Monday

September 22, 2014

September 22, 2014

Posted by **Brian** on Sunday, February 24, 2013 at 10:14pm.

I found N'(t)=(ae^-(bx))+axe^-(bx)

and i figure a=6M*.05=300000, but i cannot find the %population after 5 days. any help would be awesome, Thanks in advance!

- Calculus I, differentiation -
**Steve**, Monday, February 25, 2013 at 3:18pmN(t) = a(1 - e^-kt)

After 1 day, t=1, and we have

.05a = (1 - e^-k)

.05 = 1 - e^-k

0.95 = e^-k

k = -ln 0.95 = 0.0513

so,

N(t) = a(1 - e^-0.0513t)

So, by the end of day 5,

N(5) = a(1 - e^-.2565) = .226a = 22.6%

have heard the rumor

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