Math 30 Project Part A 1. *Mean = 3.8 Standard of deviation = 2.19 2. *7-10 misses = 4+3+2+1 (10 total) There are 81 athletes together 10 / 81 = 0.123 *binomcdf(n, p, k) ….(n=10, p=0.038, k=6) 1-binomcdf(10, 0.038, 6) = 0.04 *The first probability is about 3 times larger than the second *The second probability is assuming that the data is distributed equally, but it’s not, because the chart shows more athletes missing less targets than more. This is why the results are so different. 3. *probability of hitting = 1-0.18 = 0.82 P(hitting 16) = binompdf(20, 0.82, 16) = 0.213 4. *21 minutes and 38 seconds – 23 seconds = 21 minutes and 15 seconds this time is good enough to put him in first place if it takes him only 23 seconds per penalty lap *they all need to brush up on their aiming skills and their speed, whatever they’re lacking most in Part B 1. *5! = 5 x 4 x 3 x 2 x 1 = 120 orders 2. *2 x 2 x 2 x 2 x 2 = 32 total shadings 3. *10P10 = 3,628,800 *7P7 x 3P3 = 30240 *if order isn’t important = 10C4 = 210 if order is important = 10P4 = 5040 *3C2 x 7C2 = 63 Part C 1. *radius = 11.5cm / 2 radius = 5.75cm center is at (0,0) Standard = x^2 + y^2 = 5.75^2 General = x^2 + y^2 – 5.75^2 = 0 2. *radius = 4.50cm / 2 radius = 2.25cm center is at (0,0) Standard = x^2 + y^2 = 2.25^2 General = x^2 + y^2 – 2.25^2 = 0 3. *The small circle would have to be stretched the same in the vertical and horizontal directions at the same rate to transform into the bigger circle. Rate = radius1 / radius 2 = (23/9) 4. *Area(small) = pie(2.25)^2 = 15.9043 Area(large) = 6.53 = 7 times larger Area(large) = pie(5.75)^2 = 103.8689 Area(small) 5. *the ‘a’ value for the ellipse is the same as the ‘a’ value in the equation for the small circle. The ‘ ‘b’ value has 1.25cm added to it, which equals 3.50cm. The equation is: (x^2 / 2.25^2) + (y^2 / 3.50^2) = 1