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andreyandreev [35.5K]
3 years ago
11

Work to be performed can come from the work package level of the work breakdown structure as well as other sources. Which of the

se is NOT a source of authorized work to be performed?
a. Scope creep
b. Defect repairs
c. Preventive actions
d. Corrective actions
Engineering
1 answer:
denis-greek [22]3 years ago
8 0

A project is meant to achieve clearly defined goals within present constraints

The correct option for the source of work in a project which is NOT a source of authorized work to be performed is option (a)

a. Scope creep

The reason for the selecting the above option is as follows;

In project management, alongside work planned in a work package, authorized work to be performed can come from;

  • The repairing of defects found in completed work and which is a requirement for the project to be approve
  • Preventive actions, that are meant to mitigate factors that affect the project timeline and cost
  • Corrective actions which include actions meant to rectify parts of the project that produce an error outcome

A scope creep, however, is generally is due to the inclusion of the end

users of the project at a time that is behind schedule, insufficient analysis of

the requirements, and project complexity underestimation. and consists of

changes made after a project has commenced

A scope creep therefore, changes the amount of work to be done which

therefore alters the project timeline, and due to project, budgetary, and

end user constraints, CANNOT be a source of authorized work to be

performed

The correct option for the source that is NOT a authorized work to be performed is option a. Scope creep

Learn more about scope creep here:

brainly.com/question/14061926

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Answer:

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#include<stdlib.h>

#include<stdio.h>

#include<string.h>

void merge(int arr[], int l, int m, int r)

{

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

 

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = arr[l + i];

for (j = 0; j < n2; j++)

R[j] = arr[m + 1+ j];

i = 0;

j = 0;

k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

void mergeSort(int arr[], int l, int r)

{

if (l < r)

{

int m = l+(r-l)/2;

mergeSort(arr, l, m);

mergeSort(arr, m+1, r);

merge(arr, l, m, r);

}

}

void printArray(int A[], int size)

{

int i;

for (i=0; i < size; i++)

printf("%d ", A[i]);

printf("\n");

}

int main()

{

int arr[1000] = {0};

int arr_size =0;

int data;

char file1[20];

strcpy(file1,"data.txt");

FILE *fp;

fp = fopen(file1,"r+");

if (fp == NULL) // if file not opened return error

{

perror("Unable to open file");

return -1;

}

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{

fscanf (fp, "%d", &data);    

arr[arr_size]=data;

arr_size++;

while (!feof (fp))

{  

fscanf (fp, "%d", &data);  

arr[arr_size]=data;

arr_size++;    

}

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printf("Given array is \n");

printArray(arr, arr_size);

mergeSort(arr, 0, arr_size - 1);

printf("\nSorted array Using MERGE SORT is \n");

printArray(arr, arr_size);

return 0;

}

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