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vazorg [7]
3 years ago
12

Can u tell me the answers to this table ​

Computers and Technology
1 answer:
Sholpan [36]3 years ago
6 0

: im sorry what dose it say it is too blurry

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Depending on the manufacturer of the bios, what two different post beep code sequences indicate a memory error?
Luda [366]

<span>1.       </span>Continuous short beeps

<span>2.       </span>Continuous 3 short beeps and a pause


Almost every computer company publishes their beep code list according to their user guides and thus, the beeps may vary. For beep codes that belong to AMI BIOS, 1 beep means a DRAM refresh failure. 1 long continuous, 3 short beeps may mean a conventional or extended memory failure.






8 0
4 years ago
Obtain the 10’s complement of the following six-digit decimal numbers:<br><br> 123900<br><br> 980657
zysi [14]

Answer:

876100

019343

Explanation:

10s complement of a decimal number is obtained by the following process:

- Obtain 9s complement ( Subtract each digit by 9)

- Add 1 to the result

1) 123900

9s complement => (9-1)(9-2)(9-3)(9-9)(9-0)(9-0)

= 876099

Adding 1 , 10s complement of 123900 = 876100

2) 980657

9s complement = (9-9)(9-8)(9-0)(9-6)(9-5)(9-7)

= 019342

Adding 1 , 10s complement of 980657 = 019343

7 0
4 years ago
Write a code that calculates the Greatest Common Divisor (GCD) of two positive integers (user-defined inputs). Include an except
STALIN [3.7K]

Answer:

<em>This program is written using Java programming language</em>

import java.util.*;

public class calcGcd

{

   public static void main (String [] args)

   {

       int num1, num2;

       Scanner input = new Scanner(System.in);

       //Input two integers

       num1 = input.nextInt();

       num2 = input.nextInt();

       //Get least of the two integers

       int least = num1;

       if(num1 > num2)

       {

           least = num2;

       }

       //Initialize gcd to 1

       int gcd = 1;

       //Calculate gcd using for loop

       for(int i=1;i<=least;i++)

       {

           if(num1%i == 0 && num2%i == 0)

           {

               gcd = i;

           }

       }

       if(gcd == 1)

       {

           System.out.print("GCD is 1");

       }

       else

       {

           System.out.print("GCD is "+gcd);

       }

   }

}

Explanation:

To calculate the GCD, the program uses a for loop that iterates from 1 to the smaller number of the user input.

Within this iteration, the program checks for a common divisor of the two user inputs by the iterating element

The GCD is then displayed afterwards;

However, if the GCD is 1; the program prints the message "GCD is 1"

<em>Line by Line Explanation</em>

This line declares two integer numbers

       int num1, num2;

This line allows user the program to accept user defined inputs        

Scanner input = new Scanner(System.in);

The next two line allows gets inputs from the user

<em>        num1 = input.nextInt();</em>

<em>        num2 = input.nextInt();</em>

<em />

To calculate the GCD, the smaller of the two numbers is needed. The smaller number is derived using the following if statement

<em>        int least = num1;</em>

<em>        if(num1 > num2)</em>

<em>        {</em>

<em>            least = num2;</em>

<em>        }</em>

The next line initializes GCD to 1

       int gcd = 1;

The GCD is calculated using the following for loop

The GCD is the highest number that can divide both numbers

<em>        for(int i=1;i<=least;i++)</em>

<em>        {</em>

<em>            if(num1%i == 0 && num2%i == 0)</em>

<em>            {</em>

<em>                gcd = i;</em>

<em>            }</em>

<em>        }</em>

The following is printed if the calculated GCD is 1

       if(gcd == 1)

       {

           System.out.print("GCD is 1");

       }

Otherwise, the following is printed

       else

       {

           System.out.print("GCD is "+gcd);

       }

8 0
3 years ago
16. (PPT) You can use features on the Video Tools Playback tab to adjust how and when the video plays during the slide
Nezavi [6.7K]

Answer:

a. True

Explanation:

5 0
2 years ago
Write a function to sum the following series:
Phoenix [80]

Answer:

<u>C program to find the sum of the series( 1/2 + 2/3 + ... + i/i+1)</u>

#include <stdio.h>

double m(int i);//function declaration  

//driver function

int main() {

int i;

printf("Enter number of item in the series-\n");//Taking input from user

scanf("%d",&i);

double a= m(i);//Calling function

printf("sum=%lf",a);

return 0;

}

double m(int i)//Defining function

{

double j,k;

double sum=0;

for(j=1;j<i+1;j++)//Loop for the sum

{

k=j+1;

sum=sum+(j/k);

}

return sum;

}

<u>Output:</u>

Enter number of item in the series-5

sum=3.550000

5 0
3 years ago
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