1+5 = 6Solution1+5 = 6.pdf
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The document shows a simple math equation being solved: 1 + 5 equals 6. The equation 1 + 5 is written, followed by the solution, which is that 1 + 5 equals 6.
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A sustainable transportation system is one that • allows the basic .pdf
A sustainable transportation system is one that: • allows the basic access needs of individuals and
societies to be met safely and in a manner consistent with human and ecosystem health, and with
equity within and between generations; • is affordable, operates efficiently, offers choice of
transport mode, and supports a vibrant economy; • limits emissions and waste within the planet’s
ability to absorb them, minimizes consumption of non-renewable resources to the sustainable
yield level, reuses and recycles its components, and minimizes the use of land and the production
of noise. Such sustainability improvements around transportation may not be achieved quickly.
However, using a Triple Bottom Line framework to guide planning, policy decisions, and
implementation can provide steps toward developing a sustainable outcome. When thinking
about delivering transportation solutions and infrastructure in a more sustainable manner,
agencies and companies are considering each of the bottom line elements:
• Economy — Support economic vitality while developing infrastructure in a cost-efficient
manner. Costs of infrastructure must be within a society’s ability and willingness to pay. User
costs, including private costs, need to be within the ability of people and households to pay for
success. •
Social — Meet social needs by making transportation accessible, safe, and secure; include
provision of mobility choices for all people (including people with economic disadvantages); and
develop infrastructure that is an asset to communities. •
Environment — Create solutions that are compatible with - and that can be an enhancement to -
the natural environment, reduce emissions and pollution from the transportation system, and
reduce the material resources required to support transportation.
Following is a discussion of sustainability issues and illustrative best practice examples in: (1)
Planning and System-wide Perspective, (2) Design and Construction, (3) Operations and
Maintenance, and (4) Performance Measures.
1 )Planning and System-wide Perspective Planning and system-wide practices feed into the work
of multiple DOT divisions and can be organized into the following broad categories: Climate
Change Policies, Pricing and Revenue Policies, Land Use Strategies, Freight Strategies and
Sustainability Plans.
2)Design and Construction Perspective Design and construction practices are critical to the role
of DOTs. DOTs are responsible for designing and constructing needed facility improvements,
and redesigning and reconstructing infrastructure or facilities that are aging (for example,
bridges, highways, or facilities).
3) Operations and Maintenance Perspective Operations and maintenance are one of the key
components of the DOT organization. Most DOT employees are in the field, and DOTs have a
mission to maintain and operate the existing built systems.
4 )Use of Performance Measures Performance measures or indicators are used to link DOT
actions to their overall m.
2)The 2 strongest dimensions of my personality area)direction It.pdf
2)The 2 strongest dimensions of my personality are
a)direction : It speaks about the person\'s attitudes, motives, intentions and actions.Whether the
above factors are positively or negatively related to the welfare and well-being of other
individuals and society. In simple words whether the person is \'good\' or \'bad\'..
b)
Consciousness:
Consciousness refers to the relative development and predominance of the physical, vital and
mental aspects
2)
The people who are co-operating,keeping positive attitude in mind,open to work as a team,as a
strong passion for work
3)
The people who dont co-operate andd having head weight,dont adjust to situation.with these
people,it is difficult to work.
4)
with this self assesment,my communication has improved.It is more open to recieve things cleary
and positively.
Solution
2)The 2 strongest dimensions of my personality are
a)direction : It speaks about the person\'s attitudes, motives, intentions and actions.Whether the
above factors are positively or negatively related to the welfare and well-being of other
individuals and society. In simple words whether the person is \'good\' or \'bad\'..
b)
Consciousness:
Consciousness refers to the relative development and predominance of the physical, vital and
mental aspects
2)
The people who are co-operating,keeping positive attitude in mind,open to work as a team,as a
strong passion for work
3)
The people who dont co-operate andd having head weight,dont adjust to situation.with these
people,it is difficult to work.
4)
with this self assesment,my communication has improved.It is more open to recieve things cleary
and positively..
ReverseList.javaimport java.util.ArrayList;public class Rever.pdf
// ReverseList.java
import java.util.ArrayList;
public class ReverseList {
// function to reverse arrylist
public static void reverse(ArrayList list, int size)
{
for (int i = 0; i < size/2 ;i++ )
{
Double temp = list.get(i);
list.set(i, list.get(size-i-1));
list.set(size-i-1,temp);
}
}
public static void main(String args[]){
ArrayList list = new ArrayList();
list.add(4.8);
list.add(5.1);
list.add(9.1);
list.add(1.5);
list.add(0.8);
list.add(11.8);
list.add(5.0);
list.add(3.8);
int size = list.size();
System.out.println(\"Original list:\");
for(Double number: list){
System.out.print (number + \" \");
}
// reverse list
reverse(list,size);
System.out.println(\"\ \ After reversing the list:\");
for(Double number: list){
System.out.print (number + \" \");
}
}
}
/*
output:
Original list:
4.8 5.1 9.1 1.5 0.8 11.8 5.0 3.8
After reversing the list:
3.8 5.0 11.8 0.8 1.5 9.1 5.1 4.8
*/
Solution
// ReverseList.java
import java.util.ArrayList;
public class ReverseList {
// function to reverse arrylist
public static void reverse(ArrayList list, int size)
{
for (int i = 0; i < size/2 ;i++ )
{
Double temp = list.get(i);
list.set(i, list.get(size-i-1));
list.set(size-i-1,temp);
}
}
public static void main(String args[]){
ArrayList list = new ArrayList();
list.add(4.8);
list.add(5.1);
list.add(9.1);
list.add(1.5);
list.add(0.8);
list.add(11.8);
list.add(5.0);
list.add(3.8);
int size = list.size();
System.out.println(\"Original list:\");
for(Double number: list){
System.out.print (number + \" \");
}
// reverse list
reverse(list,size);
System.out.println(\"\ \ After reversing the list:\");
for(Double number: list){
System.out.print (number + \" \");
}
}
}
/*
output:
Original list:
4.8 5.1 9.1 1.5 0.8 11.8 5.0 3.8
After reversing the list:
3.8 5.0 11.8 0.8 1.5 9.1 5.1 4.8
*/.
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M.pdf
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M
Solution
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M.
moles of NaOH = 0.250 1.2 10^-3 = 3 10^-4 M.pdf
moles of NaOH = 0.250 * 1.2 * 10^-3 = 3 * 10^-4 M [NaOH] = [OH-] = 3 * 10^-4
/ 1.5 = 2 *10^-4 M pH = - log [H+] [H+] = 10^-14 / [OH-] = 5 * 10^-11 pH = 10.3
Solution
moles of NaOH = 0.250 * 1.2 * 10^-3 = 3 * 10^-4 M [NaOH] = [OH-] = 3 * 10^-4
/ 1.5 = 2 *10^-4 M pH = - log [H+] [H+] = 10^-14 / [OH-] = 5 * 10^-11 pH = 10.3.
molarity = no of moles volume no of moles = mas.pdf
molarity = no of moles / volume no of moles = mass / molar mass = 1.126/204.22 =
0.0055 volume = 0.002019 l molarity = 0.0055/0.002019 = 2.73
Solution
molarity = no of moles / volume no of moles = mass / molar mass = 1.126/204.22 =
0.0055 volume = 0.002019 l molarity = 0.0055/0.002019 = 2.73.
Pharmacology Stimulates alpha and beta receptors.pdf
Pharmacology Stimulates alpha and beta receptors (alpha receptors at high doses;
beta- 1 and beta- 2 receptors at moderate doses) within the sympathetic nervous system. Relaxes
smooth muscle of bronchi and iris, and is an antagonist of histamine. Pharmacokinetics
Metabolism Inactivated by enzymatic transformation to metabephrine or normetanephrine; these
are subsequently conjugated and excreted in the urine. Elimination Mostly excreted in urine as
inactive metabolites; remainder is excreted as unchanged drug or is conjugated. Onset 5 to 10
min (subcutaneous), 1 to 5 min (inhalation). Duration 4 to 6?h (subcutaneous), 1 to 4 h (IM), 1
to 3 h (inhalation). Indications and Usage Epinephrine 1:1,000 injection Relief of respiratory
distress due to bronchospasm; to provide rapid relief of hypersensitivity reactions to drugs and
other allergens (eg, anaphylactic reactions to drugs, animal serums, insect stings); to prolong the
action of local and regional anesthetics; restore cardiac rhythm in cardiac arrest due to various
causes; treatment of mucosal congestion of hay fever, rhinitis, and acute sinusitis; relieve
bronchial asthmatic paroxysms; symptomatic relief of serum sickness, urticaria, angioneurotic
edema; for relaxation of uterine musculature and to inhibit uterine contractions; epinephrine
injection can be used as a hemostatic agent; in syncope due to complete heart block or carotid
sinus hypersensitivity; for resuscitation in cardiac arrest following anesthetic accidents; used in
open-angle glaucoma. Epinephrine 1:1,000 (auto-injector) and 1:2,000 (auto-injector), Prefilled
syringe Emergency treatment of allergic reactions (type I) including anaphylaxis to insect stings
(eg, bees, fire ants, hornets, yellow jackets, wasps) and biting insects (eg, mosquitoes), allergen
immunotherapy, foods, drugs, diagnostic testing substances (eg, radiocontrast media), and other
allergens, as well as idiopathic anaphylaxis or exercise-induced anaphylaxis. The auto-injectors
and prefilled syringes are intended for immediate self-administration in patients who are at
increased risk for anaphylaxis, including individuals with a history of anaphylactic reactions.
Auto-injectors and prefilled syringes are for immediate use and are not a substitute for immediate
medical attention. Epinephrine 1:10,000 injection Treatment and prophylaxis of cardiac arrest in
the absence of ventricular fibrillation and attacks of transitory atrioventricular heart block with
syncopal seizures; to stimulate the heart in syncope due to complete heart block or carotid sinus
hypersensitivity; for resuscitation in cardiac arrest following anesthetic accidents; in
cardiopulmonary resuscitation, intracardiac puncture and intramyocardial injection of
epinephrine may be effective when external cardiac compression and attempts to restore the
circulation by electrical defibrillation or use of pacemaker fail; seldom used as a vasopressor
except in the treatment of anaphylactic shock and under .
Recommended
A sustainable transportation system is one that • allows the basic .pdf
A sustainable transportation system is one that: • allows the basic access needs of individuals and
societies to be met safely and in a manner consistent with human and ecosystem health, and with
equity within and between generations; • is affordable, operates efficiently, offers choice of
transport mode, and supports a vibrant economy; • limits emissions and waste within the planet’s
ability to absorb them, minimizes consumption of non-renewable resources to the sustainable
yield level, reuses and recycles its components, and minimizes the use of land and the production
of noise. Such sustainability improvements around transportation may not be achieved quickly.
However, using a Triple Bottom Line framework to guide planning, policy decisions, and
implementation can provide steps toward developing a sustainable outcome. When thinking
about delivering transportation solutions and infrastructure in a more sustainable manner,
agencies and companies are considering each of the bottom line elements:
• Economy — Support economic vitality while developing infrastructure in a cost-efficient
manner. Costs of infrastructure must be within a society’s ability and willingness to pay. User
costs, including private costs, need to be within the ability of people and households to pay for
success. •
Social — Meet social needs by making transportation accessible, safe, and secure; include
provision of mobility choices for all people (including people with economic disadvantages); and
develop infrastructure that is an asset to communities. •
Environment — Create solutions that are compatible with - and that can be an enhancement to -
the natural environment, reduce emissions and pollution from the transportation system, and
reduce the material resources required to support transportation.
Following is a discussion of sustainability issues and illustrative best practice examples in: (1)
Planning and System-wide Perspective, (2) Design and Construction, (3) Operations and
Maintenance, and (4) Performance Measures.
1 )Planning and System-wide Perspective Planning and system-wide practices feed into the work
of multiple DOT divisions and can be organized into the following broad categories: Climate
Change Policies, Pricing and Revenue Policies, Land Use Strategies, Freight Strategies and
Sustainability Plans.
2)Design and Construction Perspective Design and construction practices are critical to the role
of DOTs. DOTs are responsible for designing and constructing needed facility improvements,
and redesigning and reconstructing infrastructure or facilities that are aging (for example,
bridges, highways, or facilities).
3) Operations and Maintenance Perspective Operations and maintenance are one of the key
components of the DOT organization. Most DOT employees are in the field, and DOTs have a
mission to maintain and operate the existing built systems.
4 )Use of Performance Measures Performance measures or indicators are used to link DOT
actions to their overall m.
2)The 2 strongest dimensions of my personality area)direction It.pdf
2)The 2 strongest dimensions of my personality are
a)direction : It speaks about the person\'s attitudes, motives, intentions and actions.Whether the
above factors are positively or negatively related to the welfare and well-being of other
individuals and society. In simple words whether the person is \'good\' or \'bad\'..
b)
Consciousness:
Consciousness refers to the relative development and predominance of the physical, vital and
mental aspects
2)
The people who are co-operating,keeping positive attitude in mind,open to work as a team,as a
strong passion for work
3)
The people who dont co-operate andd having head weight,dont adjust to situation.with these
people,it is difficult to work.
4)
with this self assesment,my communication has improved.It is more open to recieve things cleary
and positively.
Solution
2)The 2 strongest dimensions of my personality are
a)direction : It speaks about the person\'s attitudes, motives, intentions and actions.Whether the
above factors are positively or negatively related to the welfare and well-being of other
individuals and society. In simple words whether the person is \'good\' or \'bad\'..
b)
Consciousness:
Consciousness refers to the relative development and predominance of the physical, vital and
mental aspects
2)
The people who are co-operating,keeping positive attitude in mind,open to work as a team,as a
strong passion for work
3)
The people who dont co-operate andd having head weight,dont adjust to situation.with these
people,it is difficult to work.
4)
with this self assesment,my communication has improved.It is more open to recieve things cleary
and positively..
ReverseList.javaimport java.util.ArrayList;public class Rever.pdf
// ReverseList.java
import java.util.ArrayList;
public class ReverseList {
// function to reverse arrylist
public static void reverse(ArrayList list, int size)
{
for (int i = 0; i < size/2 ;i++ )
{
Double temp = list.get(i);
list.set(i, list.get(size-i-1));
list.set(size-i-1,temp);
}
}
public static void main(String args[]){
ArrayList list = new ArrayList();
list.add(4.8);
list.add(5.1);
list.add(9.1);
list.add(1.5);
list.add(0.8);
list.add(11.8);
list.add(5.0);
list.add(3.8);
int size = list.size();
System.out.println(\"Original list:\");
for(Double number: list){
System.out.print (number + \" \");
}
// reverse list
reverse(list,size);
System.out.println(\"\ \ After reversing the list:\");
for(Double number: list){
System.out.print (number + \" \");
}
}
}
/*
output:
Original list:
4.8 5.1 9.1 1.5 0.8 11.8 5.0 3.8
After reversing the list:
3.8 5.0 11.8 0.8 1.5 9.1 5.1 4.8
*/
Solution
// ReverseList.java
import java.util.ArrayList;
public class ReverseList {
// function to reverse arrylist
public static void reverse(ArrayList list, int size)
{
for (int i = 0; i < size/2 ;i++ )
{
Double temp = list.get(i);
list.set(i, list.get(size-i-1));
list.set(size-i-1,temp);
}
}
public static void main(String args[]){
ArrayList list = new ArrayList();
list.add(4.8);
list.add(5.1);
list.add(9.1);
list.add(1.5);
list.add(0.8);
list.add(11.8);
list.add(5.0);
list.add(3.8);
int size = list.size();
System.out.println(\"Original list:\");
for(Double number: list){
System.out.print (number + \" \");
}
// reverse list
reverse(list,size);
System.out.println(\"\ \ After reversing the list:\");
for(Double number: list){
System.out.print (number + \" \");
}
}
}
/*
output:
Original list:
4.8 5.1 9.1 1.5 0.8 11.8 5.0 3.8
After reversing the list:
3.8 5.0 11.8 0.8 1.5 9.1 5.1 4.8
*/.
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M.pdf
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M
Solution
pOH = 14 - 11.5 = 2.5 [OH-]= 10^-2.5 = 0.003162 M.
moles of NaOH = 0.250 1.2 10^-3 = 3 10^-4 M.pdf
moles of NaOH = 0.250 * 1.2 * 10^-3 = 3 * 10^-4 M [NaOH] = [OH-] = 3 * 10^-4
/ 1.5 = 2 *10^-4 M pH = - log [H+] [H+] = 10^-14 / [OH-] = 5 * 10^-11 pH = 10.3
Solution
moles of NaOH = 0.250 * 1.2 * 10^-3 = 3 * 10^-4 M [NaOH] = [OH-] = 3 * 10^-4
/ 1.5 = 2 *10^-4 M pH = - log [H+] [H+] = 10^-14 / [OH-] = 5 * 10^-11 pH = 10.3.
molarity = no of moles volume no of moles = mas.pdf
molarity = no of moles / volume no of moles = mass / molar mass = 1.126/204.22 =
0.0055 volume = 0.002019 l molarity = 0.0055/0.002019 = 2.73
Solution
molarity = no of moles / volume no of moles = mass / molar mass = 1.126/204.22 =
0.0055 volume = 0.002019 l molarity = 0.0055/0.002019 = 2.73.
Pharmacology Stimulates alpha and beta receptors.pdf
Pharmacology Stimulates alpha and beta receptors (alpha receptors at high doses;
beta- 1 and beta- 2 receptors at moderate doses) within the sympathetic nervous system. Relaxes
smooth muscle of bronchi and iris, and is an antagonist of histamine. Pharmacokinetics
Metabolism Inactivated by enzymatic transformation to metabephrine or normetanephrine; these
are subsequently conjugated and excreted in the urine. Elimination Mostly excreted in urine as
inactive metabolites; remainder is excreted as unchanged drug or is conjugated. Onset 5 to 10
min (subcutaneous), 1 to 5 min (inhalation). Duration 4 to 6?h (subcutaneous), 1 to 4 h (IM), 1
to 3 h (inhalation). Indications and Usage Epinephrine 1:1,000 injection Relief of respiratory
distress due to bronchospasm; to provide rapid relief of hypersensitivity reactions to drugs and
other allergens (eg, anaphylactic reactions to drugs, animal serums, insect stings); to prolong the
action of local and regional anesthetics; restore cardiac rhythm in cardiac arrest due to various
causes; treatment of mucosal congestion of hay fever, rhinitis, and acute sinusitis; relieve
bronchial asthmatic paroxysms; symptomatic relief of serum sickness, urticaria, angioneurotic
edema; for relaxation of uterine musculature and to inhibit uterine contractions; epinephrine
injection can be used as a hemostatic agent; in syncope due to complete heart block or carotid
sinus hypersensitivity; for resuscitation in cardiac arrest following anesthetic accidents; used in
open-angle glaucoma. Epinephrine 1:1,000 (auto-injector) and 1:2,000 (auto-injector), Prefilled
syringe Emergency treatment of allergic reactions (type I) including anaphylaxis to insect stings
(eg, bees, fire ants, hornets, yellow jackets, wasps) and biting insects (eg, mosquitoes), allergen
immunotherapy, foods, drugs, diagnostic testing substances (eg, radiocontrast media), and other
allergens, as well as idiopathic anaphylaxis or exercise-induced anaphylaxis. The auto-injectors
and prefilled syringes are intended for immediate self-administration in patients who are at
increased risk for anaphylaxis, including individuals with a history of anaphylactic reactions.
Auto-injectors and prefilled syringes are for immediate use and are not a substitute for immediate
medical attention. Epinephrine 1:10,000 injection Treatment and prophylaxis of cardiac arrest in
the absence of ventricular fibrillation and attacks of transitory atrioventricular heart block with
syncopal seizures; to stimulate the heart in syncope due to complete heart block or carotid sinus
hypersensitivity; for resuscitation in cardiac arrest following anesthetic accidents; in
cardiopulmonary resuscitation, intracardiac puncture and intramyocardial injection of
epinephrine may be effective when external cardiac compression and attempts to restore the
circulation by electrical defibrillation or use of pacemaker fail; seldom used as a vasopressor
except in the treatment of anaphylactic shock and under .
step1 moles of FeS = 25000x1000Molar mass of FeS.pdf
step1 moles of FeS = 25000x1000/Molar mass of FeS Step2 moles of H2SO4=
moles of FeS Step3 as n factor of H2SO4 is 2; its molarity = 1/2 Normality = 36/2 =18 Step4
Volume of 18 M H2SO4 = Moles of H2SO4 / Molarity
Solution
step1 moles of FeS = 25000x1000/Molar mass of FeS Step2 moles of H2SO4=
moles of FeS Step3 as n factor of H2SO4 is 2; its molarity = 1/2 Normality = 36/2 =18 Step4
Volume of 18 M H2SO4 = Moles of H2SO4 / Molarity.
React with water Li, Ca, Ba, Na .pdf
React with water: Li, Ca, Ba, Na
Solution
React with water: Li, Ca, Ba, Na.
The Schiff test invented [1] and named after Hugo.pdf
The Schiff test invented [1] and named after Hugo Schiff is a chemical test for the
detection of aldehydes.[2] An unknown sample is added to the decolorized Schiff reagent and
when aldehyde is present a characteristic magenta or purple color develops. The Schiff reagent is
the reaction product of Fuchsine or the closely related Pararosaniline (lacks a methyl group) and
sodium bisulfite. Human skin also contains aldehydes and gets stained as well. Schiff reagents
are used for various staining methods, e.g. Feulgen stain and periodic acid-Schiff stain. [edit]
Mechanism Fuchsin itself is colored due to its quinoid structure (see also for example viologen)
but is discolorized by sulfonation at the central carbon atom which disrupts the favorable
resonance in the molecule. More bisulphite (depending on stoichiometry ) reacts with available
amine groups to so-called N-sulfinic acid groups. The reaction of the Schiff reagent with
aldehydes is complex with several research groups reporting multiple reaction products with
model compounds. Two different mechanisms appear in the literature [3] The first mechanism
explaining its action with aldehydes was proposed by H. Wieland in 1935. In it the aldehyde
groups react with the sulfinic acid groups forming an sulfonamide. Schiff reagent sulfonamide
mechanism In the second mechanism, the pararosanilin, sulfurous acid and aldehyde combine to
form an alkyl sulfonic acid with sulfur bonded to carbon rather than to nitrogen. This mechanism
was first proposed by P. Rumpf in 1935 and experimental evidence was obtained in 1964 by
Hardonk and van Duijn.[4] They noted that when the aldehyde is reacted first with sulfurous acid
and then with the Schiff compound the UV spectrum is identical to one obtained in a standard
test. This particular logic was disputed by Stoward in 1966 although on the whole he agreed with
the new mechanism.[5] In 1980 NMR spectroscopy (Robins, Abrams, Pincock) provided further
support for the sulfonic acid mechanism.[6] Schiff reagent sulfonic acid mechanism
Solution
The Schiff test invented [1] and named after Hugo Schiff is a chemical test for the
detection of aldehydes.[2] An unknown sample is added to the decolorized Schiff reagent and
when aldehyde is present a characteristic magenta or purple color develops. The Schiff reagent is
the reaction product of Fuchsine or the closely related Pararosaniline (lacks a methyl group) and
sodium bisulfite. Human skin also contains aldehydes and gets stained as well. Schiff reagents
are used for various staining methods, e.g. Feulgen stain and periodic acid-Schiff stain. [edit]
Mechanism Fuchsin itself is colored due to its quinoid structure (see also for example viologen)
but is discolorized by sulfonation at the central carbon atom which disrupts the favorable
resonance in the molecule. More bisulphite (depending on stoichiometry ) reacts with available
amine groups to so-called N-sulfinic acid groups. The reaction of the Schiff reagent with
aldeh.
CDE are already in cyclic form A is long enough t.pdf
CDE are already in cyclic form A is long enough to form cyclic The answer is B
Solution
CDE are already in cyclic form A is long enough to form cyclic The answer is B.
C) HF is the strongest acid among them. H2O and .pdf
C) HF is the strongest acid among them. H2O and CH4 is neutral. NH3 is a base.
Ch3OH is a very weak acid.
Solution
C) HF is the strongest acid among them. H2O and CH4 is neutral. NH3 is a base.
Ch3OH is a very weak acid..
broken images, cant view, cant help Re post i.pdf
broken images, can\'t view, can\'t help Re post it
Solution
broken images, can\'t view, can\'t help Re post it.
Yes. Provided that the test is actually performed fairly (true doubl.pdf
Yes. Provided that the test is actually performed fairly (true double-blind) and also because of
the small sample size
Solution
Yes. Provided that the test is actually performed fairly (true double-blind) and also because of
the small sample size.
The studies below provide evidence linking genes and behavior1. F.pdf
The studies below provide evidence linking genes and behavior:
1. Fruit flies that are mutant for the gene period have impaired circadian rhytms
2.Twin studies comparing monozygotic and dizygotic twins have found significant heritability
for aggression.
Solution
The studies below provide evidence linking genes and behavior:
1. Fruit flies that are mutant for the gene period have impaired circadian rhytms
2.Twin studies comparing monozygotic and dizygotic twins have found significant heritability
for aggression..
the matrix does not have Enough Rational Eigenvalues to diagonalise.pdf
the matrix does not have Enough Rational Eigenvalues to diagonalise
Solution
the matrix does not have Enough Rational Eigenvalues to diagonalise.
The differences in activities between system development and system .pdf
The differences in activities between system development and system integration:
System Development: There are different activities in System Development.System study: In
order to develop any system system study is very important. System analyst is responsible for
this system study in order to propose any system, Feasibility : This activity is next after the
system study which test the proposed system. In this activioty it consists of ecnomic,
Operational, Organizational,social feasibility.Detailed study,System Analyisis, Coding, Testing,
Implementation, Maintanence.are the activities in the system development.
System Integration: It is defined as logical process of devoleping or creating new system by
increasing its performances and efficiency. system integration consists of different activities they
are system planning which starts and initiate the system and gather data and give the
requirements in this activity., project initiation which initiates the project, analysis, architecture
definition, design, devolepment, implementation, post implementation are some of the activities.
Solution
The differences in activities between system development and system integration:
System Development: There are different activities in System Development.System study: In
order to develop any system system study is very important. System analyst is responsible for
this system study in order to propose any system, Feasibility : This activity is next after the
system study which test the proposed system. In this activioty it consists of ecnomic,
Operational, Organizational,social feasibility.Detailed study,System Analyisis, Coding, Testing,
Implementation, Maintanence.are the activities in the system development.
System Integration: It is defined as logical process of devoleping or creating new system by
increasing its performances and efficiency. system integration consists of different activities they
are system planning which starts and initiate the system and gather data and give the
requirements in this activity., project initiation which initiates the project, analysis, architecture
definition, design, devolepment, implementation, post implementation are some of the activities..
The nomenclature of bacteria follows certain set of rules. During ea.pdf
The nomenclature of bacteria follows certain set of rules. During early days of nomenclature,
most bacteria were named with the following rules:
* The genus name generally indicates the scientist who discovered the bacteria, or the place
where the bacterium is discovered (ecological niche), or the nature of the bacteris (for example,
Thiobacillus, which indicates it is bacillus and is related to sulfur metabolism). The names
derived from must be in the female nominative case, either by ending with \'a\' or \'ella.\' Hence
the name Shigella.
* The species name indicates the disease or condition it causes
The names are translated to Latin in general.
The bacterium Shigella was discovered by Japanese scientist Kiyoshi Shiga. The genus name
Shigella indicates the name of this scientist who discovered it. The species name dysenterie
indicates the disease it casues.
Solution
The nomenclature of bacteria follows certain set of rules. During early days of nomenclature,
most bacteria were named with the following rules:
* The genus name generally indicates the scientist who discovered the bacteria, or the place
where the bacterium is discovered (ecological niche), or the nature of the bacteris (for example,
Thiobacillus, which indicates it is bacillus and is related to sulfur metabolism). The names
derived from must be in the female nominative case, either by ending with \'a\' or \'ella.\' Hence
the name Shigella.
* The species name indicates the disease or condition it causes
The names are translated to Latin in general.
The bacterium Shigella was discovered by Japanese scientist Kiyoshi Shiga. The genus name
Shigella indicates the name of this scientist who discovered it. The species name dysenterie
indicates the disease it casues..
answer is steam is condensed to liquid water .pdf
answer is steam is condensed to liquid water
Solution
answer is steam is condensed to liquid water.
Placement AlgorithmIt decides which program to load into the memo.pdf
Placement Algorithm:
It decides which program to load into the memory. Common placement algorithms are first-fit,
next-fit, best-fit and priority fit.
Next-fit often leads to allocation of the largest block at the end of memory
First-fit favors allocation near the beginning: tends to create less fragmentation then Next-fit
Best-fit searches for smallest block: the fragment left behind is small as possible
Main memory quickly forms holes too small to hold any process: compaction generally needs to
be done more often
The following jobs are loaded into memory using fixed partition following a certain memory
allocation method (best-fit, first-fit and worst-fit).
List of Jobs
Size
Turnaround
Job 1
100k
3
Job 2
10k
1
Job 3
35k
2
Job 4
15k
1
Job 5
23k
2
Job 6
6k
1
Job 7
25k
1
Job 8
55k
2
Job 9
88k
3
Job 10
100k
3
Memory Block Size
Block 1 50k
Block 2 200k
Block 3 70k
Block 4 115k
Block 5 15k
BEST FIT
Best-fit memory allocation makes the best use of memory space but slower in making
allocation. In the illustration below, on the first processing cycle, jobs 1 to 5 are submitted and be
processed first. After the first cycle, job 2 and 4 located on block 5 and block 3 respectively and
both having one turnaround are replace by job 6 and 7 while job 1, job 3 and job 5 remain on
their designated block. In the third cycle, job 1 remain on block 4, while job 8 and job 9 replace
job 7 and job 5 respectively (both having 2 turnaround). On the next cycle, job 9 and job 8
remain on their block while job 10 replace job 1 (having 3 turnaround). On the fifth cycle only
job 9 and 10 are the remaining jobs to be process and there are 3 free memory blocks for the
incoming jobs. But since there are only 10 jobs, so it will remain free. On the sixth cycle, job 10
is the only remaining job to be process and finally on the seventh cycle, all jobs are successfully
process and executed and all the memory blocks are now free.
FIRST- FIT
First-fit memory allocation is faster in making allocation but leads to memory waste. The
illustration below shows that on the first cycle, job 1 to job 4 are submitted first while job 6
occupied block 5 because the remaining memory space is enough to its required memory size to
be process. While job 5 is in waiting queue because the memory size in block 5 is not enough for
the job 5 to be process. Then on the next cycle, job 5 replace job 2 on block 1 and job 7 replace
job 4 on block 4 after both job 2 and job 4 finish their process. Job 8 is in waiting queue because
the remaining block is not enough to accommodate the memory size of job 8. On the third cycle,
job 8 replace job 3 and job 9 occupies block 4 after processing job 7. While Job 1 and job 5
remain on its designated block. After the third cycle block 1 and block 5 are free to serve the
incoming jobs but since there are 10 jobs so it will remain free. And job 10 occupies block 2
after job 1 finish its turns. On the other hand, job 8 and job 9 remain on their block. Then on the
fifth.
mport java.io.; class Student { int rollno; String name; .pdf
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
System.out.println(“Enter How many Students:=> “);
nostud=Integer.parseInt(br.readLine());
Student s[]=new Student[nostud];
for(int i=0;i “);
rno=Integer.parseInt(br.readLine());
System.out.println(“Enter Name:=> “);
name=br.readLine();
System.out.println(“Enter No of Subject:=> “);
no=Integer.parseInt(br.readLine());
s[i]=new Student(rno,name,no);
}
for(int i=0;i
Solution
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStre.
Matrix A A =1 5 -2 8 0 7 2 1 3 5 7 9 11 13 7 6 5 4 3 2 1.pdf
Matrix A:
A =
1 5 -2 8 0 7 2
1 3 5 7 9 11 13
7 6 5 4 3 2 1
Matrix B:
B =
2 4 6 8 10 12 14
0 1 2 -3 5 2 9
3 5 7 9 11 13 15
Matrix X:
X =
4 5 6 7 8 9 10
_____________________________________________________________________________
______________________________________
(a) Statement Y = [A;X]
Single row Matrix X is appended as last row to Matrix A and assigns to Y.
Therefore Matrix Y is:
Y =
1 5 -2 8 0 7 2
1 3 5 7 9 11 13
7 6 5 4 3 2 1
4 5 6 7 8 9 10
_____________________________________________________________________________
______________________________________
(b) Statement Z = [A,B]
Matrix B is appended as Column to Matrix A and assigns to Z.
Therefore Matrix Z is:
Z =
1 5 -2 8 0 7 2 2 4 6 8 10 12 14
1 3 5 7 9 11 13 0 1 2 -3 5 2 9
7 6 5 4 3 2 1 3 5 7 9 11 13 15
_____________________________________________________________________________
______________________________________
(c) Statement Y(3,5)
Matrix Y contains 4 rows and 7 columns.
Statement Y(3,5) represents the element present at 3rd Row, 5th Column.
Therefore, Y(3,5) = 3
_____________________________________________________________________________
______________________________________
(d) Statement Z(2,4)
Matrix Z contains 3 rows and 14 columns.
Statement Z(2,4) represents the element present at 2nd Row, 4th Column.
Therefore, Z(2,4) = 7
_____________________________________________________________________________
______________________________________
(e) Statement Y(1:2:5)
Let the denotion be Y(p:q:r).
Resultant is the same as [p,p+q,p+2q, ...,p+m*q], where m = fix((r-p)/q)
=> p = 1, q = 2, r = 5
=> m = fix((5-1)/2) = 2
=> p+m*q = 1+(2*2) = 5
=> [1, 3, 5]
So elements present at above locations:
Element at location 1: 1
Element at location 3: 7
Element at location 5: 5
Therefore Y(1:2:5) = 1 7 5
_____________________________________________________________________________
______________________________________
(f) Statement A(2:3, 2:2:6)
2:3 denotes rows 2,3
2:2:6 denotes values 2 4 6 (values 2 to 6 with a step value of 2)
A(2:3, 2:2:6) represents:
A(2,2) A(2,4) A(2,6)
A(3,2) A(3,4) A(3,6)
Therefore A(2:3, 2:2:6):
3 7 11
6 4 2
_____________________________________________________________________________
______________________________________
(g) Statement A([1,2,6,7])
Location specified as single value are represented for 3x4 matrix is as follows:
1 4 7 10
2 5 8 11
3 6 9 12
So elements present at given statement:
Element at location 1: 1
Element at location 2: 1
Element at location 6: 6
Element at location 7: -2
Therefore A([1,2,6,7]): 1 1 6 -2
_____________________________________________________________________________
______________________________________
(h) A > B
Elements present at the corresponding locations of matrices A & B are compared.
If element in matrix A is greater than element in matrix B, resultant matrix will have a value 1,
else value 0.
For example A(2,3) = 5 ; B(2,3) = 2
Since 5 > 2, resultant matrix (2,3) = 1
Result of statement A > B:
0 1 0 0 0 0 0
1 1 1 1.
Implementation of Banker’s Algorithm using java is given below .pdf
Implementation of Banker’s Algorithm using java is given below:
package bankersalgorithm;
import javax.swing.JOptionPane;
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int n = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Process:\"));
int m = Integer.parseInt(JOptionPane.showInputDialog(\"Resource Type Number:\"));
int available[] = new int[m];
int max[][] = new int[n][m];
int allocation[][] = new int[n][m];
int need[][] = new int[n][m];
String sequence = \"\";
for(int i = 0; i < m; i++)
{
available[i] = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Available Resource
\" + (i) + \":\"));
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
allocation[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"Allocation P \" + (i) + \" for
R \" + (j) + \":\"));
}
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
max[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"MAX P \" + (i) + \" for R \" + (j) +
\":\"));
need[i][j] = max[i][j] - allocation[i][j];
}
}
int work[] = available;
boolean finish[] = new boolean[n];
for(int i = 0; i < n; i++)
{
finish[i] = false;
}
boolean check = true;
while(check)
{
check = false;
for(int i = 0; i < n; i++)
{
if(!finish[i])
{
int j;
for(j = 0; j < m; j++)
{
if(need[i][j] > work[j])
{
break;
}
}
if(j == m)
{
for(j=0; j < m; j++)
{
work[j] = work[j] + allocation[i][j];
}
finish[i] = true;
check = true;
sequence += i + \", \";
}
}
}
}
int i;
for(i = 0; i < n; i++)
{
if(!finish[i])
break;
}
if(i==n)
{
JOptionPane.showMessageDialog(null, \"SAFE And Sequence is:\"+sequence);
}
else
{
JOptionPane.showMessageDialog(null, \"DEAD LOCK\");
}
}
}
Solution
Implementation of Banker’s Algorithm using java is given below:
package bankersalgorithm;
import javax.swing.JOptionPane;
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int n = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Process:\"));
int m = Integer.parseInt(JOptionPane.showInputDialog(\"Resource Type Number:\"));
int available[] = new int[m];
int max[][] = new int[n][m];
int allocation[][] = new int[n][m];
int need[][] = new int[n][m];
String sequence = \"\";
for(int i = 0; i < m; i++)
{
available[i] = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Available Resource
\" + (i) + \":\"));
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
allocation[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"Allocation P \" + (i) + \" for
R \" + (j) + \":\"));
}
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
max[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"MAX P \" + (i) + \" for R \" + (j) +
\":\"));
need[i][j] = max[i][j] - allocation[i][j];
}
}
int work[] = available;
boolean finish[] = new boolean[n];
for(int i = 0; i < n; i++)
{
finish[i] = false;
}
boolean check = true;
while(check)
{
check = false;
for(int i = 0; i < n; i++)
{
if(!finish[i])
{
int j;
f.
I think it would be 7. Because the base and acid are the same concen.pdf
I think it would be 7. Because the base and acid are the same concentration.
Solution
I think it would be 7. Because the base and acid are the same concentration..
Heat removed from the electronic parts = Heat gained by the waterN.pdf
Heat removed from the electronic parts = Heat gained by the water
N*me*Ce*(Ti - Tf) = mwater * Cwater * ( Twf - Twi)
N* 0.1 * 0.14 * (300 - 80) = (10 *10-3*1000)* 4.186 * (60-20)
3.08*N = 1674.4
N = 543.6 or 543
Solution
Heat removed from the electronic parts = Heat gained by the water
N*me*Ce*(Ti - Tf) = mwater * Cwater * ( Twf - Twi)
N* 0.1 * 0.14 * (300 - 80) = (10 *10-3*1000)* 4.186 * (60-20)
3.08*N = 1674.4
N = 543.6 or 543.
Hi,you covered mostly things.there are issue to point and link poi.pdf
Hi,
you covered mostly things.there are issue to point and link pointer.I will try my best to solve
your issue.
I have added below functions which are wrong or missing.
city* addCity(city *head, city *previous, string cityName )
{
if (head == NULL)
{
head = new city;
head->name = cityName;
return head;
}
city *newNode = new city;
newNode->name = cityName;
if (previous == NULL)
{
newNode->next = head;
head = newNode;
return head;
}
newNode->next = previous->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
return head;
}
newNode->name = cityName;
city *current = new city;
current = searchNetwork(head, previous->name);
newNode->next = current->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
city *searchNetwork(city *ptr, string cityName)
{
city *current = ptr;
while (current->name != cityName)
{
if (current->name != cityName && current->next == NULL)
return NULL;
current = current->next;
}
return current;
}
city* deleteCity(city *head, string cityName)
{
city *current = head;
cout << \"Checkpoint 1\" << endl;
while (current->next->name != cityName)
{
if (current->next == NULL)
{
cout << \"City does not exist.\" << endl;
return head;
}
current = current->next;
}
cout << \"Checkpoint 2\" << endl;
current->next = current->next->next;
return head;
}
void printPath(city *ptr)
{
cout << \"== CURRENT PATH ==\" << endl;
// If the head is NULL
if (ptr == NULL)
cout << \"nothing in path\" << endl;
else
{
city *temp2 = ptr;
while( temp2!=NULL )
{
cout << temp2->name << \" -> \";// show the data in the linked list
temp2 = temp2->next; // tranfer the address of \'temp->next\' to \'temp\'
}
cout << \"NULL\" << endl;
}
cout << \"===\" << endl;
}
city* loadDefaultSetup(city *head)
{
head = deleteEntireNetwork(head);
head = addCity(head,NULL,\"Los Angeles\");
city *two = new city;
city *three = new city;
city *four = new city;
city *five = new city;
city *six = new city;
two->name = \"Phoenix\";
three->name = \"Denver\";
four->name = \"Dallas\";
five->name = \"Atlanta\";
six->name = \"New York\";
head->next = two;
two->next = three;
three->next = four;
four->next = five;
five->next = six;
six->next = NULL;
return head;
}
Solution
Hi,
you covered mostly things.there are issue to point and link pointer.I will try my best to solve
your issue.
I have added below functions which are wrong or missing.
city* addCity(city *head, city *previous, string cityName )
{
if (head == NULL)
{
head = new city;
head->name = cityName;
return head;
}
city *newNode = new city;
newNode->name = cityName;
if (previous == NULL)
{
newNode->next = head;
head = newNode;
return head;
}
newNode->next = previous->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
return head;
}
newNode->name = cityName;
city *current = new city;
current = searchNetwork(head, previous->name);
newNode->next = current->next;
previous->next = newNode;
c.
Palestine last event orientationfvgnh .pptx
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
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step1 moles of FeS = 25000x1000Molar mass of FeS.pdf
step1 moles of FeS = 25000x1000/Molar mass of FeS Step2 moles of H2SO4=
moles of FeS Step3 as n factor of H2SO4 is 2; its molarity = 1/2 Normality = 36/2 =18 Step4
Volume of 18 M H2SO4 = Moles of H2SO4 / Molarity
Solution
step1 moles of FeS = 25000x1000/Molar mass of FeS Step2 moles of H2SO4=
moles of FeS Step3 as n factor of H2SO4 is 2; its molarity = 1/2 Normality = 36/2 =18 Step4
Volume of 18 M H2SO4 = Moles of H2SO4 / Molarity.
React with water Li, Ca, Ba, Na .pdf
React with water: Li, Ca, Ba, Na
Solution
React with water: Li, Ca, Ba, Na.
The Schiff test invented [1] and named after Hugo.pdf
The Schiff test invented [1] and named after Hugo Schiff is a chemical test for the
detection of aldehydes.[2] An unknown sample is added to the decolorized Schiff reagent and
when aldehyde is present a characteristic magenta or purple color develops. The Schiff reagent is
the reaction product of Fuchsine or the closely related Pararosaniline (lacks a methyl group) and
sodium bisulfite. Human skin also contains aldehydes and gets stained as well. Schiff reagents
are used for various staining methods, e.g. Feulgen stain and periodic acid-Schiff stain. [edit]
Mechanism Fuchsin itself is colored due to its quinoid structure (see also for example viologen)
but is discolorized by sulfonation at the central carbon atom which disrupts the favorable
resonance in the molecule. More bisulphite (depending on stoichiometry ) reacts with available
amine groups to so-called N-sulfinic acid groups. The reaction of the Schiff reagent with
aldehydes is complex with several research groups reporting multiple reaction products with
model compounds. Two different mechanisms appear in the literature [3] The first mechanism
explaining its action with aldehydes was proposed by H. Wieland in 1935. In it the aldehyde
groups react with the sulfinic acid groups forming an sulfonamide. Schiff reagent sulfonamide
mechanism In the second mechanism, the pararosanilin, sulfurous acid and aldehyde combine to
form an alkyl sulfonic acid with sulfur bonded to carbon rather than to nitrogen. This mechanism
was first proposed by P. Rumpf in 1935 and experimental evidence was obtained in 1964 by
Hardonk and van Duijn.[4] They noted that when the aldehyde is reacted first with sulfurous acid
and then with the Schiff compound the UV spectrum is identical to one obtained in a standard
test. This particular logic was disputed by Stoward in 1966 although on the whole he agreed with
the new mechanism.[5] In 1980 NMR spectroscopy (Robins, Abrams, Pincock) provided further
support for the sulfonic acid mechanism.[6] Schiff reagent sulfonic acid mechanism
Solution
The Schiff test invented [1] and named after Hugo Schiff is a chemical test for the
detection of aldehydes.[2] An unknown sample is added to the decolorized Schiff reagent and
when aldehyde is present a characteristic magenta or purple color develops. The Schiff reagent is
the reaction product of Fuchsine or the closely related Pararosaniline (lacks a methyl group) and
sodium bisulfite. Human skin also contains aldehydes and gets stained as well. Schiff reagents
are used for various staining methods, e.g. Feulgen stain and periodic acid-Schiff stain. [edit]
Mechanism Fuchsin itself is colored due to its quinoid structure (see also for example viologen)
but is discolorized by sulfonation at the central carbon atom which disrupts the favorable
resonance in the molecule. More bisulphite (depending on stoichiometry ) reacts with available
amine groups to so-called N-sulfinic acid groups. The reaction of the Schiff reagent with
aldeh.
CDE are already in cyclic form A is long enough t.pdf
CDE are already in cyclic form A is long enough to form cyclic The answer is B
Solution
CDE are already in cyclic form A is long enough to form cyclic The answer is B.
C) HF is the strongest acid among them. H2O and .pdf
C) HF is the strongest acid among them. H2O and CH4 is neutral. NH3 is a base.
Ch3OH is a very weak acid.
Solution
C) HF is the strongest acid among them. H2O and CH4 is neutral. NH3 is a base.
Ch3OH is a very weak acid..
broken images, cant view, cant help Re post i.pdf
broken images, can\'t view, can\'t help Re post it
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broken images, can\'t view, can\'t help Re post it.
Yes. Provided that the test is actually performed fairly (true doubl.pdf
Yes. Provided that the test is actually performed fairly (true double-blind) and also because of
the small sample size
Solution
Yes. Provided that the test is actually performed fairly (true double-blind) and also because of
the small sample size.
The studies below provide evidence linking genes and behavior1. F.pdf
The studies below provide evidence linking genes and behavior:
1. Fruit flies that are mutant for the gene period have impaired circadian rhytms
2.Twin studies comparing monozygotic and dizygotic twins have found significant heritability
for aggression.
Solution
The studies below provide evidence linking genes and behavior:
1. Fruit flies that are mutant for the gene period have impaired circadian rhytms
2.Twin studies comparing monozygotic and dizygotic twins have found significant heritability
for aggression..
the matrix does not have Enough Rational Eigenvalues to diagonalise.pdf
the matrix does not have Enough Rational Eigenvalues to diagonalise
Solution
the matrix does not have Enough Rational Eigenvalues to diagonalise.
The differences in activities between system development and system .pdf
The differences in activities between system development and system integration:
System Development: There are different activities in System Development.System study: In
order to develop any system system study is very important. System analyst is responsible for
this system study in order to propose any system, Feasibility : This activity is next after the
system study which test the proposed system. In this activioty it consists of ecnomic,
Operational, Organizational,social feasibility.Detailed study,System Analyisis, Coding, Testing,
Implementation, Maintanence.are the activities in the system development.
System Integration: It is defined as logical process of devoleping or creating new system by
increasing its performances and efficiency. system integration consists of different activities they
are system planning which starts and initiate the system and gather data and give the
requirements in this activity., project initiation which initiates the project, analysis, architecture
definition, design, devolepment, implementation, post implementation are some of the activities.
Solution
The differences in activities between system development and system integration:
System Development: There are different activities in System Development.System study: In
order to develop any system system study is very important. System analyst is responsible for
this system study in order to propose any system, Feasibility : This activity is next after the
system study which test the proposed system. In this activioty it consists of ecnomic,
Operational, Organizational,social feasibility.Detailed study,System Analyisis, Coding, Testing,
Implementation, Maintanence.are the activities in the system development.
System Integration: It is defined as logical process of devoleping or creating new system by
increasing its performances and efficiency. system integration consists of different activities they
are system planning which starts and initiate the system and gather data and give the
requirements in this activity., project initiation which initiates the project, analysis, architecture
definition, design, devolepment, implementation, post implementation are some of the activities..
The nomenclature of bacteria follows certain set of rules. During ea.pdf
The nomenclature of bacteria follows certain set of rules. During early days of nomenclature,
most bacteria were named with the following rules:
* The genus name generally indicates the scientist who discovered the bacteria, or the place
where the bacterium is discovered (ecological niche), or the nature of the bacteris (for example,
Thiobacillus, which indicates it is bacillus and is related to sulfur metabolism). The names
derived from must be in the female nominative case, either by ending with \'a\' or \'ella.\' Hence
the name Shigella.
* The species name indicates the disease or condition it causes
The names are translated to Latin in general.
The bacterium Shigella was discovered by Japanese scientist Kiyoshi Shiga. The genus name
Shigella indicates the name of this scientist who discovered it. The species name dysenterie
indicates the disease it casues.
Solution
The nomenclature of bacteria follows certain set of rules. During early days of nomenclature,
most bacteria were named with the following rules:
* The genus name generally indicates the scientist who discovered the bacteria, or the place
where the bacterium is discovered (ecological niche), or the nature of the bacteris (for example,
Thiobacillus, which indicates it is bacillus and is related to sulfur metabolism). The names
derived from must be in the female nominative case, either by ending with \'a\' or \'ella.\' Hence
the name Shigella.
* The species name indicates the disease or condition it causes
The names are translated to Latin in general.
The bacterium Shigella was discovered by Japanese scientist Kiyoshi Shiga. The genus name
Shigella indicates the name of this scientist who discovered it. The species name dysenterie
indicates the disease it casues..
answer is steam is condensed to liquid water .pdf
answer is steam is condensed to liquid water
Solution
answer is steam is condensed to liquid water.
Placement AlgorithmIt decides which program to load into the memo.pdf
Placement Algorithm:
It decides which program to load into the memory. Common placement algorithms are first-fit,
next-fit, best-fit and priority fit.
Next-fit often leads to allocation of the largest block at the end of memory
First-fit favors allocation near the beginning: tends to create less fragmentation then Next-fit
Best-fit searches for smallest block: the fragment left behind is small as possible
Main memory quickly forms holes too small to hold any process: compaction generally needs to
be done more often
The following jobs are loaded into memory using fixed partition following a certain memory
allocation method (best-fit, first-fit and worst-fit).
List of Jobs
Size
Turnaround
Job 1
100k
3
Job 2
10k
1
Job 3
35k
2
Job 4
15k
1
Job 5
23k
2
Job 6
6k
1
Job 7
25k
1
Job 8
55k
2
Job 9
88k
3
Job 10
100k
3
Memory Block Size
Block 1 50k
Block 2 200k
Block 3 70k
Block 4 115k
Block 5 15k
BEST FIT
Best-fit memory allocation makes the best use of memory space but slower in making
allocation. In the illustration below, on the first processing cycle, jobs 1 to 5 are submitted and be
processed first. After the first cycle, job 2 and 4 located on block 5 and block 3 respectively and
both having one turnaround are replace by job 6 and 7 while job 1, job 3 and job 5 remain on
their designated block. In the third cycle, job 1 remain on block 4, while job 8 and job 9 replace
job 7 and job 5 respectively (both having 2 turnaround). On the next cycle, job 9 and job 8
remain on their block while job 10 replace job 1 (having 3 turnaround). On the fifth cycle only
job 9 and 10 are the remaining jobs to be process and there are 3 free memory blocks for the
incoming jobs. But since there are only 10 jobs, so it will remain free. On the sixth cycle, job 10
is the only remaining job to be process and finally on the seventh cycle, all jobs are successfully
process and executed and all the memory blocks are now free.
FIRST- FIT
First-fit memory allocation is faster in making allocation but leads to memory waste. The
illustration below shows that on the first cycle, job 1 to job 4 are submitted first while job 6
occupied block 5 because the remaining memory space is enough to its required memory size to
be process. While job 5 is in waiting queue because the memory size in block 5 is not enough for
the job 5 to be process. Then on the next cycle, job 5 replace job 2 on block 1 and job 7 replace
job 4 on block 4 after both job 2 and job 4 finish their process. Job 8 is in waiting queue because
the remaining block is not enough to accommodate the memory size of job 8. On the third cycle,
job 8 replace job 3 and job 9 occupies block 4 after processing job 7. While Job 1 and job 5
remain on its designated block. After the third cycle block 1 and block 5 are free to serve the
incoming jobs but since there are 10 jobs so it will remain free. And job 10 occupies block 2
after job 1 finish its turns. On the other hand, job 8 and job 9 remain on their block. Then on the
fifth.
mport java.io.; class Student { int rollno; String name; .pdf
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
System.out.println(“Enter How many Students:=> “);
nostud=Integer.parseInt(br.readLine());
Student s[]=new Student[nostud];
for(int i=0;i “);
rno=Integer.parseInt(br.readLine());
System.out.println(“Enter Name:=> “);
name=br.readLine();
System.out.println(“Enter No of Subject:=> “);
no=Integer.parseInt(br.readLine());
s[i]=new Student(rno,name,no);
}
for(int i=0;i
Solution
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStre.
Matrix A A =1 5 -2 8 0 7 2 1 3 5 7 9 11 13 7 6 5 4 3 2 1.pdf
Matrix A:
A =
1 5 -2 8 0 7 2
1 3 5 7 9 11 13
7 6 5 4 3 2 1
Matrix B:
B =
2 4 6 8 10 12 14
0 1 2 -3 5 2 9
3 5 7 9 11 13 15
Matrix X:
X =
4 5 6 7 8 9 10
_____________________________________________________________________________
______________________________________
(a) Statement Y = [A;X]
Single row Matrix X is appended as last row to Matrix A and assigns to Y.
Therefore Matrix Y is:
Y =
1 5 -2 8 0 7 2
1 3 5 7 9 11 13
7 6 5 4 3 2 1
4 5 6 7 8 9 10
_____________________________________________________________________________
______________________________________
(b) Statement Z = [A,B]
Matrix B is appended as Column to Matrix A and assigns to Z.
Therefore Matrix Z is:
Z =
1 5 -2 8 0 7 2 2 4 6 8 10 12 14
1 3 5 7 9 11 13 0 1 2 -3 5 2 9
7 6 5 4 3 2 1 3 5 7 9 11 13 15
_____________________________________________________________________________
______________________________________
(c) Statement Y(3,5)
Matrix Y contains 4 rows and 7 columns.
Statement Y(3,5) represents the element present at 3rd Row, 5th Column.
Therefore, Y(3,5) = 3
_____________________________________________________________________________
______________________________________
(d) Statement Z(2,4)
Matrix Z contains 3 rows and 14 columns.
Statement Z(2,4) represents the element present at 2nd Row, 4th Column.
Therefore, Z(2,4) = 7
_____________________________________________________________________________
______________________________________
(e) Statement Y(1:2:5)
Let the denotion be Y(p:q:r).
Resultant is the same as [p,p+q,p+2q, ...,p+m*q], where m = fix((r-p)/q)
=> p = 1, q = 2, r = 5
=> m = fix((5-1)/2) = 2
=> p+m*q = 1+(2*2) = 5
=> [1, 3, 5]
So elements present at above locations:
Element at location 1: 1
Element at location 3: 7
Element at location 5: 5
Therefore Y(1:2:5) = 1 7 5
_____________________________________________________________________________
______________________________________
(f) Statement A(2:3, 2:2:6)
2:3 denotes rows 2,3
2:2:6 denotes values 2 4 6 (values 2 to 6 with a step value of 2)
A(2:3, 2:2:6) represents:
A(2,2) A(2,4) A(2,6)
A(3,2) A(3,4) A(3,6)
Therefore A(2:3, 2:2:6):
3 7 11
6 4 2
_____________________________________________________________________________
______________________________________
(g) Statement A([1,2,6,7])
Location specified as single value are represented for 3x4 matrix is as follows:
1 4 7 10
2 5 8 11
3 6 9 12
So elements present at given statement:
Element at location 1: 1
Element at location 2: 1
Element at location 6: 6
Element at location 7: -2
Therefore A([1,2,6,7]): 1 1 6 -2
_____________________________________________________________________________
______________________________________
(h) A > B
Elements present at the corresponding locations of matrices A & B are compared.
If element in matrix A is greater than element in matrix B, resultant matrix will have a value 1,
else value 0.
For example A(2,3) = 5 ; B(2,3) = 2
Since 5 > 2, resultant matrix (2,3) = 1
Result of statement A > B:
0 1 0 0 0 0 0
1 1 1 1.
Implementation of Banker’s Algorithm using java is given below .pdf
Implementation of Banker’s Algorithm using java is given below:
package bankersalgorithm;
import javax.swing.JOptionPane;
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int n = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Process:\"));
int m = Integer.parseInt(JOptionPane.showInputDialog(\"Resource Type Number:\"));
int available[] = new int[m];
int max[][] = new int[n][m];
int allocation[][] = new int[n][m];
int need[][] = new int[n][m];
String sequence = \"\";
for(int i = 0; i < m; i++)
{
available[i] = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Available Resource
\" + (i) + \":\"));
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
allocation[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"Allocation P \" + (i) + \" for
R \" + (j) + \":\"));
}
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
max[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"MAX P \" + (i) + \" for R \" + (j) +
\":\"));
need[i][j] = max[i][j] - allocation[i][j];
}
}
int work[] = available;
boolean finish[] = new boolean[n];
for(int i = 0; i < n; i++)
{
finish[i] = false;
}
boolean check = true;
while(check)
{
check = false;
for(int i = 0; i < n; i++)
{
if(!finish[i])
{
int j;
for(j = 0; j < m; j++)
{
if(need[i][j] > work[j])
{
break;
}
}
if(j == m)
{
for(j=0; j < m; j++)
{
work[j] = work[j] + allocation[i][j];
}
finish[i] = true;
check = true;
sequence += i + \", \";
}
}
}
}
int i;
for(i = 0; i < n; i++)
{
if(!finish[i])
break;
}
if(i==n)
{
JOptionPane.showMessageDialog(null, \"SAFE And Sequence is:\"+sequence);
}
else
{
JOptionPane.showMessageDialog(null, \"DEAD LOCK\");
}
}
}
Solution
Implementation of Banker’s Algorithm using java is given below:
package bankersalgorithm;
import javax.swing.JOptionPane;
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int n = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Process:\"));
int m = Integer.parseInt(JOptionPane.showInputDialog(\"Resource Type Number:\"));
int available[] = new int[m];
int max[][] = new int[n][m];
int allocation[][] = new int[n][m];
int need[][] = new int[n][m];
String sequence = \"\";
for(int i = 0; i < m; i++)
{
available[i] = Integer.parseInt(JOptionPane.showInputDialog(\"Number Of Available Resource
\" + (i) + \":\"));
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
allocation[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"Allocation P \" + (i) + \" for
R \" + (j) + \":\"));
}
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
max[i][j] = Integer.parseInt(JOptionPane.showInputDialog(\"MAX P \" + (i) + \" for R \" + (j) +
\":\"));
need[i][j] = max[i][j] - allocation[i][j];
}
}
int work[] = available;
boolean finish[] = new boolean[n];
for(int i = 0; i < n; i++)
{
finish[i] = false;
}
boolean check = true;
while(check)
{
check = false;
for(int i = 0; i < n; i++)
{
if(!finish[i])
{
int j;
f.
I think it would be 7. Because the base and acid are the same concen.pdf
I think it would be 7. Because the base and acid are the same concentration.
Solution
I think it would be 7. Because the base and acid are the same concentration..
Heat removed from the electronic parts = Heat gained by the waterN.pdf
Heat removed from the electronic parts = Heat gained by the water
N*me*Ce*(Ti - Tf) = mwater * Cwater * ( Twf - Twi)
N* 0.1 * 0.14 * (300 - 80) = (10 *10-3*1000)* 4.186 * (60-20)
3.08*N = 1674.4
N = 543.6 or 543
Solution
Heat removed from the electronic parts = Heat gained by the water
N*me*Ce*(Ti - Tf) = mwater * Cwater * ( Twf - Twi)
N* 0.1 * 0.14 * (300 - 80) = (10 *10-3*1000)* 4.186 * (60-20)
3.08*N = 1674.4
N = 543.6 or 543.
Hi,you covered mostly things.there are issue to point and link poi.pdf
Hi,
you covered mostly things.there are issue to point and link pointer.I will try my best to solve
your issue.
I have added below functions which are wrong or missing.
city* addCity(city *head, city *previous, string cityName )
{
if (head == NULL)
{
head = new city;
head->name = cityName;
return head;
}
city *newNode = new city;
newNode->name = cityName;
if (previous == NULL)
{
newNode->next = head;
head = newNode;
return head;
}
newNode->next = previous->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
return head;
}
newNode->name = cityName;
city *current = new city;
current = searchNetwork(head, previous->name);
newNode->next = current->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
city *searchNetwork(city *ptr, string cityName)
{
city *current = ptr;
while (current->name != cityName)
{
if (current->name != cityName && current->next == NULL)
return NULL;
current = current->next;
}
return current;
}
city* deleteCity(city *head, string cityName)
{
city *current = head;
cout << \"Checkpoint 1\" << endl;
while (current->next->name != cityName)
{
if (current->next == NULL)
{
cout << \"City does not exist.\" << endl;
return head;
}
current = current->next;
}
cout << \"Checkpoint 2\" << endl;
current->next = current->next->next;
return head;
}
void printPath(city *ptr)
{
cout << \"== CURRENT PATH ==\" << endl;
// If the head is NULL
if (ptr == NULL)
cout << \"nothing in path\" << endl;
else
{
city *temp2 = ptr;
while( temp2!=NULL )
{
cout << temp2->name << \" -> \";// show the data in the linked list
temp2 = temp2->next; // tranfer the address of \'temp->next\' to \'temp\'
}
cout << \"NULL\" << endl;
}
cout << \"===\" << endl;
}
city* loadDefaultSetup(city *head)
{
head = deleteEntireNetwork(head);
head = addCity(head,NULL,\"Los Angeles\");
city *two = new city;
city *three = new city;
city *four = new city;
city *five = new city;
city *six = new city;
two->name = \"Phoenix\";
three->name = \"Denver\";
four->name = \"Dallas\";
five->name = \"Atlanta\";
six->name = \"New York\";
head->next = two;
two->next = three;
three->next = four;
four->next = five;
five->next = six;
six->next = NULL;
return head;
}
Solution
Hi,
you covered mostly things.there are issue to point and link pointer.I will try my best to solve
your issue.
I have added below functions which are wrong or missing.
city* addCity(city *head, city *previous, string cityName )
{
if (head == NULL)
{
head = new city;
head->name = cityName;
return head;
}
city *newNode = new city;
newNode->name = cityName;
if (previous == NULL)
{
newNode->next = head;
head = newNode;
return head;
}
newNode->next = previous->next;
previous->next = newNode;
cout << \"prev: \" << previous->name << \" new: \" << cityName << endl;
return head;
}
newNode->name = cityName;
city *current = new city;
current = searchNetwork(head, previous->name);
newNode->next = current->next;
previous->next = newNode;
c.
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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.
678020731-Sumas-y-Restas-Para-Colorear.pdf
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Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
special B.ed 2nd year old paper_20240531.pdf
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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.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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Home assignment II on Spectroscopy 2024 Answers.pdf
Home assignment II on Spectroscopy 2024 Answers.pdf
Instructions for Submissions thorugh G- Classroom.pptx
Instructions for Submissions thorugh G- Classroom.pptx
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI fund...
The French Revolution Class 9 Study Material pdf free download
The French Revolution Class 9 Study Material pdf free download
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