Suppose that we define an individual who has automobile insurance from the company as a "success" if they have no moving violations. calculate the probability that, in a random sample of 10 individuals, exactly 4 will have no moving violations. [3 points] (b) suppose that we define an individual as a "success" if they have at least two moving violations. calculate the probability that at most 3 out of 10 people will each have at least two moving violations. [3 points] (c) calculate the (i) mean and the (ii) standard deviation of the number of individuals out of 10 who have exactly one moving violation. [2 points each]

c

step-by-step explanation: easy math

there still no table

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16