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. 225
C. 371
D. 3520

Answer :

Sure, let's solve the problem step by step.

We are given a function [tex]\( f(t) = P \cdot e^{rt} \)[/tex], where the values for [tex]\( f(5) \)[/tex], [tex]\( r \)[/tex], and [tex]\( t \)[/tex] are provided. We need to find the approximate value of [tex]\( P \)[/tex].

Here's what we know:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]

The function we have is:

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

Substituting the given values into the function gives us:

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

First, let's calculate the exponent:

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

Next, find [tex]\( e^{0.25} \)[/tex].

Using the calculated value for the exponential part:

[tex]\[ e^{0.25} \approx 1.284 \][/tex]

Now, we substitute this back into the equation:

[tex]\[ 288.9 = P \cdot 1.284 \][/tex]

To solve for [tex]\( P \)[/tex], divide both sides of the equation by 1.284:

[tex]\[ P = \frac{288.9}{1.284} \][/tex]

[tex]\[ P \approx 225 \][/tex]

Therefore, the approximate value of [tex]\( P \)[/tex] is 225.

So the correct answer is B. 225.