3. 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 found to be given by
S(t) = 68 − 20 log (t + 1), t ≥ 0.
What was the average score when they initially took the test, t = 0?
Round your answer to a whole percent, if necessary.
a)
Answer:
Show your work in this space:
What was the average score after 4 months? after 24 months?
Round your answers to two decimal places.
b)
Answer:
Show your work in this space:
After what time t was the average score 50%?
Round your answe

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- anonymous

at t=0
S(0)=68-20log(1)=68-20*0=68
at t=4
S(4)=68-20log(5)=68-20*.699=54.02
S(24)=68-20log(25)=40.04
b)
50=68-20log(t+1) we need to find t.
18=20log(t+1)
9/10=log(t+1)
9/10= ln(t+1)/(ln(10))
.9ln(10)=ln(t+1)
e^(.9ln(10))=t+1
(e^(.9ln(10)))-1=t
t=6.94 months
check:
S(6.94)=68-20*log(7.94)=50 this is right

- anonymous

Thanks..

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- anonymous

youre welcome

- anonymous

Hey, is this a b and c broke down?

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