Answer:
The student has total average of 83.89.
Step-by-step explanation:
The student has 72 out of 100 in the attendance and with a weightage of 4%, the average will be = [tex]72 \times \frac{4}{100} = 2.88[/tex]
The student has 96 out of 100 in the first paper and with a weightage of 7%, the average will be = [tex]96 \times \frac{7}{100} = 6.72[/tex]
The student has 91 out of 100 in the second paper and with a weightage of 7%, the average will be = [tex]91 \times \frac{7}{100} = 6.37[/tex].
The student has 94 out of 100 in the test 1 and with a weightage of 12%, the average will be = [tex]94 \times \frac{12}{100} = 11.28[/tex]
The student has 73 out of 100 in the test 2 and with a weightage of 12%, the average will be = [tex]73 \times \frac{12}{100} = 8.76[/tex].
The student has 100 out of 100 in the test 3 and with a weightage of 12%, the average will be = [tex]100 \times \frac{12}{100} = 12[/tex].
The student has 78 out of 100 in the final exam and with a weightage of 46%, the average will be = [tex]78 \times \frac{46}{100} = 35.88[/tex]
Therefore, the student has the total average of (2.88 + 6.72 + 6.37 + 11.28 + 8.76 + 12 + 35.88) = 83.89. (Answer)