KJ4UTS
  • KJ4UTS
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.)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KJ4UTS
  • KJ4UTS
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?
anonymous
  • anonymous
For initial result put t=0
Michele_Laino
  • Michele_Laino
question #1 you have to evaluate N(0)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yup and for #2 put t=5
KJ4UTS
  • KJ4UTS
1. 40 2. 109 3. ?
anonymous
  • anonymous
as change is n(5) - n(10) put t= 5 and t=10 and subtract
Michele_Laino
  • Michele_Laino
yes! for third part we have to compute this: \[{N\left( 5 \right) - N\left( 0 \right) = ...?}\]
KJ4UTS
  • KJ4UTS
I am not sure exactly how to compute that what do I plug in?
Michele_Laino
  • Michele_Laino
is your formula like this: \[N\left( t \right) = \frac{{54}}{{0.35 + 0.68t}}\]
anonymous
  • anonymous
from 10 to 5 n(10) = 54/(.35+(.68^(10)))
KJ4UTS
  • KJ4UTS
@Michele_Laino that's how my formula is.
Michele_Laino
  • Michele_Laino
sorry ! Is it like below: \[\Large N\left( t \right) = \frac{{54}}{{0.35 + {{0.68}^t}}}\]
KJ4UTS
  • KJ4UTS
yes
anonymous
  • anonymous
yup ^ this power operator.
Michele_Laino
  • Michele_Laino
then if t=0, we have: \[\Large N\left( 0 \right) = \frac{{54}}{{0.35 + 1}} = ...?\]
KJ4UTS
  • KJ4UTS
40
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
if t=5, then we have: \[\Large N\left( 5 \right) = \frac{{54}}{{0.35 + {{0.68}^5}}} \cong 109\]
KJ4UTS
  • KJ4UTS
so do I just go 109-40=69?
anonymous
  • anonymous
nope i dont think so
Michele_Laino
  • Michele_Laino
for t=10, we have: \[\Large N\left( {10} \right) = \frac{{54}}{{0.35 + {{0.68}^{10}}}} \cong 145.5\]
anonymous
  • anonymous
3. Determine the change in the beaver population between t = 5 and t = 10. here t is 10 and 5 implies 145-109 =?
Michele_Laino
  • Michele_Laino
so you have to do this: \[\Large N\left( {10} \right) - N\left( 5 \right) = 145.5 - 109 = ...?\]
KJ4UTS
  • KJ4UTS
145.5-109=36.5
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
yup in whole no.
Michele_Laino
  • Michele_Laino
yes! @AdityaOS is right!
KJ4UTS
  • KJ4UTS
So the answers are: 1. 40 2. 109 3. 36.5 but isn't 36.5 not in whole number
Michele_Laino
  • Michele_Laino
you can round off 36.5 to 37
KJ4UTS
  • KJ4UTS
oh ok thank you @Michele_Laino and @AdityaOS
Michele_Laino
  • Michele_Laino
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.