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NARA [144]
3 years ago
13

"in troubleshooting a boot problem, what is the advantage of restoring all uefi/bios settings to their default values?"

Computers and Technology
1 answer:
Alex17521 [72]3 years ago
3 0
When this case would appear, one thing that I would do personally would first, go to the settings, in then, after having this done, I would then "scroll down" to where ti would say "restore (uefi/bios) files, and from there, you would get every value that would would have from the beginning in your chip.

And also, what is truly unique would be the fact that you would be able to choose the "restore point" that you would like for it to appear.
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The _____ layer addresses how the software will execute on specific computers and networks.
lidiya [134]

The problem domain layer addresses how the software will execute on specific computers and networks.

A problem domain is a software engineering term that refers to all information that defines a problem and compels the solution.

A problem domain basically looks at only the area you are interested in and excludes the rest. It includes the goals the problem owner wishes to have and the context in which the problem exists. For example, if you are creating a website for selling artwork online, the problem domain will be artwork and eCommerce.

Therefore, in order to identify a problem domain, you need to know the relevant user requirements. This can be easy when you start by finding out what people (users) think the problem is.

#SPJ4

3 0
2 years ago
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Write a program that has the user input how many classes they are taking this semester and then output how many hours they will
NeTakaya

wait i don't understand: "Write a program that has the user input how many classes they are taking this semester and then output how many hours they will need to study each week."

8 0
3 years ago
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Write a program to read as many test scores as the user wants from the keyboard (assuming at most 50 scores). Print the scores i
Oksana_A [137]

Answer: Provided in the explanation segment

Explanation:

Below is the code to carry out this program;

/* C++ program helps prompts user to enter the size of the array. To display the array elements, sorts the data from highest to lowest, print the lowest, highest and average value. */

//main.cpp

//include header files

#include<iostream>

#include<iomanip>

using namespace std;

//function prototypes

void displayArray(int arr[], int size);

void selectionSort(int arr[], int size);

int findMax(int arr[], int size);

int findMin(int arr[], int size);

double findAvg(int arr[], int size) ;

//main function

int main()

{

  const int max=50;

  int size;

  int data[max];

  cout<<"Enter # of scores :";

  //Read size

  cin>>size;

  /*Read user data values from user*/

  for(int index=0;index<size;index++)

  {

      cout<<"Score ["<<(index+1)<<"]: ";

      cin>>data[index];

  }

  cout<<"(1) original order"<<endl;

  displayArray(data,size);

  cout<<"(2) sorted from high to low"<<endl;

  selectionSort(data,size);

  displayArray(data,size);

  cout<<"(3) Highest score : ";

  cout<<findMax(data,size)<<endl;

  cout<<"(4) Lowest score : ";

  cout<<findMin(data,size)<<endl;

  cout<<"(5) Lowest scoreAverage score : ";

  cout<<findAvg(data,size)<<endl;

  //pause program on console output

  system("pause");

  return 0;

}

 

/*Function findAvg that takes array and size and returns the average of the array.*/

double findAvg(int arr[], int size)

{

  double total=0;

  for(int index=0;index<size;index++)

  {

      total=total+arr[index];

  }

  return total/size;

}

/*Function that sorts the array from high to low order*/

void selectionSort(int arr[], int size)

{

  int n = size;

  for (int i = 0; i < n-1; i++)

  {

      int minIndex = i;

      for (int j = i+1; j < n; j++)

          if (arr[j] > arr[minIndex])

              minIndex = j;

      int temp = arr[minIndex];

      arr[minIndex] = arr[i];

      arr[i] = temp;

  }

}

/*Function that display the array values */

void displayArray(int arr[], int size)

{

  for(int index=0;index<size;index++)

  {

      cout<<setw(4)<<arr[index];

  }

  cout<<endl;

}

/*Function that finds the maximum array elements */

int findMax(int arr[], int size)

{

  int max=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]>max)

          max=arr[index];

  return max;

}

/*Function that finds the minimum array elements */

int findMin(int arr[], int size)

{

  int min=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]<min)

          min=arr[index];

  return min;

}

cheers i hope this help!!!

8 0
3 years ago
Suppose users share a 1-Gbps link. Also, suppose each user requires 200 Mbps when transmitting, but each user only transmits 30
V125BC [204]

Answer:

The answer is "5 users and 1 block".

Explanation:

In Option a:

Bandwidth total = 1-Gbps \times 1000

                          = 1,000 \ Mbps      

Any User Requirement = 200 \ Mbps

The method for calculating the number of approved users also is:  

Now, calculate the price of each person for overall bandwidth and demands,  

\text{Sponsored user amount} = \frac{\text{Bandwidth total}}{\text{Each user's requirement}}

                                     =\frac{1000}{200}\\\\=\frac{10}{2}\\\\= 5 \ users

In Option b:

\text{blocking probability} = \frac{link}{\text{1-Gbps} = 10^9 \frac{bits}{sec}}

\ let = 0.05 = \frac{100}{20} \\\\\text{blocking probability} = \frac{ 200 \times 10^6}{\frac{100}{20}}

                                = \frac{ 200 \times 10^6 \times 20 }{100}\\\\= \frac{ 2 \times 10^6 \times 20 }{1}\\\\= 40 \times 10^6 \\\\

mean user = 25 \times \frac{1}{20} \\\\

                  = 1.25

max user = \frac{10^9}{40 \times 10^6} \\\\

                = \frac{10^9}{4 \times 10^7} \\\\ = \frac{10^2}{4} \\\\ = \frac{100}{4} \\\\= 25 \\\\ =\ \ 1 \ \ block \\\\

3 0
3 years ago
A digital designer might do computer animation or video games true or false
pav-90 [236]

your answer should be TRUE if i’m correct..

MARK ME BRAINLIEST PLEASE IF IM CORRECT

3 0
3 years ago
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