Answer :
We are given the function
[tex]$$
f(t) = P \cdot e^{r t}
$$[/tex]
with the values [tex]\( f(4) = 246.4 \)[/tex] and [tex]\( r = 0.04 \)[/tex].
1. Substitute [tex]\( t = 4 \)[/tex] into the function:
[tex]$$
f(4) = P \cdot e^{0.04 \times 4} = P \cdot e^{0.16}.
$$[/tex]
2. Since [tex]\( f(4) = 246.4 \)[/tex], we have:
[tex]$$
246.4 = P \cdot e^{0.16}.
$$[/tex]
3. Solve for [tex]\( P \)[/tex] by dividing both sides by [tex]\( e^{0.16} \)[/tex]:
[tex]$$
P = \frac{246.4}{e^{0.16}}.
$$[/tex]
4. Evaluating the exponential value [tex]\( e^{0.16} \)[/tex] (approximately 1.17351) gives:
[tex]$$
P \approx \frac{246.4}{1.17351} \approx 209.97.
$$[/tex]
Thus, the approximate value of [tex]\( P \)[/tex] is 210.
[tex]$$
f(t) = P \cdot e^{r t}
$$[/tex]
with the values [tex]\( f(4) = 246.4 \)[/tex] and [tex]\( r = 0.04 \)[/tex].
1. Substitute [tex]\( t = 4 \)[/tex] into the function:
[tex]$$
f(4) = P \cdot e^{0.04 \times 4} = P \cdot e^{0.16}.
$$[/tex]
2. Since [tex]\( f(4) = 246.4 \)[/tex], we have:
[tex]$$
246.4 = P \cdot e^{0.16}.
$$[/tex]
3. Solve for [tex]\( P \)[/tex] by dividing both sides by [tex]\( e^{0.16} \)[/tex]:
[tex]$$
P = \frac{246.4}{e^{0.16}}.
$$[/tex]
4. Evaluating the exponential value [tex]\( e^{0.16} \)[/tex] (approximately 1.17351) gives:
[tex]$$
P \approx \frac{246.4}{1.17351} \approx 209.97.
$$[/tex]
Thus, the approximate value of [tex]\( P \)[/tex] is 210.
