Let's first make a list of what is given in the scenario:
a.) Dakota puts 400.00 into an account to use for school expenses.
Principal Amount = 400
b.) The account earns 13% interest.
Interest rate = 13%
c.) It is compounded monthly.
n = 12
What is asked: How much will be in the account after 6 years? t = 6 years
For this type of situation, let's use the Compound Interest Formula:

Where,
A = amount after being compounded a certain interest rate in a given time.
r = interest rate, in decimal form = 13% = 13%/100% = 0.13
n = number of times the interest is being compounded = 12
t = time = 6 years
P = Principal/Initial Amount = 400.00
Let's plug in the values in the formula to be able to determine how much will be in the account after 6 years:




Therefore, Dakota's account will become 868.92 in 6 years at 13% interest compounded monthly.