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malfutka [58]
2 years ago
9

suppose that you have been given the task of writing an unloader - that is, a piece of software that can take the image of a pro

gram that has been loaded and write out an object program that could later be loaded and executed. the computer system uses a relocating loader, so the object program you produce must be capable of being loaded at a location in memory that is different from where your unloader took it. what problems do you see that would prevent you from accomplishing this task?
Computers and Technology
1 answer:
Brums [2.3K]2 years ago
8 0
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(copied from google)
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When workers demonstrate patience, are able to manage there emotions, and get along with other employees, which skills are being
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Program MATH_SCORES: Your math instructor gives three tests worth 50 points each. You can drop one of the test scores. The final
Simora [160]

Answer:

import java.util.Scanner;

public class num5 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       //Prompt and receive the three Scores

       int score1;

       int score2;

       int score3;

       do {

           System.out.println("Enter first Score Enter score between 1 -50");

           score1 = in.nextInt();

       } while(score1>50 || score1<0);

       do {

           System.out.println("Enter second Score.The second score must be between 1 -50");

           score2 = in.nextInt();

       } while(score2>50 || score2<0);

       do {

           System.out.println("Enter Third Score Third score must between 1 -50");

           score3 = in.nextInt();

       } while(score3>50 || score3<0);

       //Find the minimum of the three to drop

       int min, min2, max;

       if(score1<score2 && score1<score3){

           min = score1;

           min2 = score2;

           max = score3;

       }

       else if(score2 < score1 && score2<score3){

           min = score2;

           min2 = score1;

           max = score3;

       }

       else{

           min = score3;

           min2 = score1;

           max = score2;

       }

       System.out.println("your entered "+max+", "+min2+" and "+min+" the min is");

       int total = max+min2;

       System.out.println("Total of the two highest is "+total);

       //Finding the grade based on the cut-off points given

       if(total>=90){

           System.out.println("Grade is A");

       }

       else if(total>=80){

           System.out.println("Grade is B");

       }

       else if(total>=70){

           System.out.println("Grade is C");

       }

       else if(total>=60){

           System.out.println("Grade is D");

       }

       else{

           System.out.println("Grade is F");

       }

   }

}

Explanation:

  • Implemented with Java
  • Use the scanner class to receive user input
  • Use a do.....while loop to validate user input for each of the variables. A valid score must be between 0 and 50 while(score>50 || score<0);  
  • Use if and else to find the minimum of the three values and drop
  • Add the two highest numbers
  • use if/else if /else statements to print the corresponding grade
8 0
2 years ago
A computer has 4 GB of RAM of which the operating system occupies 512 MB. The processes are all 256 MB (for simplicity) and have
Afina-wow [57]

Answer:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

Explanation:

Previous concepts

Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"

Solution to the problem

For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:

Available = 4096 MB - 512 MB = 3584 MB

Now we can find the maximum simultaneous process than can use with this:

\frac{3584 MB}{256 MB/proc}= 14 processes

And then we can find the maximum wait I/O that can be tolerated with the following formula:

1- p^{14}= rate

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

3 0
2 years ago
Write a function PrintShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, pr
Finger [1]

Answer:

public static void PrintShampooInstructions(int numberOfCycles){

       if (numberOfCycles<1){

           System.out.println("Too Few");

       }

       else if(numberOfCycles>4){

           System.out.println("Too many");

       }

       else

           for(int i = 1; i<=numberOfCycles; i++){

               System.out.println(i +": Lather and rinse");

           }

       System.out.println("Done");

   }

Explanation:

I have used Java Programming language to solve this

Use if...elseif and else statement to determine and print "Too Few" or "Too Many".

If within range use a for loop to print the number of times

8 0
2 years ago
Read 2 more answers
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