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Keith_Richards [23]
2 years ago
7

JAVAplzz help...............​

Computers and Technology
1 answer:
patriot [66]2 years ago
7 0

Answer:

import java.util.Scanner;

import java.util.Arrays;

public class Main {

 public static void main(String[] args) {

   Scanner scan = new Scanner(System.in);

   int numbers[] = new int[3];

   System.out.print("Enter first number: ");

   numbers[0] = scan.nextInt();

   System.out.print("Enter second number: ");

   numbers[1] = scan.nextInt();

   System.out.print("Enter third number: ");

   numbers[2] = scan.nextInt();

   scan.close();

   Arrays.sort(numbers);

   if (numbers[0] == numbers[2]) {

     System.out.println("All the numbers are equal");

   } else {

     System.out.printf("Greatest number: %d\n", numbers[2]);

   }

 }

}

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Write a python application that allows a user to enter any number of student test scores until the user enters 999. If the score
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Answer:

This program is as follows

<em>total = 0; count = 0</em>

<em>testscore = int(input("Score: "))</em>

<em>while testscore != 999:</em>

<em>    if testscore < 0 or testscore > 100:</em>

<em>        print("Out of range")</em>

<em>    else:</em>

<em>        total+=testscore</em>

<em>        count+=1</em>

<em>    testscore= int(input("Score: "))</em>

<em>print(count,"scores entered")</em>

<em>print("Arithmetic Average:",total/count)</em>

Explanation:

This initializes total and count to 0

total = 0; count = 0

This gets input for score

testscore = int(input("Score: "))

The following iteration stop when 999 is entered

while testscore != 999:

This prints out of range for scores outside 0 - 100

   if testscore < 0 or testscore > 100:

<em>        print("Out of range")</em>

Otherwise

   else:

The total score is calculated

       total+=testscore

The number of score is calculated

       count+=1

Get another input

   testscore = int(input("Score: "))

The number of score is printed

print(count,"scores entered")

The average of score is printed

print("Arithmetic Average:",total/count)

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