Kathleen Carty

verified

This problem has been solved

by verified expert Kathleen Carty

100% free to try

Question

The population of a fish farm in t years is modeled by the equation P(t) = (1300)/(1 + 4e^(-0.65t)). Answer the following: 1. What is the initial population of fish? 2. To the nearest tenth, what is the doubling time for the fish population? 3. To the nearest whole number, what will the fish population be after 4 years? 4. To the nearest tenth, how long will it take for the population to reach 1050? 5. What is the carrying capacity for the fish population? (HINT: Try graphing the function with technology.)

          The population of a fish farm in t years is modeled by the equation P(t) = (1300)/(1 + 4e^(-0.65t)).

Answer the following:
1. What is the initial population of fish?
2. To the nearest tenth, what is the doubling time for the fish population?
3. To the nearest whole number, what will the fish population be after 4 years?
4. To the nearest tenth, how long will it take for the population to reach 1050?
5. What is the carrying capacity for the fish population? (HINT: Try graphing the function with technology.)

Show more…

Added by Margarita D.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Kathleen Carty

07/22/2022

Step 1

65t)): \[ P(0) = \frac{1300}{1 + 4e^{0}} = \frac{1300}{5} = 260 \] Show more…

Show all steps

lock

Thanks for your feedback!

The population of a fish farm in t years is modeled by the equation P(t) = (1300)/(1 + 4e^(-0.65t)). Answer the following: 1. What is the initial population of fish? 2. To the nearest tenth, what is the doubling time for the fish population? 3. To the nearest whole number, what will the fish population be after 4 years? 4. To the nearest tenth, how long will it take for the population to reach 1050? 5. What is the carrying capacity for the fish population? (HINT: Try graphing the function with technology.)

Play audio

Feedback

Powered by NumerAI

Kathleen Carty and 81 other subject Algebra educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:02 Okay, the initial population is population at time zero.

00:09 So just let t equal zero.

00:17 So that gives me 1 ,300 divided by 1 plus 4 or 1 ,300 divided by 5.

00:28 My initial population was 260 fish.

00:36 What is the doubling time? so the doubling time is the amount of time it takes for my initial population to double.

00:51 So i want to set my function equal to 260 times 2.

01:06 Now, 260 times 2 is just 520.

01:11 So i'm going to have 1 plus 4e to the negative 0 .65t.

01:18 Equal to 1 ,300 divided by 520.

01:23 1 ,300 divided by 520 is 2 .5.

01:36 Subtract 1, and then divide both sides by 4.

01:48 So i'm going to have e to the negative.

01:51 0 .65t equals 1 .5 divided by 4, which is 0 .375.

01:59 Take the natural log of both sides.

02:05 Drop your negative exponent down.

02:08 Ln of e is just equal to 1.

02:12 Divide both sides by negative .6 .5.

02:16 And we'll have our doubling time.

02:18 Ln of 0 .375 divided by negative .6.

02:24 It's going to give me a doubling time of 1 .509.

02:30 So about 1 .5 years...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace