Answer:

Step-by-step explanation:
The number of parrots in t years after 2010 can be modeled by the following function:

In which P(0) is the number of parrots in 2010 and r is the growth rate, as a decimal.
608 parrots in the forest in 2010.
This means that 
Then

When the scientists went back 5 years later, they found 4617 parrots.
This means that 
We use this to find 1 + r. So



![1 + r = \sqrt[5]{\frac{4617}{608}}](https://tex.z-dn.net/?f=1%20%2B%20r%20%3D%20%5Csqrt%5B5%5D%7B%5Cfrac%7B4617%7D%7B608%7D%7D)

So

THE ANSWER IS D
ABCDEFGHIJKLMNOPQRSTUVWXYZ THANKS YOU !!!
the answer is 360 because if you take a third you will have 120 then if you take half of that then you get 60
Answer: 21
Step-by-step explanation: