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anonymous
 5 years ago
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
anonymous
 5 years ago
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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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.1Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at t=0 S(0)=6820log(1)=6820*0=68 at t=4 S(4)=6820log(5)=6820*.699=54.02 S(24)=6820log(25)=40.04 b) 50=6820log(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)=6820*log(7.94)=50 this is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, is this a b and c broke down?
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