SOLUTION: Supppose 2000 bacteria are present at time t=0. Then t minutes later, the number of bacteria present will be (N)t=2000(2)^t/40 Find the number of bacteria present after 20 and 1

Question 58433: Supppose 2000 bacteria are present at time t=0. Then t minutes later, the number of bacteria present will be (N)t=2000(2)^t/40
Find the number of bacteria present after 20 and 120 minutes.
Round to the nearest whole number for each one.
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
(N)t = 50(2)^t
After 20 minutes:
(N)t = 50(2)^20 = 52,428,800
After 120 minutes:
(N)t = 50(2)^t = 6.6461399789245793645190353014017 x 10^37
or: 66,461,399,789,245,793,645,190,353,014,017,000,000
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