Question

the function ( f(x)=90 e^{-0.4 x}+10 ) 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 % ) of the information is remembered.
(round to one decimal place as needed.)
b. after one week, ( 70.3 % ) of the information is remembered.
(round to one decimal place as needed.)
c. after four weeks, ( square % ) of the information is remembered.
(round to one decimal place as needed.)

Explanation:

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

Given \(f(x)=90e^{-0.4x}+10\), when \(x = 4\), we have \(f(4)=90e^{-0.4\times4}+10\).
First, calculate the exponent: \(-0.4\times4=-1.6\). So the function becomes \(f(4)=90e^{-1.6}+10\).

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

We know that \(e^{-a}=\frac{1}{e^{a}}\). Using a calculator (since \(e\approx2.71828\)), \(e^{1.6}\approx4.95303\), then \(e^{-1.6}=\frac{1}{e^{1.6}}\approx\frac{1}{4.95303}\approx0.202\).

Step3: Calculate \(90e^{-1.6}\)

\(90\times0.202 = 18.18\).

Step4: Calculate \(f(4)\)

\(f(4)=18.18 + 10=28.18\approx28.2\)

Answer:

\(28.2\)