Question

In 2009, there were 1570 bears in a wildlife refuge. In 2010, the population had increased to

approximately 1884 bears. If this trend continues and the bear population is increasing exponentially,
how many bears will there be in 2018?

Answer 1

Explanation:

Question 15

1 / 1 pts

In 2009, there were 1570 bears in a wildlife refuge. In 2010, the population had increased to approximately 1884 bears. If this trend continues and the bear population is increasing exponentially, how many bears will there be in 2018?

ANSWER = 8,101 GOT IT RIGHT ON TEST

Answer 2

`8,101\ bears`

Explanation:

we know that

In this problem we have a exponential function of the form

`y=a(b)^(x)`

where

x ----> is the number of years since 2009

y ----> is the population of bears

a ----> is the initial value

b ---> is the base

step 1

Find the value of a

For x=0 (year 2009)

y=1,570 bears

substitute

`1.570=a(b)^(0)`

`a=1.570\ bears`

so

`y=1.570(b)^(x)`

step 2

Find the value of b

For x=1 (year 2010)

y=1,884 bears

substitute

`1,884=1.570(b)^(1)`

`b=1,884/1.570`

`b=1.2`

The exponential function is equal to

`y=1.570(1.2)^(x)`

step 3

How many bears will there be in 2018?

2018-2009=9 years

so

For x=9 years

substitute in the equation

`y=1.570(1.2)^(9)`

`y=8,101\ bears`