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Goryan [66]
2 years ago
14

) A byte is used to represent a single character in the computer ______ true or false?

Computers and Technology
2 answers:
sdas [7]2 years ago
7 0
Answer: false
Explanation:
topjm [15]2 years ago
7 0

Answer:

True!

hope this helps you..!

Explanation:

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How to code 2.9.5: Four colored triangles {Code HS}
Nitella [24]

Answer: penup()

backward(100)

for i in range(4):

pensize(5)

pendown()

left(60)

color("green")

forward(50)

right(120)

color("blue")

forward(50)

color("red")

right(120)

forward(50)

penup()

left(180)

forward(50)

Explanation:

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6 0
2 years ago
Write a class with a constructor that accepts a String object as its argument. The class should have a method that returns the n
mote1985 [20]

Answer:

import java.util.Scanner;

public class VowelsAndConsonantsDemo {

   /**

   The printMenu methods displays a menu to the user

    */

   public static void printMenu() {

       System.out.println("Please select an option: ");

       System.out.println();

       System.out.print("a. Count the number of vowels in the string.\n"

               + "b. Count the number of consonants in the string.\n"

               + "c. Count both the vowels and consonants in the string.\n"

               + "d. Enter another string.\n"

               + "e. Exit the program\n");

   }

   public static void main(String[] args) {

       String input;      // to hold the user's input

       String option;     // to hold the user's input

       char choice;       // to hold a single character

       char exit;         // user chooses 'e' to exit the program

       char letter;       //the Y or N from the user's decision to exit

       // create a Scanner object to read keyboard input.

       //  Scanner keyboard = new Scanner(System.in);

       Scanner keyboard;

       do {

           keyboard = new Scanner(System.in);

           // ask user to enter string

           System.out.print("Enter a string: ");

           input = keyboard.nextLine();

           input = input.toLowerCase();

           System.out.println();

           printMenu();

           option = keyboard.nextLine();

           choice = option.charAt(0);

           VowelsAndConsonants words = new VowelsAndConsonants(input);

           switch (choice) {

               case 'a':

               case 'A':

                   System.out.println("Number of Vowels: " + words.getVowels());

                   break;

               case 'b':

               case 'B':

                   System.out.println("Number of Consonants: " + words.getConsonants());

                   break;

               case 'c':

               case 'C':

                   System.out.println("Number of Vowels & Consonants: " + words.getConsonants()

                           + words.getVowels());

                   break;

               case 'd':

               case 'D':

                   System.out.println("Enter a string: ");

                   break;

               case 'e':

               case 'E':

                   System.exit(0);

                   break;

               default:

                   System.out.println("You did not enter a valid choice.");

           }

           //

           // keyboard.nextLine();    //consumes the new line character after the choice

           // String answer = keyboard.nextLine();

           // letter = answer.charAt(0);

       } while (true);

   }

}

4 0
2 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
2 years ago
If there are three classes, Shape, Circle and Square, what is the most likely relationship among them?
xz_007 [3.2K]

Answer: B) Shape is a base class, and circle and square are derived classes of Shape.

Explanation:

Shape is a base class because circle and squares are the shapes so these are the derived class of the shape, which is inherited by the shape like circle and square. As, the base class (shape) is the class which are derived from the other classes like circle and square and it facilitates other class which can simplified the code re-usability that is inherited from the base class. Base class is also known as parent class and the super class.  

8 0
2 years ago
Suppose that you need to maintain a collection of data whose contents are fixed- i.e., you need to search for and retrieve exist
Keith_Richards [23]

Answer:

Option a: a sorted array

Explanation:

Since the expectation on the data structure never need to add or delete items, a sorted array is the most desirable option. An array is known for its difficulty to modify the size (either by adding item or removing item). However, this disadvantage would no longer be a concern for this task. This is also the reason, linked list, binary search tree and queue is not a better option here although they offer much greater efficiency to add and remove item from collection.

On another hand,  any existing items from the sorted array can be easily retrieved using address indexing and therefore the data query process can be very fast and efficient.

4 0
2 years ago
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