Albert
Compound interest formula:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where:
A: final amount
P: principal
r: annual interest rate, as a decimal
t: time in years
n: number of times interest applied per year
Substituting with P = $1000, r = 0.012 (= 1.2/100), n = 12 (interest is compounded monthly), t = 10 years, we get:
[tex]\begin{gathered} A=1000(1+\frac{0.012}{12})^{12\cdot10} \\ A=1000(1.001)^{120} \\ A=1127.43\text{ \$} \end{gathered}[/tex]
If $500 lost 2%, then it keeps 98% of its original value, that is,
$500x98% = $490
Continuous compound formula:
[tex]A=Pe^{rt}[/tex]
where the variables have the same meaning as before.
Substituting with P = $500, r = 0.008 ( = 0.8/100), and t = 10 years, we get:
[tex]\begin{gathered} A=500\cdot e^{0.008\cdot10} \\ A=541.64\text{ \$} \end{gathered}[/tex]
The balance of Albert’s $2000 after 10 years is:
$1127.43 + $490 + $541.64 = $2159.07
Marie
Substituting in the compound interest formula with P = $1500, r = 0.014 (= 1.4/100), n = 4 (interest is compounded quartely), t = 10 years, we get:
