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tiny-mole [99]
3 years ago
7

After unit and integration testing are completed, _________ testing ensures that all hardware and software components work toget

her.?
Computers and Technology
1 answer:
raketka [301]3 years ago
5 0
Testing ensures that all hardware and software components work together
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When you select an object in the Visual Studio Designer, the object's size, color, text, and other characteristics are displayed
harina [27]

Answer:

the main window if that`s not correct sorry

Explanation:

3 0
3 years ago
Which tcp/ip troubleshooting command should you use to determine whether a client and server are communicating with each other?
sesenic [268]
The answer is <span>The ping command.   The </span><span>tcp/ip troubleshooting command you should  use to determine whether a client and server are communicating with each other is The ping command.  </span><span>The </span>ping command<span> is used to verify that a device can communicate with another on a network.</span>
5 0
3 years ago
Write a program that accepts an integer value called multiplier as user input. Create an array of integers with ARRAY_SIZE eleme
Tasya [4]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

void PrintForward(int myarray[], int size){

   for(int i = 0; i<size;i++){        cout<<myarray[i]<<" ";    }

}

void PrintBackward(int myarray[], int size){

   for(int i = size-1; i>=0;i--){        cout<<myarray[i]<<" ";    }

}

int main(){

   const int ARRAY_SIZE = 12;

   int multiplier;

   cout<<"Multiplier: ";

   cin>>multiplier;

   int myarray [ARRAY_SIZE];

   for(int i = 0; i<ARRAY_SIZE;i++){        myarray[i] = i * multiplier;    }

   PrintForward(myarray,ARRAY_SIZE);

   PrintBackward(myarray,ARRAY_SIZE);

   return 0;}

Explanation:

The PrintForward function begins here

void PrintForward(int myarray[], int size){

This iterates through the array in ascending order and print each array element

<em>    for(int i = 0; i<size;i++){        cout<<myarray[i]<<" ";    }</em>

}

The PrintBackward function begins here

void PrintBackward(int myarray[], int size){

This iterates through the array in descending order and print each array element

<em>    for(int i = size-1; i>=0;i--){        cout<<myarray[i]<<" ";    }</em>

}

The main begins here

int main(){

This declares and initializes the array size

   const int ARRAY_SIZE = 12;

This declares the multiplier as an integer

   int multiplier;

This gets input for the multiplier

   cout<<"Multiplier: ";    cin>>multiplier;

This declares the array

   int myarray [ARRAY_SIZE];

This iterates through the array and populate the array by i * multiplier

<em>    for(int i = 0; i<ARRAY_SIZE;i++){        myarray[i] = i * multiplier;    }</em>

This calls the PrintForward method

   PrintForward(myarray,ARRAY_SIZE);

This calls the PrintBackward method

   PrintBackward(myarray,ARRAY_SIZE);

   return 0;}

6 0
3 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
"while executing programs from the command line, most operating systems also allow the user to specify one or more ____________
Rudik [331]
The answer could possibly be "parameters"
6 0
3 years ago
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