Router Switeh Routen pe Ping pe Wireless. 7W aicess point DMZ Zone au.pdf
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Router Switeh Routen pe Ping pe Wireless. 7W aicess point DMZ Zone auess point out ey acess poinf Broadbond wrreless access autess point Servers Solution The logic behind IDS sensors is that they actually try to sniff the route in which your network traffic flows. whereas the IPS(intrusion protection system) is more effective.The IDS has the ability to only detect the uneven activity but can\'t do with the help of IPS feature. if a user want to check or sniff all the traffic when the user is connected to the outside world(internet) just set the IDS outside the firewal. if a user want to check the traffic which is flowing inside the network whether malicious or not the best idea would be to install IDS inside the firewall. the DMZ zone is also said as(demilitarized zone) is a sub-network where the internal network( internal LAN) is seperated from the other untrusted network like the internet. the external servers,or the resources and services are located in the DMZ so they are easily accessible from the Internet but the other part of the internal LAN remains safe as it is not reacheble.it restricts the ability of hackers to directly access the internal servers present in a network as well as the data also with the help of accessing Internet. the best way to put IPS/IDS is near the firewal(outside the firewal) so can manage every traffic comming from outside like internet and if the traffic seems suspicious can simply block the packet or traffic outside the network only. IDS is expensive so it can\'t be use at many areas but the best to use is near firewall only..
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operations for adding and deleting items. Use doubly linked list for your implementation. Call
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Solution
//package com.java2novice.ds.queue;
public class DequeDblLinkedListImpl {
private LinkedNode front;
private LinkedNode rear;
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int i=0;
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{
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i++;
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}
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}
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if(tmpRear != null) tmpRear.setNext(null);
if(tmpRear == null) front = null;
System.out.println(\"removed from rear: \"+rear.getValue());
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}
public static void main(String a[]){
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deque.insertFront(34);
deque.insertFront(67);
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deque.insertFront(29);
deque.insertFront(765);
deque.removeFront();
deque.removeFront();
deque.removeFront();
deque.insertRear(43);
deque.insertRear(83);
deque.insertRear(84);
deque.insertRear(546);
deque.insertRear(356);
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeFront();
deque.removeFront();
deque.removeFront();
}
}
class LinkedNode{
private LinkedNode prev;
private LinkedNode next;
private T value;
public LinkedNode getPrev() {
return prev;
}
public void setPrev(LinkedNode prev) {
this.prev = prev;
}
public LinkedNode getNext() {
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public void setNext(L.
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Let f(z) be an entire function such that fn+1(z) equivalent 0, Prove that f is a polynomial of
degree at most n?
Solution
We can prove the above formula using mathematical induction.
So lets us take multiple cases.
For f(x) = x, f(x)\'=1, f(x)\'\'=0. so for f(x)^2 the polynomial power is 1,
for f(x)=x^2, f(x)\'=2x, f(x)\'\'=2, f(x)\'\'\'=0, so for f(x)\'\'\' the polynomial is 2.
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Intrusion Detection System: Security Monitoring System
Intrusion Detection System: Security Monitoring System
Securing E-commerce networks in MIS and E-Commerce
Securing E-commerce networks in MIS and E-Commerce
More from amrishinda
JAVA A double-ended queue is a list that allows the addition and.pdf
**JAVA** A double-ended queue is a list that allows the addition and removal of items from
either end. One end is arbitrarily called the front and other the rear, but the two ends behave
identically. Specify, design and implement a class for a double-ended queue. Include operations
to check if it is empty and to return the number of items in the list. For each end, include
operations for adding and deleting items. Use doubly linked list for your implementation. Call
your class Deque (pronounced \"deck\").
