Present and explain all the elements in Freud theory.SolutionI.pdf
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Present and explain all the elements in Freud theory. Solution In the year 1905, Freud developed a model of the mind, which he used to describe the features of the mind’s structure and function. He stated that on the surface of the mind is consciousness, which consists of those thoughts that are the focus of our attention now. The preconscious consists of all which can be retrieved from memory. The third and most significant region is the unconscious. Here lie the processes that are the real cause of most behaviour. In 1923, Freud developed a structural model of mind. Mind has entities like id, ego and superego. Freud assumed the id operated at an unconscious level according to the pleasure principle. Eros, or life instinct, helps the individual to survive; it directs life-sustaining activities such as respiration, eating. The ego develops from the id during infancy. The ego\'s goal is to satisfy the demands of the id in a safe a socially acceptable way. The superego develops during early childhood (when the child identifies with the same sex parent) and is responsible for ensuring moral standards are followed. Freud\'s theory is very good at explaining behavior. His theory of cognitive psychology has identified unconscious processes, such as procedural memory and automatic processing..
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Clearly instant gratification of these desires is not always possible and thus psychological tension is created
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Process. The Id uses Primary Process to fulfill the need to act on an urge that is dangerous or unacceptable
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}
void display()
{
//it displays queue data with characters of queue if invalid choice is entered
st *tmp;
int j = 0;
cout << \"\ queue is:\ character \\tyear\";
for(tmp = front; tmp != rear; tmp = tmp->next, j++)
{
cout << \"\ \" << chr[j] << \" \" << tmp->year;
}
cout << \"\ \"<year;
}
private:
st *front, *rear;
};
int main()
{
//main method to call queue functions by choice
queue s1;
int ch,yr;
char el;
bool s;
while(1)
{
cout << \"\ 1.ENTERQUEUE\ 2.DELETEQUEUE\ 3.ISEMPTY\ 4.ISFULL\ 5.EXIT\ enter
your choice: \";
cin >> ch;
switch(ch)
{
case 1:
if(!s1.isfull())
{
cout << \"\ enter a year : \";
cin >> yr;
cout << \"\ enter a character : \";
cin >> el;
s1.enterqueue(yr, el);
}
else
cout << \"insertion not possible\";
break;
case 2:
s = s1.isempty();
if(!s)
{
el = s1.deletequeue();
cout << \"\ The character of queue \" << el <<\" is deleted\" << endl;
}
else
cout << \"\ queue is empty.deletion not possible\"<< endl;
break;
case 3:
s = s1.isempty();
if(s)
cout << \"\ queue is empty\" << endl;
else
cout << \"\ queue is not empty\" << endl;
break;
case 4:
s=s1.isfull();
if(s)
cout << \"\ queue is full\" << endl;
else
cout << \"\ queue is not full\" << endl;
break;
case 5:
exit(0);
default : cout << \"invalid choice\" << endl;
if(!s1.isempty())
s1.display();
}
}
return (0);
}
Solution
#include
#include
using namespace std;
struct st
{
int year;
st *next;
};
class queue
{
public:
//int sz,i=0;//modified //code modified here// everything is fine with the code except this...
queue is implemented correctly
//you can .
Count Red Nodes. Write a program that computes the percentage of red.pdf
Count Red Nodes. Write a program that computes the percentage of red nodes in a given red-
black BST. Test your program by running at least 100 trials of the experiment of inserting N
random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis.
*It is experiment from textbook Algorithms, 3.3.42
Does anyone know the answer?
Solution
/*
* Java Program to Implement Red Black Tree
*/
import java.util.Scanner;
/* Class Node */
class RedBlackNode
{
RedBlackNode left, right;
int element;
int color;
/* Constructor */
public RedBlackNode(int theElement)
{
this( theElement, null, null );
}
/* Constructor */
public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt)
{
left = lt;
right = rt;
element = theElement;
color = 1;
}
}
/* Class RBTree */
class RBTree
{
private RedBlackNode current;
private RedBlackNode parent;
private RedBlackNode grand;
private RedBlackNode great;
private RedBlackNode header;
private static RedBlackNode nullNode;
/* static initializer for nullNode */
static
{
nullNode = new RedBlackNode(0);
nullNode.left = nullNode;
nullNode.right = nullNode;
}
/* Black - 1 RED - 0 */
static final int BLACK = 1;
static final int RED = 0;
/* Constructor */
public RBTree(int negInf)
{
header = new RedBlackNode(negInf);
header.left = nullNode;
header.right = nullNode;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return header.right == nullNode;
}
/* Make the tree logically empty */
public void makeEmpty()
{
header.right = nullNode;
}
/* Function to insert item */
public void insert(int item )
{
current = parent = grand = header;
nullNode.element = item;
while (current.element != item)
{
great = grand;
grand = parent;
parent = current;
current = item < current.element ? current.left : current.right;
// Check if two red children and fix if so
if (current.left.color == RED && current.right.color == RED)
handleReorient( item );
}
// Insertion fails if already present
if (current != nullNode)
return;
current = new RedBlackNode(item, nullNode, nullNode);
// Attach to parent
if (item < parent.element)
parent.left = current;
else
parent.right = current;
handleReorient( item );
}
private void handleReorient(int item)
{
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if (parent.color == RED)
{
// Have to rotate
grand.color = RED;
if (item < grand.element != item < parent.element)
parent = rotate( item, grand ); // Start dbl rotate
current = rotate(item, great );
current.color = BLACK;
}
// Make root black
header.right.color = BLACK;
}
private RedBlackNode rotate(int item, RedBlackNode parent)
{
if(item < parent.element)
return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) :
rotateWithRightChild(parent.left) ;
else
return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) :
rotateWithRightChild(parent.right);
}
/* Rotate binary tree node with left child */
private RedBlackNode rot.
apple 2008 historically what were Apples major competitive advanta.pdf
apple 2008
historically what were Apples major competitive advantages and why did they fail to build on
these advantages to lead the industry?
Solution
Apples major competitive advantages:
Apple II and Macintosh were the machines that made Apple. They both had to following
features.
Brand loyalty among important segments
Reasons for why did Apple fail to build on these advantages to lead the industry:
Apple’s missteps:
Entry of Rivals:.
A partial results from a PERT analysis for a project with 50 activit.pdf
A partial results from a PERT analysis for a project with 50 activities shows that path A-K-Q
Solution
There are 50 activities shows the path A-K-Q.
With the help of PERT (Program Evaluation and Review Technique), and CPM (Critical Path
Method), the project manager can
1. Plan the project ahead of time and foresee possible sources of troubles and delays in
completion.
2. Schedule the project activities at the appropriate times to conform with proper job sequence so
that the project is completed as soon as possible.
3. Coordinate and control the projects activities so as to stay on schedule in completing the
project.
Thus PERT incorporates uncertainties in activity times in its analysis. It determines the
probabilities of completing various stages of the project by specified deadlines. It also calculates
the expected time to complete the project. An important and extremely useful by product of
PERT analysis is its identification of various “bottlenecks” in a project.
Application of PERT and CPM
A partial list of applications of PERT and CPM techniques is
1. Construction projects (e.g., buildings, highways, houses, and bridges)
2. Preparation of bids and proposals for large projects.
3. Maintenance planning of oil refineries, ship repairs, and other large operations.
4. Development of new weapons systems and new manufactured products.
5. Manufacture and assembly of large items such as airplanes, ships, and computers. 6. Simple
projects.
Six steps :
1. Define the projects and all of its significant activities or tasks.
2. Develop the relationships among the activities. Decide which activities must precede and
which must follow others.
3. Draw the network connecting all of the activities.
4. Assign time and/or cost estimates to each activity.
5. Compute the longest time path through the network; this is called the critical path.
6. Use the network to help plan, schedule, monitor and control the project..
briefly write about the variability of natural systems and the signi.pdf
briefly write about the variability of natural systems and the significance of the rate of change
and predictability to the stability and functioning of ecological systems
Solution
Answer:- The natural ecological systems are usually variable in physical and biological
characteristics that are always in dynamic equilibrium with their biota. The ecological systems
which diverse both in eveness and richness of species are usually stable and less prone to
extinction and resist the disturbances created by any external agency. The keystone species play
an important role in the ecosystem functioning and overall community stability. Thus variability
and stability are essential for functioning of ecological systems..
A graph is symmetric with respect to the vertical line corresponding.pdf
A graph is symmetric with respect to the vertical line corresponding to t= pi/2 and r > 0. What
does this say about the algebraic symmetry of the function?
A. r(t)=r(pi-t)
B. r(t)=r(-t)
C. r(t)=r(pi/2-t)
D. r(t)=-r(t)
Solution
Symmetrical wrt to vertical line: t=pi/2 and r>0
which above the horizontal axisi.e. 0--- pi rad
So, it means r(t) = r(180 -t) = r(pi -t)
Option A.
X-Linked Recessive Write three rules to keep in mind when counseling .pdf
X-Linked Recessive Write three rules to keep in mind when counseling someone about the
likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About
parents\' phenotype. About the probability of the offspring showing the trait. If instead of having
Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked
recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More
information is required to answer this question.
Solution
About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of
her X chromosome and the father do not have the recessive allele in his genotype, then none of
the daughters will suffer from the disease (since they will inherit only one recessive allele from
the mother) but sons will have a 50% probability of suffering from the disease as they will
inherit the recessive allele from the mother and the Y chromosome from the father, although
there is a 50% probability of inheriting the wild type allele from the mother.
Now if the mother is suffering from the disease i.e. has the recessive allele on both the X
chromosome while the father is healthy having no allele for the disease, the daughters will
always be a carrier but never suffer from the disease as they eill inherit the healthy wild type
allele from the father as one X chromosome comes from the father and the other X chromosome
from the mother. The sons will however all suffer from the disease as they inherit the Y
chromosome from the father and the X chromosome having the recessive allele causing the
disease will be inherited from the mother.
Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome
but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele
carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the
daughters will also have a chance of suffering from the disease.
If both the parents do not have a single recessive X linked disease gene on any of their X
chromosome then none of the children will suffer from the disease.
About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X
Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier
for a X linked recessive disease as males have only one X chromosome. If the diseased allele in
present on that X chromosome in males it will be expressed as there is no dominant healthy wild
type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on
X chromosome, the father\'s phenotype will be expresding the disease while if the father do not
have the allele he will be healthy. On the other hand for the mother to express the disease the
recessive allele has to be present on bith her X chromosome.
Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.
Write a Java Class to Implement a Generic Linked ListYour list mus.pdf
Write a Java Class to Implement a Generic Linked List
Your list must be implemented as a singly-linked list of generic nodes, where each Node object
has two instance variables: an object of the “type variable” class, and a pointer to the next node
on the list.
Your class will contain separate methods to handle each of the operations read from a data file
(see II., below)
Your class will also override toString() to return the objects on the list in the order in which they
occur.
Write a Test Class for Your LinkedList Class
Your main method will read list operation instructions from a data file, until end-of-file and call
the appropriate LinkedList method to execute each one.
After each operation is executed, print out the operation and the updated list.
The data file to be used is on the class web site and the operations are:
APPEND X - Append object X to the end of the list
ADD N X - Insert object X as the new Nth element in the list, increasing the size of the list by 1
E.g. Suppose the list is:
head -> 1 -> 2 -> 3 -> 4->null
After ADD 3 7 it would be:
head -> 1 -> 2 -> 7 -> 3 -> 4->null
DELETE N – Remove the Nth object from the list
SWAP M N - Interchange the positions of the Mth and Nth
objects on the list
For credit, the two nodes must actually \"trade places\" in the list, and not merely swap their data
values
REVERSE - Reverse the order of the objects on the list
This must be done by reversing the order of the nodes themselves, rather than by swapping the
data stored
To get credit for your reverse() method, it must use either one of these 2 algorithms:
For each node on the list except the current “head”
node, delete the node and insert it as the new head
Use your swap() method
6. CLEAR – Clear the list (make it empty)
No credit will be given for programs that use any additional data structures – either from the Java
API or programmer defined -, other than your own LinkedList class
txt file:
Solution
//Java Program to Implement Singly Linked List
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
/* Class Node */
class Nodes
{
protected Object data;
protected Nodes next;
/* Constructor */
public Nodes()
{
next = null;
data = null;
}
/* Constructor */
public Nodes(Object d,Nodes n)
{
data = d;
next = n;
}
/* Function to set link to next Node */
public void setLink(Nodes n)
{
next = n;
}
/* Function to set data to current Node */
public void setData(Object d)
{
data = d;
}
/* Function to get link to next node */
public Nodes getLink()
{
return next;
}
/* Function to get data from current Node */
public Object getData()
{
return data;
}
}
/* Class linkedList */
class linkList
{
protected Nodes start;
protected Nodes end ;
public int size ;
/* Constructor */
public linkList()
{
start = null;
end = null;
size = 0;
}
/* Function to check if list is empty */
public boolean isEmpty()
{
return start == null;
}
/* Function to get size of list */
public int getSize()
{
return .
Write a class (BasketballTeam) encapsulating the concept of a tea.pdf
Write a class (BasketballTeam) encapsulating the concept of a team of basketball players. This
method should have a single instance variable which is an array of basketball player objects. In
addition to that class, you will need to design and a code a Playerclass to encapsulate the concept
of a basketball player, assuming a basketball player has the following attributes: a name, a
position, number of shots taken, and number of shots made.
In your Player class you should write the following methods:
A constructor taking a String for the name, a String for the position, an int for the number of
shots taken, and an int for the number of shots made.
Accessor, mutator, toString, and equals methods.
A method returning the individual shooting percentage. Avg = Shots Made / Shots Taken. Hint:
This method should return a double!!!
In your BasketballTeam class you should write the following methods:
A constructor taking an array of Player objects as its only parameter and assigning that array to
the array data member of the class, its only instance variable.
Accessor, mutator, toString, and equals methods.
A method checking if all positions are different, returning true if they are, false if they are not.
A method returning the shooting percentage of the team. Team Shooting Percentage = Total
Shots Made / Total Shots Taken Hint: This should return a double!!! Also, this is not the average
of the the averages.
A method checking that we have a center (that is, the name of the position) on the team. If we
do not have any return false; otherwise, it returns true;
A method returning void that sorts the array of Player objects in ascending order using the
number of shots taken as the sorting key.
A method returning the array of Player objects sorted in ascending alphabetical order by name.
Hint: Use compareTo method of the String class.
A method returning the name of the player with the most shots made.
A method returning the name of the player with the highest shooting average.
In your BasketballTeamClient class, when you test all your methods, you can hard-code five
basketball Player objects. This client should call all 12 methods including the constructor for full
credit.
When writing the Player class you should write a PlayerClient class that tests all its methods
before integrating into BasketballTeam class.
Solution
public class Player {
//Attributes
private String name;
private String position;
private int shotsTaken;
private int shotsMade;
/**
* Constructor
* @param name
* @param position
* @param shotsTaken
* @param shotsMade
*/
public Player(String name, String position, int shotsTaken, int shotsMade) {
this.name = name;
this.position = position;
this.shotsTaken = shotsTaken;
this.shotsMade = shotsMade;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the position
*/
public String getPosition() {
return position;
}
/**
* @p.
Which of the following would you expect to happen in a plant that doe.pdf
Which of the following would you expect to happen in a plant that does not have enough water
(multiple answers are correct) Without water, the plant cannot make the sugar it needs to grow
the stem begins to dry out growth would be stunted because it is difficult to move nutrients
around Without water, the plant would become limp.
Solution
Answer: All the choices given are correct! If Plant that does not have enough water;
(a) Without water, plants cannot make the sugar it needs to grow.
Plants make their own food by the process of photosynthesis.The amount of water, sunlight and
carbon dioxide directly influences the process of food production for plant growth. Thus the
process of photosynthesis slows down and even stops when water is absent, which causes
internal food supplies to diminish.
(b) the stem begins to dry out
Scarcity of water leads to loss of turgor pressure. Stomata on the stem surface close as a result of
less water (for water conservation) and as a result stem dries out.
(c) the growth would be stunted because it is difficult to move nutrients around.
As water moves above the plant system it carries vital minerals and nutrients to different areas of
the plant. The stomas on the leaf and stem surface release waste products (oxygen) into the
environment and takes CO2 in. This process of transpiration maintains the turgor pressure and
also cools the plant in sunlight. In scarcity of water, transpiration process slows and, the plant
show stunded growth due to lack of nutrients supply.
(d) Without water the plant would become limp
Turgor pressure holds the plant cells and keeps them erect. This pressure is lost without water
and plants begin to collapse (wilting!) or give a limp appearance..
What is the value of including personal information about yourself a.pdf
What is the value of including personal information about yourself and your lifestyle when
researchers are looking at your DNA? How does study different than many other similar sets of
data gathered from GWAS and SNP Haplotypes?What do you think about the scope and purpose
of this type of genetics data study.
Solution
Answer to including personal information and lifestyle lies in phenotype differences. Phenotype
is influenced by genotype and environment. Hence although scientists and researchers look for
DNA or genetic material of an individual but also take a look on the environmental conditions or
life style which influences the phenotype of an organism.
GWAS (Genome wide association study) is study of genetic variants which are associated with a
trait like any major disease. By using SNPs (Short nuclear polymorphisms) short interspersed
regions can be identified to locate the traits of interest. The environmental factors like sunlight
can differ the response of an individual to it. One individual can be sensitiove to it but the other
could be diseased on exposure to sunlight, thus having a knowledge about individual\'s lifestyle
helps in tracking previous information.
The scope of GWAS has wider implications. Disease dignosis using genetics data helps in
developing drugs relating to it. It can be used to identify genes from different organisms.
Ancestral hierarchy can be established..
What is a voluntary response sampleSolutionVoluntary response.pdf
What is a voluntary response sample?
Solution
Voluntary response sampling is heavily biased because it focuses on volunteered survey answers
rather than a random sampling. What makes this biased is because the only people who volunteer
are those who have a particular interest in the topic of the survey. They only include people who
choose volunteer, whereas a random sample would need to include people whether or not they
choose to volunteer. Often, voluntary response samples oversample people who have strong
opinions and undersample people who don\'t care much about the topic of the survey. Thus
inferences from a voluntary response sample are not as trustworthy as conclusions based on a
random sample of the entire population under consideration..
What adaptive benefit do these abilities give wolbachia In oth.pdf
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
Solution
Manipulating the host reproduction help its survival as parasite.The improvement of complex
lifestyles cycles can growth parasite health. Some parasites have, via host manipulation, further
improved their reproductive success. Host manipulation in a few parasites, where they
effectively can change the behavior or morphology of their intermediate host in a way that
increases the parasite\'s trophic transmission
It is a method of parasite transmission wherein the parasite manipulates the behavior of its host
to growth transmission success to better trophic levels.
Think about autumn (Fall season) in places like New England (Pennsyl.pdf
Think about autumn (Fall season) in places like New England (Pennsylvania, New York,
Vermont, New Hampshire), where the leaves change color with the seasons. During the fall,
which pigments are the first to be broken down in the leaves? How does that explain the fall
colors we see? Why does the green color disappears during autumn (Fall season)? Please fully
and clearly explain why the green color disappeares in trees located in New England during the
autumn.
Solution
A number of factors play crucial role in changing the color of leaves in winter. Some are listed
below:
Similar is the reason with the New England trees..
The orange is which type of fruit Simple - Legume Simple - Drupe .pdf
The orange is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple
Aggregate The peas is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry
Multiple Aggregate The pistachio is which type of fruit? Simple - Legume Simple - Drupe
Simple - Berry Multiple Aggregate
Solution
Please find the answers below and reasons in parentheses:
Answer 22: Chocie 3 (Botanically, orange fruit is considered as a multiple aggregate of berry and
the juicy part or pulp is considered as succulent testa of the ovary. The peel of orange is
considered as ovarian wall.)
Answer 23: Choice 1 (A pea pod is considered as a simple legume by botanical nature. It can
open across both walls which eliminates the seeds inside the pod)
Answer 24: Choice2 (Pistachios belong to family of cashew or family Anacardiaceae
botanically. The fruits of this family are simple drupes in nature. These fruits contain the ovarian
wall as the hard covering of the fruit and the ovary itself modifies into edible seed).
The _________ is the protective chamber that houses the ovule and Lat.pdf
The _________ is the protective chamber that houses the ovule and Later matures to become the
fruit. A. Petal B. Ovary C. Carpel D. Stigma E. Sepals. What adaptation enhances bird\'s
flying ability? A. hollow bones B. sternum. C. air sacs D. homeothermy E. all of the above.
What plant characteristics are adaptations to land? A. internal fertilizations B. vascular tissues.
C. seeds with waterproof coats. D. waxy leaves. E. all of the above. Which of which is
characteristic of alternation of generation in land plants? A. haploid sporophytes make haploid
spores. B. gametophytes produce spores that develop into gametes C. sporophytes and
gametophytes are typically similar in appearance. D. meiosis in sporophytes produces haploid
spores. E. either the gametophyte or the sporophyte is unicellular. The haploid stage in a plant
life in a plant life cycle is called A. embryo B. zygote C. gametophyte D. sporophyte E.
conspicuous Animals in which the blastopo
Solution
31. ovary is the protective chamber that houses the ovule and later matures the fruit
32. tha hollow bones, air sacs , the sternum and homeothermy helps the birds to fly. the hollow
bones helps in reducing the body weight which helps in flying. the air sacs are responsible for
extracting more oxygen which hepls them to frly high. the sternum bones are present in the
wings which are strong and flexible for the movement of the wings
33. all of the above are the plants characteristics for the plant to adapt land
34.
35. the haploid stage of plant life in plant life cycle is known as gametophyte.
36.in protostomes the blastopore becomes the mouth.
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http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
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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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The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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