The baggage limit for an airplane is set at 100 pounds per passenger. thus, for an airplane with 200 passenger seats there would be a limit of 20,000 pounds. the weight of the baggage of an individual passenger is a random variable with a mean of 95 pounds and a standard deviation of 35 pounds. if all 200 seats are sold for a particular flight, what is the probability that the total weight of the passengers' baggage will exceed the 20,000-pound limit?

At a mean of 95 pounds per passenger and standard deviation of 35 pounds, multiplying this with the total number of passengers = 200, results in:

absolute mean = 95 * 200 = 19,000

absolute std dev = 35 * 200 = 7,000

Calculating for the z score:

z = (x – u) / s

where x is sample value = more than 20,000; u is the sample mean = 19,000; s is std dev = 7,000

z = (20,000 – 19,000) / 7,000

z = 0.143

From the distribution tables,

P (z = 0.14) = 0.5557

Therefore a 55.57% chance that it will be more than 20,000 pound limit

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: