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4vir4ik [10]
3 years ago
5

Assume you have a byte-addressable machine that uses 32-bit word (integers are 32-bit) and you are storing the following informa

tion: the string Inventory followed by a value (569)10 followed by another string Price followed by a value (49.75)10. Also, assume that floating point numbers are stored as single-precision IEEE 754 with 127 excess. a) Show (as hex numbers with word boundary) how this is stored on a big-endian machine. b) Show (as hex numbers with word boundary) how this is stored on a little-endian machine.
Computers and Technology
1 answer:
Tresset [83]3 years ago
7 0

Answer:

Internet protocol, IP address are 32- bit binary numbers.

Explanation: Each 32- bit is represented in 4- 8 bits segments

a) 10,11,13,14,  0,0,10,13,  0,11,12,13,  11,12,13,14,  0,0,0,11,  0,0,13,14,  10,12,13,14, O,10,13,14

b)

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tatiyna

Answer:

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3 years ago
Helen is filling her persönal details on her résumé. Which detail should she include in this section?
svetoff [14.1K]

I would think option A.

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4 0
3 years ago
write a java program that prompts the user to enter scores (each number an integer from 0 to 10) and prints the following output
Vladimir [108]

Answer:

Explanation:

The following is written in Java. It continues asking the user for inputs until they enter a -1. Then it saves all the values into an array and calculates the number of values entered, the highest, and lowest, and prints all the variables to the screen. The code was tested and the output can be seen in the attached image below.

import java.util.ArrayList;

import java.util.Arrays;

import java.util.Scanner;

class Brainly {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       int count = 0;

       int highest, lowest;

       ArrayList<Integer> myArr = new ArrayList<>();

       while (true) {

           System.out.println("Enter a number [0-10] or -1 to exit");

           int num = in.nextInt();

           if (num != -1) {

               if ((num >= 0) && (num <= 10)) {

                   count+= 1;

                   myArr.add(num);

               } else {

                   System.out.println("Wrong Value");

               }

           } else {

               break;

           }

       }

       if (myArr.size() > 3) {

           highest = myArr.get(0);

           lowest = myArr.get(0);

           for (int x: myArr) {

               if (x > highest) {

                   highest = x;

               }

               if (x < lowest) {

                   lowest = x;

               }

           }

           System.out.println("Number of Elements: " + count);

           System.out.println("Highest: " + highest);

           System.out.println("Lowest : " + lowest);

       } else {

           System.out.println("Number of Elements: " + count);

           System.out.println("No Highest or Lowest Elements");

       }

   }

}

4 0
2 years ago
Which header will be the largest?<br><br> Hello<br><br> Hello<br><br> Hello<br><br> Hello
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Answer:  

Bye

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Explanation:

4 0
3 years ago
Read 2 more answers
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balandron [24]

Answer:

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Explanation:

<h3>#CarryOnLearning</h3>

{hope it helps}}

4 0
2 years ago
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