Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Answer:
C6 greeting cards for $23.40
Step-by-step explanation:
A. 13.5/3=4.5
B. 38.25/9=4.25
C. 23.4/6=3.9
d. 32.8/8=4.1
The coordinates of the vertices of the parallelogram, given that s is a units from the origin, Z is b units from the origin, and then length of the base is c units could be the following:
W(b+c, 0), Z(b, 0), S(0, a), T(c,a)
900 divided by 3 is 300 so 300 students bring their lunch opposed to 600 students who do not.
Tip:
draw a diagonal and compare triangles ;)