Acar dealer who sells only late-model luxury cars recently hired a new salesperson and believes that this salesperson is selling at lower markups. he knows that the long-run average markup in his lot is $5,600. he takes a random sample of 16 of the new salesperson’s sales and finds an average markup of $5,000 and a standard deviation of $800. assume the markups are normally distributed. what is the value of an appropriate test statistic for the car dealer to use to test his claim? multiple choice1.t15 = –0.752.z = –3.003.z = –0.754.t15 = –3.00

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.

1/4 because there is 4 grades.

step-by-step explanation:

in which figure is point g a centroid

ok then what happened.