Martian Cubics and Unit TestingYou must implement all of the data .pdf
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Martian Cubics and Unit Testing You must implement all of the data members and methods described below. RAM is more limited on NASA missions than on your home computer, so you may NOT add any instance variables or static variables to this class other than those described below. You may add methods of your own, as long as they are private. The data type class you are writing is a very general class that could be of use in a wide variety of projects at NASA in their Mars research, so take care in your work! Implement eval -- this method takes one parameter (DoubleWithAppx), evaluates the Martian cubic at the point represented by the parameter and returns a DoubleWithAppx representing the result of that evaluation. (i.e., if your Martian cubic is 5x^3-3x^2+2x+4, and you call eval(5), it should return 564.) My Code: public class MartianCubic { private final DoubleWithAppx a ; private final DoubleWithAppx b; private final DoubleWithAppx c; private final DoubleWithAppx d; public MartianCubic() { a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); d = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx dIn) { d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx cIn, DoubleWithAppx dIn) { c = cIn; d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.b = bIn; this.c = cIn; this.d = dIn; a = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx aIn, DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.a = aIn; this.b = bIn; this.c = cIn; this.d = dIn; } public MartianCubic(MartianCubic other) { this( other.getA(), other.getB(), other.getC(), other.getD() ); } public DoubleWithAppx getA() { return a; } public DoubleWithAppx getB() { return b; } public DoubleWithAppx getC() { return c; } public DoubleWithAppx getD() { return d; } public DoubleWithAppx eval(DoubleWithAppx x) { //HINT: Think about how to chain method calls to make this compact. } public boolean equals (Object other) { if (other == null) { return false; } else if (this.getClass()!=other.getClass()) { return false; } else { MartianCubic casted = (MartianCubic)other; return ( a.equals(casted.a) && b.equals(casted.b) && c.equals(casted.c) && d.equals(casted.d) ); } } Solution //solution package ex; public class MartianCubic { private final DoubleWithAppx a ; private final DoubleWithAppx b; private final DoubleWithAppx c; private final DoubleWithAppx d; public MartianCubic() { a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); d = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx dIn) { d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx cIn, DoubleWithAppx dIn) { c = cIn; d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); } public MartianCubic.
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Can you please help me the JAVA program? Let\'s write a program that will produce a file that
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type of csv (say MyGrades.csv, for example). Write the to String value for each SSRow to the
file. Finally, write a row \"average\".=(D1+D2+D3+D4+D5+D6+D7) Open MyGrades.csv
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Solution
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Scanner;
class SSRow{
String name;
int score;
int weight;
String cal;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getScore() {
return score;
}
public void setScore(int score) {
this.score = score;
}
public int getWeight() {
return weight;
}
public void setWeight(int weight) {
this.weight = weight;
}
public String getCal() {
return cal;
}
public void setCal(String cal) {
this.cal = cal;
}
@Override
public String toString() {
return \"\\\"\" + name + \"\\\", \\\"\" + score + \"\\\", \\\"\" + weight + \"\\\", \\\"\" + cal + \"\\\"\";
}
}
public class SSWriter {
// Maintain a static map b/w workType and weights
private static final Map WORK_TYPE_TO_WEIGHT_MAP = new LinkedHashMap<>();
static {
WORK_TYPE_TO_WEIGHT_MAP.put(\"Lab\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Reading Quizzes\", 10);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Group Work\", 10);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Midterm\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Project\", 15);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Final\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Instructor Discretion\", 5);
}
public static void main(String[] args) throws IOException {
Scanner scanner = new Scanner(System.in);
ArrayList list = new ArrayList<>(); // List to maintain the rows
int index = 1;
String avgString = \"\";
for(String workType : WORK_TYPE_.
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1
Part I: written exercises.
1. Using the Red-Black Tree applet, you input the following numbers into the Red-Black tree, keep all red-black rules ( see the red-black rule on the bottom of the exercise ) either using color changes or rotation. The result graph should be height balance and red-black correct.
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a. Insert 50, 70, 20, 90
b. Insert 50, 70, 20, 90, 99
c. Insert 50, 70, 20, 90, 80
d. Insert 50, 70, 20, 80, 60, 90, 75, 95
e. Insert 50, 70, 20, 90, 60, 55, 65, 68
Red-Black Rules
When inserting (or deleting) a new node, certain rules, which we call the red-black rules, must be followed. If they’re followed, the tree will be balanced. Let’s lookbriefly at these rules:
1. Every node is either red or black.
2. The root is always black.
3. If a node is red, its children must be black (although the converse isn’t necessarily true).
4. Every path from the root to a leaf, or to a null child, must contain the same
number of black nodes.
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a. Separate chaining hash table
b. Open addressing hash table using linear probing.
c. Open addressing hash table using quadratic probing.
d. Open addressing hash table with second hash function h2(x) = 7- (x %7)
3. Draw the figures to show steps how to convert the binary tree to a heap using the Heapify algorithm.
(
4
5
2
6
7
3
1
)
4. Using the diagrams (see the attached example of heap sort) to trace the action of heapsort on the list of 1, 7, 2, 6, 3, 5, 4, give every state of your tree such as finishing a heapify or removing the root.
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Implement the PriorityQ class in the priorityQ.java program (Listing 4.6) using
a heap instead of an array. You should be able to use the Heap class in the
heap.java program (Listing 12.1) without modification. Make it a descending
queue (largest item is removed).
Listing 4.6
The priorityQ.java Program
// priorityQ.java
// demonstrates priority queue
// to run this program: C>java PriorityQApp
////////////////////////////////////////////////////////////////
class PriorityQ
{
// array in sorted order, from max at 0 to min at size-1
private int maxSize;
private long[] queArray;
private int nItems;
//-------------------------------------------------------------
public PriorityQ(int s) // constructor
{
maxSize = s;
queArray = new long[maxSize];
nItems = 0;
}
//-------------------------------------------------------------
public void insert(long item) // insert item
{
int j;
if(nItems==0) // if no items,
queArray[nItems++] = item; // insert at 0
else // if items,
{
for(j=nItems-1; j>=0; j--) // start at end,
{
if( item > queArray[j] ) // if new item larger,
queArray[j+1] = queArray[ ...
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weights used for this course. As an example, we\'ll use the following table, but you should use
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and create the following SSRow objects: \"Lab\", : 20. \"=B1Times C1/100\" \"Reading
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file. Finally, write a row \"average\".=(D1+D2+D3+D4+D5+D6+D7) Open MyGrades.csv
with Excel or LibreOffice Calc.
Solution
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Scanner;
class SSRow{
String name;
int score;
int weight;
String cal;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getScore() {
return score;
}
public void setScore(int score) {
this.score = score;
}
public int getWeight() {
return weight;
}
public void setWeight(int weight) {
this.weight = weight;
}
public String getCal() {
return cal;
}
public void setCal(String cal) {
this.cal = cal;
}
@Override
public String toString() {
return \"\\\"\" + name + \"\\\", \\\"\" + score + \"\\\", \\\"\" + weight + \"\\\", \\\"\" + cal + \"\\\"\";
}
}
public class SSWriter {
// Maintain a static map b/w workType and weights
private static final Map WORK_TYPE_TO_WEIGHT_MAP = new LinkedHashMap<>();
static {
WORK_TYPE_TO_WEIGHT_MAP.put(\"Lab\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Reading Quizzes\", 10);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Group Work\", 10);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Midterm\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Project\", 15);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Final\", 20);
WORK_TYPE_TO_WEIGHT_MAP.put(\"Instructor Discretion\", 5);
}
public static void main(String[] args) throws IOException {
Scanner scanner = new Scanner(System.in);
ArrayList list = new ArrayList<>(); // List to maintain the rows
int index = 1;
String avgString = \"\";
for(String workType : WORK_TYPE_.
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1
Part I: written exercises.
1. Using the Red-Black Tree applet, you input the following numbers into the Red-Black tree, keep all red-black rules ( see the red-black rule on the bottom of the exercise ) either using color changes or rotation. The result graph should be height balance and red-black correct.
You need to follow the rules, randomly color change and rotation will result in zero credit even the result could be red-black correct.
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a. Insert 50, 70, 20, 90
b. Insert 50, 70, 20, 90, 99
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e. Insert 50, 70, 20, 90, 60, 55, 65, 68
Red-Black Rules
When inserting (or deleting) a new node, certain rules, which we call the red-black rules, must be followed. If they’re followed, the tree will be balanced. Let’s lookbriefly at these rules:
1. Every node is either red or black.
2. The root is always black.
3. If a node is red, its children must be black (although the converse isn’t necessarily true).
4. Every path from the root to a leaf, or to a null child, must contain the same
number of black nodes.
2. Given input {4371, 1323, 6173, 4199, 4344, 9679, 1989} and a hash function h(x) = x %10 (table size is 10), showing the resulting
a. Separate chaining hash table
b. Open addressing hash table using linear probing.
c. Open addressing hash table using quadratic probing.
d. Open addressing hash table with second hash function h2(x) = 7- (x %7)
3. Draw the figures to show steps how to convert the binary tree to a heap using the Heapify algorithm.
(
4
5
2
6
7
3
1
)
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Listing 4.6
The priorityQ.java Program
// priorityQ.java
// demonstrates priority queue
// to run this program: C>java PriorityQApp
////////////////////////////////////////////////////////////////
class PriorityQ
{
// array in sorted order, from max at 0 to min at size-1
private int maxSize;
private long[] queArray;
private int nItems;
//-------------------------------------------------------------
public PriorityQ(int s) // constructor
{
maxSize = s;
queArray = new long[maxSize];
nItems = 0;
}
//-------------------------------------------------------------
public void insert(long item) // insert item
{
int j;
if(nItems==0) // if no items,
queArray[nItems++] = item; // insert at 0
else // if items,
{
for(j=nItems-1; j>=0; j--) // start at end,
{
if( item > queArray[j] ) // if new item larger,
queArray[j+1] = queArray[ ...
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public int x;
public int y;
public Coordinate( int x, int y ) {
this.x = x;
this.y = y;
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}
}
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private char[][] maze;
private int height;
private int width;
/**
* Create a new Maze of the specified height and width, initializing every
* location as empty, with a \' \'.
**/
public Maze( int width, int height ) {
// ADD STATEMENTS HERE
}
/**
* Mutator to allow us to set the specified Coordinate as blocked,
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**/
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// ADD STATEMENTS HERE
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**/
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/**
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**/
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// ADD STATEMENTS HERE
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/**
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* the specified coordinate.
**/
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Vector vector = new Vector();
// ADD STATEMENTS HERE
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}
/**
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**/
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return new Coordinate( 0, 1 );
}
/**
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**/
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// ADD STATEMENTS HERE
}
/**
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* to show the Coordinates of each cell in the maze.
**/
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StringBuilder buffer =.
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Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
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}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
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{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}
Solution
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
@Test public void testXY()
{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}.
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Listings for BinaryHeap.java and BinaryHeapTest.java are shown in th.pdf
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Everything in the BinaryHeap class has been written except the add and remove methods, which
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testAdd and testRemove should both run without errors
Note:
A comp object implementing the Comparator interface is passed to the binary heap\'s
constructor. You will need to use the comp object\'s compare method when making comparisons
in the add and remove operations.
