Astudent's course grade is based on one midterm that counts as 15% of his final grade, one class project that counts as 15% of his final grade, a set of homework assignments that counts as 35% of his final grade, and a final exam that counts as 35% of his final grade. his midterm score is 83, his project score is 97, his homework score is 82, and his final exam score is 63. what is his overall final score? what letter grade did he earn (a, b, c, d, or f)? assume that a mean of 90 or above is an a, a mean of at least 80 but < 90 is a b, and so on.

The student gets the Grade 'C' because the aggregate percentage is greater than 70 and less than 80 and this can be determined by using the given data.

Given :

A student's project marks are 97 out of 100 but the weightage of the project marks is 15%. That is:

So, the project marks are 14.55 out of 15.

A student's homework assignments marks are 82 out of 100 but the weightage of the project marks is 35%. That is:

= 28.7

So, homework assignments marks are 28.7 out of 35.

A student's midterm marks are 83 out of 100 but the weightage of the project marks is 15%. That is:

= 12.45

So, midterm marks are 12.45 out of 15.

A student's final exam marks are 63 out of 100 but the weightage of the project marks is 35%. That is:

= 22.05

So, final exam marks are 22.05 out of 35.

So, the aggregate percentage is given by:

Aggregate Percentage = 14.55 + 12.45 + 28.7 + 22.05 = 77.75%

The student gets the Grade 'C' because the aggregate percentage is greater than 70 and less than 80.

For more information, refer to the link given below:

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Overall final score = 77.75% ; Grade = C.

Step-by-step explanation:

The approach to solve this question is to realize that the marks have to be converted into the respective percentages of the whole course. This means that the marks of all the components have to be normalized according to the grading breakdown.

Project Marks = 97/100. Weightage = 15%. So 97*15/100 = 14.55/15.

This means that the student received 14.55 marks in the project out of 15.

Similarly for other components:

Mid-Term Marks = 83/100. Weightage = 15%. So 83*15/100 = 12.45/15.

Homework Marks = 82/100. Weightage = 35%. So 82*35/100 = 28.7/35.

Finals Marks = 63/100. Weightage = 35%. So 63*35/100 = 22.05/35.

After the conversion process, add up the normalized marks, which are now acting as the percentages earned in all the components.

Aggregate Percentage = 14.55 + 12.45 + 28.7 + 22.05 = 77.75%.

According to the grade scale, the student receives a C because 70 is less than 77.75 and 77.75 is less than 80.

Summarizing, the student receives a C at 77.75%!!!

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )