easoning (math - 1043 - 020)\nquestion 16, 4.c.78 - t\npart 3 of 3\nhw score: 68.75%, 11 of 16…

easoning (math - 1043 - 020)\nquestion 16, 4.c.78 - t\npart 3 of 3\nhw score: 68.75%, 11 of 16 points\npoints: 0 of 1\nabe deposits $50 each month for 30 years in an account with an apr of 3.5%. beatrice deposits $100 each month for 15 years in an account with an apr of 3.5%.\na. verify that abe and beatrice deposit the same amount of money during the stated periods of time. how much do they deposit?\nb. use technology to compute the accumulated balance in each account and explain the results.\na. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. abe and beatrice both deposit $18000\nb. abe deposits $ and beatrice deposits $, different from abes deposit.\nb. the accumulated balance in abes account is $31770.64\n(round to the nearest cent as needed.)\nthe accumulated balance in beatrices account is $ \n(round to the nearest cent as needed.)

easoning (math - 1043 - 020)\nquestion 16, 4.c.78 - t\npart 3 of 3\nhw score: 68.75%, 11 of 16 points\npoints: 0 of 1\nabe deposits $50 each month for 30 years in an account with an apr of 3.5%. beatrice deposits $100 each month for 15 years in an account with an apr of 3.5%.\na. verify that abe and beatrice deposit the same amount of money during the stated periods of time. how much do they deposit?\nb. use technology to compute the accumulated balance in each account and explain the results.\na. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. abe and beatrice both deposit $18000\nb. abe deposits $ and beatrice deposits $, different from abes deposit.\nb. the accumulated balance in abes account is $31770.64\n(round to the nearest cent as needed.)\nthe accumulated balance in beatrices account is $ \n(round to the nearest cent as needed.)

Answer

Explanation:

Step1: Calculate the total deposits for Abe

Abe deposits $50$ per month for $30$ years. There are $30\times12 = 360$ months. So the total deposit amount for Abe is $50\times360=18000$.

Step2: Calculate the total deposits for Beatrice

Beatrice deposits $100$ per month for $15$ years. There are $15\times 12=180$ months. So the total deposit amount for Beatrice is $100\times180 = 18000$.

Step3: Use 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 amount of each payment, $r$ is the interest rate per period, and $n$ is the number of periods. The annual percentage rate (APR) is $3.5%$, so the monthly interest rate $r=\frac{0.035}{12}$. For Abe: $A = 50$, $n = 360$, $r=\frac{0.035}{12}$. [ \begin{align*} F_{Abe}&=50\times\frac{(1+\frac{0.035}{12})^{360}-1}{\frac{0.035}{12}}\ &\approx50\times\frac{(1 + 0.0029167)^{360}-1}{0.0029167}\ &\approx50\times\frac{(1.0029167)^{360}-1}{0.0029167}\ &\approx50\times\frac{2.83748 - 1}{0.0029167}\ &\approx50\times\frac{1.83748}{0.0029167}\ &\approx31770.64 \end{align*} ] For Beatrice: $A = 100$, $n = 180$, $r=\frac{0.035}{12}$. [ \begin{align*} F_{Beatrice}&=100\times\frac{(1+\frac{0.035}{12})^{180}-1}{\frac{0.035}{12}}\ &\approx100\times\frac{(1 + 0.0029167)^{180}-1}{0.0029167}\ &\approx100\times\frac{(1.0029167)^{180}-1}{0.0029167}\ &\approx100\times\frac{1.69092 - 1}{0.0029167}\ &\approx100\times\frac{0.69092}{0.0029167}\ &\approx23687.77 \end{align*} ]

Answer:

a. A. Abe and Beatrice both deposit $18000$. b. The accumulated balance in Abe's account is $$31770.64$. The accumulated balance in Beatrice's account is $$23687.77$. Although they both deposit the same total principal amount of $$18000$, Abe's account has a higher accumulated balance because his money is in the account for a longer time, allowing more time for compound - interest to work.