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. 371
B. 24
C. 225
D. 3520

Answer :

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

1. Start with the given function:

[tex]\( f(t) = P e^{rt} \)[/tex]

2. Substitute the known values:

We know that [tex]\( f(5) = 288.9 \)[/tex], [tex]\( r = 0.05 \)[/tex], and [tex]\( t = 5 \)[/tex]. So, the equation becomes:

[tex]\[
288.9 = P e^{0.05 \times 5}
\][/tex]

3. Simplify the exponent:

Calculate the exponent:

[tex]\[
0.05 \times 5 = 0.25
\][/tex]

Therefore, the equation becomes:

[tex]\[
288.9 = P e^{0.25}
\][/tex]

4. Solve for [tex]\( P \)[/tex]:

Rearrange the equation to solve for [tex]\( P \)[/tex]:

[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]

5. Calculate [tex]\( e^{0.25} \)[/tex]:

Using a calculator, we find:

[tex]\[
e^{0.25} \approx 1.284025
\][/tex]

6. Calculate [tex]\( P \)[/tex]:

Plug the value of [tex]\( e^{0.25} \)[/tex] into the equation:

[tex]\[
P = \frac{288.9}{1.284025} \approx 224.995
\][/tex]

7. Choose the closest value:

Compare the calculated value of [tex]\( P \approx 224.995 \)[/tex] with the given options:

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

The closest value to 224.995 is [tex]\( C. 225 \)[/tex].

Thus, the approximate value of [tex]\( P \)[/tex] is [tex]\(\boxed{225}\)[/tex].