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anonymous

  • 5 years ago

A population grows according to the equation P(t) =6000-5500e^(-0.159t) for t>0 or t=0, t measured in years. This population will approach a limiting value as time goes on. During which year will the population reach HALF of this limiting value?

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  1. anonymous
    • 5 years ago
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    What is the limit as t goes to infinity of e^-0.159t?

  2. anonymous
    • 5 years ago
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    Isn't it Zero?

  3. anonymous
    • 5 years ago
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    So what does P(t) go to as t goes to infinity?

  4. anonymous
    • 5 years ago
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    I'm not quite sure?

  5. anonymous
    • 5 years ago
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    Since e^-0.159t goes to 0, 5500e^(-0.159t) goes to zero as well, right. It's just 5500(0).

  6. anonymous
    • 5 years ago
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    What happens to the 6000?

  7. anonymous
    • 5 years ago
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    Since that second part goes to zero, you have 6000-0 which is just 6000. That's the limiting value

  8. anonymous
    • 5 years ago
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    Okay so I basically just take the limit of each parts 6000 and -5500e^(-0.159t) and I end up with 6000 for the limiting value.

  9. anonymous
    • 5 years ago
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    Yeah.

  10. anonymous
    • 5 years ago
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    Thank you! Do you know how to do the second part?

  11. anonymous
    • 5 years ago
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    You know the limiting value. So half of that is 3000. So P(t)=3000, you need t. So 3000=6000-5500e^(-0.159t). Isolate e^(-0.159t) then take the natural log of both sides.

  12. anonymous
    • 5 years ago
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    So just simply solve for t?

  13. anonymous
    • 5 years ago
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    Yeah. They are asking for the time the population will be 3000.

  14. anonymous
    • 5 years ago
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    I know how to do it know! Thanks a lot!

  15. anonymous
    • 5 years ago
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    Np

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