Answer :
We are given the function
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with [tex]$r = 0.05$[/tex], and it is known that [tex]$f(5) = 288.9$[/tex]. To find [tex]$P$[/tex], we first substitute [tex]$t = 5$[/tex] and [tex]$r = 0.05$[/tex] into the function:
[tex]$$
f(5) = P e^{0.05 \cdot 5} = P e^{0.25}.
$$[/tex]
Since [tex]$f(5) = 288.9$[/tex], we have:
[tex]$$
288.9 = P e^{0.25}.
$$[/tex]
To solve for [tex]$P$[/tex], divide both sides of the equation by [tex]$e^{0.25}$[/tex]:
[tex]$$
P = \frac{288.9}{e^{0.25}}.
$$[/tex]
Using the approximation [tex]$e^{0.25} \approx 1.2840254166877414$[/tex], we find
[tex]$$
P \approx \frac{288.9}{1.2840254166877414} \approx 224.9955.
$$[/tex]
Rounding to the nearest whole number gives
[tex]$$
P \approx 225.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{225}$[/tex], which corresponds to option C.
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with [tex]$r = 0.05$[/tex], and it is known that [tex]$f(5) = 288.9$[/tex]. To find [tex]$P$[/tex], we first substitute [tex]$t = 5$[/tex] and [tex]$r = 0.05$[/tex] into the function:
[tex]$$
f(5) = P e^{0.05 \cdot 5} = P e^{0.25}.
$$[/tex]
Since [tex]$f(5) = 288.9$[/tex], we have:
[tex]$$
288.9 = P e^{0.25}.
$$[/tex]
To solve for [tex]$P$[/tex], divide both sides of the equation by [tex]$e^{0.25}$[/tex]:
[tex]$$
P = \frac{288.9}{e^{0.25}}.
$$[/tex]
Using the approximation [tex]$e^{0.25} \approx 1.2840254166877414$[/tex], we find
[tex]$$
P \approx \frac{288.9}{1.2840254166877414} \approx 224.9955.
$$[/tex]
Rounding to the nearest whole number gives
[tex]$$
P \approx 225.
$$[/tex]
Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{225}$[/tex], which corresponds to option C.
