A population of deer inside a park has a carrying capacity of 200 and a growth rate of 3%. If the initial population is 80 deer, what is the population of deer at any given time?

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))

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

step-by-step explanation: