## Intermediate Algebra: Connecting Concepts through Application

a) $184,320$ bacteria b) $\approx1.67$ hours
a) $4$ hours $= 240$ minutes $= 12$ 20-minute intervals Therefore $n = 12$ $B(n) = 45(2^{n})$ $B(12) = 45(2^{12})$ $= 45(4096)$ $= 184,320$ bacteria b) $1440 = 45(2^{n})$ $\frac{1440}{45} =2^{n}$ $32 = 2^{n}$ $2^{5} = 2^{n}$ $n = 5$ $5 \times 20 = 100$ minutes $\approx 1.67$ hours