Answer
a) $y=(10000)(0.9)^t$
b) $5905$
c) $21.854$ years
Work Step by Step
(a) The decay factor is $1-0.1= 0.90$ In general, the number of cases after $t$ years is $y=(10000)(0.9)^t$.
b) Setting $t=5$, we obtain the number of cases 5 years from the start:
$$
y=(10000) \cdot(0.9)^5=5904.9 \approx 5905 \text { cases }
$$
c) If we $y=(10000) \cdot(0.9)^t$ , the approximating the value of $t$ for which $y=1000$, is about $t = 21.854$ years