Solution
//package com.java2novice.ds.queue;
public class DequeDblLinkedListImpl {
private LinkedNode front;
private LinkedNode rear;
public void insertFront(T item){
//add element at the beginning of the queue
System.out.println(\"adding at front: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setNext(front);
if(front != null) front.setPrev(nd);
if(front == null) rear = nd;
front = nd;
}
public void insertRear(T item){
//add element at the end of the queue
System.out.println(\"adding at rear: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setPrev(rear);
if(rear != null) rear.setNext(nd);
if(rear == null) front = nd;
rear = nd;
}
public void removeFront(){
if(front == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front.getNext();
if(tmpFront != null) tmpFront.setPrev(null);
if(tmpFront == null) rear = null;
System.out.println(\"removed from front: \"+front.getValue());
front = tmpFront;
}
public int size(){
if(front == null && rear ==null){
return 0;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front;
LinkedNode tmpRear = rear;
int i=0;
while(tmpFront != rear)
{
i++;
tmpFront=tmpFront.getNext();
}
i++;
return i;
}
public void removeRear(){
if(rear == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpRear = rear.getPrev();
if(tmpRear != null) tmpRear.setNext(null);
if(tmpRear == null) front = null;
System.out.println(\"removed from rear: \"+rear.getValue());
rear = tmpRear;
}
public static void main(String a[]){
DequeDblLinkedListImpl deque = new DequeDblLinkedListImpl();
deque.insertFront(34);
deque.insertFront(67);
System.out.println(\"Size of the queue : \"+deque.size());
deque.insertFront(29);
deque.insertFront(765);
deque.removeFront();
deque.removeFront();
deque.removeFront();
deque.insertRear(43);
deque.insertRear(83);
deque.insertRear(84);
deque.insertRear(546);
deque.insertRear(356);
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeFront();
deque.removeFront();
deque.removeFront();
}
}
class LinkedNode{
private LinkedNode prev;
private LinkedNode next;
private T value;
public LinkedNode getPrev() {
return prev;
}
public void setPrev(LinkedNode prev) {
this.prev = prev;
}
public LinkedNode getNext() {
return next;
}
public void setNext(L.
Let f(z) be an entire function such that fn+1(z) equivalent 0, Prove.pdf
Let f(z) be an entire function such that fn+1(z) equivalent 0, Prove that f is a polynomial of
degree at most n?
Solution
We can prove the above formula using mathematical induction.
So lets us take multiple cases.
For f(x) = x, f(x)\'=1, f(x)\'\'=0. so for f(x)^2 the polynomial power is 1,
for f(x)=x^2, f(x)\'=2x, f(x)\'\'=2, f(x)\'\'\'=0, so for f(x)\'\'\' the polynomial is 2.
So we can deduce that for a polynomial with n power it should be differentiated n+1 times to get
fn+1(z)=0,
So it could be stated otherwise if fn+1(z)=0 then f(z) has a power of atmost x^n..
Implement threads and a GUI interface using advanced Java Swing clas.pdf
Implement threads and a GUI interface using advanced Java Swing classes.
Required data structure - the advanced data structure uses implementation of a multi-tree with
the following levels: Cave - level 0 Party - Level 1 Creature - Level 2 Artifacts - Level 3
Treasures - Level 3 Jobs
Use the Swing class JTree effectively to display the contents of the data file. Implement a JTable
to also show the contents of the data file.
Threads: Implement a thread for each job representing a task that creature will perform. Use the
synchronize directive to avoid race conditions and insure that a creature is performing only one
job at a time. The thread for each job should be started as the job is read in from the data file.
Only one job should be progressing for each creature at any moment. Use delays to show the
creature doing the task.
Use a JProgressBar for each creature to show the creature performing the task. Use JButton\'s on
the Job panel to allow the task to be suspended and cancelled. The GUI elements should be
distinct from the other classes in the program.
Solution
Please find the code below for the above problem :
import java.util.ArrayList;
import java.util.Scanner;
import java.util.HashMap;
import java.util.Random;
import java.io.File;
import java.io.FileNotFoundException;
import javax.swing.JProgressBar;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.JButton;
import javax.swing.JTextArea;
import javax.swing.JScrollPane;
import javax.swing.SwingConstants;
import java.awt.GridLayout;
import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.Font;
import java.awt.event.ActionListener;
import java.awt.event.ActionEvent;
public class Cave2 {
ArrayList parties = new ArrayList (); // the parties
ArrayList loners = new ArrayList (); // creatures not in parties
ArrayList drops = new ArrayList (); // Treasures not carried
ArrayList magics = new ArrayList (); // Artifacts not carried
ArrayList jobs = new ArrayList (); // Jobs to be performed
JFrame jf = new JFrame ();
JPanel jrun = new JPanel ();
JTextArea jta = new JTextArea (10, 10);
// hashmap of everything by index
HashMap hmElements = new HashMap ();
public Cave2 () {
Scanner sf = null;
String fileName = \"dataC.txt\";
try {
// sf = new Scanner (new File (fileName));
// getResourceAsStream is the way to access a data file in a jar package file
// this works in general, so leave it here as is.
