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KJ4UTS

  • one year ago

A breeding group of beavers is introduced into a protected area. After t years the number of beavers in the area is modeled by the function N(t)= 54/0.35+0.68^t 1. How many beavers were initially introduced? 2. Estimate the number of beavers after 5 years. 3. Determine the change in the beaver population between t = 5 and t = 10. (Note: All answers are whole numbers.)

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  1. KJ4UTS
    • one year ago
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    The answers that I have so far are (I am not sure if I did them right): 1. N(t)= 54/0.35+0.68^0 = 40 2. N(t)= 54/0.35+0.68^5 = 109 3. I am not sure?

  2. anonymous
    • one year ago
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    For initial result put t=0

  3. Michele_Laino
    • one year ago
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    question #1 you have to evaluate N(0)

  4. anonymous
    • one year ago
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    yup and for #2 put t=5

  5. KJ4UTS
    • one year ago
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    1. 40 2. 109 3. ?

  6. anonymous
    • one year ago
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    as change is n(5) - n(10) put t= 5 and t=10 and subtract

  7. Michele_Laino
    • one year ago
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    yes! for third part we have to compute this: \[{N\left( 5 \right) - N\left( 0 \right) = ...?}\]

  8. KJ4UTS
    • one year ago
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    I am not sure exactly how to compute that what do I plug in?

  9. Michele_Laino
    • one year ago
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    is your formula like this: \[N\left( t \right) = \frac{{54}}{{0.35 + 0.68t}}\]

  10. anonymous
    • one year ago
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    from 10 to 5 n(10) = 54/(.35+(.68^(10)))

  11. KJ4UTS
    • one year ago
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    @Michele_Laino that's how my formula is.

  12. Michele_Laino
    • one year ago
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    sorry ! Is it like below: \[\Large N\left( t \right) = \frac{{54}}{{0.35 + {{0.68}^t}}}\]

  13. KJ4UTS
    • one year ago
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    yes

  14. anonymous
    • one year ago
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    yup ^ this power operator.

  15. Michele_Laino
    • one year ago
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    then if t=0, we have: \[\Large N\left( 0 \right) = \frac{{54}}{{0.35 + 1}} = ...?\]

  16. KJ4UTS
    • one year ago
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    40

  17. Michele_Laino
    • one year ago
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    correct!

  18. Michele_Laino
    • one year ago
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    if t=5, then we have: \[\Large N\left( 5 \right) = \frac{{54}}{{0.35 + {{0.68}^5}}} \cong 109\]

  19. KJ4UTS
    • one year ago
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    so do I just go 109-40=69?

  20. anonymous
    • one year ago
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    nope i dont think so

  21. Michele_Laino
    • one year ago
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    for t=10, we have: \[\Large N\left( {10} \right) = \frac{{54}}{{0.35 + {{0.68}^{10}}}} \cong 145.5\]

  22. anonymous
    • one year ago
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    3. Determine the change in the beaver population between t = 5 and t = 10. here t is 10 and 5 implies 145-109 =?

  23. Michele_Laino
    • one year ago
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    so you have to do this: \[\Large N\left( {10} \right) - N\left( 5 \right) = 145.5 - 109 = ...?\]

  24. KJ4UTS
    • one year ago
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    145.5-109=36.5

  25. Michele_Laino
    • one year ago
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    that's right!

  26. anonymous
    • one year ago
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    yup in whole no.

  27. Michele_Laino
    • one year ago
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    yes! @AdityaOS is right!

  28. KJ4UTS
    • one year ago
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    So the answers are: 1. 40 2. 109 3. 36.5 but isn't 36.5 not in whole number

  29. Michele_Laino
    • one year ago
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    you can round off 36.5 to 37

  30. KJ4UTS
    • one year ago
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    oh ok thank you @Michele_Laino and @AdityaOS

  31. Michele_Laino
    • one year ago
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    :)

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