public class LinkedHashEntry { private int key; private int valu.pdf

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anjanaarts2014

public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table; HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext(); if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } } public void remove(int key) { int hash = (key % TABLE_SIZE); if (table[hash] != null) { LinkedHashEntry prevEntry = null; LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) { prevEntry = entry; entry = entry.getNext(); } if (entry.getKey() == key) { if (prevEntry == null) table[hash] = entry.getNext(); else prevEntry.setNext(entry.getNext()); } } } } Solution public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table; HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext(); if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } } public void remove(.

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$public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table;$

$HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext(); if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } }$

$public void remove(int key) { int hash = (key % TABLE_SIZE); if (table[hash] != null) { LinkedHashEntry prevEntry = null; LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) { prevEntry = entry; entry = entry.getNext(); } if (entry.getKey() == key) { if (prevEntry == null) table[hash] = entry.getNext(); else prevEntry.setNext(entry.getNext()); } } } } Solution public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; }$

$public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table; HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext();$

$if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } } public void remove(int key) { int hash = (key % TABLE_SIZE); if (table[hash] != null) { LinkedHashEntry prevEntry = null; LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) { prevEntry = entry; entry = entry.getNext(); } if (entry.getKey() == key) { if (prevEntry == null) table[hash] = entry.getNext(); else prevEntry.setNext(entry.getNext());$

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How do you update an address according to this code? Currently, I figured out up to (the bolded line) How do I update address, display all entries and quit. Add a name (n) Look up a name (l) Update address (u) Delete an entry (d) Display all entries (a) Quit (q) -> n import java.util.Scanner; public class AddressBook { public static void main(String[] args) { Tablea addressBook = new Tablea(); Scanner scanner = new Scanner(System. in ); String key, value; // Insert a new entry into the address book System. out .print("Enter a name: "); key = scanner.nextLine(); System. out .print("Enter a address: "); value = scanner.nextLine(); addressBook.insert(key, value); // Lookup an entry in the address book System. out .print("Enter a name to look up: "); key = scanner.nextLine(); String result = addressBook.lookUp(key); if (result != null) { System. out .println("address: " + result); } else { System. out .println("name not found"); } // Delete an entry from the address book System. out .print("Enter a name to delete: "); key = scanner.nextLine(); boolean deleted = addressBook.delete(key); if (deleted) { System. out .println("address deleted"); } else { System. out .println("name not found"); } } } public class Node { private String key; private String value; private Node next; Node() { // add here .. } Node(String key, String value) { // add here .. this.key = key; this.value = value; this.next = null; } public String getKey() { // add here .. return this.key; } public void setKey(String key) { // add here .. this.key = key; } public String getValue() { // add here .. return this.value; } public void setValue(String value) { // add here .. this.value = value; } public Node getNext() { // add here .. return this.next; } public void setNext(Node next) { // add here .. this.next = next; } } import java.util.Scanner; public class Tablea { private Node mark; public Node getMark() { return this.mark; } public void setMark(Node mark) { this.mark = mark; } public boolean insert(String key, String value) { Node newNode= new Node(key,value); if (this.mark== null) { this.mark = newNode; }else { newNode.setNext(this.mark.getNext()); this.mark.setNext(newNode); } return true; } public String lookUp(String key) { Node current = this.mark; while(current!=null) { if (current.getKey().equals(key)) { return current.getValue(); } current = current.getNext(); } return null; } public boolean delete(String key) { Node current = this.mark; Node prev = null; while (current!= null) { if (current.getKey().equals(key)) { if (prev == null) { this.mark = current.getNext(); }else { prev.setNext(current.getNext()); } return true; } prev = current; current = current.getNext(); } return false; } public boolean update(String key, String newValue) { Node current = this.mark; while (current != null) { if (current.getKey().equals(key)) { current.setValue(newValue); return true; } current = current.getNext(); } return false; } public boolean markToStart() { if (this.mark== null) { retur.

