Answer :
We are given the function
$$
f(t) = P e^{rt},
$$
with $r = 0.03$ and the condition $f(3) = 191.5$. Substituting $t=3$ and $r=0.03$ into the function, we get
$$
f(3) = P e^{0.03 \times 3} = P e^{0.09} = 191.5.
$$
To solve for $P$, divide both sides of the equation by $e^{0.09}$:
$$
P = \frac{191.5}{e^{0.09}}.
$$
Evaluating the denominator, we find that
$$
e^{0.09} \approx 1.09417.
$$
Thus, the value of $P$ is approximately
$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$
Therefore, the approximate value of $P$ is $\boxed{175}$, which corresponds to option C.
$$
f(t) = P e^{rt},
$$
with $r = 0.03$ and the condition $f(3) = 191.5$. Substituting $t=3$ and $r=0.03$ into the function, we get
$$
f(3) = P e^{0.03 \times 3} = P e^{0.09} = 191.5.
$$
To solve for $P$, divide both sides of the equation by $e^{0.09}$:
$$
P = \frac{191.5}{e^{0.09}}.
$$
Evaluating the denominator, we find that
$$
e^{0.09} \approx 1.09417.
$$
Thus, the value of $P$ is approximately
$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$
Therefore, the approximate value of $P$ is $\boxed{175}$, which corresponds to option C.
