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mixer [17]
3 years ago
15

The __________ endian storage format places the __________ byte of a word in the lowest memory address. The __________ endian st

orage format places the __________ byte of a word in the highest memory address.
Computers and Technology
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

i. Big

ii. most significant

iii. little

iv. most significant

Explanation:

You might be interested in
In JAVA please:
blagie [28]

Answer:

import java.util.Scanner;

public class ArraysKeyValue {

public static void main (String [] args) {

final int SIZE_LIST = 4;

int[] keysList = new int[SIZE_LIST];

int[] itemsList = new int[SIZE_LIST];

int i;

keysList[0] = 13;

keysList[1] = 47;

keysList[2] = 71;

keysList[3] = 59;

itemsList[0] = 12;

itemsList[1] = 36;

itemsList[2] = 72;

itemsList[3] = 54;  

/* Your solution goes here */

for ( i = 0; i < SIZE_LIST; i++){

 if (keysList[i]>50){

  System.out.println(itemsList[i] + " ");  }  }

System.out.println("");

}

}

Explanation:

I will explain the whole program flow.

  • There are two arrays here
  • keysList and itemsList
  • The first list (keysList) contains the following elements:

13 element at first position of the array (0th index)

47 element at second  position of the array (1st index)

71 element at third position of the array (2nd index)

59 element at fourth position of the array (3rd index)

  • The other list (itemsList) contains the following elements:

12 element at first position of the array (0th index)

36 element at second  position of the array (1st index)

72 element at third position of the array (2nd index)

54 element at fourth position of the array (3rd index)

  • The size of the array elements is fixed which is 4 and is stored in the variable SIZE_LIST.
  • Then the loop starts. The loop contains a variable i which is initialized to 0. First it checks if the value of i is less than the size of the list. It is true as SIZE_LIST=4 and i=0.
  • So the program control enters the body of the loop.
  • In first iteration, IF condition checks if the i-th element of the keysList is greater than 50. As i=0 So the element at 0th index of the keysList is 13 which is not greater than 50 so the body of IF statement will not execute as the condition evaluates to false. The value of i increments by 1 so now i becomes 1.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=1 So the element at 1st index of the keysList is 47 which is not greater than 50 so the body of IF statement will not execute as the condition evaluates to false. The value of i is incremented by 1 so now i becomes 2.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again as i= 2 which is less than SIZE_LIST so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=2 So the element at 2nd index of the keysList is 71 which is greater than 50 so the body of IF statement is executed as the condition evaluates to true. So in the body of the IF statement there is a print statement which prints the i-th element of the itemsList. As i = 2 so the value at the index 2 of the itemsList is displayed in the output which is 72. Next value of i is incremented by 1 so now i becomes 3.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again as i= 3 which is less than SIZE_LIST so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=3 So the element at 3rd index of the keysList is 59 which is greater than 50 so the body of IF statement is executed as the condition evaluates to true. So in the body of the IF statement there is a print statement which prints the i-th element of the itemsList. As i = 3 so the value at the index 3 of the itemsList is displayed in the output which is 54. Next value of i is incremented by 1 so now i becomes 4.
  • In next iteration loop again checks if the value of i is less than the size of the list which is now false as i=4 which is equal to the SIZE_LIST= 4. So the loop breaks.
  • So the output of the above program is:

72

54

5 0
3 years ago
Phases of systems development life cycle do developers identify the particular features nd functions of a new system
ki77a [65]

The phase where developers identify the particular features and functions of a new system is the requirement and analysis stage.

<h3>What is system development?</h3>

System development is the process of defining, designing, testing, and implementing a new software application or program.

System development has a life cycle. The primary stages are as follows;

  • Planning stage
  • Requirement and Analysis
  • Design
  • Development
  • Testing
  • Implementation.
  • Operation and maintenance.

