Which account below is not a subdivision of owners equityAccumu.pdf
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Which account below is not a subdivision of owner\'s equity? Accumulated Comprehensive Income. Bond Sinking fund. Addition Paid-in-Capital. Retained Earnings. Accumulated Comprehensive Income. Bond Sinking fund. Addition Paid-in-Capital. Retained Earnings. Solution Bonk Sinking Fund not a subdivision of owner\'s equity Parts of Stockholder Equity are Capital Stock, Addition Paid-in-Capital, Retained Earnings., Accumulated Comprehensive Income, Treasury Stock, NONCONTROLLING INTEREST (MINORITY INTEREST).
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Describe 3 functions of the cellular membrane and how are these func.pdf
Describe 3 functions of the cellular membrane and how are these functions accomplished by the
chemical make up of the membrane.
Solution
The main functions of the cell membrane are:to maintain the physical integrity of the cell,that is
to mechanically enclose the contents of the cell,also to control the movement of particles e.g.
ions or molecules,into and out of the cell.It also isolates the cytoplasm from the external
environment.
The hydrophilic head and hydrophobic tails of the phospholipid molecules that form the
phospholipid bilayer of the cell membrane forming its physical/mechanical structure.The ion
channels responsible for the selective permeability of the cell membrane.Membrane pumps that
enable active transport of specific molecules across the cell membrane.Carrier proteins and
receptor proteins that transport specific substance across the cell membrane and receive
(chemical) signals from outside the cell that tell the cell to take a specific
action,respectively.Cholesterol molecules and carbohydrate chains that support the structure of
the membrane and its attachment to other tissues,respectively.Surface protein markers that
identify the cell.Membrane proteins acts as a recptor to combine with specific substance such as
hormones while some form narrow passage ways or channels through which various molecules
and ions can cross the cell membrane.Enzymatic proteins carry out metabolic reactions..
Discuss the three main types of intellectual capitalSolutionth.pdf
Discuss the three main types of intellectual capital
Solution
the main types of intellectual capitals are: Copy rights, patents, trade marks and trade secrets.
copy rights: gives the right to the inventor or author or a book or artistick work to his invention.
book publishing and movie distribution comes under this example.
patents: gives the inventor to sell the right to use his invention for commercial and for that he
will charge something from the user. the user will get licence to use it and will be benefited.
manufacturing of McDonald and Dominos pizzas comes under this action.
trademark: is facilitate the seller or producer to differentitate his prodcut from his competitors
product and will be benefited. is like giving licence to sell the products in a region for a
particular period. the producer will allows the seller to sell the product and will get some benefit..
Describer in detail the innate immune response to both an intercellu.pdf
Describer in detail the innate immune response to both an intercellular infection and an
extracellular infection.
Solution
Innate immunity is the type of immunity which is developed by birth..! The defense against the
extracellur infection are from the barrier which are present on the body like skin, hairs in the
nose, tears which contain lisozymes in them, phagocyotic cells like neutrophils etc which kills
the outer cells..!
Then the intra cellular immunity like cytokines, inflammatory barriers like histamine which
produces inflammation due to the infection, complement protein...
Describe the difference between tropic and non-tropic hormones.S.pdf
Describe the difference between tropic and non-tropic hormones.
Solution
TROPIC HORMONES : Tropic hormones are those hormones which act on other endocrine
hormonesthey are named as respective tropin.
Ex: adreno cartico tropic hormone (acth) is a tropic hormone which acts on the cells of adrenal
cortex and stimulate the synthesis of glucocorticoid hormones.similarly,Thyroid stimulating
Hormone (TSH) and Leutinising hormone(LH) and Follicle stimulating hormone (FSH) also act
on specific tissues or hormones and stimulate the synthesis of other hormones.
NON-TROPIC HORMONES : these are the hormones that act directly on the specialised tissues
or cells and induce the effect.they play a vital role in many metabolic functions.
Ex: cartisol is a non topic hormone which is a result of many chain reactions that are induced due
to ACTH (tropic hormone).testosterone,estrogen,vassopressin,epinep[hrine and non-epinephrine
are some of the examples of non-tropic hormones.
both tropic and non-tropic hormones are crucial in proper endocrine functions.
for example if ACTH is inhibited, release of cortisol is inhibited due to interruption of chain
reactions.
Most of the hormones released from endocrine glands like pancreas,gonads,pituatary galnds are
non-tropic..
C++ ProgrammingGivenWrite a closest Pair FunctionDeliverable.pdf
C++ Programming
Given:
Write a closest Pair Function:
Deliverables
ClosestPair.h
ClosestPair.cpp
Solution
code:
Pair Closest_Pair( Point P[ ], int n)
{
float min = FLT_MAX;
int indexone=0;
int indextwo=0;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
{
min = dist(P[i], P[j]);
indexone=i;
indextwo=j;
}
Struct Pair st ;
st.indexPointOne = indexone;
st.indexPointTwo = indextwo;
return min;
}
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}.
Can someone explain to me what the learning curve equation (Tn=T1n^.pdf
Can someone explain to me what the learning curve equation (Tn=T1*n^b) is and how you get
this equation? What is the relationship between the variables?
Solution
The learning curve equation is defined as follows : -
Tn = T1 nb
Tn is the production time for the nth unit
T1 is the production time for the first unit
n = number of units
b is slope of the learning curve and is defined as b = log(learning rate) / log2
Lr , the learning rate , is expressed as a decimal , eg ., Lr = 80% = 0.8.
Both influenza virus and (IFV), and Measles Virus (MV) are envel.pdf
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Solution
Influenza virus is an enveloped, single-stranded negative-sense RNA virus belonging to the
Orthomyxoviridae family. Influenza A virions have three integral membrane proteins: an ion
channel protein (M2) and two glycoproteins, hemagglutinin (HA), required for entry into host
cells, and neuraminidase (NA), involved in the release of progeny virions from the host cell.
Underneath the membrane, most virions have a layer of M1 matrix protein enclosing the viral
ribonucleoprotein complexes (RNPs). On exposure to low pH of around 5 the viral RNP\'s are
released. It is followed by the M2-mediated pumping of protons into the virion, this induces the
change in the intermolecular interactions of the M1 matrix protien and its binding to the viral
membrane, this leads to membrane fusion and conformational changes in the hemagglutinin
causing them to be inactive because this conformational change causes exposure of a
hydrophobic part of the molecule. Fusion of the virion with target membrane occurs at low pH
where a transition from a relaxed conformation to an active conformation of HA occurs, this
transition at low pH in absence of a target membrane leads to inactivation.
The measles virus is a paramyxovirus, is 120–250 nm in diameter, with a core of single-stranded
RNA. It has Two membrane envelope proteins are important in pathogenesis. They are the F
(fusion) protein, responsible for fusion of virus and host cell membranes, viral penetration, and
hemolysis, and the H (hemagglutinin) protein, which is responsible for adsorption of virus to
cells. measles virus could not be inactivated at pH range of 5.5 to 9.0. It is inactivated at pH of
<3 or >7Attachmentnt to host cell happens through binding of virus to the cellular receptor and
fusion of the viral and cellular membranes. MV enters cell in a pH independent manner and fuses
directly to the plasma membrane. At around pH 5 there is no conformational change that occurs
in the multifaceted strucuture of MV..
All of the following are characteristic to fatigue fracture surfaces.pdf
All of the following are characteristic to fatigue fracture surfaces, except A. Clamshell marks. B.
Beach marks. C. Dimples. D. Striations.
Solution
STRIATIONS
Striations means a series of ridges, furrows or linear marks, and is used in several ways: Glacial
striation. Striation (geology), a striation as a result of a geological fault..
