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bulgar [2K]
3 years ago
10

For each of the threats and vulnerabilities from the Identifying Threats and Vulnerabilities in an IT Infrastructure lab in this

lab manual (list at least three and no more than five) that you have remediated, what must you assess as part of your overall COBIT P09 risk management approach for your IT infrastructure?
Computers and Technology
1 answer:
tia_tia [17]3 years ago
5 0

COBIT P09 is a management tool  or approach for your IT infrastructure  where THREATS  and VULNERABILITIES  affected IT infrastructure lab

<u>Explanation</u>

1. The first process should update antivirus each pc or workstation on laptop or  workstation

2. Update operating system updates if any pending for each pc or workstation or laptop

3. Each PC or desktop or workstation to be scanned and check any threats or Vulnerabilities

4. Once the scanning is completed check still Threats and Vulnerabilities  still exists physically check schedule task any new task created if so disable or delete it  

5. If Windows operating system kindly check user\appdata\local folder etc and delete unwanted folders

6. Repeats the process of step 5 till threats and Vulnerabilities is removed In each desktop or workstation  or laptop.

COBIT P09 is assets the risk process for any infrastructure lab, It is a strategic process that is applied with another process to get rid of Threats and Vulnerabilities.

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olga2289 [7]

Binary search is used similarly. K'th element is found in more productive way.

<u>Explanation:</u>

arr1 and arr2 are the middle elements which are compared, let us compute as mid1 and mid2. Let us predict arr1 [mid1] k, the elements are not needed after mid2 elements.  arr2 are set as last element for arr2 [mid2].  New sub problems are defined. When one array is half the size.

C++ code for the algorithm.

#include <iostream>

using namespace std;    

int kth(int *arr1, int *arr2, int *end1, int *end2, int k)

{

if (arr1 == end1)

return arr2[k];

if (arr2 == end2)

return arr1[k];

int mid1 = (end1 - arr1) / 2;

int mid2 = (end2 - arr2) / 2;

if (mid1 + mid2 < k)

{

if (arr1[mid1] > arr2[mid2])

return kth(arr1, arr2 + mid2 + 1, end1, end2,

k - mid2 - 1);

else

return kth(arr1 + mid1 + 1, arr2, end1, end2,

k - mid1 - 1);

}

else

{

if (arr1[mid1] > arr2[mid2])

return kth(arr1, arr2, arr1 + mid1, end2, k);

else

return kth(arr1, arr2, end1, arr2 + mid2, k);

}    

int main()

{

int arr1[5] = {2, 3, 6, 7, 9};

int arr2[4] = {1, 4, 8, 10};    

int k = 5;

cout << kth(arr1, arr2, arr1 + 5, arr2 + 4, k - 1);

return 0;

}

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julsineya [31]

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