A six-week fitness program was designed to decrease the time it takes retired individuals to walk one mile. At the beginning of the program, 20 randomly selected retired individuals were invited to participate, and their times to walk a mile were recorded. After the six-week program, their times to walk a mile were again recorded. Most participants saw little to no improvement in their times to walk one mile; however, a few participants saw drastic improvements in their times to walk one mile. The program director would like to perform a hypothesis test to determine if the program reduces the mean time for retired individuals to walk a mile. Which of the following statements is true? a. Because there are 40 observations in total, the sample size is large enough to assume that the sampling distribution of the sample mean differences is approximately normal.
b. Because the walk times before and after the program are paired, a hypothesis test is not appropriate.
c. The sampling distribution of sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed.
d. Because the sample size of 20 is less than 30 and the improvements in walk times in the sample data appear to be skewed, the distribution of sample means should not be assumed to be approximately normal.
e. Because the program recruited only retired individuals, the distribution of sample means should not be assumed to be approximately normal.

c. The sampling distribution of the sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed

Step-by-step explanation:

From the given data, we have;

The category of the sample = Retired individuals

The number of participants in the sample = 20

The duration of program = six-weeks

The improvement seen by most participants = Little to no improvement

The improvement seen by few participants = Drastic improvement

Therefore, given that the participants are randomly selected and the majority of the participants make the same observation of improvement in the time to walk a mile, we have that, the majority of the outcomes show little difference in walk times after the program, therefore, the distribution of the sample data is not skewed and can be assumed to be approximately normal

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 graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)