Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, it decided to collect some data. Specifically, a random sample of 1,000 passengers was selected. The passengers were observed, and the number of bags carried on the plane was noted. Out of the 1,000 passengers, 345 had more than one bag.
The domestic version of Boeing's 747 has a capacity for 568 passengers. Determine an interval estimate of the number of passengers that you would expect to carry more than one piece of luggage on the plane. Assume the plane is at its passenger capacity.
a) (171.651, 216.214)
b) (181.514, 208.313)
c) (174.412, 217.218)
d) (179.20, 212.716)

d) (179.20, 212.716)

Step-by-step explanation:

1000 : 345

569 : 195.96

Midpoint of (179.20, 212.716) is 195.958 which is the closest to 195.96

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

the answer is two. hope this !