#### Answer

$ \$ 1246.08$

#### Work Step by Step

Recall:
Continuous Compounding Formula
$$A = P e^{r t}$$
where
$P:$ The principal amount
$r:$ Annual interest rate
$t:$ Number of years
$A:$ Amount after $t$ years
The given problem has:
$P= \$ 1000, r = 0.11, t=2$
Thus, using these values and the formula above gives:
$A = 1000 \times e^{0.11 \times 2}$
$A \approx \$ 1246.08$