if one can understand a few things it is easier to solve these kind .pdf
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if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive expression but Wild type (I+) is dominant over I- . so it is inducible operon. Solution if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation .
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Step1 First compound has 2 double bonds ; Second .pdf
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Solution
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This above plant name is Solidago sempervirens, belong to genus Soli.pdf
This above plant name is Solidago sempervirens, belong to genus Solidago. Species
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Solution
This above plant name is Solidago sempervirens, belong to genus Solidago. Species
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Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120-
180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of
most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the
top of the plant, often with branches that bend backwards towards the base. This species blooms
in late summer and well into the fall, later in the season than most of its relatives. Its fruits are
wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short
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angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed
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flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed
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The Java Program for the above given isimport java.io.File;impo.pdf
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Solution
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This above plant name is Solidago sempervirens, belong to genus Solidago. Species
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Solution
The Java Program for the above given is:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.util.Scanner;
import java.util.NoSuchElementException;
import java.io.PrintWriter;
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Scanner in = new Scanner(System.in);
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String inputFileName = in.next();
System.out.print(\"Output File Name: \");
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double total= 0;
while (in.hasNextDouble())
{
double value = in.nextDouble();
out.printf(\"%15.2f\ \", value);
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The fidelity of DNA replication determines the genome stability and .pdf
The fidelity of DNA replication determines the genome stability and it plays a main role in
evolution of species. The DNA replication fidelity is a key factor of mutations which in turn
leads to many human diseases.
The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III
holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication
of both leading and lagging DNA strand with high speed and high fidelity. There are also four
other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V.
Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA
strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is
responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is
responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a
high constitutive intracellular concentration. Pol V is induced constitutively in the absence of
exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA
primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand.
In eukaryotes, the replication of leading and lagging strands takes place by Pol and Pol
respectively, which leads to fidelity differences. Whereas in E.coli the replication of both leading
and lagging strands takes place by same enzyme polymerase III. This is also one of the reason
why E.coli has high fidelity of DNA replication.
The tau subunit of Pol III plays a central role in DNA replication fidelity, as it interacts with both
leading and lagging strand. This subunit also contributes for polymerase switching during
replication.
Solution
The fidelity of DNA replication determines the genome stability and it plays a main role in
evolution of species. The DNA replication fidelity is a key factor of mutations which in turn
leads to many human diseases.
The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III
holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication
of both leading and lagging DNA strand with high speed and high fidelity. There are also four
other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V.
Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA
strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is
responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is
responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a
high constitutive intracellular concentration. Pol V is induced constitutively in the absence of
exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA
primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand.
In eukaryotes, the replication of leading .
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2.Addressing scheme:
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2. The adress of ipv6 is 128 bit.
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7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
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Solution
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2.extension header is present with length 40 bytes in ipv6.
2.Addressing scheme:
1. A 32-bit permiting a amont of 2^32 in ipv4.
2. The adress of ipv6 is 128 bit.
3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6.
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5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6
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7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
spot..
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TFTP - Same as FTP but not connection oriented.
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IMAP - Like POP3, Internet Message Access Protocol is a standard protocol for accessing e-
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HTTP - The Hypertext Transfer Protocol is the set of rules for exchanging files (text, graphic
images, sound, video, and other multimedia files) on the World Wide Web. It is the protocol
controlling the transfer and addressing of HTTP requests and responses.
HTTPS - Signifies that a web page is using the Secure Sockets Layer (SSL) protocol and is
providing a secure connection. This is used for secure internet business transactions.
NTP - Network Time Protocol is a protocol that is used to synchronize computer clock times in
a network of computers.
SNMP - Stands for Simple Network Management Protocol and is used for monitoring and status
information on a network. SNMP can be used to monitor any device that is SNMP capable and
this can include computers, printers, routers, servers, gateways and many more using agents on
the target systems. The agents report information back to the management systems by the use of
“traps” which capture snapshot data of the system. This trap information could be system errors,
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Static arrays are structures whose size is fixed at compile time and therefore cannot be extended
or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is
overhead associated with the operation of copying old data and freeing the memory associated
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be accessed very efficiently using an index. However, for applications that can be better
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A linked list is a collection of objects linked together by references from one object to another
object. By convention these objects are named as nodes. So the basic linked list is collection of
nodes where each node contains one or more data fields AND a reference to the next node. The
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Linked lists are widely used in many applications because of the flexibility it provides. Unlike
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not random accessed like arrays. To find information in a linked list one must start from the head
of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage
of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust”
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There are few different types of linked lists. A singly linked list as described above provides
access to the list from the head node. Traversal is allowed only one way and there is no going
back. A doubly linked list is a list that has two references, one to the next node and another to
previous node. Doubly linked list also starts from head node, but provide access both ways. That
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Linked list is a collection of linked nodes. A node is a struct with at least a.
D) II, III, IV, and V only .pdf
This document provides the answer choice "D) II, III, IV, and V only" as the solution to a problem or question. No other context or details are given about the problem or question being answered. The answer simply states that choices II, III, IV, and V are the correct options.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.next;
}
}
return false;
}
public void printList(){
Node l = first;
while(l.next!=null){
System.out.print(l.number+\" \");
l = l.next;
}
System.out.println();
}
}
Node.java (implements a single node in the linked list; stores number and pointers)
public class Node{
String number;
Node previous;
Node next;
public Node(String num){
number = num;
previous = null;
next = null;
}
public Node(String num, Node p, Node n){
number = num;
previous = p;
next = n;
}
}
Main.java (contains the main method and the helper methods to solve the questions given)
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
static boolean checkHappiness(String num){
Node current = new Node(num, null, null);
NumberList numberList = new NumberList(current);
int len = num.length();
int resultant = 0;
for (int i=0; i happyNumbersfrom1to10000(){
ArrayList numbers = new ArrayList();
for (int j=0; j<10000; j++){
if (checkHappiness(String.valueOf(j+1))){
numbers.add(String.valueOf(j+1));
}
}
return numbers;
}
static void happyNumbersfrom9001to10000(){
for (int j=9001; j<=10000; j++){
//System.out.println(j);
if (checkHappiness(String.valueOf(j))){
System.out.println(String.valueOf(j));
}
}
}
static String getLargeHappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
static String getLargeUnhappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (!checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
public static void main(String[] args){
happyNumbersfrom9001to10000();
System.out.println(happyNumbersfrom1to10000());
}
}
Solution
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation
Solution
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation.
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i
Solution
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i.
Interphase.This is the phase where the cell prepares for the next .pdf
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle.
Solution
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle..
B.the rotation will be zero as L and D will cance.pdf
B.the rotation will be zero as L and D will cancel out each other..
Solution
B.the rotation will be zero as L and D will cancel out each other...
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
In the above conversation it is belonging to stereotypes.
Stereotypes are a form of generalization. When we generalize, we group or classify people,
places, or things according to the traits they have in common. For example, we may say most
Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations
are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood
hasn\'t produced any quality movies in the past fifteen years.\"
But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending
generalizations about people, places, or things. When a stereotype is challenged, the person who
holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As
Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that
sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full
of on those subjects.\"75
The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning,
clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse,
and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy
liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these
stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles
brave, the Irish devout, blacks athletic.)
Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film
directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics,
atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers,
psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society
matrons. There are stereotypes of institutions as well: marriage, the church, government, the
military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would
include even God and mother.)
FACTS DON\'T MATTER
It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is
seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice,
pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People
who stereotype don\'t dust accept the facts that are offered to them. They measure those facts
against what they already \"know.\" That is, they measure them against the stereotype itself.
Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar
facts.
People who think in terms of stereotypes tend to be selective in their perceptions. They reject
conditions that challenge their preformed judgment and retain those that reinforce it. Thus a
person can notice the Jewish employer promoting another Jew but ign.
In general, the objective of an audit is to assess the risk of mater.pdf
In general, the objective of an audit is to assess the risk of material misstatements in the financial
statements. Material misstatements can arise from inadequacies in internal controls and from
inaccurate management assertions. Thus, testing the validity of the various implicit managerial
assertions is a key objective of an auditor.
Existence and Completeness
Auditing standards require that auditors test basic underlying management assertions implicit in
the financial statements. Key among these various assertions are existence or occurrence, which
describe a singular concept: Journal entries are not fiction. As the name implies, an auditor will
conduct various procedures to verify that assets do in fact exist and that recorded transactions did
in fact occur. Additionally, an auditor will seek evidence of completeness, so that the financial
statements include all material transactions that occurred, and so the records do not omit material
transactions for any reason.
Rights and Obligations
The various rights and obligations of the company are important management assertions inherent
in the financial statements. Thus, an auditor will obtain evidence regarding a company\'s rights,
such as proper title to assets and status of intellectual property. An auditor will be concerned
with assertions relating to the company\'s obligations, such as accounts payable balances, long-
term debts and tax liabilities. Thus, the audit objectives will be fulfilled upon validating these
specific assertions.
Valuation or Allocation
Valuation or allocation are managerial assertions which are often material to the financial
statements; thus, an auditor will diligently conduct audit procedures relating to these objectives.
Generally accepted accounting principles, or GAAP, require that certain balance sheet items be
presented using different valuation methodologies. Meeting these standards is a key audit
objective, as the risk of material misstatement is low in probability, but high in magnitude. Thus,
among other things, the historical cost of assets is verified, depreciation methods are scrutinized
and the fair value of investments are calculated to satisfy this objective.
Presentation and Disclosure
Another specific audit objective is validating the presentation of the financial statements and the
adequacy of the disclosures therein. Financial statements should conform to certain requirements
and expectations, and should include the balance sheet, income statement, statement of cash
flows and the statement of owner\'s equity. Relating to disclosure, the auditor will consider the
sufficiency and clarity of footnotes and the transparency in management discussion and analysis,
so he can assess the risk of material misstatement and fulfill the audit objective.
Specific audit objectives are the application of the general audit objectives to a given class of
transactions, account balance, or presentation and disclosure. There must be at least one specific
audit object.
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy
Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl
Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl-
Spectator ions are : Cl - &H+
Spectator ions are : Cl - &H+
Net ionic Equation is : Cu2+ + SO42- --->CuSiO4
Solution
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy
Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl
Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl-
Spectator ions are : Cl - &H+
Spectator ions are : Cl - &H+
Net ionic Equation is : Cu2+ + SO42- --->CuSiO4.
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
fixed 3 issues. I am working on remaining.
package chegg;
public class HugeIntegerApp {
public static void main(String args[]) {
String[][] testInputs = { { \"987654321\", \"234567890\" },
{ \"987654321\", \"-234567890\" }, { \"-987654321\", \"234567890\" },
{ \"-987654321\", \"-234567890\" }, { \"234567890\", \"987654321\" },
{ \"234567890\", \"-987654321\" }, { \"-234567890\", \"987654321\" },
{ \"-234567890\", \"-987654321\" } };
for (String[] ints : testInputs) {
HugeInteger h1 = new HugeInteger(ints[0]);
HugeInteger h2 = new HugeInteger(ints[1]);
HugeInteger hcopy=new HugeInteger(ints[0]);
HugeInteger hcopy1=new HugeInteger(ints[0]);
HugeInteger hcopy2=new HugeInteger(ints[0]);
System.out.println(\"h1=\" + h1);
System.out.println(\"h2=\" + h2);
if (h1.isEqualTo(h2)) {
System.out.println(\"h1 is equal to h2.\");
}
if (h1.isNotEqualTo(h2)) {
System.out.println(\"h1 is not equal to h2.\");
}
if (h1.isGreaterThan(h2)) {
System.out.println(\"h1 is greater than h2.\");
}
if (h1.isLessThan(h2)) {
System.out.println(\"h1 is less than to h2.\");
}
if (h1.isGreaterThanOrEqualTo(h2)) {
System.out.println(\"h1 is greater than or equal to h2.\");
}
if (h1.isLessThanOrEqualTo(h2)) {
System.out.println(\"h1 is less than or equal to h2.\");
}
h1.add(h2); // h1 += h2
System.out.println(\"h1.add(h2);\");
System.out.printf(\"h1=%s\ \", h1);
hcopy.subtract(h2); // h1 -= h2
System.out.println(\"h1.subtract(h2);\");
System.out.println(\"h1.subtract(h2);\");
System.out.printf(\"h1=%s\ \", hcopy);
//h1.subtract(h2); // h1 -= h2
hcopy1.add(h2); // h1 += h2
System.out.println(\"h1.add(h2);\");
hcopy2.multiply(h2); // h1 *= h2
System.out.println(\"h1.multiply(h2);\");
System.out.printf(\"h1=%s\ \ \", h1);
}
}
}
package chegg;
public class HugeInteger {
private static final int NUM_DIGITS = 40;
private int digits[] = new int[NUM_DIGITS];
private boolean positive;
public HugeInteger(String num) {
int len = num.length();
if (num.indexOf(0) == \'-\') {
len--;
positive = false;
} else {
positive = true;
}
for (int i = 0; i < len; i++) {
digits[NUM_DIGITS - 1 - i] = num.charAt(len - 1 - i) - (int) \'0\';
digits.toString();
}
}
public void multiply(HugeInteger hi) {
this.digits = multiplyArraydigits(this.digits, hi.digits);
}
private int[] multiplyArraydigits(int[] array1, int[] array2) {
int[] result = new int[array1.length + array2.length];
int carry = 0;
int set = 0;
for (int i = array1.length - 1; i >= 0; i--) {
int temp = result.length - 1 - set;
for (int j = array2.length - 1; j >= 0; j--) {
int sum = result[temp] + (array1[i] * array2[j]) + carry;
result[temp] = sum % 10;
carry = sum / 10;
temp--;
}
if (carry > 0) {
result[temp] += carry;
while (result[temp] >= 10) {
carry = result[temp] / 10;
result[temp] = result[temp] % 10;
result[temp--] += carry;
}
carry = 0;
}
set++;
}
return result;
}
public void subtract(HugeInteger hi) {
// if the signs of the two numbers are not the same
if (this.positive != hi.positive) {
if (this.positive) { // \'this\' is positive...
.
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
class Mesh{
public:
/*Mesh data*?
vectorvertices;
vector indices;
vectortextures;
?*functions*?
Mesh(vectorvertices,vectorindices,vector textures);
void draw(shader shader);
private:
/* Render data*/
gluint vao,vbo,ebo;
/* functions */
void setupmesh();
};
Solution
class Mesh{
public:
/*Mesh data*?
vectorvertices;
vector indices;
vectortextures;
?*functions*?
Mesh(vectorvertices,vectorindices,vector textures);
void draw(shader shader);
private:
/* Render data*/
gluint vao,vbo,ebo;
/* functions */
void setupmesh();
};.
Bacteria was used in transformation experiment, to prove Dna was the.pdf
Bacteria was used in transformation experiment, to prove Dna was the genetic material and
replicated semi conservatively, used in gene expression and regulation, preferred host for cloning
genes, and in manipulation of DNA
Solution
Bacteria was used in transformation experiment, to prove Dna was the genetic material and
replicated semi conservatively, used in gene expression and regulation, preferred host for cloning
genes, and in manipulation of DNA.
c) 11Solutionc) 11.pdf
El documento presenta la solución a una ecuación matemática. La respuesta es el número 11.
assembly mips subroutineSolutionassembly mips subroutine.pdf
This document discusses assembly language code for MIPS subroutines. MIPS subroutines allow breaking programs into smaller reusable parts. To call a subroutine, the address of the subroutine is placed in register $25 and execution is transferred there using the JAL instruction. When the subroutine finishes, it returns using the JR instruction.
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using
tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is
responsible for browning of fruits and cut vegetables.
2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate
specificity for phenolic compounds.
3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol
group.
4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs
that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active
site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor.
Phenylthiourea and resorcinol is a non-competitive inhibitor.
6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction
is best at the optimum temperature or physiological temperature.
Solution
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using
tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is
responsible for browning of fruits and cut vegetables.
2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate
specificity for phenolic compounds.
3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol
group.
4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs
that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active
site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor.
Phenylthiourea and resorcinol is a non-competitive inhibitor.
6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction
is best at the optimum temperature or physiological temperature..
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.)
Solution
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.).
Accounting for Restricted Grants When and How To Record Properly
In this webinar, member learned how to stay in compliance with generally accepted accounting principles (GAAP) for restricted grants.
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This short document discusses a chemical reaction with an electrode potential (E) of 0.28 volts. At this electrode potential, the reactants of the reaction are favored to form, rather than the products. The document simply states that with an E value of 0.28V, the reactants are favored in the chemical reaction.
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
The differences between ipv4 and ipv6 are as below
1. Header:
1.Checksum and options feild are present and the length of header with 20 bytes in ipv4.
2.extension header is present with length 40 bytes in ipv6.
2.Addressing scheme:
1. A 32-bit permiting a amont of 2^32 in ipv4.
2. The adress of ipv6 is 128 bit.
3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6.
4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is
present to describe packet flow for QOS handling.
5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6
SLAAC +DHPCPV6 is used.
