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

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.