Answer :
We are given the function
[tex]$$
f(t)=P e^{rt},
$$[/tex]
with [tex]$f(3)=191.5$[/tex] and [tex]$r=0.03$[/tex]. The function at [tex]$t=3$[/tex] is
[tex]$$
f(3)=P e^{0.03 \times 3}=P e^{0.09}.
$$[/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]
Next, we compute [tex]$e^{0.09}$[/tex]. The value is approximately
[tex]$$
e^{0.09}\approx1.0941742837052104.
$$[/tex]
Now substitute this value back into the equation for [tex]$P$[/tex]:
[tex]$$
P \approx \frac{191.5}{1.0941742837052104} \approx 175.02.
$$[/tex]
Rounding to the nearest whole number, we find
[tex]$$
P \approx 175.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option D.
[tex]$$
f(t)=P e^{rt},
$$[/tex]
with [tex]$f(3)=191.5$[/tex] and [tex]$r=0.03$[/tex]. The function at [tex]$t=3$[/tex] is
[tex]$$
f(3)=P e^{0.03 \times 3}=P e^{0.09}.
$$[/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]
Next, we compute [tex]$e^{0.09}$[/tex]. The value is approximately
[tex]$$
e^{0.09}\approx1.0941742837052104.
$$[/tex]
Now substitute this value back into the equation for [tex]$P$[/tex]:
[tex]$$
P \approx \frac{191.5}{1.0941742837052104} \approx 175.02.
$$[/tex]
Rounding to the nearest whole number, we find
[tex]$$
P \approx 175.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex], which corresponds to option D.
