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Romashka-Z-Leto [24]
3 years ago
6

Complete the sentence with the correct response. In a two-way selection, if the initial statement is evaluated as , the code ski

ps to the ELSE command.
Computers and Technology
1 answer:
mestny [16]3 years ago
8 0

Answer:

True.

The code skips the else command

Explanation:

I will answer this question with the following code segment

<em>n = 1</em>

<em>If n > 0:</em>

<em>   Print("greater than 0")</em>

<em>else:</em>

<em>   Print("not greater than 0")</em>

<em />

From the code segment above

<em>Print("greater than 0")</em> will be executed because the condition <em>If n > 0 </em>is true

Since the if condition is true, the else condition will not be executed

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Array Challenge Have the function ArrayChallenge (arr) take the array of numbers stored in arr and return the string true if any
vampirchik [111]

Answer:

The code is given as follows,

Explanation:

Code:

#include <stdio.h>

#include <string.h>  

int n; //to store size of array  

char* ArrayChallenge(int arr[]) //function returns string true or false

{

  int i, j;

  int sum = 0;

  for(i = 0; i < n; i++)

  {

      sum = sum + arr[i]; //count sum

  }  

  for(i = 0; i < n; i++)

  {

      for(j = i+1; j < n; j++) //loop for every two elements in array

      {

          if(arr[i]*arr[j] > 2*sum) //check if proudct of two elements > 2 times sum

          {

              printf("\n%d x %d = %d, 2xSum = %d\n", arr[i], arr[j], arr[i]*arr[j], 2*sum);

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          }

      }

  }

  return "false"; // If proudct of two elements < 2 times sum. return false

}  

int main()

{  

  printf("\nEnter size of array: ");

  scanf("%d", &n); //read size of array

  int A[n]; //array of size n

  printf("\nEnter array elements: ");

  int i;

  for(i = 0; i < n; i++)

  {

      scanf("%d", &A[i]); //read array from stdin

  }

  printf("%s\n",ArrayChallenge(A)); //ccall function and print answer

 

  return 0;

}

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