The number line M-N is from 0 - 9.
2/3 of the way will be the same as 9 x 2/3
That equals 18/3 which reduces to 6.
6 is the coord 2/3 across the numberline from 0 to 9.
Answer:
mulitply 2nd equation by 2
we will get
6x+6y=-10
6x-10y=-22
Subtract 2nd from 1 you will get value of y and out the value of y in 1st equation you'll get the value of x
Answer:
--- Mean absolute deviation of educational documentary
--- Mean absolute deviation of prime time drama
Step-by-step explanation:
Given
![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
See attachment for table
Required
Determine the mean absolute deviation of each
Mean absolute deviation m(x) is calculated as:
![m(x) = \frac{1}{n}\sum |x - \mu|](https://tex.z-dn.net/?f=m%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5Csum%20%7Cx%20-%20%5Cmu%7C)
For Educational Documentary
First, calculate the mean
![\mu = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
So:
![\mu_A = \frac{26 + 28 + 30 + 18+27 + 18 + 20 +31+17+17}{10}](https://tex.z-dn.net/?f=%5Cmu_A%20%3D%20%5Cfrac%7B26%20%2B%2028%20%2B%2030%20%2B%2018%2B27%20%2B%2018%20%2B%2020%20%2B31%2B17%2B17%7D%7B10%7D)
![\mu_A = \frac{232}{10}](https://tex.z-dn.net/?f=%5Cmu_A%20%3D%20%5Cfrac%7B232%7D%7B10%7D)
![\mu_A = 23.2](https://tex.z-dn.net/?f=%5Cmu_A%20%3D%2023.2)
The mean absolute deviation is then calculated as:
![m(x)_A = \frac{1}{n}\sum |x - \mu_A|](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5Csum%20%7Cx%20-%20%5Cmu_A%7C)
![m(x)_A = \frac{1}{10}(|26 -23.2| +|28 -23.2| +|30 -23.2| +|18 -23.2| +|27 -23.2| +|18 -23.2| +|20 -23.2| +|31 -23.2| +|17 -23.2| +|17 -23.2|)](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%20%5Cfrac%7B1%7D%7B10%7D%28%7C26%20-23.2%7C%20%2B%7C28%20%20-23.2%7C%20%2B%7C30%20%20-23.2%7C%20%2B%7C18%20-23.2%7C%20%2B%7C27%20-23.2%7C%20%2B%7C18%20%20-23.2%7C%20%2B%7C20%20-23.2%7C%20%2B%7C31%20-23.2%7C%20%2B%7C17%20-23.2%7C%20%2B%7C17%20-23.2%7C%29)
![m(x)_A = \frac{1}{10}(|2.8| +|4.8| +|6.8| +|-5.2| +|3.8| +|-5.2| +|-3.2| +|7.8| +|-6.2| +|-6.2|)](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%20%5Cfrac%7B1%7D%7B10%7D%28%7C2.8%7C%20%2B%7C4.8%7C%20%2B%7C6.8%7C%20%2B%7C-5.2%7C%20%2B%7C3.8%7C%20%2B%7C-5.2%7C%20%2B%7C-3.2%7C%20%2B%7C7.8%7C%20%2B%7C-6.2%7C%20%2B%7C-6.2%7C%29)
![m(x)_A = \frac{1}{10}(2.8 +4.8 +6.8 +5.2 +3.8 +5.2 +3.2 +7.8 +6.2 +6.2)](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%20%5Cfrac%7B1%7D%7B10%7D%282.8%20%2B4.8%20%2B6.8%20%2B5.2%20%2B3.8%20%2B5.2%20%2B3.2%20%2B7.8%20%2B6.2%20%2B6.2%29)
![m(x)_A = \frac{1}{10}*52](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%20%5Cfrac%7B1%7D%7B10%7D%2A52)
![m(x)_A = 5.2](https://tex.z-dn.net/?f=m%28x%29_A%20%3D%205.2)
For Prime time Drama
First, calculate the mean
![\mu = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
So:
![\mu_B = \frac{39+39+35+29+40+27+41+29+32+30}{10}](https://tex.z-dn.net/?f=%5Cmu_B%20%3D%20%5Cfrac%7B39%2B39%2B35%2B29%2B40%2B27%2B41%2B29%2B32%2B30%7D%7B10%7D)
![\mu_B = \frac{341}{10}](https://tex.z-dn.net/?f=%5Cmu_B%20%3D%20%5Cfrac%7B341%7D%7B10%7D)
![\mu_B = 34.1](https://tex.z-dn.net/?f=%5Cmu_B%20%3D%2034.1)
The mean absolute deviation is then calculated as:
![m(x)_B = \frac{1}{n}\sum |x - \mu_B|](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5Csum%20%7Cx%20-%20%5Cmu_B%7C)
![m(x)_B = \frac{1}{10}(|39-34.1|+|39-34.1|+|35-34.1|+|29-34.1|+|40-34.1|+|27-34.1|+|41-34.1|+|29-34.1|+|32-34.1|+|30-34.1|)](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%20%5Cfrac%7B1%7D%7B10%7D%28%7C39-34.1%7C%2B%7C39-34.1%7C%2B%7C35-34.1%7C%2B%7C29-34.1%7C%2B%7C40-34.1%7C%2B%7C27-34.1%7C%2B%7C41-34.1%7C%2B%7C29-34.1%7C%2B%7C32-34.1%7C%2B%7C30-34.1%7C%29)
![m(x)_B = \frac{1}{10}(|4.9|+|4.9|+|0.9|+|-5.1|+|5.9|+|-7.1|+|6.9|+|-5.1|+|-2.1|+|-4.1|)](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%20%5Cfrac%7B1%7D%7B10%7D%28%7C4.9%7C%2B%7C4.9%7C%2B%7C0.9%7C%2B%7C-5.1%7C%2B%7C5.9%7C%2B%7C-7.1%7C%2B%7C6.9%7C%2B%7C-5.1%7C%2B%7C-2.1%7C%2B%7C-4.1%7C%29)
![m(x)_B = \frac{1}{10}(4.9+4.9+0.9+5.1+5.9+7.1+6.9+5.1+2.1+4.1)](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%20%5Cfrac%7B1%7D%7B10%7D%284.9%2B4.9%2B0.9%2B5.1%2B5.9%2B7.1%2B6.9%2B5.1%2B2.1%2B4.1%29)
![m(x)_B = \frac{1}{10}*47](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%20%5Cfrac%7B1%7D%7B10%7D%2A47)
![m(x)_B = 4.7](https://tex.z-dn.net/?f=m%28x%29_B%20%3D%204.7)
Answer:
(7,4)
Step-by-step explanation:
The midpoint formula is:
![M=(\frac{x_1+x_1}{2},\frac{y_1+y_2}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7Bx_1%2Bx_1%7D%7B2%7D%2C%5Cfrac%7By_1%2By_2%7D%7B2%7D%29)
Let (5,6) be x₁ and y₁ and let's let (9,2) be x₂ and y₂. Thus, the midpoint is:
![M=(\frac{5+9}{2},\frac{6+2}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7B5%2B9%7D%7B2%7D%2C%5Cfrac%7B6%2B2%7D%7B2%7D%29)
Add:
![M=(\frac{14}{2},\frac{8}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7B14%7D%7B2%7D%2C%5Cfrac%7B8%7D%7B2%7D%29)
Divide by 2:
![M=(7,4)](https://tex.z-dn.net/?f=M%3D%287%2C4%29)
So, our midpoint is (7,4).
And we're done!
Answer:
use the area formulae for trapezium