Answer:
Option B, 218 deer per year
Explanation:
As we know
[tex]\frac{dP}{dt} = kP (1- \frac{P}{K})[/tex]
Where K is the carrying capacity
K is the rate constant
dP/dt is the change in population per unit time and
P is the population
Substituting the given values we get
[tex]\frac{dP}{dt} = k * 420 * (1-\frac{420}{875})\\= 1 *420 * (1-0.48)\\= 218[/tex]
Option B is correct
