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If [tex]f(4) = 246.4[/tex] when [tex]r = 0.04[/tex] for the function [tex]f(t) = P e^t[/tex], then what is the approximate value of [tex]P[/tex]?

A. 50
B. 210
C. 1220
D. 289

Answer :

We are given the function

[tex]$$
f(t)=P e^{rt},
$$[/tex]

with [tex]$r=0.04$[/tex] and [tex]$t=4$[/tex]. This means

[tex]$$
f(4)=P e^{0.04\times4}=P e^{0.16}.
$$[/tex]

Since [tex]$f(4)=246.4$[/tex], we have:

[tex]$$
246.4=P e^{0.16}.
$$[/tex]

To solve for [tex]$P$[/tex], divide both sides of the equation by [tex]$e^{0.16}$[/tex]:

[tex]$$
P=\frac{246.4}{e^{0.16}}.
$$[/tex]

Evaluating the value, we have that [tex]$e^{0.16}\approx1.17351$[/tex]. Therefore,

[tex]$$
P\approx\frac{246.4}{1.17351}\approx209.97.
$$[/tex]

This value is approximately [tex]$210$[/tex], which corresponds to option B.

Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{210}$[/tex].

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