A species with an initial population of 120 is growing in an environment where the carrying capacity is 750 . After 3 years the population is up to 450. Use a logistic function to estimate the population after 5 years. Round to the nearest whole number. Population = [? ] individuals.

We can model the population using a logistic growth function of the form   P(t) = K / [1 + A·e^(–r·t)], where   K = carrying capacity,   A is a constant determined by the initial condition, and   r is the growth rate. Step 1. Find A using the initial condition P(0) = 120.   At t = 0:    P(0) = 750 / (1 + A) = 120.   Solve for A:    1 + A = 750/120 = 6.25  ⟹  A = 6.25 – 1 = 5.25. Step 2. Determine r using the condition P(3) = 450.   At t = 3:    450 = 750 / [1 + 5.25·e^(–3r)].   Solve for e^(–3r):    1 + 5.25·e^(–3r) = 750/450 = 5/3.    5.25·e^(–3r) = (5/3) – 1 = 2/3.   Thus, e^(–3r) = (2/3) / 5.25.   Since 5.25 = 21/4, we have:    e^(–3r) = (2/3)·(4/21) = 8/63.   Taking natural logs:    –3r = ln(8/63),   so r = –(1/3)·ln(8/63) = (1/3)·ln(63/8). Step 3. Estimate the population after 5 years, P(5).   P(5) = 750 / [1 + 5.25·e^(–r·5)].   We already have r = (1/3)·ln(63/8). Compute the exponent:    r·5 = (5/3)·ln(63/8).   Thus, e^(–5r) = e^(–(5/3)·ln(63/8)) = (63/8)^(–5/3) = (8/63)^(5/3). You can now compute the approximate value. First, calculate r numerically:   r = (1/3)·ln(63/8).   Calculate 63/8 = 7.875.   ln(7.875) ≈ 2.066, so r ≈ 2.066/3 ≈ 0.6887 per year. Then compute the exponent for t = 5:   5r ≈ 5 × 0.6887 = 3.4435.   Thus, e^(–3.4435) ≈ 0.03205. Now find the denominator:   Denom = 1 + 5.25·(0.03205) ≈ 1 + 0.16826 = 1.16826. Finally, the population at 5 years is:   P(5) ≈ 750 / 1.16826 ≈ 642.22. Rounded to the nearest whole number, the population after 5 years is approximately 642 individuals. Population = 642 individuals.