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Kipish [7]
3 years ago
10

Take the average of some numbers. Show all the numbers that are below average. You can assume that there will not be more than 2

0 numbers entered. For example: Enter a weight (0 to stop): 1 Enter a weight (0 to stop): 2 Enter a weight (0 to stop): 3 Enter a weight (0 to stop): 4 Enter a weight (0 to stop): 5 Enter a weight (0 to stop): 6 Enter a weight (0 to stop): 7 Enter a weight (0 to stop): 8 Enter a weight (0 to stop): 9 Enter a weight (0 to stop): 10 Enter a weight (0 to stop): 0 The total was 55 The average is 5.5 Here are all the numbers less than the average: 1 2 3 4 5
Computers and Technology
1 answer:
olchik [2.2K]3 years ago
6 0

Here is code in java.

import java.util.*; // import package

class Main //  creating class

{

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

{

Scanner br=new Scanner(System.in); // scanner class for input

       int []a=new int[20]; // declaring array

       int count=0,x,j; // creating variable

       int sum =0;

       System.out.println("Enter a weight (0 to stop):");

       x=br.nextInt(); // taking input

while(x!=0) // iterating over the loop

       {

           a[count] = x;

           count++;

           System.out.println("Enter a weight (0 to stop):");

           x=br.nextInt();

}

       for( j=0;j<count;j++) // iterating over the loop

       {

       sum+=a[j];

       }

       double avg = sum/(double)count;

       System.out.println("The total is: "+sum);

       System.out.println("The average is: "+avg);

       System.out.println("Here are all the numbers less than the average: ");

     for( j=0;j<count;j++) //  // iterating over the loop

       {

           if(a[j]<avg)

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

       }

}

}

Explanation:

First create an object of "Scanner" class to read input from the user.

Create an array of size 20, which store the number given by user.Ask user

to give input and keep the count of number in the variable "count". If user

give input 0 then it will stop taking input.And then calculate sum of all the

input.It will calculate average by dividing the sum with count and print it.

It will compare the all elements of array with average value, if the element

is less than average, it will print that element.

Output:

Enter a weight (0 to stop):

1

Enter a weight (0 to stop):

2

Enter a weight (0 to stop):

3

Enter a weight (0 to stop):

4

Enter a weight (0 to stop):

5

Enter a weight (0 to stop):

6

Enter a weight (0 to stop):

7

Enter a weight (0 to stop):

8

Enter a weight (0 to stop):

9

Enter a weight (0 to stop):

10

Enter a weight (0 to stop):

0

The total is: 55

The average is: 5.5

Here are all the numbers less than the average:

1 2 3 4 5

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mart [117]

Answer:Following is the C program:-

#include <stdio.h>

int fun()//function fun of return type int and it returns value 6.

{

   return 6;

}

int main() {

  int a, b;

  a = 10;

  b = a + fun();//adds 6 to a.

  printf("With the function call on the right, ");

  printf("\n%d ",b);//printing b..

return 0;

}

Output:-

With the function call on the right,  

16

Explanation:

The function fun return the value 6 so it adds 6 to a and stores the result in b.

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3 years ago
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Answer:

Written in Python

name = input("Name: ")

wageHours = int(input("Hours: "))

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if wageHours >= 60:

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print(name)

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print(total)

Explanation:

The program is self-explanatory.

However,

On line 4, the program checks if wageHours is greater than 60.

If yes, the corresponding wage is calculated.

On line 6, if workHours is not up to 60, the total wages is calculated by multiplying workHours by regPay, since there's no provision for how to calculate total wages for hours less than 60

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1. How many bits would you need to address a 2M × 32 memory if:
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Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

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hence ; L = 2^21 ; w = 32

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w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

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b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

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L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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