public static class LinkedDequeT implements DequeADTT{ priv.pdf

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The LinkedDeque class implements a doubly-linked deque (dequeue) data structure. It contains methods for checking if the deque is empty, getting the size, adding/removing elements from the front and back of the deque, and checking for null elements. The class uses first and last nodes and a size variable to track the front, back, and count of elements in the deque.

Education

public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null
tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();

Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list
public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}

private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}
Solution
public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null

tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();
Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list

public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}
private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}

Similar to public static class LinkedDequeT implements DequeADTT{ priv.pdf

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The LinkedList1 class implements a Linked list. class.pdf

malavshah9013

**JAVA** A double-ended queue is a list that allows the addition and removal of items from either end. One end is arbitrarily called the front and other the rear, but the two ends behave identically. Specify, design and implement a class for a double-ended queue. Include operations to check if it is empty and to return the number of items in the list. For each end, include operations for adding and deleting items. Use doubly linked list for your implementation. Call your class Deque (pronounced \"deck\"). Solution //package com.java2novice.ds.queue; public class DequeDblLinkedListImpl { private LinkedNode front; private LinkedNode rear; public void insertFront(T item){ //add element at the beginning of the queue System.out.println(\"adding at front: \"+item); LinkedNode nd = new LinkedNode(); nd.setValue(item); nd.setNext(front); if(front != null) front.setPrev(nd); if(front == null) rear = nd; front = nd; } public void insertRear(T item){ //add element at the end of the queue System.out.println(\"adding at rear: \"+item); LinkedNode nd = new LinkedNode(); nd.setValue(item); nd.setPrev(rear); if(rear != null) rear.setNext(nd); if(rear == null) front = nd; rear = nd; } public void removeFront(){ if(front == null){ System.out.println(\"Deque underflow!! unable to remove.\"); return; } //remove an item from the beginning of the queue LinkedNode tmpFront = front.getNext(); if(tmpFront != null) tmpFront.setPrev(null); if(tmpFront == null) rear = null; System.out.println(\"removed from front: \"+front.getValue()); front = tmpFront; } public int size(){ if(front == null && rear ==null){ return 0; } //remove an item from the beginning of the queue LinkedNode tmpFront = front; LinkedNode tmpRear = rear; int i=0; while(tmpFront != rear) { i++; tmpFront=tmpFront.getNext(); } i++; return i; } public void removeRear(){ if(rear == null){ System.out.println(\"Deque underflow!! unable to remove.\"); return; } //remove an item from the beginning of the queue LinkedNode tmpRear = rear.getPrev(); if(tmpRear != null) tmpRear.setNext(null); if(tmpRear == null) front = null; System.out.println(\"removed from rear: \"+rear.getValue()); rear = tmpRear; } public static void main(String a[]){ DequeDblLinkedListImpl deque = new DequeDblLinkedListImpl(); deque.insertFront(34); deque.insertFront(67); System.out.println(\"Size of the queue : \"+deque.size()); deque.insertFront(29); deque.insertFront(765); deque.removeFront(); deque.removeFront(); deque.removeFront(); deque.insertRear(43); deque.insertRear(83); deque.insertRear(84); deque.insertRear(546); deque.insertRear(356); deque.removeRear(); deque.removeRear(); deque.removeRear(); deque.removeRear(); deque.removeFront(); deque.removeFront(); deque.removeFront(); } } class LinkedNode{ private LinkedNode prev; private LinkedNode next; private T value; public LinkedNode getPrev() { return prev; } public void setPrev(LinkedNode prev) { this.prev = prev; } public LinkedNode getNext() { return next; } public void setNext(L.

JAVA A double-ended queue is a list that allows the addition and.pdf

amrishinda

How do I fix it in java? LinkedList.java: /** * Defines a doubly-linked list class * @author * @author */ import java.util.NoSuchElementException; public class LinkedList { private class Node { private T data; private Node next; private Node prev; public Node(T data) { this.data = data; this.next = null; this.prev = null; } } private int length; private Node first; private Node last; private Node iterator; /**** CONSTRUCTORS ****/ /** * Instantiates a new LinkedList with default values * @postcondition */ public LinkedList() { first = null; last = null; iterator = null; length = 0; } /** * Converts the given array into a LinkedList * @param array the array of values to insert into this LinkedList * @postcondition */ public LinkedList(T[] array) { } /** * Instantiates a new LinkedList by copying another List * @param original the LinkedList to copy * @postcondition a new List object, which is an identical, * but separate, copy of the LinkedList original */ public LinkedList(LinkedList original) { } /**** ACCESSORS ****/ public T getFirst() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return first.data; } public T getLast() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return last.data; } /** * Returns the data stored in the iterator node * @precondition * @return the data stored in the iterator node * @throw NullPointerException */ public T getIterator() throws NullPointerException { if (iterator != null){ return iterator.data; }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } /** * Returns the current length of the LinkedList * @return the length of the LinkedList from 0 to n */ public int getLength() { return length; } /** * Returns whether the LinkedList is currently empty * @return whether the LinkedList is empty */ public boolean isEmpty() { return length == 0; } /** * Returns whether the iterator is offEnd, i.e. null * @return whether the iterator is null */ public boolean offEnd() { return iterator == null; } /**** MUTATORS ****/ public void addFirst(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ newNode.next = first; first.prev = newNode; first = newNode; } length++; } public void addLast(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ last.next = newNode; newNode.prev = last; last = newNode; } length++; } /** * Inserts a new element after the iterator * @param data the data to insert * @precondition * @throws NullPointerException */ public void addIterator(T data) throws NullPointerException{ if(iterator != null){ Node newNode = new Node(data); newNode.next = iterator.next; iterator.next = newNode; if (iterator == first){ first = newNode; } }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } / public void removeFirst() throws NoSuchElementException { if(is.

How do I fix it in javaLinkedList.java Defines a doubl.pdf

fmac5

How do I fix it in LinkedList.java? LinkedList.java: /** * Defines a doubly-linked list class * @author * @author */ import java.util.NoSuchElementException; public class LinkedList { private class Node { private T data; private Node next; private Node prev; public Node(T data) { this.data = data; this.next = null; this.prev = null; } } private int length; private Node first; private Node last; private Node iterator; /**** CONSTRUCTORS ****/ /** * Instantiates a new LinkedList with default values * @postcondition */ public LinkedList() { first = null; last = null; iterator = null; length = 0; } /** * Converts the given array into a LinkedList * @param array the array of values to insert into this LinkedList * @postcondition */ public LinkedList(T[] array) { } /** * Instantiates a new LinkedList by copying another List * @param original the LinkedList to copy * @postcondition a new List object, which is an identical, * but separate, copy of the LinkedList original */ public LinkedList(LinkedList original) { } /**** ACCESSORS ****/ public T getFirst() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return first.data; } public T getLast() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return last.data; } /** * Returns the data stored in the iterator node * @precondition * @return the data stored in the iterator node * @throw NullPointerException */ public T getIterator() throws NullPointerException { if (iterator != null){ return iterator.data; }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } /** * Returns the current length of the LinkedList * @return the length of the LinkedList from 0 to n */ public int getLength() { return length; } /** * Returns whether the LinkedList is currently empty * @return whether the LinkedList is empty */ public boolean isEmpty() { return length == 0; } /** * Returns whether the iterator is offEnd, i.e. null * @return whether the iterator is null */ public boolean offEnd() { return iterator == null; } /**** MUTATORS ****/ public void addFirst(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ newNode.next = first; first.prev = newNode; first = newNode; } length++; } public void addLast(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ last.next = newNode; newNode.prev = last; last = newNode; } length++; } /** * Inserts a new element after the iterator * @param data the data to insert * @precondition * @throws NullPointerException */ public void addIterator(T data) throws NullPointerException{ if(iterator != null){ Node newNode = new Node(data); newNode.next = iterator.next; iterator.next = newNode; if (iterator == first){ first = newNode; } }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } / public void removeFirst() throws NoSuchElementExcept.

How do I fix it in LinkedList.javaLinkedList.java Define.pdf

mail931892

Linked Stack program.docx

kudikalakalabharathi

In the class we extensively discussed a generic singly linked list i.pdf

birajdar2

How do I fix it in LinkedList.java? this is what i did LabProgram.java LinkedList.java: /** * Defines a doubly-linked list class * @author * @author */ import java.util.NoSuchElementException; public class LinkedList { private class Node { private T data; private Node next; private Node prev; public Node(T data) { this.data = data; this.next = null; this.prev = null; } } private int length; private Node first; private Node last; private Node iterator; /**** CONSTRUCTORS ****/ /** * Instantiates a new LinkedList with default values * @postcondition */ public LinkedList() { first = null; last = null; iterator = null; length = 0; } /** * Converts the given array into a LinkedList * @param array the array of values to insert into this LinkedList * @postcondition */ public LinkedList(T[] array) { } /** * Instantiates a new LinkedList by copying another List * @param original the LinkedList to copy * @postcondition a new List object, which is an identical, * but separate, copy of the LinkedList original */ public LinkedList(LinkedList original) { } /**** ACCESSORS ****/ public T getFirst() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return first.data; } public T getLast() throws NoSuchElementException { if (isEmpty()){ throw new NoSuchElementException("The list is empty"); } return last.data; } /** * Returns the data stored in the iterator node * @precondition * @return the data stored in the iterator node * @throw NullPointerException */ public T getIterator() throws NullPointerException { if (iterator != null){ return iterator.data; }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } /** * Returns the current length of the LinkedList * @return the length of the LinkedList from 0 to n */ public int getLength() { return length; } /** * Returns whether the LinkedList is currently empty * @return whether the LinkedList is empty */ public boolean isEmpty() { return length == 0; } /** * Returns whether the iterator is offEnd, i.e. null * @return whether the iterator is null */ public boolean offEnd() { return iterator == null; } /**** MUTATORS ****/ public void addFirst(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ newNode.next = first; first.prev = newNode; first = newNode; } length++; } public void addLast(T data) { Node newNode = new Node(data); if(isEmpty()){ first = newNode; last = newNode; } else{ last.next = newNode; newNode.prev = last; last = newNode; } length++; } /** * Inserts a new element after the iterator * @param data the data to insert * @precondition * @throws NullPointerException */ public void addIterator(T data) throws NullPointerException{ if(iterator != null){ Node newNode = new Node(data); newNode.next = iterator.next; iterator.next = newNode; if (iterator == first){ first = newNode; } }else{ throw new NullPointerException("Iterator is off the end opf the list."); } } / public void remov.

How do I fix it in LinkedList.javathis is what i didLabProgra.pdf

mail931892

Add functions push(int n, Deque &dq) and pop(Deque &dq). Functions push(int n, Deque &dq) and pop(Deque &dq) should call functions of the Deque class to simulate a stack's push and pop functions. The Deque class should not be touched . I will use the original unmodified Deque class when testing your code. Demonstrate to convert a base 10 number to base 2. Ask the user to enter a base 10 number, output the equivalent base 2 number. The main() function would create a Deque object and pass it by reference when calling the new push(int n, Deque &dq) and pop(Deque &dq) functions simulating stack operations. keep it simple and to the instructions and code provided please! and thank you! // C++ implementation of De-queue using circular // array #include<iostream> using namespace std; // Maximum size of array or Dequeue #define MAX 100 // A structure to represent a Deque class Deque { int arr[MAX]; int front; int rear; int size; public : Deque(int size) { front = -1; rear = 0; this->size = size; } // Operations on Deque: void insertfront(int key); void insertrear(int key); void deletefront(); void deleterear(); bool isFull(); bool isEmpty(); int getFront(); int getRear(); }; // Checks whether Deque is full or not. bool Deque::isFull() { return ((front == 0 && rear == size-1)|| front == rear+1); } // Checks whether Deque is empty or not. bool Deque::isEmpty () { return (front == -1); } // Inserts an element at front void Deque::insertfront(int key) { // check whether Deque if full or not if (isFull()) { cout << "Overflow\n" << endl; return; } // If queue is initially empty if (front == -1) { front = 0; rear = 0; } // front is at first position of queue else if (front == 0) front = size - 1 ; else // decrement front end by '1' front = front-1; // insert current element into Deque arr[front] = key ; } // function to inset element at rear end // of Deque. void Deque ::insertrear(int key) { if (isFull()) { cout << " Overflow\n " << endl; return; } // If queue is initially empty if (front == -1) { front = 0; rear = 0; } // rear is at last position of queue else if (rear == size-1) rear = 0; // increment rear end by '1' else rear = rear+1; // insert current element into Deque arr[rear] = key ; } // Deletes element at front end of Deque void Deque ::deletefront() { // check whether Deque is empty or not if (isEmpty()) { cout << "Queue Underflow\n" << endl; return ; } // Deque has only one element if (front == rear) { front = -1; rear = -1; } else // back to initial position if (front == size -1) front = 0; else // increment front by '1' to remove current // front value from Deque front = front+1; } // Delete element at rear end of Deque void Deque::deleterear() { if (isEmpty()) { cout << " Underflow\n" << endl ; return ; } // Deque has only one element if (front == rear) { front = -1; rear = -1; } else if (rear == 0) rear = size-1; else rear = rear-1; } // Returns front element of Deque int Deque::getFront() { // check whether Deque is empty or not if (isEmp.

Add functions push(int n- Deque &dq) and pop(Deque &dq)- Functions pus.docx

WilliamZnlMarshallc

Link list part 2

Anaya Zafar

Please help write BinaryTree.java Thank you! Create a class BinaryTree. A class that implements the ADT binary tree. import java.util.Iterator; import java.util.NoSuchElementException; import StackAndQueuePackage.*; public class BinaryTree <T> implements BinaryTreeInterface<T> { private BinaryNode<T> root; public BinaryTree() { root = null; } // end default constructor public BinaryTree(T rootData) { root = new BinaryNode<>(rootData); } // end constructor public BinaryTree(T rootData, BinaryTree<T> leftTree, BinaryTree<T> rightTree) { privateSetTree(rootData, leftTree, rightTree); } // end constructor public void setTree(T rootData) { root = new BinaryNode<>(rootData); } // end setTree public void setTree(T rootData, BinaryTreeInterface<T> leftTree, BinaryTreeInterface<T> rightTree) { privateSetTree(rootData, (BinaryTree<T>)leftTree, (BinaryTree<T>)rightTree); } // end setTree private void privateSetTree(T rootData, BinaryTree<T> leftTree, BinaryTree<T> rightTree) { root = new BinaryNode<>(rootData); if ((leftTree != null) && !leftTree.isEmpty()) root.setLeftChild(leftTree.root); if ((rightTree != null) && !rightTree.isEmpty()) { if (rightTree != leftTree) root.setRightChild(rightTree.root); else root.setRightChild(rightTree.root.copy()); } // end if if ((leftTree != null) && (leftTree != this)) leftTree.clear(); if ((rightTree != null) && (rightTree != this)) rightTree.clear(); } // end privateSetTree //... private class PreorderIterator implements Iterator<T> { private StackInterface<BinaryNode<T>> nodeStack; public PreorderIterator() { nodeStack = new LinkedStack<>(); if (root != null) nodeStack.push(root); } // end default constructor public boolean hasNext() { return !nodeStack.isEmpty(); } // end hasNext public T next() { BinaryNode<T> nextNode; if (hasNext()) { nextNode = nodeStack.pop(); BinaryNode<T> leftChild = nextNode.getLeftChild(); BinaryNode<T> rightChild = nextNode.getRightChild(); // Push into stack in reverse order of recursive calls if (rightChild != null) nodeStack.push(rightChild); if (leftChild != null) nodeStack.push(leftChild); } else { throw new NoSuchElementException(); } return nextNode.getData(); } // end next public void remove() { throw new UnsupportedOperationException(); } // end remove } // end PreorderIterator public void iterativePreorderTraverse () { StackInterface<BinaryNode<T>> nodeStack = new LinkedStack<>(); if (root != null) nodeStack.push(root); BinaryNode<T> nextNode; while (!nodeStack.isEmpty()) { nextNode = nodeStack.pop(); BinaryNode<T> leftChild = nextNode.getLeftChild(); BinaryNode<T> rightChild = nextNode.getRightChild(); // Push into stack in reverse order of recursive calls if (rightChild != null) nodeStack.push(rightChild); if (leftChild != null) nodeStack.push(leftChild); System.out.print(nextNode.getData() + " "); } // end while } // end iterativePreorderTraverse private class InorderIterator implements Iterator<T> { private StackInterface<BinaryNode<T>> nodeStack; private BinaryNode<T> cu.

Please help write BinaryTree-java Thank you! Create a class BinaryTr.pdf

info750646

Java Concurrency Gotchas

Alex Miller

More from amaresh6333

a) It is given E is the event of getting 2. We can observe that 2 is an even number. It is given F is the event of getting an even number. Event E is giving an even number and event F is also giving an even number.That means both are occuring at the same time. so they are not mutually exclusive. b)The events E and F to be complementary their probabilities should add up to 1. P(E) = 1/6 P(F) = 3/6 P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3 2/3 is not equal to one. so events E and F are not complementary. Solution a) It is given E is the event of getting 2. We can observe that 2 is an even number. It is given F is the event of getting an even number. Event E is giving an even number and event F is also giving an even number.That means both are occuring at the same time. so they are not mutually exclusive. b)The events E and F to be complementary their probabilities should add up to 1. P(E) = 1/6 P(F) = 3/6 P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3 2/3 is not equal to one. so events E and F are not complementary..

a) It is given E is the event of getting 2. We can observe that 2 is.pdf

amaresh6333

4.1 (a) Softwoods These generally come from the gymnosperm trees which have needles and cones or simply called the coniferous trees. These trees have needles in place of leaves and remain green throughout the year. All the softwoods are not softer than the hard woods, there are certain hardwoods that are softer. They grow primarly in pacific northwest and south of united states. Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire resistant compared to the hardwoods. Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock. Uses: Although they are classified as softwoods, but they are sufficiently dense and hard to be used for various construction purposes. They are generally used for making the framework of houses. About 80% of the timber comes from the softwoods. Various building components such as windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and many more things use softwoods. 4.1 (b) Hardwoods Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots. These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have higher density than the softwoods. They grow slower than the softwoods, so they are expensive. examples of hardwoods: oak, maple, mahogany, cherry, walnut, rosewood, teak Uses: mostly used to manufacture the high quality furniture,decks, flooring and construction that needs to last for longer periods. 4.1 (c) Most of the structural timber (80%) comes from the softwoods, but the hardwoods are used for the high quality furniture, decks, flooring and constructions which needs to last for longer periods. Solution 4.1 (a) Softwoods These generally come from the gymnosperm trees which have needles and cones or simply called the coniferous trees. These trees have needles in place of leaves and remain green throughout the year. All the softwoods are not softer than the hard woods, there are certain hardwoods that are softer. They grow primarly in pacific northwest and south of united states. Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire resistant compared to the hardwoods. Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock. Uses: Although they are classified as softwoods, but they are sufficiently dense and hard to be used for various construction purposes. They are generally used for making the framework of houses. About 80% of the timber comes from the softwoods. Various building components such as windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and many more things use softwoods. 4.1 (b) Hardwoods Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots. These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have higher density than the softwoods. They grow slower than the softwoods, so they are expensive. .

4.1 (a) SoftwoodsThese generally come from the gymnosperm trees wh.pdf

amaresh6333

A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast involves delivering a message from one sender to many recipients. Senders direct an IP broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick up that message. This broadcast is \'limited in that it does not reach every node on the Internet, only nodes on the LAN Solution A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast involves delivering a message from one sender to many recipients. Senders direct an IP broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick up that message. This broadcast is \'limited in that it does not reach every node on the Internet, only nodes on the LAN.

A special type of IP address is the limited broadcast address 255.25.pdf

amaresh6333

10 leading causes of death in the United States : In the recent statistics the causes of death in the US is heart disease and cancer. 1. HEART DISEASE (614,348) : Heart disease is the leading causes of death for both men and women in US.Major symptoms are chest pain,breathlessness,nausea,cold sweats etc.Causes of heart disease are high cholesterol, high blood pressure,obesity,smoking,lack of physical activity,sleep apnea,stress. 2. CANCER (591,699 ) : Cancer effects people of all gender,ages.Estimated cancer related deaths (male) in 2015 are lung and bronchus, prostate,colon and rectum,pancreas,liver,leukemia etc and for female lung and bronchus,breast,colon and rectum,pancreas,ovary,leukemia,uterine corpous. 3. CHRONIC LOWER RESPIRATORY DISEASE (147,101) :Causes of chronic lower repiratory disease is mainly smoking,air pollution and dust,genetics.It also includes the deaths from chronic bronchitis,emphysema and asthma. 4. ACCIDENTS (136,053) : Motor vehicle accidents,accidental falls,accidental drowning,accidental exposure to fire etc.The raod vehicle accidents can be avoided by not drinking and driving. 5. STROKE (133,103) : Stroke is not considerd by heart disease.Cerebrocvascular disease is developes with the complication of blood vessels.The type of cerebrovascular disease are Stroke,Transient ischemic attack,Vascular dementia.Symptoms are numbness on face and arm,confusion,trouble in speaking,vision problem,headache.Stroke can be prevented by eating healthy diet,maintaining weight,physical activity and exercise,avoid smoking and alcohol,maintain cholesterol and blood pressure level. 6. ALZHEIMER\'S DISEASE ( 93,514) : This disease develops by the progressive mental deteriorataion in the old age.5.3 millions Americans may have alzheimer\'s disease in 2015.Symptoms are memory loss,difficulty in family tasks and solving problems,confusion with time and place,changes mood and personality. 7. DIABETES/DIABETES MELLITUS (76,488) : Diabetes caused by the deficiency of insulin and failure to metabolize sugar and starch in the body,diabetes can cause serious helath complications like heart disease,kidney failure,vision problem etc.There are two types of diabetes a) type 1: Insulin-depedent diabetes mellitus and b) type 2 : Non-insulin-dependent diabetes mellitus. Symptoms are frequent urination,excessive thirst,extreme hunger,vision change,feeling tired,dry skin. 8. INFLUENZA AND PNEUMONIA (55,227) : Influenza is highly contagious viral infection and most severe in winter season.The cause of Pneumonia are virus,chemicals,mycoplasmas and infectious fungi.Symptoms are fever,headache,cough,chills,sore throat,nasal congestion,wheezing. 9. KIDNEY DISEASE (48,146) : Nephritis,nephrotic syndrome,nephrosis are the conditions of kidney diseases.Symptoms are loss of appetite,fatigue,headache,nausea etc. 10. SUICIDE AND SELF-HARM (42,773) : Causes of suicude are mental depression and stress,substance abuse,family violence including sexual abuse,prior suici.

10 leading causes of death in the United States In the recent sta.pdf

amaresh6333

// C Program to shuffle a given Aay #include #include #include // A utility function to print an Aay void printArray (int A[], int n) { int i; for (i = 0; i < n; i++) printf(\"%d \", A[i]); printf(\"\ \"); } int* shuffle(int A[], int B[]) { int n1 = 5; int n2 = 2; int i = 0, j = 0, k = 0; int *C; C = (int*)malloc(7 * sizeof(int)); // Merging starts while (i < n1 && j < n2) { if (k%2 == 0) { C[k] = A[i]; i++; k++; } else { C[k] = B[j]; k++; j++; } } /* elements in array \'A\' are still remaining where as the array \'B\' is exhausted */ while (i < n1) { C[k] = A[i]; i++; k++; } /* elements in array \'B\' are still remaining where as the array \'A\' is exhausted */ while (j < n2) { C[k] = B[j]; k++; j++; } return C; } int main() { int A[] = {1, 2, 3, 4, 5}; int B[] = {10, 20}; int n1 = 5; int n2 = 2; int nC = 7; int *C; C = (int*)malloc(7 * sizeof(int)); printf(\"Array A: \"); printArray(A, n1); printf(\"Array B: \"); printArray(B, n2); C = shuffle (A, B); printf(\"\ Shuffled: \"); printArray(C,nC); return 0; } /* output: Array A: 1 2 3 4 5 Array B: 10 20 Shuffled: 1 10 2 20 3 4 5 */ Solution // C Program to shuffle a given Aay #include #include #include // A utility function to print an Aay void printArray (int A[], int n) { int i; for (i = 0; i < n; i++) printf(\"%d \", A[i]); printf(\"\ \"); } int* shuffle(int A[], int B[]) { int n1 = 5; int n2 = 2; int i = 0, j = 0, k = 0; int *C; C = (int*)malloc(7 * sizeof(int)); // Merging starts while (i < n1 && j < n2) { if (k%2 == 0) { C[k] = A[i]; i++; k++; } else { C[k] = B[j]; k++; j++; } } /* elements in array \'A\' are still remaining where as the array \'B\' is exhausted */ while (i < n1) { C[k] = A[i]; i++; k++; } /* elements in array \'B\' are still remaining where as the array \'A\' is exhausted */ while (j < n2) { C[k] = B[j]; k++; j++; } return C; } int main() { int A[] = {1, 2, 3, 4, 5}; int B[] = {10, 20}; int n1 = 5; int n2 = 2; int nC = 7; int *C; C = (int*)malloc(7 * sizeof(int)); printf(\"Array A: \"); printArray(A, n1); printf(\"Array B: \"); printArray(B, n2); C = shuffle (A, B); printf(\"\ Shuffled: \"); printArray(C,nC); return 0; } /* output: Array A: 1 2 3 4 5 Array B: 10 20 Shuffled: 1 10 2 20 3 4 5 */.

C Program to shuffle a given Aay#include stdio.h #include .pdf

amaresh6333

1.Router: 1.It is a machine which aheads datagrams over the networks. 2.This router is present at a node which connects two or more networks. 3.It avails of headers and tables to establish the path. 4.We can filter the data through this router in very small. 5.in order to communicate one another this router uses a protocol. 2.Hubs: 1.a hub is a frequent connection point . 2.It contains no of ports. 3.The data type in hub is frames. 4.hubs are present as an association for the networks. 5.It is some what slow. 6.we can use it as a stand alone device. 3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices. 2.generally this servers sre many types as follows 1.Database server:A system which operate through queries. 2.Print server :A system which directs one or more printer. 3.File server and network server which directs the network traffic. 4.It offers many functionality frequently called \"servers.\" 5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a simple computer. 4.Network cabling: 1.It is used to join the one network device to another network device. 2.It the agency to travel from one netork to another network 3.Depending on their sizeand topology the cables are different as folllows. 1 .co-axial cable 2.optical fibre cable 3.twisted pair cacle 4.For acheving the good network we need to know the charecteristics of a cable at first. 5.Sound masking upgrades the isolation. Solution 1.Router: 1.It is a machine which aheads datagrams over the networks. 2.This router is present at a node which connects two or more networks. 3.It avails of headers and tables to establish the path. 4.We can filter the data through this router in very small. 5.in order to communicate one another this router uses a protocol. 2.Hubs: 1.a hub is a frequent connection point . 2.It contains no of ports. 3.The data type in hub is frames. 4.hubs are present as an association for the networks. 5.It is some what slow. 6.we can use it as a stand alone device. 3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices. 2.generally this servers sre many types as follows 1.Database server:A system which operate through queries. 2.Print server :A system which directs one or more printer. 3.File server and network server which directs the network traffic. 4.It offers many functionality frequently called \"servers.\" 5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a simple computer. 4.Network cabling: 1.It is used to join the one network device to another network device. 2.It the agency to travel from one netork to another network 3.Depending on their sizeand topology the cables are different as folllows. 1 .co-axial cable 2.optical fibre cable 3.twisted pair cacle 4.For acheving the good network we need to know the charecteristics of a cable at first. 5.Sound masking upgrades the isolation..

