Question

After 10 years, a 2000-dollar investment compounded annually has grown to 3600. What is the interest rate to the nearest whole-number percent?

Answer 1

In this question we need to use the compound interest formula, which is

$A= P(1+r)^n$

A represents the amount after n years and P is the principal.

And in this question, P = 2000, n =10 years, r is unknown and A = 3600.

Substituting these values in the formula

$3600 = 2000(1+r)^{10}$

$\frac{9}{5} = (1+r)^{10}$

$1+r = ( \frac{9}{5})^{1/10}$

1+r = 1.06054

r=0.06054 or approx 6%

Answer 2

Answer:

6%

Step-by-step explanation:

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