In figuring nine-week grades, a social stidies teacher wants homework to count for 20%, quizzes to count for 20%, class participation to count for 10%, and exams to count for 50% of the grade.
Question: a) what average would a student get if her scores in those categories were 80,70,90, and 75 respectively?
b) If a student had a homework score of 90, a quiz score of 80, and a class participation score of 100, what exam score will give a nine-week average of at least 80?
2007-04-29 14:36:26 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
a) 80*0.2 + 70*0.2 + 90*0.1 + 75*0.5
=16 + 14 + 9 + 37.5
= 76.5% which would probably be rounded up to 77%
b) 90*0.2 + 80*0.2 + 100*.1 + x*0.5 = 80
18 + 16 + 10 + x/2 = 80
x/2 = 36
x=72%
2007-04-29 14:42:15 · answer #1 · answered by gudspeling 7 · 1⤊ 0⤋
A. 80 X .20 = 16
70 X .20 = 14
90 X .10 = 9
75 X .50 = 37.5
16 + 14 + 9 + 37.5 = FINAL GRADE
B. 90 X .20 = 18
80 X .20 = 16
100 X .10 = 10
18 + 16 + 10 + 44
80 - 44 = 36
36 X 2 = REQUIRED EXAM SCORE
2007-04-29 14:51:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
H = 20%
Q = 20%
CP = 10%
E = 50%
a) 80*20/100 + 70*20/100 + 90*10/100 + 75*50/100 = 76,5
b) 90*20/100 + 80*20/100 + 100*10/100 + x*50/100 > or = 80
44 + x/2 > or = 80
x/2 > or = 34
x > or = 68
2007-04-29 14:52:47 · answer #3 · answered by Anonymous · 0⤊ 0⤋
a) .20(80)+.20(70)+.10(90)+.50(75) = 76.5
b) .20(90)+.20(80)+.10(100)+.50(x) = 80
.50x = 80 -(.20(90)+.20(80)+.10(100)
.50x = 80-44=36
x= 72
The student would have to make at least 72 on the exam.
.
2007-04-29 14:53:03 · answer #4 · answered by Robert L 7 · 0⤊ 0⤋
(2*80 + 2*70 + 1*90 + 5*75) / 10
(160 + 140 + 90 + 375) / 10
765 / 10 = 76.5
(2*90 + 2*80 + 1*100 + 5 * N) / 10 = 80
(180 + 160 + 100 + 5*N) = 800
5* N = 360
N = 72
2007-04-29 14:45:01 · answer #5 · answered by bob h 3 · 0⤊ 0⤋