Question

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

Answer 1

Answer:

$8,101\ bears$

Step-by-step 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

$y=1.570(1.2)^{9}$

$y=8,101\ bears$