Therefore, the phase where developers identify the particular features and functions of a new system is the requirement and analysis stage.

learn more on system development here: brainly.com/question/13042526

#SPJ12

7 0
2 years ago
Read 2 more answers
What will be displayed if code corresponding to the following pseudocode is executed? Set Number = 4 Repeat Write 2 * Number Set
Elis [28]

Answer:

8

12

Explanation:

I made the code a bit easier to understand then worked out how it would go. Here's what I did.

number = 4

repeat until number = 8:

   write 2 * number

   number = number + 2

Following this itenary, we have, the system first writes "8" as it multipled 4 by 2. Number is now equal to 6.

Next repetition, the system writes "12" as it multipled 6 by 2. Now, number = 8. The proccess now stops as number is now equal to 8.

5 0
3 years ago
Convert the following Base 10 (decimal) numbers to base 2(binary):<br> 107<br> 200
melamori03 [73]

<u>Answer:</u>

107₁₀ - 1101011

₂ (Binary representation)

200₁₀- 11001000₂ (Binary representation)

<u>Explanation:</u>

Converting from Decimal to binary:

Procedure -

1. Divide the number by 2, write the reminder separately.

2. Divide the divisor by 2, write the reminder before the previously written reminders.

Keep doing this till you get your divisor as 1.

Then we will write the divisor  before the written reminders and that will be the binary representation.

   

 For 107 :

       

          We divide 107 by 2 ,we get the divisor 53 and remainder 1

                                                  ( 1 )

Then,we will divide this divisor i.e 53 by 2,we get divisor 26 and remainder 1 ,we will put this remainder before the previous one.

                                                ( 1 1 )      

Then,we will divide this divisor i.e 26 by 2,we get divisor 13 and remainder 0 ,we will put this remainder before the previous one.    

                                                 ( 0 1 1 )

Then,we will divide this divisor i.e 13 by 2,we get divisor 6 and remainder 1 ,we will put this remainder before the previous one.    

                                                 ( 1 0 1 1 )

Then,we will divide this divisor i.e 6 by 2,we get divisor 3 and remainder 0,we will put this remainder before the previous one.    

                                                 ( 0 1 0 1 1 )

Then,we will divide this divisor i.e 3 by 2,we get divisor 1 and remainder 1  ,we will put this remainder before the previous one.    

                                                 ( 1 0 1 0 1 1 )

As we get the divisor=1,then we will stop.we will write the divisor before the written reminders.

                                              ( 1 1 0 1 0 1 1 )

This will be the binary representation of 107.

For 200 :

         We divide 200 by 2 ,we get the divisor 100 and remainder 0

                                                  ( 0 )

Then,we will divide this divisor i.e 100 by 2,we get divisor 50 and remainder 0 ,we will put this remainder before the previous one.

                                                ( 0 0 )      

Then,we will divide this divisor i.e 50 by 2,we get divisor 25 and remainder 0 ,we will put this remainder before the previous one.    

                                                 ( 0 0 0 )

Then,we will divide this divisor i.e 25 by 2,we get divisor 12 and remainder 1 ,we will put this remainder before the previous one.    

                                                 ( 1 0 0 0 )

Then,we will divide this divisor i.e 12 by 2,we get divisor 6 and remainder 0 ,we will put this remainder before the previous one.    

                                                 ( 0 1 0 0 0 )

Then,we will divide this divisor i.e 6 by 2,we get divisor 3 and remainder 0  ,we will put this remainder before the previous one.    

                                                 ( 0 0 1 0 0 0 )

Then,we will divide this divisor i.e 3 by 2,we get divisor 1 and remainder 1,we will put this remainder before the previous one.    

                                                 ( 1 0 0 1 0 0 0 )

As we get the divisor=1,then we will stop.we will write the divisor before the written reminders.

                                                 ( 1 1 0 0 1 0 0 0 )

This will be the binary representation of 200.

7 0
3 years ago
Big data are used to _____. Select 3 options.
Anon25 [30]

Answer:

Big data are used to _____. Select 3 options.

understand people’s hobbies

analyze people’s interests

make decisions about marketing

Explanation:

100%

8 0
3 years ago
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