Respuesta :
Answer:
n = 1, A = $1,343.92
n = 2, A = $1,346.86
n = 4, A = $1,348.35
n = 12, A = $1,349.35
n = 365, A = $1,349.84
Continuous compounding, A = $1,349.86
Step-by-step explanation:
We are given the following in the question:
P = $1000
r = 3% = 0.03
t = 10 years
Formula:
The compound interest is given by
[tex]A = P\bigg(1 + \displaystyle\frac{r}{n}\bigg)^{nt}[/tex]
where P is the principal, r is the interest rate, t is the time, n is the nature of compound interest and A is the final amount.
For n = 1
[tex]A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{10}\\\\A = \$1,343.92[/tex]
For n = 2
[tex]A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{20}\\\\A = \$1,346.86[/tex]
For n = 4
[tex]A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{40}\\\\A = \$1,348.35[/tex]
For n = 12
[tex]A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{120}\\\\A = \$1,349.35[/tex]
For n = 365
[tex]A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{3650}\\\\A = \$1,349.84[/tex]
Continuous compounding:
[tex]A = Pe^{rt}\\A = 1000e^{0.03\times 10}\\A = \$1,349.86[/tex]
