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You invested $2,500\$2,500 in a savings account, and after 66 years, the balance grew to $3,397.50\$3,397.50. The interest is compounded annually. What is the annual interest rate? Use the formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}, where AA is the balance (final amount), PP is the principal (starting amount), rr is the interest rate expressed as a decimal, nn is the number of times per year that the interest is compounded, and tt is the time in years. Round your answer to the nearest tenth.

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Q. You invested $2,500\$2,500 in a savings account, and after 66 years, the balance grew to $3,397.50\$3,397.50. The interest is compounded annually. What is the annual interest rate? Use the formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}, where AA is the balance (final amount), PP is the principal (starting amount), rr is the interest rate expressed as a decimal, nn is the number of times per year that the interest is compounded, and tt is the time in years. Round your answer to the nearest tenth.
  1. Identify Given Values: Identify the given values. A = $3,397.50\$3,397.50, P = $2,500\$2,500, t = 66 \text{ years}\, n = 11 \text{ (compounded annually)}
  2. Use Formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}: Use the formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}. 3,397.50=2,500(1+r)63,397.50 = 2,500\left(1 + r\right)^6
  3. Divide to Isolate (1+r)6(1 + r)^6: Divide both sides by 2,5002,500 to isolate (1+r)6(1 + r)^6. 3,397.502,500=(1+r)6\frac{3,397.50}{2,500} = (1 + r)^6 1.359=(1+r)61.359 = (1 + r)^6
  4. Take 66th Root: Take the 66th root of both sides to solve for (1+r) (1 + r) . (1.359)(1/6)=1+r (1.359)^{(1/6)} = 1 + r 1.050=1+r 1.050 = 1 + r
  5. Subtract to Solve for rr: Subtract 11 from both sides to solve for rr. 1.0501=r1.050 - 1 = r r=0.050r = 0.050
  6. Convert to Percentage: Convert rr to a percentage. r=0.050imes100extextpercentr = 0.050 imes 100 ext{ extpercent} r=5.0extextpercentr = 5.0 ext{ extpercent}

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