Question

NO LINKS!!!
Complete the table by finding the balance A when P dollars is invested at rate r for t years and compounded n times per year. (Round your answers to the nearest cent)

P = $2100, r= 8.5%, t = 9 years

n A
1 $
2 $
4 $
12 $
365 $
Continuous $

Answer 1

Answer:

$\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} n&A \\\cline{1-2}\vphantom{\dfrac12} 1& \ 4376.10\\\vphantom{\dfrac12} 2& \ 4442.10\\\vphantom{\dfrac12} 4& \ 4476.86\\\vphantom{\dfrac12} 12& \ 4500.73\\\vphantom{\dfrac12} 365& \ 4512.49\\\vphantom{\dfrac12} \sf Continuous& \ 4512.89 \\\cline{1-2} \end{array}$

Step-by-step explanation:

$\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\ \phantom{ww} \bullet P = principal amount \\ \phantom{ww} \bullet r = interest rate (in decimal form) \\ \phantom{ww} \bullet n = number of times interest is applied per year \\ \phantom{ww} \bullet t = time (in years) \\ \end{minipage}}$

Given:

• P = $2100
• r = 8.5% = 0.085
• t = 9 years

Substitute the given values into the formula to create an equation for A in terms of n:

$\implies A=2100\left(1+\dfrac{0.085}{n}\right)^{9n}$

Substitute each value of n into the equation:

$\begin{aligned}n=1 \implies A&=2100\left(1+\dfrac{0.085}{1}\right)^{9 \times 1}\\&=2100\left(1.085\right)^{9}\\&=\ 4376.10\end{aligned}$

$\begin{aligned}n=2 \implies A&=2100\left(1+\dfrac{0.085}{2}\right)^{9 \times 2}\\&=2100\left(1.0425\right)^{18}\\&=\ 4442.10\end{aligned}$

$\begin{aligned}n=4 \implies A&=2100\left(1+\dfrac{0.085}{4}\right)^{9 \times 4}\\&=2100\left(1.02125\right)^{36}\\&=\ 4476.86\end{aligned}$

$\begin{aligned}n=12 \implies A&=2100\left(1+\dfrac{0.085}{12}\right)^{9 \times 12}\\&=2100\left(1.00708333...\right)^{108}\\&=\ 4500.73\end{aligned}$

$\begin{aligned}n=365 \implies A&=2100\left(1+\dfrac{0.085}{365}\right)^{9 \times 365}\\&=2100\left(1.00023287...\right)^{3285}\\&=\ 4512.49\end{aligned}$

$\boxed{\begin{minipage}{8.5 cm}\underline{Continuous Compounding Formula}\\\\ A=Pe^{rt} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\\phantom{ww} \bullet P = principal amount \\\phantom{ww} \bullet e = Euler's number (constant) \\\phantom{ww} \bullet r = annual interest rate (in decimal form) \\\phantom{ww} \bullet t = time (in years) \\\end{minipage}}$

$\implies A=2100e^{0.085 \times 9}$

$\implies A=2100e^{0.765}$

$\implies A=2100(2.14899437...)$

$\implies A=\ 4512.89$

Input the calculated values into the table:

$\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} n&A \\\cline{1-2}\vphantom{\dfrac12} 1& \ 4376.10\\\vphantom{\dfrac12} 2& \ 4442.10\\\vphantom{\dfrac12} 4& \ 4476.86\\\vphantom{\dfrac12} 12& \ 4500.73\\\vphantom{\dfrac12} 365& \ 4512.49\\\vphantom{\dfrac12} \sf Continuous& \ 4512.89 \\\cline{1-2} \end{array}$