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Masteriza [31]
4 years ago
5

Now that the classList Array has been implemented, we need to create methods to access the list items.

Computers and Technology
1 answer:
natima [27]4 years ago
7 0

The following code will be applied to Create the following static methods for the Student class

<u>Explanation:</u>

/* Note: Array index starts from 0 in java so ,, user ask for 1 index then the position will be i-1 , below program is based on this concept if you dont want this way just remove -1 from classList.get(i-1).getName(); ,and classList.add(i-1,student); */

/* ClassListTester.java */

public class ClassListTester

{

public static void main(String[] args)

{

//You don't need to change anything here, but feel free to add more Students!

Student alan = new Student("Alan", 11);

Student kevin = new Student("Kevin", 10);

Student annie = new Student("Annie", 12);

System.out.println(Student.printClassList());

System.out.println(Student.getLastStudent());

System.out.println(Student.getStudent(1));

Student.addStudent(2, new Student("Trevor", 12));

System.out.println(Student.printClassList());

System.out.println(Student.getClassSize());

}

}

/* Student.java */

import java.util.ArrayList;

public class Student

{

private String name;

private int grade;

//Implement classList here:

private static ArrayList<Student> classList = new ArrayList<Student>();

public Student(String name, int grade)

{

this.name = name;

this.grade = grade;

classList.add(this);

}

public String getName()

{

return this.name;

}

//Add the static methods here:

public static String printClassList()

{

String names = "";

for(Student name: classList)

{

names+= name.getName() + "\n";

}

return "Student Class List:\n" + names;

}  

public static String getLastStudent() {

// index run from 0 so last student will be size -1

return classList.get(classList.size()-1).getName();

}

public static String getStudent(int i) {

// array starts from 0 so i-1 will be the student

return classList.get(i-1).getName();

}

public static void addStudent(int i, Student student) {

// array starts from 0 so, we add at i-1 position

classList.add(i-1,student);

// remove extra student

classList.remove(classList.size()-1);

}  

public static int getClassSize() {

return classList.size();

}

}

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Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working​ alone, one machine would take one hour long
Alborosie

<em><u>Answer</u></em>

5 hours

<em><u>Explanation</u></em>

The two working together can finish a job in

\frac{20}{9}  \: hours

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

\frac{1}{x}    +  \frac{1}{x - 1}

This should be equal to 20/9 hours.

\frac{1}{x}  +  \frac{1}{x - 1}  =  \frac{9}{20}

Multiply through by;

20x(x - 1) \times \frac{1}{x}  +20x(x - 1) \times   \frac{1}{x - 1}  =  20x(x - 1) \times \frac{9}{0}

20(x - 1)  +20x = 9x(x - 1)

20x - 20+20x = 9{x}^{2}  - 9x

9{x}^{2}  - 9x - 20x - 20x + 20= 0

9{x}^{2}  - 49x  + 20= 0

Factor to get:

(9x - 4)(x - 5) = 0

x =  \frac{4}{9}  \: or \: x = 5

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

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Answer:

Here is the JAVA program:

import java.util.Scanner;  //to accept input from user

public class Main { // class name

 public static void main(String [] args) {  // start of main method

   Scanner input = new Scanner(System.in);  // creates Scanner class object to take input from user

   System.out.println("Please Enter the Cost of the Item: ");  // prompts user to enter the cost of item

   double itemCost = input.nextDouble();  //scans and reads the input value of item cost

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   double amount = input.nextDouble();  //scans and reads the value of amount from user

   int change = (int)(amount * 100 - itemCost * 100);  //compute the remaining amount i.e. change

   System.out.println("Change Owed: " + change / 100.0);  // displays the owed change

   int quarters = change / 25;  // computes the value for quarters

   change = change % 25;  // compute the quarters remaining

   int dimes = change / 10;  //  computes dimes

   change = change % 10;  //  computes dimes remaining

   int nickels = change / 5;  // computes nickels

   change = change % 5;  // compute nickels remaining

   int pennies = change;  // computes pennies

   System.out.println("Quarters: " + quarters);  // displays computed value of quarters

   System.out.println("Dimes: " + dimes);  // displays value of dimes

   System.out.println("Nickels: " + nickels);  // displays value of nickels

   System.out.println("Pennies: " + pennies);   }} //displays value of pennies

Explanation:

I will explain the program with an examples.

Suppose the user enters 4.57 as cost of the item and 5.00 as amount paid. Then the program works as follows:

Change is computed as

change = (int)(amount * 100 - itemCost * 100);

This becomes;

change = (int)(5.00 * 100 - 4.57 * 100)

            = 500 - 457

change = 43

Now the change owed is computed as:

change / 100.0 = 43/100.0 = 0.43

Hence change owed = 0.43

Next quarters are computed as:

quarters = change / 25;

This becomes:

quarters = 43/25

quarters = 1

Now the remaining is computed as:

   change = change % 25;

This becomes:

   change = 43 % 25;

  change = 18

Next the dimes are computed from remaining value of change as:

dimes = change / 10;

dimes = 18 / 10

dimes = 1

Now the remaining is computed as:

   change = change % 10;

This becomes:

   change = 18 % 10

  change = 8

Next the nickels are computed from remaining value of change as:

nickels = change / 5;

nickels = 8 / 5

nickels = 1

Now the remaining is computed as:

   change = change % 5;

This becomes:

   change = 8 % 5

  change = 3

At last the pennies are computed as:

pennies = change;

pennies = 3

So the output of the entire program is:

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The screenshot of the output is attached.

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