# The rate of growth of the fish population was modeled by the equation G(t) = (60,000e^-0.6t)/(1+5e^-0.6t)^2, where t is measured in years since 2000 and G in kilograms per year. If the biomass was 25,000 kg in the year 2000, what is the predicted biomass for the year 2020?

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The rate of growth of the fish population was modeled by the equation G(t) = (60,000e^-0.6t)/(1+5e^-0.6t)^2, where t is measured in years since 2000 and G in kilograms per year. If the biomass was 25,000 kg in the year 2000, what is the predicted biomass for the year 2020?

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Step 1

Given: help_outlineImage Transcriptionclose-0.6f G(t)= (1+ Se*«) -0.61 where t is measured in years since 2000. And the biomass was 25,000kg in the year 2000. fullscreen
Step 2

The net change from 2000 to 2020 is, help_outlineImage Transcriptionclose60,000e-0.61 dt (1+5eo) 20 20 0.6f (1+5e061 Let u =1+5e -0.61 0.61 dt → du = -3e Then t =0 implies, u =1+ 5(1) = 6 t = 20 implies, u=1+5(e1?) =1+3.0721061×10¯³ z 1 fullscreen
Step 3

Therefore... help_outlineImage Transcriptionclose20 (60,000e -0s") -0.61 60,000 dt du 3 = + Se-0.61 60,000 iu²du 3 6. = -20,000|u du = 20,000[ 1" ), J6 = 20,000 6. 1 z 16,666 kg fullscreen

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### Calculus 