sf = new Scanner (getClass().getResourceAsStream (fileName));
}
catch (Exception e) {
System.out.println (e + \"\ File bummer: \" + fileName);
} // end open file try/catch block
if (sf == null)
return;
readFile (sf);
jta.setText (toString ());
jf.setTitle (\"Cave2\");
jf.add (jrun, BorderLayout.PAGE_END);
jrun.setLayout (new GridLayout (0, 5, 2, 5));
jf.add (new JScrollPane (jta), BorderLayout.CENTER);
jta.setFont (new Font (\"Monospaced\", 0, 12));
jf.pack ();
jf.setVisible (true);
jf.setLocationRelativeTo (null);
} // end no-parameter constru.
If a cells contents are greatly hypo-osmotic to the surrounding env.pdf
If a cell\'s contents are greatly hypo-osmotic to the surrounding environment, the cell will
probably A. burst divide shrivel expand
Solution
hypo-osmotic cell means the cell contains less amount of sloute than surrounding ..indirectly
saying cell has more water content..surrounndings are hyperosmotic ..water always moves from
hypo osmotic to hyperosmotic so here water moves out of the cell to surroundings thats leading
dehydration of cells results in shrinkage of cell answer is C shrivel.
How are the forelimbs of humans, the wings of birds, the wings of ba.pdf
How are the forelimbs of humans, the wings of birds, the wings of bats, the wings of insects
similar, how are they different, and what do we think happened in the past evolution of these
structures ?
Solution
All these structures mentioned, that is, the forelimbs of humans, the wings of birds, the wings of
bats, the wings of insects have a similarfunction because they all are used for movement. But
they are structurally very dissimilar. Hence they are also known as analogous characters.
The independent evolution of characters in members of different species with similar function
but differences in structure and ancestral origin is known as convergent evolution. This type of
evolution leads to the formation of analogous characters or traits. The convergent evolution of
flight of birds, bats and insects is known as functional convergence, as their function is similar.
In the very past, there must have been a common ancestor, from whom the wings and forelimbs
were evolved and structurally became different in the course of evolution..
Fom 1120, Sched Mark for follow up Question 15 of 105. A corporat.pdf
Fom 1120, Sched Mark for follow up Question 15 of 105. A corporation is required to file an
annual Form 1120 if all of the following are true EXCEPT The corporation was dissolved in the
prior year O The corporation retains no assets rts a loss The corporation no longer carries on
substantial activities Mark for follow up « Back Save /Return Later Summary Next >»
Solution
Answer to Question 1:
The correct option is (a) , a corporation was dissolved in the prior year..
Germ cells are set aside during animal development and give rise to .pdf
Germ cells are set aside during animal development and give rise to the gonads during
embryogensis. Discuss the implications of this with regards to inheritance of the genetic
mutations inherited from a parent vs. mutations acquired during the lifespan of the organism.
Solution
Inheritance of the genetic mutations inherited from a parent:
Genetic mutations inherited from a parent are also called germline mutations because they are
present in the parent’s germ cells i.e. their egg or sperm cells. These cells give rise to the gonads
during embryogenesis. When embryogenesis takes place the resulting fertilized egg cell receives
DNA from both parents. The mutated germ cells thus result in formation of mutated DNA
making it an inherited mutation.
Mutations acquired during the lifespan of the organism:
When mutation takes place during lifespan and in some cells of the body, it results in acquired
mutation. These can be caused by environmental factors such as ultraviolet radiation from the
sun, or can occur if a mistake in DNA copieng during cell division. Acquired mutations takes
place in somatic cells which are other than the germ cells. Such mutations which are not
inherited do not take place in germ cells..
