t0 = 80000 * (1/2)^9
t180 = 80000 + 2^9
The doubling period of a bacterial population is 10 minutes. At time t=90
minutes, the bacterial population was 80000.
What was the initial population at time t=0?
Incorrect
Find the size of the bacterial population after 3 hours
3 answers
count = initial * 2^(t/10) , where t is in minutes
80000 = initial * (2^9)
initial = 8000/2^9 = 156.25
(mathematically possible, but we can't have partial bacteria, but anyway.....)
count after 3 hours, t = 180
count = 156.25 (2^18) = 40,960,000
80000 = initial * (2^9)
initial = 8000/2^9 = 156.25
(mathematically possible, but we can't have partial bacteria, but anyway.....)
count after 3 hours, t = 180
count = 156.25 (2^18) = 40,960,000
correction...hit the wrong key
t180 = 80000 × 2^9
t180 = 80000 × 2^9