A radioactive substance decays according to A(t) = 800(0.85)^t, where t is in years. Which value is closest to the amount remaining after…

Substitute $t=4$: $A(4)=800(0.85)^4$. Since $0.85^2=0.7225$ and $0.7225^2\approx0.522$, we get $A(4)\approx800(0.522)\approx417.6$, which is closest to $418$, choice F. Choice J ($680$) applies the decay factor once, $800(0.85)$, instead of four times. Choices G and H are values that do not come from $800(0.85)^t$ for any whole number of years near $4$. Because the substance loses $15\%$ each year, after four years about half should remain, and $418$ is just over half of $800$.