9. Online students require tutors with Academic Technology training. Academic Technology training is offered to staff with more than 2 years of service at the academic support center. Luke wants to identify the staff members eligible for Academic Technology training in the table. In cell M2, enter a formula using a nested IF function and structured references to determine first if a staff member already has completed Academic Technology training, and if not, whether that staff member is eligible for Academic Technology training. a. If the value of the Academic Technology Training column is equal to "Yes", the formula should return the text Completed. Remember to use a structured reference to the Academic Technology Training column. b. If the value of the Academic Technology Training column is equal to No, the formula should determine if the value in the Service Years column is greater than 2. c. The formula should return the text Yes if the staff member’s Service Years value is greater than 2. d. The formula should return the text No if the staff member’s Service Years value is not greater than 2.

The java program is given below

import java.util.Comparator;

import java.util.PriorityQueue;

public class PriorityQueueTest {

static class PQsort implements Comparator<Integer> {

 public int compare(Integer one, Integer two) {

 return two - one;

 }

}

public static void main(String[] args) {

 int[] ia = { 1, 10, 5, 3, 4, 7, 6, 9, 8 };

 PriorityQueue<Integer> pq1 = new PriorityQueue<Integer>();

 // use offer() method to add elements to the PriorityQueue pq1

 for (int x : ia) {

 pq1.offer(x);

 }

 System.out.println("pq1: " + pq1);

 PQsort pqs = new PQsort();

 PriorityQueue<Integer> pq2 = new PriorityQueue<Integer>(10, pqs);

 // In this particular case, we can simply use Collections.reverseOrder()

 // instead of self-defined comparator

 for (int x : ia) {

 pq2.offer(x);

 }

 System.out.println("pq2: " + pq2);

 // print size

 System.out.println("size: " + pq2.size());

 // return highest priority element in the queue without removing it

 System.out.println("peek: " + pq2.peek());

 // print size

 System.out.println("size: " + pq2.size());

 // return highest priority element and removes it from the queue

 System.out.println("poll: " + pq2.poll());

 // print size

 System.out.println("size: " + pq2.size());

 System.out.print("pq2: " + pq2);

}

}

answer is c. complication