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ELEN [110]
3 years ago
13

Exploring Arraylists. Remember that arraylists are good for adding and removing things, and that they are clock-cycle friendlier

in this case than resizing arrays as we discussed in lecture. This lab allows you to get some hands-on using an arraylist so you will remember it for those times it would be a good fit for an application. a) Create an Arraylist of type String and prompt the user for three names and add these names to your ArrayList. b) Print a message with the number of elements in the ArrayList for the user using the size() method (to find the number to print.) c) Prompt the user for two more names and add them to the ArrayList and once again print a message with the number of elements in the ArrayList for the user. d) Use a loop to print all of the names in the ArrayList for the user. e. Ask the user for a name to remove, remove the value the user provides, and then use a loop to print all of the names in the ArrayList for the user. (25 points)
Computers and Technology
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

Code is given below:

Explanation:

import java.io.*;

import java.util.*;

import java.util.Scanner;

class Solution {

  public static void main(String args[]) {

      Scanner in = new Scanner(System.in);

     

     

      ArrayList<String> ans = new ArrayList();

     

      System.out.println("(a) ADD THREE NAMES: ");

     

      for(int i=0;i<3;i++)

      {

          ans.add(in.next());

         

      }

      System.out.println("(b) NUMBER OF ELEMENTS: " + ans.size());

     

      System.out.println("(c) ADD TWO MORE NAMES : ");

      for(int i=0;i<2;i++)

      {

          ans.add(in.next());

         

         

      }

      System.out.println("(d) PRINT ALL NAMES: ");

      for(int i=0;i<ans.size();i++)

      {

          System.out.println(ans.get(i));

      }

     

      System.out.println("(e) ENTER A NAME TO REMOVE: ");

      ans.remove(in.next());

      for(int i=0;i<ans.size();i++)

      {

          System.out.println(ans.get(i));

      }

     

     

  }

}

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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

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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:

A

Explanation:

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Answer:

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