The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an exponential decay function that models this situation. Then find the population in 2010.
Accepted Solution
A:
Answer:Part 1) The exponential function is equal to [tex]y=1,350(0.95)^{x}[/tex]Part 2) The population in 2010 was [tex]992\ fish[/tex] Step-by-step explanation:Part 1) Write an exponential decay function that models this situationwe know thatIn this problem we have a exponential function of the form[tex]y=a(b)^{x}[/tex]wherey ----> the fish population of Lake Collins since 2004x ----> the time in yearsa is the initial valueb is the basewe have[tex]a=1,350\ fish[/tex][tex]b=(100\%-5\%)=95\%=0.95[/tex]substitute[tex]y=1,350(0.95)^{x}[/tex] ----> exponential function that represent this scenarioPart 2) Find the population in 2010we have[tex]y=1,350(0.95)^{x}[/tex] soFor [tex]x=(2010-2004)=6\ years[/tex]substitute[tex]y=1,350(0.95)^{6}=992\ fish[/tex]