Answer:
a) [tex] N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684[/tex]
b) [tex] N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639[/tex]
c) [tex] 70 =\frac{95}{1+8.5 e^{-0.12t}}[/tex]
[tex] 1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}[/tex]
[tex]8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}[/tex]
[tex] e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}[/tex]
[tex] ln e^{-0.12t} = ln (\frac{5}{119})[/tex]
[tex] -0.12 t = ln(\frac{5}{119})[/tex]
[tex] t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks[/tex]
d) If we find the limit when t tend to infinity for the function we have this:
[tex] lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95[/tex]
So then the number of words per minute have a limit and is 95 as t increases without bound.
Step-by-step explanation:
For this case we have the following expression for the average number of words per minutes typed adter t weeks:
[tex] N(t) = \frac{95}{1+8.5 e^{-0.12t}}[/tex]
Part a
For this case we just need to replace the value of t=10 in order to see what we got:
[tex] N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684[/tex]
So the number of words per minute typed after 10 weeks are approximately 27.
Part b
For this case we just need to replace the value of t=20 in order to see what we got:
[tex] N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639[/tex]
So the number of words per minute typed after 20 weeks are approximately 54.
Part c
For this case we want to solve the following equation:
[tex] 70 =\frac{95}{1+8.5 e^{-0.12t}}[/tex]
And we can rewrite this expression like this:
[tex] 1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}[/tex]
[tex]8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}[/tex]
Now we can divide both sides by 8.5 and we got:
[tex] e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln e^{-0.12t} = ln (\frac{5}{119})[/tex]
[tex] -0.12 t = ln(\frac{5}{119})[/tex]
And then if we solve for t we got:
[tex] t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks[/tex]
And we can see this on the plot 1 attached.
Part d
If we find the limit when t tend to infinity for the function we have this:
[tex] lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95[/tex]
So then the number of words per minute have a limit and is 95 as t increases without bound.
