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

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

Answer :

To solve the problem, we need to find the approximate value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex].

### Step-by-Step Solution:

1. Identify the Given Values:
- We know that [tex]\( f(4) = 246.4 \)[/tex].
- The rate [tex]\( r = 0.04 \)[/tex].
- The time [tex]\( t = 4 \)[/tex].

2. Understand the Function:
- We are given the function [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex].
- We need to find the value of [tex]\( P \)[/tex].

3. Rearrange the Function to Solve for [tex]\( P \)[/tex]:
- Start by setting up the equation with the known values:
[tex]\[
246.4 = P \cdot e^{0.04 \cdot 4}
\][/tex]
- Simplify the exponent:
[tex]\[
246.4 = P \cdot e^{0.16}
\][/tex]

4. Solve for [tex]\( P \)[/tex]:
- Divide both sides by [tex]\( e^{0.16} \)[/tex] to isolate [tex]\( P \)[/tex]:
[tex]\[
P = \frac{246.4}{e^{0.16}}
\][/tex]

5. Approximate Calculation:
- Calculating [tex]\( e^{0.16} \)[/tex], and then [tex]\(\frac{246.4}{e^{0.16}}\)[/tex], we find the approximate value of [tex]\( P \)[/tex].

6. Select the Closest Approximate Value:
- The numerical result is approximately 210.

Therefore, the approximate value of [tex]\( P \)[/tex] is C. 210.

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