Suppose that the variables grade1, grade2, and grade3 are declared as follows: int grade1 = 88; int grade2 = 92; int grade3 = 83; which of these expressions correctly calculates the average of grade1, grade2, and grade3? a. (grade1 + grade2 + grade3) / 3 b. (grade1 + grade2 + grade3) / 3.0 c. ( + grade2 + grade3) / 3) d. (double)grade1 + grade2 + grade3 / 3.0 e. (double)grade1 + (double)grade2 + (double)grade3 / 3.0

Option B (grade1 + grade2 + grade3) / 3.0

Explanation:

The most accurate result is calculated by the expression given in the option B.  This is because all the grade1, grade2 and grade3 variables are integer type and therefore we need to add a decimal point after divisor (e.g. 3.0) so that the decimal points of the result can be preserved.

For example:

(88 + 92 + 83) / 3.0  should result in 87.66666666666667

If we do  (88 + 92 + 83) / 3, the decimal points are truncated to 87.

The expression in option c, (double)((grade1 + grade2 + grade3) / 3), will result in 87 as the  (grade1 + grade2 + grade3) / 3 will be evaluated first prior to converting into double type.

The expression in option d and e are inaccurate as they don't calculate average of the three variables.  

the formula for work is

f*d

therefore work=2.0n*3.0=6n*m