Question

the function $f(x)=60e^{-0.4x}+40$ describes the percentage of information, $f(x)$, that a particular person remembers x weeks after learning the information.
a. substitute 0 for x and, without using a calculator, find the percentage of information remembered at the moment it is first learned.
b. substitute 1 for x and find the percentage of information that is remembered after 1 week.
c. find the percentage of information that is remembered after 4 weeks.
d. find the percentage of information that is remembered after one year (52 weeks).
a. at the moment it is first learned, 100.0% of the information is remembered.
(round to one decimal place as needed.)
b. after one week, 80.2% of the information is remembered.
(round to one decimal place as needed.)
c. after four weeks, % of the information is remembered.
(round to one decimal place as needed.)

Explanation:

Step1: Substitute \(x = 4\) into the function

Given \(f(x)=60e^{-0.4x}+40\), when \(x = 4\), we have \(f(4)=60e^{-0.4\times4}+40\).
First, calculate the exponent: \(-0.4\times4=-1.6\). So the function becomes \(f(4)=60e^{- 1.6}+40\).

Step2: Evaluate \(e^{-1.6}\)

We know that \(e^{-a}=\frac{1}{e^{a}}\). Using a calculator (since the problem doesn't restrict for this part as it did for part a), \(e^{1.6}\approx4.953\), so \(e^{-1.6}=\frac{1}{e^{1.6}}\approx\frac{1}{4.953}\approx0.202\).

Step3: Calculate \(60e^{-1.6}\)

\(60\times0.202 = 12.12\).

Step4: Calculate \(f(4)\)

\(f(4)=12.12 + 40=52.12\approx52.1\)

Answer:

\(52.1\)