Question

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected to grow rapidly at first but to level out when the population grows to near the maximum that the environment can support. Such growth is known as logistic population growth, and ecologists sometimes use a formula to describe it. The number N of deer present at time t (measured in years since the herd was introduced) on a certain wildlife reserve has been determined by ecologists to be given by the function N= 12.36 (12.36 divided by 0.03 + 0.55^t) ------------ 0.03 + 0.55^t (t is an exponent here) questions: i have the answers but i need help getting there... a. How many deer were initially on the reserve? 12 b. Calculate N (10) and explain the meaning of the number you have calculated. N(10)= 380 deer c. Express the number of deer present after 15 years using functional notation, and then calculate it. N(15)= 410 deer d. How much increase in the deer population do you expect from the 10th to the 15th year? 30 deer

Answer 1

Answer:

A. Initially, there were 12 deer.

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.