Question

Learning Theory In a typing class,the averege number N of words per minutes typed after t weeks of lessons can be modeled by N = 95/(1+8.5e- 0.12t).
(a) Find the averege number of words per minute typed after 10 weeks.
(b) Find the averege number of words per minute typed after 20 weeks.
(c) Use a graphing utility to graph the model. Find the number of weeks required to achieve an averege of 70 words per minute.
(d) Does the number of words per minute have a limit at t increases without bound?Explain your reasoning.

Answer 1

Answer:

a) $N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

b) $N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

c) $70 =\frac{95}{1+8.5 e^{-0.12t}}$

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

d) If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

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:

$N(t) = \frac{95}{1+8.5 e^{-0.12t}}$

Part a

For this case we just need to replace the value of t=10 in order to see what we got:

$N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

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:

$N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

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:

$70 =\frac{95}{1+8.5 e^{-0.12t}}$

And we can rewrite this expression like this:

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

Now we can divide both sides by 8.5 and we got:

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

And then if we solve for t we got:

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

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:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.