a) The language is syntactically invalid. There are 7 syntax rules b.pdf
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a) The language is syntactically invalid. There are 7 syntax rules being violated. b) The language is syntactically invalid. There are 4 syntax rules being violated. c) The language is syntactically invalid. There are 18 syntax rules being violated. d) The language is syntactically valid. Solution a) The language is syntactically invalid. There are 7 syntax rules being violated. b) The language is syntactically invalid. There are 4 syntax rules being violated. c) The language is syntactically invalid. There are 18 syntax rules being violated. d) The language is syntactically valid..
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1. How new modern materials prompted changes in architecture in the .pdf
1. How new modern materials prompted changes in architecture in the late nineteenth century in
reference to the construction of the Crystal Palace or Eiffel Tower.
After the Baroque faded slowly away, eighteenth-century architecture consisted primarily of
revivals of previous periods. This time was to be the calm before the storm, for the approaching
Industrial Revolution was to change everything about the world as it was then, including
architecture. Previously, building materials had been restricted to a few manmade materials
along with those available in nature: timber, stone, timber, lime mortar, and concrete. Metals
were not available in sufficient quantity or consistent quality to be used as anything more than
ornamentation. Structure was limited by the capabilities of natural materials. The Industrial
Revolution changed this situation dramatically.
In 1800, the worldwide tonnage of iron produced was 825,000 tons. By 1900, with the Industrial
Revolution in full swing, worldwide production stood at 40 million tons, almost 50 times as
much. Iron was available in three forms. The least processed form, cast iron, was brittle due to a
high percentage of impurities. It still displayed impressive compressive strength, however.
Wrought iron was a more refined form of iron, malleable, though with low tensile strength. Steel
was the strongest, most versatile form of iron. Through a conversion process, all of the impurities
were burned out of the iron ore, then precise amounts of carbon were added for hardness. Steel
had tensile and compressive strength greater than any material previously available, and its
capabilities would revolutionize architecture.
This change did not happen over night. Prior to the introduction of bulk iron, architecture relied
on compressive strength to hold buildings up. Even great structures like the Chartres Cathedral
or the Parthenon were essentially orderly piles of stone. Architects were accustomed to thinking
of certain ways of creating structure, and though they glimpsed some of the possibilities of the
new materials, the first applications were made using the old ideas.
The explosion in the development of iron and steel structures was driven initially by the advance
of the railroads. Bridges were required to span gorges and rivers. In 1779, the first iron bridge
was built across the Severn River in Coalescence, England. It was not an iron bridge as we might
conceive of it today, but rather a traditional arch made of iron instead of stone. The compressive
strength of limestone is 20 tons per square foot. The compressive strength of cast iron is 10 tons
per square inch, 72 times as high, permitting significantly larger spans. Later, the truss, long used
in timber roofs, became the primary element of bridge building. A triangle is the strongest
structural element known, and applied force only makes it more stable. When a diagonal is added
to a square, the form can be viewed as two triangles sharing a side, the fundam.
c++ code for a Median of Integer Stream from Text File program#.pdf
// c++ code for a Median of Integer Stream from Text File program
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#include
double streamMedian(vector v, int size)
{
double* s = new double[size];
for (int i = 0; i < size; ++i)
{
s[i] = v[i];
}
for (int i = size - 1; i > 0; --i)
{
for (int j = 0; j < i; ++j)
{
if (s[j] > s[j+1])
{
double t = s[j];
s[j] = s[j+1];
s[j+1] = t;
}
}
}
double median = 0.0;
if ((size % 2) == 0)
{
median = (s[size/2] + s[(size/2) - 1])/2.0;
}
else
{
median = s[size/2];
}
return median;
}
int main()
{
int number;
double median;
std::vector v;
int count = 1;
ifstream myfile (\"input.txt\");
if (myfile.is_open())
{
while ( myfile >> number )
{
cout << \"After reading \" << count << \" element of stream - \";
v.push_back(number);
for (int i = 0; i < v.size(); ++i)
{
cout << v[i] << \" \";
}
cout << \"-> \";
sort(v.begin(), v.end());
cout << \"median - \" << streamMedian(v,v.size()) << endl;
}
myfile.close();
}
return 0;
}
/*
input.txt
5
15
1
3
output:
After reading 1 element of stream - 5 -> median - 5
After reading 1 element of stream - 5 15 -> median - 10
After reading 1 element of stream - 5 15 1 -> median - 5
After reading 1 element of stream - 1 5 15 3 -> median - 4
*/
Solution
// c++ code for a Median of Integer Stream from Text File program
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#include
double streamMedian(vector v, int size)
{
double* s = new double[size];
for (int i = 0; i < size; ++i)
{
s[i] = v[i];
}
for (int i = size - 1; i > 0; --i)
{
for (int j = 0; j < i; ++j)
{
if (s[j] > s[j+1])
{
double t = s[j];
s[j] = s[j+1];
s[j+1] = t;
}
}
}
double median = 0.0;
if ((size % 2) == 0)
{
median = (s[size/2] + s[(size/2) - 1])/2.0;
}
else
{
median = s[size/2];
}
return median;
}
int main()
{
int number;
double median;
std::vector v;
int count = 1;
ifstream myfile (\"input.txt\");
if (myfile.is_open())
{
while ( myfile >> number )
{
cout << \"After reading \" << count << \" element of stream - \";
v.push_back(number);
for (int i = 0; i < v.size(); ++i)
{
cout << v[i] << \" \";
}
cout << \"-> \";
sort(v.begin(), v.end());
cout << \"median - \" << streamMedian(v,v.size()) << endl;
}
myfile.close();
}
return 0;
}
/*
input.txt
5
15
1
3
output:
After reading 1 element of stream - 5 -> median - 5
After reading 1 element of stream - 5 15 -> median - 10
After reading 1 element of stream - 5 15 1 -> median - 5
After reading 1 element of stream - 1 5 15 3 -> median - 4
*/.
1) cobalt iodide has an intense color, thus this colorless salt cann.pdf
1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled.
2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains
waters of crystallization.
Solution
1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled.
2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains
waters of crystallization..
the hybridisation is sp2 since the electron cloud.pdf
the hybridisation is sp2 since the electron cloud is lost due to positive charge and
the three( rest of the lobes) sprea out forming sp2 hybridisation.
Solution
the hybridisation is sp2 since the electron cloud is lost due to positive charge and
the three( rest of the lobes) sprea out forming sp2 hybridisation..
Sodium hydroxide, also known as lye and caustic s.pdf
Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular
formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes,
granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This
alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide
is used in many industries, mostly as a strong chemical base in the manufacture of pulp and
paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide
production in 2004 was approximately 60 million tonnes, while demand was 51 million
tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high
temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad
experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original
sample.
Solution
Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular
formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes,
granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This
alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide
is used in many industries, mostly as a strong chemical base in the manufacture of pulp and
paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide
production in 2004 was approximately 60 million tonnes, while demand was 51 million
tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high
temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad
experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original
sample..
Hydrocrbon is the organic compound which containing only the atoms .pdf
Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon.
Based on the saturation of hydrogens with the carbons they are classified as mainly three types .
1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn
H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double
bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds
have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2
Solution
Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon.
Based on the saturation of hydrogens with the carbons they are classified as mainly three types .
1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn
H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double
bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds
have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2.
No. In order for a substance to conduct electrici.pdf
No. In order for a substance to conduct electricity it must have charged particles
that can move freely. Not all solids have that facility such as ionic compounds in solid state.
Solution
No. In order for a substance to conduct electricity it must have charged particles
that can move freely. Not all solids have that facility such as ionic compounds in solid state..
HCHO OS for H is +1, O is -2 there are two Hs o.pdf
HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (-
2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H
+ 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo.
Solution
HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (-
2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H
+ 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo..
Recommended
1. How new modern materials prompted changes in architecture in the .pdf
1. How new modern materials prompted changes in architecture in the late nineteenth century in
reference to the construction of the Crystal Palace or Eiffel Tower.
After the Baroque faded slowly away, eighteenth-century architecture consisted primarily of
revivals of previous periods. This time was to be the calm before the storm, for the approaching
Industrial Revolution was to change everything about the world as it was then, including
architecture. Previously, building materials had been restricted to a few manmade materials
along with those available in nature: timber, stone, timber, lime mortar, and concrete. Metals
were not available in sufficient quantity or consistent quality to be used as anything more than
ornamentation. Structure was limited by the capabilities of natural materials. The Industrial
Revolution changed this situation dramatically.
In 1800, the worldwide tonnage of iron produced was 825,000 tons. By 1900, with the Industrial
Revolution in full swing, worldwide production stood at 40 million tons, almost 50 times as
much. Iron was available in three forms. The least processed form, cast iron, was brittle due to a
high percentage of impurities. It still displayed impressive compressive strength, however.
Wrought iron was a more refined form of iron, malleable, though with low tensile strength. Steel
was the strongest, most versatile form of iron. Through a conversion process, all of the impurities
were burned out of the iron ore, then precise amounts of carbon were added for hardness. Steel
had tensile and compressive strength greater than any material previously available, and its
capabilities would revolutionize architecture.
This change did not happen over night. Prior to the introduction of bulk iron, architecture relied
on compressive strength to hold buildings up. Even great structures like the Chartres Cathedral
or the Parthenon were essentially orderly piles of stone. Architects were accustomed to thinking
of certain ways of creating structure, and though they glimpsed some of the possibilities of the
new materials, the first applications were made using the old ideas.
The explosion in the development of iron and steel structures was driven initially by the advance
of the railroads. Bridges were required to span gorges and rivers. In 1779, the first iron bridge
was built across the Severn River in Coalescence, England. It was not an iron bridge as we might
conceive of it today, but rather a traditional arch made of iron instead of stone. The compressive
strength of limestone is 20 tons per square foot. The compressive strength of cast iron is 10 tons
per square inch, 72 times as high, permitting significantly larger spans. Later, the truss, long used
in timber roofs, became the primary element of bridge building. A triangle is the strongest
structural element known, and applied force only makes it more stable. When a diagonal is added
to a square, the form can be viewed as two triangles sharing a side, the fundam.
c++ code for a Median of Integer Stream from Text File program#.pdf
// c++ code for a Median of Integer Stream from Text File program
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#include
double streamMedian(vector v, int size)
{
double* s = new double[size];
for (int i = 0; i < size; ++i)
{
s[i] = v[i];
}
for (int i = size - 1; i > 0; --i)
{
for (int j = 0; j < i; ++j)
{
if (s[j] > s[j+1])
{
double t = s[j];
s[j] = s[j+1];
s[j+1] = t;
}
}
}
double median = 0.0;
if ((size % 2) == 0)
{
median = (s[size/2] + s[(size/2) - 1])/2.0;
}
else
{
median = s[size/2];
}
return median;
}
int main()
{
int number;
double median;
std::vector v;
int count = 1;
ifstream myfile (\"input.txt\");
if (myfile.is_open())
{
while ( myfile >> number )
{
cout << \"After reading \" << count << \" element of stream - \";
v.push_back(number);
for (int i = 0; i < v.size(); ++i)
{
cout << v[i] << \" \";
}
cout << \"-> \";
sort(v.begin(), v.end());
cout << \"median - \" << streamMedian(v,v.size()) << endl;
}
myfile.close();
}
return 0;
}
/*
input.txt
5
15
1
3
output:
After reading 1 element of stream - 5 -> median - 5
After reading 1 element of stream - 5 15 -> median - 10
After reading 1 element of stream - 5 15 1 -> median - 5
After reading 1 element of stream - 1 5 15 3 -> median - 4
*/
Solution
// c++ code for a Median of Integer Stream from Text File program
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#include
double streamMedian(vector v, int size)
{
double* s = new double[size];
for (int i = 0; i < size; ++i)
{
s[i] = v[i];
}
for (int i = size - 1; i > 0; --i)
{
for (int j = 0; j < i; ++j)
{
if (s[j] > s[j+1])
{
double t = s[j];
s[j] = s[j+1];
s[j+1] = t;
}
}
}
double median = 0.0;
if ((size % 2) == 0)
{
median = (s[size/2] + s[(size/2) - 1])/2.0;
}
else
{
median = s[size/2];
}
return median;
}
int main()
{
int number;
double median;
std::vector v;
int count = 1;
ifstream myfile (\"input.txt\");
if (myfile.is_open())
{
while ( myfile >> number )
{
cout << \"After reading \" << count << \" element of stream - \";
v.push_back(number);
for (int i = 0; i < v.size(); ++i)
{
cout << v[i] << \" \";
}
cout << \"-> \";
sort(v.begin(), v.end());
cout << \"median - \" << streamMedian(v,v.size()) << endl;
}
myfile.close();
}
return 0;
}
/*
input.txt
5
15
1
3
output:
After reading 1 element of stream - 5 -> median - 5
After reading 1 element of stream - 5 15 -> median - 10
After reading 1 element of stream - 5 15 1 -> median - 5
After reading 1 element of stream - 1 5 15 3 -> median - 4
*/.
1) cobalt iodide has an intense color, thus this colorless salt cann.pdf
1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled.
2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains
waters of crystallization.
Solution
1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled.
2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains
waters of crystallization..
the hybridisation is sp2 since the electron cloud.pdf
the hybridisation is sp2 since the electron cloud is lost due to positive charge and
the three( rest of the lobes) sprea out forming sp2 hybridisation.
Solution
the hybridisation is sp2 since the electron cloud is lost due to positive charge and
the three( rest of the lobes) sprea out forming sp2 hybridisation..
Sodium hydroxide, also known as lye and caustic s.pdf
Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular
formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes,
granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This
alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide
is used in many industries, mostly as a strong chemical base in the manufacture of pulp and
paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide
production in 2004 was approximately 60 million tonnes, while demand was 51 million
tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high
temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad
experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original
sample.
Solution
Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular
formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes,
granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This
alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide
is used in many industries, mostly as a strong chemical base in the manufacture of pulp and
paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide
production in 2004 was approximately 60 million tonnes, while demand was 51 million
tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high
temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad
experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original
sample..
Hydrocrbon is the organic compound which containing only the atoms .pdf
Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon.
Based on the saturation of hydrogens with the carbons they are classified as mainly three types .
1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn
H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double
bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds
have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2
Solution
Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon.
Based on the saturation of hydrogens with the carbons they are classified as mainly three types .
1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn
H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double
bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds
have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2.
No. In order for a substance to conduct electrici.pdf
No. In order for a substance to conduct electricity it must have charged particles
that can move freely. Not all solids have that facility such as ionic compounds in solid state.
Solution
No. In order for a substance to conduct electricity it must have charged particles
that can move freely. Not all solids have that facility such as ionic compounds in solid state..
HCHO OS for H is +1, O is -2 there are two Hs o.pdf
HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (-
2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H
+ 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo.
Solution
HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (-
2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H
+ 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo..
The molecular orbital (MO) theory is a way of loo.pdf
The molecular orbital (MO) theory is a way of looking at the structure of a
molecule by using molecular orbitals that belong to the molecule as a whole Figure 1.
Combination of two 1s atomic orbitals to form a sigma bonding orbital or a sigma-starred
antibonding orbital. rather than to the individual atoms. When simple bonding occurs between
two atoms, the pair of electrons forming the bond occupies an MO that is a mathematical
combination of the wave functions of the atomic orbitals of the two atoms involved. The MO
method originated in the work of Friedrich Hund and Robert S. Mulliken. When atoms combine
to form a molecule, the number of orbitals in the molecule equals the number of orbitals in the
combining atoms. When two very simple atoms, each with one atomic orbital, are combined, two
molecular orbitals are formed. One is a bonding orbital, lower in energy than the atomic orbitals,
and derived from their sum. It is called sigma. The other is an antibonding orbital, higher in
energy than the atomic orbitals, and resulting from their difference. It is called sigma-starred ( s
*). (See the diagram in Figure 1.) The basic idea might best be illustrated by considering
diatomic molecules of hydrogen and helium. The energy diagrams are shown in Figure 2. Each
hydrogen atom has one 1 s electron. In the H 2 molecule the two hydrogen electrons go into the
lowest energy MO available, the sigma orbital. In the case of helium, each helium atom has two
electrons, so the He 2 molecule would have four. Two would go into the lower energy bonding
orbital, but the other two would have to go into the higher energy sigma-starred orbital. Figure
2. Molecular orbital energy diagrams for (a) H 2 showing both electrons in the bonding sigma
MO; and (b) He 2 in which two of the electrons are in the antibonding sigma-starred MO Figure
3. Combination of p atomic orbitals to form (a) sigma MOs by end-to-end interactions or (b) pi
MOs by sideways interaction. Bond Order The bond order for a molecule can be determined as
follows: bond order = ½ (bonding electrons - antibonding electrons). Therefore, the H 2
molecule has a bond order of ½ (2 - 0) = 1. In other words, there is a single bond connecting the
two H atoms in the H 2 molecule. In the case of He 2 , on the other hand, the bond order is ½ (2 -
2) = 0. This means that He 2 is not a stable molecule. Multiple Bonds Double or triple bonds
involve two or three pairs of bonding electrons. Single bonds are always sigma bonds, but in
multiple bonds the first bond is sigma, while any second or third bonds are pi bonds. The overlap
of p orbitals can yield either pi or sigma MOs, as shown in Figure 3. When they overlap end to
end, they form sigma orbitals, but when they overlap side to side, they form pi orbitals. Consider
now the oxygen molecule. The Lewis structure for oxygen is :Ö::Ö: The double bond is
necessary in order to satisfy the octet rule for both oxygen atoms. The measured bond length for
oxygen supports the.
The electrons reach or give off a certain amount of energy and diffe.pdf
The electrons reach or give off a certain amount of energy and different amounts of energy in
the electron changes the color because it jumps to a differrent energy level Hope this helps! rate
lifesaver :)
Solution
The electrons reach or give off a certain amount of energy and different amounts of energy in
the electron changes the color because it jumps to a differrent energy level Hope this helps! rate
lifesaver :).
Weak London disperision and dipole- dipoleModerate Hydrogen bond.pdf
Weak: London disperision and dipole- dipole
Moderate: Hydrogen bonding and ion-dipole
Trust, got these wrong due to inputting the wrong answer above.
Solution
Weak: London disperision and dipole- dipole
Moderate: Hydrogen bonding and ion-dipole
Trust, got these wrong due to inputting the wrong answer above..
var min =0; var max=10; var tab = {}; var name; var score;.pdf
var min =0;
var max=10;
var tab = {};
var name;
var score;
var sum=0;
while(name=prompt(\"Enter name:\")){ // Get userinput for name
do{
score = prompt(\"Enter score\"); // Get Score
tab[name] = score; // Create object with name as key and score as value.
}while(isNaN(score) || !(score>min) || !(score\";
for(item in tab){ // Create table
ui = ui + (\"\"+item+\"\"+tab[item]+\"\");
sum = sum + parseInt(tab[item]);
}
ui = ui + \"\";
document.write(ui); // Display table
document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and
displaying on the page
Solution
var min =0;
var max=10;
var tab = {};
var name;
var score;
var sum=0;
while(name=prompt(\"Enter name:\")){ // Get userinput for name
do{
score = prompt(\"Enter score\"); // Get Score
tab[name] = score; // Create object with name as key and score as value.
}while(isNaN(score) || !(score>min) || !(score\";
for(item in tab){ // Create table
ui = ui + (\"\"+item+\"\"+tab[item]+\"\");
sum = sum + parseInt(tab[item]);
}
ui = ui + \"\";
document.write(ui); // Display table
document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and
displaying on the page.
Unfortunately several cancers are not predictable with simple tests .pdf
Unfortunately several cancers are not predictable with simple tests rather required much
intensive diagnosis. The whole genome sequence in determining the entire DNA in normal cells
as well as in tumor cells provide the facility to compare and pin point the mutations occurring in
the oncogenes or tumor suppressor genes responsible for causing the cancer seems to be a better
approach. But such approaches are highly expensive and require expertise.
Though, the whole genome sequencing is carried out pin pointing of specific driver mutations
responsible for cancer phenotype would be a question. Some critical mutations can be identified
in some cancer genome sequences while others are hard to find among the thousands of
possibilities. The mutations like missense, nonsense frameshift, rearrangements may alter the
coding sequences of any gene must be checked. The regulatory mutations that increase the
transcription of oncogenes or decrease that of tumor suppressor genes are potentially important
but are very difficult to find in a whole genome sequence.
The catalogs of regulatory elements in human genomes provide list of certain potential
sequences to be targeted. But the catalogs are rudimentary or incomplete. Therefore, the whole
genome sequencing is not a panacea that will lead to immediate cures of all cancers. Recent
studies have shown that the whole genome sequences of cells derived from different regions of
same tumor have suggested an important reason for cancer reference. Certain cancers are
heterogenous and cells within the tumor have different genomes. Certain mutations in oncogens
or tumor suppressor genes are common to all cells in the tumor but some are not.
These findings make a sense that accumulation of mutations in clone of cells cause cancer.
Therefore, designing drug targeting specific cells protecting adult stem cells cannot be
accomplished. Thus, effective cancer treatments would be directed against the common
mutations but is difficult to identify the specific mutations without sequencing of genomes of
many cells throughout the tumor. Cancer landscape is a large scale cancer genome sequencing
project (cancer genome atlas) funded by National Institute of Health (NIH). In this project the
whole genomes or exomex of several hundreds of cancer are being characterized. The recurrent
patterns of mutations are subdivided cancers in to groups with probable clinical relevance.
For example, 4 groups of breast cancers have been identified. Almost 178 lung squamous cell
carcinoma’s genomes have been characterized and matched with DNA of normal cells from
same patients. Mutations in certain tumor suppressor genes like p53 and certain oncogenes have
shown relatively high frequencies among these cancers. The mutations of certain cancers
resemble cancers in some other organs. For example, pattern of mutation in breast cancer
resembles many ovarian cancers more than other types of breast cancers. These findings may
help in designing the drugs .
The van t Hoff factor i (named after J. H. van t Hoff) is a meas.pdf
The van \'t Hoff factor i (named after J. H. van \'t Hoff) is a measure of the effect of a solute
upon colligative properties, such as vapor pressure, osmotic pressure and freezing point
depression. The van \'t Hoff factor is the ratio between the actual concentration of particles
produced when the substance is dissolved, and the concentration of a substance as calculated
from its mass. For most non-electrolytes dissolved in water, the van\' t Hoff factor is essentially
1. For most ionic compounds dissolved in water, the van \'t Hoff factor is equal to the number of
discrete ions in a formula unit of the substance. This is true for ideal solutions only, as
occasionally ion pairing occurs in solution. At a given instant a small percentage of the ions are
paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte
solutions. This causes deviation from the van \'t Hoff factor. The deviation for the van \'t Hoff
factor tends to be greatest where the ions have multiple charges.
Relationship with the dissociation degree
The dissociation degree is the fraction of original solute molecules that have dissociated. It is
usually indicated by the Greek symbol . There is a simple relationship between this parameter
and the van \'t Hoff factor. If the solute substance dissociates into n ions, then
i = 1 + (n 1)
For instance, for the following dissociation
KCl K+ + Cl-
As n = 2, we would have that i = 1 +
Physical significance of i
When solute particles associate in solution, i is less than 1. (e.g. ethanoic acid in benzene,
benzoic acid in benzene)
When solute particles dissociate in solution, i is greater than 1. (e.g. sodium chloride in water,
potassium chloride in water, magnesium chloride in water)
When solute particles neither dissociate nor associate in solution, i equals 1. (e.g. Glucose in
water)
The value of i is ; i = the actual number of particles in solution after dissociation ÷ the number
of formula units initially dissolved in solution. Means the number of particles per formula unit of
the solute when a solution is dilute.
Solution
The van \'t Hoff factor i (named after J. H. van \'t Hoff) is a measure of the effect of a solute
upon colligative properties, such as vapor pressure, osmotic pressure and freezing point
depression. The van \'t Hoff factor is the ratio between the actual concentration of particles
produced when the substance is dissolved, and the concentration of a substance as calculated
from its mass. For most non-electrolytes dissolved in water, the van\' t Hoff factor is essentially
1. For most ionic compounds dissolved in water, the van \'t Hoff factor is equal to the number of
discrete ions in a formula unit of the substance. This is true for ideal solutions only, as
occasionally ion pairing occurs in solution. At a given instant a small percentage of the ions are
paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte
solutions. This causes deviation from th.
Polar Bonds and Molecular Shape A polar molecule.pdf
Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net
dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into
determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non-
polar, you must determine both factors. The polarity of the individual bonds in the molecule. The
shape or geometry of the molecule. First, to determine if a given individual bond is polar, you
need to know the electronegativity of the two atoms involved in that bond. To find the
electronegativities of all the elements, look at the periodic table (follow the link below this
answer under Web Links). If the electronegativity of the two atoms has a difference of 0.3 or
less, then the bond is non-polar. If the electronegativity difference is greater that 0.3 but less than
1.7, then the bond is polar. If the two values have a difference greater than 1.7, then the bond is
ionic, which is just very very polar. Once you know which bonds in the molecule are polar and
which are non-polar, you must use the shape of the molecule. You need the shape because two
polar bonds, if oriented correctly can cancel each other out (like two equally strong people
pulling in opposite directions on a rope -- nobody moves). The three possible outcomes: If all
bonds are non-polar, then the whole molecule is non-polar regardless of its shape. If there is
symmetry in the molecule so that the polarity of the bonds cancels out, then the molecule is non-
polar. A common example of this is carbon dioxide, or CO2. The molecule is linear, and its
Lewis dot structure is like this: O=C=O (this doesn\'t include two sets of lone pairs on each
oxygen). The carbon-oxygen bond is a polar bond, but because they are exactly opposed to each
other, the molecule is overall non-polar. Another example of this is CCl4, where each carbon-
chlorine bond is polar, but the molecule is non-polar. Here, how they cancel out isn\'t as obvious,
but they do. CCl4 is a tetrahedral molecule, and the 4 C-Cl polar bonds cancel each other out. If
there are polar bonds but there is no symmetry such that they cancel each other out, the overall
molecule is polar. Water is a typical example of this. The two O-H bonds are oriented in a V-
shape, and so the don\'t cancel out. Similarly, CH3Cl is also polar. It is the same shape as CCl4
(see above), but now it doesn\'t have the same symmetry because there is only one C-Cl bond
and the bonds don\'t cancel out
Solution
Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net
dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into
determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non-
polar, you must determine both factors. The polarity of the individual bonds in the molecule. The
shape or geometry of the molecule. First, to determine if a given individual bond is polar, you
need to know the electronegativ.
The good functioning of an economy depends on the proper functioning.pdf
The good functioning of an economy depends on the proper functioning of the country\'s
financial market and financial institutions. Both these are involved in the financial transactions
that take place in the economy. Financial transactions take place through exchange of funds.
There are two ways in which the exchange of funds can take place. They are: Direct exchange
and Indirect exchange.
Direct exchange takes place when the buyers and the sellers meet each other directly and
exchange their securities.For example: Stocks or bonds
Indirect exchange takes place when the buyers and sellers do not meet each other directly. They
carry out their financial transaction through a financial institution. For example: Bank
The financial market and financial institution have the main function of transferring the funds
from people who have excess funds to people who have shortage of funds. Some functions are
common for both financial markets as well as financial institutions. They both are involved in the
borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller
to the buyer. They both help in providing information to the buyers and sellers about the
financial assets.
However, there are also some functions that are unique to each. Financial markets determine the
price of the financial asset based on the demand and supply forces. Financial institutions help to
lower the transaction costs and also reduce the assymetry of information about financial assets.
The financial institutions also provide specialized services to th buyers and sellers.
Thus, it can be said that in the financial system, the financial market and financial institutions
have an important role to play. Direct exchange takes place through financial market while
indirect exchnage takes place through financial institutions.
Solution
The good functioning of an economy depends on the proper functioning of the country\'s
financial market and financial institutions. Both these are involved in the financial transactions
that take place in the economy. Financial transactions take place through exchange of funds.
There are two ways in which the exchange of funds can take place. They are: Direct exchange
and Indirect exchange.
Direct exchange takes place when the buyers and the sellers meet each other directly and
exchange their securities.For example: Stocks or bonds
Indirect exchange takes place when the buyers and sellers do not meet each other directly. They
carry out their financial transaction through a financial institution. For example: Bank
The financial market and financial institution have the main function of transferring the funds
from people who have excess funds to people who have shortage of funds. Some functions are
common for both financial markets as well as financial institutions. They both are involved in the
borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller
to the buyer. They both help.
The answer is Yes, it is a reduction-oxidation reaction2 HNO2 + 2.pdf
The answer is: Yes, it is a reduction-oxidation reaction
2 HNO2 + 2 H+ + 3 I- => 2 NO + I3- + 2 H2O
Oxidation state of N in HNO2 = +3
Oxidation state of N in NO = +2
Thus HNO2 is reduced to NO as oxidation state of N decreases.
Oxidation state of I in I- = -1
Oxidation state of I in I3- = -1/3
Thus I- is oxidized to I3- as oxidation state of I increases.
Solution
The answer is: Yes, it is a reduction-oxidation reaction
2 HNO2 + 2 H+ + 3 I- => 2 NO + I3- + 2 H2O
Oxidation state of N in HNO2 = +3
Oxidation state of N in NO = +2
Thus HNO2 is reduced to NO as oxidation state of N decreases.
Oxidation state of I in I- = -1
Oxidation state of I in I3- = -1/3
Thus I- is oxidized to I3- as oxidation state of I increases..
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmer.pdf
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a
large quadriped with unusual posture, heavy back, short forelimbs, spiked tail.
Integumentary system is that which is composed of skin and its appendages. Its primary function
is to protect the body from external injury, temperature regulation, cushioning and self defense.
Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them
measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in
selfdefense and regulation of body temperature. The plates were supplied with blood through the
blood vessels and the air flow through the plates is hypothesized to have regulated the body
temperature. The plates were in double row as per findings in late 1960\'s.
Lymphatic system: There is little evidence to clearly understand the lymphatic system although it
is hypothesized that they had a 4 chambered heart and were warmblooded with a well
functioning circulatory system. They had higher aerobic capacity and higher pressure or higher
gas exchange circulatory system.
Solution
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a
large quadriped with unusual posture, heavy back, short forelimbs, spiked tail.
Integumentary system is that which is composed of skin and its appendages. Its primary function
is to protect the body from external injury, temperature regulation, cushioning and self defense.
Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them
measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in
selfdefense and regulation of body temperature. The plates were supplied with blood through the
blood vessels and the air flow through the plates is hypothesized to have regulated the body
temperature. The plates were in double row as per findings in late 1960\'s.
Lymphatic system: There is little evidence to clearly understand the lymphatic system although it
is hypothesized that they had a 4 chambered heart and were warmblooded with a well
functioning circulatory system. They had higher aerobic capacity and higher pressure or higher
gas exchange circulatory system..
The answer is d- the hydrogen bonds in iceThe high heat of fusion.pdf
The answer is: d- the hydrogen bonds in ice
The high heat of fusion indicates that strong intermolecular forces are present between the H2O
molecules in the form of hydrogen bonds.
Solution
The answer is: d- the hydrogen bonds in ice
The high heat of fusion indicates that strong intermolecular forces are present between the H2O
molecules in the form of hydrogen bonds..
SolutionOption (a) will qualify as a database as dictionary is a .pdf
Solution
:
Option (a) will qualify as a database as dictionary is a database of different words with their
meanings stored in it.
Option (b) will qualify as a database as telephone directory contains data regarding people like
their name, address, phone number.
Option (c) will qualify as a database as rolodex is a file that contains cards holding information
about people like names, addresses and phone numbers.
Option (d) will not qualify as a database as textbook index only points to the topics in the book,
it itself does not store any data in it related to the topic or anything..
Tests of Controls in an Audit of Internal Control. The objective of .pdf
Tests of Controls in an Audit of Internal Control. The objective of the tests of controls in an audit
of internal control over financial reporting is to obtain evidence about the effectiveness of
controls to support the auditor\'s opinion on the company\'s internal control over financial
reporting. The auditor\'s opinion relates to the effectiveness of the company\'s internal control
over financial reporting as of a point in time and taken as a whole.
The general standards are applicable to an audit of internal control over financial reporting.
Those standards require technical training and proficiency as an auditor, independence, and the
exercise of due professional care, including professional skepticism. This standard establishes
the fieldwork and reporting standards applicable to an audit of internal control over financial
reporting. The auditor should use the same suitable, recognized control framework to perform his
or her audit of internal control over financial reporting as management uses for its annual
evaluation of the effectiveness of the company\'s internal control over financial reporting.
To express an opinion on internal control over financial reporting as of a point in time, the
auditor should obtain evidence that internal control over financial reporting has operated
effectively for a sufficient period of time, which may be less than the entire period (ordinarily
one year) covered by the company\'s financial statements. To express an opinion on internal
control over financial reporting taken as a whole, the auditor must obtain evidence about the
effectiveness of selected controls over all relevant assertions. This requires that the auditor test
the design and operating effectiveness of controls
When concluding on the effectiveness of internal control over financial reporting for purposes of
expressing an opinion on internal control over financial reporting, the auditor should incorporate
the results of any additional tests of controls performed to achieve the objective related to
expressing an opinion on the financial statements, as discussed in the following section.
Tests of Controls in an Audit of Financial Statements.
To express an opinion on the financial statements, the auditor ordinarily performs tests of
controls and substantive procedures. The objective of the tests of controls the auditor performs
for this purpose is to assess control risk. To assess control risk for specific financial statement
assertions at less than the maximum, the auditor is required to obtain evidence that the relevant
controls operated effectively during the entire period upon which the auditor plans to place
reliance on those controls. However, the auditor is not required to assess control risk at less than
the maximum for all relevant assertions and, for a variety of reasons, the auditor may choose not
to do so.
When concluding on the effectiveness of controls for the purpose of assessing control risk, the
auditor also should evaluate the .
Software Quality Assurance is important because Quality in the softw.pdf
Software Quality Assurance is important because Quality in the software is one of the pivotal
aspects of a software development company. quality assurance starts from the beginning of a
software project, right from the analysis phase. Software Quality Assurance (SQA) is defined as
a well planned and systematic approach to evaluate the quality of software. It checks the
adherence to software product standards, processes, and procedures. SQA includes the
systematic process of assuring that standards and procedures are established and are followed
throughout the software development life cycle and test cycle as well. The compliance of the
built with agreed-upon standards and procedures is evaluated through process monitoring,
product evaluation, project management etc.
Software quality assurance methods
The main reason of involving quality in software assurance in the process of software product
development is to make sure that the final product built is as per the requirement specification
and comply with the standards.
SQA encompasses different activities, like :
1) Requirement analysis and definition
2) Design architecture and description
3) Coding and logic analysis
4) Change and configuration management
5) Testing and standard compliance
6) Release management and Release Control
Different Functionalities of Quality Assurance are the following:
1. Maintaining the quality of the project as per business requirements and Specifications.
2. Defect Prevention. And formal methods for other defect prevention techniques.
3. Defect Reduction
4. Inspection, formal and informal reviews: Direct fault detection and removal without
executing the project scenario.
5. Testing the project for Failure observation and bug removal.
6. Risk identification.
7. Defect tracking techniques and methods
8. Software fault tolerance.
9. Concluding Remarks and maintaining reports.
Software quality can be quantified into two major groups such as:
1. Software Functional Quality: It basically shows how well the software product conforms to
the basic design, based on functional requirements. The attribute can also be described as the
fitness for purpose of a software.
2. Software Structural Quality: It reflects to how well the project meets the non-functional
requirements such as usability, accessibility and security that helps in proper the delivery of the
functional requirements. It basically defines the degree of correctness of the product.
.
Solution
Software Quality Assurance is important because Quality in the software is one of the pivotal
aspects of a software development company. quality assurance starts from the beginning of a
software project, right from the analysis phase. Software Quality Assurance (SQA) is defined as
a well planned and systematic approach to evaluate the quality of software. It checks the
adherence to software product standards, processes, and procedures. SQA includes the
systematic process of assuring that standards and procedures are establis.
Oxygen rich blood leaves the lungs. It travels via the pulmonary vei.pdf
Oxygen rich blood leaves the lungs. It travels via the pulmonary vein to the heart.
It travels through the aorta to the iliac artery.
It then travels through the femoral artery to lower leg capillaries.
It leaves the lower leg capillaries through the femoral vein
It travels through the iliac vein to the inferior vena cava to the heart
Carbon di oxide rich blood reaches the lungs through the pulmonary artery.
Solution
Oxygen rich blood leaves the lungs. It travels via the pulmonary vein to the heart.
It travels through the aorta to the iliac artery.
It then travels through the femoral artery to lower leg capillaries.
It leaves the lower leg capillaries through the femoral vein
It travels through the iliac vein to the inferior vena cava to the heart
Carbon di oxide rich blood reaches the lungs through the pulmonary artery..
Normal Concept For colour vision, our normal human eyes have 3 con.pdf
Normal Concept : For colour vision, our normal human eyes have 3 cones and photoreceptors.
L cone - sensitive to long wavelength of yellowish green light
M cone - middle wavelength of green light
s cone - short wavelength to bluish violet light
Protonapia - complete absence of red cone photoreceptors.He felt difficulty in identifying the red
colour
Deuteranopia - complete absence of green cone photoreceptors.He felt difficulty in identifying
green colour
Red green colour discrimination is called red- green dichromat vision blindness.He can able to
see yellow circles sometimes in place of red or green colour, it shows the unstable working
colour vision receptors.
If one colour dominates the vision, the other will be absent,for instance, if red hue is dominant, it
strikes the receptors of green colour and vice versa.
Red green dichromat friend will have reduction view in the normally appearing red - green
colour.The hues, he perceives was not like normal, it looks fade olive in colour.This is due to
specific absence of photoreceptors.
If he able to perceive colours, either one of the phoreceptors work in exitatory state and the other
in inhibitory state.This is called opponent colour theory.
Of all that, he might have some degeneration or improper formation issues with myelin
sheath(covering the axon of a neuron) which is responsible for colour determination.
Solution
Normal Concept : For colour vision, our normal human eyes have 3 cones and photoreceptors.
L cone - sensitive to long wavelength of yellowish green light
M cone - middle wavelength of green light
s cone - short wavelength to bluish violet light
Protonapia - complete absence of red cone photoreceptors.He felt difficulty in identifying the red
colour
Deuteranopia - complete absence of green cone photoreceptors.He felt difficulty in identifying
green colour
Red green colour discrimination is called red- green dichromat vision blindness.He can able to
see yellow circles sometimes in place of red or green colour, it shows the unstable working
colour vision receptors.
If one colour dominates the vision, the other will be absent,for instance, if red hue is dominant, it
strikes the receptors of green colour and vice versa.
Red green dichromat friend will have reduction view in the normally appearing red - green
colour.The hues, he perceives was not like normal, it looks fade olive in colour.This is due to
specific absence of photoreceptors.
If he able to perceive colours, either one of the phoreceptors work in exitatory state and the other
in inhibitory state.This is called opponent colour theory.
Of all that, he might have some degeneration or improper formation issues with myelin
sheath(covering the axon of a neuron) which is responsible for colour determination..
LinkedListOperations.java-class Node{ public int item; .pdf
LinkedListOperations.java:-
class Node
{
public int item;
public Node next;
public Node(int val)
{
item = val;
}
public void displayNode()
{
System.out.println(\"[\" + item + \"] \");
}
}
class LinkedList
{
private Node first;
public LinkedList()
{
first = null;
}
public boolean isEmpty()
{
return (first==null);
}
public void insert(int val)//inserts at beginning of list
{
Node newNode = new Node(val);
newNode.next = first;
first = newNode;
}
public Node delete()//deletes at beginning of list
{
Node temp = first;
first = first.next;
return temp;
}
public void display()
{
System.out.println(\"List items from first to last :\");
Node current = first;
while(current != null)
{
current.displayNode();
current = current.next;
}
System.out.println(\"\");
}
public Node search(int val)
{
Node current = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
current = current.next;
}
return current;
}
public Node delete(int val)
{
Node current = first;
Node previous = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
{
previous = current;
current = current.next;
}
}
if(current == first)
first = first.next;
else
previous.next = current.next;
return current;
}
}
class LinkedListOperations
{
public static void main(String[] args)
{
LinkedList object = new LinkedList();
object.insert(10);
object.insert(20);
object.insert(30);
object.display();
while( !object.isEmpty() )
{
Node member = object.delete();
System.out.print(\"Deleted \");
member.displayNode();
System.out.println(\"\");
}
object.display();
object.insert(40);
object.insert(50);
object.insert(60);
object.display();
Node objecttosearch = object.search(50);
if( objecttosearch != null)
System.out.println(\"Found Node : \" + objecttosearch.item);
else
System.out.println(\"Cannot locate the node\");
Node objecttodelete = object.delete(50);
if( objecttodelete != null )
System.out.println(\"Deleted node : \" + objecttodelete.item);
else
System.out.println(\"Cannot delete the node\");
object.display();
}
}
Solution
LinkedListOperations.java:-
class Node
{
public int item;
public Node next;
public Node(int val)
{
item = val;
}
public void displayNode()
{
System.out.println(\"[\" + item + \"] \");
}
}
class LinkedList
{
private Node first;
public LinkedList()
{
first = null;
}
public boolean isEmpty()
{
return (first==null);
}
public void insert(int val)//inserts at beginning of list
{
Node newNode = new Node(val);
newNode.next = first;
first = newNode;
}
public Node delete()//deletes at beginning of list
{
Node temp = first;
first = first.next;
return temp;
}
public void display()
{
System.out.println(\"List items from first to last :\");
Node current = first;
while(current != null)
{
current.displayNode();
current = current.next;
}
System.out.println(\"\");
}
public Node search(int val)
{
Node current = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
current = current.next;
}
return current;
}
public Node delete(int val)
{
N.
In Keynesian economic theory, a factor that quantifies the change in.pdf
In Keynesian economic theory, a factor that quantifies the change in total income as compared to
the injection of capital deposits or investments which originally fueled the growth. It is usually
used as a measurement of the effects of government spending on income, and it can be calculated
as one divided by the marginal propensity to save.
Average propensity to consume, or APC, is the ratio of total consumption to total disposable
income, whereas marginal propensity to consume, or MPC, is the ratio of change in consumption
to the change in disposable income.
Y= C + S
That is, disposable income (income after tax) = Consumption + Savings
Think of Bob the Builder who earns a gross income of $1000 per week. After tax (which he
legally has no choice but to pay) he brings in $850 dollars. His DISPOSABLE Income is $850.
He then has two very simple choices- to save it all (i.e. place it in a financial institution such as a
credit union, bank, super fund) or to spend it (consume as economists would say). Realistically,
he will probably save a fraction and spend a fraction.
The fraction of his disposable income (Y) in which he decides to save (S) is his APS- Average
Propensity to Save
Therefore- Amount Saved/Disposable Income or S/Y.
Very simple maths.
Similarly, the amount of disposable income that he uses on consumption (spends) is his Average
Propensity to Consume-
Consumption/Disposable Income or C/Y.
The APC and APS are expressed as fractions of Disposable Income- for example 0.3 and 0.7.
As he only has two choices, whether to save or spend, the APC and APS must always sum to
give 1.
They vary with individuals, particularly with those of different disposable incomes.
Surprisingly, the wealthy has a higher APS and a lower APC compared to the poor. This is
because APS/APC are proportional things; poor people will spend a larger amount of their
income on bills and just general living expenses leaving little to no room for saving. As you get
wealthy, your consumption will rise however you can afford to live comfortably plus have a lot
more money left over for savings, hence their APS is higher.
Extending this, economists have come up with the Marginal Propensity (Tendency) to
Save/Consume. This looks at the tendency to save/spend each extra dollar of income that is
earned.
MPC is the proportion of the next dollar in which someone earns that is spent, and similarly
MPS is the proportion that is saved. Eg- I give you one dollar, you spend 80 cents and keep 20
cents, your MPC is 0.8 and your MPS is 0.2. As you can see-
MPC + MPS also add to 1.
As pointed out above, this concept must be clearly understood, as it is the fundamental of
Keynes demand theory and the multiplier which you look at to explain economic growth in year
12.
As for the difference between savings and investment. Savings are placed away in a financial
institution, the incentive to do so is that they earn interest and hence ‘grow’. If you look at the
circular flow of income model (if yo.
import java.util.LinkedList; import java.util.Random; import jav.pdf
import java.util.LinkedList;
import java.util.Random;
import java.util.Scanner;
/**
* author
*
*/
public class LinkedListTest {
/**
* param args
*/
public static void main(String[] args) {
int M, N, Lower, Upper;
Scanner scanner = null;
try {
LinkedList list1 = new LinkedList();
LinkedList list2 = new LinkedList();
LinkedList list3 = new LinkedList();
Random random = new Random();
scanner = new Scanner(System.in);
System.out.print(\"Enter the Lower and Upper bounds:\");
Lower = scanner.nextInt();
Upper = scanner.nextInt();
System.out.print(\"Enter the N value to Insert list1:\");
N = scanner.nextInt();
System.out.print(\"Enter the M value to Insert list2:\");
M = scanner.nextInt();
for (int i = 1; i <= N; i++) {
list1.add(random.nextInt((Upper - Lower) + 1) + Lower);
}
for (int i = 1; i <= M; i++) {
list2.add(random.nextInt((Upper - Lower) + 1) + Lower);
}
System.out.println(\"list1: \" + list1);
System.out.println(\"list2: \" + list2);
System.out.println(\"Minimum elelment in list1 :\" + getMin(list1));
System.out.println(\"Minimum elelment in list2 :\" + getMin(list2));
System.out
.println(\"Count Divisibles by 3 and 5 elelment in list1 :\"
+ countDivisibles(list1));
System.out
.println(\"Count Divisibles by 3 and 5 elelment in list2 :\"
+ countDivisibles(list2));
System.out.println(\"Common Elements From list1 and list2:\"
+ getCommonList(list1, list2));
} catch (Exception e) {
// TODO: handle exception
}
}
/**
* param list
* return
*/
public static int getMin(LinkedList list) {
int min = list.get(0);
for (Integer integer : list) {
if (min > integer) {
min = integer;
}
}
return min;
}
/**
* param list
* return
*/
public static int countDivisibles(LinkedList list) {
int count = 0;
for (Integer integer : list) {
if (integer % 3 == 0 && integer % 5 == 0) {
count++;
}
}
return count;
}
/**
* param list1
* param list2
* return
*/
public static LinkedList getCommonList(LinkedList list1,
LinkedList list2) {
LinkedList list = new LinkedList();
LinkedList listCommon = new LinkedList();
list = (list1.size() > list2.size()) ? list1 : list2;
for (int i = 0; i < list.size(); i++) {
Integer number = list.get(i);
if (list2.contains(number) && list1.contains(number)) {
listCommon.add(number);
}
}
return listCommon;
}
}
OUTPUT:
Enter the Lower and Upper bounds:1 10
Enter the N value to Insert list1:10
Enter the M value to Insert list2:20
list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5]
list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2]
Minimum elelment in list1 :1
Minimum elelment in list2 :1
Count Divisibles by 3 and 5 elelment in list1 :0
Count Divisibles by 3 and 5 elelment in list2 :0
Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6]
Solution
import java.util.LinkedList;
import java.util.Random;
import java.util.Scanner;
/**
* author
*
*/
public class LinkedListTest {
/**
* param args
*/
public static void main(String[] args) {
int M, N, Lower, Upper;
Scanner scanner = null;
try {
LinkedList list1 = new LinkedList();
L.
Hi,Please find the answers as followsPart AExpec.pdf
Hi,
Please find the answers as follows:
Part A:
Expected Return = .1*-10 + .2*0 + .4*10 + .2*20 + .1*30 = 10%
Part B:
Variance = .1*(-10 - 10)^2 + .2*(0 - 10)^2 + .4*(10 - 10)^2 + .2*(20 - 10)^2 + .1*(30 -10)^2 =
120
Standard Deviation = (120)^1/2 = 10.95%
Part C:
Coeffcient of Variation = Standard Deviation/Expected Return = 10.95/10 = 1.095
Part D:
Stand Alone Risk is measured by th Standard Deviation or Coefficient of variation (CV) of
returns.
Part E:
Stand Alone Risk reflects the uncertainty about the cash flows. It is relevant when the the
undiversified risk of an individual asset is required to be measured.
Thanks.
Solution
Hi,
Please find the answers as follows:
Part A:
Expected Return = .1*-10 + .2*0 + .4*10 + .2*20 + .1*30 = 10%
Part B:
Variance = .1*(-10 - 10)^2 + .2*(0 - 10)^2 + .4*(10 - 10)^2 + .2*(20 - 10)^2 + .1*(30 -10)^2 =
120
Standard Deviation = (120)^1/2 = 10.95%
Part C:
Coeffcient of Variation = Standard Deviation/Expected Return = 10.95/10 = 1.095
Part D:
Stand Alone Risk is measured by th Standard Deviation or Coefficient of variation (CV) of
returns.
Part E:
Stand Alone Risk reflects the uncertainty about the cash flows. It is relevant when the the
undiversified risk of an individual asset is required to be measured.
Thanks..
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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The molecular orbital (MO) theory is a way of loo.pdf
The molecular orbital (MO) theory is a way of looking at the structure of a
molecule by using molecular orbitals that belong to the molecule as a whole Figure 1.
Combination of two 1s atomic orbitals to form a sigma bonding orbital or a sigma-starred
antibonding orbital. rather than to the individual atoms. When simple bonding occurs between
two atoms, the pair of electrons forming the bond occupies an MO that is a mathematical
combination of the wave functions of the atomic orbitals of the two atoms involved. The MO
method originated in the work of Friedrich Hund and Robert S. Mulliken. When atoms combine
to form a molecule, the number of orbitals in the molecule equals the number of orbitals in the
combining atoms. When two very simple atoms, each with one atomic orbital, are combined, two
molecular orbitals are formed. One is a bonding orbital, lower in energy than the atomic orbitals,
and derived from their sum. It is called sigma. The other is an antibonding orbital, higher in
energy than the atomic orbitals, and resulting from their difference. It is called sigma-starred ( s
*). (See the diagram in Figure 1.) The basic idea might best be illustrated by considering
diatomic molecules of hydrogen and helium. The energy diagrams are shown in Figure 2. Each
hydrogen atom has one 1 s electron. In the H 2 molecule the two hydrogen electrons go into the
lowest energy MO available, the sigma orbital. In the case of helium, each helium atom has two
electrons, so the He 2 molecule would have four. Two would go into the lower energy bonding
orbital, but the other two would have to go into the higher energy sigma-starred orbital. Figure
2. Molecular orbital energy diagrams for (a) H 2 showing both electrons in the bonding sigma
MO; and (b) He 2 in which two of the electrons are in the antibonding sigma-starred MO Figure
3. Combination of p atomic orbitals to form (a) sigma MOs by end-to-end interactions or (b) pi
MOs by sideways interaction. Bond Order The bond order for a molecule can be determined as
follows: bond order = ½ (bonding electrons - antibonding electrons). Therefore, the H 2
molecule has a bond order of ½ (2 - 0) = 1. In other words, there is a single bond connecting the
two H atoms in the H 2 molecule. In the case of He 2 , on the other hand, the bond order is ½ (2 -
2) = 0. This means that He 2 is not a stable molecule. Multiple Bonds Double or triple bonds
involve two or three pairs of bonding electrons. Single bonds are always sigma bonds, but in
multiple bonds the first bond is sigma, while any second or third bonds are pi bonds. The overlap
of p orbitals can yield either pi or sigma MOs, as shown in Figure 3. When they overlap end to
end, they form sigma orbitals, but when they overlap side to side, they form pi orbitals. Consider
now the oxygen molecule. The Lewis structure for oxygen is :Ö::Ö: The double bond is
necessary in order to satisfy the octet rule for both oxygen atoms. The measured bond length for
oxygen supports the.
The electrons reach or give off a certain amount of energy and diffe.pdf
The electrons reach or give off a certain amount of energy and different amounts of energy in
the electron changes the color because it jumps to a differrent energy level Hope this helps! rate
lifesaver :)
Solution
The electrons reach or give off a certain amount of energy and different amounts of energy in
the electron changes the color because it jumps to a differrent energy level Hope this helps! rate
lifesaver :).
