The number of total rushing yards for 23 football players for the 2014 and 2015 seasons are listed below.


2014 Total Yards 2015 Total Yards
37 11
76 56
4 4
5 3
50 87
1 2
45 31
82 55
21 3
2 1
35 34
17 22
50 25
8 42
146 147
4 1
51 25
4 5
1 17
23 4
14 10
9 3
42 67

Part A: On your own paper, create a back-to-back stemplot. Submit a list of the stems you used in order from least to greatest. Justify your reasoning for split or nonsplit stems. (4 points)
Part B: Compare and contrast the two data sets. Justify your answer using key features of the data. (6 points)

Answers

Answer 1

Answer:

the totalyatds of 2015 is76 56

Answer 2

A stem and leaf plot presents quantitative data graphically

The given data is presented as follows;

[tex]\begin{array}{ccc}2014 \ Total \ Yards&&2015 \ Total \ Yards\\37&&11\\76&&56\\4&&4\\5&&3\\50&&87\\1&&2\\45&&31\\82&&55\\21&&3\\2&&1\\35&&34\\17&&22\\50&&25\\17&&22\\50&&25\\8&&42\\146&&147\\4&&1\\51&&25\\4&&5\\1&&17\\23&&4\\14&&10\\9&&3\\42&&67\end{array}[/tex]

Part A

The back to back stem plot is presented as follows;

[tex]\begin{array}{r|c|l}2014 \ Total \ Yards&&2015 \ Total \ Yards\\9 \ 8 \ 5 \ 4 \ 4 \ 4 \ 2 \ 1 \ 1&0&1 \ 1 \ 2 \ 3 \ 3 \ 3 \ 4 \ 4 \ 5\\7 \ 4&1&0 \ 1 \ 7\\3 \ 1 &2&2 \ 5 \ 5\\7 \ 5&3&1 \ 4\\5 \ 2&4&2\\1 \ 0 \ 0&5&5 \ 6\\&6&7\\6&7&\\2&8&7\\6&14&7\end{array}[/tex]

The list of stem used which is non split stem is; 0, 1, 2, 3, 4, 5, 6, 7, 8, and 14

This is so because, the values of the data are spread across the tens unit position from 1 to 8

Part B;

The similarities are;

Both dataset are skewed towards the lower (unit) total yards valuesThe number of players that have a unit total yards in 2014 and 2015 are equal to 9The number of players in 30s, 80s and 140s are equal in 2014 an 2015

The differences are;

The number of players in the 10s, 20s and 60s are more in 2015 than in 2014, while those 50s and 70s are more in 2014 than in2015

From the data sets, we have that the sum of the total rushing yards for 2014 (727) is more than the sum of the total rushing yards for 2015 (655)

The average for 2014 (31.6087), is therefore, larger for 2014 than 2015 (28.4783)

The standard deviation which is a measure of dispersion is smaller for 2014 (33.9434) is lesser than the standard deviation for 2015 (34.5088)

The median for 2014 (21) is larger than the median for 2015 (17)

Learn more about stem and leaf plot here:

https://brainly.com/question/17491250


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PLEASE HELP ME OUT
I've already filled out the first few blanks, but I really need your help on the rest of this problem
PLEASE SHOW YOUR STEPS
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Answer:

Explanation:

Part D

For d, the very first thing you need to do is figure out which one of the steps you are going to use. You have 2 in b^2 + 2, so even if b = 0 the two still matters. It means that you use f(x) = -x + 3 because that's what you use when you have 2 or above.

The second thing you have to realize is that f(x) = -x + 3 has the meaning of what ever you see on the left in the place of x, you put on the right wherever there is an x.

In this case f(b^2 + 2) = -b^2 - 2 + 3 = 1 - b^2

I'm not sure enough to give you an answer for the domain and range, not this time of night.

9514 1404 393

Answer:

  a) 9

  b) 1

  c) -1

  d) -b² +1

  domain: (-∞, ∞)

  range: (-∞, 1] ∪ [3, ∞)

Explanation:

When you evaluate a piecewise function, the first step is to determine what piece is applicable. Then, you fill in the argument value and do the arithmetic.

a) -6 < 2, so the first piece applies. f(-6) = |-6| +3 = 9

b) 2 = 2, so the second piece applies. f(2) = -2 +3 = 1

c) 4 > 2, so the second piece applies. f(4) = -4 +3 = -1

d) b² +2 ≥ 2, so the second piece applies. f(b² +2) = -(b² +2) +3 = -b² +1

__

The function is defined for all values of x, so the domain is (-∞, ∞).

The minimum value of the first piece is +3. The maximum value of the second piece is f(2) = 1. So, values of y between 1 and 3 are not part of the range.

The range is (-∞, 1] ∪ [3, ∞).

_____

Additional comment

The square of a real number can never be negative, so b²+2 cannot be less than 2.

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