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Nadya [2.5K]
3 years ago
10

The process by which information gets into memory storage is

Computers and Technology
1 answer:
Kay [80]3 years ago
7 0

Answer:

That is the first step and it is called encoding process.

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What type of pointing device is often used by artists, and why is it ideal for artists?​
devlian [24]

Answer:

A drawing/graphics tablet

Explanation:

It is ideal for artists, due to it being very similar to as if you were to draw on paper. The stylus replicates a pencil or pen.

4 0
3 years ago
Which of these is an on-site metric for social media marketing?
love history [14]
The answer to you question is C
3 0
3 years ago
What is the best definition of the 7x7 rule for maximizing audience comprehension
olganol [36]

The 7x7 Rule states that a PowerPoint slide should have no more than seven lines of text and no more than seven words in each of those lines.

7 0
3 years ago
Write a program consisting of: a. A function named right Triangle() that accepts the lengths of two sides of a right triangle as
Lunna [17]

Answer:

The java program is as follows.

import java.lang.*;

public class Triangle

{

   //variables to hold sides of a triangle

   static double height;

   static double base;

   static double hypo;

   //method to compute hypotenuse

   static double rightTriangle(double h, double b)

   {

       return Math.sqrt(h*h + b*b);

   }

public static void main(String[] args) {

    height = 4;

    base = 3;

    hypo = rightTriangle(height, base);

 System.out.printf("The hypotenuse of the right-angled triangle is %.4f", hypo);

}

}

OUTPUT

The hypotenuse of the right-angled triangle is 5.0000

Explanation:

1. The variables to hold all the three sides of a triangle are declared as double. The variables are declared at class level and hence, declared with keyword static.

2. The method, rightTriangle() takes the height and base of a triangle and computes and returns the value of the hypotenuse. The square root of the sum of both the sides is obtained using Math.sqrt() method.

3. The method, rightTriangle(), is also declared static since it is called inside the main() method which is a static method.

4. Inside main(), the method, rightTriangle() is called and takes the height and base variables are parameters. These variables are initialized inside main().

5. The value returned by the method, rightTriangle(), is assigned to the variable, hypo.

6. The value of the hypotenuse of the triangle which is stored in the variable, hypo, is displayed to the user.

7. The value of the hypotenuse is displayed with 4 decimal places which is done using printf() method and %.4f format specifier. The number 4 can be changed to any number, depending upon the decimal places required.

8. In java, all the code is written inside a class.

9. The name of the program is same as the name of the class having the main() method.

10. The class having the main() method is declared public.

11. All the variables declared outside main() and inside another method, are local to that particular method. While the variables declared outside main() and inside class are always declared static in java.

4 0
3 years ago
How long does it take to send a 15 MiB file from Host A to Host B over a circuit-switched network, assuming: Total link transmis
alexdok [17]

Answer:

The answer is "102.2 milliseconds".

Explanation:

Given:

Size of file = 15 MiB

The transmission rate of the total link= 49.7 Gbps

User = 7

Time of setup = 84.5 ms

calculation:

1\ MiB = 2^{20} = 1048576\ bytes\\\\1\ MiB = 2^{20 \times 8}= 8388608\  bits\\\\15\ MiB= 8388608 \times 15 = 125829120\ bits\\\\

So,

Total Number of bits = 125829120 \ bits

Now

The transmission rate of the total link= 49.7 Gbps

1\ Gbps = 1000000000\ bps\\\\49.7 \ Gbps = 49.7 \times 1000000000 =49700000000\ bps\\\\FDM \ \ network

\text{Calculating the transmission rate for 1 time slot:}

=\frac{ 49700000000}{7} \ bits / second\\\\= 7100000000 \ bits / second\\\\ = \frac{49700000000}{(10^{3\times 7})} \ in\  milliseconds\\\\ =7100000 \ bits / millisecond

Now,

\text{Total time taken to transmit 15 MiB of file} = \frac{\text{Total number of bits}}{\text{Transmission rate}}

= \frac{125829120}{7100000}\\\\= 17.72\\\\

\text{Total time = Setup time + Transmission Time}\\\\  

                 = 84.5+ 17.72\\\\= 102.2 \ milliseconds

7 0
3 years ago
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