consider the two savings plans below. compare the balances in each plan after 10 years. which person…

consider the two savings plans below. compare the balances in each plan after 10 years. which person deposited more money in the plan? which of the two investment strategies is better? yolanda deposits $200 per month in an account with an apr of 6%, while zach deposits $2000 at the end of each year in an account with an apr of 6.5%. the balance in yolandas saving plan after 10 years was $32775.87 (round the final answer to the nearest cent as needed. round all intermediate values to seven decimal places as needed.) the balance in zachs saving plan after 10 years was $ (round the final answer to the nearest cent as needed. round all intermediate values to seven decimal places as needed.)

consider the two savings plans below. compare the balances in each plan after 10 years. which person deposited more money in the plan? which of the two investment strategies is better? yolanda deposits $200 per month in an account with an apr of 6%, while zach deposits $2000 at the end of each year in an account with an apr of 6.5%. the balance in yolandas saving plan after 10 years was $32775.87 (round the final answer to the nearest cent as needed. round all intermediate values to seven decimal places as needed.) the balance in zachs saving plan after 10 years was $ (round the final answer to the nearest cent as needed. round all intermediate values to seven decimal places as needed.)

Answer

Explanation:

Step1: Recall the future - value of an ordinary annuity formula

The formula for the future - value of an ordinary annuity is $F = A\times\frac{(1 + r)^{n}-1}{r}$, where $A$ is the annual payment, $r$ is the interest rate per period, and $n$ is the number of periods. For Zach, $A = 2000$, the annual percentage rate (APR) is $6.5%=0.065$, and $n = 10$.

Step2: Substitute the values into the formula

$F=2000\times\frac{(1 + 0.065)^{10}-1}{0.065}$. First, calculate $(1 + 0.065)^{10}$. Using the formula $a^{b}$, where $a = 1.065$ and $b = 10$, we have $(1.065)^{10}\approx1.8771374$. Then, $(1.065)^{10}-1\approx1.8771374 - 1=0.8771374$. $\frac{(1.065)^{10}-1}{0.065}\approx\frac{0.8771374}{0.065}\approx13.4944215$. $F = 2000\times13.4944215=26988.843\approx26988.84$.

Answer:

$26988.84$