You have to leave solids and liquids out of the equation. This lea.pdf
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You have to leave solids and liquids out of the equation. This leaves only product B for the product, nothing for the reactants. Initially there was no product B. At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 = which rounds to answer E) 1.3e-3 it was a trick question. Solution You have to leave solids and liquids out of the equation. This leaves only product B for the product, nothing for the reactants. Initially there was no product B. At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 = which rounds to answer E) 1.3e-3 it was a trick question..
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Tungsten is a body centered cubic latticeSolutionTungsten is a.pdf
Tungsten is a body centered cubic lattice
Solution
Tungsten is a body centered cubic lattice.
This is the final code which meets your requirement.Than youCard.j.pdf
This is the final code which meets your requirement.Than you
Card.java
import java.util.Random;
public class Card {
//declaring instance variables
private String suit;
private int face_value;
public static int count=52;
//Creating an instance of Random class reference
Random rand;
//Default Constructor
public Card() {
super();
//Passing random class object to the Random class reference
rand = new Random();
}
//Getters and setters
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
public int getFace_value() {
return face_value;
}
public void setFace_value(int face_value) {
this.face_value = face_value;
}
public int getCount() {
return count;
}
//This shuffle() method will shuffle the deck
public void shuffle()
{
//Picking a card from the deck randomly and find its face value and suit
this.face_value = 1 + rand.nextInt((9) + 1);
int suit_random_value = 1 + rand.nextInt((3) + 1);
if(suit_random_value==1)
{
this.suit=\"Clubs\";
}
else if(suit_random_value==2)
{
this.suit=\"hearts\";
}
else if(suit_random_value==3)
{
this.suit=\"Spades\";
}
else if(suit_random_value==4)
{
this.suit=\"diamonds\";
}
this.count=count-1;
}
}
________________________________________________________________________
DriverClass.java
public class DriverClass {
public static void main(String[] args) {
//creating an array which can hold 52 Card class Objects
Card c[]=new Card[52];
//Creating 52 card class objects and storing them into an array
for(int i=0;i<52;i++)
{
//Creating Card class Object
c[i]=new Card();
}
//Displaying each card face value and its suit.
for(int i=0;i<52;i++)
{
//Shuffling the deck before every card picking from the deck
c[i].shuffle();
//Displaying randomly picked card face value and its suit
System.out.println(c[i].getFace_value()+\" of \"+c[i].getSuit()+\" Cards remaining
\"+c[i].getCount());
}
}
}
_________________________________________________________________________
Output:
8 of diamonds
Cards remaining 51
4 of Clubs
Cards remaining 50
9 of Clubs
Cards remaining 49
7 of Spades
Cards remaining 48
9 of diamonds
Cards remaining 47
7 of diamonds
Cards remaining 46
4 of diamonds
Cards remaining 45
9 of Clubs
Cards remaining 44
5 of hearts
Cards remaining 43
7 of Spades
Cards remaining 42
10 of Clubs
Cards remaining 41
7 of diamonds
Cards remaining 40
8 of diamonds
Cards remaining 39
6 of hearts
Cards remaining 38
4 of hearts
Cards remaining 37
6 of Spades
Cards remaining 36
4 of Clubs
Cards remaining 35
3 of diamonds
Cards remaining 34
9 of diamonds
Cards remaining 33
3 of diamonds
Cards remaining 32
5 of Spades
Cards remaining 31
4 of Clubs
Cards remaining 30
3 of Spades
Cards remaining 29
6 of diamonds
Cards remaining 28
3 of hearts
Cards remaining 27
9 of hearts
Cards remaining 26
7 of hearts
Cards remaining 25
3 of hearts
Cards remaining 24
10 of Clubs
Cards remaining 23
2 of hearts
Cards remaining 22
9 of Clubs
Cards remaining 21
8 of diamonds
Cards remaining 20
8 of Clubs
Cards remaining 19
2 of h.
Statistics is the mathematical science involving the collection, ana.pdf
Statistics is the mathematical science involving the collection, analysis and interpretation of data.
A number of specialties have evolved to apply statistical theory and methods to various
disciplines. Certain topics have \"statistical\" in their name but relate to manipulations of
probability distributions rather than to statistical analysis.
· Actuarial science is the discipline that applies mathematical and statistical methods to
assess risk in the insurance and financeindustries.
· Astrostatistics is the discipline that applies statistical analysis to the understanding of
astronomical data.
· Biostatistics is a branch of biology that studies biological phenomena and observations by
means of statistical analysis, and includes medical statistics.
· Business analytics is a rapidly developing business process that applies statistical methods
to data sets (often very large) to develop new insights and understanding of business
performance & opportunities
· Chemometrics is the science of relating measurements made on a chemical system or
process to the state of the system via application of mathematical or statistical methods.
· Demography is the statistical study of all populations. It can be a very general science that
can be applied to any kind of dynamic population, that is, one that changes over time or space.
· Econometrics is a branch of economics that applies statistical methods to the empirical
study of economic theories and relationships.
· Environmental statistics is the application of statistical methods to environmental science.
Weather, climate, air and water quality are included, as are studies of plant and animal
populations.
· Epidemiology is the study of factors affecting the health and illness of populations, and
serves as the foundation and logic of interventions made in the interest of public health and
preventive medicine.
· Geostatistics is a branch of geography that deals with the analysis of data from disciplines
such as petroleum geology, hydrogeology, hydrology, meteorology,oceanography, geochemistry,
geography.
· Machine Learning
· Operations research (or Operational Research) is an interdisciplinary branch of applied
mathematics and formal science that uses methods such as mathematical modeling, statistics, and
algorithms to arrive at optimal or near optimal solutions to complex problems.
· Population ecology is a sub-field of ecology that deals with the dynamics of species
populations and how these populations interact with the environment.
· Psychometric is the theory and technique of educational and psychological measurement
of knowledge, abilities, attitudes, and personality traits.
· Quality control reviews the factors involved in manufacturing and production; it can make
use of statistical sampling of product items to aid decisions in process control or in accepting
deliveries.
· Quantitative psychology is the science of statistically explaining and changing mental
processes and behaviors in humans.
· Reliabi.
Secondary phloemApples (Malus communis, M. pumila, & M. sylvestr.pdf
Secondary phloem
Apples (Malus communis, M. pumila, & M. sylvestris), pears (Pyrus communis) and quince
(Cydonia oblonga) belong to the rose family (Rosaceae), and include literally hundreds of
cultivated varieties. In the apple, the original ancestral species is obscured by so many cultivated
variations throughout the centuries that some authors lump them all into one species, Malus
domestica. They all originated in western Asia (or Eurasia) and are characterized by fleshy fruits
called pomes. In the pome, a thick, fleshy hypanthium layer (also called the floral cup or calyx
tube) surrounds (and is fused with) the seed-bearing ovary or core. The sepals, petals and
stamens arise from the rim of the hypanthium. Since the ovary is situated below the attachment
of the sepals, petals and stamens, it is termed \"inferior\" in technical plant taxonomy books. The
fleshy hypanthium of a rose (Rosa) surrounds a cluster of small one-seeded achenes. Since the
achenes represent separate ripened ovaries all derived from a single flower, the entire structure
(called a rose hip) can be considered an aggregate fruit or etaerio. Rose hips are eaten raw and
are ground up as a supplemental source of vitamin C (ascorbic acid).
Bing cherries (Prunus avium) showing the long stalk (pedicel) and fleshy drupe containng a hard,
stony, seed-bearing endocarp. Sweet cherries such as these are usually considered to belong to P.
avium, while sour cherries belong to the P. cerasus group. There are literally hundreds of
varieties of cherries. Maraschino cherries are made from sweet cherries which have been
bleached, deseeded, and soaked in a sugar solution to which red food coloring and flavoring have
been added. Maraschino cherries are commonly covered with chocolate, placed as a decorative
topping in ice cream sundies, and in mixed drinks.
The raspberry or hindberry is the edible fruit of a multitude of plant species in the genus Rubus,
most of which are in the subgenusIdaeobatus; the name also applies to these plants themselves.
Raspberries are perennial, with woody stems.
Raspberries are grown for the fresh fruit market and for commercial processing into individually
quick frozen (IQF) fruit, purée, juice, or as dried fruit used in a variety of grocery products.
Traditionally, raspberries were a mid-summer crop, but with new technology, cultivars, and
transportation, they can now be obtained year-round. Raspberries need ample sun and water for
optimal development. Raspberries thrive in well-drained soil with a pH of between 6 and 7 with
ample organic matter to assist in retaining water. While moisture is essential, wet and heavy soils
or excess irrigation can bring on Phytophthora root rot which is one of the most serious pest
problems facing red raspberry. As a cultivated plant in moist temperate regions, it is easy to grow
and has a tendency to spread unless pruned. Escaped raspberries frequently appear as garden
weeds, spread by seeds found in bird droppings.
An indiv.
Polygon .java package chegg2;public class Polygon { Var.pdf
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------------
Enter Number of Sides:9
setSides() results: 9
Polygons
--------------------------------------------------
setSides() results: 3
setSides() results: 9
BUILD SUCCESSFUL (total time: 1 second)
Solution
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------.
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2).pdf
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134
Solution
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134.
OutputFibonacci till 20 0 1 1 2 3 5 8 13 21 .pdf
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster.
Solution
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster..
Matter is made up of electrically charge particle but the Constituen.pdf
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms
Solution
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms.
Recommended
Tungsten is a body centered cubic latticeSolutionTungsten is a.pdf
Tungsten is a body centered cubic lattice
Solution
Tungsten is a body centered cubic lattice.
