Respuesta :
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.
[tex]A(t) = Pe^{rt}[/tex]
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 [tex]P = 330[/tex]
$590 in the year 2007
2007 is 8 years after 1999, so P(8) = 590.
We use this to find r.
[tex]A(t) = Pe^{rt}[/tex]
[tex]590 = 330e^{8r}[/tex]
[tex]e^{8r} = \frac{590}{330}[/tex]
[tex]e^{8r} = 1.79[/tex]
Applying ln to both sides:
[tex]\ln{e^{8r}} = \ln{1.79}[/tex]
[tex]8r = \ln{1.79}[/tex]
[tex]r = \frac{\ln{1.79}}{8}[/tex]
[tex]r = 0.0726[/tex]
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).
[tex]A(t) = 330e^{0.0726t}[/tex]
[tex]A(10) = 330e^{0.0726*10} = 682[/tex]
The value of the account in the year 2009 will be $682.
