a) The carrying capacity is the maximum number of elephants that the conservation area can support. In this case, we can find the carrying capacity by taking the limit of the population function as t approaches infinity.

Taking the limit as t approaches infinity, we have:

lim(t->∞) P(t) = lim(t->∞) 1900/(1+0.9e^(-0.14t))

Since the exponent -0.14t approaches negative infinity as t approaches infinity, the e^(-0.14t) term approaches 0. Therefore, the limiting value of the population function is:

lim(t->∞) P(t) = 1900/(1+0.9(0)) = 1900

So, the carrying capacity is 1900 elephants.

b) The growth rate is the rate at which the population is increasing per unit of time. In this case, we can find the growth rate by taking the derivative of the population function with respect to t.

P'(t) = dP/dt = (1900(0.9)(-0.14)e^(-0.14t))/((1+0.9e^(-0.14t))^2)

Simplifying, we have:

P'(t) = -0.2394e^(-0.14t)/(1+0.9e^(-0.14t))^2

At any given time t, the growth rate is given by P'(t).

c) The initial population is the population at t=0. Plugging in t=0 into the population function, we have:

P(0) = 1900/(1+0.9e^(-0.14(0)))
= 1900/(1+0.9e^0)
= 1900/(1+0.9(1))
= 1900/(1+0.9)
= 1900/1.9
= 1000

So, the initial population is 1000 elephants.

d) To find how long it will take to reach 1250 elephants, we can set the population function equal to 1250 and solve for t.

1250 = 1900/(1+0.9e^(-0.14t))

Multiply both sides by (1+0.9e^(-0.14t)):

1250(1+0.9e^(-0.14t)) = 1900

1250 + 1125e^(-0.14t) = 1900

1125e^(-0.14t) = 650

Divide both sides by 1125:

e^(-0.14t) = 650/1125

Take the natural logarithm of both sides:

-0.14t = ln(650/1125)

Solve for t:

t = (ln(650/1125)) / -0.14

Using a calculator, we find:

t ≈ 5 years

So, it will take approximately 5 years to reach 1250 elephants.

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