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Elodia [21]
3 years ago
8

HELP PLEASE !!!!!!!!!!!! Amy needs to ensure that she can enter a parameter that will match the values End, deadend, and flatend

. Which option will she use?
a. “end*”
b. “*end”
c. “&end”
d. “end&”
Computers and Technology
1 answer:
ELEN [110]3 years ago
6 0

Answer: it’s d ynnn

Explanation:

It’s d

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D.Limited Access Highway
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Before you insert a page break, what should you do? select the font you want to use for the text that comes after the page break
Triss [41]
Put the insertion point where you want the page to break is what I would write.
7 0
3 years ago
Start
Crank

Answer:

N=5  key task decision start and end

Explanation:

8 0
3 years ago
(1) Prompt the user for a string that contains two strings separated by a comma. (1 pt) Examples of strings that can be accepted
denis-greek [22]

Answer:

Check the explanation

Explanation:

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#define MAX_LIMIT 50

int checkComma(char *input)

{

int flag = 0;

for(int i = 0; i < strlen(input); i++)

{

if(input[i] == ',')

{

flag = 1;

break;

}

}

return flag;

}

int main(void)

{

char input[MAX_LIMIT];

char *words[2];

char delim[] = ", ";

printf("\n");

do

{

printf("Enter input string: ");

fgets(input, MAX_LIMIT, stdin);

size_t ln = strlen(input) - 1;

if (*input && input[ln] == '\n')

input[ln] = '\0';

if(strcmp(input, "q") == 0)

{

printf("Thank you...Exiting\n\n");

exit(1);

}

else

{

if(checkComma(input) == 0)

{

printf("No comma in string.\n\n");

}

else

{

char *ptr = strtok(input, delim);

int count = 0;

while(ptr != NULL)

{

words[count++] = ptr;

ptr = strtok(NULL, delim);

}

printf("First word: %s\n", words[0]);

printf("Second word: %s\n\n", words[1]);

}

}

}while(strcmp(input, "q") != 0);

return 0;

}

Kindly check the attached image below for the output.

4 0
3 years ago
Write a computer program that determines how many grades are between 0 and 19.
True [87]

Answer:

public class nnn {

   public static void main(String[] args) {

       int [] examScores = {31, 70, 92, 5, 47, 88, 81, 73, 51, 76, 80, 90, 55, 23, 43,98,36,87,22,61, 19,69,26,82,89,99, 71,59,49,64};

       int zeroTo19 = 0;

       int nineteenTo39 = 0;

       int fortyTo59 = 0;

       int sixtyTo79 = 0;

       int eightyTo100 = 0;

       for(int i =0; i<examScores.length; i++){

           if(examScores[i]<=19){

               zeroTo19++;

           }

           else if(examScores[i]>19&&examScores[i]<=39){

               nineteenTo39++;

           }

           else if(examScores[i]>39&&examScores[i]<=59){

               fortyTo59++;

           }

           else if(examScores[i]>59&&examScores[i]<=79){

               sixtyTo79++;

           }

           else {

               eightyTo100++;

           }

       }

       System.out.println("0 - 19 is "+zeroTo19);

       System.out.println("20 - 39 is "+nineteenTo39);

       System.out.println("40 - 59 is "+fortyTo59);

       System.out.println("60 - 79 is "+sixtyTo79);

       System.out.println("80 - 100 is "+eightyTo100);

   }

}

Explanation:

  • This has been solved with Java
  • Create an array of the exam scores
  • Create new variables for each of the score range
  • Use multiple if statements as u loop through the array to determine the range of scores
  • Finally outside the loop print them variables out
7 0
3 years ago
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