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balandron [24]
2 years ago
11

Is the type of vision useful in sending motion and objects outside normal vision

Computers and Technology
1 answer:
nlexa [21]2 years ago
8 0
Yes it is in fact it is usful
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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
2 years ago
What are the answers to everfi
FinnZ [79.3K]
FutureSmart focuses on the important Middle School years by empowering students to become the stewards of their financial futures. This three hour web-based resource educates students on the practicalities of daily financial decisions and the payoffs of long-term planning. Since Middle School is an important period for positive habits
to take form and grow, this course is particularly meaningful.
Through a compelling narrative in which students play the Mayor of a town, local citizens are helped with real-life decisions. From weighing opportunity costs, to delaying instant gratification for long- term gain, students face important questions on their way to becoming FutureSmart. At the end of the course, students compose their own blueprint for the future. FutureSmart c
6 0
2 years ago
How long is the bachelor's program at Eth Zurich? 3 or 4 years?
natulia [17]
The bachelor's program at Eth Zurich is 3 years.
I hope this helps!
:-)
4 0
3 years ago
Read one positive integer n. Then create an n X n two-dimensional array and write the code that stores integers from 1 to n2 as
marysya [2.9K]

Answer:

The program in Java is as follows:

import java.util.*;

public class Main{

public static void main(String[] args) {

    int n;

    Scanner input = new Scanner(System.in);

 System.out.print("Size of array: ");

 n = input.nextInt();

 int count = 1;

 int[][] arr = new int[n][n];

 for(int i = 0; i<n;i++){

     for(int j = 0; j<n;j++){

         arr[i][j] = count;

         count++;      }  }

 for(int i = 0; i<n;i++){

     for(int j = 0; j<n; j++){

         System.out.printf(arr[i][j]+" ");      }

     System.out.println();  } }}

Explanation:

This declares the size of the array

    int n;

    Scanner input = new Scanner(System.in);

This prompts the user for size of array

 System.out.print("Size of array: ");

This gets input for size of array

 n = input.nextInt();

This initializes the array element to 1

 int count = 1;

This creates a 2d array

 int[][] arr = new int[n][n];

This iterates through the rows

 for(int i = 0; i<n;i++){

This iterates through the columns

     for(int j = 0; j<n;j++){

This populates the array

         arr[i][j] = count;

         count++;      }  }

The following nested loop prints the array elements

 for(int i = 0; i<n;i++){

     for(int j = 0; j<n; j++){

         System.out.printf(arr[i][j]+" ");      }

     System.out.println();  } }}

8 0
2 years ago
After reading through the code, what will happen when you click run?​
AnnyKZ [126]

Answer:

B.) Laurel will try to collect treasure when there is not any and it causes an error

Explanation:

Laurel is at the top right corner of the map. When you click run, laurel moves 1 step ahead. This spot is right above the coordinate of the very first diamond. There is no treasure at this spot.

Therefore, when the next command calls for collecting, there will be an error, as there is no treasure at the spot right above the first diamond.

4 0
2 years ago
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