Question
The function f left parenthesis x right parenthesis equals 70 e Superscript negative 0.6 x Baseline plus 30 describes the percentage of information, f(x), that a particular person remembers x weeks after learning the information.
Bold a. nbsp
Substitute 0 for x and, without using a calculator, find the percentage of information remembered at the moment it is first learned.
nbsp Bold b. nbsp
Substitute 1 for x and find the percentage of information that is remembered after 1 week.
nbsp Bold c.
Find the percentage of information that is remembered after 8 weeks.
Bold d. nbsp
Find the percentage of information that is remembered after one year (52 weeks).
Bold a. nbsp
Substitute 0 for x and, without using a calculator, find the percentage of information remembered at the moment it is first learned.
nbsp Bold b. nbsp
Substitute 1 for x and find the percentage of information that is remembered after 1 week.
nbsp Bold c.
Find the percentage of information that is remembered after 8 weeks.
Bold d. nbsp
Find the percentage of information that is remembered after one year (52 weeks).
Answers
Answered by
GPT-4o mini
To solve the problem, we will substitute the values of \( x \) into the function \( f(x) = 70 e^{-0.6x} + 30 \) and calculate the answers for each part.
### Part a
Substitute \( x = 0 \):
\[
f(0) = 70 e^{-0.6 \cdot 0} + 30
\]
Since \( e^0 = 1 \):
\[
f(0) = 70 \cdot 1 + 30 = 70 + 30 = 100
\]
**So, the percentage of information remembered at the moment it is first learned is 100%.**
### Part b
Substitute \( x = 1 \):
\[
f(1) = 70 e^{-0.6 \cdot 1} + 30
\]
Calculating \( e^{-0.6} \):
Using the approximation \( e^{-0.6} \approx 0.5488 \):
\[
f(1) = 70 \cdot 0.5488 + 30 \approx 38.416 + 30 \approx 68.416
\]
**Thus, the percentage of information that is remembered after 1 week is approximately 68.42%.**
### Part c
Substitute \( x = 8 \):
\[
f(8) = 70 e^{-0.6 \cdot 8} + 30
\]
Calculating \( e^{-4.8} \):
Using the approximation \( e^{-4.8} \approx 0.008229 \):
\[
f(8) = 70 \cdot 0.008229 + 30 \approx 0.575 + 30 \approx 30.575
\]
**So, the percentage of information that is remembered after 8 weeks is approximately 30.58%.**
### Part d
Substitute \( x = 52 \) (for one year):
\[
f(52) = 70 e^{-0.6 \cdot 52} + 30
\]
Calculating \( e^{-31.2} \):
Using the approximation \( e^{-31.2} \approx 8.5 \times 10^{-15} \) (which is a very small value):
\[
f(52) = 70 \cdot 8.5 \times 10^{-15} + 30 \approx 0 + 30 \approx 30
\]
**Therefore, the percentage of information that is remembered after one year is approximately 30%.**
### Summary of Results:
- a. 100%
- b. Approximately 68.42%
- c. Approximately 30.58%
- d. Approximately 30%
### Part a
Substitute \( x = 0 \):
\[
f(0) = 70 e^{-0.6 \cdot 0} + 30
\]
Since \( e^0 = 1 \):
\[
f(0) = 70 \cdot 1 + 30 = 70 + 30 = 100
\]
**So, the percentage of information remembered at the moment it is first learned is 100%.**
### Part b
Substitute \( x = 1 \):
\[
f(1) = 70 e^{-0.6 \cdot 1} + 30
\]
Calculating \( e^{-0.6} \):
Using the approximation \( e^{-0.6} \approx 0.5488 \):
\[
f(1) = 70 \cdot 0.5488 + 30 \approx 38.416 + 30 \approx 68.416
\]
**Thus, the percentage of information that is remembered after 1 week is approximately 68.42%.**
### Part c
Substitute \( x = 8 \):
\[
f(8) = 70 e^{-0.6 \cdot 8} + 30
\]
Calculating \( e^{-4.8} \):
Using the approximation \( e^{-4.8} \approx 0.008229 \):
\[
f(8) = 70 \cdot 0.008229 + 30 \approx 0.575 + 30 \approx 30.575
\]
**So, the percentage of information that is remembered after 8 weeks is approximately 30.58%.**
### Part d
Substitute \( x = 52 \) (for one year):
\[
f(52) = 70 e^{-0.6 \cdot 52} + 30
\]
Calculating \( e^{-31.2} \):
Using the approximation \( e^{-31.2} \approx 8.5 \times 10^{-15} \) (which is a very small value):
\[
f(52) = 70 \cdot 8.5 \times 10^{-15} + 30 \approx 0 + 30 \approx 30
\]
**Therefore, the percentage of information that is remembered after one year is approximately 30%.**
### Summary of Results:
- a. 100%
- b. Approximately 68.42%
- c. Approximately 30.58%
- d. Approximately 30%
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