Question

Students in a mathematics class were given an exam and then tested monthly with an equivalent exam. The average scores for the class are given by the human memory model f(t) = 76 − 18 log10(t + 1), 0 ≤ t ≤ 12 where t is the time in months. Verify your answers in parts (a), (b), and (c) using a graphing utility.
(a) What was the average score on the original exam (t = 0)? f(0) =
(b) What was the average score after 2 months? (Round your answer to one decimal place.) f(2) =
(c) What was the average score after 11 months? (Round your answer to one decimal place.) f(11) =

Answer 1

Answer:

a) $f(t) = 76-18 log_{10} (0+1)= 76 - 18 log_{10} 1= 76$

b) $f(t) = 76-18 log_{10} (2+1)= 76 - 18 log_{10} 3= 76-8.588=67.41$

c) $f(t) = 76-18 log_{10} (11+1)= 76 - 18 log_{10} 12= 76-19.425=56.574$

Step-by-step explanation:

For this case we know that the average scores for the class are given by the following model:

$f(t) = 76-18 log_{10} (t+1) , 0 \leq t \leq 12$

Where t is in months. The graph attached illustrate the function for this case

And for this case we can answer the questions like this:

Part a

We just need to replace t =0 into the model and we got:

$f(t) = 76-18 log_{10} (0+1)= 76 - 18 log_{10} 1= 76$

Part b

We just need to replace t =2 into the model and we got:

$f(t) = 76-18 log_{10} (2+1)= 76 - 18 log_{10} 3= 76-8.588=67.41$

Part c

We just need to replace t =11 into the model and we got:

$f(t) = 76-18 log_{10} (11+1)= 76 - 18 log_{10} 12= 76-19.425=56.574$