In the following code we would like to compute a grade as a percentage #include

Question

In the following code we would like to compute a grade as a percentage #include int main (void) 2 3 5 6 7 int ptotal-80; int ppossible-100; float score; ore=ptotal/Ppossible*100 printf ("The grade is %4.1f%% " , score) ; return 0; 9 10 The student should get an 80%. But this is the output. The grade is 0.0% we can fix line 7, as numbered above, to generate the correct output of 80%. There are many ways. which of the following alternatives to line 7 will show the output at 80%? Please select ALL correct answers score-ptotal/ppossible 100.; score- 100. *ptotal/ppossible score-100 (float)ptotal/ppossible; score-ptotal/ppossible*100.; score-100 ptotal/ppossible score-(float) 100*ptotal/ppossible;

Explanation / Answer

1. score-ptotal/ppossible 100.;

2. score- 100. *ptotal/ppossible;

3. score-100 (float)ptotal/ppossible;

4. score-ptotal/ppossible*100.;

5. score-100 ptotal/ppossible;

6. score-(float) 100*ptotal/ppossible;

The code prints 0.0% as ptotal/ppossible, which are both integers outputs an integer = 0 due to the / operator.

#2 score=100.*ptotal/ppossible outputs 80% //100 is written as 100. which makes it a floating point and outputs a floating point

#3 score=100 * (float) ptotal/ppossible ouputs 80% //we explicitly type cast the integer to a floating point

#6 score=(float) 100*ptotal/ppossible outputs 80% //we explicitly type cast the integer to a floating point by writing (float)

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