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

a)
`N(t=10) = (95)/(1+8.5 e^(-0.12(10)))= (95)/(1+ 8.5 e^(-1.2)) = 26.684`

b)
`N(t=20) = (95)/(1+8.5 e^(-0.12(20)))= (95)/(1+ 8.5 e^(-2.4)) = 53.639`

c)
`70 =(95)/(1+8.5 e^(-0.12t))`

`1+ 8.5 e^(-0.12t) = (95)/(70)= (19)/(14)`

`8.5 e^(-0.12t) = (19)/(14)-1= (5)/(14)`

`e^(-0.12t) = ((5)/(14))/(8.5)= (5)/(119)`

`ln e^(-0.12t) = ln ((5)/(119))`

`-0.12 t = ln((5)/(119))`

`t = (ln((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) (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.

Explanation:

For this case we have the following expression for the average number of words per minutes typed adter t weeks:

`N(t) = (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) = (95)/(1+8.5 e^(-0.12(10)))= (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) = (95)/(1+8.5 e^(-0.12(20)))= (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 =(95)/(1+8.5 e^(-0.12t))`

And we can rewrite this expression like this:

`1+ 8.5 e^(-0.12t) = (95)/(70)= (19)/(14)`

`8.5 e^(-0.12t) = (19)/(14)-1= (5)/(14)`

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

`e^(-0.12t) = ((5)/(14))/(8.5)= (5)/(119)`

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

`ln e^(-0.12t) = ln ((5)/(119))`

`-0.12 t = ln((5)/(119))`

And then if we solve for t we got:

`t = (ln((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) (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.