Explanation
Albert) Let's define
[tex]\begin{cases}A=\text{ money earned from 1000 dollars and 1.2\% of annual interest compounded monthly,} \\ L=\text{ 2\% of 500 dollars, lost over the course of the ten years,} \\ B=\text{ money earned from 500 dollars growing compounded continuously at a rate of 0.8\% annually.}\end{cases}[/tex]
Then,
[tex]M(\text{Albert})=A+(500-L)+B.[/tex]
To calculate A, we have the following compound interest formula:
[tex]A=1000\cdot(1+\frac{0.012}{12})^{12\cdot10}\approx1127.43[/tex]
L is easy to calculate:
[tex]L=0.02\cdot500=10.[/tex]
To calculate B, we have a formula as well:
[tex]B=500\cdot e^{0.008\cdot10}\approx541.64.[/tex]
Then,
[tex]M(\text{Albert})\approx1127.43+(500-10)+541.64=2159.07.[/tex]
Answer
The balance of Albert's $2000 after ten years is $2159.07.
