MATH SOLVE

4 months ago

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

Accepted Solution

A:

Answer:[tex]8,101\ bears[/tex]Step-by-step explanation:we know thatIn this problem we have a exponential function of the form[tex]y=a(b)^{x}[/tex]wherex ----> is the number of years since 2009y ----> is the population of bearsa ----> is the initial value b ---> is the basestep 1Find the value of aFor x=0 (year 2009)y=1,570 bearssubstitute[tex]1.570=a(b)^{0}[/tex][tex]a=1.570\ bears[/tex]so[tex]y=1.570(b)^{x}[/tex]step 2Find the value of bFor x=1 (year 2010)y=1,884 bearssubstitute[tex]1,884=1.570(b)^{1}[/tex][tex]b=1,884/1.570[/tex][tex]b=1.2[/tex]The exponential function is equal to[tex]y=1.570(1.2)^{x}[/tex]step 3How many bears will there be in 2018?2018-2009=9 yearssoFor x=9 yearssubstitute in the equation[tex]y=1.570(1.2)^{9}[/tex][tex]y=8,101\ bears[/tex]