package ex2- public class Exercise2-E- { private static class N.pdf

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package ex2; public class Exercise2<E> { private static class Node<E> { private E element; private Node<E> prev; private Node<E> next; public Node(E e, Node<E> p, Node<E> n) { element = e; prev = p; next = n; } public E getElement() { return element; } public Node<E> getPrev() { return prev; } public Node<E> getNext() { return next; } public void setPrev(Node<E> p) { prev = p; } public void setNext(Node<E> n) { next = n; } } private Node<E> header; private Node<E> trailer; private int size = 0; public Exercise2() { header = new Node<>(null, null, null); trailer = new Node<>(null, header, null); header.setNext(trailer); } public int size() { return size; } public boolean isEmpty() { return size == 0; } public E first() { if (isEmpty()) return null; return header.getNext().getElement(); } public E last() { if (isEmpty()) return null; return trailer.getPrev().getElement(); } public void addFirst(E e) { addBetween(e, header, header.getNext()); } public void addLast(E e) { addBetween(e, trailer.getPrev(), trailer); } public E removeFirst() { if (isEmpty()) return null; return remove(header.getNext()); } public E removeLast() { if (isEmpty()) return null; return remove(trailer.getPrev()); } private void addBetween(E e, Node<E> predecessor, Node<E> successor) { if (predecessor == header && successor == trailer) { // skip creation of new node for header/trailer header.setNext(new Node<>(e, header, trailer)); trailer.setPrev(header.getNext()); } else { Node<E> newest = new Node<>(e, predecessor, successor); predecessor.setNext(newest); successor.setPrev(newest); } size++; } private E remove(Node<E> node) { Node<E> predecessor = node.getPrev(); Node<E> successor = node.getNext(); predecessor.setNext(successor); successor.setPrev(predecessor); size--; return node.getElement(); } public String toString() { StringBuilder sb = new StringBuilder("("); Node<E> walk = header; while (walk != null) { sb.append(walk.getElement()); walk = walk.getNext(); if (walk != null) { sb.append(", "); } } sb.append(")"); return sb.toString(); } public void reverseList() { Node<E> current = header; header = trailer.getPrev(); trailer = current.getNext(); while (current != null) { Node<E> temp = current.getNext(); current.setNext(current.getPrev()); current.setPrev(temp); current = temp; } } public static void main(String[] args) { Exercise2<String> list = new Exercise2<String>(); list.addFirst("MSP"); list.addLast("ATL"); list.addLast("BOS"); list.addFirst("LAX"); System.out.println(list); list.reverseList(); System.out.println(list); } } Here's my code, the output should reverse the list, but I'm keep getting errors, anyone mind to help me to get a proper output? .

Education

public void addFirst(E e) {
addBetween(e, header, header.getNext());
}
public void addLast(E e) {
addBetween(e, trailer.getPrev(), trailer);
}
public E removeFirst() {
if (isEmpty()) return null;
return remove(header.getNext());
}
public E removeLast() {
if (isEmpty()) return null;
return remove(trailer.getPrev());
}
private void addBetween(E e, Node<E> predecessor, Node<E> successor) {
if (predecessor == header && successor == trailer) {
// skip creation of new node for header/trailer
header.setNext(new Node<>(e, header, trailer));
trailer.setPrev(header.getNext());
} else {
Node<E> newest = new Node<>(e, predecessor, successor);
predecessor.setNext(newest);
successor.setPrev(newest);
}
size++;
}
private E remove(Node<E> node) {
Node<E> predecessor = node.getPrev();
Node<E> successor = node.getNext();
predecessor.setNext(successor);
successor.setPrev(predecessor);
size--;
return node.getElement();
}
public String toString() {
StringBuilder sb = new StringBuilder("(");
Node<E> walk = header;
while (walk != null) {
sb.append(walk.getElement());
walk = walk.getNext();

if (walk != null) {
sb.append(", ");
}
}
sb.append(")");
return sb.toString();
}
public void reverseList() {
Node<E> current = header;
header = trailer.getPrev();
trailer = current.getNext();
while (current != null) {
Node<E> temp = current.getNext();
current.setNext(current.getPrev());
current.setPrev(temp);
current = temp;
}
}
public static void main(String[] args)
{
Exercise2<String> list = new Exercise2<String>();
list.addFirst("MSP");
list.addLast("ATL");
list.addLast("BOS");
list.addFirst("LAX");
System.out.println(list);
list.reverseList();
System.out.println(list);
}
}
Here's my code, the output should reverse the list, but I'm keep getting errors, anyone mind to
help me to get a proper output?

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package com.java2novice.ds.linkedlist; import java.util.NoSuchElementException; public class DoublyLinkedListImpl { private Node head; private Node tail; private int size; public DoublyLinkedListImpl() { size = 0; } /** * this class keeps track of each element information * @author java2novice * */ private class Node { E element; Node next; Node prev; public Node(E element, Node next, Node prev) { this.element = element; this.next = next; this.prev = prev; } } /** * returns the size of the linked list * @return */ public int size() { return size; } /** * return whether the list is empty or not * @return */ public boolean isEmpty() { return size == 0; } /** * adds element at the starting of the linked list * @param element */ public void addFirst(E element) { Node tmp = new Node(element, head, null); if(head != null ) {head.prev = tmp;} head = tmp; if(tail == null) { tail = tmp;} size++; System.out.println(\"adding: \"+element); } /** * adds element at the end of the linked list * @param element */ public void addLast(E element) { Node tmp = new Node(element, null, tail); if(tail != null) {tail.next = tmp;} tail = tmp; if(head == null) { head = tmp;} size++; System.out.println(\"adding: \"+element); } /** * this method walks forward through the linked list */ public void iterateForward(){ System.out.println(\"iterating forward..\"); Node tmp = head; while(tmp != null){ System.out.println(tmp.element); tmp = tmp.next; } } /** * this method walks backward through the linked list */ public void iterateBackward(){ System.out.println(\"iterating backword..\"); Node tmp = tail; while(tmp != null){ System.out.println(tmp.element); tmp = tmp.prev; } } /** * this method removes element from the start of the linked list * @return */ public E removeFirst() { if (size == 0) throw new NoSuchElementException(); Node tmp = head; head = head.next; head.prev = null; size--; System.out.println(\"deleted: \"+tmp.element); return tmp.element; } /** * this method removes element from the end of the linked list * @return */ public E removeLast() { if (size == 0) throw new NoSuchElementException(); Node tmp = tail; tail = tail.prev; tail.next = null; size--; System.out.println(\"deleted: \"+tmp.element); return tmp.element; } public static void main(String a[]){ DoublyLinkedListImpl dll = new DoublyLinkedListImpl(); dll.addFirst(10); dll.addFirst(34); dll.addLast(56); dll.addLast(364); dll.iterateForward(); dll.removeFirst(); dll.removeLast(); dll.iterateBackward(); } } Solution package com.java2novice.ds.linkedlist; import java.util.NoSuchElementException; public class DoublyLinkedListImpl { private Node head; private Node tail; private int size; public DoublyLinkedListImpl() { size = 0; } /** * this class keeps track of each element information * @author java2novice * */ private class Node { E element; Node next; Node prev; public Node(E element, Node next, Node prev) { this.element = element; this.next = next; this.prev = prev; } } /** * returns the size of the l.

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Complete the class ArraySet-1.java which implements the SetADT.java interface. Implement only the methods public Set<E> union(Set<E> set) public Set<E> intersection(Set<E> set) ArraySet-1.java public class ArraySet<E> implements Set<E>, Iterable<E> { private static Random rand = new Random(); private final int DEFAULT_CAPACITY = 100; private final int NOT_FOUND = -1; private int count; // the current number of elements in the set private E[] contents; private int modCount; public ArraySet() { count = 0; modCount = 0; contents = (E[])(new Object[DEFAULT_CAPACITY]); } public ArraySet(int initialCapacity) { count = 0; contents = (E[])(new Object[initialCapacity]); } public boolean add(E element) { boolean result = false; if (!(contains(element))) { if (size() == contents.length) { expandCapacity(); } contents[count] = element; count += 1; result = true; } modCount += 1; return result; } public void addAll(Set<E> set) { Iterator<E> scan = set.iterator(); while (scan.hasNext()) { add (scan.next()); } modCount += 1; } public E removeRandom() throws NoSuchElementException { if (isEmpty()) { throw new NoSuchElementException("ArraySet"); } int choice = rand.nextInt(count); E result = contents[choice]; contents[choice] = contents[count-1]; // fill the gap contents[count-1] = null; count--; modCount += 1; return result; } public E remove(E target) throws NoSuchElementException, NoSuchElementException { int search = NOT_FOUND; if (isEmpty()) { throw new NoSuchElementException("ArraySet"); } for (int index=0; index < count && search == NOT_FOUND; index += 1) { if (contents[index].equals(target)) { search = index; } } if (search == NOT_FOUND) { throw new NoSuchElementException(); } E result = contents[search]; contents[search] = contents[count-1]; contents[count-1] = null; count--; modCount += 1; return result; } public Set<E> union(Set<E> set) { // Implement this method return null; // Replace when implementinging this method } public Set<E> difference(Set<E> set) { ArraySet<E> diff = new ArraySet<E>(); for (int index = 0; index < count; index += 1) { if (!(set.contains(contents[index]))) { diff.add (contents[index]); } } return diff; } public Set<E> intersection(Set<E> set) { // Implement this method return null; // Replace when implmenting this method } public boolean contains (E target) { int search = NOT_FOUND; for (int index=0; index < count && search == NOT_FOUND; index += 1) { if (contents[index].equals(target)) { search = index; } } return (search != NOT_FOUND); } public boolean equals(Set<E> set) { boolean result = false; ArraySet<E> temp1 = new ArraySet<E>(); ArraySet<E> temp2 = new ArraySet<E>(); E element; if (size() == set.size()) { temp1.addAll(this); temp2.addAll(set); Iterator<E> scan = set.iterator(); while (scan.hasNext()) { element = scan.next(); if (temp1.contains(element)) { temp1.remove(element); temp2.remove(element); } } result = (temp1.isEmpty() && temp2.isEmpty()); } return result; } public boolean isEmpty() { return (count == 0).

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Please finish everything marked TODO BinaryNode.java public class BinaryNode<E extends Comparable<E>> implements TreePrinter.PrintableNode{ private E data; private BinaryNode<E> left; private BinaryNode<E> right; private int height; private int size; private BinaryNode<E> parent; public BinaryNode(E data){ this.data = data; this.left = null; this.right = null; this.parent = null; this.height = 1; this.size = 1; } // TODO: Set up the BinaryNode public BinaryNode(E data, BinaryNode<E> left, BinaryNode<E> right, BinaryNode<E> parent){ } // Access fields E data() { return this.data; }; BinaryNode<E> left() { return this.left; } BinaryNode<E> right() { return this.right; } BinaryNode<E> parent() { return this.parent; } // Setter fields void setLeft(BinaryNode<E> left) { this.left = left; if(left != null) left.setParent(this); } void setRight(BinaryNode<E> right) { this.right = right; if(right != null) right.setParent(this); } void setParent(BinaryNode<E> parent) { this.parent = parent; } void setData(E data) { this.data = data; } void setHeight(int h){ this.height = h; } // Basic properties int height() { return this.height; } int size() { return this.size; } boolean isBalanced() { int leftHeight = (hasLeft()) ? left.height() : 0; int rightHeight = (hasRight()) ? right().height() : 0; return Math.abs(leftHeight - rightHeight) < 2; } boolean hasLeft(){ return left == null ? false : true; } boolean hasRight(){ return right == null ? false :true; } boolean hasParent(){ return parent == null ? false :true; } public boolean equals(BinaryNode<E> other){ if(other == null) return false; return other.data.equals(this.data); } // Can use these to help debug public TreePrinter.PrintableNode getLeft() { return left == null ? null : left; } public TreePrinter.PrintableNode getRight() { return right == null ? null : right; } public String getText() { return String.valueOf(data); } public String toString(){ String ret = ""; return "root " + this.data + " Left: " +(hasLeft() ? left.data : null) + " Right: " +(hasRight() ? right.data : null) + " parent: " + (hasParent() ? parent.data : null) ; } } BST.java import java.util.ArrayList; import java.util.Collection; import java.util.List; public class BST<E extends Comparable<E>> implements Tree<E> { private int height; private int size; private BinaryNode<E> root; public BST(){ this.root = null; this.height = 0; this.size = 0; } // TODO: BST public BST(BinaryNode<E> root){ } // Access field public BinaryNode<E> root() { return this.root; } // Basic properties public int height() { return this.height; } public int size() { return this.size; } public boolean isBalanced() { return root.isBalanced(); } // TODO: updateHeight - Update the root height to reflect any changes public void updateHeight() { } // Traversals that return lists // TODO: Preorder traversal public List<E> preOrderList() { return new ArrayList<>(); } // .

