College

If [tex]$f(5)=288.9$[/tex] when [tex]$r=0.05$[/tex] for the function [tex]$f(t)=P e^{rt}$[/tex], then what is the approximate value of [tex]$P$[/tex]?

A. 24
B. 3520
C. 225
D. 371

Answer :

To find the approximate value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = P e^{rt} \)[/tex] given that [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex], we can follow these steps:

1. Understand the function:
The function given is [tex]\( f(t) = P e^{rt} \)[/tex]. For our specific problem, we know:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]

2. Set up the equation:
We substitute the known values into the equation:
[tex]\[
288.9 = P \times e^{0.05 \times 5}
\][/tex]
Simplify the exponent:
[tex]\[
e^{0.05 \times 5} = e^{0.25}
\][/tex]

3. Calculate [tex]\( e^{0.25} \)[/tex]:
The approximate value of [tex]\( e^{0.25} \)[/tex] is around 1.284.

4. Solve for [tex]\( P \)[/tex]:
Use the equation to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{288.9}{e^{0.25}} \approx \frac{288.9}{1.284}
\][/tex]
[tex]\[
P \approx 225
\][/tex]

5. Choose the closest answer:
By solving the above, we find [tex]\( P \)[/tex] is approximately 225. Therefore, the correct answer is:

- C. 225

This is how you determine the approximate value of [tex]\( P \)[/tex] using the given function and parameters.

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