Determine the missing amounts. (Hint For example, to solve for (a), .pdf
Determine the missing amounts. (Hint: For example, to solve for (a), Assets Liabilities
Stockholders Equity - $28,730.) anuary 1, 2015 Assets Liabilities Stockholders\' equity $ 86,190
$126,412 s 137904 (g) 86,190 51,714 $172,380 57,460 57460 (d) 68,952 57460 0) 114,920
28730 (a) December 31, 2015 Assets Liabilities Stockholders\' equity 109174 (b) 63,206 45,968
157,440 229,840 252824 (k) 91,936 86,190 71250 (e) 17,238 80444 (h) 149,396 160,888
Stockholders\' equity changes in year Additional investment Dividends Total revenues Total
expenses 17,238 11,492 574,600 11,492 11,492 402,220 379,236 16,089 482,664 442,442
393,026
Solution
Donatello Company:
January 1, 2015:
Stockholders’ Equity = Assets - Liabilities
Stockholders’ Equity = $86,190 - $57,460
Stockholders’ Equity = $28,730
December 31, 2015:
Assets = Liabilities + Stockholders’ Equity
Assets = $63,206 + $45,968
Assets = $109,174
Change in Stockholders’ Equity = Stockholders’ Equity, December 31, 2015 - Stockholders’
Equity, January 1, 2015
Change in Stockholders’ Equity = $45,968 - $28,730
Change in Stockholders’ Equity = $17,238
Change in Stockholders’ Equity = Additional Investment + Total Revenues - Total Expenses -
Dividends
$17,238 = Additional Investment + $402,220 - $379,236 - $11,492
Additional Investment = $5,746
Leonardo Company:
January 1, 2015:
Assets = Liabilities + Stockholders’ Equity
$126,412 = Liabilities + $68,952
Liabilities = $57,460
December 31, 2015:
Assets = Liabilities + Stockholders’ Equity
$157,440 = $86,190 + Stockholders’ Equity
Stockholders’ Equity = $71,250
Change in Stockholders’ Equity = Stockholders’ Equity, December 31, 2015 - Stockholders’
Equity, January 1, 2015
Change in Stockholders’ Equity = $71,250 - $68,952
Change in Stockholders’ Equity = $2,298
Change in Stockholders’ Equity = Additional Investment + Total Revenues - Total Expenses -
Dividends
$2,298 = $17,238 + $482,664 - $442,442 - Dividends
Dividends = $55,162
Michelangelo Company:
January 1, 2015:
Assets = Liabilities + Stockholders’ Equity
Assets = $86,190 + $51,714
Assets = $137,904
December 31, 2015:
Assets = Liabilities + Stockholders’ Equity
$229,840 = Liabilities + $149,396
Liabilities = $80,444
Change in Stockholders’ Equity = Stockholders’ Equity, December 31, 2015 - Stockholders’
Equity, January 1, 2015
Change in Stockholders’ Equity = $149,396 - $51,714
Change in Stockholders’ Equity = $97,682
Change in Stockholders’ Equity = Additional Investment + Total Revenues - Total Expenses -
Dividends
$97,682 = $11,492 + Total Revenues - $393,026 - $16,089
Total Revenues = $495,305
Raphael Company:
January 1, 2015:
Assets = Liabilities + Stockholders’ Equity
$172,380 = Liabilities + $114,920
Liabilities = $57,460
December 31, 2015:
Assets = Liabilities + Stockholders’ Equity
Assets = $91,936 + $160,888
Assets = $252,824
Change in Stockholders’ Equity = Stockholders’ Equity, December 31, 2015 - Stockholders’
Equity, January 1, 2015
Change in Stockholders’ Equity = $160,888 - $114,920
Cha.
Discuss the various methods of intellectual property protection avai.pdf
Discuss the various methods of intellectual property protection available to businesses and
individuals.
Solution
The World Trade organisation (WTO) divides intellectual property in two main areas:
(i) Copyright and rights related to copyright. back to top
The rights authors have over literary and artistic works of their creation (such as books and other
types of creations, like paintings, sculpture, computer programs, music and films), are protected
by copyright, for a minimum time of 50 years after the author’s death.
Also are protected by copyright (neighbouring) the rights of performers such as actors, singers,
musicians, producers of phongrams and broadcasting agencies. The main purpose of copyright
laws is to protect and encourage authors and creators to keep working on their artistic and
creative products, having a legal base to cover them as owners of their work.
