When you don't know the answer to a question, a good incident management strategy is to tell the user that you'll research the question and get back to him or her.
<h3>What is
incident management?</h3>
Incident management can be defined as a strategic process through which a business organization or company identifies, analyzes, and correct hazards, so as to ensure that normal service operation is restored as quickly as possible to end users after a disruption, as well as to prevent a re-occurrence of these hazards in the future.
As a support agent, if you don't know the answer to a question, a good incident management strategy is to tell the user that you'll research the question and get back to him or her at a latter time.
Read more on incident management here: brainly.com/question/11595883
Answer:
// A optimized school method based C++ program to check
// if a number is composite.
#include <bits/stdc++.h>
using namespace std;
bool isComposite(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return true;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return true;
return false;
}
// Driver Program to test above function
int main()
{
isComposite(11)? cout << " true\n": cout << " false\n";
isComposite(15)? cout << " true\n": cout << " false\n";
return 0;
}
Explanation:
Answer:
B. {1, 2, 2, 3, 3, 4, 5}
Explanation:
Given
The above code segment
Required
Determine which list does not work
The list that didn't work is 
Considering options (A) to (E), we notice that only list B has consecutive duplicate numbers i.e. 2,2 and 3,3
All other list do not have consecutive duplicate numbers
Option B can be represented as:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 2](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%202)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 3](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%203)
![nums[5] = 4](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%204)
![nums[6] = 5](https://tex.z-dn.net/?f=nums%5B6%5D%20%3D%205)
if (nums.get(j).equals(nums.get(j + 1)))
The above if condition checks for duplicate numbers.
In (B), when the elements at index 1 and 2 (i.e. 2 and 2) are compared, one of the 2's is removed and the Arraylist becomes:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 3](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%203)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 4](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%204)
![nums[5] = 5](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%205)
The next comparison is: index 3 and 4. Meaning that comparison of index 2 and 3 has been skipped.
<em>This is so because of the way the if statement is constructed.</em>