Abraham and blaise each have 10$. they repeatedly flip a fair coin. if it comes up heads, abraham gives blaise 1$. if it comes up tails, blaise gives abraham 1$. what is the expected number of flips until one of them runs out of money?

it's 100

step-by-step explanation:

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

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answer is  

a.

p = 78

step-by-step explanation: