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leva [86]
4 years ago
5

Define a JavaScript function named showGrades which does not have any parameters. Your function should create and return an arra

y containing 3 Numbers. The values for each of these entries should be: get the value from the div whose id is "GradeA" and store it at index 0 of your array; get the value from the div whose id is "Passing" and store it at index 1 of your array; and get the value from the div whose id is "Learning" and store it at index 2 of your array. Your must then encode the array as a JSON blob and return that JSON blob.
Computers and Technology
1 answer:
grandymaker [24]4 years ago
8 0

Answer:

see explaination

Explanation:

//selective dev elements by id name

var gradeA = document.querySelector("#GradeA");

var passing = document.querySelector("#Passing");

var learning = document.querySelector("#Learning");

//function showGrades

function showGrades() {

var arr = [];

//converting string to int and inserting into array

arr[0] = parseInt(gradeA.textContent);

arr[1] = parseInt(passing.textContent);

arr[2] = parseInt(learning.textContent);

//creating json blob

var blob = new Blob(new Array(arr), {type:"text/json"});

return blob;

}

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g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
grin007 [14]

Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

3 0
3 years ago
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