1. ## problem solving

P(t)=0.8/(1+1.67e^-0.16t)
The model represents the proportion of new computers sold that use windows 98. Let t=0 represent 1998, t=1 represent 1999, and so on.

1)determine the maximum proportion of new computers sold that will use win 98. That is, determine the limit of P(t) as t -> infinity

2)using a graphing utlility graph p(t)

3)When will 75% of new computers sold use win 98?

2. Originally Posted by dmak263
P(t)=0.8/(1+1.67e^-0.16t)
The model represents the proportion of new computers sold that use windows 98. Let t=0 represent 1998, t=1 represent 1999, and so on.

1)determine the maximum proportion of new computers sold that will use win 98. That is, determine the limit of P(t) as t -> infinity

2)using a graphing utlility graph p(t)

3)When will 75% of new computers sold use win 98?
Someone may want to double check this >.<

1. Think about what happens to e^x for a large negative x.

Spoiler:
$\text {as} t \rightarrow \infty , e^{-0.16t} \rightarrow 0$

$
P_{\infty} = \frac{0.8}{1} = \frac{4}{5}$

2. Plug in the values and away you go. I'll edit in my graph later

3. As P(t) = 0.75, solve for t. (Isolate e^-0.16t and take logs)

Spoiler:
$0.75 =\frac{0.8}{1+1.67e^{-0.16t}}$

Take the reciprocal of both sides:

$\frac{4}{3} = \frac{1+1.67e^{-0.16t}}{0.8}$

$0.8\frac{4}{3} = 1+1.67e^{-0.16t}$

$\frac{16}{15} - 1 = 1.67e^{-0.16t}$

$e^{-0.16t} = \frac{15}{16} \times \frac{1}{1.67} = \frac{375}{668}$

$t = -\frac{1}{0.16}ln(\frac{375}{668}) = 3.61 years$

3. I don't understand the steps for #1...#3, I comprehend completely.
There was another question... what proportion of new computers sold in 1998 use windows 98?