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arsen [322]
3 years ago
9

Write steps: To open an old documents​

Computers and Technology
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

1.Click.

2.Choose “All Programs”

3.Then select, the “Microsoft Office” folder.

4.Now, open your desired Office application. 5.for e.g Microsoft Word 2010.

6.Once the application opens, select.

Click.

7.Now, select the document you wish to open in Office 2010 and click.

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What is the size of the key space if all 8 characters are randomly chosen 8-bit ascii characters?
Slav-nsk [51]
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4 years ago
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
how many bits must be flipped (i.e. changed from 0 to 1 or from 1 to 0) in order to capitalize a lowercase 'a' that’s represente
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Lowercase a is decimal 97 ; upper case is decimal 65

It's easier to think of them in octal, however: a = octal 141, and A is octal 101

octal to binary is easy, each digit is three bits. 
141 = 001 100 001 
101 = 001 000 001

So, how many bits are changed above? 
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Artificial intelligence can be used to help science students research and verify their research. It is also a very interesting topic to study in general.

Hope that helped!!! k

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State one way the projector can be integrated into teaching and learning ​
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Answer:

The projector con be used to show videos or pictures on a particular subject or topic to gain more understanding

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