Answer :
We are given the function
$$
f(t) = P e^{rt}
$$
with $r = 0.05$ and $f(5) = 288.9$. This means that when $t = 5$, the equation becomes
$$
P e^{0.05 \times 5} = 288.9.
$$
Simplify the exponent:
$$
0.05 \times 5 = 0.25,
$$
so we have
$$
P e^{0.25} = 288.9.
$$
To solve for $P$, divide both sides of the equation by $e^{0.25}$:
$$
P = \frac{288.9}{e^{0.25}}.
$$
Evaluating $e^{0.25}$ gives approximately $1.2840254166877414$. Therefore,
$$
P \approx \frac{288.9}{1.2840254166877414} \approx 224.99554622932885.
$$
Rounding this to the nearest whole number yields $225$.
Thus, the approximate value of $P$ is $\boxed{225}$.
$$
f(t) = P e^{rt}
$$
with $r = 0.05$ and $f(5) = 288.9$. This means that when $t = 5$, the equation becomes
$$
P e^{0.05 \times 5} = 288.9.
$$
Simplify the exponent:
$$
0.05 \times 5 = 0.25,
$$
so we have
$$
P e^{0.25} = 288.9.
$$
To solve for $P$, divide both sides of the equation by $e^{0.25}$:
$$
P = \frac{288.9}{e^{0.25}}.
$$
Evaluating $e^{0.25}$ gives approximately $1.2840254166877414$. Therefore,
$$
P \approx \frac{288.9}{1.2840254166877414} \approx 224.99554622932885.
$$
Rounding this to the nearest whole number yields $225$.
Thus, the approximate value of $P$ is $\boxed{225}$.
