Resolving Probabilities of Exactly One and At Least One: Suppose the US Transportation Security Administration (TSA) requires 10% of all airplane passengers be randomly selected for a full screening before boarding a plane. Now, suppose you will travel on a plane 8 times in the coming year. (a) What is the probability that you will get screened exactly once?
(b) What is the probability that you will get screened at least once?
(c) Is the probability in favor (> 50%) of you getting screened exactly once? Is the probability in favor (> 50%) of you getting screened at least once?

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.

slope intercept form is y = mx + b

a

step-by-step explanation:

given the 2 equations

- x + y = - 2 → (1)

2x - 3y = 4 → (2)

multiplying (1) by 2 and adding to (2) will eliminate the x- term

- 2x + 2y = - 4 → (3)

add (2) and (3) term by term

(2x - 2x) + (- 3y + 2y) = (4 - 4), simplifying gives

- y = 0 ⇒ y = 0

substitute y = 0 in either of the 2 equations for corresponding value of x

substituting in (2) gives

2x = 4 ( divide both sides by 2 )

x = 2

solution is (2, 0 ) → a