For a recent season in college football, the total number of rushing yards for that season is recorded for each running back. Th
1 answer:
Answer:
The standard deviation of the number of rushing yards for the running backs that season is 350.
Step-by-step explanation:
Consider the provided information.
The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards.
Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.
Use the formula:
Here z is 2.42 and μ is 790, substitute the respective values as shown.
Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.
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Solution:
Keep in mind ,
Equation of line joining two points (a,b) and (p,q) is given by :
Equation of line P which is obtained by joining (- 8,15) and (6, - 12) is given by:
Equation of line Q which is obtained by joining (4,16) and (-9, 10) is given by:
Equation of line P and Q are
27 x+14 y+6=0-------(1)× 2
6 x-13 y+184=0-------(2)× 9
54 x + 2 8 y+12=0---(1)
54 x -117 y +1656=0----(2)
(1) - (2)
145 y= 1644
y=11.33,
27 x+14 y+6=0-------(1)×13
6 x-13 y+184=0-------(1)×14
351 x + 182 y + 78=0-----(1)
84 x - 182 y +2576=0----(2)
(1) + (2)
435 x + 2654=0
x = - 6.11
So, solution set is (-6.11, 11.33)
Option (C) is true. (−2, 4), because this point makes both the equations incorrect.
Answer:
Step-by-step explanation:
Given
4, 17, 12, 9, 6, 10, 1, 5, 9, 3
Calculating the range;
From the given data;
Highest = 17 and Lowest = 1
Hence;
Calculating the Inter-quartile Range
Inter quartile range (IQR) is calculates as thus
Where
Q3 = Upper Quartile and Q1 = Lower Quartile
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<em>Start by arranging the data in ascending order</em>
1, 3, 4, 5, 6, 9, 9, 10, 12, 17
N = Number of data; N = 10
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Calculating Q3
<em>Substitute 10 for N</em>
Express 8.25 as 8 + 0.25
Express 0.25 as fraction
From the arranged data;
and
Calculating Q1
<em>Substitute 10 for N</em>
Express 2.75 as 2 + 0.75
Express 0.75 as fraction
From the arranged data;
and
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Recall that
Calculating Variance
Start by calculating the mean
Subtract the mean from each data, then square the result
Sum the result
Divide by number of observation;
Calculating Standard Deviation (SD)
<em>(Approximated)</em>