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Akimi4 [234]
4 years ago
11

A student'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.
Mathematics
1 answer:
kow [346]4 years ago
7 0

Answer:

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%!!!

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The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

Step-by-step explanation:

<u><em>Step(i):</em></u>-

Given  

                cos( 3x - \frac{\pi }{3} )  = \frac{\sqrt{3} }{2}

                  cos( 3x - \frac{\pi }{3} )  = cos (\frac{\pi }{6} )

                      3x - \frac{\pi }{3}  =  \frac{\pi }{6}

                      3x - \frac{\pi }{3  } + \frac{\pi }{3}   =  \frac{\pi }{6} + \frac{\pi }{3}

                      3x = \frac{2\pi +\pi }{6} = \frac{3\pi }{6} = \frac{\pi }{2}

                     x = \frac{\pi }{6}

<u><em>Step(ii)</em></u>:-

The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

<u><em>verification </em></u>:-

      cos( 3x - \frac{\pi }{3} )  = \frac{\sqrt{3} }{2}

put  x = \frac{\pi }{6}

    cos( 3(\frac{\pi }{6})  - \frac{\pi }{3} )  = \frac{\sqrt{3} }{2}

    cos (\frac{\pi }{6} ) = \frac{\sqrt{3} }{2} \\\\\frac{\sqrt{3} }{2} =  \frac{\sqrt{3} }{2}

Both are equal

∴The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

                     

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