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mars1129 [50]
3 years ago
5

Assume you have an XML Tree variable called quiz initialized with the following valid XML document. You are not required to draw

the tree, but you might find it helpful. <?xml version="1.0" encoding="UTF-8"?> 155 What is the size of this tree? What would be the value returned by quiz. child(0).child(1).child(0).label() What would be the value returned by quiz. child(1).attributeValue("number")

Computers and Technology
1 answer:
blagie [28]3 years ago
8 0

Answer:

See explaination.

Explanation:

question is the root element which have two children one is M-C and other is Coding.

Again M-C have two sub children Points and parts

Note: The tree will be as shown in the attachment. kindly refer to attachment.

Here if we see the child of quiz(questions) at 0 position is number of type M-C and another child at location 1 is number of type Coding.

Now quiz.child(0) is number of type M-C which has two child and child at 0 is Points and child at 1 is Parts

quiz.child(0).child(1) is Points and now further points doesn't have any children hence going further to quiz.child(0).child(1).child(0) is nothing hence it will not return anything.

Next quiz.child(1) is number of type coding and value is 5.

Size of the tree is (2^depth)-1

and here depth of tree is 3 hence size is (2^3)-1 i.e. 7

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Read 2 more answers
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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