1.Router 1.It is a machine which aheads datagrams over the networks.pdf

amaresh6333

1) List currently running jobs ANS) see currently running command: ps examples: $ps 2.Kill one of the running jobs ANS) a.kill any process by PID command: kill {PID} example: $kill 1012 3.Run the ls command with –help option ANS) ls command: ls command is used to list the contents of a directory Syntax: ls [-a] [-A] [-b] [-c] [-C] [-d] [-f] [-F] [-g] [-i] [-l] [-L] [-m] [-o] [-p][-q] [-r] [-R] [-s] [-t] [-u] [-x] [pathnames] Parameters OR Switches • -a Shows All The Files, Including Hidden File Which Begin With ADot. • -A Shows All The Files Including Hidden Files. Does Not Displays The Working Directory (.) OR The Parent Directory (..) • -b Force Printing of Non-Printable Characters To Be in Octal \\ddd Notation. • -c Use Time of Last Modification of The i-Node For Sorting (-t) OR For Printing (-l OR -n) • -C Multi-Column Output With Entries Sorted Down The Columns. Which is Generally The Default Option. • -d Lists Only The Directory Names But Not its Contents. • -f Force Each Argument To Be Interpreted As A Directory And List The Name Found in Each Slot. -f Option Turns off -l, -t, -s, And -r, And Turns on –a, The List Order is The Order in Which Entries Appear in The Directory. • -F Mark Directories With A Trailing Slash (/), DOORS With A Trailing Greater-Than Sign (>), Executable Files With A Trailing Asterisk (*), FIFOs With A Trailing Vertical Bar (|), Symbolic Links With A Trailing Equals Sign (=). • -g Same As -l Parameter, Except The Owner is Not Printed. • -i For Each File, Prints The i-Node Number in The First Column. • -l Shows The User With Huge Amounts of Information With Permissions, Owners, Size, And When Last Modified.) • -L If An Argument is A Symbolic Link, List The File OR Directory To Which The Link References Rather Than The Link Itself. • -m Stream Output Format, In This Files Are Listed Across The Page, Separated By Commas. • -n Same As –l Parameter, Except That The Owner\'s UID And Group\'s GID Numbers Are Printed, Rather Than The Associated Character Strings. -o The Same As -l, Except That The Group is Not Printed. -p Displays A Forward Slash (/) in Front of All Directories. -q Force Printing of Non-Printable Characters in File Names As The Character Question Mark (?). -r Reverses The Order of How The Files Are Displayed. -R Includes The Contents of Subdirectories. -s Give Size in Blocks, Including Indirect Blocks, For Each Entry. -t Shows The End User The Files Modification Time. -u Use Time of Last Access Instead of Last Modification For Sorting Use With The -t Option OR For Printing Use With The -l Option. -x Displays Files in Columns. -1 Prints One Entry Per Line of Output. pathnames File OR Directory To List. 4.Run the help command on echo ANS) echo command display to the Screen. “echo” is Useful For • Producing Diagnostics in Command Files • For Sending Known Data into A Pipe • For Displaying The Contents of Environment Variables syntax: echo [-n] text 5.Cat an existing file using a command that misspells one c.

1) List currently running jobsANS) see currently runningcommand.pdf

amaresh6333

Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at 295.3 K Step3 = 767.0- 20.2 = 746.8 torr Solution Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at 295.3 K Step3 = 767.0- 20.2 = 746.8 torr.

Step1 Total pressure = partial pressure of Hydrog.pdf

amaresh6333

I would say very many elements exista as monoatomic molecules.All the metals,for example.Don\'t get confused by the fact that if metals have a crystaline structure,then the molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an element is any material that is composed of only a single kind of atom (ie: all its atoms are identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another Compounds... A molecule is a group of atoms that are bonded together, but they do not form a compound. A compound is is something whose molecules contain more than one kind of atom. Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of another element you get a compound. In Short Element Is A Single Atom While Molecule Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element and a molecule. You can compare an element and a compound. An element is made of only a single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does not matter. A compound is made up of molecules (or crystals) with more than one kind of atom (NH3, H2O, NaCl) Solution I would say very many elements exista as monoatomic molecules.All the metals,for example.Don\'t get confused by the fact that if metals have a crystaline structure,then the molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an element is any material that is composed of only a single kind of atom (ie: all its atoms are identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another Compounds... A molecule is a group of atoms that are bonded together, but they do not form a compound. A compound is is something whose molecules contain more than one kind of atom. Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of another element you get a compound. In Short Element Is A Single Atom While Molecule Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element and a molecule. You can compare an element and a compound. An element is made of only a single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does not matter. A compound is made up of .

I would say very many elements exista as monoatom.pdf

amaresh6333

E is simply hexane...no polarity at all...nothing to speak of has extra electron density. A is a benzene ring, and if you play with the resonance you will see that sometimes the electrons can be in many different places which will make it polar at those times. :) nate Solution E is simply hexane...no polarity at all...nothing to speak of has extra electron density. A is a benzene ring, and if you play with the resonance you will see that sometimes the electrons can be in many different places which will make it polar at those times. :) nate.

E is simply hexane...no polarity at all...nothing.pdf

amaresh6333

Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation that leads to rust. Same thing basically for the OH. Solution Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation that leads to rust. Same thing basically for the OH..

Ferroxyl contains the chemical compound Potassium.pdf

amaresh6333

There are two different types of mechanisms. There are elimination and substitution reactions. The elimination reaction is when there is a proton transfer and loss of leaving group taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is unclear. Solution There are two different types of mechanisms. There are elimination and substitution reactions. The elimination reaction is when there is a proton transfer and loss of leaving group taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is unclear..

There are two different types of mechanisms. Ther.pdf

amaresh6333

What must be true about a set of data when its standard deviation is.pdf

amaresh6333

When one drives from sea level to trailhead in high Sierras, altitude increases. At higher altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This is detrimental as it can even stop our brain due to lack of oxygen. These symptoms are collectively called Altitude sickness which reduces, as one spends some time in such region as our body gets acclimatized to such environment. Solution When one drives from sea level to trailhead in high Sierras, altitude increases. At higher altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This is detrimental as it can even stop our brain due to lack of oxygen. These symptoms are collectively called Altitude sickness which reduces, as one spends some time in such region as our body gets acclimatized to such environment..

When one drives from sea level to trailhead in high Sierras, altitud.pdf

amaresh6333

the reaction as follows- 14N7 + 0n1 -- 4He2 + 1.pdf

amaresh6333

The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps! Solution The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps!.

The important thing here is to understand the Standard Electrode Pot.pdf

amaresh6333

The pathway would be overactive.The GTPase activity of the alpha s.pdf

amaresh6333

The metal ion is Pb2+.The white precipitate PbCl2 is insoluble in .pdf

amaresh6333

The electron carriers that feed into the ETC are NADH and FADH2.NA.pdf

amaresh6333

The developmental potential, or potency of a cell, describes the range of different cell types it can become. Therefore it involves. The determination means a cell restricts itself choosing a particular fate. Differentiation follows determination and the cell becomes a specific cell type ultimately. Solution The developmental potential, or potency of a cell, describes the range of different cell types it can become. Therefore it involves. The determination means a cell restricts itself choosing a particular fate. Differentiation follows determination and the cell becomes a specific cell type ultimately..

The developmental potential, or potency of a cell, describes the ran.pdf

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public static class LinkedDequeT implements DequeADTT{ priv.pdf

1. public static class LinkedDeque implements DequeADT{ private int size; private Node first; private Node last; public LinkedDeque() { } // checking empty method or not public boolean isEmpty() { return first == null; } // checking size public int size() { return size; } // enqueing front public void enqueueFront(T num) { checkNull(num); Node tempNode = new Node(); tempNode.num = num; if (first != null) { // check root is not null tempNode.next = first; first.previous = tempNode; } first = tempNode; if (last == null) last = first; size++; } // removing front public Item dequeueFront() { CheckEmpty();

2. Node oldFirst = first; first = first.next; if (first == null) last = null; else first.previous = null; size--; return oldFirst.item; } // inseeting element at end of list public void enqueueBack(Item item) { checkNull(item); Node newLast = new Node(); newLast.item = item; if (last != null) { newLast.previous = last; last.next = newLast; } last = newLast; if (first == null) first = last; size++; } // removing element from end of the list public Item dequeueBack() { CheckEmpty(); Node oldLast = last; last = oldLast.previous; if (last == null) first = null; else last.next = null; size--; return oldLast.item; }

3. private void CheckEmpty() { if (first == null) throw new NoSuchElementException(); } private void checkNull(Item item) { if (item == null) throw new NullPointerException(); } } Solution public static class LinkedDeque implements DequeADT{ private int size; private Node first; private Node last; public LinkedDeque() { } // checking empty method or not public boolean isEmpty() { return first == null; } // checking size public int size() { return size; } // enqueing front public void enqueueFront(T num) { checkNull(num); Node tempNode = new Node(); tempNode.num = num; if (first != null) { // check root is not null

4. tempNode.next = first; first.previous = tempNode; } first = tempNode; if (last == null) last = first; size++; } // removing front public Item dequeueFront() { CheckEmpty(); Node oldFirst = first; first = first.next; if (first == null) last = null; else first.previous = null; size--; return oldFirst.item; } // inseeting element at end of list public void enqueueBack(Item item) { checkNull(item); Node newLast = new Node(); newLast.item = item; if (last != null) { newLast.previous = last; last.next = newLast; } last = newLast; if (first == null) first = last; size++; } // removing element from end of the list

5. public Item dequeueBack() { CheckEmpty(); Node oldLast = last; last = oldLast.previous; if (last == null) first = null; else last.next = null; size--; return oldLast.item; } private void CheckEmpty() { if (first == null) throw new NoSuchElementException(); } private void checkNull(Item item) { if (item == null) throw new NullPointerException(); } }

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