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Fiesta28 [93]
3 years ago
6

3.2 lesson practice edhesive ​

Computers and Technology
1 answer:
Nookie1986 [14]3 years ago
6 0
Ummm what is the question
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3 years ago
Write a user-defined function that calculates the GPA. For the function name and arguments, use the following: The input argumen
QveST [7]

Answer:

#include <iostream>

#include <vector>

using namespace std;

void calGPA();

vector<int> g;

vector<int> h;

int main(){

   char pushMore = 'y';

   int fg, fh;

   for (;;){

       if (pushMore == 'n'){

           break;

       } else{

           cout<< "Enter integer for grade: ";

           cin>> fg;

           cout<< "Enter integer for credit hours: ";

           cin>> fh;

           g.push_back(fg);

           h.push_back(fh);

           cout<< "Do you want to add more grade and credit hours? y/n: ";

           cin>> pushMore;

       }

   }

   calGPA();

}

void calGPA(){

   double total = 0, GPA;

   for (int i = 0; i < g.size(); ++i) {

        total +=  g.at(i) * h.at(i) ;

   }

   cout<< "The GPA is : "<< total/g.size();

}

Explanation:

The C++ source code above defines two vectors 'g' and 'h'. The latter holds the grades of a student while the former holds the credit hours for the subject taken. The void 'calGPA' function calculates and prints the student's GPA.

7 0
2 years ago
In terms of their eligibility for legal protection, how do ideas differ from song lyrics?
4vir4ik [10]

Answer:

D

Explanation:

3 0
3 years ago
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What are three procedures used by browsers
ra1l [238]
I believe that the three procedures are...

1.  Sub - Standard sub routine

2.  Function - a routine that returns an answer

3.  Property - reserved for <span> Class Modules</span>

<span>Hope this helps, please mark brainliest!</span>

5 0
3 years ago
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Problem 4 (25 points)Consider a byte addressing architecture with 64-bit memory addresses.(a)Which bits of the address would be
zysi [14]

Answer:

Following are the solution to the given points:

Explanation:

The Memory address value = 64 bit

The Size of the word = \frac{64}{8} =8 \  Byte

In point a:

The offset size = 3\ bits ( in 1-word block size)  

The Index size = 9 \  bits (as block number =512)  

Tag size = 64 - 12  = 52\  bits

In point b:

The offset size = 8 \times 8 \ bytes = 2^6 = 6 \ bits.

The Index size = 64 \ bits = 2^6 \ =6 \ bits

Tag size  = 64 - 12  = 52\  bits

In point c:

The Ratio at point a  

\to 3:64

The Ratio at point b

\to 6:64

5 0
2 years ago
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