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vivado [14]
2 years ago
10

What are the first and the last physical memory addressesaccessible using

Computers and Technology
1 answer:
Misha Larkins [42]2 years ago
6 0

Answer

For First physical memory address,we add 00000 in segment values.

For Last physical memory address,we add 0FFFF in segment values.

<u>NOTE</u>-For addition of hexadecimal numbers ,you first have to convert it into binary then add them,after this convert back it in hexadecimal.

a)1000

For First physical memory address, we add 00000 in segment value

We add 0 at the least significant bit while calculating.

          1000<u>0</u> +00000 = 1000<u>0</u> (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1000<u>0</u> +0FFFF =1FFFF    (from note)

b)0FFF

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0FFF<u>0</u> +00000=0FFF0 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0FFF<u>0</u> +0FFFF =1FFEF (from note)

c)0001

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0001<u>0</u> +00000=00010 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0001<u>0</u> +0FFFF =1000F (from note)

d) E000

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    E000<u>0</u> +00000=E0000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           E000<u>0</u> +0FFFF =EFFFF (from note)

e) 1002

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    1002<u>0</u> +00000=10020  (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1002<u>0</u> +0FFFF =2001F (from note)

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Object-Oriented Programming (Using Java Language)
Delicious77 [7]

Answer:

import java.util.Scanner;

class Main {  

 public static void main(String args[]) {

       Scanner scan = new Scanner(System.in);

       System.out.print("Enter a decimal value (0 to 15): ");

       int num = scan.nextInt();

       scan.close();

       

       if (num < 0 || num >15) {

           System.out.printf("%d is an invalid input\n", num);

       } else {

           System.out.printf("The hex value is %X\n", num);

       }

 }

}

Explanation:

Hopefully this example will get you going for the other assignments.

3 0
2 years ago
Is recursion ever required to solve a problem? What other approach can you use to solve a problem that is repetitive in nature?
Setler79 [48]

Answer:

No you can not tell that recursion is ever required to solve a problem.

Recursion is required when in the problem, the solution of the input depends on the solution of the subsets of the input.

Iteration is also another form of repetitive approach we follow to solve that kind of problems.

But the difference between recursion and iteration is :

  • In recursion we call the function repeatedly to return the result to next level.
  • In iteration certain bunch of instructions in a loop are executed until certain conditions met.

Explanation:

For example in the Fibonacci sequence problem, to find f_{n}, we need to compute f_{n-1} and f_{n-2} before that.

  • In case of recursion we just call the method Fibonacci(n) repeatedly only changing the parameter Fibonacci(n-1), that calculates the value and return it.

                           Fibonacci(n)

                           1. if(n==0 or n==1)

                           2.       return 1.

                           3.else

                           4.      return( Fibonacci(n-1)+Fibonacci(n-1) )

  • But in case of iteration we run a loop for i=2 to n, within which we add the value of current f_{i-1} and f_{i-2} to find the value of f_{i}

                           Fibonacci(n)

                           1. if(n<=2)

                           2.    result = 1

                           3.     else

                           4. result1 =1 and result2=1.

                           5.      {  result = result1 +result2.

                           6.         result1= result2.

                           7.         result2 = result.

                           8.      }

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8 0
3 years ago
Working with text in presentation programs is similar to using text in other applications. True False
Yuliya22 [10]
I would say true because text is really important so you eant to make it look as good as possible.
8 0
3 years ago
Create an application containing an array that stores eight integers. The application should call five methods that in turn (1)
Butoxors [25]

Answer:

package b4;

public class ArrayMethodDemo {

public static void main(String[] args) {

 // Create an array to store 8 random integers.

 int[] integernumbers = { 3, 4, 6, 9, 5, 6, 7, 2 };

 // Call the display method to show the elements in the array.

 display(integernumbers);

 // Call the method to display the elements in reverse order.

 displayReverse(integernumbers);

 // Call the method to find the sum of the elements.

 sum(integernumbers);

 // Call the method to display all elements less than a limiting value, say 5.

 lessThan(integernumbers, 5);

 // Call the method to display all elements greater than the average of the array

 // elements.

 higherThan(integernumbers);

}

// Method to display the elements in the array.

// The method receives only a single argument which is the array.

// Loop through the array and at every cycle, print out each element

public static void display(int[] arr) {

 System.out.print("All integers in the array : ");

 for (int i = 0; i < arr.length; i++) {

  System.out.print(arr[i] + " ");

 }

 System.out.println();

}

// Method to display the elements in the array in reverse order

// The method receives only a single parameter which is the array

// Loop through the array starting at the last index

// At every cycle, print out the elements in the array

public static void displayReverse(int[] arr) {

 System.out.print("All integers in the array in reverse : ");

 for (int i = arr.length - 1; i >= 0; i--) {

  System.out.print(arr[i] + " ");

 }

 System.out.println();

}

// Method to print out the sum of the elements in the array.

// The method receives only a single parameter which is the array.

// Declare a variable called sum to hold the sum of the elements

// Initialize sum to zero

// Loop through the array and cumulatively add each element to sum

// Print out the sum at the end of the loop

public static void sum(int[] arr) {

 int sum = 0;

 for (int i = 0; i < arr.length; i++) {

  sum += arr[i];

 }

 System.out.println("The sum of the integers in the array is " + sum);

}

// Method to print all values in the array that are less than a limiting

// argument.

// The method has two parameters - the array and the limiting argument.

// Loop through the array and at every cycle,

// check if the current array element is less than the limiting argument.

// If it is, print it out. Else continue

public static void lessThan(int[] arr, int limitingargument) {

 System.out.print("All values less than the limiting argument (" + limitingargument + ") are: ");

 for (int i = 0; i < arr.length; i++) {

  if (arr[i] < limitingargument) {

   System.out.print(arr[i] + " ");

  }

 }

 System.out.println();

}

// Method to print all values in the array that are higher than the average of

// the array.

// The method has one parameter - the array.

// First, calculate the average of the array elements.

// Loop through the array and at every cycle,

// check if the current array element is greater than the average.

// If it is, print it out. Else continue

public static void higherThan(int[] arr) {

 int sum = 0;

 for (int i = 0; i < arr.length; i++) {

  sum += arr[i];

 }

 double average = sum / arr.length;

 System.out.print("All values higher than the calculate average (" + average + ") is ");

 for (int i = 0; i < arr.length; i++) {

  if (arr[i] > average) {

   System.out.print(arr[i] + " ");

  }

 }

 System.out.println();

}

}

Explanation:

The program has been written in Java.

Please go through the comments in the code for explanation.

The source code has also been attached to this response.

Hope this helps!

Download java
3 0
3 years ago
Uses computer resources from multiple locations to solve a common problem.
Lynna [10]

Answer:

Distributed Computing is used to solve common problem from multiple locations at different locations.

Explanation:

It is the architecture of processor, that is used to combine the different resources of computer from different domain to solve the common problem.

It allows us resource sharing such as hardware and software that are connected to the computer.

Examples

Following are the examples of Distributed computing.

  1. Telecommunication Network
  2. sensor network
  3. banking network

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