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))
Y=-2/1+0, or just y=-2/1 because it goes through 0 so it really makes no difference
Answer:
plz mark brainliest if it helps
Answer:
,‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️☹️☹️☹️☹️☹️☹️☹️☹️☹️☹️☹️☹️☹️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️☹️☹️⏹️☹️‼️☹️‼️☹️‼️☹️‼️☹️‼️‼️☹️‼️☹️‼️☹️‼️☹️☺️‼️☺️☹️☺️☹️‼️☺️☹️☹️‼️☺️‼️☹️☹️‼️☹️‼️☹️☹️‼️‼️‼️‼️‼️☹️☹️☹️☹️‼️‼️☹️☹️☹️‼️☹️‼️☹️‼️☹️
Answer: b i just took the test
Step-by-step explanation:
