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nikklg [1K]
3 years ago
15

a marketing company is setting up a new office in the city. which type of contract should they sign for periodic maintenance of

their computers and their peripherals
Computers and Technology
1 answer:
True [87]3 years ago
8 0

Answer:

The contract they had to sign was a computer maintenance contract with computer service providers.

Explanation:

<em>Why </em><em>do companies have to </em><em>maintain computers</em><em> regularly? </em>

Because of all the activities of the company at this time on average use applications from computers. With the maintenance of computers, the activities of the company will continue to run well.

There are two types of maintenance that can be carried out:

  • Hardware Maintenance
  • Software Maintenance

#AnswerForTrees

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Output each floating-point value with two digits after the decimal point, which can be achieved by executing cout &lt;&lt; fixed
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Answer:

In C++:

#include <iomanip>

#include <iostream>

#include <vector>

using namespace std;

int main(){

vector<double> myvector;

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cout<<"Enter 5 numbers: ";

for (int i = 1; i <= 5; i++){

 cin>>userinput;

    myvector.push_back(userinput);

}

for (int i = 0; i < 5; i++){

    cout << fixed << setprecision(2)<< myvector.at(i) <<' ';

}

return 0;

}

Explanation:

This declares the vector as double

 vector<double> myvector;

This declares a variable for input as double

double userinput;

This prompts the user for input of 5 numbers

cout<<"Enter 5 numbers: ";

This iterates from 1 to 5

for (int i = 1; i <= 5; i++){

This gets each integer input

 cin>>userinput;

This pushes each input into the vector

    myvector.push_back(userinput); }

This iterates through the vector

for (int i = 0; i < 5; i++){

This prints each vector element separated by space using fixed setprecision

cout << fixed << setprecision(2)<< myvector.at(i) <<' ';

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3 years ago
Selection Sort List the resulting array after each iteration of the outer loop of the selection sort algorithm. Indicate the num
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Solution :

Initial array = $\text{C,Q,S,A,X,B,T}$

$n= 7$(length of the array)

$\text{1st}$ Iteration:

i = 1

  j = 1

  $\text{a[j-1]}$ = C

  a[j] = Q

  since $\text{a[j-1]}$ < a[j] , break from inner loop

 

Number of comparisons in 1st Iteration = 1

After 1st Iteration:

Array : C,Q,S,A,X,B,T

2nd Iteration:

i = 2

  j = 2

  a[j-1] = Q

  a[j] = S

  since a[j-1] < a[j], break from inner loop

 

Number of comparisons in 2nd Iteration = 1

After 2nd Iteration:

Array : C,Q,S,A,X,B,T

3rd Iteration:

i = 3

  j = 3

  a[j-1] = S

  a[j] = A

  since a[j-1] > a[j], exchange a[2] with a[3]

  Array : C,Q,A,S,X,B,T

 

  j = 2

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  Array : C,A,Q,S,X,B,T

 

  j = 1

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  since a[j-1] > a[j], exchange a[0] with a[1]

  Array : A,C,Q,S,X,B,T

 

  j = 0, break from inner loop

Number of comparisons in 3rd Iteration = 3

After 3rd Iteration:

Array : A,C,Q,S,X,B,T

4th Iteration:

i = 4

  j = 4

  a[j-1] = S

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  since a[j-1] < a[j], break from inner loop

 

Number of comparisons in 4th Iteration = 1

After 4th Iteration:

Array : A,C,Q,S,X,B,T

5th Iteration:

i = 5

  j = 5

  a[j-1] = X

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  since a[j-1] > a[j], exchange a[4] with a[5]

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  j = 4

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  Array : A,C,Q,B,S,X,T

 

  j = 3

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  j = 2

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  since a[j-1] > a[j], exchange a[1] with a[2]

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  j = 1

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  a[j] = B

  since a[j-1] < a[j], break from inner loop

 

Number of comparisons in 5th Iteration = 5

After 5th Iteration:

Array : A,B,C,Q,S,X,T

6th Iteration:

i = 6

  j = 6

  a[j-1] = X

  a[j] = T

  since a[j-1] > a[j], exchange a[5] with a[6]

  Array : A,B,C,Q,S,T,X

 

  j = 5

  a[j-1] = S

  a[j] = T

  since a[j-1] < a[j], break from inner loop

 

Number of comparisons in 6th Iteration = 2

After 6th Iteration:

Array : A,B,C,Q,S,T,X

Sorted Array : A B C Q S T X  

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