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Norma-Jean [14]
3 years ago
11

Determine the number of bytes necessary to store an uncompressed RGB color image of size 640 × 480 pixels using 8, 10, 12 and 14

bits per color channel.
Computers and Technology
1 answer:
dexar [7]3 years ago
8 0

Answer:

For 8 bits / Color Channel: 921, 600 bytes.

For 10 bits / Color Channel: 1,152,000 bytes.

For 12 bits / Color Channel: 1,382,400 bytes.

For 14 bits / Color Channel : 1,612, 800 bytes.

Explanation:

Un uncompressed RGB color image of 640 pixels wide by 480 pixels height (VGA resolution) uses 3 samples of x bits fin order to represent the quantity of each primary color (red, green and blue) in a given pixel of the image.

So, if we use 8 bits/ sample, we can calculate the bytes needed to store a complete image as follows:

640*480*3*8 bits = 7, 372,800 bits. (1)

As, by definition, 1 byte= 8 bits, we have:

Nº of bytes = 7,372,800 / 8 = 921,600 bytes. (2)

In order to get the bytes needed to store the same image, using 10, 12 and 14 bits/ sample, all we need is replacing 8 by 10, 12 and 14 in (1), and then dividing by 8 as in (2).

Proceeding in this way we obtain:

640*480*3*10 = 9,216,000 bits ⇒    1,152,000 bytes.

640*480*3*12 = 11,059,200 bits ⇒   1,382,400 bytes.

640*480*3*14=   12,902,400 bits ⇒  1,612,800 bytes.

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import java.util.*;

public class Main

{

public static void main(String[] args) {

    Scanner input = new Scanner(System.in);

    System.out.print("Enter the first number: ");

    int n1 = input.nextInt();

    System.out.print("Enter the second number: ");

    int n2 = input.nextInt();

    System.out.print("Enter the third number: ");

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    return smallestNum;

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}

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