Question

students in an english class took a final exam. they took equivalent forms of the exam at monthly intervals thereafter. the average score S(t), in percent , after t months was foundto be given by:
S(t)=68-20log(t+1), t is greater than 0.

1a). what was the average score when they initially took the test, t=0

1b) what was the average score after 14 months

1c) after what time was the average score 40%

Answer 1

since S(t) is the average score for t months

a) S(0) : plug in 0 for every t in S(t). Should give S(t) = 68

b) S(14): plug in 14 for every t in S(t) [only one t in this case]

c) S(t) = 0.4. Find t. lmk if u need help with this one after doing the 2 first

Answer 2

i dont really understand

Answer 3

which one?

Answer 4

heres what i have
1a) 68

Answer 5

the other ones are confusing

Answer 6

b). plug in 14 for every t

S(14) = 68 -20log(14+1) = 68 - 20(1.176) = 44.478

the value of log15 was obtained with my calculator.

note: i used log base 10 [log button] and not the natural logarithm [ln button]

Answer 7

and 1c.

Answer 8

0.4 = 68 - 20log(t+1)
re-arranged
20log(t+1) = 67.6
log(t+1) = 3.38

assuming log is log base 10 and not log base e, doing it's inverse operation:

(t+1) = 10^(3.38)

t = 2397.83

which seems like a very high number.

Answer 9

thank you.