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. 225
B. 3520
C. 24
D. 371

Answer :

Sure, I'll help you solve this problem step-by-step.

We are given the function:

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

where:

- [tex]\( f(t) = 288.9 \)[/tex] when [tex]\( t = 5 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]

We need to find the value of [tex]\( P \)[/tex].

### Step-by-Step Solution:

1. Plug in the given values into the function:

[tex]\[ f(5) = P e^{0.05 \cdot 5} \][/tex]

2. Substitute [tex]\( f(5) \)[/tex] with 288.9:

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

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

Using a calculator or referencing the numerical result, we have:

[tex]\[ e^{0.25} \approx 1.2840254166877414 \][/tex]

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

[tex]\[ P = \frac{288.9}{1.2840254166877414} \][/tex]

5. Calculate the value of [tex]\( P \)[/tex]:

[tex]\[ P \approx 224.99554622932885 \][/tex]

The approximate value of [tex]\( P \)[/tex] is very close to 225.

### Final Answer:

[tex]\[ \boxed{225} \][/tex]

So, the correct answer is A. 225.