On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the e-ticket check-in kiosk to obtain boarding passes. In a random sample of eight passengers, what is the probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Answer :

0.2322

Step-by-step explanation:

Proportion, p of those who used e - ticket to get boarding pass = 112 / 280 = 0.4

p = 0.4

1 - p = 1 - 0.4 = 0.6

Number of samples, n = 8

P(x = 4)

Using the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x = 4) = 8C4 * 0.4^4 * 0.6^4

P(x = 4) = 70 * 0.0256 * 0.1296

P(x = 4) = 0.2322432

P(x = 4) = 0.2322

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

d. 145*2 is 290 than the total angle is 360 then minus 290 then divide by two