Question

An initial population of 810 fish is introduced into a take. This fish population grows according to a continuous exponential growth model. There are 972 fish in the lake after 11 years. (a) Let t be the time (in years) since the inltial population is introduced, and let y be the number of fish at time t. Write a formula relating y to t. Use exact expresslons to fil in the missing parts of the formula. Do not use approximations. y = __e⁻⁻t (b) How many fish are there 21 years atter the initial population is introduced? Do not raund amy intermedste computations, and round your answer to the nearest whole number.

Answer 1

The formula relating the number of fish to the time is y = 810 * e^((1/11) * ln(972/810) * t). Plugging in t = 21 will give you the number of fish after 21 years.

Step-by-step explanation:

To write a formula relating the number of fish, y, to the time, t, we can use the exponential growth model which is given by the formula: y = A * e^(kt), where A is the initial population and k is the growth rate. In this case, the initial population is 810 fish and the number of fish after 11 years is 972. Plugging these values into the formula, we get: 972 = 810 * e^(11k). To find the value of k, we need to solve this equation for k:

972/810 = e^(11k)

e^(11k) = 972/810

11k = ln(972/810)

k = (1/11) * ln(972/810)

So the formula relating y to t is: y = 810 * e^((1/11) * ln(972/810) * t)

(b) To find the number of fish after 21 years, we can plug t = 21 into the formula:

y = 810 * e^((1/11) * ln(972/810) * 21)

Calculating this value will give you the approximate number of fish after 21 years.

Learn more about Exponential growth