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Sagot :
Using compound interest, it is found that:
a) A(8) = 2389.66
b) t = 31.15
c) P = 1870.85
Compound interest:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
- A(t) is the amount of money after t years.
- P is the principal(the initial sum of money).
- r is the interest rate(as a decimal value).
- n is the number of times that interest is compounded per year.
- t is the time in years for which the money is invested or borrowed.
In this problem:
- The APR is of 2.25%, hence [tex]r = 0.0225[/tex].
- No information about the number of compounding per year, hence [tex]n = 1[/tex].
Item a:
[tex]P = 2000[/tex], hence:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}[/tex]
[tex]A(8) = 2389.66[/tex]
Item b:
[tex]A(t) = 4000[/tex], hence:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]4000 = 2000\left(1 + \frac{0.0225}{1}\right)^{t}[/tex]
[tex](1.0225)^t = 2[/tex]
[tex]\log{(1.0225)^t} = \log{2}[/tex]
[tex]t\log{1.0225} = \log{2}[/tex]
[tex]t = \frac{\log{2}}{\log{1.0225}}[/tex]
[tex]t = 31.15[/tex]
Item c:
[tex]A(3) = 2000[/tex], hence:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}[/tex]
[tex]P = \frac{2000}{(1.0225)^3}[/tex]
[tex]P = 1870.85[/tex]
A similar problem is given at https://brainly.com/question/24850750
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