Please finish everything marked TODO BinaryNode-java public class Bi.docx

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Java весна 2013 лекция 2

Technopark

To complete the task, you need to fill in the missing code. I’ve included code to create an Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked list. You can use the .next() method to advance the Iterator to the next item (the first time you call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==> C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling .remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove() once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator() and after you’ve added all the nodes to the list, where l is your CircularLinkedList Solution import java.util.Iterator; class CircularLinkedList implements Iterable { // Your variables // You can include a reference to a tail if you want Node head; int size; // BE SURE TO KEEP TRACK OF THE SIZE // implement this constructor public CircularLinkedList() { head=null; } // writing helper functions for add and remove, like the book did can help // but remember, the last element\'s next node will be the head! // attach a node to the end of the list // Be sure to handle the adding to an empty list // always returns true public boolean add(E e) { Node newNode=new Node(e); if(size==0){ head=newNode; } else{ Node last=getNode(size-1); last.next=newNode; } newNode.next=head; //last element node is set to head size++; return true; } // need to handle // out of bounds // empty list // adding to front // adding to middle // adding to \"end\" // REMEMBER TO INCREMENT THE SIZE public boolean add(int index, E e){ if(index>size) return false; Node tmp=new Node(e); if(index==0){ tmp.next=head; Node last=getNode(size-1); head=tmp; last.next=head; } else { Node curr=getNode(index-1); tmp.next=curr.next; curr.next=tmp; } size++; return true; } // I highly recommend using this helper method // Return Node found at the specified index // be sure to handle out of bounds cases private Node getNode(int index ) { Node prev=head; for(int i=0;isize) { e=null; } else if(index==0){ e = head.getElement(); Node last=getNode(size-1); head=head.next; last.next=head; size--; } else{ Node prev=getNode(index-1); Node curr=getNode(index); e=curr.getElement(); prev.next=curr.next; size--; } return e; } // Turns your list into a string // Useful for debugging public String toString(){ Node current = head; StringBuilder result = new StringBuilder(); if(size == 0){ return \"\"; } if(size == 1) { return head.getElement().toString(); } else{ do{ result.append(current.getElement()); result.append(\" ==> \"); current = current.next; } while(current != head); } return result.toString(); } public Iterator iterator() { return new ListIterator(); } /.

To complete the task, you need to fill in the missing code. I’ve inc.pdf

ezycolours78

STAGE 2: The Methods (65 points) Implement all the methods that are marked as Left as Exercise. The Driver program only works when you have done all the methods. I have added the template of all the methods with a placeholder value as return types of the methods. For example, return 0 if method returns an int value. List of the methods left as exercise are as follows: /** Create a list from an array of objects */ @SuppressWarnings("unchecked") public MyLinkedList(E[] objects) /** Add a new element at the specified index in this list in ascending order */ public void addInOrder(E e) /** Check to see if this list contains element e */ public boolean contains(E e) /** Remove all the occurrences of the element e * from this list. Return true if the element is * removed. */ public boolean removeAll(E e) /** Remove the first occurrence of the element e * from this list. Return true if the element is * removed. */ public boolean remove(E e) /** Return the element at the specified index */ public E get(int index) /** Return the index of the head matching element in * this list. Return -1 if no match. */ public int indexOf(Object e) /** Return the index of the last matching element in * this list. Return -1 if no match. */ public int lastIndexOf(E e) /** Replace the element at the specified position * in this list with the specified element. */ public E set(int index, E e) /** Print this list using recursion */ public void printList() /** Returns an array containing all of the elements * in this collection; * the runtime type of the returned array is that of * the specified array. */ public <E> E[] toArray(E[] array) /** Split the original list in half. The original * list will continue to reference the * front half of the original list and the method * returns a reference to a new list that stores the * back half of the original list. public class MyLinkedList<E extends Comparable<E>> { private Node<E> head, tail; private int size = 0; // Number of elements in the list public MyLinkedList() { head=tail=null; size=0; } /** Return the head element in the list */ public E getFirst() { if (size == 0) { return null; } else { return head.element; } } /** Return the last element in the list */ public E getLast() { if (size == 0) { return null; } else { return tail.element; } } /** Add an element to the beginning of the list */ public void addFirst(E e) { Node<E> newNode = new Node<>(e); // Create a new node newNode.next = head; // link the new node with the head head = newNode; // head points to the new node size++; // Increase list size if (tail == null) // the new node is the only node in list tail = head; } /** Add an element to the end of the list */ public void addLast(E e) { Node<E> newNode = new Node<>(e); // Create a new for element e if (tail == null) { head = tail = newNode; // The new node is the only node in list } else { tail.next = newNode; // Link the new with the last node tail = newNode; // tail now points to the last node } size++; // Incr.

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babitasingh698417

There is a host of sociological and cultural research that paints a robust picture of the effects of globalization on culture. This Application focuses on the flows of culture between Western and developing nations. As a practitioner in this global environment, you should be familiar with these culture effects. Use newspapers, magazines, and the Internet to research cultural changes in both Western and developing countries due to globalization. Then perform the following tasks: Outline the cultural aspects of globalization. Take an anthropological, rather than a business, perspective. Explain how you think understanding culture helps in doing business in today\'s global economy. Cite resources to justify your response. Solution Information technology has penetrated almost every aspect of our lives, “shrinking” our world into a global village. Economies and cultures have come closer. People are now aware of the cultures, traditions, lifestyle, living conditions prevailing in almost every corner of the world. Interestingly, this is going beyond awareness and into a state of integration that is a result of cross-pollinated views, ideologies, products and services. This evolution is termed as “globalization.” Culture has many definitions. My own definition is that culture is our collective experience as a society, and its impact on our reaction and decision-making relative to every-day facts and circumstances. Why is cross-cultural competence critical to your professional future and the viability of your company? It’s omnipresent in every business interaction and strategic decision.It is not feasible to be an expert on all the world’s cultures. It is possible, however, to incorporate a cross-cultural framework that improves cross-cultural understanding and interactions. Multinational firms whose operations are borderless have to consider the cultural variability of different regions of the world and develop cultural understanding. Major cultural constraints encountered by businesses include local attitudes, taste preferences, language, religion, management style, gender discrimination, skills, personalities, education, etc. To be successful, they need to mold their business actions in accordance with the local cultural models, they need to establish a global mindset. Let’s consider an example of the food giant, McDonald’s. The company enjoys a global presence; operating in more than 100 countries serving 70 million people every day. Their headquarters and senior management are U.S.-based but they entrust their local operations to local managers of the countries they operate in. Operations in more than 50 percent of their outlets are franchised. Furthermore, their menus are customized according to cultural habits and local taste preferences in every country. It is without a doubt that global thinking and cultural understanding are both powerful business tools which allow multinational firms to dominate the local markets and establish a global pres.

This is problem is same problem which i submitted on 22017, I just.pdf

fcaindore

code: #include using namespace std; class Point { public: int x,y; Point() { x=0; y=0; } Point(int xi,int yi) { x=xi; y=yi; } int getx() { return x; } int gety() { return y; } void setx(int xi) { x=xi; } void sety(int yi) { y=yi; } }; class Line { public: Point* p1; Point* p2; Line() { } Line(Point* pi1,Point* pi2) { p1= pi1; p2 = pi2; } Point* getP1() { return p1; } Point* getP2() { return p2; } void setp1(Point* pi1) { p1=pi1; } void setp2(Point* pi2) { p2=pi2; } float getslope() { return (p2->y - p1->y)/ (p2->x - p1->x); } }; class Plane { public: Line* l1; Line* l2; Plane() { } Plane(Line* li1,Line* li2) { l1= li1; l2 = li2; } Line* getl1() { return l1; } Line* getl2() { return l2; } void setl1(Line* li1) { l1 = li1; } void setl2(Line* li2) { l2 = li2; } }; int main() { Point* p1 = new Point(0,0); Point* p2 = new Point(0,1); Point* p3 = new Point(1,0); Point* p4 = new Point(1,1); Line* l1 = new Line(p1,p2); Line* l2 = new Line(p3,p4); float m1 = l1->getslope();//finding slope of line 1 float m2 = l2->getslope();//finding slope of line 2 if(m1==m2) //if slope of both lines are equal then they are paralled otherwise not { cout<<\"lines are parallel\"< Solution code: #include using namespace std; class Point { public: int x,y; Point() { x=0; y=0; } Point(int xi,int yi) { x=xi; y=yi; } int getx() { return x; } int gety() { return y; } void setx(int xi) { x=xi; } void sety(int yi) { y=yi; } }; class Line { public: Point* p1; Point* p2; Line() { } Line(Point* pi1,Point* pi2) { p1= pi1; p2 = pi2; } Point* getP1() { return p1; } Point* getP2() { return p2; } void setp1(Point* pi1) { p1=pi1; } void setp2(Point* pi2) { p2=pi2; } float getslope() { return (p2->y - p1->y)/ (p2->x - p1->x); } }; class Plane { public: Line* l1; Line* l2; Plane() { } Plane(Line* li1,Line* li2) { l1= li1; l2 = li2; } Line* getl1() { return l1; } Line* getl2() { return l2; } void setl1(Line* li1) { l1 = li1; } void setl2(Line* li2) { l2 = li2; } }; int main() { Point* p1 = new Point(0,0); Point* p2 = new Point(0,1); Point* p3 = new Point(1,0); Point* p4 = new Point(1,1); Line* l1 = new Line(p1,p2); Line* l2 = new Line(p3,p4); float m1 = l1->getslope();//finding slope of line 1 float m2 = l2->getslope();//finding slope of line 2 if(m1==m2) //if slope of both lines are equal then they are paralled otherwise not { cout<<\"lines are parallel\"<.

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package hw1; public class Runner { public static void main(String[] args) { /*print top header ID Name Gender Age CH Academic Year */ Student.printHeader(); /* Creates a student and prints his info. * Parameters here are: Name, age, number of courses enrolled in, * academic year, and gender. * Pay attention that the ID is auto generated and does not have a setter. * valid age is [1-100] otherwise, it is set to 1 * valid academicYear is [1-10] otherwise, it is set to 1 * valid numEnrolledCourses is [1-9] otherwise, it is set to 0 */ //You must make this constructor call the empty constructor which sets the value of the id Student s1 = new Student("Ahemd Ali", 20, 6, 2); /*printStudentInfo() prints the ID, name, Age, Credit Hours, Academic year, and gender. * credit hours is the number of courses the student is enrolled in * 3 */ s1.printStudentInfo(); //Creates a second student and prints the students' info //You must make this constructor call the constructor that takes all attributes but not the gender Student s2 = new Student("Radwan Ameer", 23, 5, 6, Gender.MALE); s2.printStudentInfo(); //Creates a third student and prints the students' info Student s3 = new Student(); //sets the name of s3 to Aisha Fawaz System.out.println("Setting the name of p3 to \"Aisha Fawaz\""); s3.setName("Aisha Fawaz"); //Set number of enrolled courses for Aisha to 4 System.out.println("Setting number of enrolled courses of Aisha to 4"); s3.setNumEnrolledCourses(4); /*printStudentInfo() for s3 */ s3.printStudentInfo(); //Set number of enrolled courses of Aisha to 10! System.out.println("Set enrolled courses to 10!"); s3.setNumEnrolledCourses(10); /*printStudentInfo() for s3 */ s3.printStudentInfo(); //Sets the age of Aisha to 19 System.out.println("Setting Aisha's age to 19"); s3.setAge(19); //sets the gender of Aisha to Female s3.setGender(Gender.FEMALE); // printStudentInfo() for s3 s3.printStudentInfo(); /*checks if s1 and s3 having the same values for all of their corresponding attributes*/ System.out.println("s1.equals(s3)? "+s1.equals(s3)); //creates student object s4 Student s4 = new Student(); /*copies data of s1 to s4*/ s1.copy(s4); System.out.println("s4 is a copy of s1"); /*checks if s1 and s4 having the same values for all of their corresponding attributes*/ System.out.println("s1.equals(s4)? "+s1.equals(s4)); /*creates a clone of s3 and returns its reference to be saved in s5. * s5 is a clone of s3 */ Student s5 = s3.clone(); System.out.println("s5 is a clone of s3"); /*checks if s3 and s5 having the same values for all of their corresponding attributes*/ System.out.println("s3.equals(s5)? "+s3.equals(s5)); /*compares s1 to s2 based on the value of the attribute id. * if the value of id of s1 is less than the value of id of s2 the method returns -1, * if the value of id of s1 is greater than the value of id of s2 the method returns 1, * otherwise the two values of id are equal, it returns 0. */ switch(s1.compareTo(s2)) { case -1: System.out.println("s.

package hw1-public class Runner { public static void main(String-- arg (1).pdf

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P1) Given the Von Neumann architecture- answer the following questions.pdf

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P1) Answer the following questions (a) We discussed about bottleneck issues in Neumann architecture. Explain how the introduction of cache memory can resolve this issue. (b) Explain the two main advantages of using Interpreter based Languages. (c) Alias is presence of two or more distinct referencing methods for the same memory location. What is the main risk of Alias? (c) Alias is presence of two or more distinct referencing methods for the same memory location. What is the main risk of Alias? (d) Explain the benefits of using JIT (Just in Time Compiler) in Java..

P1) Answer the following questions (a) We discussed about bottleneck i.pdf

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P1) Answer the following questions ( 40 points) (a) We discussed about bottleneck issues in Neumann architecture. Explain how the introduction of cache memory can resolve this issue. (b) Explain the two main advantages of using Interpreter based Languages. (c) Alias is presence of two or more distinct referencing methods for the same memory location. What is the main risk of Alias? (d) Explain the benefits of using JIT (Just in Time Compiler) in Java..

P1) Answer the following questions ( 40 points) (a) We discussed about.pdf

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Our Space is a social media site that is growing in popularity. The firm has been around for a few years and has created a nice market niche for itself. In fact, it actually turned a profit last year, albeit a fairly small one. After doing some basic research on the company, you've decided to take a closer look. You plan to use the price-to-sales ratio to value the stock, and you have collected PIS multiples on the following social media companies: Find the average PiS ratio for these three firms. Given that Our Space is expected to generate $42 million in sales next year and will have 14 million shares of stock outstanding, use the average P/S ratio you computed above to put a value on Our Space's stock The average PIS ratio for the 3 firms is (Round to two decimal places) The value of Our Space's stock is $ (Round to the nearest cent) Data table (Click on the icon here 1 in order to copy the contents of the data table below into a spreadsheet.).

Our Space is a social media site that is growing in popularity- The fi.pdf

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P(Zn-kZn1-j)-k!(j)kej (5 points) Let T be the minimal value of n such.pdf

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P(Zn-kZn1-j)-k!(j)kejfn-P(T-10Z1-n)gn-P(T-20Z1-n).pdf

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