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Eva8 [605]
3 years ago
7

Changing how information is represented so that it can be read by a computer is called

Computers and Technology
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

Your answer is D) Encode

Explanation:

Definition of Encode: To change how information is represented so that it can be read by a computer.

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The following checksum formula is widely used by banks and credit card companies to validate legal account numbers: d0 + f(d1) +
Aleksandr [31]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to import Scanner class

public class ISBN

{   public static void main(String[] args)  { // start of main() function body

   Scanner s = new Scanner(System.in); // creates Scanner object

//prompts the user to enter 10 digit integer

   System.out.println("Enter the digits of an ISBN as integer: ");    

   String number = s.next(); // reads the number from the user

   int sum = 0; // stores the sum of the digits

   for (int i = 2; i <= number.length(); i++) {

//loop starts and continues till the end of the number is reached by i

          sum += (i * number.charAt(i - 1) ); }

/*this statement multiplies each digit of the number with i and adds the value of sum to the product result and stores in the sum variable*/

          int remainder = (sum % 11);  // take mod of sum by 11 to get checksum  

   if (remainder == 10)

/*if remainder is equal to 10 adds X at the end of given isbn number as checksum value */

  { System.out.println("The ISBN number is " + number + "X"); }

  else

// displays input number with the checksum value computed

 {System.out.println("The ISBN number is " + number + remainder); }  }  }  

Explanation:

This program takes a 10-digit integer as a command line argument and uses Scanner class to accept input from the user.

The for loop has a variable i that starts from 2 and the loop terminates when the value of i exceeds 10 and this loop multiplies each digit of the input number with the i and this product is added and stored in variable sum. charAt() function is used to return a char value at i-1.

This is done in the following way: suppose d represents each digit:

sum=d1 * 1 + d2 * 2 + d3 * 3 + d4 * 4 + d5 * 5 + d6 * 6 + d7 * 7 + d8 * 8 + d9 * 9

Next the mod operator is used to get the remainder by dividing the value of sum with 11 in order to find the checksum and stores the result in remainder variable.

If the value of remainder is equal to 10 then use X for 10 and the output will be the 10 digits and the 11th digit checksum (last digit) is X.

If the value of remainder is not equal to 10, then it prints a valid 11-digit number with the given integer as its first 10 digits and the checksum computed by sum % 11 as the last digit.  

8 0
3 years ago
What doe the &amp; operator do in python programming software
ArbitrLikvidat [17]
Hardware software is the answer
5 0
3 years ago
Due to a fire at ABC Software Solutions, all
zaharov [31]

Answer:

chaos among people who worked in the company.

6 0
2 years ago
It is a function that removes an existing file from the server.
kondaur [170]

Answer: A) remove()

Explanation:

 The remove() function removes an existing file from the server but it does not affect the existing directory and file. We can also ease and remove files in the file handling by using the remove() function. And it is built-in function that removes any type of the data from the function by taking values in the parameter whose values are equal with the passing value in the parameter.

7 0
3 years ago
In the following four questions, we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the pa
pantera1 [17]

Answer:

1)we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Initially suppose there is only one link between source and destination. Also suppose that the entire MP3 file is sent as one packet. The TRANSMISSION DELAY is:

3 Seconds

2)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. The END TO END DELAY(transmission delay plus propagation delay) is

3.05 seconds

3)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. how many bits will the source have transmitted when the first bit arrives at the destination.

500,000 bits

4)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP3 file is sent as one packet. Suppose there is no congestion, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. The end-to-end delay is

6.1 seconds

5)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose that the MP3 file is broken into 3 packets, each of 10 Mbits. Ignore headers that may be added to these packets. Also ignore router processing delays. Assuming store and forward packet switching at the router, the total delay is

4.05 seconds

6)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 TDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

7)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

6 0
3 years ago
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