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Given below is the completed implementation of MyLinkedList class. Other classes are not modified and remain same as in question - MyList, MyAbstractList, TestMyLinkedList. The output of TestMyLinkedList is shown below. Please don\'t forget to rate the answer if it helped. Thank you very much. Since the MyList is not completely pasted in the question ... giving the complete class file below MyList.java public interface MyList> extends java.lang.Iterable { // Add a new element at the end of this list public void add (E e); // Add a new element at specified index in this list public void add (int index, E e); // Return true if this list contains the element public boolean contains (E e); // Return the element at the specified index public E get (int index); // Return the index of the first matching element // Return -1 if no match. public int indexOf (E e); // Return the index of the last matching element // Return -1 if no match. public int lastIndexOf (E e); E remove(int index); void clear(); E set(int index, E e); } MyLinkedList.java public class MyLinkedList> extends MyAbstractList implements Comparable> { private Node head, tail; // head and tail pointers // Create a default list public MyLinkedList() { } public int size() { return size; } // Need this for implementing Comparable public int compareTo (MyLinkedList e) { return(size() - e.size()); } // Create a list from an array of objects public MyLinkedList (E[] objects) { super(objects); // Passes the array up to abstract parent } // 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) { // Create a new node for element e Node newNode = new Node(e); 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) // new node is only node { tail = head; } } // Add an element to the end of the list public void addLast (E e) { // Create a new node for element e Node newNode = new Node(e); if (tail == null) { head = tail = newNode; // new node is only node } else { tail.next = newNode; // Link the new with the last node tail = tail.next; // tail now points to the last node } size++; // Increase size } // Remove the head node and // return the object from the removed node public E removeFirst() { if (size == 0) { return null; } else { Node temp = head; head = head.next; size--; if (head == null) { tail = null; } return temp.element; } } // Remove the last node and return the object from it public E removeLast() { if (size == 0) { return null; } else if (size == 1) { Node temp = head; head = tail = null; size = 0; return temp.element; } else { Node current = head; for (int i = 0; i < size - 2; i++) { current = current.next; } Node temp = tail; tai.

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info430661

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How do I fix this error " Exception in thread "main" java.lang.NullPointerException: Cannot invoke "String.hashCode()" because "action" is null at AddressBook.main(AddressBook.java:25)" import java.util.Scanner; public class AddressBook { public static void main(String[] args) { Tablea addressBook = new Tablea(); Scanner scanner = new Scanner(System.in); String name, address; String action = null; boolean moreEntries = true; while (moreEntries) { System.out.println("/nMenu"); System.out.println("n.Add Entry"); System.out.println("d. Delete Entry"); System.out.println("u. Update Entry"); System.out.println("l. Search"); System.out.println("a. View All"); System.out.println("q. Exit"); // Insert a new entry into the address book switch (action) { case "n": System.out.print("Enter a name: "); name = scanner.nextLine(); System.out.print("Enter an address: "); address = scanner.nextLine(); addressBook.insert(name, address); System.out.println("New Contact has been added,"); System.out.print("Add another entry? (y/n): "); String answer = scanner.nextLine().toLowerCase(); moreEntries = answer.equals("y") || answer.equals("yes"); break; // Delete an entry from the address book case "d": System.out.println("Enter a name to delete: "); name = scanner.nextLine(); boolean deleted = addressBook.delete(name); if (deleted) { System.out.println("Address deleted"); } else { System.out.println("Name not found"); } break; // Update an entry in the address book case "u": System.out.print("Enter a name to update: "); name = scanner.nextLine(); System.out.print("Enter a new address: "); address = scanner.nextLine(); boolean updated = addressBook.update(name, address); if (updated) { System.out.println("Address updated"); } else { System.out.println("Name not found"); } break; // Lookup an entry in the address book case "l": System.out.print("Enter a name to look up: "); name = scanner.nextLine(); String result = addressBook.lookUp(name); if (result != null) { System.out.println("Address: " + result); } else { System.out.println("Name not found"); } break; // Display all entries in the address book case "a": System.out.println("All entries:"); addressBook.displayAll(); break; // Quit the program case"q": System.out.println("Quitting..."); } } } } Other files that go with it public class Node { private String name; private String address; private Node next; Node() { // add here .. } Node(String name, String address) { // add here .. this.name = name; this.address = address; this.next = null; } public String getKey() { // add here .. return this.name; } public void setKey(String name) { // add here .. this.name = name; } public String getValue() { // add here .. return this.address; } public void setValue(String address) { // add here .. this.address = address; } public Node getNext() { // add here .. return this.next; } public void setNext(Node next) { // add here .. this.next = next; } } import java.util.Scanner; public class Tablea { private Node mark; public Node.

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A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode Solution A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode.

A is correct. The ntp server 10.0.0.5 command is entered in Global C.pdf

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A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent. Solution A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent..

A. Gene expression.B. Transcription factor.C. Gene expression..pdf

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All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y) Solution All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y).

