This problem involves calculating the total balance $B(t)$ in two accounts:

1. An investment account that compounds monthly with a 6% annual interest rate.
2. A savings account with a constant balance of $800.

Breaking Down the Problem:

• The formula for compound interest is:
  $A = P(1 + \frac{r}{n})^{nt}$
  where:
  $P$ = principal amount ($200),
  $r$ = annual interest rate (6% = 0.06),
  $n$ = number of compounding periods per year (12 for monthly),
  $t$ = number of years.

Thus, the balance of the investment account after $t$ years is:
$200 \left(1 + \frac{0.06}{12}\right)^{12t} = 200(1.005)^{12t}.$

• Adding the savings account balance of $800, the total balance $B(t)$ becomes:
  $B(t) = 200(1.005)^{12t} + 800.$

Correct Answer:

The correct choice is: $\text{(B)} \quad B(t) = 200(1.005)^{12t} + 800.$

Let me know if you want further clarification!

Here are 5 related questions to expand your understanding:

1. What happens to the total balance if the savings account increases by $100 annually?
2. How does the compounding frequency affect the total balance in the investment account?
3. If the interest rate is increased to 8%, how would that change the formula?
4. How can you graph $B(t)$ to visualize the growth over time?
5. What is the balance after 5 years?

Tip: Always confirm the compounding frequency when working with compound interest problems—it directly impacts the formula and the results.