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anygoal [31]
3 years ago
6

Assume your computer clock runs at the speed of 2.5 GHz (G=10^9).

Computers and Technology
1 answer:
azamat3 years ago
7 0

Answer:

a) T=4*10^{-4}μs

b) N=2.5*10^6 cycles

c) 10000 programs.

Explanation:

a) We know that the frequency is the inverse of the period, so:

f=\frac{1}{T}\\\\T=\frac{1}{f}\\T=4*10^{-10}s

1μs is equal to 1*10^{-6}s

so T=4*10^{-4}us

b) If in a second there are 2.5*10^9 cycles:

N=2.5*10^9*(1*10^{-3})=2.5*10^6 cycles

c) we have to make a conversion, we know that a program takes 100*10^(-3) milliseconds, that is, 1*10^(-4) seconds so in 1 second we can execute:

P=\frac{1s}{1*10^{-4}s}=10000

10000 programs.

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3 years ago
Write a program that will open the file random.txt and calculate and display the following: A. The number of numbers in the file
Rama09 [41]

Answer:

Here is the C++ program:

#include <iostream>  //to use input output functions

#include <fstream>  //to manipulate files

using namespace std;  //to identify objects like cin cout

int main(){  //start of main function

  ifstream file;   //creates an object of ifstream

   file.open("random.txt"); //open method to open random.txt file using object file of ifstream

       

    int numCount = 0;  //to store the number of all numbers in the file          

    double sum = 0;   //to store the sum of all numbers in the file

    double average = 0.0;   //to store the average of all numbers in the file

    int number ; //stores numbers in a file

                 

        if(!file){   //if file could not be opened

           cout<<"Error opening file!\n";    }   //displays this error message

       

       while(file>>number){   //reads each number from the file till the end of the file and stores into number variable

           numCount++; //adds 1 to the count of numCount each time a number is read from the file          

           sum += number;  }  //adds all the numbers and stores the result in sum variable

           average = sum/numCount;  //divides the computed sum of all numbers by the number of numbers in the file

     

       cout<<"The number of numbers in the file: "<<numCount<<endl;  //displays the number of numbers

       cout<<"The sum of all the numbers in the file: "<<sum<<endl;  //displays the sum of all numbers

       cout<<"The average of all the numbers in the file: "<< average<<endl;  //displays the average of all numbers

       file.close();     }  //closes the file    

   

Explanation:

Since the random.txt is not given to check the working of the above program, random.txt is created and some numbers are added to it:

35

48

21

56

74

93

88

109

150

16

while(file>>number) statement reads each number and stores it into number variable.

At first iteration:

35 is read and stored to number

numCount++;  becomes

numCount = numCount + 1

numCount = 1      

sum += number; this becomes:

sum = sum + number

sum = 0 + 35

sum = 35

At second iteration:

48 is read and stored to number

numCount++;  becomes

numCount = 1+ 1

numCount = 2    

sum += number; this becomes:

sum = sum + number

sum = 35 + 48

sum = 83

So at each iteration a number is read from file, the numCount increments to 1 at each iteration and the number is added to the sum.

At last iteration:

16 is read and stored to number

numCount++;  becomes

numCount = 9 + 1

numCount = 10    

sum += number; this becomes:

sum = sum + number

sum = 674 + 16

sum = 690

Now the loop breaks and the program moves to the statement:

       average = sum/numCount;  this becomes:

       average = 690/10;

       average = 69

So the entire output of the program is:

The number of numbers in the file: 10                                                                                                           The sum of all the numbers in the file: 690                                                                                                     The average of all the numbers in the file: 69

The screenshot of the program and its output is attached.

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Answer:

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