All are capable to grow in lactose medium . In presence of lactose t.pdf

anjanaarts2014

1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .hta have extenstion ) to their targets. When the user run that program,hackers trigger a Throjan horse or Remote Administration Tool(RAT) by this hackers can remotely acces users machine. SSL protocal is used for internet banking.By using Man-in-middle software which employs the ARP posioning technique to can sniff the password. So be alert while using ATM,Credit card and check the statements carefully. 4)Insider threats or organised crime more than 50 percent cyber security threats are caused by internal figures means computer crime may be carried out by one person by hacking prepaid phone card system.That is \"Organised crime\" china,Brazil,Russia cooperating with internal employees in criminal model CaaS. 5)These internal attacks are most dangerous threat to data and systems.Rogue employees,who are members of the IT team with knowledge of and access to networks,data centers and admin accounts, can cause serious damage. 6)Misunderstanding GRC:So many managements dont\'t understand the concept of GRC(Governance,Risk Management,Compliance) for IT governance,information security governance and corporate governance.To deal with training in IT governance,corporate governance,information security governance are best practice. ITIL and CobiT helps to deal with this problem. Solution 1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .h.

1)The viruse which is the most dangerous threat.In DTI survey,72 of.pdf

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HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf

anjanaarts2014

The -OCl ion. .pdf

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I have already answered it. Check. .pdf

anjanaarts2014

Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane Solution Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane.

Iodide is a strong nucleophile but a weak base, s.pdf

anjanaarts2014

NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for the positive and negative ions between the other ions in the crystal and the water molecules. Several water molecules are attracted to each of the ions. Whether the crystal dissolves is determined by which attractive force is stronger. If the internal ionic forces in the crystal are the strongest, the crystal does not dissolve. This is the situation in reactions where precipitates form. If the attractions for the ions by the polar water molecules are the strongest, the crystal will dissolve. This is the situation in sodium chloride. Once the ions are released from the crystals, the ions are completely surrounded by water molecules. Note that the proper atom in the water molecule must \"point\" toward the correct ion. The charge principle and the partial charges in the polar molecule determine the correct orientation. Partially negative oxygen atoms in the water molecule interact with the positive sodium ion. Partially positive hydrogen atoms in the water molecule interact with the negative chloride ion. Sugar : The interaction between sugar molecule and water is greater than between sugar molecules. So When sugar is put into water, it likes to form hydrogen bonds with water. So it breaks glycosidic linkages and forms hydrogen bonds with water. Initially it separates into polymer sugar chains, but later it dissociates into individual molecules. Solution NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for.

NaCl Its the coulombic interaction between Na.pdf

anjanaarts2014

Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes referred to as the Jovian or outer planets. The planet was known by astronomers of ancient times and was associated with the mythology and religious beliefs of many cultures. The Romans named the planet after the Roman god Jupiter.[14] When viewed from Earth, Jupiter can reach an apparent magnitude of -2.94, making it on average the third-brightest object in the night sky after the Moon and Venus. (Mars can briefly match Jupiter\'s brightness at certain points in its orbit.) Jupiter is primarily composed of hydrogen with a quarter of its mass being helium; it may also have a rocky core of heavier elements. Because of its rapid rotation, Jupiter\'s shape is that of an oblate spheroid (it possesses a slight but noticeable bulge around the equator). The outer atmosphere is visibly segregated into several bands at different latitudes, resulting in turbulence and storms along their interacting boundaries. A prominent result is the Great Red Spot, a giant storm that is known to have existed since at least the 17th century when it was first seen by telescope. Surrounding the planet is a faint planetary ring system and a powerful magnetosphere. There are also at least 63 moons, including the four large moons called the Galilean moons that were first discovered by Galileo Galilei in 1610. Ganymede, the largest of these moons, has a diameter greater than that of the planet Mercury. Jupiter has been explored on several occasions by robotic spacecraft, most notably during the early Pioneer and Voyager flyby missions and later by the Galileo orbiter. The most recent probe to visit Jupiter was the Pluto-bound New Horizons spacecraft in late February 2007. The probe used the gravity from Jupiter to increase its speed. Future targets for exploration in the Jovian system include the possible ice-covered liquid ocean on the moon Europa. Jupiter\'s structure: Jupiter is composed primarily of gaseous and liquid matter. It is the largest of four gas giants as well as the largest planet in the solar system with a diameter of 142,984 km at its equator. The density of Jupiter, 1.326 g/cm³, is the second highest of the gas giant planets. However, the density is lower than any of the four terrestrial planets. Solution Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes refe.

Here Jupiter is the fifth planet from the Sun a.pdf

anjanaarts2014

moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf

anjanaarts2014

Lithium is group 1 metal, while nitrogen is group.pdf

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there is some special character in ur question li.pdf

anjanaarts2014

a. Times Interest Earned Ratio = EBIT / interest Expense For 2014: $8,799 / $600 = 14.7 times For 2015: $9,454 / $838 = 11.3 times For 2016: $10,338 / $1,058 = 9.8 times The TImes Interest Earned (TIE) Ratio measures the operating income level that can be used for interest payments in the future, i.e. how many times the Company can make the interest payments. It is a solvency ratio since it measures the Company\'s ability to service it\'s debt interest obligations. Hence, the higher the ratio, the more favorable is the company\'s liquidity position. In the given question, we can observe that the ratio has consistently decreased year on year. This implies that the Company\'s debt has increased while the increase in operating income has not been at the same pace. The Company can afford to service interest fewer times each year from 2014 to 2016. This implies that the Company has become comparatively more risky than before from the Creditor\'s standpoint. b. Current Ratio = Current Assets / Current Liabilities For 2014: $25,983 / $19,027 = 1.4 For 2015: $29,158 / $23,169 = 1.3 For 2016: $31,042 / $26,250 = 1.2 The Current Ratio is an indicator of how efficiently the Company manages it\'s operating cycle. A higher Current Ratio implies that the COmpany is in a better position to discharge all it\'s short term obligations without affecting or jeopardising it\'s working capital requirements. On the flipside, a ratio that is too high indicates that the Company\'s current assets are lying idle or that it is not efficiently using short term funds. In the Given case, the Company\'s current ratio shows a steady decline. However, the lowest is 1.2 times, which implies that the Company is still sufficiently liquid. This is becase even at the lowest, the Company\'s Current Assets can pay off the Current Liabilities 1.2 times without affecting working capital cycle for regular business needs. Hence, as per the options given, we can say that CVS’s current ratio has decreased over the past three years, however, it is greater than 1 indicating CVS is liquid. c. Total Liabilities to Equity Ratio = Total Liabilities / Total Equity For 2014: $36,224 / $37,963 = 0.95 For 2015: $55,234 / $37,203 = 1.5 For 2016: $57,628 / $36,834 = 1.6 Total Liabilities include all the claims, long term and short term, against the Company\'s Assets. Total Liabilities to Equity Ratio indicates how much of the Company\'s Assets are funded by creditors versus how much is funded by the equity holders. Broadly, a 1:1 ratio is considered acceptable for most industries. Creditors would prefer a lower ratio as this implies their money is at less risk. However, equity holders would like to benefit from the risk taken by them due to the Company using debt, and would prefer a higher ratio. In the given case, this ratio is increasing year on year. This implies that the COmpany is taking on more debt financing. This implies that the Company has more risk overall, although it benefits the equity .

a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf

anjanaarts2014

45 yrs.Solution45 yrs..pdf

anjanaarts2014

the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions. Solution the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions..

the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf

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1. public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table;

2. HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext(); if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } }

3. public void remove(int key) { int hash = (key % TABLE_SIZE); if (table[hash] != null) { LinkedHashEntry prevEntry = null; LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) { prevEntry = entry; entry = entry.getNext(); } if (entry.getKey() == key) { if (prevEntry == null) table[hash] = entry.getNext(); else prevEntry.setNext(entry.getNext()); } } } } Solution public class LinkedHashEntry { private int key; private int value; private LinkedHashEntry next; LinkedHashEntry(int key, int value) { this.key = key; this.value = value; this.next = null; } public int getValue() { return value; }

4. public void setValue(int value) { this.value = value; } public int getKey() { return key; } public LinkedHashEntry getNext() { return next; } public void setNext(LinkedHashEntry next) { this.next = next; } } public class HashMap { private final static int TABLE_SIZE = 128; LinkedHashEntry[] table; HashMap() { table = new LinkedHashEntry[TABLE_SIZE]; for (int i = 0; i < TABLE_SIZE; i++) table[i] = null; } public int get(int key) { int hash = (key % TABLE_SIZE); if (table[hash] == null) return -1; else { LinkedHashEntry entry = table[hash]; while (entry != null && entry.getKey() != key) entry = entry.getNext();

5. if (entry == null) return -1; else return entry.getValue(); } } public void put(int key, int value) { int hash = (key % TABLE_SIZE); if (table[hash] == null) table[hash] = new LinkedHashEntry(key, value); else { LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) entry = entry.getNext(); if (entry.getKey() == key) entry.setValue(value); else entry.setNext(new LinkedHashEntry(key, value)); } } public void remove(int key) { int hash = (key % TABLE_SIZE); if (table[hash] != null) { LinkedHashEntry prevEntry = null; LinkedHashEntry entry = table[hash]; while (entry.getNext() != null && entry.getKey() != key) { prevEntry = entry; entry = entry.getNext(); } if (entry.getKey() == key) { if (prevEntry == null) table[hash] = entry.getNext(); else prevEntry.setNext(entry.getNext());

6. } } } }

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