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Math Help - Sea turtles!!!!

  1. #1
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    Sea turtles!!!!

    Suppose a species of sea turtles is confined to an island and has a current population of 70 turtles. If the population, N, is modeled by N=70(5+2t)/(5+.04t) , t≥0 where t is the time in years, how many turtles are predicated to be on the island in 38 years?

    ok well i thought this was pretty simple, but it seemed that i was wrong. i plugged 38 into the formula like so N=70(5+2(38)/(5+.04(38) and got 1520.4 which did not match the answer in the back of the book. what did i do wrong???
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  2. #2
    Super Member Bacterius's Avatar
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    N=\frac{70(5+2t)}{5+0.04t}

    t \geq 0

    We want N when t = 38 :

    N=\frac{70(5+2 \times 38)}{5+0.04 \times 38}

    N=\frac{70 \times 81}{6,52}

    N=\frac{5670}{6,52}

    N \approx 869,6 (2 dp)

    Is that right ? What was the answer in the book anyway, knowing it should help ...
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  3. #3
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    Quote Originally Posted by Bacterius View Post


    N=\frac{70(5+2t)}{5+0.04t}

    t \geq 0

    We want N when t = 38 :

    N=\frac{70(5+2 \times 38)}{5+0.04 \times 38}

    N=\frac{70 \times 81}{6,52}

    N=\frac{5670}{6,52}

    N \approx 869,6 (2 dp)

    Is that right ? What was the answer in the book anyway, knowing it should help ...
    yes that is the answer in the back. but i wonder what i did wrong?? i guess i was the problem. thank you.
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  4. #4
    Super Member Bacterius's Avatar
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    Quote Originally Posted by VNVeteran View Post
    yes that is the answer in the back. but i wonder what i did wrong?? i guess i was the problem. thank you.
    I believe you did a calculation error, maybe because of misplaced brackets. It happens sometimes.
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