You are a security professional. You have been asked to develop strategies to protect data and file systems in a small Linux environment. The environment contains a Linux application server, a Linux web server, and 30 workstations and laptops. Several users connect to the network from their personal mobile devices. a. Develop layered access security strategies that protect data at rest, data in motion, and file systems.
b. Prepare an elaborated brief summary report describing an access control plan for implementing the strategies. The report should include the purpose of the plan, steps in the creation of the plan, and brief implementation procedures.

The answer should be menu

No. you would need some type of program downloaded to open the special file

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.