Answered

Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2000 there were about
1,150 fish. Write an exponential decay function to model this situation. Then find the population in
2006.

Sagot :

Answer:

Number of fish in 2006 = 845 fishes (Approx.)

Step-by-step explanation:

Given;

Number of fish in 2000 = 1,150 fishes

Decreasing rate = 5% per year

Find:

Number of fish in 2006

Computation:

Exponential decay function:

F = P[1-r]ⁿ

Number of year = 2006 - 2000

Number of year =  6 year

Number of fish in 2006 = 1,150[1-5%]⁶

Number of fish in 2006 = 1,150[1-0.05]⁶

Number of fish in 2006 = 1,150[0.95]⁶

Number of fish in 2006 = 1,150[0.735]

Number of fish in 2006 = 845.25

Number of fish in 2006 = 845 fishes (Approx.)

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.