Answer :
We are given the function
[tex]$$
f(t) = P e^{rt}
$$[/tex]
with [tex]$r = 0.03$[/tex]. Since [tex]$f(3) = 191.5$[/tex], we substitute into the equation:
[tex]$$
191.5 = P e^{0.03 \times 3}.
$$[/tex]
First, compute the exponent:
[tex]$$
0.03 \times 3 = 0.09.
$$[/tex]
Now the equation becomes:
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], divide both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Using the approximate value:
[tex]$$
e^{0.09} \approx 1.0941742837052104,
$$[/tex]
we substitute to get:
[tex]$$
P \approx \frac{191.5}{1.0941742837052104} \approx 175.01782197944019.
$$[/tex]
Thus, the value of [tex]$P$[/tex] is approximately [tex]$175$[/tex], which corresponds to option A.
[tex]$$
f(t) = P e^{rt}
$$[/tex]
with [tex]$r = 0.03$[/tex]. Since [tex]$f(3) = 191.5$[/tex], we substitute into the equation:
[tex]$$
191.5 = P e^{0.03 \times 3}.
$$[/tex]
First, compute the exponent:
[tex]$$
0.03 \times 3 = 0.09.
$$[/tex]
Now the equation becomes:
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], divide both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Using the approximate value:
[tex]$$
e^{0.09} \approx 1.0941742837052104,
$$[/tex]
we substitute to get:
[tex]$$
P \approx \frac{191.5}{1.0941742837052104} \approx 175.01782197944019.
$$[/tex]
Thus, the value of [tex]$P$[/tex] is approximately [tex]$175$[/tex], which corresponds to option A.
