Write a function, average_peer_pressure, which takes in a month, and returns a vector of length two. For the first element, consider doctors who began prescribing in that month, then return the average proportion of prescribers among their contacts. For the second element, consider doctors who began prescribing after that month (or never did), then return the average proportion of prescribers among their contacts. Call your functions from (2) and (5); use a function from the apply family rather than a loop if at all possible. Check that month 5 returns 0.2698052 and 0.525822. (Hint: mean has an na. rm option)

Iwould say d run disk cleanup on the hard drive

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.

Ineed with this one!