6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4
with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes.
7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
spot.
Solution
The differences between ipv4 and ipv6 are as below
1. Header:
1.Checksum and options feild are present and the length of header with 20 bytes in ipv4.
2.extension header is present with length 40 bytes in ipv6.
2.Addressing scheme:
1. A 32-bit permiting a amont of 2^32 in ipv4.
2. The adress of ipv6 is 128 bit.
3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6.
4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is
present to describe packet flow for QOS handling.
5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6
SLAAC +DHPCPV6 is used.
6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4
with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes.
7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
spot..
TCP - TCP breaks data into manageable packets and tracks information.pdf
TCP - TCP breaks data into manageable packets and tracks information such as source and
destination of packets. It is able to reroute packets and is responsible for guaranteed delivery of
the data.
IP - This is a connectionless protocol, which means that a session is not created before sending
data. IP is responsible for addressing and routing of packets between computers. It does not
guarantee delivery and does not give acknowledgement of packets that are lost or sent out of
order as this is the responsibility of higher layer protocols such as TCP.
UDP - A connectionless, datagram service that provides an unreliable, best-effort delivery.
ICMP - Internet Control Message Protocol enables systems on a TCP/IP network to share status
and error information such as with the use of PING and TRACERT utilities.
SMTP - Used to reliably send and receive mail over the Internet.
FTP - File transfer protocol is used for transferring files between remote systems. Must resolve
host name to IP address to establish communication. It is connection oriented (i.e. verifies that
packets reach destination).
TFTP - Same as FTP but not connection oriented.
ARP - provides IP-address to MAC address resolution for IP packets. A MAC address is your
computer\'s unique hardware number and appears in the form 00-A0-F1-27-64-E1 (for example).
Each computer stores an ARP cache of other computers ARP-IP combinations.
POP3 - Post Office Protocol. A POP3 mail server holds mail until the workstation is ready to
receive it.
IMAP - Like POP3, Internet Message Access Protocol is a standard protocol for accessing e-
mail from your local server. IMAP (the latest version is IMAP4) is a client/server protocol in
which e-mail is received and held for you by your Internet server.
TELNET - Provides a virtual terminal or remote login across the network that is connection-
based. The remote server must be running a Telnet service for clients to connect.
HTTP - The Hypertext Transfer Protocol is the set of rules for exchanging files (text, graphic
images, sound, video, and other multimedia files) on the World Wide Web. It is the protocol
controlling the transfer and addressing of HTTP requests and responses.
HTTPS - Signifies that a web page is using the Secure Sockets Layer (SSL) protocol and is
providing a secure connection. This is used for secure internet business transactions.
NTP - Network Time Protocol is a protocol that is used to synchronize computer clock times in
a network of computers.
SNMP - Stands for Simple Network Management Protocol and is used for monitoring and status
information on a network. SNMP can be used to monitor any device that is SNMP capable and
this can include computers, printers, routers, servers, gateways and many more using agents on
the target systems. The agents report information back to the management systems by the use of
“traps” which capture snapshot data of the system. This trap information could be system errors,
resource information, or other info.
Static arrays are structures whose size is fixed at compile time and.pdf
Static arrays are structures whose size is fixed at compile time and therefore cannot be extended
or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is
overhead associated with the operation of copying old data and freeing the memory associated
with the old data structure. One potential problem of using arrays for storing data is that arrays
require a contiguous block of memory which may not be available, if the requested contiguous
block is too large. However the advantages of using arrays are that each element in the array can
be accessed very efficiently using an index. However, for applications that can be better
managed without using contiguous memory we define a concept called “linked lists”.
A linked list is a collection of objects linked together by references from one object to another
object. By convention these objects are named as nodes. So the basic linked list is collection of
nodes where each node contains one or more data fields AND a reference to the next node. The
last node points to a NULL reference to indicate the end of the list.
Types of Linked Lists
Linked lists are widely used in many applications because of the flexibility it provides. Unlike
arrays that are dynamically assigned, linked lists do not require memory from a contiguous
block. This makes it very appealing to store data in a linked list, when the data set is large or
device (eg: PDA) has limited memory. One of the disadvantages of linked lists is that they are
not random accessed like arrays. To find information in a linked list one must start from the head
of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage
of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust”
the array.
There are few different types of linked lists. A singly linked list as described above provides
access to the list from the head node. Traversal is allowed only one way and there is no going
back. A doubly linked list is a list that has two references, one to the next node and another to
previous node. Doubly linked list also starts from head node, but provide access both ways. That
is one can traverse forward or backward from any node. A multilinked list (see figures 1 & 2) is
a more general linked list with multiple links from nodes. For examples, we can define a Node
that has two references, age pointer and a name pointer. With this structure it is possible to
maintain a single list, where if we follow the name pointer we can traverse the list in alphabetical
order of names and if we traverse the age pointer, we can traverse the list sorted by ages. This
type of node organization may be useful for maintaining a customer list in a bank where same
list can be traversed in any order (name, age, or any other criteria) based on the need.
Designing the Node of a Linked List
Linked list is a collection of linked nodes. A node is a struct with at least a.
D) II, III, IV, and V only .pdf
This document provides the answer choice "D) II, III, IV, and V only" as the solution to a problem or question. No other context or details are given about the problem or question being answered. The answer simply states that choices II, III, IV, and V are the correct options.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.next;
}
}
return false;
}
public void printList(){
Node l = first;
while(l.next!=null){
System.out.print(l.number+\" \");
l = l.next;
}
System.out.println();
}
}
Node.java (implements a single node in the linked list; stores number and pointers)
public class Node{
String number;
Node previous;
Node next;
public Node(String num){
number = num;
previous = null;
next = null;
}
public Node(String num, Node p, Node n){
number = num;
previous = p;
next = n;
}
}
Main.java (contains the main method and the helper methods to solve the questions given)
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
static boolean checkHappiness(String num){
Node current = new Node(num, null, null);
NumberList numberList = new NumberList(current);
int len = num.length();
int resultant = 0;
for (int i=0; i happyNumbersfrom1to10000(){
ArrayList numbers = new ArrayList();
for (int j=0; j<10000; j++){
if (checkHappiness(String.valueOf(j+1))){
numbers.add(String.valueOf(j+1));
}
}
return numbers;
}
static void happyNumbersfrom9001to10000(){
for (int j=9001; j<=10000; j++){
//System.out.println(j);
if (checkHappiness(String.valueOf(j))){
System.out.println(String.valueOf(j));
}
}
}
static String getLargeHappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
static String getLargeUnhappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (!checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
public static void main(String[] args){
happyNumbersfrom9001to10000();
System.out.println(happyNumbersfrom1to10000());
}
}
Solution
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation
Solution
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation.
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i
Solution
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i.
Interphase.This is the phase where the cell prepares for the next .pdf
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle.
Solution
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle..
B.the rotation will be zero as L and D will cance.pdf
B.the rotation will be zero as L and D will cancel out each other..
Solution
B.the rotation will be zero as L and D will cancel out each other...
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
In the above conversation it is belonging to stereotypes.
Stereotypes are a form of generalization. When we generalize, we group or classify people,
places, or things according to the traits they have in common. For example, we may say most
Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations
are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood
hasn\'t produced any quality movies in the past fifteen years.\"
But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending
generalizations about people, places, or things. When a stereotype is challenged, the person who
holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As
Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that
sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full
of on those subjects.\"75
The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning,
clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse,
and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy
liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these
stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles
brave, the Irish devout, blacks athletic.)
Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film
directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics,
atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers,
psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society
matrons. There are stereotypes of institutions as well: marriage, the church, government, the
military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would
include even God and mother.)
FACTS DON\'T MATTER
It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is
seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice,
pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People
who stereotype don\'t dust accept the facts that are offered to them. They measure those facts
against what they already \"know.\" That is, they measure them against the stereotype itself.
Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar
facts.
People who think in terms of stereotypes tend to be selective in their perceptions. They reject
conditions that challenge their preformed judgment and retain those that reinforce it. Thus a
person can notice the Jewish employer promoting another Jew but ign.
In general, the objective of an audit is to assess the risk of mater.pdf
In general, the objective of an audit is to assess the risk of material misstatements in the financial
statements. Material misstatements can arise from inadequacies in internal controls and from
inaccurate management assertions. Thus, testing the validity of the various implicit managerial
assertions is a key objective of an auditor.
Existence and Completeness
Auditing standards require that auditors test basic underlying management assertions implicit in
the financial statements. Key among these various assertions are existence or occurrence, which
describe a singular concept: Journal entries are not fiction. As the name implies, an auditor will
conduct various procedures to verify that assets do in fact exist and that recorded transactions did
in fact occur. Additionally, an auditor will seek evidence of completeness, so that the financial
statements include all material transactions that occurred, and so the records do not omit material
transactions for any reason.
Rights and Obligations
The various rights and obligations of the company are important management assertions inherent
in the financial statements. Thus, an auditor will obtain evidence regarding a company\'s rights,
such as proper title to assets and status of intellectual property. An auditor will be concerned
with assertions relating to the company\'s obligations, such as accounts payable balances, long-
term debts and tax liabilities. Thus, the audit objectives will be fulfilled upon validating these
specific assertions.
Valuation or Allocation
Valuation or allocation are managerial assertions which are often material to the financial
statements; thus, an auditor will diligently conduct audit procedures relating to these objectives.
Generally accepted accounting principles, or GAAP, require that certain balance sheet items be
presented using different valuation methodologies. Meeting these standards is a key audit
objective, as the risk of material misstatement is low in probability, but high in magnitude. Thus,
among other things, the historical cost of assets is verified, depreciation methods are scrutinized
and the fair value of investments are calculated to satisfy this objective.
Presentation and Disclosure
Another specific audit objective is validating the presentation of the financial statements and the
adequacy of the disclosures therein. Financial statements should conform to certain requirements
and expectations, and should include the balance sheet, income statement, statement of cash
flows and the statement of owner\'s equity. Relating to disclosure, the auditor will consider the
sufficiency and clarity of footnotes and the transparency in management discussion and analysis,
so he can assess the risk of material misstatement and fulfill the audit objective.
Specific audit objectives are the application of the general audit objectives to a given class of
transactions, account balance, or presentation and disclosure. There must be at least one specific
audit object.
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy
Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl
Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl-
Spectator ions are : Cl - &H+
Spectator ions are : Cl - &H+
Net ionic Equation is : Cu2+ + SO42- --->CuSiO4
Solution
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy
Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl
Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl-
Spectator ions are : Cl - &H+
Spectator ions are : Cl - &H+
Net ionic Equation is : Cu2+ + SO42- --->CuSiO4.
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
fixed 3 issues. I am working on remaining.
package chegg;
public class HugeIntegerApp {
public static void main(String args[]) {
String[][] testInputs = { { \"987654321\", \"234567890\" },
{ \"987654321\", \"-234567890\" }, { \"-987654321\", \"234567890\" },
{ \"-987654321\", \"-234567890\" }, { \"234567890\", \"987654321\" },
{ \"234567890\", \"-987654321\" }, { \"-234567890\", \"987654321\" },
{ \"-234567890\", \"-987654321\" } };
for (String[] ints : testInputs) {
HugeInteger h1 = new HugeInteger(ints[0]);
HugeInteger h2 = new HugeInteger(ints[1]);
HugeInteger hcopy=new HugeInteger(ints[0]);
HugeInteger hcopy1=new HugeInteger(ints[0]);
HugeInteger hcopy2=new HugeInteger(ints[0]);
System.out.println(\"h1=\" + h1);
System.out.println(\"h2=\" + h2);
if (h1.isEqualTo(h2)) {
System.out.println(\"h1 is equal to h2.\");
}
if (h1.isNotEqualTo(h2)) {
System.out.println(\"h1 is not equal to h2.\");
}
if (h1.isGreaterThan(h2)) {
System.out.println(\"h1 is greater than h2.\");
}
if (h1.isLessThan(h2)) {
System.out.println(\"h1 is less than to h2.\");
}
if (h1.isGreaterThanOrEqualTo(h2)) {
System.out.println(\"h1 is greater than or equal to h2.\");
}
if (h1.isLessThanOrEqualTo(h2)) {
System.out.println(\"h1 is less than or equal to h2.\");
}
h1.add(h2); // h1 += h2
System.out.println(\"h1.add(h2);\");
System.out.printf(\"h1=%s\ \", h1);
hcopy.subtract(h2); // h1 -= h2
System.out.println(\"h1.subtract(h2);\");
System.out.println(\"h1.subtract(h2);\");
System.out.printf(\"h1=%s\ \", hcopy);
//h1.subtract(h2); // h1 -= h2
hcopy1.add(h2); // h1 += h2
System.out.println(\"h1.add(h2);\");
hcopy2.multiply(h2); // h1 *= h2
System.out.println(\"h1.multiply(h2);\");
System.out.printf(\"h1=%s\ \ \", h1);
}
}
}
package chegg;
public class HugeInteger {
private static final int NUM_DIGITS = 40;
private int digits[] = new int[NUM_DIGITS];
private boolean positive;
public HugeInteger(String num) {
int len = num.length();
if (num.indexOf(0) == \'-\') {
len--;
positive = false;
} else {
positive = true;
}
for (int i = 0; i < len; i++) {
digits[NUM_DIGITS - 1 - i] = num.charAt(len - 1 - i) - (int) \'0\';
digits.toString();
}
}
public void multiply(HugeInteger hi) {
this.digits = multiplyArraydigits(this.digits, hi.digits);
}
private int[] multiplyArraydigits(int[] array1, int[] array2) {
int[] result = new int[array1.length + array2.length];
int carry = 0;
int set = 0;
for (int i = array1.length - 1; i >= 0; i--) {
int temp = result.length - 1 - set;
for (int j = array2.length - 1; j >= 0; j--) {
int sum = result[temp] + (array1[i] * array2[j]) + carry;
result[temp] = sum % 10;
carry = sum / 10;
temp--;
}
if (carry > 0) {
result[temp] += carry;
while (result[temp] >= 10) {
carry = result[temp] / 10;
result[temp] = result[temp] % 10;
result[temp--] += carry;
}
carry = 0;
}
set++;
}
return result;
}
public void subtract(HugeInteger hi) {
// if the signs of the two numbers are not the same
if (this.positive != hi.positive) {
if (this.positive) { // \'this\' is positive...
.
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
class Mesh{
public:
/*Mesh data*?
vectorvertices;
vector indices;
vectortextures;
?*functions*?
Mesh(vectorvertices,vectorindices,vector textures);
void draw(shader shader);
private:
/* Render data*/
gluint vao,vbo,ebo;
/* functions */
void setupmesh();
};
Solution
class Mesh{
public:
/*Mesh data*?
vectorvertices;
vector indices;
vectortextures;
?*functions*?
Mesh(vectorvertices,vectorindices,vector textures);
void draw(shader shader);
private:
/* Render data*/
gluint vao,vbo,ebo;
/* functions */
void setupmesh();
};.
Bacteria was used in transformation experiment, to prove Dna was the.pdf
Bacteria was used in transformation experiment, to prove Dna was the genetic material and
replicated semi conservatively, used in gene expression and regulation, preferred host for cloning
genes, and in manipulation of DNA
Solution
Bacteria was used in transformation experiment, to prove Dna was the genetic material and
replicated semi conservatively, used in gene expression and regulation, preferred host for cloning
genes, and in manipulation of DNA.
c) 11Solutionc) 11.pdf
El documento presenta la solución a una ecuación matemática. La respuesta es el número 11.
assembly mips subroutineSolutionassembly mips subroutine.pdf
This document discusses assembly language code for MIPS subroutines. MIPS subroutines allow breaking programs into smaller reusable parts. To call a subroutine, the address of the subroutine is placed in register $25 and execution is transferred there using the JAL instruction. When the subroutine finishes, it returns using the JR instruction.
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using
tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is
responsible for browning of fruits and cut vegetables.
2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate
specificity for phenolic compounds.
3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol
group.
4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs
that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active
site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor.
Phenylthiourea and resorcinol is a non-competitive inhibitor.
6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction
is best at the optimum temperature or physiological temperature.
Solution
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using
tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is
responsible for browning of fruits and cut vegetables.
2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate
specificity for phenolic compounds.
3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol
group.
4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs
that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active
site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor.
Phenylthiourea and resorcinol is a non-competitive inhibitor.
6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction
is best at the optimum temperature or physiological temperature..
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.)
Solution
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.).
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Static arrays are structures whose size is fixed at compile time and.pdf
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c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
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In general, the objective of an audit is to assess the risk of mater.pdf
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I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
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fixed 3 issues. I am working on remaining.package chegg;public c.pdf
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Bacteria was used in transformation experiment, to prove Dna was the.pdf
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if one can understand a few things it is easier to solve these kind .pdf
- 1. if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive
- 2. expression but Wild type (I+) is dominant over I- . so it is inducible operon. Solution if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to
- 3. grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive expression but Wild type (I+) is dominant over I- . so it is inducible operon.