Now, we are ready to create our modelwe just need some data! In March 2014, the recognized start of the Ebola outbreak, there were 49 cases Ebola in West Africa. One month later, there were 253 cases. This is enough information to build our linear model. 3. We will use the information above to determine the values to use for P0 and d in our linear growth model for the Ebola outbreak. 1. Since March 2014 is the start of the outbreak, we let n = 0 represent March 2014. We will also choose to define n to be the number of months since March 2014. How much time passes from n = 0 to n = 1? 1 month 2. What month and year does n = 1 correspond to on the calendar? April 2014. 3. How many cases of Ebola were there when n = 0? What about when n = 1? At n+0 march 2014 there was 49 cases at n =1 april 2014 there were 253 cases 4. Use your answer from part (c) to compute the common difference, or slope, for our linear growth model. d=Cases at n=1Cases at n=0 d=25349=204d=25349=204 Thus, d = __204________. 4. What is the initial number or starting number of cases of Ebola for our linear growth equation? Ebola at n=0n=0 (March 2014) is 49. Thus, P0 = _49________. 4. Using the values from the above parts, write a linear growth model for the number of cases of Ebola in West Africa during the outbreak in 2014: Pn = 49+204n 5. Use your model from Question 2 to predict the number of cases of Ebola in West Africa 1. after 2 months. 2. in June 2014. 3