Answer:
The total balance on December 1, 2016 is B = $7356.28
Explanation:
I am going to call B1 the balance for the first deposit, B2 the balance for the second deposit and B3 for the third deposit.
This is a compound interest problem.
The compound interest formula is given by:
![A = P(1 + (r)/(n))^(nt)](https://img.qammunity.org/2020/formulas/mathematics/college/dsad63du8aukkfd64adgjgs94f0mgywaeq.png)
A: Amount of money(Balance)
P: Principal(Initial deposit)
r: interest rate(as a decimal value)
n: number of times that interest is compounded per unit t
t: time the money is invested or borrowed for.
First step: Balance B1 on December 1, 2016
A = amount of money = B1
P = 1,700
r = 0.056
n = 12(compounded mothly, and t is in years)
t = 8 years
![B1 = 1700(1 + (0.056)/(12))^(96)](https://img.qammunity.org/2020/formulas/mathematics/college/9ail5zz7nzjkg835pd686t6xtq0nc1jcsu.png)
![B1 = $2658.17](https://img.qammunity.org/2020/formulas/mathematics/college/9hx32y3nndvwiex85dpl17id2th7zyjxxj.png)
Second step: Balance B2 on December 1, 2016
A = B3
P = 1,700
r = 0.056
n = 12(compounded mothly, and t is in years)
t: From January 1, 2010 to December 1, 2016 there are 6 years and 11 months. 11 months is 0.92 of a year. So
t = 6.92
![B2 = 1700(1 + (0.056)/(12))^(83.04)](https://img.qammunity.org/2020/formulas/mathematics/college/v6qlwgmzugkh5fgm79qrndz42n6sp1tsw2.png)
![B2 = $2502.39](https://img.qammunity.org/2020/formulas/mathematics/college/4b0hx5ttjaqp583ib7cs10s56ohc4ek4p0.png)
Third step: Balance B3 on December 1, 2016
A = B3
P = 1700
r = 0.056
n = 12
t: 4 years and 7 months. 7 months is 7/12 = 0.58. So t = 4.58 years.
![B3 = 1700(1 + (0.056)/(12))^(54.96)](https://img.qammunity.org/2020/formulas/mathematics/college/s4c8jus65exnitcq6uarv88wzzrl87mik1.png)
![B3 = $2195.72](https://img.qammunity.org/2020/formulas/mathematics/college/27xgtooojjg1zo55fnhmi4bolvrlly24hq.png)
Final step: Total balance on December 1, 2016
B = 2195.72 + 2502.39 + 2658.17
B = $7356.28
So the total balance on December 1, 2016 is B = $7356.28