Complete the do-while loop to output 0 to the value of countLimit using printVal. Assume the user will only input a positive number. For example, if countLimit is 5 the expected output will be

Question not complete

Complete the do-while loop to output every number form 0 to countLimit using printVal. Assume the user will only input a positive number. For example, if countLimit is 5 the expected output will be 0 1 2 3 4 5

import java.util.Scanner;

public class CountToLimit {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

int countLimit = 0;

int printVal = 0;

// Get user input

countLimit = scnr.nextInt();

printVal = 0;

do {

System.out.print(printVal + " ");

printVal = printVal + 1;

} while ( /* Your solution goes here */ );

System.out.println("");

return;

}

}

Replace while ( /* Your solution goes here */ );

With

while ( printVal <= countLimit );

This will check if the value of variable printVal is still within print range of variable countLimit.

The full program becomes

import java.util.Scanner;

public class CountToLimit {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

int countLimit = 0;

int printVal = 0;

// Get user input

countLimit = scnr.nextInt();

printVal = 0;

do {

System.out.print(printVal + " ");

printVal = printVal + 1;

} while ( printVal <= countLimit );

System.out.println("");

return;

}

}

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

step-by-step explanation: