The processes are executed in the following manner in a system with Round Robin scheduling. a. The process is scheduled to execute until the end of the current CPU burst then process is put back into the ready queue. b. The process is executed for a pre-defined time quantum then the process is preempted and put back into the ready queue. c. The process coming first is executed first without preemption till its completion d. The process having the smallest burst time is executed first.

Idon't have a single clue tbfh

True, it makes it easier on the eyes and easier to make if it is the same color scheme

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.

Answer(cpu)