Answer :
We are given the function
[tex]$$
f(t) = P e^{r t},
$$[/tex]
and the information that when [tex]$t = 3$[/tex] and [tex]$r = 0.03$[/tex], we have
[tex]$$
f(3) = 191.5.
$$[/tex]
This means that
[tex]$$
191.5 = P e^{0.03 \cdot 3}.
$$[/tex]
Since
[tex]$$
0.03 \cdot 3 = 0.09,
$$[/tex]
the equation becomes
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], we isolate it by dividing both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Evaluating [tex]$e^{0.09}$[/tex], we find that it is approximately [tex]$1.09417$[/tex]. Thus,
[tex]$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$[/tex]
Therefore, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option C.
[tex]$$
f(t) = P e^{r t},
$$[/tex]
and the information that when [tex]$t = 3$[/tex] and [tex]$r = 0.03$[/tex], we have
[tex]$$
f(3) = 191.5.
$$[/tex]
This means that
[tex]$$
191.5 = P e^{0.03 \cdot 3}.
$$[/tex]
Since
[tex]$$
0.03 \cdot 3 = 0.09,
$$[/tex]
the equation becomes
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], we isolate it by dividing both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Evaluating [tex]$e^{0.09}$[/tex], we find that it is approximately [tex]$1.09417$[/tex]. Thus,
[tex]$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$[/tex]
Therefore, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option C.