Recommended
Describe 3 functions of the cellular membrane and how are these func.pdf
Describe 3 functions of the cellular membrane and how are these functions accomplished by the
chemical make up of the membrane.
Solution
The main functions of the cell membrane are:to maintain the physical integrity of the cell,that is
to mechanically enclose the contents of the cell,also to control the movement of particles e.g.
ions or molecules,into and out of the cell.It also isolates the cytoplasm from the external
environment.
The hydrophilic head and hydrophobic tails of the phospholipid molecules that form the
phospholipid bilayer of the cell membrane forming its physical/mechanical structure.The ion
channels responsible for the selective permeability of the cell membrane.Membrane pumps that
enable active transport of specific molecules across the cell membrane.Carrier proteins and
receptor proteins that transport specific substance across the cell membrane and receive
(chemical) signals from outside the cell that tell the cell to take a specific
action,respectively.Cholesterol molecules and carbohydrate chains that support the structure of
the membrane and its attachment to other tissues,respectively.Surface protein markers that
identify the cell.Membrane proteins acts as a recptor to combine with specific substance such as
hormones while some form narrow passage ways or channels through which various molecules
and ions can cross the cell membrane.Enzymatic proteins carry out metabolic reactions..
Discuss the three main types of intellectual capitalSolutionth.pdf
Discuss the three main types of intellectual capital
Solution
the main types of intellectual capitals are: Copy rights, patents, trade marks and trade secrets.
copy rights: gives the right to the inventor or author or a book or artistick work to his invention.
book publishing and movie distribution comes under this example.
patents: gives the inventor to sell the right to use his invention for commercial and for that he
will charge something from the user. the user will get licence to use it and will be benefited.
manufacturing of McDonald and Dominos pizzas comes under this action.
trademark: is facilitate the seller or producer to differentitate his prodcut from his competitors
product and will be benefited. is like giving licence to sell the products in a region for a
particular period. the producer will allows the seller to sell the product and will get some benefit..
Describer in detail the innate immune response to both an intercellu.pdf
Describer in detail the innate immune response to both an intercellular infection and an
extracellular infection.
Solution
Innate immunity is the type of immunity which is developed by birth..! The defense against the
extracellur infection are from the barrier which are present on the body like skin, hairs in the
nose, tears which contain lisozymes in them, phagocyotic cells like neutrophils etc which kills
the outer cells..!
Then the intra cellular immunity like cytokines, inflammatory barriers like histamine which
produces inflammation due to the infection, complement protein...
Describe the difference between tropic and non-tropic hormones.S.pdf
Describe the difference between tropic and non-tropic hormones.
Solution
TROPIC HORMONES : Tropic hormones are those hormones which act on other endocrine
hormonesthey are named as respective tropin.
Ex: adreno cartico tropic hormone (acth) is a tropic hormone which acts on the cells of adrenal
cortex and stimulate the synthesis of glucocorticoid hormones.similarly,Thyroid stimulating
Hormone (TSH) and Leutinising hormone(LH) and Follicle stimulating hormone (FSH) also act
on specific tissues or hormones and stimulate the synthesis of other hormones.
NON-TROPIC HORMONES : these are the hormones that act directly on the specialised tissues
or cells and induce the effect.they play a vital role in many metabolic functions.
Ex: cartisol is a non topic hormone which is a result of many chain reactions that are induced due
to ACTH (tropic hormone).testosterone,estrogen,vassopressin,epinep[hrine and non-epinephrine
are some of the examples of non-tropic hormones.
both tropic and non-tropic hormones are crucial in proper endocrine functions.
for example if ACTH is inhibited, release of cortisol is inhibited due to interruption of chain
reactions.
Most of the hormones released from endocrine glands like pancreas,gonads,pituatary galnds are
non-tropic..
C++ ProgrammingGivenWrite a closest Pair FunctionDeliverable.pdf
C++ Programming
Given:
Write a closest Pair Function:
Deliverables
ClosestPair.h
ClosestPair.cpp
Solution
code:
Pair Closest_Pair( Point P[ ], int n)
{
float min = FLT_MAX;
int indexone=0;
int indextwo=0;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
{
min = dist(P[i], P[j]);
indexone=i;
indextwo=j;
}
Struct Pair st ;
st.indexPointOne = indexone;
st.indexPointTwo = indextwo;
return min;
}
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}.
Can someone explain to me what the learning curve equation (Tn=T1n^.pdf
Can someone explain to me what the learning curve equation (Tn=T1*n^b) is and how you get
this equation? What is the relationship between the variables?
Solution
The learning curve equation is defined as follows : -
Tn = T1 nb
Tn is the production time for the nth unit
T1 is the production time for the first unit
n = number of units
b is slope of the learning curve and is defined as b = log(learning rate) / log2
Lr , the learning rate , is expressed as a decimal , eg ., Lr = 80% = 0.8.
Both influenza virus and (IFV), and Measles Virus (MV) are envel.pdf
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
two viruses?
Solution
Influenza virus is an enveloped, single-stranded negative-sense RNA virus belonging to the
Orthomyxoviridae family. Influenza A virions have three integral membrane proteins: an ion
channel protein (M2) and two glycoproteins, hemagglutinin (HA), required for entry into host
cells, and neuraminidase (NA), involved in the release of progeny virions from the host cell.
Underneath the membrane, most virions have a layer of M1 matrix protein enclosing the viral
ribonucleoprotein complexes (RNPs). On exposure to low pH of around 5 the viral RNP\'s are
released. It is followed by the M2-mediated pumping of protons into the virion, this induces the
change in the intermolecular interactions of the M1 matrix protien and its binding to the viral
membrane, this leads to membrane fusion and conformational changes in the hemagglutinin
causing them to be inactive because this conformational change causes exposure of a
hydrophobic part of the molecule. Fusion of the virion with target membrane occurs at low pH
where a transition from a relaxed conformation to an active conformation of HA occurs, this
transition at low pH in absence of a target membrane leads to inactivation.
The measles virus is a paramyxovirus, is 120–250 nm in diameter, with a core of single-stranded
RNA. It has Two membrane envelope proteins are important in pathogenesis. They are the F
(fusion) protein, responsible for fusion of virus and host cell membranes, viral penetration, and
hemolysis, and the H (hemagglutinin) protein, which is responsible for adsorption of virus to
cells. measles virus could not be inactivated at pH range of 5.5 to 9.0. It is inactivated at pH of
<3 or >7Attachmentnt to host cell happens through binding of virus to the cellular receptor and
fusion of the viral and cellular membranes. MV enters cell in a pH independent manner and fuses
directly to the plasma membrane. At around pH 5 there is no conformational change that occurs
in the multifaceted strucuture of MV..
All of the following are characteristic to fatigue fracture surfaces.pdf
All of the following are characteristic to fatigue fracture surfaces, except A. Clamshell marks. B.
Beach marks. C. Dimples. D. Striations.
Solution
STRIATIONS
Striations means a series of ridges, furrows or linear marks, and is used in several ways: Glacial
striation. Striation (geology), a striation as a result of a geological fault..
A cancer biology topic Describte how ChIP (chromatin immunoprecipit.pdf
A cancer biology topic: Describte how ChIP (chromatin immunoprecipitation), DNA
methylation array, and gene expression arrays are performed. What are the similarities?
Solution
DNA methylation is main epigenetic mechanism of gene expression regulation.Alteration in this
leads to tumor. Various stimulant factors stimulate DNA methylation leads to generation of
canceroua cells.
DNA methylation process do down regulation of gene.
Methylation of CpG sites blocks the binding of regulatory proteins resulting transcription
modulation.
In CHIP DNA methylation stauninig of CHIP done by various stains. Stained CHIP show
intensities of methylated and unmethylated bead types. Data analysis gives florescence intensity
ratio between methylated and unmethylated group.
Value of ratio is 0, 1, 0.5 represents non methylation, total methylation and one copy methylated
other not respectively.
DNA methylation measure by following other method also
Bisulfite treatment
Whole genomic DNA amplification
Hybridization and single base extension
In Gene expression two following procees take important part:-
TRANSCRIPTION:- DNA to RNA by RNA polymerase which use DNA as template and
produce RNA.
TRANSLATION:- RNA to Protein translation (tRNA charged with amino acid enter in ribosome
with correct mRNA triplet , ribisome then add aminoacid.
In cancer, gene promoter contain CpG island with numerous CpG sites. When CpG
sitesethylated gene become silence.
Thanks...
You get a call from a customer who reinstalled Windows on his comput.pdf
You get a call from a customer who reinstalled Windows on his computer after a crash.
Fortunately, he had recently backed up all this data to an external USB drive and was able to
restore everything on his NTFS-formatted file system. However, he had a number of encrypted
files that he had backed up along with the rest of his data that he can no longer access. What
happened?
A. NTFS-encrypted files cannot be backed up, and when the computer crashed, those files were
lost forever. When the customer tried to restore these files, only corrupted versions were
restored.
B. NTFS-encrypted files are not restored with the original encryption key used before the backup
process and thus become inaccessible.
C. When the customer reinstalled Windows, even though he used his old username, the computer
assigned him a completely different security identifier (SID). The restored NTFS-encrypted files
are looking for his original SID.
D. When the customer reinstalled Windows, he didn’t use his original username. The restored
NTFS-encrypted files are looking for the original username.
Solution
C. NTFS encryption used the customer’s SID to associate with the encrypted file. Now that the
customer reinstalled Windows, a completely new SID was assigned to the customer. Without the
original SID, he can’t access the encrypted files. A third-party decryption application may help
him access his encrypted data..
You are running the Windows installation setup for Windows 7 on a co.pdf
You are running the Windows installation setup for Windows 7 on a computer. When the
Windows Setup screen appears, what do you notice?
A. The Setup screen has been drastically redesigned from Windows Vista.
B. The Setup screen is almost exactly the same as Windows Vista.
C. The Setup screen hasn’t changed since Windows XP.
D. Unlike Windows Vista, the Setup screen immediately launches into the GUI.
Solution
B. The Windows 7 Setup screen is virtually the same as the screen in Windows Vista, both
beginning in a black character-mode screen and then launching the GUI..
Write a program to decipher messages encoded using a prefix code, gi.pdf
Write a program to decipher messages encoded using a prefix code, given the encoding tree.
Such codes are widely used in applications that compress data, including JPEG for images, MP3
for music, and DivX for video. using Huffman Trees (Prefix Binary Trees) c++
Solution
#include #include #include #include // for CHAR_BIT #include #include const int
UniqueSymbols = 1 << CHAR_BIT; const char* SampleString = \"this is an example for
huffman encoding\"; typedef std::vector HuffCode; typedef std::map HuffCodeMap; class
INode { public: const int f; virtual ~INode() {} protected: INode(int f) : f(f) {} };
class InternalNode : public INode { public: INode *const left; INode *const right;
InternalNode(INode* c0, INode* c1) : INode(c0->f + c1->f), left(c0), right(c1) {}
~InternalNode() { delete left; delete right; } }; class LeafNode :
public INode { public: const char c; LeafNode(int f, char c) : INode(f), c(c) {} };
struct NodeCmp { bool operator()(const INode* lhs, const INode* rhs) const { return lhs->f
> rhs->f; } }; INode* BuildTree(const int (&frequencies)[UniqueSymbols]) {
std::priority_queue, NodeCmp> trees; for (int i = 0; i < UniqueSymbols; ++i) {
if(frequencies[i] != 0) trees.push(new LeafNode(frequencies[i], (char)i)); }
while (trees.size() > 1) { INode* childR = trees.top(); trees.pop();
INode* childL = trees.top(); trees.pop(); INode* parent = new
InternalNode(childR, childL); trees.push(parent); } return trees.top(); } void
GenerateCodes(const INode* node, const HuffCode& prefix, HuffCodeMap& outCodes) {
if (const LeafNode* lf = dynamic_cast(node)) { outCodes[lf->c] = prefix; }
else if (const InternalNode* in = dynamic_cast(node)) { HuffCode leftPrefix =
prefix; leftPrefix.push_back(false); GenerateCodes(in->left, leftPrefix,
outCodes); HuffCode rightPrefix = prefix; rightPrefix.push_back(true);
GenerateCodes(in->right, rightPrefix, outCodes); } } int main() { // Build
frequency table int frequencies[UniqueSymbols] = {0}; const char* ptr =
SampleString; while (*ptr != \'\\0\') ++frequencies[*ptr++]; INode* root =
BuildTree(frequencies); HuffCodeMap codes; GenerateCodes(root, HuffCode(),
codes); delete root; for (HuffCodeMap::const_iterator it = codes.begin(); it !=
codes.end(); ++it) { std::cout << it->first << \" \"; std::copy(it-
>second.begin(), it->second.end(), std::ostream_iterator(std::cout)); std::cout
<< std::endl; } return 0; }.
With the biological species concept, the process of speciation is fre.pdf
With the biological species concept, the process of speciation is frequently defined as the
evolution of between populations. reproductive isolation genetic drift gene flow reproductive
inclusion geographic separation The phylogenetic species concept looks at __. environmental
adaptations and morphology species response to adverse conditions and genetic data genetic data
and environmental adaptations morphology and environmental adaptations morphology and
genetic data A phylogenetic species comprise __. interbreeding population that are
reproductively isolated individuals that share measurable traits that distinguish them populations
that share a recent evolutionary history individuals that share a recent evolutionary history one
large gene pool subdivided into local populations
Solution
Reproductive isolation
According to biological species concept, species is a group of actually or potentially
interbreeding individuals who are reproductively isolated from other such groups..
Write a class ArrayList that represents an array of integers. Init.pdf
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
However the output for the prime numbers 1 -50 is incorrect. Displaying numbers like 0, 625 for
prime numbers.
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a: Default constructor used to unitialize the array with capacity=1
ArrayList::ArrayList()
{
asize = 0;
capacity = 1;
array = new int[capacity];
}
// This function returns the actual size of array
int ArrayList::size()
{
return asize;
}
// 1.b:The overloaded [] oprtaor return element at ith position in array
int & ArrayList::operator [](unsigned int i)
{
if(i >= asize)
{
cout << \"invalid reference \";
exit(0);
}
return *(array + i);
}
//1.c: This function push m at the end of the array
void ArrayList::push_back(int m)
{
// this condition checks if size is beyond capacity or not. if so double the capacity.
if(asize == capacity)
{
capacity = 2 * capacity;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
array[asize] = m;
asize = asize + 1;
}
void ArrayList::erase(int m){
int pos = -1;
for(int i = 0; i < asize; i++){
if(array[i] == m){
pos = i;
break;
}
}
if(pos == -1){
cout << \"Element not present\ \";
}
else{
for(int i = pos; i < asize - 1; i++){
array[i] = array[i + 1];
}
}
asize--;
if(asize == capacity / 2)
{
capacity = capacity / 2;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
}
string ArrayList::toString(){
string str = \"[\";
for(int i = 0; i < asize; i++){
ostringstream oss;
oss << array[i];
str += oss.str();
if(i != asize - 1){
str += \", \";
}
}
str += \"]\";
return str;
}
//---------------------------
//You can test above class by using main() below
int main()
{
ArrayList arr; // call ro default constructor
for (int i = 1; i <= 50; i++)
{
arr.push_back(i); // call to push_back()
}
cout << \"Should contain numbers 1..50, is \";
cout << arr.toString();
cout << \"\ \";
return 0;
}
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a:.
what is the relationship between lnx and 1xSolutionThe relat.pdf
what is the relationship between lnx and 1/x?
Solution
The relation between ln x and (1/x) is explained in terms of definite and indefinte integrals.
Indefinite integral:
Int (1/x) dx + C = lnx +C.
Definite integral:
{Int (1/t) dt from t = 0 to t = x} = lnx.
Or in other words ln (x) is the area under the curve f(x) = ln(t) between the ordinates t= 0 to t =
x, which could be calculated by evaluation of the definite integral Int (1/t)dt from t = 0 to t =x..
Which of these conclusions is supported by plant phylogenies (evolut.pdf
Which of these conclusions is supported by plant phylogenies (evolutionary trees)?
Question 4 options:1)It is likely that the transition to land occurred directly from red algae.2)In
all of the trees generated so far, land plants trace back to more than one ancestral group.3)The
oldest lineages of land plants are descended from some green algae (Charophytes)4)The closest
relative to land plants is fungi
Solution
Answer is option 3) The oldest lineages of land plants are descended from some green algae
(Charophytes)
Land plants evolbed from green algae approximately 510 million years ago. The Charophytes are
still present..
Who are some other individuals who might want to use the informa.pdf
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Solution
Some Other Individuals who might use the information in financial statements
1) Suppliers - Suppliers are interested in the information of financial Statements. Suppliers get to
know about solvency and liquidity of company to pay its short term obligations.
2) Employees- Employees use the financial statement information to analyse about the
company\'s ability to pay salary and benefits. Others reasons of using financial statement
information are career development, Financial performance.
3) Customers - Customers use the financial statement information for assessing the company
expansion possibilities, its stability of operations, Its compliances review, for investment
decisions.
What is the nucleoid region of a prokaryotic cell a membrane bound .pdf
What is the nucleoid region of a prokaryotic cell? a membrane bound organelle that contains
the DNA a region within the nucleus where parts of ribosomes are made a place within the cell
where DNA is more abundant a place where you\'d expect to find a proton and a neutron
Solution
A place with the cell where DNA is more abundant
The nucleoid refers to an irregularly shaped region within a prokaryotic cell that contains the
genetic material..
What is heartwood What is sapwood What is the most abundant tissu.pdf
What is heartwood? What is sapwood? What is the most abundant tissue in the three-year-old
stem? What are three contrasting differences in structure between the sunflower stem and the
TNia stem? What are two techniques that can be used to determine the age of a woody twig?
Solution
1. Heart wood is the dead and inner wood tissue produced by cork cambium. It is dark in colr
and is distinguished from sap wood by its color and texture.
Sapwood is the outermost living wood tissue. It is also produced by cambium but is composed of
different types of cells. It is lightin color. It is involved conduction of water and minerals from
roots to all parts of tree.
2. In a three year old stem, the secondary xylem and phloem are bundant than primary xylem and
phloem. These vascular tissues conspicuously appear as annual rings by which the age of the tree
is known.
3.
4.
The age of a woody twig can be determined by counting the number of groups of bud scale scars
on it and he the number of terminal bud scale scars present.SunflowerTiliaAnnualperennialherb
exhibits no secondary growthtree exhibits secondary growthVascular bundles form a ring. Xylem
lies inside the vascular bundles. Phloem lies outside and phloem fibers are tough. The center of
stem is filled with ground tissue known as pithOne year old stem exhibits no secondary growth
and contains a layer of xylem and two later meristems called vascular and cork cambium.
During second year growth 2 rings of xylem is observed. During third year secondary xylem is
obseved and secondary phloem towards inside to xylem..
Use arrays to store data for analysis. Use functions to perform the .pdf
Use arrays to store data for analysis. Use functions to perform the analysis described below. All
output is to be written to an output file.
Solution
import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
public class DNA{
private String dna;
public DNA(){
}
public void read_store(String filename){
/*try {
Scanner in;
//in = new Scanner(new FileReader(filename));
dna = in.next();
}catch(IOException e){
System.out.println(\"file not there \" + e);
return;
}*/
}
public char compliment_base(char c){
if(c == \'A\') return \'T\';
if(c == \'T\') return \'A\';
if(c == \'G\') return \'C\';
if(c == \'C\') return \'G\';
return \' \';
}
public void DrawDNA(String seq){
System.out.println(\"---+-----+------+------+------\");
for (char c : seq.toCharArray() ) {
System.out.print(c + \" \");
}
System.out.println(\"DNA Sequence \");
System.out.println(\"................................\");
for (char c : seq.toCharArray() ) {
System.out.print(compliment_base(c) + \" \");
}
System.out.println(\" Complement \");
System.out.println(\"---+-----+------+------+------\");
System.out.println(\"\");
}
public void Prefix(int n, String base){
String output = base.substring(0,n);
DrawDNA(output);
}
public static void main(String[] args) {
DNA d = new DNA();
d.DrawDNA(\"ATGC\");
d.Prefix(4,\"ACGTTGCA\");
}
}
---+-----+------+------+------
A T G C DNA Sequence
................................
T A C G Complement
---+-----+------+------+------
---+-----+------+------+------
A C G T DNA Sequence
................................
T G C A Complement
---+-----+------+------+------.
Using a telephone survey, a team of sociologists collect data on edu.pdf
Using a telephone survey, a team of sociologists collect data on education and income for 154
individuals. Let xi be the highest grade completed by the ith respondent (example xi =14 means
that ith respondent completed 2 years of college). Let yi be the income of the ith respondent. The
sociologists have computed the following quantities after running the simple linear regression to
predict the income of the individual by the highest grade completed:
http://imgur.com/sdszLUM
Solution.
The “Tripwire” system calculates all the hashes for all executable f.pdf
The “Tripwire” system calculates all the hashes for all executable files on a clean computer and
then compares the hashes with the actual files, when the machine is scanned, to ensure nothing
has changed and no malware is hiding as legitimate programs.
1) What kind of attacks will not be detectable by such a scheme?
2) What attack will render the above strategy useless (i.e. will not find infected files, even if they
exist).
Solution
The near-ubiquitous UNIX system is an example of a file system where such monitoring is
useful. Flaws
and weaknesses in typical UNIX systems are well-documented (e.g., [8, 25, 19, 4, 9]). UNIX
file systems
are susceptible to threats in the guise of unauthorized users, intruders, viruses, worms, and logic
bombs
as well as failures and bugs. As such, UNIX system administrators are faced with prospects of
subtle,
difficult-to-detect damage to files, malicious and accidental.
Tripwire is an integrity checking tool designed for the UNIX environment to aid system
administrators to
monitor their file systems for unauthorized modifications. First made available on November 2,
1992, it has
proven to be a popular tool, being portable, configurable, scalable, flexible, manageable,
automatable, and
secure. It was written in response to repeated break-in activity on the Internet, and the difficulty
experienced
by affected administrators in finding all of the “backdoors” left by the intruders.
To aid in the detection of the appropriate threats, system administrators would use an integrity
checker to
monitor file systems for added, deleted, and changed files. Meaningfully reporting changed files
is difficult,
because most files are expected to change: system log files are written to, program sources are
updated,
and documents are revised. Typically, these changes would not concern system administrators.
However,
changes to certain files, such as system binaries, might elicit a different reaction.
The database used by the integrity checker should be protected from unauthorized
modifications; an intruder
who can change the database can subvert the entire integrity checking scheme. Although the
system administrator
can secure the database by storing it on some media inaccessible to remote intruders (e.g., paper
printout), usability is sacrificed. A database stored in some machine readable format may risk
unauthorized
modification, but allows the integrity checking process to be automated. Storing the database on
read-only
media provides the best of both approaches, allowing machine access but preventing changes.
Most available UNIX security tools fall into two categories: static audit tools and integrity
checkers.
Among the most prominent are COPS[8], TAMU[22], crc check[8], Hobgoblin[15], and
ATP[28].5 A few
commercial security tools also exist, but they are comparable to the user-community tools
mentioned here.
While many of these tools may be outstanding in their own right, most are mismatches for
integrity checking
in UNIX env.
The archaeal RNA viruses described in the introduction were single-s.pdf
The archaeal RNA viruses described in the introduction were single-stranded plus-sense viruses,
like many of the viruses described in the textbook chapter (e.g., phage MS2, poliovirus, and
coronaviruses). They had the genes for RNA replicase and some appeared to use polyproteins.
Based on your knowledge of other viral life cycles, what are the most likely lifecycles for these
archaeal RNA viruses?
Solution
RNA viruses are those viruses in which the sole nucleic acid present is RNA, mostly single
stranded or some viruses have double-stranded RNA. The single stranded RNA can be (+) sense
or (-) sense. The RNA viruses do not have DNA intermediates in their life cycle as compared to
retroviruses.
Most likely lifecycle for archaeal RNA virus.
propose 3 key word centered on the topic which will make the basis o.pdf
propose 3 key word centered on the topic which will make the basis of your poster. Write one
sentence about each key word explaining the rational behind it. The topic that I choose was Owl
migration.
Solution
Three key word on topic Owl migration
1 .Flight behaviour
2. Nocturnal field oberservation
3. Seasonal migration.
Molecular clock studies by Sarich and Wilson have suggested that .pdf
\"Molecular clock\" studies by Sarich and Wilson have suggested that humans and the African
apes diverged from a common ancestor around 2 to 3 million years ago. 5 to 8 million years
ago. 15 to 20 million years ago. 50 million year ago. 10 to 15 million years ago. Which of the
following is true of the gibbon? they are the smallest of the apes. they live mainly in central
Africa. they live in multi-male, multi-female groups in the rainforest. all of the above Old
World monkeys are characterized by all of the following EXCEPT bilophodont molars.
quadrupedal molars quadrupedal locomotion, with all four limbs around the same length. a 2-1-
2-3 dental formula. a grasping, or prehensile, tail.
Solution
Q. 19. 4 to 5 million years ago
Q. 20 they are the smallest of the apes - true
They live mainly in Central Africa - false. They live in Southeast Asia
They live in multi male and multi female groups in rainforest - false. They live in small families
in the
rainforest- (monogamous)
Q 21. They do not have a grasping and prehensile tail.
Limestone rocks, coral reefs, and forests are examples of Select one.pdf
Limestone rocks, coral reefs, and forests are examples of Select one a. carbon sources b.
carbon sinks c. carbon credits d. geo-sequestration. Differences in the climates of Venus, Earth,
and Mars can be best explained by Select one: a. the distances of each planet from the Sun b.
the chemical make-up of their atmospheres c. the percentage of carbon dioxide in their
atmospheres d. the distances of each planet from the Sun and the chemical make-up of their
atmospheres Which of the following statements is supported by the data presented in the graph?
Select one a. The planet closest to the Sun has the warmest temperature b. Mars is too far from
the Sun to support life c. Distance from the Sun alone is not sufficient to explain the variation in
temperatures of the planets d. Because it is the warmest of the planets, Venus is likely to support
life
Solution
1. carbon sinks
2. the distances of each planet from Sun and their chemical makeup of their atmospheres.
Because of its thin atmosphere and greater distance from Sun, the surface temperature is colder
for Mars. however, it experiences temperature fluctuations.
3. distance from Sun alone is not sufficient to explain the variation in temperatures of the planets.
3..
I have the first program completed (not how request, but it works) a.pdf
I have the first program completed (not how request, but it works) and I am trying to get the
second portion together.
First Program requested (complete)**********************************************
(1) Create three files to submit:
ItemToPurchase.h - Class declaration
ItemToPurchase.cpp - Class definition
main.cpp - main() function
Build the ItemToPurchase class with the following specifications:
Default constructor
Public class functions (mutators & accessors)
SetName() & GetName() (2 pts)
SetPrice() & GetPrice() (2 pts)
SetQuantity() & GetQuantity() (2 pts)
Private data members
string itemName - Initialized in default constructor to \"none\"
int itemPrice - Initialized in default constructor to 0
int itemQuantity - Initialized in default constructor to 0
(2) In main(), prompt the user for two items and create two objects of the ItemToPurchase class.
Before prompting for the second item, call cin.ignore() to allow the user to input a new string. (2
pts)
Ex:
(3) Add the costs of the two items together and output the total cost. (2 pts)
Ex:
Second Portion I am on
now**************************************************************************
*******
7.19 Program: Online shopping cart (continued) (C++)
This program extends the earlier \"Online shopping cart\" program. (Consider first saving your
earlier program).
(1) Extend the ItemToPurchase class per the following specifications:
Parameterized constructor to assign item name, item description, item price, and item quantity
(default values of 0). (1 pt)
Public member functions
SetDescription() mutator & GetDescription() accessor (2 pts)
PrintItemCost() - Outputs the item name followed by the quantity, price, and subtotal
PrintItemDescription() - Outputs the item name and description
Private data members
string itemDescription - Initialized in default constructor to \"none\"
Ex. of PrintItemCost() output:
Ex. of PrintItemDescription() output:
(2) Create three new files:
ShoppingCart.h - Class declaration
ShoppingCart.cpp - Class definition
main.cpp - main() function (Note: main()\'s functionality differs from the warm up)
Build the ShoppingCart class with the following specifications. Note: Some can be function
stubs (empty functions) initially, to be completed in later steps.
Default constructor
Parameterized constructor which takes the customer name and date as parameters (1 pt)
Private data members
string customerName - Initialized in default constructor to \"none\"
string currentDate - Initialized in default constructor to \"January 1, 2016\"
vector < ItemToPurchase > cartItems
Public member functions
GetCustomerName() accessor (1 pt)
GetDate() accessor (1 pt)
AddItem()
Adds an item to cartItems vector. Has parameter ItemToPurchase. Does not return anything.
RemoveItem()
Removes item from cartItems vector. Has a string (an item\'s name) parameter. Does not return
anything.
If item name cannot be found, output this message: Item not found in cart. Nothing removed.
ModifyItem()
Modifies an it.
How does chemotherapy function as a cancer treatmentA.It kills ce.pdf
How does chemotherapy function as a cancer treatment?
A.It kills cells in the process of dividing.
B.It attaches toxic drugs to immune system cells that can bind to cancer cells.
C.It finds oncogenes and corrects their DNA sequences.
D.It uses a virus to alert the immune system to the cancer.
E.It attacks cells with unusual proteins on their plasma membranes.
Solution.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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A cancer biology topic Describte how ChIP (chromatin immunoprecipit.pdf
A cancer biology topic: Describte how ChIP (chromatin immunoprecipitation), DNA
methylation array, and gene expression arrays are performed. What are the similarities?
Solution
DNA methylation is main epigenetic mechanism of gene expression regulation.Alteration in this
leads to tumor. Various stimulant factors stimulate DNA methylation leads to generation of
canceroua cells.
DNA methylation process do down regulation of gene.
Methylation of CpG sites blocks the binding of regulatory proteins resulting transcription
modulation.
In CHIP DNA methylation stauninig of CHIP done by various stains. Stained CHIP show
intensities of methylated and unmethylated bead types. Data analysis gives florescence intensity
ratio between methylated and unmethylated group.
Value of ratio is 0, 1, 0.5 represents non methylation, total methylation and one copy methylated
other not respectively.
DNA methylation measure by following other method also
Bisulfite treatment
Whole genomic DNA amplification
Hybridization and single base extension
In Gene expression two following procees take important part:-
TRANSCRIPTION:- DNA to RNA by RNA polymerase which use DNA as template and
produce RNA.
TRANSLATION:- RNA to Protein translation (tRNA charged with amino acid enter in ribosome
with correct mRNA triplet , ribisome then add aminoacid.
In cancer, gene promoter contain CpG island with numerous CpG sites. When CpG
sitesethylated gene become silence.
Thanks...
You get a call from a customer who reinstalled Windows on his comput.pdf
You get a call from a customer who reinstalled Windows on his computer after a crash.
Fortunately, he had recently backed up all this data to an external USB drive and was able to
restore everything on his NTFS-formatted file system. However, he had a number of encrypted
files that he had backed up along with the rest of his data that he can no longer access. What
happened?
A. NTFS-encrypted files cannot be backed up, and when the computer crashed, those files were
lost forever. When the customer tried to restore these files, only corrupted versions were
restored.
B. NTFS-encrypted files are not restored with the original encryption key used before the backup
process and thus become inaccessible.
C. When the customer reinstalled Windows, even though he used his old username, the computer
assigned him a completely different security identifier (SID). The restored NTFS-encrypted files
are looking for his original SID.
D. When the customer reinstalled Windows, he didn’t use his original username. The restored
NTFS-encrypted files are looking for the original username.
Solution
C. NTFS encryption used the customer’s SID to associate with the encrypted file. Now that the
customer reinstalled Windows, a completely new SID was assigned to the customer. Without the
original SID, he can’t access the encrypted files. A third-party decryption application may help
him access his encrypted data..
You are running the Windows installation setup for Windows 7 on a co.pdf
You are running the Windows installation setup for Windows 7 on a computer. When the
Windows Setup screen appears, what do you notice?
A. The Setup screen has been drastically redesigned from Windows Vista.
B. The Setup screen is almost exactly the same as Windows Vista.
C. The Setup screen hasn’t changed since Windows XP.
D. Unlike Windows Vista, the Setup screen immediately launches into the GUI.
Solution
B. The Windows 7 Setup screen is virtually the same as the screen in Windows Vista, both
beginning in a black character-mode screen and then launching the GUI..
Write a program to decipher messages encoded using a prefix code, gi.pdf
Write a program to decipher messages encoded using a prefix code, given the encoding tree.
Such codes are widely used in applications that compress data, including JPEG for images, MP3
for music, and DivX for video. using Huffman Trees (Prefix Binary Trees) c++
Solution
#include #include #include #include // for CHAR_BIT #include #include const int
UniqueSymbols = 1 << CHAR_BIT; const char* SampleString = \"this is an example for
huffman encoding\"; typedef std::vector HuffCode; typedef std::map HuffCodeMap; class
INode { public: const int f; virtual ~INode() {} protected: INode(int f) : f(f) {} };
class InternalNode : public INode { public: INode *const left; INode *const right;
InternalNode(INode* c0, INode* c1) : INode(c0->f + c1->f), left(c0), right(c1) {}
~InternalNode() { delete left; delete right; } }; class LeafNode :
public INode { public: const char c; LeafNode(int f, char c) : INode(f), c(c) {} };
struct NodeCmp { bool operator()(const INode* lhs, const INode* rhs) const { return lhs->f
> rhs->f; } }; INode* BuildTree(const int (&frequencies)[UniqueSymbols]) {
std::priority_queue, NodeCmp> trees; for (int i = 0; i < UniqueSymbols; ++i) {
if(frequencies[i] != 0) trees.push(new LeafNode(frequencies[i], (char)i)); }
while (trees.size() > 1) { INode* childR = trees.top(); trees.pop();
INode* childL = trees.top(); trees.pop(); INode* parent = new
InternalNode(childR, childL); trees.push(parent); } return trees.top(); } void
GenerateCodes(const INode* node, const HuffCode& prefix, HuffCodeMap& outCodes) {
if (const LeafNode* lf = dynamic_cast(node)) { outCodes[lf->c] = prefix; }
else if (const InternalNode* in = dynamic_cast(node)) { HuffCode leftPrefix =
prefix; leftPrefix.push_back(false); GenerateCodes(in->left, leftPrefix,
outCodes); HuffCode rightPrefix = prefix; rightPrefix.push_back(true);
GenerateCodes(in->right, rightPrefix, outCodes); } } int main() { // Build
frequency table int frequencies[UniqueSymbols] = {0}; const char* ptr =
SampleString; while (*ptr != \'\\0\') ++frequencies[*ptr++]; INode* root =
BuildTree(frequencies); HuffCodeMap codes; GenerateCodes(root, HuffCode(),
codes); delete root; for (HuffCodeMap::const_iterator it = codes.begin(); it !=
codes.end(); ++it) { std::cout << it->first << \" \"; std::copy(it-
>second.begin(), it->second.end(), std::ostream_iterator(std::cout)); std::cout
<< std::endl; } return 0; }.
With the biological species concept, the process of speciation is fre.pdf
With the biological species concept, the process of speciation is frequently defined as the
evolution of between populations. reproductive isolation genetic drift gene flow reproductive
inclusion geographic separation The phylogenetic species concept looks at __. environmental
adaptations and morphology species response to adverse conditions and genetic data genetic data
and environmental adaptations morphology and environmental adaptations morphology and
genetic data A phylogenetic species comprise __. interbreeding population that are
reproductively isolated individuals that share measurable traits that distinguish them populations
that share a recent evolutionary history individuals that share a recent evolutionary history one
large gene pool subdivided into local populations
Solution
Reproductive isolation
According to biological species concept, species is a group of actually or potentially
interbreeding individuals who are reproductively isolated from other such groups..
Write a class ArrayList that represents an array of integers. Init.pdf
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
However the output for the prime numbers 1 -50 is incorrect. Displaying numbers like 0, 625 for
prime numbers.
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a: Default constructor used to unitialize the array with capacity=1
ArrayList::ArrayList()
{
asize = 0;
capacity = 1;
array = new int[capacity];
}
// This function returns the actual size of array
int ArrayList::size()
{
return asize;
}
// 1.b:The overloaded [] oprtaor return element at ith position in array
int & ArrayList::operator [](unsigned int i)
{
if(i >= asize)
{
cout << \"invalid reference \";
exit(0);
}
return *(array + i);
}
//1.c: This function push m at the end of the array
void ArrayList::push_back(int m)
{
// this condition checks if size is beyond capacity or not. if so double the capacity.
if(asize == capacity)
{
capacity = 2 * capacity;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
array[asize] = m;
asize = asize + 1;
}
void ArrayList::erase(int m){
int pos = -1;
for(int i = 0; i < asize; i++){
if(array[i] == m){
pos = i;
break;
}
}
if(pos == -1){
cout << \"Element not present\ \";
}
else{
for(int i = pos; i < asize - 1; i++){
array[i] = array[i + 1];
}
}
asize--;
if(asize == capacity / 2)
{
capacity = capacity / 2;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
}
string ArrayList::toString(){
string str = \"[\";
for(int i = 0; i < asize; i++){
ostringstream oss;
oss << array[i];
str += oss.str();
if(i != asize - 1){
str += \", \";
}
}
str += \"]\";
return str;
}
//---------------------------
//You can test above class by using main() below
int main()
{
ArrayList arr; // call ro default constructor
for (int i = 1; i <= 50; i++)
{
arr.push_back(i); // call to push_back()
}
cout << \"Should contain numbers 1..50, is \";
cout << arr.toString();
cout << \"\ \";
return 0;
}
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a:.
what is the relationship between lnx and 1xSolutionThe relat.pdf
what is the relationship between lnx and 1/x?
Solution
The relation between ln x and (1/x) is explained in terms of definite and indefinte integrals.
Indefinite integral:
Int (1/x) dx + C = lnx +C.
Definite integral:
{Int (1/t) dt from t = 0 to t = x} = lnx.
Or in other words ln (x) is the area under the curve f(x) = ln(t) between the ordinates t= 0 to t =
x, which could be calculated by evaluation of the definite integral Int (1/t)dt from t = 0 to t =x..
Which of these conclusions is supported by plant phylogenies (evolut.pdf
Which of these conclusions is supported by plant phylogenies (evolutionary trees)?
Question 4 options:1)It is likely that the transition to land occurred directly from red algae.2)In
all of the trees generated so far, land plants trace back to more than one ancestral group.3)The
oldest lineages of land plants are descended from some green algae (Charophytes)4)The closest
relative to land plants is fungi
Solution
Answer is option 3) The oldest lineages of land plants are descended from some green algae
(Charophytes)
Land plants evolbed from green algae approximately 510 million years ago. The Charophytes are
still present..
Who are some other individuals who might want to use the informa.pdf
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Solution
Some Other Individuals who might use the information in financial statements
1) Suppliers - Suppliers are interested in the information of financial Statements. Suppliers get to
know about solvency and liquidity of company to pay its short term obligations.
2) Employees- Employees use the financial statement information to analyse about the
company\'s ability to pay salary and benefits. Others reasons of using financial statement
information are career development, Financial performance.
3) Customers - Customers use the financial statement information for assessing the company
expansion possibilities, its stability of operations, Its compliances review, for investment
decisions.
What is the nucleoid region of a prokaryotic cell a membrane bound .pdf
What is the nucleoid region of a prokaryotic cell? a membrane bound organelle that contains
the DNA a region within the nucleus where parts of ribosomes are made a place within the cell
where DNA is more abundant a place where you\'d expect to find a proton and a neutron
Solution
A place with the cell where DNA is more abundant
The nucleoid refers to an irregularly shaped region within a prokaryotic cell that contains the
genetic material..
What is heartwood What is sapwood What is the most abundant tissu.pdf
What is heartwood? What is sapwood? What is the most abundant tissue in the three-year-old
stem? What are three contrasting differences in structure between the sunflower stem and the
TNia stem? What are two techniques that can be used to determine the age of a woody twig?
Solution
1. Heart wood is the dead and inner wood tissue produced by cork cambium. It is dark in colr
and is distinguished from sap wood by its color and texture.
Sapwood is the outermost living wood tissue. It is also produced by cambium but is composed of
different types of cells. It is lightin color. It is involved conduction of water and minerals from
roots to all parts of tree.
2. In a three year old stem, the secondary xylem and phloem are bundant than primary xylem and
phloem. These vascular tissues conspicuously appear as annual rings by which the age of the tree
is known.
3.
4.
The age of a woody twig can be determined by counting the number of groups of bud scale scars
on it and he the number of terminal bud scale scars present.SunflowerTiliaAnnualperennialherb
exhibits no secondary growthtree exhibits secondary growthVascular bundles form a ring. Xylem
lies inside the vascular bundles. Phloem lies outside and phloem fibers are tough. The center of
stem is filled with ground tissue known as pithOne year old stem exhibits no secondary growth
and contains a layer of xylem and two later meristems called vascular and cork cambium.
During second year growth 2 rings of xylem is observed. During third year secondary xylem is
obseved and secondary phloem towards inside to xylem..
Use arrays to store data for analysis. Use functions to perform the .pdf
Use arrays to store data for analysis. Use functions to perform the analysis described below. All
output is to be written to an output file.
Solution
import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
public class DNA{
private String dna;
public DNA(){
}
public void read_store(String filename){
/*try {
Scanner in;
//in = new Scanner(new FileReader(filename));
dna = in.next();
}catch(IOException e){
System.out.println(\"file not there \" + e);
return;
}*/
}
public char compliment_base(char c){
if(c == \'A\') return \'T\';
if(c == \'T\') return \'A\';
if(c == \'G\') return \'C\';
if(c == \'C\') return \'G\';
return \' \';
}
public void DrawDNA(String seq){
System.out.println(\"---+-----+------+------+------\");
for (char c : seq.toCharArray() ) {
System.out.print(c + \" \");
}
System.out.println(\"DNA Sequence \");
System.out.println(\"................................\");
for (char c : seq.toCharArray() ) {
System.out.print(compliment_base(c) + \" \");
}
System.out.println(\" Complement \");
System.out.println(\"---+-----+------+------+------\");
System.out.println(\"\");
}
public void Prefix(int n, String base){
String output = base.substring(0,n);
DrawDNA(output);
}
public static void main(String[] args) {
DNA d = new DNA();
d.DrawDNA(\"ATGC\");
d.Prefix(4,\"ACGTTGCA\");
}
}
---+-----+------+------+------
A T G C DNA Sequence
................................
T A C G Complement
---+-----+------+------+------
---+-----+------+------+------
A C G T DNA Sequence
................................
T G C A Complement
---+-----+------+------+------.
Using a telephone survey, a team of sociologists collect data on edu.pdf
Using a telephone survey, a team of sociologists collect data on education and income for 154
individuals. Let xi be the highest grade completed by the ith respondent (example xi =14 means
that ith respondent completed 2 years of college). Let yi be the income of the ith respondent. The
sociologists have computed the following quantities after running the simple linear regression to
predict the income of the individual by the highest grade completed:
http://imgur.com/sdszLUM
Solution.
The “Tripwire” system calculates all the hashes for all executable f.pdf
The “Tripwire” system calculates all the hashes for all executable files on a clean computer and
then compares the hashes with the actual files, when the machine is scanned, to ensure nothing
has changed and no malware is hiding as legitimate programs.
1) What kind of attacks will not be detectable by such a scheme?
2) What attack will render the above strategy useless (i.e. will not find infected files, even if they
exist).
Solution
The near-ubiquitous UNIX system is an example of a file system where such monitoring is
useful. Flaws
and weaknesses in typical UNIX systems are well-documented (e.g., [8, 25, 19, 4, 9]). UNIX
file systems
are susceptible to threats in the guise of unauthorized users, intruders, viruses, worms, and logic
bombs
as well as failures and bugs. As such, UNIX system administrators are faced with prospects of
subtle,
difficult-to-detect damage to files, malicious and accidental.
Tripwire is an integrity checking tool designed for the UNIX environment to aid system
administrators to
monitor their file systems for unauthorized modifications. First made available on November 2,
1992, it has
proven to be a popular tool, being portable, configurable, scalable, flexible, manageable,
automatable, and
secure. It was written in response to repeated break-in activity on the Internet, and the difficulty
experienced
by affected administrators in finding all of the “backdoors” left by the intruders.