This is the final code which meets your requirement.Than youCard.j.pdf
This is the final code which meets your requirement.Than you
Card.java
import java.util.Random;
public class Card {
//declaring instance variables
private String suit;
private int face_value;
public static int count=52;
//Creating an instance of Random class reference
Random rand;
//Default Constructor
public Card() {
super();
//Passing random class object to the Random class reference
rand = new Random();
}
//Getters and setters
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
public int getFace_value() {
return face_value;
}
public void setFace_value(int face_value) {
this.face_value = face_value;
}
public int getCount() {
return count;
}
//This shuffle() method will shuffle the deck
public void shuffle()
{
//Picking a card from the deck randomly and find its face value and suit
this.face_value = 1 + rand.nextInt((9) + 1);
int suit_random_value = 1 + rand.nextInt((3) + 1);
if(suit_random_value==1)
{
this.suit=\"Clubs\";
}
else if(suit_random_value==2)
{
this.suit=\"hearts\";
}
else if(suit_random_value==3)
{
this.suit=\"Spades\";
}
else if(suit_random_value==4)
{
this.suit=\"diamonds\";
}
this.count=count-1;
}
}
________________________________________________________________________
DriverClass.java
public class DriverClass {
public static void main(String[] args) {
//creating an array which can hold 52 Card class Objects
Card c[]=new Card[52];
//Creating 52 card class objects and storing them into an array
for(int i=0;i<52;i++)
{
//Creating Card class Object
c[i]=new Card();
}
//Displaying each card face value and its suit.
for(int i=0;i<52;i++)
{
//Shuffling the deck before every card picking from the deck
c[i].shuffle();
//Displaying randomly picked card face value and its suit
System.out.println(c[i].getFace_value()+\" of \"+c[i].getSuit()+\" Cards remaining
\"+c[i].getCount());
}
}
}
_________________________________________________________________________
Output:
8 of diamonds
Cards remaining 51
4 of Clubs
Cards remaining 50
9 of Clubs
Cards remaining 49
7 of Spades
Cards remaining 48
9 of diamonds
Cards remaining 47
7 of diamonds
Cards remaining 46
4 of diamonds
Cards remaining 45
9 of Clubs
Cards remaining 44
5 of hearts
Cards remaining 43
7 of Spades
Cards remaining 42
10 of Clubs
Cards remaining 41
7 of diamonds
Cards remaining 40
8 of diamonds
Cards remaining 39
6 of hearts
Cards remaining 38
4 of hearts
Cards remaining 37
6 of Spades
Cards remaining 36
4 of Clubs
Cards remaining 35
3 of diamonds
Cards remaining 34
9 of diamonds
Cards remaining 33
3 of diamonds
Cards remaining 32
5 of Spades
Cards remaining 31
4 of Clubs
Cards remaining 30
3 of Spades
Cards remaining 29
6 of diamonds
Cards remaining 28
3 of hearts
Cards remaining 27
9 of hearts
Cards remaining 26
7 of hearts
Cards remaining 25
3 of hearts
Cards remaining 24
10 of Clubs
Cards remaining 23
2 of hearts
Cards remaining 22
9 of Clubs
Cards remaining 21
8 of diamonds
Cards remaining 20
8 of Clubs
Cards remaining 19
2 of h.
Statistics is the mathematical science involving the collection, ana.pdf
Statistics is the mathematical science involving the collection, analysis and interpretation of data.
A number of specialties have evolved to apply statistical theory and methods to various
disciplines. Certain topics have \"statistical\" in their name but relate to manipulations of
probability distributions rather than to statistical analysis.
· Actuarial science is the discipline that applies mathematical and statistical methods to
assess risk in the insurance and financeindustries.
· Astrostatistics is the discipline that applies statistical analysis to the understanding of
astronomical data.
· Biostatistics is a branch of biology that studies biological phenomena and observations by
means of statistical analysis, and includes medical statistics.
· Business analytics is a rapidly developing business process that applies statistical methods
to data sets (often very large) to develop new insights and understanding of business
performance & opportunities
· Chemometrics is the science of relating measurements made on a chemical system or
process to the state of the system via application of mathematical or statistical methods.
· Demography is the statistical study of all populations. It can be a very general science that
can be applied to any kind of dynamic population, that is, one that changes over time or space.
· Econometrics is a branch of economics that applies statistical methods to the empirical
study of economic theories and relationships.
· Environmental statistics is the application of statistical methods to environmental science.
Weather, climate, air and water quality are included, as are studies of plant and animal
populations.
· Epidemiology is the study of factors affecting the health and illness of populations, and
serves as the foundation and logic of interventions made in the interest of public health and
preventive medicine.
· Geostatistics is a branch of geography that deals with the analysis of data from disciplines
such as petroleum geology, hydrogeology, hydrology, meteorology,oceanography, geochemistry,
geography.
· Machine Learning
· Operations research (or Operational Research) is an interdisciplinary branch of applied
mathematics and formal science that uses methods such as mathematical modeling, statistics, and
algorithms to arrive at optimal or near optimal solutions to complex problems.
· Population ecology is a sub-field of ecology that deals with the dynamics of species
populations and how these populations interact with the environment.
· Psychometric is the theory and technique of educational and psychological measurement
of knowledge, abilities, attitudes, and personality traits.
· Quality control reviews the factors involved in manufacturing and production; it can make
use of statistical sampling of product items to aid decisions in process control or in accepting
deliveries.
· Quantitative psychology is the science of statistically explaining and changing mental
processes and behaviors in humans.
· Reliabi.
