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tester [92]
3 years ago
5

An online museum is creating a site hosting hundreds of thousands of digital representations of art from around the world. The w

ant to use a lossless compression, so that members of the museum can download and decompress the image and have an identical copy of the original file. Which of the following are true? Select two answers.
A. the museum can choose a heuristic approach which will achieve a lossless compression, but they cannot be sure that it is the most efficient compression for each image.
B. the museum can choose a heuristic approach which will achieve lossless compression for most digital representations
C. algorithms for lossless compression exist, so the museum can use those to compress the image
D. the museum should not compress digital representations of art, as there is always a danger that the compression is actual lossy
Computers and Technology
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

A. the museum can choose a heuristic approach which will achieve a lossless compression, but they cannot be sure that it is the most efficient compression for each image

C. algorithms for lossless compression exist, so the museum can use those to compress the image

Explanation:

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Tortise and hare race java g Modify the main class so it runs the race 100 times and reports how many times each runner wins. (T
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Answer:

Game.java file

import java.util.Scanner;

public class Game {

/**

* t_pos and h_pos are the current positions of the Tortoise and Hare

*/

static int t_pos,h_pos;

static Tortoise tortoise;

static Hare hare;

public static void main(String[] args) {

play(); /*starting the game*/

}

public static void play(){

/**

* the method will starts the play, loop until the game is over, displays the winner

* and prompts the user if they want to play again

*/

/**

* defining Tortoise and Hare objects

*/

tortoise=new Tortoise();

hare=new Hare();

t_pos=1;

h_pos=1;

System.out.println("The race is about to start");

tortoise.printTrack();

hare.printTrack();

while(t_pos != 50 && h_pos !=50){

System.out.println("\n\n\n"); /*printing blank lines*/

t_pos=tortoise.move(); /*moving and getting the current position of tortoise*/

h_pos=hare.move();/*moving and getting the current position of hare*/

tortoise.printTrack(); /*displaying the tracks*/

hare.printTrack();

try { /*comment this part to skip the 1s break between each round; for testing*/

Thread.sleep(1000);

} catch (InterruptedException e) {

e.printStackTrace();

}

}

System.out.println("\nRace Over");

if(t_pos==50 && h_pos==50){

System.out.println("Its a tie");

}

else if(t_pos==50){

System.out.println("Tortoise wins");

}else if(h_pos==50){

System.out.println("Hare wins");

}

System.out.println("Do you want to play again? (y/n)");

Scanner scanner=new Scanner(System.in);

String ch=scanner.next();

if(ch.equalsIgnoreCase("y")){

play();

}else if(ch.equalsIgnoreCase("n")){

System.out.println("Thanks for playing, Goodbye");

}else{

System.out.println("Invalid choice, quitting..");

}

}

}

//Tortoise.java

public class Tortoise {

/**

* the current position of the tortoise

*/

int position;

/**

* track array

*/

char[] track;

/**

* speed of tortoise

*/

int speed=1;

public Tortoise() {

position=0;

track=new char[50];

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

/**

* filling the track

*/

track[i]='-';

}

}

public int move(){

if(position<track.length){

position=position+speed;

}

return position+1;

}

public void printTrack(){

/**

* the current position of tortoise will be displayed by 'T' everything else will be '-'

*/

System.out.println();

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

if(i==position){

System.out.print('T');

}else{

System.out.print(track[i]);

}

}

}

}

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import java.util.Random;

public class Hare {

int position;

int speed=10;

char[] track;

/**

* Random object to generate a random number

*/

Random random;

/**

* resting percent denotes how much time Hare will be resting

*/

int resting_percent=90;

public Hare() {

position=0;

track=new char[50];

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

track[i]='-';

}

random=new Random();

}

public int move(){

int n=random.nextInt(100-1)+1; /*generating a random number between 1 and 100*/

if(n<=resting_percent){

/**

* at rest; will not move, returns the current position.

*/

return position;

}else{

/**

* not resting..

*/

if(position<track.length){

if(position+speed>=track.length){

position=track.length-1;

}else{

position=position+speed;

}

}

return position+1;

}

}

public void printTrack(){

System.out.println();

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

if(i==position){

System.out.print('H');

}else{

System.out.print(track[i]);

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}

}

}

/*Output (partial and random)*/

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

--T-----------------------------------------------

H-------------------------------------------------

---T----------------------------------------------

----------H---------------------------------------

.

.

.

.

-----------------------------------------------T--

-------------------------------------------------H

Race Over

Hare wins

Do you want to play again? (y/n)

y

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

.

.

.

.

-----------------------------------------------T--

--------------------H-----------------------------

------------------------------------------------T-

--------------------H-----------------------------

-------------------------------------------------T

--------------------H-----------------------------

Race Over

Tortoise wins

Do you want to play again? (y/n)

n

Explanation:

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Answer:

four heads are better than one

Explanation:

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Answer:

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Consider an error-free 64 kbps satellite channel used to send 512 byte data frames in one direction, with very short acknowledge
Vladimir [108]

Answer:

The answer is "2".

Explanation:

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1 bytes = 8 bits

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Frame length = 4096 bits

Formula

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Window size = 1+2a

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Window size = 1+0.864

Window size = 1.864

Window size = 2

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3 years ago
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