Air-usa has a policy of booking as many as 21 persons on an airplane that can seat only 19. (past studies have revealed that only 88% of the booked passengers actually arrive for the flight.)find the probability that if air-usa books 21 persons, not enough seats will be available. prob =is this probability low enough so that overbooking is not a real concern for passengers if you define unusual as 5% or less? yes, it is low enough not to be a concernno, it is not low enough to not be a concernwhat about defining unusual as 10% or less? yes, it is low enough not to be a concernno, it is not low enough to not be a concern

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

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step-by-step explanation: