Question 1014921
P(t)=Kpeřt/K+p(eřt-1) 


your equation, as best i can decipher it, can be represented as follows:


p(t) = k * p * e^(rt) / (k + p * (e^(rt) - 1)


k is the population limit.
p is the starting quantity.
e is the scientific constant of 2.718281828.....
r is the growth rate
t is the number of days.


in your problem:


k = 10,000
p = 2
e = 2.71818...
r = .2
t = number of days.


the questions are:


question a:
When will the growth of the infected population start to level off, and how many people will be infected at that point?  Explain your reasoning, and include any graphs you draw with or without technology. 


the graph of the equation is shown below:


<img src = "http://theo.x10hosting.com/2016/012004.jpg" alt="$$$" </>


from the graph, it appears that the growth rate is starting to level off around the 50th day and appears to be nearing the point of saturation at around the 70th day.


note:
the variable of t is represented by the variable of x in the graph.
the variable of p(t) is represented by the variable of y in the graph.


question b:
When will the infected population equal the uninfected population?  Explain your steps. 


the infected population will equal the uninfected population when 5000 people have been infected because 5000 people represents half the population.


from the graph it's easy to see by plotting the line y = 5000 and find the intersection of the graph of the equation with the graph of the line at y = 5000.


algebraically, your would set p(t) equal to 5000 and solve for t.


when you solve for t, you will find that t = 42.58496586 which appears to be very close to what is shown on the graph.


in fact, it should be right on except for rounding.