Weak London disperision and dipole- dipoleModerate Hydrogen bond.pdf
Weak: London disperision and dipole- dipole
Moderate: Hydrogen bonding and ion-dipole
Trust, got these wrong due to inputting the wrong answer above.
Solution
Weak: London disperision and dipole- dipole
Moderate: Hydrogen bonding and ion-dipole
Trust, got these wrong due to inputting the wrong answer above..
var min =0; var max=10; var tab = {}; var name; var score;.pdf
var min =0;
var max=10;
var tab = {};
var name;
var score;
var sum=0;
while(name=prompt(\"Enter name:\")){ // Get userinput for name
do{
score = prompt(\"Enter score\"); // Get Score
tab[name] = score; // Create object with name as key and score as value.
}while(isNaN(score) || !(score>min) || !(score\";
for(item in tab){ // Create table
ui = ui + (\"\"+item+\"\"+tab[item]+\"\");
sum = sum + parseInt(tab[item]);
}
ui = ui + \"\";
document.write(ui); // Display table
document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and
displaying on the page
Solution
var min =0;
var max=10;
var tab = {};
var name;
var score;
var sum=0;
while(name=prompt(\"Enter name:\")){ // Get userinput for name
do{
score = prompt(\"Enter score\"); // Get Score
tab[name] = score; // Create object with name as key and score as value.
}while(isNaN(score) || !(score>min) || !(score\";
for(item in tab){ // Create table
ui = ui + (\"\"+item+\"\"+tab[item]+\"\");
sum = sum + parseInt(tab[item]);
}
ui = ui + \"\";
document.write(ui); // Display table
document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and
displaying on the page.
Unfortunately several cancers are not predictable with simple tests .pdf
Unfortunately several cancers are not predictable with simple tests rather required much
intensive diagnosis. The whole genome sequence in determining the entire DNA in normal cells
as well as in tumor cells provide the facility to compare and pin point the mutations occurring in
the oncogenes or tumor suppressor genes responsible for causing the cancer seems to be a better
approach. But such approaches are highly expensive and require expertise.
Though, the whole genome sequencing is carried out pin pointing of specific driver mutations
responsible for cancer phenotype would be a question. Some critical mutations can be identified
in some cancer genome sequences while others are hard to find among the thousands of
possibilities. The mutations like missense, nonsense frameshift, rearrangements may alter the
coding sequences of any gene must be checked. The regulatory mutations that increase the
transcription of oncogenes or decrease that of tumor suppressor genes are potentially important
but are very difficult to find in a whole genome sequence.
The catalogs of regulatory elements in human genomes provide list of certain potential
sequences to be targeted. But the catalogs are rudimentary or incomplete. Therefore, the whole
genome sequencing is not a panacea that will lead to immediate cures of all cancers. Recent
studies have shown that the whole genome sequences of cells derived from different regions of
same tumor have suggested an important reason for cancer reference. Certain cancers are
heterogenous and cells within the tumor have different genomes. Certain mutations in oncogens
or tumor suppressor genes are common to all cells in the tumor but some are not.
These findings make a sense that accumulation of mutations in clone of cells cause cancer.
Therefore, designing drug targeting specific cells protecting adult stem cells cannot be
accomplished. Thus, effective cancer treatments would be directed against the common
mutations but is difficult to identify the specific mutations without sequencing of genomes of
many cells throughout the tumor. Cancer landscape is a large scale cancer genome sequencing
project (cancer genome atlas) funded by National Institute of Health (NIH). In this project the
whole genomes or exomex of several hundreds of cancer are being characterized. The recurrent
patterns of mutations are subdivided cancers in to groups with probable clinical relevance.
For example, 4 groups of breast cancers have been identified. Almost 178 lung squamous cell
carcinoma’s genomes have been characterized and matched with DNA of normal cells from
same patients. Mutations in certain tumor suppressor genes like p53 and certain oncogenes have
shown relatively high frequencies among these cancers. The mutations of certain cancers
resemble cancers in some other organs. For example, pattern of mutation in breast cancer
resembles many ovarian cancers more than other types of breast cancers. These findings may
help in designing the drugs .
The van t Hoff factor i (named after J. H. van t Hoff) is a meas.pdf
The van \'t Hoff factor i (named after J. H. van \'t Hoff) is a measure of the effect of a solute
upon colligative properties, such as vapor pressure, osmotic pressure and freezing point
depression. The van \'t Hoff factor is the ratio between the actual concentration of particles
produced when the substance is dissolved, and the concentration of a substance as calculated
from its mass. For most non-electrolytes dissolved in water, the van\' t Hoff factor is essentially
1. For most ionic compounds dissolved in water, the van \'t Hoff factor is equal to the number of
discrete ions in a formula unit of the substance. This is true for ideal solutions only, as
occasionally ion pairing occurs in solution. At a given instant a small percentage of the ions are
paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte
solutions. This causes deviation from the van \'t Hoff factor. The deviation for the van \'t Hoff
factor tends to be greatest where the ions have multiple charges.
Relationship with the dissociation degree
The dissociation degree is the fraction of original solute molecules that have dissociated. It is
usually indicated by the Greek symbol . There is a simple relationship between this parameter
and the van \'t Hoff factor. If the solute substance dissociates into n ions, then
i = 1 + (n 1)
For instance, for the following dissociation
KCl K+ + Cl-
As n = 2, we would have that i = 1 +
Physical significance of i
When solute particles associate in solution, i is less than 1. (e.g. ethanoic acid in benzene,
benzoic acid in benzene)
When solute particles dissociate in solution, i is greater than 1. (e.g. sodium chloride in water,
potassium chloride in water, magnesium chloride in water)
When solute particles neither dissociate nor associate in solution, i equals 1. (e.g. Glucose in
water)
The value of i is ; i = the actual number of particles in solution after dissociation ÷ the number
of formula units initially dissolved in solution. Means the number of particles per formula unit of
the solute when a solution is dilute.
Solution
The van \'t Hoff factor i (named after J. H. van \'t Hoff) is a measure of the effect of a solute
upon colligative properties, such as vapor pressure, osmotic pressure and freezing point
depression. The van \'t Hoff factor is the ratio between the actual concentration of particles
produced when the substance is dissolved, and the concentration of a substance as calculated
from its mass. For most non-electrolytes dissolved in water, the van\' t Hoff factor is essentially
1. For most ionic compounds dissolved in water, the van \'t Hoff factor is equal to the number of
discrete ions in a formula unit of the substance. This is true for ideal solutions only, as
occasionally ion pairing occurs in solution. At a given instant a small percentage of the ions are
paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte
solutions. This causes deviation from th.
Polar Bonds and Molecular Shape A polar molecule.pdf
Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net
dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into
determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non-
polar, you must determine both factors. The polarity of the individual bonds in the molecule. The
shape or geometry of the molecule. First, to determine if a given individual bond is polar, you
need to know the electronegativity of the two atoms involved in that bond. To find the
electronegativities of all the elements, look at the periodic table (follow the link below this
answer under Web Links). If the electronegativity of the two atoms has a difference of 0.3 or
less, then the bond is non-polar. If the electronegativity difference is greater that 0.3 but less than
1.7, then the bond is polar. If the two values have a difference greater than 1.7, then the bond is
ionic, which is just very very polar. Once you know which bonds in the molecule are polar and
which are non-polar, you must use the shape of the molecule. You need the shape because two
polar bonds, if oriented correctly can cancel each other out (like two equally strong people
pulling in opposite directions on a rope -- nobody moves). The three possible outcomes: If all
bonds are non-polar, then the whole molecule is non-polar regardless of its shape. If there is
symmetry in the molecule so that the polarity of the bonds cancels out, then the molecule is non-
polar. A common example of this is carbon dioxide, or CO2. The molecule is linear, and its
Lewis dot structure is like this: O=C=O (this doesn\'t include two sets of lone pairs on each
oxygen). The carbon-oxygen bond is a polar bond, but because they are exactly opposed to each
other, the molecule is overall non-polar. Another example of this is CCl4, where each carbon-
chlorine bond is polar, but the molecule is non-polar. Here, how they cancel out isn\'t as obvious,
but they do. CCl4 is a tetrahedral molecule, and the 4 C-Cl polar bonds cancel each other out. If
there are polar bonds but there is no symmetry such that they cancel each other out, the overall
molecule is polar. Water is a typical example of this. The two O-H bonds are oriented in a V-
shape, and so the don\'t cancel out. Similarly, CH3Cl is also polar. It is the same shape as CCl4
(see above), but now it doesn\'t have the same symmetry because there is only one C-Cl bond
and the bonds don\'t cancel out
Solution
Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net
dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into
determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non-
polar, you must determine both factors. The polarity of the individual bonds in the molecule. The
shape or geometry of the molecule. First, to determine if a given individual bond is polar, you
need to know the electronegativ.
The good functioning of an economy depends on the proper functioning.pdf
The good functioning of an economy depends on the proper functioning of the country\'s
financial market and financial institutions. Both these are involved in the financial transactions
that take place in the economy. Financial transactions take place through exchange of funds.
There are two ways in which the exchange of funds can take place. They are: Direct exchange
and Indirect exchange.
Direct exchange takes place when the buyers and the sellers meet each other directly and
exchange their securities.For example: Stocks or bonds
Indirect exchange takes place when the buyers and sellers do not meet each other directly. They
carry out their financial transaction through a financial institution. For example: Bank
The financial market and financial institution have the main function of transferring the funds
from people who have excess funds to people who have shortage of funds. Some functions are
common for both financial markets as well as financial institutions. They both are involved in the
borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller
to the buyer. They both help in providing information to the buyers and sellers about the
financial assets.
However, there are also some functions that are unique to each. Financial markets determine the
price of the financial asset based on the demand and supply forces. Financial institutions help to
lower the transaction costs and also reduce the assymetry of information about financial assets.
The financial institutions also provide specialized services to th buyers and sellers.
Thus, it can be said that in the financial system, the financial market and financial institutions
have an important role to play. Direct exchange takes place through financial market while
indirect exchnage takes place through financial institutions.
Solution
The good functioning of an economy depends on the proper functioning of the country\'s
financial market and financial institutions. Both these are involved in the financial transactions
that take place in the economy. Financial transactions take place through exchange of funds.
There are two ways in which the exchange of funds can take place. They are: Direct exchange
and Indirect exchange.
Direct exchange takes place when the buyers and the sellers meet each other directly and
exchange their securities.For example: Stocks or bonds
Indirect exchange takes place when the buyers and sellers do not meet each other directly. They
carry out their financial transaction through a financial institution. For example: Bank
The financial market and financial institution have the main function of transferring the funds
from people who have excess funds to people who have shortage of funds. Some functions are
common for both financial markets as well as financial institutions. They both are involved in the
borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller
to the buyer. They both help.
