What is the result when the following code is run? double x = 1; double y = 1; int i = 0; do { y = x / 2; x = x + y; i = i + 1; } while (x < 2.5); System. out. print(i + " ");

The answer is: 3

Step-by-step explanation:

1. At the begining of the program we start by declairing the variables:

 double x=1, double y=1 and int i=0.

2. The structure do...while is used to defined the loop. x<2.5 is the finalization condition of the loop. i is the counter of the loop.

y=x/2 is the first calculation

x=x+y is the second one. Here is where the values of the variable x changes.

a) for the first iteration, the values of y and x are shown below:

The variable x is minor to 2.5 so the loop will continue computing.

b)   the second iteration, the values of y and x are shown below::

The variable x is still minor to 2.5 so the loop will continue computing.

c) third iteration:

The condition x<2.5 is not true so the loop ends.

3. System.out.print(i + " "); displays the value of the variable i wich value is 3.

Therefore the number 3 is display.

answer: answer

step-by-step explanation:

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong