Question 1171737
P(t)=L/(1+Be^-(kt))
when t=0, P(t)=10
The denominator is 1+B, and L/(1+B)=10
L=10+10B
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L=200, the horizontal asymptote.  When k is large, Be^-(kt) becomes close to 0 and the denominator 1.
Since L=10+10B, B=19
P(t)=200/(1+19 e^(-kt))
The inflection point is where the 2nd derivative goes from positive to negative and is at L/2 or 100 cases.
The horizontal asymptote is at P(t)=200. 


{{{graph(300,300,-10,40,-10,200,200/(1+19e^(-0.25x)),100)}}}