Secondary phloemApples (Malus communis, M. pumila, & M. sylvestr.pdf
Secondary phloem
Apples (Malus communis, M. pumila, & M. sylvestris), pears (Pyrus communis) and quince
(Cydonia oblonga) belong to the rose family (Rosaceae), and include literally hundreds of
cultivated varieties. In the apple, the original ancestral species is obscured by so many cultivated
variations throughout the centuries that some authors lump them all into one species, Malus
domestica. They all originated in western Asia (or Eurasia) and are characterized by fleshy fruits
called pomes. In the pome, a thick, fleshy hypanthium layer (also called the floral cup or calyx
tube) surrounds (and is fused with) the seed-bearing ovary or core. The sepals, petals and
stamens arise from the rim of the hypanthium. Since the ovary is situated below the attachment
of the sepals, petals and stamens, it is termed \"inferior\" in technical plant taxonomy books. The
fleshy hypanthium of a rose (Rosa) surrounds a cluster of small one-seeded achenes. Since the
achenes represent separate ripened ovaries all derived from a single flower, the entire structure
(called a rose hip) can be considered an aggregate fruit or etaerio. Rose hips are eaten raw and
are ground up as a supplemental source of vitamin C (ascorbic acid).
Bing cherries (Prunus avium) showing the long stalk (pedicel) and fleshy drupe containng a hard,
stony, seed-bearing endocarp. Sweet cherries such as these are usually considered to belong to P.
avium, while sour cherries belong to the P. cerasus group. There are literally hundreds of
varieties of cherries. Maraschino cherries are made from sweet cherries which have been
bleached, deseeded, and soaked in a sugar solution to which red food coloring and flavoring have
been added. Maraschino cherries are commonly covered with chocolate, placed as a decorative
topping in ice cream sundies, and in mixed drinks.
The raspberry or hindberry is the edible fruit of a multitude of plant species in the genus Rubus,
most of which are in the subgenusIdaeobatus; the name also applies to these plants themselves.
Raspberries are perennial, with woody stems.
Raspberries are grown for the fresh fruit market and for commercial processing into individually
quick frozen (IQF) fruit, purée, juice, or as dried fruit used in a variety of grocery products.
Traditionally, raspberries were a mid-summer crop, but with new technology, cultivars, and
transportation, they can now be obtained year-round. Raspberries need ample sun and water for
optimal development. Raspberries thrive in well-drained soil with a pH of between 6 and 7 with
ample organic matter to assist in retaining water. While moisture is essential, wet and heavy soils
or excess irrigation can bring on Phytophthora root rot which is one of the most serious pest
problems facing red raspberry. As a cultivated plant in moist temperate regions, it is easy to grow
and has a tendency to spread unless pruned. Escaped raspberries frequently appear as garden
weeds, spread by seeds found in bird droppings.
An indiv.
Polygon .java package chegg2;public class Polygon { Var.pdf
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------------
Enter Number of Sides:9
setSides() results: 9
Polygons
--------------------------------------------------
setSides() results: 3
setSides() results: 9
BUILD SUCCESSFUL (total time: 1 second)
Solution
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------.
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2).pdf
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134
Solution
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134.
OutputFibonacci till 20 0 1 1 2 3 5 8 13 21 .pdf
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster.
Solution
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster..
Matter is made up of electrically charge particle but the Constituen.pdf
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms
Solution
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms.
it is a polynomial of degree 3Solutionit is a polynomial of de.pdf
it is a polynomial of degree 3
Solution
it is a polynomial of degree 3.
JUnit is a Regression Testing Framework used by developers to implem.pdf
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case.
Solution
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case..
Intitially take a graph sheet and plot the points and the whole figu.pdf
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6.
Solution
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6..
Incapsula Enterprise is the best mitigation service provider with th.pdf
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacities equip it with the size needed to
prevent large-scale volumetric attacks,27 data centers are connected which are located all over
the world, the service is prepared for even the most advanced DDoS attacks. In addition to
enormous network capacity and a high number of data centers, Incapsula hosts a variety of
mitigation techniques that intercept application layer, volumetric and protocol attacks.
With the recent acquisition of Defense.net, F5 has adjusted its service to offer cloud-based
mitigation technique, on-site and hybrid deployment methods,both hardware and the cloud is
used, F5 Silverline Protection is equipped for sophisticated DDoS attacks. This service surpasses
Incapsula in scrubbing capacity, it provides 2TB of bandwidth to filter, clean and remove
infected traffic. While the service does not offer unlimited mitigation, this protection service can
be used on-demand or always on, customizing it to fit the needs of any company.
Third one comes Arbor Cloud which offers a large network and scrubbing capacity and have
1.14TB of bandwidth, this server prepare the system for specialised volumetric attacks.variety of
protection techniques are provided by this service including IP blocking, rate limiting, automatic
bot discernment and more. While the cloud-based mitigation techniques are impressive, Arbor is
known for its hardware. The service is in a unique position, it\'s hardware is the challenge for it\'s
competitors
similar mitigation techniques is provided by all DDOS mitigation service providers to keep an
eye on any attacks and constant monitoring, they differ in price, deployment methods and
network size.It is not necessary that bigger mean better, the services that offer larger networks
and more mitigation technologies allow for more customization that you employ the best
protection service for your IT infrastructure.
Solution
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacit.
Go with the most obvious one first...Carboxylic Acids will have a.pdf
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A
Solution
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A.
Flexible benefit plans gives employees a choice between qualified be.pdf
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan.
Solution
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan..
EnvironmentLet assume the environment is a grid of colored tiles(.pdf
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<
Solution
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<.
Classification of organisms is based upon a number of physical and p.pdf
Classification of organisms is based upon a number of physical and physiological features. The
structure and organization of reproductive organs remains in important feature usually used to
distinguish organisms. Some of the features can be discussed as below:
1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body
is present or not. Presence of a fleshy or modified cover around the seed is a characteristic
feature of angiosperms whereas a naked seed located directly above a vegetative structure of a
plant is a feature of gymnosperms.
2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the
leaf blades of lower and higher ferns. Their arrangement and mode of release determines the
nature, frequency and ease of sexual reproduction in these lower plants.
3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for
sexual reproduction. The mode of transfer of pollen determines the geographical range of
pollination in plants. Sulphur shower is a critical example for sexual reproduction in pines
(gymnosperms) which distinguishes them from other trees.
4. Sporocarp: A sporocarp or a fruiting body is a bulging of the vegetative extension from the
main structure of a lower fungi which contains a group of cells which either follow meiotic
division or mitotic division and serve as the reproductive structures. Their growth and nature can
be monitored microscopically and the fungus can be thus classified appropriately. Some algae
also contain a unique localization and arrangement of haploid cells called spores which can be
present in singles or doublets and thus represent a very unique discrimination from other
organisms.
These, these set of information provide an insight into usability of features of a reproductive
structure for classification of organisms.
Solution
Classification of organisms is based upon a number of physical and physiological features. The
structure and organization of reproductive organs remains in important feature usually used to
distinguish organisms. Some of the features can be discussed as below:
1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body
is present or not. Presence of a fleshy or modified cover around the seed is a characteristic
feature of angiosperms whereas a naked seed located directly above a vegetative structure of a
plant is a feature of gymnosperms.
2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the
leaf blades of lower and higher ferns. Their arrangement and mode of release determines the
nature, frequency and ease of sexual reproduction in these lower plants.
3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for
sexual reproduction. The mode of transfer of pollen determines the geographical range of
pollination in plants. Sulphur shower is a critical example for.
Assets=Liabilities+Paid in capital+Retained earnings1..pdf
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Solution
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450.
Hypothesis Test for population variance If we have a sample fro.pdf
Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?
Solution
Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?.
ANSWERFundamental configurations of Windows printer Deployments.pdf
ANSWER:
Fundamental configurations of Windows printer Deployments:
The Following sections describe four fundamental configuraions that are the basis of most
Windows Printer Deployments
1.Direct Printing
2.Locally Attached printer Sharing
3.Network-attached Printing
4.Network-Attached printer sharing
you can scale these configurations up to accommodate a network of virtually any size.
1.DIRECT PRINTING:
The simplest print architecture consists of one print device connected to one computer, also
known as a locally attached print device, as shown in Figure 2-13. When you connect a print
device directly to a Windows Server 2012 R2 computer and print from an application running on
that system, the computer supplies the printer, printer driver, and print server functions.
2.LOCALLY ATTACHED PRINTER SHARING:
In addition to printing from an application running on that computer, you can also share the
printer (and the print device) with other users on the same network. In this arrangement, the
computer with the locally attached print device functions as a print server. Figure 2-14 shows the
other computers on the network, which are known as the print clients.
In the default Windows Server 2012 R2 printer-sharing configuration, each client uses its own
printer and printer driver. As before, the application running on the client computer sends the
print job to the printer and the printer driver renders the job, based on the capabilities of the print
device.
The main advantage of this printing arrangement is that multiple users, located anywhere on the
network, can send jobs to a single print device connected to a computer functioning as a print
server. The downside is that processing the print jobs for many users can impose a significant
burden on the print server. Although any Windows computer can function as a print server, you
should use a workstation for this purpose only when you have no more than a handful of print
clients to support or you have a very light printing volume.
3.NETWORK-ATTACHED PRINTING:
The printing solutions discussed thus far involve print devices connected directly to a computer
using a USB or other port. Print devices do not necessarily have to be attached to computers,
however. You can connect a print device directly to the network instead. Many print device
models are equipped with network interface adapters, enabling you to attach a standard network
cable. Some print devices have expansion slots into which you can install a network printing
adapter you have purchased separately. Finally, for print devices with no networking capabilities,
standalone network print servers are available, which connect to the network and enable you to
attach one or more print devices. Print devices so equipped have their own IP addresses and
typically have an embedded web-based configuration interface.
With network-attached print devices, the primary deployment decision the administrator must
make is to decide which computer will function as th.
Since HCl is strong acid , it dissociates fully t.pdf
Since HCl is strong acid , it dissociates fully to give H+ and Cl - so the
concentration of H+ = 2 M so p H = - log[H+] = - log (2) = - 0.301
Solution
Since HCl is strong acid , it dissociates fully to give H+ and Cl - so the
concentration of H+ = 2 M so p H = - log[H+] = - log (2) = - 0.301.
@author public class Person{ String sname, .pdf
/**
* @author
*
*/
public class Person
{
String sname, iname;
int fid, sid;
}
/**
* @author
*
*/
public class Student extends Person
{
String sname;
int sid;
int no_of_credits, tot_grade_pts;
/**
* @param sname
* @param sid
*/
public Student(String sname, int sid)
{
this.sname = sname;
this.sid = sid;
}
/**
* @return
*/
public String get_name()
{
return sname;
}
public int get_sid()
{
return sid;
}
public boolean two_equal(int sid)
{
if ((this.sid) == sid)
return true;
else
return false;
}
public void set_credits(int credits)
{
no_of_credits = credits;
}
public int get_credits()
{
return no_of_credits;
}
public void set_tot_gpts(int gpts)
{
tot_grade_pts = gpts;
}
public int get_tgpts()
{
return tot_grade_pts;
}
public float get_GPA()
{
return ((tot_grade_pts) / (no_of_credits));
}
}
/**
* @author
*
*/
public class Instructor extends Person
{
String iname;
int fid;
String dept;
/**
* @param iname
* @param fid
*/
public Instructor(String iname, int fid)
{
this.iname = iname;
this.fid = fid;
}
/**
* @param dept
*/
void set_dept(String dept)
{
this.dept = dept;
}
/**
* @return
*/
public String get_dept()
{
return dept;
}
}
import java.util.ArrayList;
/**
* @author
*
*/
public class Course
{
String cname, instructor_name;
int creg_code, max_students, no_of_students;
ArrayList reg_students = new ArrayList();
/**
* @param cname
* @param creg_code
* @param max_students
*/
Course(String cname, int creg_code, int max_students)
{
this.cname = cname;
this.creg_code = creg_code;
this.max_students = max_students;
}
/**
* @param instructor_name
*/
void set_instructor(String instructor_name)
{
this.instructor_name = instructor_name;
}
/**
* method to search student id
*
* @param sid
* @return
*/
public boolean search_student(int sid)
{
if (reg_students.contains(sid))
return true;
else
return false;
}
/**
* method to add student id
*
* @param sid
*/
public void add_student(int sid)
{
try {
if ((reg_students.size()) == max_students)
{
throw new MyException();
}
else
reg_students.add(sid);
}
catch (Exception e)
{
System.out.println(\"Course has maximum students\");
}
}
/**
* method to remove student id
*
* @param sid
*/
public void rem_student(int sid)
{
try {
if (reg_students.contains(sid))
reg_students.remove(new Integer(sid));
else
throw new MyException();
} catch (Exception e)
{
System.out.println(\"NO STUDENT FOUND WITH THE GIVEN id\");
}
}
}
/**
* @author
*
*/
public class MyException extends Exception
{
public MyException() {
// TODO Auto-generated constructor stub
super(\" Invalid Data\");
}
}
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Registrar
{
/**
* @param args
* @throws NumberFormatException
* @throws IOException
*/
public static void main(String args[]) throws NumberFormatException,
IOException
{
Course c1 = new Course(\"default_course\", 123, 10);
int ch, regcode, max_stu, sid;
String cname, sname;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
do {
System..
no even covalent coupunds also relese heat thos.pdf
no even covalent coupunds also relese heat those reaction are called combustion
reactions
Solution
no even covalent coupunds also relese heat those reaction are called combustion
reactions.
organic layer note this ammonium salt (itself is.pdf
organic layer note: this ammonium salt (itself is water soluble) reacts with NaOH to
form neutral amine compound that is an organic compound and thus more soluble in organic
solvent, Et2O.
Solution
organic layer note: this ammonium salt (itself is water soluble) reacts with NaOH to
form neutral amine compound that is an organic compound and thus more soluble in organic
solvent, Et2O..
Glucose is more sluble in water because it has mo.pdf
Glucose is more sluble in water because it has moreno . of -OH groups on it that
can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound
it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose.
Solution
Glucose is more sluble in water because it has moreno . of -OH groups on it that
can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound
it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose..
compounds 1 and 3 are the same be cause they are .pdf
compounds 1 and 3 are the same be cause they are both the same compounds
written differently
Solution
compounds 1 and 3 are the same be cause they are both the same compounds
written differently.
C) Its a double displacement reaction .pdf
C) It\'s a double displacement reaction
Solution
C) It\'s a double displacement reaction.
B)light interacting with a metal and producing a .pdf
B)light interacting with a metal and producing a current.
Solution
B)light interacting with a metal and producing a current..
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
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it is a polynomial of degree 3Solutionit is a polynomial of de.pdf
it is a polynomial of degree 3
Solution
it is a polynomial of degree 3.
JUnit is a Regression Testing Framework used by developers to implem.pdf
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case.
Solution
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case..
Intitially take a graph sheet and plot the points and the whole figu.pdf
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6.
Solution
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6..
Incapsula Enterprise is the best mitigation service provider with th.pdf
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacities equip it with the size needed to
prevent large-scale volumetric attacks,27 data centers are connected which are located all over
the world, the service is prepared for even the most advanced DDoS attacks. In addition to
enormous network capacity and a high number of data centers, Incapsula hosts a variety of
mitigation techniques that intercept application layer, volumetric and protocol attacks.
With the recent acquisition of Defense.net, F5 has adjusted its service to offer cloud-based
mitigation technique, on-site and hybrid deployment methods,both hardware and the cloud is
used, F5 Silverline Protection is equipped for sophisticated DDoS attacks. This service surpasses
Incapsula in scrubbing capacity, it provides 2TB of bandwidth to filter, clean and remove
infected traffic. While the service does not offer unlimited mitigation, this protection service can
be used on-demand or always on, customizing it to fit the needs of any company.
Third one comes Arbor Cloud which offers a large network and scrubbing capacity and have
1.14TB of bandwidth, this server prepare the system for specialised volumetric attacks.variety of
protection techniques are provided by this service including IP blocking, rate limiting, automatic
bot discernment and more. While the cloud-based mitigation techniques are impressive, Arbor is
known for its hardware. The service is in a unique position, it\'s hardware is the challenge for it\'s
competitors
similar mitigation techniques is provided by all DDOS mitigation service providers to keep an
eye on any attacks and constant monitoring, they differ in price, deployment methods and
network size.It is not necessary that bigger mean better, the services that offer larger networks
and more mitigation technologies allow for more customization that you employ the best
protection service for your IT infrastructure.
Solution
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacit.
Go with the most obvious one first...Carboxylic Acids will have a.pdf
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A
Solution
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A.
Flexible benefit plans gives employees a choice between qualified be.pdf
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan.
Solution
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan..
EnvironmentLet assume the environment is a grid of colored tiles(.pdf
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<
Solution
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<.
Classification of organisms is based upon a number of physical and p.pdf
Classification of organisms is based upon a number of physical and physiological features. The
structure and organization of reproductive organs remains in important feature usually used to
distinguish organisms. Some of the features can be discussed as below:
1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body
is present or not. Presence of a fleshy or modified cover around the seed is a characteristic
feature of angiosperms whereas a naked seed located directly above a vegetative structure of a
plant is a feature of gymnosperms.
2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the
leaf blades of lower and higher ferns. Their arrangement and mode of release determines the
nature, frequency and ease of sexual reproduction in these lower plants.
3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for
sexual reproduction. The mode of transfer of pollen determines the geographical range of
pollination in plants. Sulphur shower is a critical example for sexual reproduction in pines
(gymnosperms) which distinguishes them from other trees.
4. Sporocarp: A sporocarp or a fruiting body is a bulging of the vegetative extension from the
main structure of a lower fungi which contains a group of cells which either follow meiotic
division or mitotic division and serve as the reproductive structures. Their growth and nature can
be monitored microscopically and the fungus can be thus classified appropriately. Some algae
also contain a unique localization and arrangement of haploid cells called spores which can be
present in singles or doublets and thus represent a very unique discrimination from other
organisms.
These, these set of information provide an insight into usability of features of a reproductive
structure for classification of organisms.
Solution
Classification of organisms is based upon a number of physical and physiological features. The
structure and organization of reproductive organs remains in important feature usually used to
distinguish organisms. Some of the features can be discussed as below:
1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body
is present or not. Presence of a fleshy or modified cover around the seed is a characteristic
feature of angiosperms whereas a naked seed located directly above a vegetative structure of a
plant is a feature of gymnosperms.
2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the
leaf blades of lower and higher ferns. Their arrangement and mode of release determines the
nature, frequency and ease of sexual reproduction in these lower plants.
3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for
sexual reproduction. The mode of transfer of pollen determines the geographical range of
pollination in plants. Sulphur shower is a critical example for.
Assets=Liabilities+Paid in capital+Retained earnings1..pdf
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Solution
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450.
Hypothesis Test for population variance If we have a sample fro.pdf
Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?
Solution
Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?.
ANSWERFundamental configurations of Windows printer Deployments.pdf
ANSWER:
Fundamental configurations of Windows printer Deployments:
The Following sections describe four fundamental configuraions that are the basis of most
Windows Printer Deployments
1.Direct Printing
2.Locally Attached printer Sharing
3.Network-attached Printing
4.Network-Attached printer sharing
you can scale these configurations up to accommodate a network of virtually any size.
1.DIRECT PRINTING:
The simplest print architecture consists of one print device connected to one computer, also
known as a locally attached print device, as shown in Figure 2-13. When you connect a print
device directly to a Windows Server 2012 R2 computer and print from an application running on
that system, the computer supplies the printer, printer driver, and print server functions.
2.LOCALLY ATTACHED PRINTER SHARING:
In addition to printing from an application running on that computer, you can also share the
printer (and the print device) with other users on the same network. In this arrangement, the
computer with the locally attached print device functions as a print server. Figure 2-14 shows the
other computers on the network, which are known as the print clients.
In the default Windows Server 2012 R2 printer-sharing configuration, each client uses its own
printer and printer driver. As before, the application running on the client computer sends the
print job to the printer and the printer driver renders the job, based on the capabilities of the print
device.
The main advantage of this printing arrangement is that multiple users, located anywhere on the
network, can send jobs to a single print device connected to a computer functioning as a print
server. The downside is that processing the print jobs for many users can impose a significant
burden on the print server. Although any Windows computer can function as a print server, you
should use a workstation for this purpose only when you have no more than a handful of print
clients to support or you have a very light printing volume.
3.NETWORK-ATTACHED PRINTING:
The printing solutions discussed thus far involve print devices connected directly to a computer
using a USB or other port. Print devices do not necessarily have to be attached to computers,
however. You can connect a print device directly to the network instead. Many print device
models are equipped with network interface adapters, enabling you to attach a standard network
cable. Some print devices have expansion slots into which you can install a network printing
adapter you have purchased separately. Finally, for print devices with no networking capabilities,
standalone network print servers are available, which connect to the network and enable you to
attach one or more print devices. Print devices so equipped have their own IP addresses and
typically have an embedded web-based configuration interface.
With network-attached print devices, the primary deployment decision the administrator must
make is to decide which computer will function as th.
Since HCl is strong acid , it dissociates fully t.pdf
Since HCl is strong acid , it dissociates fully to give H+ and Cl - so the
concentration of H+ = 2 M so p H = - log[H+] = - log (2) = - 0.301
Solution
Since HCl is strong acid , it dissociates fully to give H+ and Cl - so the
concentration of H+ = 2 M so p H = - log[H+] = - log (2) = - 0.301.
@author public class Person{ String sname, .pdf
/**
* @author
*
*/
public class Person
{
String sname, iname;
int fid, sid;
}
/**
* @author
*
*/
public class Student extends Person
{
String sname;
int sid;
int no_of_credits, tot_grade_pts;
/**
* @param sname
* @param sid
*/
public Student(String sname, int sid)
{
this.sname = sname;
this.sid = sid;
}
/**
* @return
*/
public String get_name()
{
return sname;
}
public int get_sid()
{
return sid;
}
public boolean two_equal(int sid)
{
if ((this.sid) == sid)
return true;
else
return false;
}
public void set_credits(int credits)
{
no_of_credits = credits;
}
public int get_credits()
{
return no_of_credits;
}
public void set_tot_gpts(int gpts)
{
tot_grade_pts = gpts;
}
public int get_tgpts()
{
return tot_grade_pts;
}
public float get_GPA()
{
return ((tot_grade_pts) / (no_of_credits));
}
}
/**
* @author
*
*/
public class Instructor extends Person
{
String iname;
int fid;
String dept;
/**
* @param iname
* @param fid
*/
public Instructor(String iname, int fid)
{
this.iname = iname;
this.fid = fid;
}
/**
* @param dept
*/
void set_dept(String dept)
{
this.dept = dept;
}
/**
* @return
*/
public String get_dept()
{
return dept;
}
}
import java.util.ArrayList;
/**
* @author
*
*/
public class Course
{
String cname, instructor_name;
int creg_code, max_students, no_of_students;
ArrayList reg_students = new ArrayList();
/**
* @param cname
* @param creg_code
* @param max_students
*/
Course(String cname, int creg_code, int max_students)
{
this.cname = cname;
this.creg_code = creg_code;
this.max_students = max_students;
}
/**
* @param instructor_name
*/
void set_instructor(String instructor_name)
{
this.instructor_name = instructor_name;
}
/**
* method to search student id
*
* @param sid
* @return
*/
public boolean search_student(int sid)
{
if (reg_students.contains(sid))
return true;
else
return false;
}
/**
* method to add student id
*
* @param sid
*/
public void add_student(int sid)
{
try {
if ((reg_students.size()) == max_students)
{
throw new MyException();
}
else
reg_students.add(sid);
}
catch (Exception e)
{
System.out.println(\"Course has maximum students\");
}
}
/**
* method to remove student id
*
* @param sid
*/
public void rem_student(int sid)
{
try {
if (reg_students.contains(sid))
reg_students.remove(new Integer(sid));
else
throw new MyException();
} catch (Exception e)
{
System.out.println(\"NO STUDENT FOUND WITH THE GIVEN id\");
}
}
}
/**
* @author
*
*/
public class MyException extends Exception
{
public MyException() {
// TODO Auto-generated constructor stub
super(\" Invalid Data\");
}
}
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Registrar
{
/**
* @param args
* @throws NumberFormatException
* @throws IOException
*/
public static void main(String args[]) throws NumberFormatException,
IOException
{
Course c1 = new Course(\"default_course\", 123, 10);
int ch, regcode, max_stu, sid;
String cname, sname;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
do {
System..
no even covalent coupunds also relese heat thos.pdf
no even covalent coupunds also relese heat those reaction are called combustion
reactions
Solution
no even covalent coupunds also relese heat those reaction are called combustion
reactions.
organic layer note this ammonium salt (itself is.pdf
organic layer note: this ammonium salt (itself is water soluble) reacts with NaOH to
form neutral amine compound that is an organic compound and thus more soluble in organic
solvent, Et2O.
Solution
organic layer note: this ammonium salt (itself is water soluble) reacts with NaOH to
form neutral amine compound that is an organic compound and thus more soluble in organic
solvent, Et2O..
Glucose is more sluble in water because it has mo.pdf
Glucose is more sluble in water because it has moreno . of -OH groups on it that
can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound
it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose.
Solution
Glucose is more sluble in water because it has moreno . of -OH groups on it that
can hydrogen bond withwater molecules. C-OH-----------H-O-H But NaCl is an ionic compound
it can dissociates to Na+ & Cl-ions only, it will not form hydrogen bonding like glucose..
compounds 1 and 3 are the same be cause they are .pdf
compounds 1 and 3 are the same be cause they are both the same compounds
written differently
Solution
compounds 1 and 3 are the same be cause they are both the same compounds
written differently.
C) Its a double displacement reaction .pdf
C) It\'s a double displacement reaction
Solution
C) It\'s a double displacement reaction.
B)light interacting with a metal and producing a .pdf
B)light interacting with a metal and producing a current.
Solution
B)light interacting with a metal and producing a current..
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