The parent, leftChild, and rightChild methods in BinaryHeapTest are for testing purposes only,
and should not be used to implement add and remove
Solution
BinaryHeapTest.java
import java.util.Comparator;
import org.junit.Test;
import static org.junit.Assert.*;
public class BinaryHeapTest {
/**
* A binary heap of Integer objects
*/
private BinaryHeap bh;
/**
* Some Integer nodes for testing
*/
private Integer n1, n2, n3, n4, n5, n6, n7, n8, n9, n10;
/**
* An object for comparing Integer nodes
*/
private Comparator comp;
public BinaryHeapTest() {
comp = new Comparator() {
public int compare(Integer t, Integer t1) {
return t.compareTo(t1);
}
};
bh = new BinaryHeap(10, comp);
n1 = new Integer(1);
n2 = new Integer(2);
n3 = new Integer(3);
n4 = new Integer(4);
n5 = new Integer(5);
n6 = new Integer(6);
n7 = new Integer(7);
n8 = new Integer(8);
n9 = new Integer(9);
n10 = new Integer(10);
}
@Test
public void testAdd() {
buildHeap();
assertTrue(leftChild(n1) == n2);
assertTrue(rightChild(n1) == n5);
assertTrue(leftChild(n2) == n4);
assertTrue(rightChild(n2) == n3);
assertTrue(leftChild(n5) == n9);
assertTrue(rightChild(n5) == n6);
assertTrue(leftChild(n4) == n10);
assertTrue(rightChild(n4) == n7);
assertTrue(leftChild(n3) == n8);
assertTrue(rightChild(n3) == null);
assertTrue(leftChild(n9) == null);
assertTrue(rightChild(n9) == null);
assertTrue(leftChild(n6) == null);
assertTrue(rightChild(n6) == null);
assertTrue(parent(n10) == n4);
assertTrue(parent(n7) == n4);
assertTrue(parent(n8) == n3);
assertTrue(parent(n4) == n2);
assertTrue(parent(n3) == n2);
assertTrue(parent(n9) == n5);
assertTrue(parent(n6) == n5);
assertTrue(parent(n2) == n1);
assertTrue(parent(n5) == n1);
assertTrue(parent(n1) == null);
}
/**
* An auxiliary method that builds a heap that should look like:
*
* 1
* / \\
* 2 5
* / \\ / \\
* 4 3 9 6
* / \\ /
* 10 7 8
*/
private void buildHeap() {
bh.clear();
assertTrue(bh.isEmpty());
bh.add(n10);
bh.add(n9);
bh.add(n8);
bh.add(n7);
bh.add(n6);
bh.add(n5);
bh.add(n4);
bh.add(n3);
bh.add(n2);
bh.add(n1);
assertTrue(bh.size() == 10);
}
@Test
public void testRemove() {
buildHeap();
assertTrue(bh.remove() == n1);
assertTrue(bh.remove() == n2);
assertTrue(bh.remove() == n3);
assertTrue(bh.remove() == n4);
assertTrue(bh.remove() == n5);
assertTrue(bh.remove() == n6);
assertTrue(bh.remove() == n7);
assertTrue(bh.remove() == n8);
assertTru.
How to Read the City Evaluating evidence to generate informed urban .pdf
How to Read the City? Evaluating evidence to generate informed urban perspectives. People
understand cities through recognizable elements that provide legibility and emotional security of
being in familiar settings. Please describe two recognizable elements that you associate with
Baltimore as a City. 3.
Solution
Answer:- We can read a city by evaluating health of people and development of city . Traffic of
road is also a element of read of city. Crime evaluation is also very important in read of city.
Evaluation means the systematic investigation of the merit , worth or significance of an object or
effort.
Evaluation of community health and development program is very necessary.
There are two major element for evaluate
(i) health
(ii) Development of infrastructure
Evaluation of health of people of Baltimore city is very necessary. Evaluation of development of
infrastructure of city is very necessary for city. By help of evaluation we can read condition of
city.
Pollution in city is main cause of health problem in city. Quality of water is very important for
evaluating health condition of people of city.
By development of infrastructure we can improve condition of city. Development is very
necessary for Baltimore city..
How do transpiration, cohesion and adhesion all work together to pro.pdf
How do transpiration, cohesion and adhesion all work together to provide the forces necessary to
overcome gravity to transport water from the roots to the leaves without expending energy?
Include definitions of the three terms as they relate to the transport of water up the plant.
HINT: Begin with transpiration (define); then explain how cohesion (define) and adhesion
(define) work together to get water up the plant. Don\'t be too long
Solution
Transpiration is the process of water movement through a plant and evaporates from aerial parts
- leaves, stems and flowers. Water is necessary for plants but only a small amount of water is
used for growth and metabolism, above 97% of water is lost by transpiration and guttation.
Transpiration occurs through the stomata cells during photosynthesis. Transpiration also cools
plants, changes osmotic pressure of cells, and enables mass flow of mineral nutrients and water
from roots to shoots.
Cohesion: It is the phenomenon attraction between the similar molecules. The water molecules
remain attracted by cohesive forces and cannot be separated from one another.
Adhesion: The ability of a substance to stick to an unlike molecules (attraction between unlike
molecules)
Cohesion Tension theory is a theory of ascent of sap. It was proposed by Dixon and Joly in
1894. The theory describes the movement of water from roots to the leaves of a plant. Water
molecules are bonded to each other by hydrogen bonding, hence water form a string (linear) of
molecules during its movement toward xylem.
The water is filled inside the xylem capillaries and due to the cohesion and adhesion properties
of water it forms a continuous water column from roots to leaves.
Cohesive and adhesive properties of water molecules to form an unbroken continuous water
column in the xylem. Transpiration pull exerted on this water column. The water column could
not be pulled away from the xylem walls because of cohesion-adhesion properties of water..
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public class Coordinate {
public int x;
public int y;
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public class Maze {
private char[][] maze;
private int height;
private int width;
/**
* Create a new Maze of the specified height and width, initializing every
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**/
public Maze( int width, int height ) {
// ADD STATEMENTS HERE
}
/**
* Mutator to allow us to set the specified Coordinate as blocked,
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**/
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// ADD STATEMENTS HERE
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// ADD STATEMENTS HERE
}
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// ADD STATEMENTS HERE
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StringBuilder buffer =.
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NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
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Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
Hi,I have modified the Point.java file as per your requirement.P.pdf
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
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Point p = new Point();
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Solution
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
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{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}.
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Listings for BinaryHeap.java and BinaryHeapTest.java are shown in th.pdf
Listings for BinaryHeap.java and BinaryHeapTest.java are shown in the menu at left
Everything in the BinaryHeap class has been written except the add and remove methods, which
you must write:
First, implement add, then run the test class and observe the testAdd results
Next, implement remove, then run the test class and observe the testRemove results
testAdd and testRemove should both run without errors
Note:
A comp object implementing the Comparator interface is passed to the binary heap\'s
constructor. You will need to use the comp object\'s compare method when making comparisons
in the add and remove operations.
The parent, leftChild, and rightChild methods in BinaryHeapTest are for testing purposes only,
and should not be used to implement add and remove
Solution
BinaryHeapTest.java
import java.util.Comparator;
import org.junit.Test;
import static org.junit.Assert.*;
public class BinaryHeapTest {
/**
* A binary heap of Integer objects
*/
private BinaryHeap bh;
/**
* Some Integer nodes for testing
*/
private Integer n1, n2, n3, n4, n5, n6, n7, n8, n9, n10;
/**
* An object for comparing Integer nodes
*/
private Comparator comp;
public BinaryHeapTest() {
comp = new Comparator() {
public int compare(Integer t, Integer t1) {
return t.compareTo(t1);
}
};
bh = new BinaryHeap(10, comp);
n1 = new Integer(1);
n2 = new Integer(2);
n3 = new Integer(3);
n4 = new Integer(4);
n5 = new Integer(5);
n6 = new Integer(6);
n7 = new Integer(7);
n8 = new Integer(8);
n9 = new Integer(9);
n10 = new Integer(10);
}
@Test
public void testAdd() {
buildHeap();
assertTrue(leftChild(n1) == n2);
assertTrue(rightChild(n1) == n5);
assertTrue(leftChild(n2) == n4);
assertTrue(rightChild(n2) == n3);
assertTrue(leftChild(n5) == n9);
assertTrue(rightChild(n5) == n6);
assertTrue(leftChild(n4) == n10);
assertTrue(rightChild(n4) == n7);
assertTrue(leftChild(n3) == n8);
assertTrue(rightChild(n3) == null);
assertTrue(leftChild(n9) == null);
assertTrue(rightChild(n9) == null);
assertTrue(leftChild(n6) == null);
assertTrue(rightChild(n6) == null);
assertTrue(parent(n10) == n4);
assertTrue(parent(n7) == n4);
assertTrue(parent(n8) == n3);
assertTrue(parent(n4) == n2);
assertTrue(parent(n3) == n2);
assertTrue(parent(n9) == n5);
assertTrue(parent(n6) == n5);
assertTrue(parent(n2) == n1);
assertTrue(parent(n5) == n1);
assertTrue(parent(n1) == null);
}
/**
* An auxiliary method that builds a heap that should look like:
*
* 1
* / \\
* 2 5
* / \\ / \\
* 4 3 9 6
* / \\ /
* 10 7 8
*/
private void buildHeap() {
bh.clear();
assertTrue(bh.isEmpty());
bh.add(n10);
bh.add(n9);
bh.add(n8);
bh.add(n7);
bh.add(n6);
bh.add(n5);
bh.add(n4);
bh.add(n3);
bh.add(n2);
bh.add(n1);
assertTrue(bh.size() == 10);
}
@Test
public void testRemove() {
buildHeap();
assertTrue(bh.remove() == n1);
assertTrue(bh.remove() == n2);
assertTrue(bh.remove() == n3);
assertTrue(bh.remove() == n4);
assertTrue(bh.remove() == n5);
assertTrue(bh.remove() == n6);
assertTrue(bh.remove() == n7);
assertTrue(bh.remove() == n8);
assertTru.
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How to Read the City Evaluating evidence to generate informed urban .pdf
How to Read the City? Evaluating evidence to generate informed urban perspectives. People
understand cities through recognizable elements that provide legibility and emotional security of
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Solution
Answer:- We can read a city by evaluating health of people and development of city . Traffic of
road is also a element of read of city. Crime evaluation is also very important in read of city.
Evaluation means the systematic investigation of the merit , worth or significance of an object or
effort.
Evaluation of community health and development program is very necessary.
There are two major element for evaluate
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Evaluation of health of people of Baltimore city is very necessary. Evaluation of development of
infrastructure of city is very necessary for city. By help of evaluation we can read condition of
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evaluating health condition of people of city.
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How do transpiration, cohesion and adhesion all work together to pro.pdf
How do transpiration, cohesion and adhesion all work together to provide the forces necessary to
overcome gravity to transport water from the roots to the leaves without expending energy?
Include definitions of the three terms as they relate to the transport of water up the plant.
HINT: Begin with transpiration (define); then explain how cohesion (define) and adhesion
(define) work together to get water up the plant. Don\'t be too long
Solution
Transpiration is the process of water movement through a plant and evaporates from aerial parts
- leaves, stems and flowers. Water is necessary for plants but only a small amount of water is
used for growth and metabolism, above 97% of water is lost by transpiration and guttation.
Transpiration occurs through the stomata cells during photosynthesis. Transpiration also cools
plants, changes osmotic pressure of cells, and enables mass flow of mineral nutrients and water
from roots to shoots.
Cohesion: It is the phenomenon attraction between the similar molecules. The water molecules
remain attracted by cohesive forces and cannot be separated from one another.
Adhesion: The ability of a substance to stick to an unlike molecules (attraction between unlike
molecules)
Cohesion Tension theory is a theory of ascent of sap. It was proposed by Dixon and Joly in
1894. The theory describes the movement of water from roots to the leaves of a plant. Water
molecules are bonded to each other by hydrogen bonding, hence water form a string (linear) of
molecules during its movement toward xylem.
The water is filled inside the xylem capillaries and due to the cohesion and adhesion properties
of water it forms a continuous water column from roots to leaves.
Cohesive and adhesive properties of water molecules to form an unbroken continuous water
column in the xylem. Transpiration pull exerted on this water column. The water column could
not be pulled away from the xylem walls because of cohesion-adhesion properties of water..
How can the Internet be used for competitive advantage in business .pdf
How can the Internet be used for competitive advantage in business? Describe three firms which
are using the Internet for conducting business, and speculate as to the underlying technologies
which might be in use in these firms. Will the Internet become an essential business tool like the
telephone or will it go the way of the dinosaurs?
Solution
a) Usage of Internet for Competitive Advantage in Business
At its most fundamental level, competitive advantage is a measure of the value that a firm
is able to create and transfer to its customers. Essentially, a firm can create value for its
customers in one of three ways:
1.By providing their products and/or services at a lower price than its competitors;
2. By differentiating its products and/or services from competitors through the
provision of unique benefits which offset a premium price;
3. Through a combination of both a low price and differentiation strategy which is
focused on a particular market or segment .
Information and networking technologies are creating a new marketing paradigm, this new
paradigm is described as \'one-to-one marketing\', where greater emphasis is being placed on
individuality and customisation.
The Internet, as an interactive medium, offers the capability to support the emerging business
ethos more completely than traditional communication media. Empirical surveys show that the
majority of businesses using the Internet have a passive presence, suggesting that strategic
business use of the World Wide Web is still in its infancy.
Information technology will render rivalry between competitors more intense, yet also create
additional opportunities to develop competitive advantage such as ,
i) the creation of new business interrelationships within the value system,
ii) the expansion of industry scope, and
iii) the increased ability to co-ordinate value activities regionally, nationally, and globally.
Porter and Miller suggest that IT will create competitive advantage by providing
companies with new ways to outperform their competitors.
IT can have a powerful effect on the cost structure of a business, can increase opportunities for
differentiation, and can alter a firm\'s competitive scope.
Examining the impact of the Internet in relation to this proposition, there are a number of
competitive advantages to be realised at each stage of the value system. These include,
i) process inputs from suppliers (upstream value),
ii) internal operations (value chain), and
iii) customer relations (downstream value).
b) Firms that use Internet
Travel Industry:
Travel and Tourism is one of the highest revenue-generating sectors of the internet, The rapid
growth of the internet is having an impact on the distribution of travel services and has
heightened speculation about the potential for disintermediation of the travel agent.
Banking Sector:
Nearly 57 percent of the Indian commercial banks are providing transactional Internet banking
services. The univariate analysis indicates that Int.
give an example of an qualitative variablegive an example of.pdf
give an example of an qualitative variable
give an example of an qualitative variable
give an example of an qualitative variable
Solution
A qualitative variable does not contain numbers, it only desribes in terms of \"qualities\" . They
are also called categorical variable, and have no natural sense of ordering.
Examples: First Name of a person, hair color, zodiac sign etc.
Dynamic biological process influence population density, dispersion a.pdf
Dynamic biological process influence population density, dispersion and demographics
Distinguish between density and dispersion of population. Describe conditions that may result in
clumped dispersion, uniform dispersion, and random dispersion of individuals in a population.
What is a life table and why is it a useful tool for ecologists interested in population ecology?
Distinguish between a life table and a reproductive table. What does a reproductive table tell us?
Describe the characteristics of populations that exhibit Type I, Type II, and Type III survivorship
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environment and the elastic model describes how a population grows more slowly as it nears its
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What expectations do we have about exponential growth in natural populations? Under what
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populations Life history traits are products of natural selection What are life history traits?
What defines an organisms life history? Define and distinguish between semelparity and
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Solution
Density of a population means the number of individuals present in one unit area. Dispersion
means in what manner they are dispersed or spread out in that area whether randomly distributed,
or in groups or evenly dispersed. In clumped dispersion,animals make groups to move and live
together. Often when there is shortage and scarcity of resources in summers, animals tend to
make groups to share resources.These groups can be safeguards for juveniles and can protect
them. Preying skills of few animals can benefit rest of the group in getting the food. In Uniform
dispersion all the animals are evenly distributed on the area and mostly independent to prey and
fight for their own food. They find mate according to their own choice and compatibility and
protect themselves alone. In random dispersion, the population is distributed randomly in the
area where there is mix of all three dispersion patterns. There is enough food in the viscinity,
animals can roam and search freely their food and also can form group. Life table is a table that
has data about death and birth rates of a population under study, that can give ecologists
prediction about probable growth or shrink of a population. It gives them the pattern of .
Discuss some disadvantages of viable counts of microorganisms.So.pdf
Discuss some disadvantages of viable counts of microorganisms.
Solution
Disadvantages of viable counts of microorganisms:-
1. The assumption that each colony is arsing from single cell- Staphylococcus is an example in
this regard, which form clumps of microorganisms where each clump is counted as one colony.
2. Variation in the rate at which bacteria give rise to an observable colony- In case of a short
incubation period, some colonies may be skipped and hence, the optomization of temperature
and medium condition is required.
3. Difficult in distinguishing living and non living cells- In case of a direct microscopic count,
this becomes an issue.
4. High sensitivity- Sensitivity in case of a viable count is a disadvantage if plating conditions
are optimal..
Describe the pathophysiology, clinical manifestations, diagnostic te.pdf
Describe the pathophysiology, clinical manifestations, diagnostic tests and nursing management
for patients with abnormal cell growth of the head, neck and chest.
Solution
The abnormal cell growth of the head, neck and chest may be due to neuroblastoma or the head
and neck cancer. The treatment and manifestations depend on type and location of growing cells,
tumor size, how far it has spread, and if the tumor cells secrete hormones.
PATHOPHYSIOLOGY
Cancers of the head and neck are recognized by using the area wherein they begin: the oral
cavity, salivary glands, paranasal sinuses and nasal cavity, pharynx, larynx, and lymph nodes
within the neck. Tumors inside the neck can often be visible or felt as a hard, painless lump. The
main reasons of head and neck cancers include tobacco (smoked or chewed), and alcohol. other
threat factors are solar exposure, human papillomavirus (HPV) infection, radiation to the top and
neck, ancestry, Epstein-Barr virus infection, wooden dust or airborne asbestos, intake of certain
preservatives or salted meals, terrible oral hygiene, and Plummer-Vinson syndrome.
CLINICAL MENIFESTATION
If the tumor is in the chest, it would press at the superior vena cava. This will motive swelling
inside the face, neck, hands, and upper chest and occasionally with a bluish-crimson skin shade.
it can additionally cause headaches, dizziness, and a alternate in awareness if it influences the
mind. The tumor may press at the throat or windpipe, which could purpose coughing and
problem respiration or swallowing. Neuroblastomas that press on nerves within the chest or neck
can sometimes purpose other symptoms, together with a drooping eyelid and a small pupil.
Stress on different nerves near the backbone might have an effect on the potential to experience
or move their arms or legs. Symptoms of numerous head and neck cancer sites include a lump or
sore that does not heal, a sore throat that does not leave, issue swallowing, and a change or
hoarseness inside the voice. Neuroblastomas sometimes launch hormones that may reason
problems with tissues and organs, those troubles are known as paraneoplastic syndromes which
consist of steady diarrhea, Fever, excessive blood pressure causing irritability, speedy heartbeat,
Reddening of the pores and skin, Sweating etc.
DIAGNOSIS:
Tests conducted to diagnose head and neck most cancers range relying on the symptoms and can
include a biopsy, endoscopy, blood or urine checks, X-rays, CT scans, MRIs, and pet scans.
• Physical examination may encompass visible inspection of the oral and nasal cavities, neck,
throat, and tongue the usage of a small mirror and/or lights. The physician may feel for lumps on
the neck, lips, gums, and cheeks.
• Endoscopy examines regions within the body. The kind of endoscope the doctor uses depends
at the vicinity being examined.
• Laboratory exams test the samples of blood, urine, or other substances from the body.
If the analysis is cancer, the doctor will need to learn.
Could you please do it in R program 1. Assume that we draw n data v.pdf
Could you please do it in R program? 1. Assume that we draw n data values independently from
an exponential distribution with rate parameter , ie. (1, n ~ Exp(A). The exponential distribution
is a positive and right skewed distribution function (see the Wikipedia entry here and the R help
output here) and the probability distribution function is given by It is well known that the mean
of the exponential distribution is EX- and the variance is Var(X] 2 . The median of the
exponential distribution is given by log(2) Now a statistician is considering using two statistics to
estimate the mean of the distribution . The two statistics are: . The sample mean T1 = X = M 72
. The sample median divided by log(2), i.e. T2-X/log(2) The statistician knows that because Xi,...
,Xn are independent and identically dis- nVar(X tributed (iid.), Var(T) = (a) Using the central
limit theorem, what is the asymptotic distribution of V n(-1)? (b) Using the lecture notes, what is
the asymptotic distribution of Vn(X - lo2)? (c) Using the property that Var(cT-cyar(T) for any
statistic T and scalar c, along with the answer in part a, derive the formula for the asymptotic
variance Var(T2) (d) If we were to compare the two statistics, what is the asymptotic relative
efficiency ar(T (e) Based upon part c, which statistic is the most efficient statistic to estimate 1?
Broadly speaking, discuss the advantages and disadvantages of both statistics? (f) Generate 1000
Exp(1) random variables using
Solution
X1,X2,…..,Xn is a random sample from Exp() distribution.
E(Xi)=1/ for i=1(1)n
V(Xi)=1/2 for i=1(1)n
(a)
Given T1= Xbar = (X1+X2+…..+Xn)/n
Then E(T1 ) =1/ and V(T1) = 1/n2
Using Lindeberg Levy Central limit theorem (clt) we get
(T1-1/)/(1/sqrt(n)) ~ N(0,1) asymptotically
=> sqrt(n)(T1-1/ ) ~ N(0,1) asymptotically
=> sqrt(n)( T1-1/ ) ~ N(0,1/2) asymptotically
(b)
The asymptotic distribution of sample median med(X) is normal distribution with mean =
population median and variance = 1/4n{f(pop med)}2
The population median is given as
med = log(2)/
Then f(log(2)/) = e-*log(2)/ = e-log(2) = /2
Then var=1/(4n*2/4) = 1/n2
Then med(X) ~ N(log(2)/,1/n 2) asymptotically.
=> med(X) – log(2)/ ~ N(0,1/n 2) asymptotically
=> sqrt(n){med(X)-log(2)/ ) ~ N(0, 1/2) asymptotically
(c)
V(T2) = V(med(X)/log(2))
= V(med(X))/{log(2)}2
= 1/n 2 {log(2)}2 asymptotically
(d)
ARE(T1,T2) = V(T1)/V(T2)
= {1/n2}/{1/n2(log(2))2}
= {log(2)}2
(e)
Note that E(T1)=1/ for all >0
But E(T2)= 1/ asymptotically
Also note that
ARE(T1,T2) = V(T1)/V(T2) < 1
=> V(T1) < V(T2) asymptotically
Though both T1 and T2 are unbiased for asymptotically, asymptotic variance of T1 is lesser
than that of T2 implying that T1 is better estimator of 1/ than T2 .
(f)
Using the following codes in R we get
Code:
x<-rexp(1000,1);summary(x)
The output is
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.001265 0.271700 0.746000 1.087000 1.509000 6.689000
Here we observe that sample mean is closer to the original mean value (1 in this case) than
median. .
Class diagrams are used to define the dynamic behavior of an executi.pdf
Class diagrams are used to define the dynamic behavior of an executing system. Provide
rationale.
Solution
System development focuses on three overall different system models.
1.Functional:
These are Use Case diagrams, which describe system functionality from the point of view of
the user.
2.Object:
These are Class Diagrams, which describe the structure of the system in terms of objects,
attributes,
associations, and operations.
3.Dynamic:
Interaction Diagrams, State Machine Diagrams, and Activity Diagrams are used to describe
the internal
behavior of the system.
Rational:
-->Identifiable object-oriented modeling languages began to appear between mid-1970 and the
late 1980s
as various methodologists experimented with different approaches to object-oriented analysis
and design.
-->Software engineers and other experts needed a simplified and consistent way to model
systems,
but new languages were cropping up at a rapid pace.
-->UML itself was developed in late 1994, when the two creators of popular methodologies
teamed up.
-->Grady Booch and Jim Rumbaugh of Rational Software Corporation worked to unify the
Booch and OMT
(Object Modeling Technique) methods, and in 1995, Ivar Jacobson joined their endeavor.
-->Their combined method was known as OOSE Method 1—Object-Oriented Software
Engineering.
-->‘The Three Amigos’ of software engineering as they were known, had evolved other
methodologies.
They teamed up to provide clarity for programmers by creating new standards.
-->The collaboration between Grady, Booch, and Rumbaugh made all three methods stronger
and improved the final product..
Can someone please help me with my ICD 10 II assignment...A 47-yea.pdf
Can someone please help me with my ICD 10 II assignment...
A 47-year-old obese male was admitted with chest pain and difficulty swallowing, resulting in
regurgitation. The patient has a history of coronary artery disease (CAD), but tests ruled out
angina. Based on negative test results and the patient’s history, it was established that the chest
pain was probably related to a heavy meal.
Final diagnoses: (1) Probable gastroesophageal reflux disease (GERD), (2) CAD, (3) Obesity
Assign the appropriate codes.
Solution
It would be (1) Probable GERD-Gastroesophageal reflux disease.
Reason-
-Angina has been ruled out as cause of chest pain after taking into consideration of history of
CAD.
-Since patient is having difficulty swallowing and results in regurgitation.
-Obesity is not the sole reason in such cases always.
-Only best probabalistics cause is GERD.
CAD is a critical technology for an integrated product development p.pdf
CAD is a critical technology for an integrated product development process in a CIM
organization. t or f
Solution
yes its true.
cim ia the higher degree of the automation
it is requries to store the large and critical integrity of data to mrun and mentain the machines
these data wil store by using CAD and other aided techniqes like CAM, CAE and etc..
Byzantine influence can be seen in Russian literature because the Ru.pdf
Byzantine influence can be seen in Russian literature because the Russians allowed only
literature from the Byzantine Empire into the country. Educated Russians read all the Greek
classics. The Russians translated Greek and religious texts into Russian. Russians following the
Eastern Orthodox religion read only religious texts.
Solution
Option (A)
The Byzantine Empire saw its greatest success in the region of Southern Russia. It was only the
byzantine that used Cyril-made an alphabet so that the Russians could read the Bible. Before that
the general population of Russia could not connect to the church ceremonies and rituals but the
change made it easy for the general people to understand the literature..
balance sheet Google Search MC Qu. 122 An example of an operating... .pdf
balance sheet Google Search MC Qu. 122 An example of an operating... An example of an
operating activity is: Multiple Cholce Paying off a loan Borrowing money from a bank Seling
stock Purchasing office equipment 80
Solution
Selling Stock is a operating activity because it is a business function.All other activities are not
operating activities..
Write a 1-2 page essay researching biological science and its famous.pdf
Write a 1-2 page essay researching biological science and its famous female scientists and the
issues they encounter in this area of science including the following:
What is your understanding of biological science? What kind of research goes on in this area of
science and is it as male dominant as other areas of science?
Explore the critical issues faced by female scientists in this area of study and has the situation
changed in the last 50 years?
Solution
Biological Science: it is also termed as Biology, it is study of living organisms such as plants,
animals. Biological science aim to understand at molecular level biochemical level to
organismal and population levels. It include branches like Biochemistry, cell biology,
computational Biology, Biophysics, Development biology, evolution, ecology, microbiology,
Molecular biology etc.,
Beginning in the 1960, when the women movement in 20th century was nearing its peak, woman
compaigned for and finally won equal right in education and employment . Instead of few
woman training to be teacher, nurse, female student in increase in number were now choosing
career in physics, chemistry, maths
Famous female scientists: Some greatest scientist of all the time were females who have made
important discoveries in a variety fields in science. Several of their contributions throughout
history have even surpassed those their male counterparts. List of famous female scientists below
are organized in order of popularity as given below:
1. Marie Curie (1867-1934)
Famous for: Work on radioactivity
Marie Curie was the first woman to won Noble prize and the only woman win this award in two
categories: Physics and Chemistry. She discovered polonium and radium and her work help
creation of X-rays
2. Rita Levi-Montalcini (1909-2012)
Famous for : Nerve growth studies
Rita levi-montalcini was neurobiologist who won noble prize in physiology in 1986 for her
findings in Nerve Growth Factor. She was first Noble laureate to live past her 100th birthday.
3. Lise Meitner ( 1878-1968)
Famous for : Work on Nuclear physics
Lise Meitner was a key member of discoveries of nuclear fission.
4. Maria Mitchell (1818 - 1889)
Famous for : Work on astronomy
She was the very first american woman to become professional astronomer. She discovered a
comet in 1847 that was named as Miss Mitchell Comet.
5. Elizabeth Blackburn (1948)
Famous for : Telomere (present at end of chromosomes)
Elizabeth Black won the noble prize in 2009 for her discovery of telomerase which is enzyme
which replenishes telomere
6. Gertrude B. Elion ( 1918-1919)
Famous for : Development of new drugs
She won joint winner of noble prize in physiology and medicine in 1988 , one of her notable
creations was the development of AIDS drug AZT.
7.Rosalind Franklin (1920-1958)
Famous for : Research on DNA, RNA, graphite
She was a X-ray crystallographer and biophysictist whose work greatly contributed to revolved
x-ray diffraction images of DNA. Her worked result in finding DNA.
Why do plants in early stage exhibit high levels of tolerance Early .pdf
Why do plants in early stage exhibit high levels of tolerance? Early plants cannot inhibit the
growth of plants. Early deal with harsh conditions, such as fluctuating temperatures plants a and
thus must other plants. plants have small and light seeds that germinate well under shade. Late
succession species are likely to have speared seeds. large seeds. many seeds. fast growth rate A
mechanism of succession in which the presence of one species increases the probability that a
second species can become established is facilitation tolerance. inhibition priority effect
Inhibition is a mechanism of succession in which the probability that a species can become
established depends on the presence of another the arrival of one species at a site affects the
subsequent colonization of other one species decreases the probability that a second species will
become established. one species increases the probability that a second species will become
established. Garlic mustard Alliaria petiolate is a common invasive species in the eastern United
States. One mechanism that allows it to dominate in some areas is the production of chemicals
that interfere with the ability of other plants to gain nutrients through mutualistic fungi. This
mechanism is an example of tolerance. inhibition. the priority effect. facilitation.
Solution
14. plants in early seral stage show high leel of tolerance because early plant must deal with
harsh conditions as fluctuating temperature. it may be a effect of global warming. as temparature
change can cause molecular changes in the cell.
15. late succession species are also known as climax species. they show fast growth rate. on the
result of ecological homoeostasis they show maximum permitted biodiversity.
16. a process in succession where the presence of one species increases the probablity of second
species is known as faciliation. it can be of two types mutulism where both the species are
benefited, commensalism where one is benifited and other is uneffected.
17. inhibition is a mechanism of succession in which one species decreases the probablity that
the other species will establish. in this mechanism one plant growth prevents the growth of
others..
Which researcher(s) paralyzed subjects using Scoline in a classical .pdf
Which researcher(s) paralyzed subjects using Scoline in a classical conditioning study that
predated modern ethical guidelines for research?
Bramel
Campbell, Sanderson, and Laverty
Donnerstein, Linz, and Penrod
Milgram
Solution
Campbell, Sanderson, and Laverty
paralyzed subjects using Scoline in a classical conditioning study that predated modern ethical
guidelines for research.
Which of the following statements is NOT accurate about Mycobacteriu.pdf
Which of the following statements is NOT accurate about Mycobacterium tuberculosis?
QUESTION 1 Mycobacterium tuberculosis Which of the following statements is NOT accurate
about M. tuberculosis contains mycolic acid in it\'s cell wall. M. tuberculosis is very slow
growing and may take 2-6 weeks to culture visible colonies. M. tuberculosis generally colonizes
the lungs but may spread to other organs M. tuberculosis is the only bacteria known to stain by
the Acid Fast Staining method. QUESTION 2 Which of the following risk factors is NOT
responsible for causing a latent TB infection to become active TB disease? RapidM. tuberculosis
replication and multiplication in the anaerobic granuloma Aging HIV co-infection Poor living
conditions such as malnutrition, bad air quality, and lack of health care
Solution
Mycobacteriym tuberculosis is the causative agent for tuberculosis.
Answer for Question 1
M. tuberculosis is the only bacteria known to stain by Acid Fast Staining method.
Answer for Question 2
Rapid M. tuberculosis replication and multiplication in the anaerobic granuloma
Answer for Question 3:
M. tuberculosis can remain latent and undetected in the Macrophage..
Which of the following is likely to increase the electrical conductiv.pdf
Which of the following is likely to increase the electrical conductivity of Si? Increasing the
temperature Doping with an n-type impurity atom Doping with a p-type impurity atom All of
the above
Solution
answer is d)
a beacause as temperature increases electrons get activated even bad gap is more so electrical
conductivity increase
b because doping with n-type electrons is more in number due which electrical conductivity
increases
c because doping with p-type holes participate in electric current passage.
What is the difference between endogenous and exogenous infections .pdf
What is the difference between endogenous and exogenous infections? What is a protein
allergen?
Solution
Endogenous infections are defined as those infections that are caused by microorganisms that
are already in the body whereas exogenous infections are defined as those which are caused by
microorganisms from the external environment. Usually in endogenous infection, microbes are
already present in the body but in dormant state but are stimulated by some factors, tuberculosis
may be taken as a case. However, in exogenous infection microorganism gain their entry through
respiratory, gastrointestinal, vaginal, urinal or through injury. In case of exogenous infection,
pathogenic bacteria entering the body always cause disease. However, in case of endogenous
infection, some bacteria normally residing in the gut usually don’t cause any infection; however,
if the same bacterium enters the other sterile organ may cause infection there. As an instance, the
residential bacterium E. coli of the GI tract enters the urinary tract. This causes a urinary tract
infection. Sometimes the same type of infection occurs through both the exogenous and
endogenous ways. For example, the exogenous eczema occurs by an external irritant while as
endogenous eczema of which one type is infant eczema occurs by a stimulant already present
within the body. Examples of exogenous infections include Salmonella enterocolitis, Cholera,
Anthrax, Botulism etc. and examples of endogenous infections include tuberculosis etc..
What is a scattergramscatterdotscatterplot (different names for sa.pdf
What is a scattergram/scatterdot/scatterplot (different names for same plot)? What statistic is it
commonly used with?
Solution
What is a scattergram/scatterdot/scatterplot (different names for same plot)? What statistic is it
commonly used with?
A scatterplot is a graphic tool used to display the relationship between two quantitative variables.
A scatterplot is a useful plot of two quantitative variables, usually drawn before working out a
linear correlation coefficient or fitting a regression line.
Correlation coefficient is the statistic , it is commonly used with..
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- 1. Martian Cubics and Unit Testing You must implement all of the data members and methods described below. RAM is more limited on NASA missions than on your home computer, so you may NOT add any instance variables or static variables to this class other than those described below. You may add methods of your own, as long as they are private. The data type class you are writing is a very general class that could be of use in a wide variety of projects at NASA in their Mars research, so take care in your work! Implement eval -- this method takes one parameter (DoubleWithAppx), evaluates the Martian cubic at the point represented by the parameter and returns a DoubleWithAppx representing the result of that evaluation. (i.e., if your Martian cubic is 5x^3-3x^2+2x+4, and you call eval(5), it should return 564.) My Code: public class MartianCubic { private final DoubleWithAppx a ; private final DoubleWithAppx b; private final DoubleWithAppx c; private final DoubleWithAppx d; public MartianCubic() { a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); d = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx dIn) { d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx cIn, DoubleWithAppx dIn) { c = cIn; d = dIn;
- 2. a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.b = bIn; this.c = cIn; this.d = dIn; a = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx aIn, DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.a = aIn; this.b = bIn; this.c = cIn; this.d = dIn; } public MartianCubic(MartianCubic other) { this( other.getA(), other.getB(), other.getC(), other.getD() ); } public DoubleWithAppx getA() { return a; } public DoubleWithAppx getB() { return b; }
- 3. public DoubleWithAppx getC() { return c; } public DoubleWithAppx getD() { return d; } public DoubleWithAppx eval(DoubleWithAppx x) { //HINT: Think about how to chain method calls to make this compact. } public boolean equals (Object other) { if (other == null) { return false; } else if (this.getClass()!=other.getClass()) { return false; } else { MartianCubic casted = (MartianCubic)other; return ( a.equals(casted.a) && b.equals(casted.b) && c.equals(casted.c) && d.equals(casted.d) ); } } Solution
- 4. //solution package ex; public class MartianCubic { private final DoubleWithAppx a ; private final DoubleWithAppx b; private final DoubleWithAppx c; private final DoubleWithAppx d; public MartianCubic() { a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); d = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx dIn) { d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); c = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx cIn, DoubleWithAppx dIn) { c = cIn; d = dIn; a = new DoubleWithAppx(0); b = new DoubleWithAppx(0); } public MartianCubic(DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.b = bIn; this.c = cIn; this.d = dIn; a = new DoubleWithAppx(0); }
- 5. public MartianCubic(DoubleWithAppx aIn, DoubleWithAppx bIn, DoubleWithAppx cIn, DoubleWithAppx dIn) { this.a = aIn; this.b = bIn; this.c = cIn; this.d = dIn; } public MartianCubic(MartianCubic other) { this( other.getA(), other.getB(), other.getC(), other.getD() ); } public DoubleWithAppx getA() { return a; } public DoubleWithAppx getB() { return b; } public DoubleWithAppx getC() { return c; } public DoubleWithAppx getD() { return d; }
- 6. public DoubleWithAppx eval(DoubleWithAppx x) { //HINT: Think about how to chain method calls to make this compact. double ans; ans= getA().getValue()* Math.pow(x.getValue(),3)- getB().getValue()*Math.pow(x.getValue(),2)+getC().getValue()*x.getValue()+getD().getValue() ; return new DoubleWithAppx((int) ans); } public boolean equals (Object other) { if (other == null) { return false; } else if (this.getClass()!=other.getClass()) { return false; } else { MartianCubic casted = (MartianCubic)other; return ( a.equals(casted.a) && b.equals(casted.b) && c.equals(casted.c) && d.equals(casted.d) ); } } }