Sure, I'll help you complete the table with a detailed solution based on the given interest rate, principal, and time period. Let's fill in the values one by one.
### Given Values:
- Principal ([tex]\(P\)[/tex]) = \[tex]$14,000
- Annual interest rate (\(r\)) = 4% or 0.04
- Time (\(t\)) = 8 years
### Formulas:
1. Compounded annually:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where \( n = 1 \)
2. Compounded quarterly:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where \( n = 4 \)
3. Compounded monthly:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where \( n = 12 \)
4. Compounded daily:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where \( n = 365 \)
5. Compounded continuously:
\[ A = Pe^{rt} \]
Where \( e \approx 2.718 \)
### Calculations:
1. Annually:
- \( n = 1 \)
- Result: \( \$[/tex] 19,160 \)
2. Quarterly:
- [tex]\( n = 4 \)[/tex]
- Result: [tex]\( \$ 19,249 \)[/tex]
3. Monthly:
- [tex]\( n = 12 \)[/tex]
- Result: [tex]\( \$ 19,270 \)[/tex]
4. Daily:
- [tex]\( n = 365 \)[/tex]
- Result: [tex]\( \$ 19,279 \)[/tex]
5. Continuously:
- Result: [tex]\( \$ 19,279.79 \)[/tex]
Now, let's complete the table with these values:
\begin{tabular}{|l|l|c|c|}
\hline
& Compounding Option & [tex]$n$[/tex] Value & Result \\ \hline
(a) & Annually & [tex]$n=1$[/tex] & \[tex]$19,160 \\ \hline
(b) & Quarterly & $[/tex]n=4[tex]$ & \$[/tex]19,249 \\ \hline
(c) & Monthly & [tex]$n=12$[/tex] & \[tex]$19,270 \\ \hline
(d) & Daily & $[/tex]n=365[tex]$ & \$[/tex]19,279 \\ \hline
(e) & Continuously & Not Applicable & \[tex]$19,279.79 \\ \hline
\end{tabular}
Thus, the table shows the effect of the number of compounding periods on the final amount after 8 years given an initial investment of \$[/tex]14,000 at a 4% interest rate.
