1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
padilas [110]
3 years ago
10

Discuss the five challenges that limit the use of computer in remote places in Africa​

Computers and Technology
1 answer:
Alchen [17]3 years ago
3 0

The first challenge is that there is not electric conection.So the computer won't be able to function.Also there will be noone who will be able to use it due to the lack of education.Plus there will not be internet connection , limitting the abilities of the computer to the very basics.In addition there will not be any tools or spare parts in order to fix a computer.Finally even if thete were spare parts there woul not be someone able to gix the computer.

I hop I was helpful.Let me know if you need anything else.

You might be interested in
Write a C++ program that reads students' names followed by their test scores. The program should output each students' name foll
Mashutka [201]

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

struct studentdata{

char Fname[50];

char Lname[50];

int marks;

char grade;

};

main()

{

studentdata s[20];

for (int i=0; i<20;i++)

   {

cout<<"\nenter the First name of student";

cin>>s[i].Fname;

cout<<"\nenter the Last name of student";

cin>>s[i].Lname;

cout<<"\nenter the marks of student";

cin>>s[i].marks;

}  

 

for (int j=0;j<20;j++)

{

if (s[j].marks>90)

{

 s[j].grade ='A';

}

else if (s[j].marks>75 && s[j].marks<90)

{

   s[j].grade ='B';

}

else if (s[j].marks>60 && s[j].marks<75)

{

 s[j].grade ='C';

}

else

{

 s[j].grade ='F';

}

}

int highest=0;

int z=0;

for (int k=0;k<20; k++)  

{

if (highest<s[k].marks)

{

 highest = s[k].marks;

 z=k;

}

 

}

cout<<"\nStudents having highest marks"<<endl;

 

cout<<"Student Name"<< s[z].Fname<<s[z].Lname<<endl;

cout<<"Marks"<<s[z].marks<<endl;

cout<<"Grade"<<s[z].grade;

getch();  

}

Explanation:

This program is used to enter the information of 20 students that includes, their first name, last name and marks obtained out of 100.

The program will compute the grades of the students that may be A,B, C, and F. If marks are greater than 90, grade is A, If marks are greater than 75 and less than 90, grade is B. For Mark from 60 to 75 the grade is C and below 60 grade is F.

The program will further found the highest marks and than display the First name, last name, marks and grade of student who have highest marks.

6 0
3 years ago
A flat panel detector’s detector element size determines what?
Nastasia [14]

Answer:

A flat panel detector’s detector element size determines Spatial resolution.

7 0
3 years ago
Define a python functions that draw a pinwheel using the pinwheel function that has input paramiters that allows you to change t
QveST [7]

Start with your draw_pinwheel() function

During each iteration of the loop decide which color to set the turtle to using the color() function

On even iterations, use color1

On odd iterations, use color2

Use an if/else statement to do the decision making

After deciding the color, surround a call to draw_triangle() with begin_fill() and end_fill() so that drawing the triangle creates a colored triangle.

If you have have forgotten, you can use an if/else to check for even/oddness in python as follows.

my_number = 3

if(my_number % 2 == 0):     # the remainder of dividing by 2 is 0

   print("The number is "+str(my_number))

   print("The number is EVEN")

else:                       # the remainder must be 1

   print("The number is "+str(my_number))

   print("The number is ODD")

Good luck <3

8 0
3 years ago
Competition between individuals in a population is due to (2 points)
madam [21]
Competition between individuals in a population is due to (2 points)
limited resources
5 0
4 years ago
Write a C program to multiply matrix A and matrix B of variable. For both A and B, the size of the matrix will be given along wi
victus00 [196]

Answer:

<u><em>Code:</em></u>

    #include <iostream>

    #include <fstream>

    #include <cctype>

    #include <cmath>

    #include <vector>

    #include <float.h>

   

    using namespace std;

   

    typedef vector<vector<double> > C2DArray;

   

    void Matrix_Mult(bool* error_flag, const int mRows, const int nCols, const int pCols, const C2DArray A_Matrix, const C2DArray B_Matrix, C2DArray& C_Matrix) {      

     int i = A_Matrix.size(), j = A_Matrix[0].size(), k = B_Matrix[0].size();

     double dummy;

     if ((j != B_Matrix.size()) || (i != mRows) || (j != nCols) || (k != pCols))  *error_flag = true;

   

     if ((C_Matrix.size() != i) || (k != C_Matrix[0].size()))  *error_flag = true;

   

     if (!(*error_flag)){

     for (k = 0; k < mRows; k++) { // Do each row of A with each column of B

      for (i = 0; i < pCols; i++) {

       dummy = 0.0;

       for (j = 0; j < nCols; j++) {

        dummy += A_Matrix[k][j]*B_Matrix[j][i];

          } // End for j

          C_Matrix[k][i] = dummy;

      } // End for i

     } // End for k

     }

   

        return;

    } // End Matrix_Mult

   

    int main()

    {char rflag = 0; //Readiness flag  

   

    cout << "                 Mat_Mul1 (7 June 2013)\n";

    cout << "=========================================================================== \n";

    cout << "This program multiplies two matrices, [A] and [B]: [A][B] = [C].\n";

    cout << "[A] is M X N, [B] is N X P, and [C] is M X P\n";

    cout << "The dimensions of the [A] and [B] matrices, M, N, and P, should \n";

    cout << "have been saved beforehand in a file named mulmatdat.\n";

    cout << "mulmatdat should be in the same folder as the Mat_Mul1 executable.\n";

    cout << "The next values of mulmatdat should be the entries for matrix [A],\n";

    cout << "with data for row 1 first, then row 2, then row 3, etc.\n";

    cout << "The entries for the [B] matrix should follow, in the same order.\n";

    cout << "\nThe data is assumed to be of type double. Variables used within this program\n";

    cout << "are type double.\n";

    cout << "\nThe output is written to the file mulmatout.txt.\n";

   

    cout << "\nIs everything ready (are you ready to continue?)? If yes, Enter y. \n";

    cout << "Otherwise Enter any other key. \n";

    cin >> rflag;

   

    if (toupper(rflag) == 'Y') {

    C2DArray A, B, C; // A, B, and C Matrices

    int i, j, mDim, nDim, pDim; // Array index variables and matrix dimensions

    bool erFlag = false;  // Error flag

   

    ifstream in("mulmatdat.txt", ios::in);

   

    if (!in) {

            cout << "\nCannot open the input file.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            return 0;

    }

   

    in >> mDim >> nDim >> pDim; //Input the Matrices' dimensions from the file

     

    if ((mDim < 1) || (nDim < 1) || (pDim < 1)){

            cout << "\nInvalid dimension entered. Program terminated. \n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

   

    ofstream out("mulmatout.txt", ios::out);

    if (!out) {

            cout << "\nCannot open the output file. Program terminated.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

   

    // Beginning of try block, if vector re-sizing unsuccessful

    try {

   

           // Resize the arrays to the appropriate sizes

            A.resize(mDim); // M rows

         B.resize(nDim); // N rows

         C.resize(mDim); // M rows

            for (i = 0; i < mDim; i++){

               A[i].resize(nDim); // M columns

            C[i].resize(pDim); // P columns

            } // End for i

   

         for (i = 0; i < nDim; i++){

               B[i].resize(pDim); // P columns

            } // End for i

   

    } // End of try block

   

    catch (bad_alloc& xa) { // Catch block, for exceptions

   

            in.close();

            out.close();

            cerr << "\nIn catch block, so an exception occurred: " << xa.what() << "\n";

            cout << "\nEnter any key to continue. \n";

            cin >> rflag;

            return 0;

   

    } // End of catch block

   

    for (i = 0; i < mDim; i++){ //Input the A Matrix from the file

            for (j = 0; j < nDim; j++){

              in >> A[i][j];

            }//End for j

    }//End for i

   

    for (i = 0; i < nDim; i++){ //Input the B Matrix from the file

            for (j = 0; j < pDim; j++){

             in >> B[i][j];

            }//End for j

    }//End for i

   

    in.close(); //Close the input file

   

    Matrix_Mult(&erFlag, mDim, nDim, pDim, A, B, C);

   

    if (erFlag){

      cout << "Inconsistent data sent to Matrix_Mult routine. Matrix multiplication NOT performed. Check data before running again.\n";

    }

    else {

   

     out.precision(DBL_DIG);

   

     out << "\nThe A matrix follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < nDim; j++){

     out << A[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

   

     out << "\nThe B matrix follows:\n";

     out << "\n";

     for (i = 0; i < nDim; i++){

      for (j = 0; j < pDim; j++){

     out << B[i][j] << " ";

      } // End for j

      out << "\n";

    }//End for i

   

     out << "\nThe product matrix, C, follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < pDim; j++){

       out << C[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

   

     cout << "\nDone! The solution is in the text file mulmatout.txt \n";

    } // End else !erFlag

    out.close();

    }

    else cout << "\nNot ready. Try again when ready with information. \n";

    cout << "\nEnter any key to continue. \n";

    cin >> rflag;

    return 0;

    }

<u><em></em></u>

4 0
4 years ago
Other questions:
  • An LPN/LVN has transferred to a nursing unit and arrives for the first day. The RN checks with the LPN/LVN often throughout the
    11·1 answer
  • How do rubrics increase learning?
    15·2 answers
  • Thomas is using a drawing program to complete a construction. Which construction could he be completing? (5 points).
    5·2 answers
  • In PHP, the only way to output text is with echo.
    10·1 answer
  • In the following data definition, assume that List2 begins at offset 2000h. What is the offset of the third value (5)?
    10·1 answer
  • Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards
    10·1 answer
  • A developer has been asked to create code that will meet the following requirements: Receives input of: Map, List Performs a pot
    13·1 answer
  • Which of the following game markets especially benefits from using vector graphics?
    8·1 answer
  • Why is it important to ensure that DNS Servers have been secured before implementing an e-mail system
    7·1 answer
  • In a computer-controlled greenhouse, a temperature sensor and a window motor are connected to the computer.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!