We are given the following model for the spread of the disease in the community.

Where A is the number of people who have contracted the flu after t days, and t is the time in days.
(a) How many people have contracted the flu after 7 days?
Let us substitute t = 7 into the above model

Therefore, 176 people have contracted the flu after 7 days.
(b) What is the carrying capacity for this model?
The carrying capacity of the model can found be by locating the point on the graph where the population line becomes horizontal.
As you can see from the graph of the model, the carrying capacity of the model is 2300
Carrying capacity = 2300
(c) How many days will it take for 625 people to contract the flu?
We need to substitute A = 625 into the model and solve for t

Take ln on both sides of the equation

Therefore, it will take 9 days for 625 people to contract the flu.