Industrial property
Industrial property is divided in two areas:
Industrial property is protected by law so as to encourage inventors, technology creators in all
areas to do their work, and to ensure they are the owners of their own work and gained
knowledge about some technical or technological solution, useful to mankind..
devise a hypothesis that explains the geographical distribution of m.pdf
devise a hypothesis that explains the geographical distribution of modern species, that if backed
by evidence would support the theory of evolution. Devise a hypothesis that explains the
geographical distribution of modern species, that is backed by evidence would refute the theory
of evolution. regarding chapter 22 descent with modification campbell 10ed
Solution
One of the difficulties in knowledge the ecological basis for species range distributions is
knowing the general significance of abiotic factors like temperature, precipitation, topography, or
earth situations (i.e., bioclimatic or potential niche) vs. biotic interactions among species like
predation, opposition or mutualisms (biotic or noticed niche) in deciding wherever species can be
found geographically. Moreover, if weather change stands to improve species geographic
locations, we however don\'t know if their biotic interdependencies, established by their
interactions, will become unravelled called community disassembly or if community can transfer
as intact units. Modern analyses of species geographic range distributions concentration on
bioclimatic niches, with restricted work to think about biotic interactions.
Classically there were two hypotheses for the progress of humans, the multiregional hypothesis -
modern Homo sapiens evolved simultaneously for the duration of the old world from archaic
Homo sapiens Homo sapiens with exchange of with exchange of genetic data by gene flow and
the out-of-Africa hypothesis - modern humans evolved in Africa and transferred out changing
previously widely dispersed archaic humans. The newest kind, the modern horse (Equus), is a lot
larger in size, is one-toed, and has teeth appropriate for grazing. The transitional forms are well
preserved as fossils, as are a great many other types of extinct horses that evolved in different
directions and remaining no living descendants..
Describe a diode array detector and CCD camera. Differentiate betwee.pdf
Describe a diode array detector and CCD camera. Differentiate between them in terms of the
output and how signal is collected.
Solution
diode array detector are linear array fabricated in single strip of semiconductor. They are used in
high level application and they are available in single or double array. typically diode array
detector cost less than CCD and have similar spectral senitivity in some range.
CCD means charge coupled device . these are also built in single semiconductor chip but they
are two dimensional. here pixels are read out sequentially. down side is CCD have long response
time than diode array..
Connect a motor, two push buttons and two LEDS to the Arduino as sho.pdf
Connect a motor, two push buttons and two LEDS to the Arduino as shown below. The fan
should operate as listed.
Begin with
int buttonPin1 = 2;
int buttonPin2 = 4;
int ledPinGreen = 11;
int ledPinRed = 6;
int motorPin = 9;
Please program in Arduino as the following:
1. Fan is started or stopped by pressing push button 1
When starting the fan, ramp up using 110ms delay from 0 to 4 volts (levels 0 to 200), fan
remains running at 4 volts.
When stopping the fan, slow down using the same values (110ms delay and level 200 to 0).
2. When operator presses the overload switch (button 2), fan starts working at full speed (255)
and the overload alarm is lit (red LED).
3. When the overload switch (button 2) is pressed again, fan returns to normal operation at 4
volts (level 200).
4. Pressing the overload button when the fan is not operating has no effect.
5. Anytime the fan is on, on/off indictor is lit, otherwise it is off.
Solution
const int buttonPin1 = 2;
const int buttonPin2 = 4;
const int ledPinGreen = 11;
const int ledPinRed = 6;
const int motorPin = 9
//state varibles
boolean B1_state = false; //button 1 state
boolean B2_state = false; //button 2 state
// function declaration
void motorRun_Stop(boolean);
void overload(boolean,boolean);
void setup()
{
Serial.begin(9600); // for debugging
pinMode(buttonPin1,INPUT); // declaring as input
digitalWrite(buttonPin1,HIGH); // activating the pullup resistor
pinMode(buttonPin2,INPUT); // declaring as input
digitalWrite(buttonPin1,HIGH); // activating the pullup resistor
pinMode(ledPinGreen,OUTPUT);
pinMode(ledPinRed,OUTPUT);
pinMode(motorPin, OUTPUT);
}
void loop()
{
if(digitalRead(buttonPin1)== 1)
{
B1_state != B1_state;
motorRun_Stop(B1_state);
}
else if(digitalRead(buttonPin2)== 1)
{
B2_state != B2_state;
overload();
}
else
{
}
}
void motorRun_Stop(boolean x)
{
byte wait = 0;
if(x == true )
{
Serial.println(\"Motor has Overloaded\");
digitalWrite(ledPinGreen,HIGH);
for (int i = 0; i<200;i++)
{
analogWrite(motorPin,i);
wait = micros();
while((micros()-wait)>550);
}
}
else if(x == false)
{
for (int i = 200; i>0;i--)
{
analogWrite(motorPin,i);
wait = micros();
while((micros()-wait)>550);
}
Serial.println(\"Motor has Stoped\");
digitalWrite(ledPinGreen,LOW);
}
else
{
}
}
void overload(boolean x,boolean y)
{
byte wait = 0;
if(x == true && y == true)
{
analogWrite(motorPin,255);
Serial.println(\"Motor has overloaded\");
digitalWrite(ledPinRed,HIGH);
}
else if(x == true && y==false)
{
analogWrite(motorPin,200);
Serial.println(\"Motor has stopped Overloading\");
digitalWrite(ledPinRed,LOW);
}
else
{
}
}.
Consider managers (supervisors) that you have worked for in the past.pdf
Consider managers (supervisors) that you have worked for in the past. What type of power did
they have? Was it effective?
Must be a good sized paragraph and site information
Solution
earlier i worked with 2 different kind of people who were my bosses. the previous one is Naveen,
HOD of management studies in the college. he is a consultative manager, he recieves the targets
and orders from the top level people and once he recieves the orders he called all of us and ask
how to implement and what are the limitations and possibilities. here he have the ultimate power
to do it and implement it. and some times he distributes the work along with power to others and
they work in their own limitations.
when ever he feels that if any activity involves much risk, then he used to call top level people
and explain them about the consequences and the decision of them. if they ask to proceed he can
move forward otherwilse they rethinks about the task and re-assign it by doing some
modifications.
the power has been enjoying by him is given by the position (HOD) and he is the answerable
person to the higher authorities. he is the whole responsible person to the failures and even for
success also. yes, it was effective and it makes sense. ofcourse some times it leads to delay the
work and once a while fail also..
Caterpillar is one of the Americas major exporters. The company sell.pdf
Caterpillar is one of the Americas major exporters. The company sells its construction
equipment, mining, equipment and engines to some 200 contries.
Discuss the following questions briefly
1.In the 1980s a stronger dollar hurt Caterpillars competitive position, but in 2008 a stronger
dollar did not seem to have the same effect. What had changed?
2. How did Caterpillar use strategy as a \"real hudge\" to reduce its exposure to foreign exchange
risk? What is the downside of its its approach?
3. Explain the difference between transaction exposure and translation exposure using the
material in the Caterpillar case to illustrate your answer.
Solution
1.In the 1980s a stronger dollar hurt Caterpillars competitive position, but in 2008 a stronger
dollar did not seem to have the same effect. What had changed?
Confidence in US Assets: The dollar started to strengthen again in 2008, foreing investment in
US assests was strong, Treasury bills was safe haven in economic stom
Deal with Unions: in the 80´s : employee layoffs and striks, outsourcing parts of production .
Mid 2000s: reached deal with union, increase of productivity.
Change in company structure: Centralized to decentralized, own pricing decisions, own product
desings, improved quality and efficiency
Improvement in product technology & energy efficiency.
2. How did Caterpillar use strategy as a \"real hudge\" to reduce its exposure to foreign exchange
risk? What is the downside of its its approach?
Caterpillar´s global foot print: \"From Calgary to Cairo and from Sydney to Salt Lake city
Caterpillar operates hundreds of offices and facilitites across the world. Caterpillar´s global
presence, product breadth and financial strenght enable us wo tin in today´s competitive market
place.
Forward Exchange: When 2 parties agree to exchange and execute the deal at some specific date
in the future.
Downside of its approach:
Currency´s value is determinated by market force, government intervention as well as demand
and supply of currency:
Globalisation: More competition within the global market that have adopted globalisation
Forward Exchange: Based on future expectation of the foreign exchange market
3. Explain the difference between transaction exposure and translation exposure using the
material in the Caterpillar case to illustrate your answer.
Transaction Exposure: The extent to which income from invididual transactions is affected by
fluctuation in foreign exchange valus
1980´s
Us dollar strenfthens
Foreign payable to Caterpillar become more expensive
Loss of foreign sales = reduced revenues + Limited means of managing transaction exposure =
Massive reduction in future cash flows
Translation Exposure: The extent to which consolidated financial results of a corporation are
affected by movements in exchange rates.
what numbers go with these descriptions.1. Uses ATP hydrolysis2..pdf
what numbers go with these descriptions.
1. Uses ATP hydrolysis
2.Has highly nonpolar structure
3.Powered by an ion gradient
4.Transports glucose across the membrane
5.Na+/glucose symport
6.Na+/K+ ATPase
7.Movement of water molecules
8.Direction of transport depends on concentration gradient
ONLY ONE ANSWER GOES WITH EACH! For each of the following, choose the item from
the list that best describes the item being described. There is only 1 best answer for each. Some
of the items from the list may be used more than once, some may not be used at all Numbe Item
Simple diffusion and facilitated transport Active transport None of the listed processes Simple
diffusion 5 Coupled transport Facilitated diffusion and active transport Facilitated diffusion
(passive transport) Active transport and coupled transport
Solution
6. Na+/K+ ATPase : Active transport
5. Na+/glucose : Active transport and coupled transport
4. Transport glucose across the membrane: Faciliated transport
7. Movement of water molecules: simple diffusion and facilitated transport.
What are the three forms of Business Organizations Answer the follow.pdf
What are the three forms of Business Organizations? Answer the following Accounts: (A)Asset
(L)Liability (SE)Stockholders\' Equity 1. Cash 2. Common Stock 3. Accounts Payable 4. Rent
Expense 5. Sales or Revenue 6. Accounts Receivable What is the first financial statement to be
prepared? What are the three Activities of a Business? Financial Accounting provides
information for decision makers primarily inside or outside the entity? What are the Rules of
Accounting called? Write the Accounting Equation? Can it be written another way?
Solution
The three forms of business organizations are Sole proprietorship, Partnership, and a
Corporation.
1. Cash – (A)
2. Common stock – (SE)
3. Accounts payable – (L)
4. Rent expense – (SE)
5. Sales or Revenue – (SE)
6. Accounts receivable – (A)
The first financial statement to be prepared is the Income Statement.
The three activities of a business are Operating activities, Investing activities, and Financing
activities.
Financial Accounting provides information for decision makers primarily outside the entity such
as shareholders, creditors, etc.
The Rules of Accounting are called the Generally Accepted Accounting Principles or GAAP for
short.
The Accounting Equation is typically written as:
Assets = Liabilities + Stockholders’ Equity
It can also be written in two more ways as under:
Liabilities = Assets – Stockholders’ Equity; or
Stockholders’ Equity = Assets - Liabilities.
Anatomy Question Use your own words.A girl with a defective heart.pdf
Anatomy Question: Use your own words.
A girl with a defective heart receives a new heart transplanted from another child who was killed
in an accident. The patient is given an antilymphocyte serum containing antibodies against her
lymphocytes. The transplanted heart is not rejected, but the patient dies of an overwhelming
bacterial infection. Why was the antilymphocyte serum given? Why was the patient so
vulnerable to infection?
Solution
Ans: antilymphocyte serum is given or suppress immune response to stop graft rejection as the
heart received is not matched with receiver due to different HLA or MHC on the donor and
recepient tissues. So that heart I.e allograft can\'t be rejected by host immune system.
But here comes a problem I.e host immune system is compromised due to antilymphocyte
serum,so no longer host immune system can fight against disease.so host is more prone to
diseases and soon dies.
Why parkinsons disease is special among the different neurological.pdf
Why parkinson\'s disease is special among the different neurological movement disorders in
terms of understanding and therapy?
Solution
Hi,
Perkinson\'s disease is neurological disorder which is degenerative and progressive.
It is due to degeneration of cell of substangia nigra which is small part of brain.
Symptoms:
Therapies :
I hope my answer is satisfactory.
Thanks!!.
Which statement is not true promoters are transcribed the Pribnow .pdf
Which statement is not true? promoters are transcribed the Pribnow box found in bacteria,
consists of TATAAT Enhancers are DNA sequences exons are spliced activators are proteins
Solution
Promoters are transcribed
This statement is false because promoters are regions which binds to transcription factors that
initiate transcription upstream. Thus they are the regions that regulate the transcription but itself
is not transcribed.
Exons are spliced
The above sequence is correct because in alternative splicing some exons may be included and
some are spliced to generate multiple RNA from single DNA sequence.
Hence option A is correct..
When extracting DNA of an fruit how does the environemnt (how hot or.pdf
When extracting DNA of an fruit how does the environemnt (how hot or cold the onion is) affect
the DNA?
Solution
The temperature of the extraction will determine the working of the enzymatic systems (internal
and external). As increasing the temperature will hamper the the enzymes, further the DNA
might undergo annealing (happens in PCR).
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