To aid in the detection of the appropriate threats, system administrators would use an integrity
checker to
monitor file systems for added, deleted, and changed files. Meaningfully reporting changed files
is difficult,
because most files are expected to change: system log files are written to, program sources are
updated,
and documents are revised. Typically, these changes would not concern system administrators.
However,
changes to certain files, such as system binaries, might elicit a different reaction.
The database used by the integrity checker should be protected from unauthorized
modifications; an intruder
who can change the database can subvert the entire integrity checking scheme. Although the
system administrator
can secure the database by storing it on some media inaccessible to remote intruders (e.g., paper
printout), usability is sacrificed. A database stored in some machine readable format may risk
unauthorized
modification, but allows the integrity checking process to be automated. Storing the database on
read-only
media provides the best of both approaches, allowing machine access but preventing changes.
Most available UNIX security tools fall into two categories: static audit tools and integrity
checkers.
Among the most prominent are COPS[8], TAMU[22], crc check[8], Hobgoblin[15], and
ATP[28].5 A few
commercial security tools also exist, but they are comparable to the user-community tools
mentioned here.
While many of these tools may be outstanding in their own right, most are mismatches for
integrity checking
in UNIX env.
The archaeal RNA viruses described in the introduction were single-s.pdf
The archaeal RNA viruses described in the introduction were single-stranded plus-sense viruses,
like many of the viruses described in the textbook chapter (e.g., phage MS2, poliovirus, and
coronaviruses). They had the genes for RNA replicase and some appeared to use polyproteins.
Based on your knowledge of other viral life cycles, what are the most likely lifecycles for these
archaeal RNA viruses?
Solution
RNA viruses are those viruses in which the sole nucleic acid present is RNA, mostly single
stranded or some viruses have double-stranded RNA. The single stranded RNA can be (+) sense
or (-) sense. The RNA viruses do not have DNA intermediates in their life cycle as compared to
retroviruses.
Most likely lifecycle for archaeal RNA virus.
propose 3 key word centered on the topic which will make the basis o.pdf
propose 3 key word centered on the topic which will make the basis of your poster. Write one
sentence about each key word explaining the rational behind it. The topic that I choose was Owl
migration.
Solution
Three key word on topic Owl migration
1 .Flight behaviour
2. Nocturnal field oberservation
3. Seasonal migration.
Molecular clock studies by Sarich and Wilson have suggested that .pdf
\"Molecular clock\" studies by Sarich and Wilson have suggested that humans and the African
apes diverged from a common ancestor around 2 to 3 million years ago. 5 to 8 million years
ago. 15 to 20 million years ago. 50 million year ago. 10 to 15 million years ago. Which of the
following is true of the gibbon? they are the smallest of the apes. they live mainly in central
Africa. they live in multi-male, multi-female groups in the rainforest. all of the above Old
World monkeys are characterized by all of the following EXCEPT bilophodont molars.
quadrupedal molars quadrupedal locomotion, with all four limbs around the same length. a 2-1-
2-3 dental formula. a grasping, or prehensile, tail.
Solution
Q. 19. 4 to 5 million years ago
Q. 20 they are the smallest of the apes - true
They live mainly in Central Africa - false. They live in Southeast Asia
They live in multi male and multi female groups in rainforest - false. They live in small families
in the
rainforest- (monogamous)
Q 21. They do not have a grasping and prehensile tail.
Limestone rocks, coral reefs, and forests are examples of Select one.pdf
Limestone rocks, coral reefs, and forests are examples of Select one a. carbon sources b.
carbon sinks c. carbon credits d. geo-sequestration. Differences in the climates of Venus, Earth,
and Mars can be best explained by Select one: a. the distances of each planet from the Sun b.
the chemical make-up of their atmospheres c. the percentage of carbon dioxide in their
atmospheres d. the distances of each planet from the Sun and the chemical make-up of their
atmospheres Which of the following statements is supported by the data presented in the graph?
Select one a. The planet closest to the Sun has the warmest temperature b. Mars is too far from
the Sun to support life c. Distance from the Sun alone is not sufficient to explain the variation in
temperatures of the planets d. Because it is the warmest of the planets, Venus is likely to support
life
Solution
1. carbon sinks
2. the distances of each planet from Sun and their chemical makeup of their atmospheres.
Because of its thin atmosphere and greater distance from Sun, the surface temperature is colder
for Mars. however, it experiences temperature fluctuations.
3. distance from Sun alone is not sufficient to explain the variation in temperatures of the planets.
3..
I have the first program completed (not how request, but it works) a.pdf
I have the first program completed (not how request, but it works) and I am trying to get the
second portion together.
First Program requested (complete)**********************************************
(1) Create three files to submit:
ItemToPurchase.h - Class declaration
ItemToPurchase.cpp - Class definition
main.cpp - main() function
Build the ItemToPurchase class with the following specifications:
Default constructor
Public class functions (mutators & accessors)
SetName() & GetName() (2 pts)
SetPrice() & GetPrice() (2 pts)
SetQuantity() & GetQuantity() (2 pts)
Private data members
string itemName - Initialized in default constructor to \"none\"
int itemPrice - Initialized in default constructor to 0
int itemQuantity - Initialized in default constructor to 0
(2) In main(), prompt the user for two items and create two objects of the ItemToPurchase class.
Before prompting for the second item, call cin.ignore() to allow the user to input a new string. (2
pts)
Ex:
(3) Add the costs of the two items together and output the total cost. (2 pts)
Ex:
Second Portion I am on
now**************************************************************************
*******
7.19 Program: Online shopping cart (continued) (C++)
This program extends the earlier \"Online shopping cart\" program. (Consider first saving your
earlier program).
(1) Extend the ItemToPurchase class per the following specifications:
Parameterized constructor to assign item name, item description, item price, and item quantity
(default values of 0). (1 pt)
Public member functions
SetDescription() mutator & GetDescription() accessor (2 pts)
PrintItemCost() - Outputs the item name followed by the quantity, price, and subtotal
PrintItemDescription() - Outputs the item name and description
Private data members
string itemDescription - Initialized in default constructor to \"none\"
Ex. of PrintItemCost() output:
Ex. of PrintItemDescription() output:
(2) Create three new files:
ShoppingCart.h - Class declaration
ShoppingCart.cpp - Class definition
main.cpp - main() function (Note: main()\'s functionality differs from the warm up)
Build the ShoppingCart class with the following specifications. Note: Some can be function
stubs (empty functions) initially, to be completed in later steps.
Default constructor
Parameterized constructor which takes the customer name and date as parameters (1 pt)
Private data members
string customerName - Initialized in default constructor to \"none\"
string currentDate - Initialized in default constructor to \"January 1, 2016\"
vector < ItemToPurchase > cartItems
Public member functions
GetCustomerName() accessor (1 pt)
GetDate() accessor (1 pt)
AddItem()
Adds an item to cartItems vector. Has parameter ItemToPurchase. Does not return anything.
RemoveItem()
Removes item from cartItems vector. Has a string (an item\'s name) parameter. Does not return
anything.
If item name cannot be found, output this message: Item not found in cart. Nothing removed.
ModifyItem()
Modifies an it.
How does chemotherapy function as a cancer treatmentA.It kills ce.pdf
How does chemotherapy function as a cancer treatment?
A.It kills cells in the process of dividing.
B.It attaches toxic drugs to immune system cells that can bind to cancer cells.
C.It finds oncogenes and corrects their DNA sequences.
D.It uses a virus to alert the immune system to the cancer.
E.It attacks cells with unusual proteins on their plasma membranes.
Solution.
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