# Thread: Sea turtles!!!!

1. ## 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???

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

$\displaystyle t \geq 0$

We want $\displaystyle N$ when $\displaystyle t = 38$ :

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

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

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

$\displaystyle N \approx 869,6$ (2 dp)

Is that right ? What was the answer in the book anyway, knowing it should help ...

3. Originally Posted by Bacterius

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

$\displaystyle t \geq 0$

We want $\displaystyle N$ when $\displaystyle t = 38$ :

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

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

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

$\displaystyle N \approx 869,6$ (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.

4. Originally Posted by VNVeteran
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.