Investing \( \$ 10,000 \) in a savings account at \( 5 \% \) annual inte money after 10 years? Use the formula: \( \Lambda=P\left(1+\frac{r}{m}\right)^{m t} \) \( \$ 13,448.89 \) \( \$ 15,000.00 \) \( \$ 16,288.95 \) \( \$ 16,386.16 \)

Given: - Principal amount, \( P = \$10,000 \) - Annual interest rate, \( r = 5\% = 0.05 \) - Time period, \( t = 10 \) years - Compounding frequency, \( m = 1 \) (compounded annually) We can use the formula for compound interest to calculate the amount of money in the savings account after 10 years: \[ A = P\left(1+\frac{r}{m}\right)^{m t} \] Substitute the given values into the formula: \[ A = 10000\left(1+\frac{0.05}{1}\right)^{1 \times 10} \] Now, we can calculate the amount of money in the savings account after 10 years. Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.05\right)^{10}\) - step1: Add the numbers: \(10000\times 1.05^{10}\) - step2: Convert the expressions: \(10000\left(\frac{21}{20}\right)^{10}\) - step3: Simplify: \(10000\times \frac{21^{10}}{20^{10}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{21^{10}}{20^{10}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{21^{10}}{20^{10}}\) - step6: Reduce the numbers: \(25\times \frac{21^{10}}{20^{8}}\) - step7: Rewrite the expression: \(25\times \frac{21^{10}}{5^{8}\times 4^{8}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{21^{10}}{5^{8}\times 4^{8}}\) - step9: Reduce the numbers: \(\frac{21^{10}}{5^{6}\times 4^{8}}\) - step10: Multiply the fractions: \(\frac{21^{10}}{15625\times 4^{8}}\) - step11: Expand the expression: \(\frac{21^{10}}{1024000000}\) The amount of money in the savings account after 10 years is approximately \$16,288.95. Therefore, the correct answer is \( \$16,288.95 \).