The answer is Yes, it is a reduction-oxidation reaction2 HNO2 + 2.pdf
The answer is: Yes, it is a reduction-oxidation reaction
2 HNO2 + 2 H+ + 3 I- => 2 NO + I3- + 2 H2O
Oxidation state of N in HNO2 = +3
Oxidation state of N in NO = +2
Thus HNO2 is reduced to NO as oxidation state of N decreases.
Oxidation state of I in I- = -1
Oxidation state of I in I3- = -1/3
Thus I- is oxidized to I3- as oxidation state of I increases.
Solution
The answer is: Yes, it is a reduction-oxidation reaction
2 HNO2 + 2 H+ + 3 I- => 2 NO + I3- + 2 H2O
Oxidation state of N in HNO2 = +3
Oxidation state of N in NO = +2
Thus HNO2 is reduced to NO as oxidation state of N decreases.
Oxidation state of I in I- = -1
Oxidation state of I in I3- = -1/3
Thus I- is oxidized to I3- as oxidation state of I increases..
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmer.pdf
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a
large quadriped with unusual posture, heavy back, short forelimbs, spiked tail.
Integumentary system is that which is composed of skin and its appendages. Its primary function
is to protect the body from external injury, temperature regulation, cushioning and self defense.
Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them
measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in
selfdefense and regulation of body temperature. The plates were supplied with blood through the
blood vessels and the air flow through the plates is hypothesized to have regulated the body
temperature. The plates were in double row as per findings in late 1960\'s.
Lymphatic system: There is little evidence to clearly understand the lymphatic system although it
is hypothesized that they had a 4 chambered heart and were warmblooded with a well
functioning circulatory system. They had higher aerobic capacity and higher pressure or higher
gas exchange circulatory system.
Solution
Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a
large quadriped with unusual posture, heavy back, short forelimbs, spiked tail.
Integumentary system is that which is composed of skin and its appendages. Its primary function
is to protect the body from external injury, temperature regulation, cushioning and self defense.
Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them
measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in
selfdefense and regulation of body temperature. The plates were supplied with blood through the
blood vessels and the air flow through the plates is hypothesized to have regulated the body
temperature. The plates were in double row as per findings in late 1960\'s.
Lymphatic system: There is little evidence to clearly understand the lymphatic system although it
is hypothesized that they had a 4 chambered heart and were warmblooded with a well
functioning circulatory system. They had higher aerobic capacity and higher pressure or higher
gas exchange circulatory system..
The answer is d- the hydrogen bonds in iceThe high heat of fusion.pdf
The answer is: d- the hydrogen bonds in ice
The high heat of fusion indicates that strong intermolecular forces are present between the H2O
molecules in the form of hydrogen bonds.
Solution
The answer is: d- the hydrogen bonds in ice
The high heat of fusion indicates that strong intermolecular forces are present between the H2O
molecules in the form of hydrogen bonds..
SolutionOption (a) will qualify as a database as dictionary is a .pdf
Solution
:
Option (a) will qualify as a database as dictionary is a database of different words with their
meanings stored in it.
Option (b) will qualify as a database as telephone directory contains data regarding people like
their name, address, phone number.
Option (c) will qualify as a database as rolodex is a file that contains cards holding information
about people like names, addresses and phone numbers.
Option (d) will not qualify as a database as textbook index only points to the topics in the book,
it itself does not store any data in it related to the topic or anything..
Tests of Controls in an Audit of Internal Control. The objective of .pdf
Tests of Controls in an Audit of Internal Control. The objective of the tests of controls in an audit
of internal control over financial reporting is to obtain evidence about the effectiveness of
controls to support the auditor\'s opinion on the company\'s internal control over financial
reporting. The auditor\'s opinion relates to the effectiveness of the company\'s internal control
over financial reporting as of a point in time and taken as a whole.
The general standards are applicable to an audit of internal control over financial reporting.
Those standards require technical training and proficiency as an auditor, independence, and the
exercise of due professional care, including professional skepticism. This standard establishes
the fieldwork and reporting standards applicable to an audit of internal control over financial
reporting. The auditor should use the same suitable, recognized control framework to perform his
or her audit of internal control over financial reporting as management uses for its annual
evaluation of the effectiveness of the company\'s internal control over financial reporting.
To express an opinion on internal control over financial reporting as of a point in time, the
auditor should obtain evidence that internal control over financial reporting has operated
effectively for a sufficient period of time, which may be less than the entire period (ordinarily
one year) covered by the company\'s financial statements. To express an opinion on internal
control over financial reporting taken as a whole, the auditor must obtain evidence about the
effectiveness of selected controls over all relevant assertions. This requires that the auditor test
the design and operating effectiveness of controls
When concluding on the effectiveness of internal control over financial reporting for purposes of
expressing an opinion on internal control over financial reporting, the auditor should incorporate
the results of any additional tests of controls performed to achieve the objective related to
expressing an opinion on the financial statements, as discussed in the following section.
Tests of Controls in an Audit of Financial Statements.
To express an opinion on the financial statements, the auditor ordinarily performs tests of
controls and substantive procedures. The objective of the tests of controls the auditor performs
for this purpose is to assess control risk. To assess control risk for specific financial statement
assertions at less than the maximum, the auditor is required to obtain evidence that the relevant
controls operated effectively during the entire period upon which the auditor plans to place
reliance on those controls. However, the auditor is not required to assess control risk at less than
the maximum for all relevant assertions and, for a variety of reasons, the auditor may choose not
to do so.
When concluding on the effectiveness of controls for the purpose of assessing control risk, the
auditor also should evaluate the .
Software Quality Assurance is important because Quality in the softw.pdf
Software Quality Assurance is important because Quality in the software is one of the pivotal
aspects of a software development company. quality assurance starts from the beginning of a
software project, right from the analysis phase. Software Quality Assurance (SQA) is defined as
a well planned and systematic approach to evaluate the quality of software. It checks the
adherence to software product standards, processes, and procedures. SQA includes the
systematic process of assuring that standards and procedures are established and are followed
throughout the software development life cycle and test cycle as well. The compliance of the
built with agreed-upon standards and procedures is evaluated through process monitoring,
product evaluation, project management etc.
Software quality assurance methods
The main reason of involving quality in software assurance in the process of software product
development is to make sure that the final product built is as per the requirement specification
and comply with the standards.
SQA encompasses different activities, like :
1) Requirement analysis and definition
2) Design architecture and description
3) Coding and logic analysis
4) Change and configuration management
5) Testing and standard compliance
6) Release management and Release Control
Different Functionalities of Quality Assurance are the following:
1. Maintaining the quality of the project as per business requirements and Specifications.
2. Defect Prevention. And formal methods for other defect prevention techniques.
3. Defect Reduction
4. Inspection, formal and informal reviews: Direct fault detection and removal without
executing the project scenario.
5. Testing the project for Failure observation and bug removal.
6. Risk identification.
7. Defect tracking techniques and methods
8. Software fault tolerance.
9. Concluding Remarks and maintaining reports.
Software quality can be quantified into two major groups such as:
1. Software Functional Quality: It basically shows how well the software product conforms to
the basic design, based on functional requirements. The attribute can also be described as the
fitness for purpose of a software.
2. Software Structural Quality: It reflects to how well the project meets the non-functional
requirements such as usability, accessibility and security that helps in proper the delivery of the
functional requirements. It basically defines the degree of correctness of the product.
.
Solution
Software Quality Assurance is important because Quality in the software is one of the pivotal
aspects of a software development company. quality assurance starts from the beginning of a
software project, right from the analysis phase. Software Quality Assurance (SQA) is defined as
a well planned and systematic approach to evaluate the quality of software. It checks the
adherence to software product standards, processes, and procedures. SQA includes the
systematic process of assuring that standards and procedures are establis.
Oxygen rich blood leaves the lungs. It travels via the pulmonary vei.pdf
Oxygen rich blood leaves the lungs. It travels via the pulmonary vein to the heart.
It travels through the aorta to the iliac artery.
It then travels through the femoral artery to lower leg capillaries.
It leaves the lower leg capillaries through the femoral vein
It travels through the iliac vein to the inferior vena cava to the heart
Carbon di oxide rich blood reaches the lungs through the pulmonary artery.
Solution
Oxygen rich blood leaves the lungs. It travels via the pulmonary vein to the heart.
It travels through the aorta to the iliac artery.
It then travels through the femoral artery to lower leg capillaries.
It leaves the lower leg capillaries through the femoral vein
It travels through the iliac vein to the inferior vena cava to the heart
Carbon di oxide rich blood reaches the lungs through the pulmonary artery..
Normal Concept For colour vision, our normal human eyes have 3 con.pdf
Normal Concept : For colour vision, our normal human eyes have 3 cones and photoreceptors.
L cone - sensitive to long wavelength of yellowish green light
M cone - middle wavelength of green light
s cone - short wavelength to bluish violet light
Protonapia - complete absence of red cone photoreceptors.He felt difficulty in identifying the red
colour
Deuteranopia - complete absence of green cone photoreceptors.He felt difficulty in identifying
green colour
Red green colour discrimination is called red- green dichromat vision blindness.He can able to
see yellow circles sometimes in place of red or green colour, it shows the unstable working
colour vision receptors.
If one colour dominates the vision, the other will be absent,for instance, if red hue is dominant, it
strikes the receptors of green colour and vice versa.
Red green dichromat friend will have reduction view in the normally appearing red - green
colour.The hues, he perceives was not like normal, it looks fade olive in colour.This is due to
specific absence of photoreceptors.
If he able to perceive colours, either one of the phoreceptors work in exitatory state and the other
in inhibitory state.This is called opponent colour theory.
Of all that, he might have some degeneration or improper formation issues with myelin
sheath(covering the axon of a neuron) which is responsible for colour determination.
Solution
Normal Concept : For colour vision, our normal human eyes have 3 cones and photoreceptors.
L cone - sensitive to long wavelength of yellowish green light
M cone - middle wavelength of green light
s cone - short wavelength to bluish violet light
Protonapia - complete absence of red cone photoreceptors.He felt difficulty in identifying the red
colour
Deuteranopia - complete absence of green cone photoreceptors.He felt difficulty in identifying
green colour
Red green colour discrimination is called red- green dichromat vision blindness.He can able to
see yellow circles sometimes in place of red or green colour, it shows the unstable working
colour vision receptors.
If one colour dominates the vision, the other will be absent,for instance, if red hue is dominant, it
strikes the receptors of green colour and vice versa.
