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 ¡Ý 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
when t = 0
S(0) = 68 - 20log(1) = 68
when t= 4
S(4) = 68 - 20log(5) = 54.02... or 54
etc.
for score to be 50
50 = 68 - 20log(t+1)
log(t+1) = 18/20 = .9
t+1 = 10^.9
t = 10^.9 - 1 months
= 6.943 or appr. 7 months
a) To find the average score when they initially took the test (t = 0), we can substitute t = 0 into the given formula:
S(t) = 68 - 20 log(t + 1)
S(0) = 68 - 20 log(0 + 1)
= 68 - 20 log(1)
= 68 - 20 * 0
= 68 - 0
= 68
Therefore, the average score when they initially took the test (t = 0) is 68%.
b) To find the average score after 4 months and after 24 months, we can substitute t = 4 and t = 24 into the given formula:
For 4 months:
S(t) = 68 - 20 log(t + 1)
S(4) = 68 - 20 log(4 + 1)
= 68 - 20 log(5)
≈ 68 - 20 * 0.69897
≈ 68 - 13.9794
≈ 54.0206
Therefore, the average score after 4 months is approximately 54.02%.
For 24 months:
S(t) = 68 - 20 log(t + 1)
S(24) = 68 - 20 log(24 + 1)
= 68 - 20 log(25)
≈ 68 - 20 * 1.39794
≈ 68 - 27.9588
≈ 40.0412
Therefore, the average score after 24 months is approximately 40.04%.
c) To find the time (t) when the average score is 50%, we need to solve the equation:
S(t) = 68 - 20 log(t + 1) = 50
Subtracting 68 from both sides:
-20 log(t + 1) = 50 - 68
-20 log(t + 1) = -18
Dividing by -20:
log(t + 1) = 18/20
log(t + 1) = 0.9
Taking the inverse logarithm:
t + 1 = 10^0.9
t + 1 ≈ 7.943282347
Subtracting 1 from both sides:
t ≈ 7.943282347 - 1
t ≈ 6.943282347
Therefore, the average score is 50% after approximately 6.94 months.