IDNLearn.com provides a seamless experience for finding and sharing answers. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
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.
### 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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.