In a lab experiment, 60 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to doub
le every 21 hours. How long would it be, to the nearest tenth of an hour, until there are 107 bacteria present?
1 answer:
Answer:
17.5 hours
Step-by-step explanation:
107 = 60(2^n)
n(lg2) = lg(107/60)
n = 0.8345763908 × 21
n = 17.52610421 hours
You might be interested in
<span>Not all functions will have intercepts</span>
Answer:
The Office
Step-by-step explanation:
Answer:
1 - 3/5 = 2/5
Step-by-step explanation:
The answer is u=4
u(2+u)-2u=16
2u+u^2-2u=16
The 2u's cancel out
u^2=16
You find the square root of u^2 and 16 which leaves you with
u=4
