Answer:THe answer is A
Step-by-step explanation:
I think it's true but I'm not 100% sure
Answer:
Step-by-step explanation:
Plug into the midpoint formula
<u>Hope this helps :-)</u>
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Answer:
The population of deer at any given time = 200(e^0.03t) ÷ (1.5 + (e^0.03t))
Step-by-step explanation:
This is an example of logistic equation on population growth
carrying capacity, k = 200
Rate, r = 3% = 0.03
Initial Population, P1 = 80
P(t) =?
P(t) = (P1 (k)(e^rt)) ÷ (k- P1 + P1(e^rt))
P(t) = (80 (200)(e^0.03t)) ÷ (200 - 80 + 80(e^0.03t))
= (16000(e^0.03t)) ÷ (120 + 80(e^0.03t))
= 200(e^0.03t) ÷ (1.5 + (e^0.03t))
Answer:
the answer is 2+_+_:$-!$:_
