Answer
The investment with $4\%$ compounded is better by $ 179$ dollars.
Work Step by Step
Step 1. For the first account, we have
$P=6000, t=5, r=0.04, n=12$
and the end balance is
$A_1=P(1+\frac{r}{n})^{nt}=6000(1+\frac{0.04}{12})^{12(5)}\approx7326$ dollars.
Step 2. For the second account, we have
$P=6000, t=5, r=0.035$
and the end balance is
$A_2=Pe^{rt}=6000(e)^{0.035(5)}\approx7147$ dollars.
Step 3. Since $A_1\gt A_2$, the first investment is better by $A_1-A_2=179$ dollars.
