Answer:
The value of the account in the year 2009 will be $682.
Step-by-step explanation:
The acount's balance, in t years after 1999, can be modeled by the following equation.

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.
$330 in an account in the year 1999
This means that 
$590 in the year 2007
2007 is 8 years after 1999, so P(8) = 590.
We use this to find r.




Applying ln to both sides:




Determine the value of the account, to the nearest dollar, in the year 2009.
2009 is 10 years after 1999, so this is A(10).


The value of the account in the year 2009 will be $682.