Answer :
To find the approximate value of [tex]\( P \)[/tex], we have the function [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex]. Given that [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex], we need to solve for [tex]\( P \)[/tex].
The equation becomes:
[tex]\[
f(5) = P \cdot e^{0.05 \cdot 5}
\][/tex]
This simplifies to:
[tex]\[
288.9 = P \cdot e^{0.25}
\][/tex]
To find [tex]\( P \)[/tex], we solve:
[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]
Here, [tex]\( e^{0.25} \approx 1.284 \)[/tex].
So, substituting the approximate value:
[tex]\[
P \approx \frac{288.9}{1.284}
\][/tex]
[tex]\[
P \approx 225
\][/tex]
Therefore, the approximate value of [tex]\( P \)[/tex] is 225.
The correct answer choice is D. 225.
The equation becomes:
[tex]\[
f(5) = P \cdot e^{0.05 \cdot 5}
\][/tex]
This simplifies to:
[tex]\[
288.9 = P \cdot e^{0.25}
\][/tex]
To find [tex]\( P \)[/tex], we solve:
[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]
Here, [tex]\( e^{0.25} \approx 1.284 \)[/tex].
So, substituting the approximate value:
[tex]\[
P \approx \frac{288.9}{1.284}
\][/tex]
[tex]\[
P \approx 225
\][/tex]
Therefore, the approximate value of [tex]\( P \)[/tex] is 225.
The correct answer choice is D. 225.
