Good afternoon,
I would like you to confirm that my result is true on contrary to the one given by the correction.
we call [imath]r[/imath] the rate (r:=5.2%) and [imath]u_n[/imath] the amount of the population after a delay of [imath]n[/imath] years.
So after n years we have [imath]u_n=u_{n-1}(1-r)[/imath], so we deduce that [math]u_n=u_0(1-r)^n[/math]
Since we work out [imath]n[/imath] so that the population is beneath 46%, once that is equated that gives [imath]u_0(1-0.46)=u_0(1-r)^n[/imath] therefore we have :[math]u_0(1-0.46)=u_0(1-r)^n\iff \ln 0.54=n\ln 0.948\iff n=\dfrac{\ln 0.54}{\ln 0.948}=11.53[/math], we finally deduce the answer : [math]\boxed{n=12}[/math]
The correction gives n=15
Thanks for your help !
The town of Wattle has a population that is decreasing steadily.
It is found that the population is decreasing at a rate of 5.2% per year.
How long will it take for the population to be beneath 46% of what it is now?
I would like you to confirm that my result is true on contrary to the one given by the correction.
we call [imath]r[/imath] the rate (r:=5.2%) and [imath]u_n[/imath] the amount of the population after a delay of [imath]n[/imath] years.
So after n years we have [imath]u_n=u_{n-1}(1-r)[/imath], so we deduce that [math]u_n=u_0(1-r)^n[/math]
Since we work out [imath]n[/imath] so that the population is beneath 46%, once that is equated that gives [imath]u_0(1-0.46)=u_0(1-r)^n[/imath] therefore we have :[math]u_0(1-0.46)=u_0(1-r)^n\iff \ln 0.54=n\ln 0.948\iff n=\dfrac{\ln 0.54}{\ln 0.948}=11.53[/math], we finally deduce the answer : [math]\boxed{n=12}[/math]
The correction gives n=15
Thanks for your help !