3. Write a project named Money that prompts for and reads a double value representing a monetary amount. Then determine the fewest number of each bill and coin needed to represent that amount, starting with the highest (assume that a ten-dollar bill is the maximum size need). For example, if the value entered is 47.63 (forty-seven dollars and sixty-three cents), then the program should print the equivalent amount as:

Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.

Explanation:

Answer:

The program in csharp for the given scenario is shown.

using System;

class ScoreGrade {

static void Main() {

//variables to store score and corresponding grade

double score;

char grade;

//user input taken for score

Console.Write("Enter the score: ");

score = Convert.ToInt32(Console.ReadLine());

//grade decided based on score

if(score>=90)

grade='A';

else if(score>=80 && score<90)

grade='B';

else if(score>=70 && score<80)

grade='C';

else if(score>=60 && score<70)

grade='D';

else

grade='F';

//score and grade displayed  

Console.WriteLine("Score: "+score+ " Grade: "+grade);

}

}

OUTPUT1

Enter the score: 76

Score: 76 Grade: C

OUTPUT2

Enter the score: 56

Score: 56 Grade: F

Explanation:

1. The variables to hold the score and grade are declared as double and char respectively.

int score;

char grade;

2. The user is prompted to enter the score. The user input is not validated and the input is stored in the variable, score.

3. Using if-else-if statements, the grade is decided based on the value of the score.

if(score>=90)

grade='A';

else if(score>=80 && score<90)

grade='B';

else if(score>=70 && score<80)

grade='C';

else if(score>=60 && score<70)

grade='D';

else

grade='F';

4. The score and the corresponding grade are displayed.

5. The program can be tested for different values of score.

6. The output for two different scores and two grades is included.

7. The program is saved using ScoreGrade.cs. The .cs extension indicates a csharp program.

8. The whole code is written inside a class since csharp is a purely object-oriented language.

9. In csharp, user input is taken using Console.ReadLine() which reads a string.

10. This string is converted into an integer using Convert.ToInt32() method.

score = Convert.ToInt32(Console.ReadLine());

11. The output is displayed using Console.WriteLine() or Console.Write() methods; the first method inserts a line after displaying the message which is not done in the second method.

12. Since the variables are declared inside Main(), they are not declared static.

13. If the variables are declared outside Main() and at the class level, it is mandatory to declare them with keyword, static.

Here is the complete question.

In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

start input testScore,

classRank if testScore >= 90 then if classRank >= 25 then output "Accept"

else output "Reject" endif else if testScore >= 80

then if classRank >= 50 then output "Accept" else output "Reject" endif

else if testScore >= 70

then if classRank >= 75 then output "Accept"

else output "Reject"

endif else output "Reject"

endif

endif

endif

stop

Study the pseudocode in picture above. Write the interactive input statements to retrieve: A student’s test score (testScore) A student's class rank (classRank) The rest of the program is written for you. Execute the program by clicking "Run Code." Enter 87 for the test score and 60 for the class rank. Execute the program by entering 60 for the test score and 87 for the class rank.

[comment]: <> (3. Write the statements to convert the string representation of a student’s test score and class rank to the integer data type (testScore and classRank, respectively).)

Function: This program determines if a student will be admitted or rejected. Input: Interactive Output: Accept or Reject

*/ #include using namespace std; int main() { // Declare variables

// Prompt for and get user input

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl; }

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl; }

else cout << "Reject" << endl; } } } //End of main() function

Answer:

Explanation:

The objective here is to use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

PROGRAM:

#include<iostream>

using namespace std;

int main(){

// Declare variables

int testScore, classRank;

// Prompt for and get user input

cout<<"Enter test score: ";

cin>>testScore;

cout<<"Enter class rank: ";

cin>>classRank;

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)  

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl;

}

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl;

}

else cout << "Reject" << endl; }

}

return 0;

} //End of main() function

OUTPUT:

See the attached file below:

Answer:

Check the explanation

Explanation:

SDLC

SDLC stands for Software Development Life Cycle

It provides steps for developing a software product

It also checks for the quality and correctness of a software

The main aim of SDLC is to develop a software, which meets a customer requirements

SDLC involves various phases to develop a high-quality product for its end user

Phases of SDLC

           There are seven phases in a software development life cycle as follows,

Requirement Analysis   Feasibility Study   Design    Coding    Testing Deployment      Maintenance

kindly check the attached image below

Requirement analysis

This is the first stage in a SDLC.

