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Anastaziya [24]
1 year ago
6

The Kansas City Chiefs are trying to figure out if there is a relationship between the number of passes thrown by Quarterback Pa

trick Mahomes and the total passing yards earned in each game.
Using your graphing calculator, analyze their data and then answer the 2 questions below.

Passes Thrown, [39] [35] [35] [37] and [40
Total Passing Yards [360] [235] [262] [249] [338]

• What is the linear regression equation that represents this data? Round to the nearest whole number. (worth 2 points)

Using your equation, calculate the number of passing yards the Chiefs can expect if Mahomes throws 38 passes in a game. (worth 2 points)

Explain how you calculated your answer for the expected passing yards if 38 passes are thrown. (worth 2 points)

Mathematics
1 answer:
stich3 [128]1 year ago
8 0

The linear regression equation that represents this data is given as follows: y = 22x - 518.

Using this equation, when 38 passes are thrown, the expected yardage is of 318 yards.

<h3>How to find the equation of linear regression?</h3>

To find the regression equation, also called line of best fit or least squares regression equation, we need to insert the points (x,y) in the calculator. These points are given on a table or in a scatter plot in the problem.

In the context of this problem, the input and output are given as follows:

  • Input: number of passes thrown by Patrick Mahomes.
  • Output: number of passing yards obtaining by Mahomes/KC.

The points to be inserted in the calculator are given as follows:

(39, 360), (35, 235), (35, 262), (37, 249), (40, 338).

Hence the equation is given by:

y = 22x - 518.

With 38 passes, the expected yardage is calculated as follows:

y = 22(38) - 518 = 318 yards.

More can be learned about linear regression at brainly.com/question/22992800

#SPJ1

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Answer:

So, the sample mean is 31.3.

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Step-by-step explanation:

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So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}

Now, we calculate the sum

\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\

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\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98

So, the sample standard deviation is 6.98.

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