Question

Complete the table to determine the effect of the number of compounding periods when computing interest. Suppose that $8000 is invested at 3.8%
interest for 12 yr under the following compounding options. Round answers in the second column to the nearest whole number. Round answers in the
last column to the nearest cent.
Compounding Option
n Value
Result
x
(a) Annually
72 =
S
(b) Quarterly
$
(c) Monthly
12 =
S
(d) Daily
n = 365
(e) Continuously
Not Applicable
SI​

Answer 1

Answer:

$\left|\begin{array}{c|c|c|c}& Compounding Option& n Value& Result(\ )\\----&----&----&----\\(a)& Annually&1&\ 12515.79\\(b)& Quarterly&4&\ 12585.91\\(c)& Monthly&12&\ 12616.91\\(d)& Daily&365&\ 12621.70\\(e)& Continuosly& Not Applicable &\ 12622.00\end{array}\right|$

Explanation:

The compound interest formula is:

$A=P(1+\frac{r}{n})^{nt}$ where:

• A = accumulated amount = ?
• P = Principal amount = $8000
• r= Rate of interest = 3.8% = 0.038
• t = number of years = 12 years
• n= Compounding Period

We are required to complete the table below:

$\left|\begin{array}{c|c|c|c}& Compounding Option& n Value& Result(\ )\\(a)& Annually&\\(b)& Quarterly&\\(c)& Monthly&\\(d)& Daily&365\\(e)& Continuosly& Not Applicable &\end{array}\right|$

(a)Annually

$n=1\\A=8000(1+\frac{0.038}{1})^{1*12}\\=\ 12515.79$

(b)Quarterly

$n=3\\A=8000(1+\frac{0.038}{3})^{3*12}\\=\ 12585.91$

(c)Monthly

$n=12\\A=8000(1+\frac{0.038}{12})^{12*12}\\=\ 12616.91$

(d)Daily

$n=365\\A=8000(1+\frac{0.038}{365})^{365*12}\\=\ 12621.70$

(e)Continuously

$P(t)=P_0e^{rt}\\=8000 \times e^{0.038 \times 12}\\=\ 12622.00$

The completed table therefore is:

$\left|\begin{array}{c|c|c|c}& Compounding Option& n Value& Result(\ )\\----&----&----&----\\(a)& Annually&1&\ 12515.79\\(b)& Quarterly&4&\ 12585.91\\(c)& Monthly&12&\ 12616.91\\(d)& Daily&365&\ 12621.70\\(e)& Continuosly& Not Applicable &\ 12622.00\end{array}\right|$