Red green dichromat friend will have reduction view in the normally appearing red - green
colour.The hues, he perceives was not like normal, it looks fade olive in colour.This is due to
specific absence of photoreceptors.
If he able to perceive colours, either one of the phoreceptors work in exitatory state and the other
in inhibitory state.This is called opponent colour theory.
Of all that, he might have some degeneration or improper formation issues with myelin
sheath(covering the axon of a neuron) which is responsible for colour determination..
LinkedListOperations.java-class Node{ public int item; .pdf
LinkedListOperations.java:-
class Node
{
public int item;
public Node next;
public Node(int val)
{
item = val;
}
public void displayNode()
{
System.out.println(\"[\" + item + \"] \");
}
}
class LinkedList
{
private Node first;
public LinkedList()
{
first = null;
}
public boolean isEmpty()
{
return (first==null);
}
public void insert(int val)//inserts at beginning of list
{
Node newNode = new Node(val);
newNode.next = first;
first = newNode;
}
public Node delete()//deletes at beginning of list
{
Node temp = first;
first = first.next;
return temp;
}
public void display()
{
System.out.println(\"List items from first to last :\");
Node current = first;
while(current != null)
{
current.displayNode();
current = current.next;
}
System.out.println(\"\");
}
public Node search(int val)
{
Node current = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
current = current.next;
}
return current;
}
public Node delete(int val)
{
Node current = first;
Node previous = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
{
previous = current;
current = current.next;
}
}
if(current == first)
first = first.next;
else
previous.next = current.next;
return current;
}
}
class LinkedListOperations
{
public static void main(String[] args)
{
LinkedList object = new LinkedList();
object.insert(10);
object.insert(20);
object.insert(30);
object.display();
while( !object.isEmpty() )
{
Node member = object.delete();
System.out.print(\"Deleted \");
member.displayNode();
System.out.println(\"\");
}
object.display();
object.insert(40);
object.insert(50);
object.insert(60);
object.display();
Node objecttosearch = object.search(50);
if( objecttosearch != null)
System.out.println(\"Found Node : \" + objecttosearch.item);
else
System.out.println(\"Cannot locate the node\");
Node objecttodelete = object.delete(50);
if( objecttodelete != null )
System.out.println(\"Deleted node : \" + objecttodelete.item);
else
System.out.println(\"Cannot delete the node\");
object.display();
}
}
Solution
LinkedListOperations.java:-
class Node
{
public int item;
public Node next;
public Node(int val)
{
item = val;
}
public void displayNode()
{
System.out.println(\"[\" + item + \"] \");
}
}
class LinkedList
{
private Node first;
public LinkedList()
{
first = null;
}
public boolean isEmpty()
{
return (first==null);
}
public void insert(int val)//inserts at beginning of list
{
Node newNode = new Node(val);
newNode.next = first;
first = newNode;
}
public Node delete()//deletes at beginning of list
{
Node temp = first;
first = first.next;
return temp;
}
public void display()
{
System.out.println(\"List items from first to last :\");
Node current = first;
while(current != null)
{
current.displayNode();
current = current.next;
}
System.out.println(\"\");
}
public Node search(int val)
{
Node current = first;
while(current.item != val)
{
if(current.next == null)
return null;
else
current = current.next;
}
return current;
}
public Node delete(int val)
{
N.
In Keynesian economic theory, a factor that quantifies the change in.pdf
In Keynesian economic theory, a factor that quantifies the change in total income as compared to
the injection of capital deposits or investments which originally fueled the growth. It is usually
used as a measurement of the effects of government spending on income, and it can be calculated
as one divided by the marginal propensity to save.
Average propensity to consume, or APC, is the ratio of total consumption to total disposable
income, whereas marginal propensity to consume, or MPC, is the ratio of change in consumption
to the change in disposable income.
Y= C + S
That is, disposable income (income after tax) = Consumption + Savings
Think of Bob the Builder who earns a gross income of $1000 per week. After tax (which he
legally has no choice but to pay) he brings in $850 dollars. His DISPOSABLE Income is $850.
He then has two very simple choices- to save it all (i.e. place it in a financial institution such as a
credit union, bank, super fund) or to spend it (consume as economists would say). Realistically,
he will probably save a fraction and spend a fraction.
The fraction of his disposable income (Y) in which he decides to save (S) is his APS- Average
Propensity to Save
Therefore- Amount Saved/Disposable Income or S/Y.
Very simple maths.
Similarly, the amount of disposable income that he uses on consumption (spends) is his Average
Propensity to Consume-
Consumption/Disposable Income or C/Y.
The APC and APS are expressed as fractions of Disposable Income- for example 0.3 and 0.7.
As he only has two choices, whether to save or spend, the APC and APS must always sum to
give 1.
They vary with individuals, particularly with those of different disposable incomes.
Surprisingly, the wealthy has a higher APS and a lower APC compared to the poor. This is
because APS/APC are proportional things; poor people will spend a larger amount of their
income on bills and just general living expenses leaving little to no room for saving. As you get
wealthy, your consumption will rise however you can afford to live comfortably plus have a lot
more money left over for savings, hence their APS is higher.
Extending this, economists have come up with the Marginal Propensity (Tendency) to
Save/Consume. This looks at the tendency to save/spend each extra dollar of income that is
earned.
MPC is the proportion of the next dollar in which someone earns that is spent, and similarly
MPS is the proportion that is saved. Eg- I give you one dollar, you spend 80 cents and keep 20
cents, your MPC is 0.8 and your MPS is 0.2. As you can see-
MPC + MPS also add to 1.
As pointed out above, this concept must be clearly understood, as it is the fundamental of
Keynes demand theory and the multiplier which you look at to explain economic growth in year
12.
As for the difference between savings and investment. Savings are placed away in a financial
institution, the incentive to do so is that they earn interest and hence ‘grow’. If you look at the
circular flow of income model (if yo.
import java.util.LinkedList; import java.util.Random; import jav.pdf
import java.util.LinkedList;
import java.util.Random;
import java.util.Scanner;
/**
* author
*
*/
public class LinkedListTest {
/**
* param args
*/
public static void main(String[] args) {
int M, N, Lower, Upper;
Scanner scanner = null;
try {
LinkedList list1 = new LinkedList();
LinkedList list2 = new LinkedList();
LinkedList list3 = new LinkedList();
Random random = new Random();
scanner = new Scanner(System.in);
System.out.print(\"Enter the Lower and Upper bounds:\");
Lower = scanner.nextInt();
Upper = scanner.nextInt();
System.out.print(\"Enter the N value to Insert list1:\");
N = scanner.nextInt();
System.out.print(\"Enter the M value to Insert list2:\");
M = scanner.nextInt();
for (int i = 1; i <= N; i++) {
list1.add(random.nextInt((Upper - Lower) + 1) + Lower);
}
for (int i = 1; i <= M; i++) {
list2.add(random.nextInt((Upper - Lower) + 1) + Lower);
}
System.out.println(\"list1: \" + list1);
System.out.println(\"list2: \" + list2);
System.out.println(\"Minimum elelment in list1 :\" + getMin(list1));
System.out.println(\"Minimum elelment in list2 :\" + getMin(list2));
System.out
.println(\"Count Divisibles by 3 and 5 elelment in list1 :\"
+ countDivisibles(list1));
System.out
.println(\"Count Divisibles by 3 and 5 elelment in list2 :\"
+ countDivisibles(list2));
System.out.println(\"Common Elements From list1 and list2:\"
+ getCommonList(list1, list2));
} catch (Exception e) {
// TODO: handle exception
}
}
/**
* param list
* return
*/
public static int getMin(LinkedList list) {
int min = list.get(0);
for (Integer integer : list) {
if (min > integer) {
min = integer;
}
}
return min;
}
/**
* param list
* return
*/
public static int countDivisibles(LinkedList list) {
int count = 0;
for (Integer integer : list) {
if (integer % 3 == 0 && integer % 5 == 0) {
count++;
}
}
return count;
}
/**
* param list1
* param list2
* return
*/
public static LinkedList getCommonList(LinkedList list1,
LinkedList list2) {
LinkedList list = new LinkedList();
LinkedList listCommon = new LinkedList();
list = (list1.size() > list2.size()) ? list1 : list2;
for (int i = 0; i < list.size(); i++) {
Integer number = list.get(i);
if (list2.contains(number) && list1.contains(number)) {
listCommon.add(number);
}
}
return listCommon;
}
}
OUTPUT:
Enter the Lower and Upper bounds:1 10
Enter the N value to Insert list1:10
Enter the M value to Insert list2:20
list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5]
list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2]
Minimum elelment in list1 :1
Minimum elelment in list2 :1
Count Divisibles by 3 and 5 elelment in list1 :0
Count Divisibles by 3 and 5 elelment in list2 :0
Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6]
Solution
import java.util.LinkedList;
import java.util.Random;
import java.util.Scanner;
/**
* author
*
*/
public class LinkedListTest {
/**
* param args
*/
public static void main(String[] args) {
int M, N, Lower, Upper;
Scanner scanner = null;
try {
LinkedList list1 = new LinkedList();
L.
Hi,Please find the answers as followsPart AExpec.pdf
Hi,
Please find the answers as follows:
Part A:
Expected Return = .1*-10 + .2*0 + .4*10 + .2*20 + .1*30 = 10%
Part B:
Variance = .1*(-10 - 10)^2 + .2*(0 - 10)^2 + .4*(10 - 10)^2 + .2*(20 - 10)^2 + .1*(30 -10)^2 =
120
Standard Deviation = (120)^1/2 = 10.95%
Part C:
Coeffcient of Variation = Standard Deviation/Expected Return = 10.95/10 = 1.095
Part D:
Stand Alone Risk is measured by th Standard Deviation or Coefficient of variation (CV) of
returns.
Part E:
Stand Alone Risk reflects the uncertainty about the cash flows. It is relevant when the the
undiversified risk of an individual asset is required to be measured.
Thanks.
Solution
Hi,
Please find the answers as follows:
Part A:
Expected Return = .1*-10 + .2*0 + .4*10 + .2*20 + .1*30 = 10%
Part B:
Variance = .1*(-10 - 10)^2 + .2*(0 - 10)^2 + .4*(10 - 10)^2 + .2*(20 - 10)^2 + .1*(30 -10)^2 =
120
Standard Deviation = (120)^1/2 = 10.95%
Part C:
Coeffcient of Variation = Standard Deviation/Expected Return = 10.95/10 = 1.095
Part D:
Stand Alone Risk is measured by th Standard Deviation or Coefficient of variation (CV) of
returns.
Part E:
Stand Alone Risk reflects the uncertainty about the cash flows. It is relevant when the the
undiversified risk of an individual asset is required to be measured.
Thanks..
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