How many ways a continous fragment of array will lead to arithematic mean?

Answer (a) with a mirror like surface;

Public static int mystery(int a[], int x) {     int c = 0;     for(int i = 0; i < a.length; i++) {        if (a[i] == x) c++;     } return c; }the method counts how many times an integer shows up in an integer array.the c variable is declared in the method to keep track of how many times the certain value has shown up.the for-loop loops through the entire array.the line    if (a[i] == x) c++;   checks if each element of the array is equal to the inputted integer to check. for each number in the array that is equal to the inputted integer x, the c variable will be incremented by 1. otherwise it will do nothing.then c is returned at the end of the loop.for example, if we have, in the main method with the mystery method coded in the class:     int arraytest[] = {2, 23, 23, 23, 3};     int testme = 23;     int count = mystery(arraytest, testme);     system.out.println(count); then the print out will be    3 because there are three 23s in that array.

Yes theater is a bird andorbs gos

The answer is c. 36======analyzing the for-loop, we see that the control variable, k, starts at 0 and increments by after every loop iteration. the condition  tells us that the loop will stop when k is above 7; in this case, when k is 8.0 1 2 3 4 5 6 7 are the values for k that will be used. each number is for each iteration; therefore, the for-loop will have 8 iterations in total.the code inside the loop itself goes like this: 1st iteration (k=0)starting with sum=0, val=1    means that add  to  .since sum is 0 and val is 1, 0 + 1 = 1. the new value of sum is 1.    means to increment  by 1.since val is 1, the new value of val of 2.end of iteration with  k=0; k is increased by one to become  k=12nd iteration (k=1)starting with sum=1, val=2    ======let us establish a pattern. this is moreso logic than programming: we can  rewrite so that we have the elements from the previous iteration. the 1 came from 0 + 1 before. so we can write it as      .and if we did the third iteration, by the end, sum would look like      notice how the number of plus signs is equal to the iteration you are on (which is just k+1), and what is being added is increasing by one each time.  for example, the third iteration (k=2)  ends up with  with 3 plus signs.the second iteration (k=1) ends up with  with two plus signs.the first iteration (k=0) ends up with  with one plus sign.this is a pattern. the nth iteration will have n plus signs.since we established before that there are 8 iterations of this loop in total, the value of sum at k=7 (the 8th and last iteration) will have sum with 8 plus signs.    so the value of sum at the end of the loop is 36.