Answer :
We are given the exponential function
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with [tex]$f(3) = 191.5$[/tex], [tex]$r = 0.03$[/tex], and [tex]$t = 3$[/tex]. Substituting these values into the function, we have
[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 the exponential term, we find that [tex]$e^{0.09} \approx 1.094174$[/tex]. Hence,
[tex]$$
P \approx \frac{191.5}{1.094174} \approx 175.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option B.
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with [tex]$f(3) = 191.5$[/tex], [tex]$r = 0.03$[/tex], and [tex]$t = 3$[/tex]. Substituting these values into the function, we have
[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 the exponential term, we find that [tex]$e^{0.09} \approx 1.094174$[/tex]. Hence,
[tex]$$
P \approx \frac{191.5}{1.094174} \approx 175.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option B.
