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-BARSIC- [3]
3 years ago
9

For an activity with more than one immediate predecessor activity, which of the following is used to compute its earliest finish

(EF) time? a. the largest EF among the immediate predecessors b. the average EF among the immediate predecessors c. the largest LF among the immediate predecessors d. the difference in EF among the immediate predecessors
Computers and Technology
1 answer:
laila [671]3 years ago
7 0

Answer:

The correct option is A

Explanation:

In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.

Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.

That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).

Cheers!

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3 years ago
QUESTION
Lady_Fox [76]

Answer:

Explanation:

#include <iostream>

using namespace std;

int main ()

{

   int responses[30],count[6];

   int score = 0;

   string resp = " ";

   for (int i = 0; i < 30; i++)

   {

    responses[i] = 0;

   }

   for (int i = 0; i < 6; i++)

   {

    count[i,1]=0;

    count[i,2]=0;

    count[i,0]=0;

   }

   while ((resp != "Y") && (resp != "y"))

   {

    for (int i = 0; i < 30; i++)

    {

        while ((score > 5) || (score < 1))

           {

            cout << "Student " << (i+1)<< " please enter a value (1-5):";

            cin >> score;

        }

           responses[i] = score;

           if((score > 5)||(score<1))

           {

               if(score==1) count[1]++;        

               if(score==2) count[2]++;        

               if(score==3) count[3]++;        

               if(score==4) count[4]++;        

               if(score==5) count[5]++;        

           }

           score = 0;

    }

    cout<< "Response               Frequency              Percentage"<<endl;;

    cout<< "    1                      "<<count[1]<<"               "<<(count[1]/30)<<"%"<<endl;

    cout<< "    2                      "<<count[2]<<"               "<<(count[2]/30)<<"%"<<endl;

    cout<< "    3                      "<<count[3]<<"               "<<(count[3]/30)<<"%"<<endl;

    cout<< "    4                      "<<count[4]<<"               "<<(count[4]/30)<<"%"<<endl;

    cout<< "    5                      "<<count[5]<<"               "<<(count[5]/30)<<"%"<<endl;

    cout<< "Do you want to exit? Press Y to exit any other key to continue: ";

    cin>> resp;

   }

   return 0;

}

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Learn more about radio buttons, here:

brainly.com/question/14316393

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