In this phase, a detailed and precise requirements are collected from various teams by senior team members

It gives a clear idea of the entire project scope, opportunities and upcoming issues in a project

This phase also involves, planning assurance requirements and analyze risk involved in the project

Here, the timeline to finish the project is finalized

Feasibility Study

In this stage, the SRS document, that is “Software Requirement Specification” document is defined, which includes everything to be designed and developed during the life cycle of a project

Design

In this phase as the name indicates, software and system design documents are prepared, based on the requirement specification document

This enable us to define the whole system architecture

Two types of design documents are prepared in this phase as follows,

High-Level Design

Low-Level Design

Coding

In this phase, the coding of project is done in a selected programming language

Tasks are divided into many units and given to various developers

It is the longest phase in the SDLC

Testing

In this stage, the developed coding is tested by the testing team

The testing team, checks the coding in the user point of view

Installation/Deployment

In this stage, a product is released by the company to the client

Maintenance

Once the product is delivered to the client, the maintenance phase work will start by fixing errors, modifying and enhancing the product.

Employee Payroll System

Requirement analysis

Gather information from various team for requirement

Analyze the number of employees to handle the project

Timeline to finish the project, eg.1 month

Feasibility Study

The system requirements such as how many systems are required

Decide which software to be installed to handle the project.

For example, if we are going to develop the software using C language, then software for that has to be installed

The SRS document has to be prepared

Design

In this the HLD and LLD is prepared, the overall software architecture is prepared

The modules involved in coding are decided

Here the modules can be employee, payroll manager, Employee Log time, account details

The input and output are designed, such as employee name is the input and output is the salary for him, based on working hours

Coding

Here the coding is divided into various sections and given to 2 or more employees

One may code employee detail, one will code working hours of employees and one may code the banking details of employee

Testing

The coding is tested for syntax, declaration, data types and various kinds of error

The testing team will test the coding with various possible values

They may even check with wrong values and analyze the output for that

Installation

Now the software is installed in the client system and the client is calculating the payroll for an employee based on working hours in a month

Maintenance

If any error occurs, the team will clear the issue.

When a new employee joins, then the employee data will added to the database and the database is updated

Also if the client asks for any new features, it will done in this phase.

Answer:

Bits

Explanation:

Answer:

There are no limitation, policies or regulations that limit these tools for use on privately owned machines or home networks. In most businesses networks, intranets or internets the use of them is illegal if used for malicious intent. Penetration testing teams sign rules of engagement before using these tools.

Explanation:

There are no limitation, limitations, policies, or regulations in their use on local machines, privately owned machines or home networks, or small business networks.

In most businesses networks, intranets or internets the use of them is often and mostly illegal especially in a situation where they are been used for malicious intent.

Penetration testing teams would often or always make sure that they sign rules of engagement before using these tools.

Answer:

Pay raise is an example of the rewards & consequences characteristic of an organization.

Answer: (a). 11.3137

(b). 22.849

Explanation:

Provided below is a step by step analysis to solving this problem

(a)

clc;close all;clear all;

a=2;x=3.5;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('2^3.5 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

2^3.5 from tailor series=11.3137 after adding n=15 terms

disp('2^3.5 using calculator =11.3137085');

Command window:

2^3.5 from tailor series=11.3137 after adding n=15 terms

2^3.5 using calculator =11.3137085

(b)

clc;close all;clear all;

a=6.3;x=1.7;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('6.3^1.7 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

disp('6.3^1.7 using calculator =22.84961748');

Command window:

6.3^1.7 from tailor series=22.8496 after adding n=16 terms

6.3^1.7 using calculator =22.84961748

cheers i hope this helped !!!

Answer:

public class RectangleArea {

   public static void main(String[] args) {

       Rectangle myRectangle = new Rectangle(15.8, 7.9);

       double calArea = myRectangle.calcArea(15.8, 7.9);

       System.out.println("The area is "+ calArea);

   }

}

class Rectangle{

   private double length;

   private double width;

   //Defined Constructor

   public Rectangle(double length, double width) {

       this.length = length;

       this.width = width;

   }

   public double calcArea(double len, double wi){

       return len* wi;

   }

}

Explanation:

Answer:

See explaination

Explanation:

/* Heap.h */

#ifndef HEAP_H

#define HEAP_H

#include<iostream>

using namespace std;

template<class T>

class Heap

{

public:

//constructor

Heap()

{

arr = nullptr;

size = maxsize = 0;

}

//parameterized constructor

Heap(const T elements[], int arraySize)

{

maxsize = arraySize;

size = 0;

arr = new T[maxsize];

for(int i=0; i<maxsize; i++)

{

add(elements[i]);

}

}

//Remove the root from the heap and maintain the heap property

T remove() //code is altered

{

T item = arr[0];

arr[0] = arr[size-1];

size--;

T x = arr[0];

int parent = 0;

while(1)

{

int child = 2*parent + 1;

if(child>=size) break;

if(child+1 < size &&arr[child+1]>arr[child])

child++;

if(x>=arr[child]) break;

arr[parent] = arr[child];

parent = child;

}

arr[parent] = x;

return item;

}

//Insert element into the heap and maintain the heap property

void add(const T&element)

{

if(size==0)

{

arr[size] = element;

size++;

return;

}

T x = element;

int child = size;

int parent = (child-1)/2;

while(child>0 && x>arr[parent])

{

arr[child] = arr[parent];

child = parent;

parent = (child-1)/2;

}

arr[child] = x;

size++;

}

//Get the number of element in the heap

int getSize() const

{

return size;

}

private:

T *arr; //code is altered

int size, maxsize;

};

#endif // HEAP_H

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

/* main.cpp */

#include<iostream>

#include "Heap.h"

using namespace std;

template <typename T>

void heapSort(T list[], int arraySize)

{

Heap<T> heap(list, arraySize);

int n = heap.getSize();

for(int i=n-1; i>0; i--)

{

list[i] = heap.remove();

}

}

int main()

{

const int SIZE = 9;

int list[] = {1, 7, 3, 4, 9, 11, 3, 1, 2};

heapSort<int>(list, SIZE);

cout << "Sorted elements using heap: \n";

for (int i = 0; i < SIZE; i++)

cout << list[i] << " ";

cout <<endl;

system("pause");

return 0;

}

Answer:

You should name it honest and sincere intention

Explanation:

That is a good name for what you are going to talk about in your slideshow.

Answer:

All the classes are mentioned with code and screenshots. . That class is shown at last.

Explanation:

Solution

Class.Java:

Code

**

* atauthor your_name

* This class denotes Class at the college.

*

*/

public class Class {

  private String className,classID;

  private Professor professor;

  /**

  * atparam className

  * atparam classID

  * atparam professor

  */

  public Class(String className, String classID, Professor professor) {

      this.classID=classID;

      this.className=className;

      this.professor=professor;

  }

     /**

  * at return classID of the class

  */

  public String getClassID() {

      return classID;

  }

    /**

  * Override toString() from Object Class

  */

  public String toString() {    

      return classID+" "+className+" "+professor.getName()+" "+professor.getEducation();        

  }  

}

Person.Java:

Code:

/**

* atauthor your_name

* This class represents Person

*

*/

public class Person {    

  protected String name;  

  /**method to fetch name

  * at return

  */

  public String getName() {

      return name;

  }

  /**method to set name

  * at param name

  */

  public void setName(String name) {

      this.name = name;

  }

Professor.java:

Code:

import java.util.ArrayList;

import java.util.List;  

/**

* at author your_name

*

*This class represents professors

*

*/

public class Professor extends Person{    

  private String professorID, education;

  private List<Class> classes=new ArrayList<Class>();    

  /**

  * at param name

  * at param professorID

  * at param education

  */

  public Professor(String name,String professorID,String education) {        

      this.name=name;

      this.professorID=professorID;

      this.education=education;        

  }

  /**

  * at return

  */

  public String getEducation() {

      return this.education;

  }    

  /**

  * at return

  */

  public String getprofessorID() {

      return this.professorID;

  }  

  /** to add classes

  * at param Class

  */

  public void addClass(Class c) {

      classes.add(c);

  }  

  /**

  * Override toString() from Object Class

  */

  public String toString() {

      String result=this.getName()+" - "+professorID+" - "+education;

      for(Class c:classes) {

          result+=c.toString()+"\n";

      }      

      return result;

  }

}

}

Student.java:

Code:

import java.util.ArrayList;

import java.util.List;  

/**

* This class represents students

*  at author your_Name

*

*/

public class Student extends Person{    

  private String studentID;

  private List<Class> classes=new ArrayList<Class>();    

  /**

  * atparam name

  * atparam studentID

  */

  public Student(String name,String studentID) {      

      this.name=name;

      this.studentID=studentID;      

  }  

  /**

  * atreturn

  */

  public String getStudentID() {

      return studentID;

  }

     /**

  * atparam c

  */

  public void addClass(Class c) {

      classes.add(c);

  }    

  /**

  * atreturn

  */

  public int getClassCount() {

      return classes.size();

  }

  /**

  * Override toString() from Object Class

  */

  public String toString() {

      String result=this.getName()+" - "+studentID+"\n";

      for(Class c:classes) {

          result+=c.toString()+"\n";

      }    

      return result;

  }  

}

NOTE: Kindly find an attached copy of screenshot of the output, which is a part of the solution to this question

Answer:

See explaination

Explanation:

#include <cstring>

#include <cstdio>

#include <string>

#include <array>

#include <random>

#include <algorithm>

struct RandomGenerator {

RandomGenerator(const size_t min, const size_t max) : dist(min, max) {}

std::random_device rd;

std::uniform_int_distribution<size_t> dist;

unsigned operator()() { return dist(rd); }

};

struct Gallow {

void Draw() const

{

std::printf(" ________\n"

"| |\n"

"| %c %c\n"

"| %c%c%c\n"

"| %c %c\n"

"|\n"

"|\n", body[0], body[1], body[2], body[3],

body[4], body[5], body[6]);

}

bool Increment()

{

switch (++errors) {

case 6: body[6] = '\\'; break;

case 5: body[5] = '/'; break;

case 4: body[4] = '\\'; break;

case 3: body[3] = '|'; break;

case 2: body[2] = '/'; break;

case 1: body[0] = '(', body[1] = ')'; break;

}

return errors < 6;

}

char body[7] { '\0' };

int errors { 0 };

};

struct Game {

void Draw() const

{

#ifdef _WIN32

std::system("cls");

#else

std::system("clear");

#endif

gallow.Draw();

std::for_each(guess.begin(), guess.end(), [](const char c) { std::printf("%c ", c); });

std::putchar('\n');

}

bool Update()

{

std::printf("Enter a letter: ");

const char letter = std::tolower(std::getchar());

while (std::getchar() != '\n') {}

bool found = false;

for (size_t i = 0; i < word.size(); ++i) {

if (word[i] == letter) {

guess[i] = letter;

found = true;

}

}

const auto end_game = [this](const char* msg) {

Draw();

std::puts(msg);

return false;

};

if (not found and not gallow.Increment())

return end_game("#### you lose! ####");

else if (found and word == guess)

return end_game("#### you win! ####");

return true;

}

RandomGenerator rand_gen { 0, words.size() - 1 };

const std::string word { words[rand_gen()] };

std::string guess { std::string().insert(0, word.size(), '_') };

Gallow gallow;

static const std::array<const std::string, 3> words;

};

const std::array<const std::string, 3> Game::words{{"control", "television", "computer"}};

int main()

{

Game game;

do {

game.Draw();

} while (game.Update());

return EXIT_SUCCESS;

}

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

Check the explanation

Explanation:

Let the population mean time on task of original website be [tex]\mu_1[/tex] and that of new prototype be [tex]\mu_2[/tex]

Here we are to test

[tex]H_0:\mu_1=\mu_2\;\;against\;\;H_1:\mu_1<\mu_2[/tex]

The data are summarized as follows:

                                                       Sample 1             Sample 2

Sample size                                        n=20                    n=20

Sample mean                                 [tex]\bar{x}_1=243.4[/tex]             [tex]\bar{x}_2=253.4[/tex]

Sample SD                                     s1=130.8901          s2=124.8199

The test statistic is obtained from online calculator as

t=-0.23

The p-value is given by 0.821

As the p-value is more than 0.05, we fail to reject the null hypothesis at 5% level of significance and hence conclude that there is no statistically significant difference between the original website and the prototype developed by the team at 5% level of significance.

Answer:

A. 0.0450

B. 4

C. 0.25

D. 37.68

E. 6Hz

F. -0.523

G. 1.5m/s

H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

I. -1.67m/s.

Explanation:

Given the equation:

y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)

Standard wave equation:

y(x, t)=Asin(kx−ωt+ϕ)

a.) Amplitude = 0.0450

b.) Wave number = 1/ λ

λ=2π/k

From the equation k = 25.12

Wavelength(λ ) = 2π/25.12 = 0.25

Wave number (1/0.25) = 4

c.) Wavelength(λ ) = 2π/25.12 = 0.25

d.) Angular frequency(ω)

ωt = 37.68t

ω = 37.68

E.) Frequency (f)

ω = 2πf

f = ω/2π

f = 37.68/6.28

f = 6Hz

f.) Phase angle(ϕ) = -0.523

g.) Wave propagation speed :

ω/k=37.68/25.12=1.5m/s

h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.

Answer:

See explaination

Explanation:

//ReadFile.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class ReadFile

{

public static void main(String[] args)

{

int elementsCount=0;

//Create a File class object

File file=null;

Scanner fileScanner=null;

String fileName="sample.txt";

try

{

//Create an instance of File class

file=new File(fileName);

//create a scanner class object

fileScanner=new Scanner(file);

//read file elements until end of file

while (fileScanner.hasNext())

{

double value=fileScanner.nextInt();

elementsCount++;

}

//print smallest value

System.out.println(elementsCount+" data values are read");

}

//Catch the exception if file not found

catch (FileNotFoundException e)

{

System.out.println("File Not Found");

}

}

}

Sample output:

sample.txt

10

20

30

40

50

output:

5 data values are read

Answer:

The formula used in this question is called the probability of combinations or combination formula.

Explanation:

Solution

Given that:

Formula applied is stated as follows:

nCr = no of ways to choose r objects from n objects

  = n!/(r!*(n-r)!)

The Data given :

Menu A : 5 appetizers and 3 main dishes

Menu B : 3 appetizers and 4 main dishes

Total appetizers - 6

Total main dishes - 5

Now,

Part A :

Total ways = No of ways to select menu A + no of ways to select menu B

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = 6C5*5C3 + 6C3*5C4

          = 6*10 + 20*5

          = 160

Part B :

Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)

Total ways = No of ways to select menu A + no of ways to select menu B  

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = (6^5)*(5^3) + (6^3)*(5^4)

          = 7776*125 + 216*625

          = 1107000

Part C :

No of ways to select same appetizers and main dish for all the options

= No of ways to select menu A + no of ways to select menu B  

= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

=(6*5) + (6*5)

= 60

Total ways = Part B - (same appetizers and main dish selected)      

= 1107000 - 60

= 1106940

=======================================================

//Class header definition

public class DisArray {

   //First method with int array as parameter

   public static void convert2D(int[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the

       //length of  the one dimensional array,

       //then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       // number of rows

       int col = ((row * row) >= arraylength) ? row : row + 1;

       //3. Create a 2D int array with the number of rows and cols

       int [ ][ ] twoD = new int [row][col];

       //4. Place the elements in the one dimensional array into

       //the two dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       // two  dimensional array.

       for (int i = 0; i < row; i++) {

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

              }

              //Otherwise, just pad the array with zeros

               else {

                   twoD[i][j] = 0;

               }

           }

       }

       //You might want to create another pair of loop to print the elements

       //in the  populated two dimensional array as follows

       for (int i = 0; i < twoD.length; i++) {

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }     //End of first method

   //Second method with String array as parameter

   public static void convert2D(String[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the length of  

       //the one dimensional array, then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       //number of rows.

       int col = (row * row >= arraylength) ? row : row + 1;

       //3. Create a 2D String array with the number of rows and cols

       String[][] twoD = new String[row][col];

       //4. Place the elements in the one dimensional array into the two

       // dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       //two  dimensional array.

       for (int i = 0; i < row; i++) {

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

               }

               //Otherwise, just pad the array with null values

               else {

                   twoD[i][j] = null;

               }

           }

       }

       //You might want to create another pair of loop to print the elements  

       //in the populated two dimensional array as follows:

       for (int i = 0; i < twoD.length; i++) {

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }    // End of the second method

   //Create the main method

   public static void main(String[] args) {

       //1. Create an arbitrary one dimensional int array

       int[] x = {23, 3, 4, 3, 2, 4, 3, 3, 5, 6, 5, 3, 5, 5, 6, 3};

       //2. Create an arbitrary two dimensional String array

       String[] names = {"abc", "john", "dow", "doe", "xyz"};

       //Call the respective methods

       convert2D(x);

       System.out.println("");

       convert2D(names);

   }         // End of the main method

}            // End of class definition

=========================================================

==========================================================

23 3 4 3  

2 4 3 3  

5 6 5 3  

5 5 6 3  

abc john dow  

doe xyz null

==========================================================

The above code has been written in Java and it contains comments explaining each line of the code. Please go through the comments. The actual executable lines of code are written in bold-face to distinguish them from the comments.

Answer:

A 512 GB Solid State Drive (SSD) will be recommended

Explanation:

Recommended hard disk for the installation of Microsoft PowerPoint application is 3 GB and since the computer is a new one it will be best to buy a hard disc with enough room for expansion, performance speed, durability and reliability

Therefore, a 512 GB Solid State Drive (SSD) is recommended as the price difference is small compared to the spinning hard drive and also there is ample space to store PowerPoint training presentation items locally.

Answer:

Check Explanation.

Explanation:

Immediately there is a connection between your laptop and the wireless hotspot, data will be transferred via the Physical layer that will be used for encoding of the network and the modulation of the network and that layer is called the OSI MODEL.

Another Important layer is the APPLICATION LAYER in which its main aim and objectives is for support that is to say it is used in software supporting.

The next layer is the PRESENTATION LAYER which is used during the process of sending the e-mail. This layer is a very important layer because it helps to back up your data and it is also been used to make connections of sessions between servers of different computers

Answer: Provided in the explanation segment

Explanation:

Below is the code to run this program.

we have that the

Program

import java.util.Scanner;

public class CharMatching {

  public static void main(String[] args) {

      //Scanner object for keyboard read

      Scanner scnr=new Scanner(System.in);

      //Variable for user input string

      String userInput;

      //Variable for firstletter input

      char firstLetter;

      //Read user input string

      userInput=scnr.nextLine();

      //Read first letter from user

      firstLetter=scnr.nextLine().charAt(0);

      //Comparison without case sensititvity and result

      if(Character.toUpperCase(firstLetter)==Character.toUpperCase(userInput.charAt(0))) {

          System.out.println("Found match: "+firstLetter);

      }

      else {

          System.out.println("No match: "+firstLetter);

      }

  }

}

Output

Hello

h

Found match: h

cheers i hope this helped !!!

Answer:

The java program for the given scenario is as follows.

import java.util.*;

//interface with method area

interface ObjectWithTwoParameters

{

double area (double d1, double d2);

}

class RectangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return d1*d2;

}

}

class TriangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return (d1*d2)/2;

}  

}

class CylinderClass implements ObjectWithTwoParameters

{

public double area (double d1, double d2)

{

return ( 2*3.14*d1*d1 + d2*(2*3.14*d1) );

}

}

public class Test

{

public static void main(String[] args)

{

//area displayed for all three shapes

ObjectWithTwoParameters r = new RectangleClass();

double arear = r.area(2, 3);

System.out.println("Area of rectangle: "+arear);

ObjectWithTwoParameters t = new TriangleClass();

double areat = t.area(4,5);

System.out.println("Area of triangle: "+areat);

ObjectWithTwoParameters c = new CylinderClass();

double areac = c.area(6,7);

System.out.println("Area of cylinder: "+areac);

}

}

OUTPUT

Area of rectangle: 6.0

Area of triangle: 10.0

Area of cylinder: 489.84

Explanation:

1. The program fulfils all the mentioned requirements.

2. The program contains one interface, ObjectWithTwoParameters, three classes which implement that interface, RectangleClass, TriangleClass and CylinderClass, and one demo class, Test, containing the main method.

3. The method in the interface has no access specifier.

4. The overridden methods in the three classes have public access specifier.

5. No additional variables have been declared.

6. The test class having the main() method is declared public.

7. The area of the rectangle, triangle and the cylinder have been computed as per the respective formulae.

8. The interface is similar to a class which can have only declarations of both, variables and methods. No method can be defined inside an interface.

9. The other classes use the methods of the interface by implementing the interface using the keyword, implements.

10. The object is created using the name of the interface as shown.

ObjectWithTwoParameters r = new RectangleClass();

   //Method definition of words_typed

   //The return type is int

   //Takes two int parameters: typingSpeed and timeInterval

   public static int words_typed(int typingSpeed, int timeInterval) {

       //Get the number of words typed by  

       //finding the product of the typing speed and the time interval

       //and then dividing the result by 60 (since the typing speed is in "words

       // per minute"  and the time interval is in "seconds")

       int numberOfWords = typingSpeed * timeInterval / 60;

       //return the number of words

       return numberOfWords;

   }        //end of method declaration

The code above has been written in Java and it contains comments explaining each of the lines of the code. Please go through the comments.

Answer:

I am writing the program in Python.

def words_typed(typing_speed,time_interval):

   typing_speed>=0

   time_interval>0

   no_of_words=int(typing_speed*(time_interval/60))

   return no_of_words

output=words_typed(20,30)

print(output)

Explanation:

I will explain the code line by line.

First the statement def words_typed(typing_speed,time_interval) is the definition of a function named words_typed which has two parameters typing_speed and time_interval.

typing_speed variable of integer type in words per minute.

time_interval variable of int type in seconds

The statements typing_speed>=0 and time_interval>0 means that value of typing_speed is greater than or equal to zero and value of time_interval is greater than zero as specified in the question.

The function words_typed is used to return the number of words that a  person with typing speed would type in that time interval. In order to compute the words_typed, the following formula is used:

   no_of_words=int(typing_speed*(time_interval/60))

The value of typing_speed is multiplied by value of time_interval in order to computer number of words and the result of the multiplication is stored in no_of_words. Here the time_interval is divided by 60 because the value of time_interval is in seconds while the value of typing_speed is in minutes. So to convert seconds into minutes the value of time_interval is divided by 60 because 1 minute = 60 seconds.

return no_of_words statement returns the number of words.

output=words_typed(20,30) statement calls the words_typed function and passed two values i.e 20 for typing_speed and 30 for time_interval.

print(output) statement displays the number of words a person with typing speed would type in that time interval, on the output screen.

Answer:

class Main {

  public static void main(String[] args) {

      char arr[] = {'T','E','D','R','W','B','S','V','A'};

      int n = arr.length;

      System.out.println("Selection Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp1 = selectionSort(arr);

      System.out.println("Total comparisons: "+comp1);

      System.out.println("\nInsertion Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp2 = insertionSort(arr);

      System.out.println("Total Comparisons: "+comp2);

      System.out.println("\nOverall Total Comparisons: "+(comp1+comp2));

  }

  static long selectionSort(char arr[]) {

      // applies selection sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      long comparisons = 0;

      // One by one move boundary of unsorted subarray

      for (int i = n-1; i>=n-n/2; i--) {

              // Find the minimum element in unsorted array

              int max_idx = i;

              for (int j = i-1; j>=0; j--) {

                      // there is a comparison everytime this loop returns

                      comparisons++;

                      if (arr[j] > arr[max_idx])

                              max_idx = j;

              }

              // Swap the found minimum element with the first

              // element

              char temp = arr[max_idx];

              arr[max_idx] = arr[i];

              arr[i] = temp;

              System.out.print(n-1-i+"\t");

              printArray(arr);

              System.out.println("\t"+comparisons);

      }

      return comparisons;

  }

  static long insertionSort(char arr[]) {

      // applies insertion sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      n = n-n/2;   // sort only the first n/2 elements

      long comparisons = 0;

      for (int i = 1; i < n; ++i) {

          char key = arr[i];

          int j = i - 1;

          /* Move elements of arr[0..i-1], that are

                  greater than key, to one position ahead

                  of their current position */

          while (j >= 0) {

              // there is a comparison everytime this loop runs

              comparisons++;

              if (arr[j] > key) {

                  arr[j + 1] = arr[j];

              } else {

                  break;

              }

              j--;

          }

          arr[j + 1] = key;

          System.out.print(i-1+"\t");

          printArray(arr);

          System.out.println("\t"+comparisons);

      }

      return comparisons;

  }  

  static void printArray(char arr[]) {

      for (int i=0; i<arr.length; i++)

          System.out.print(arr[i]+" ");

  }

}

Explanation:

Explanation is in the answer.

Answer:

How was the Aswan high dam had an adverse effect on human health? It created new ecosystems in which diseases such as malaria flourished. An environmental challenge facing the Great Man Made River is that pumping water near the coast. ... Water scarcity makes it impossible to grow enough food for the population.

Explanation: