To estimate the value of the account 12 years after it was opened, we need to fit an exponential regression model to the given data. The model can be represented by the equation:
[tex]\[ V(t) = a \cdot e^{bt} \][/tex]
where [tex]\( V(t) \)[/tex] is the value of the account at time [tex]\( t \)[/tex] years, [tex]\( a \)[/tex] is the initial amount of the account, and [tex]\( b \)[/tex] is the growth rate.
Given the account values at different times:
- At [tex]\( t = 0 \)[/tex], [tex]\( V(0) = 5000 \)[/tex]
- At [tex]\( t = 2 \)[/tex], [tex]\( V(2) = 5510 \)[/tex]
- At [tex]\( t = 5 \)[/tex], [tex]\( V(5) = 6390 \)[/tex]
- At [tex]\( t = 8 \)[/tex], [tex]\( V(8) = 7390 \)[/tex]
- At [tex]\( t = 10 \)[/tex], [tex]\( V(10) = 8150 \)[/tex]
By performing exponential regression on these data points, we find the parameters:
[tex]\[ a \approx 5000.224 \][/tex]
[tex]\[ b \approx 0.048861 \][/tex]
Now, to estimate the value of the account 12 years after it was opened, we substitute [tex]\( t = 12 \)[/tex] into the exponential model:
[tex]\[ V(12) = 5000.224 \cdot e^{0.048861 \cdot 12} \][/tex]
This calculation gives:
[tex]\[ V(12) \approx 8987.303 \][/tex]
Next, we need to match this estimated value with the closest value from the provided choices:
- [tex]\( \$8,910 \)[/tex]
- [tex]\( \$8,980 \)[/tex]
- [tex]\( \$13,660 \)[/tex]
- [tex]\( \$16,040 \)[/tex]
The closest value to [tex]\( 8987.303 \)[/tex] is [tex]\( \$8,980 \)[/tex].
Therefore, the best estimate of the value of the account 12 years after it was opened is:
[tex]\[ \boxed{8980} \][/tex]
