Math 103 Final Project (1)

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Feb 20, 2024

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～ MATH 103 FINAL PROJECT ∽ Focus Problems Pg. 15, 16 - Pg.15: a) 1450/52 = 27.88 - Mary spends $27.88 each week on gas. b) 27.88 x 3 = 83.64 - Mary should budget $83.64 for gas for the next 3 weeks. c) No, because the gas price changes over time. Pg.16: w(t) = 0.792t^3 - 4.196t^2 + 3.872t + 24.021 a) 2010 to 2012 w(21.317) - w(24.021)/2 - 0 = -2.704/2 = -1.352 b) 2012 to 2014 w(23.061) - w(21.317)/4 - 2 = 1.744/2 = 0.872 c) 2014 to 2016 w(67.269) - w(23.061)/6 - 4 = 44.28/2 = 22.104 Between 2010 and 2012 the growth of the plant yellow star thistle is decreasing, and then it goes back to increasing from 2012 to 2014. Then between 2014 and 2016 the plant is increasing even more from the previous years. This suggests that the plant is going to take over more and more areas of the park in the future. Pg. 29, 49- Pg. 29 - You might use linear functions to find the rates of change throughout the data and use that to predict the functions to represent the bear population in 2015. A better option for this data would be to use a rational function. This would not be a good way to represent the data because the rate of change is not constant. Pg. 49 - A piecewise function would not be a better way to approach the problem because the rate of change is not constant. This would make it difficult to make a piecewise function for the data. Pg. 95, 107- Problem pg. 95 - Increase growth of the Geyer willows in Yellowstone National Park. - Growth Function: A(t)= 5.1 e ^0.092(t) - A(t) is the area of a ring in mm^2 of a willow t years after 1990. In what year do you predict that the area of a ring will exceed 20 mm^2? Do you think this is reasonable? Solution - 20= 5.1 e^0.092(t) - 3.9216= e^0.092(t)

- ln3.9216= 0.092(t) - 1.3665/ 0.092 = t - t= 14.8533 years after 1990 - Yes, it is reasonable to have a tree’s area of a ring to grow to 20mm^2 in the span of about 14-15years. Problem pg. 107 - This problem is a continuation of the problem above except it includes populations for elk and wolves. - In blue is what was given from the textbook - In green is what was found using the equations given - E(t)=8e^-0.0957t {elk population t years after the wolves were reintroduced} - W(E) = 40/E {area inside a ring of a willow as a function of the elk pop} Solution - W(t) = 40/(8e^-0.0957(t)) {t is years} - So for 1990 the equation would be W(t) = 40/(8e^-0.0957(0)) = 5 Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 Ring Area (in mm^2) 5.1 7.7 6.9 7.8 6.4 7.3 6.7 9.5 9.7 13.9 12.3 Elk pop. after wolves reintroduced 5 5.5 4.1 3.4 3.1 8.07 8.88 9.77 10.75 11.83 13.02 - The predicted effect of the reintroduction of wolves on the growth of willow shows that the trees fluctuate for the first 6 years then dramatically increase for the remaining four years. The elk population matches the willow population over the years. Search and Rescue Team - Teron 1. 5 = 𝜋 R^2 R=1.263 Km R^2 = 1.5915 Km 1.5915 + h^2 = 2^2 h^2= 1.552Km The pilot should fly at a height of 1.552Km. 2. 5= 𝜋 R^2 R^2=1.5915 R= 1.261566261 Km 1.5915km + h^2 = 0.5^2 h^2 = -1.341549431 - No it wouldn't make sense because the height ends up being negative. 3. 5 = 𝜋 R^2 = 1.5915 km= R^2

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