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frosja888 [35]
2 years ago
7

If a machine cycle is 2 nanoseconds , how many machine cycles occur each second?

Computers and Technology
1 answer:
worty [1.4K]2 years ago
7 0

Answer: 5 × 10^8 cycles per second

Explanation:

First and foremost, we should note that 1 nanosecond = 1 × 10^-9 seconds

We are told that the machine cycle is 2 nanoseconds.

The number of machine cycles that occur each second will them be calculated as:

1 = 2 × 10^-9

= 5 × 10^8 cycles per second

The number of machine cycles that occur each second is 5 × 10^8 cycles per second.

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Select the correct answer. Which input device uses optical technology?
daser333 [38]

Answer:

barcode reader is the correct answer

8 0
2 years ago
Two forms of compression are lossy and lossless. State giving reasons which
Helen [10]

Answer:

(i) When transmitting a draft manuscript for a book, the lossless compression technique is most suitable because after decompression, the data is rebuilt and restored in its form as it was from where it originated

(ii) When transmitting a video recording which you have made of the school play, a lossy compression technique is most suitable because the large size of video files require the increased data carrying capacity which is provided by the lossy transmission technique. The quality of the video can be reduced without affecting the message intended to be delivered

Explanation:

4 0
2 years ago
The value of the mathematical constant e can be expressed as an infinite series: e=1+1/1!+1/2!+1/3!+... Write a program that app
LUCKY_DIMON [66]

Answer:

// here is code in c++ to find the approx value of "e".

#include <bits/stdc++.h>

using namespace std;

// function to find factorial of a number

double fact(int n){

double f =1.0;

// if n=0 then return 1

if(n==0)

return 1;

for(int a=1;a<=n;++a)

       f = f *a;

// return the factorial of number

return f;

}

// driver function

int main()

{

// variable

int n;

double sum=0;

cout<<"enter n:";

// read the value of n

cin>>n;

// Calculate the sum of the series

  for (int x = 0; x <= n; x++)

  {

     sum += 1.0/fact(x);

  }

  // print the approx value of "e"

    cout<<"Approx Value of e is: "<<sum<<endl;

return 0;

}

Explanation:

Read the value of "n" from user. Declare and initialize variable "sum" to store the sum of series.Create a function to Calculate the factorial of a given number. Calculate the sum of all the term of the series 1+1/1!+1/2!.....+1/n!.This will be the approx value of "e".

Output:

enter n:12

Approx Value of e is: 2.71828

6 0
3 years ago
Write a program that reads in an integer value for n and then sums the integers from n to 2 * n if n is nonnegative, or from 2 *
devlian [24]

Answer:

Following are the program in c++

First code when using for loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

for(i=n; i<=(2*n); i++)

{

sum1+= i;

}

}

else //  check if n  is negative

{

for(i=(2*n); i<=n; i++)

{

sum1+= i;

}

}

cout<<"The sum is:"<<sum1;

return 0;

Output

Enter the number: 1

The sum is:3

second code when using while loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

else //  check if n  is negative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

cout<<"The sum is:"<<sum1;

return 0;

}

Output

Enter the number: 1

The sum is:3

Explanation:

Here we making 2 program by using different loops but using same logic

Taking a user input in "n" variable.

if n is nonnegative then we iterate the loops  n to 2 * n and storing the sum in "sum1" variable.

Otherwise iterate a loop  2 * n to n and storing the sum in "sum1" variable.  Finally display  sum1 variable .

4 0
3 years ago
What ""old fashioned"" features of checking accounts is p2p replacing
balandron [24]

Answer:

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Explanation:

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If only one answer is allowed then I would lean towards "ATM Cash Withdrawls".

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