For functions, even means f(-x)=f(x) and odd means f(-x)=-f(x)
![\sqrt{x^2} - 9](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%7D%20-%209)
The only occurrence of x is squared so this is even.
![|x-3|](https://tex.z-dn.net/?f=%7Cx-3%7C)
This isn't even or odd, it's neither. f(1)=|1-3|=2. f(-1)=|-1-3|=4 for example.
![f(x) = \dfrac{x}{x^2-1}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cdfrac%7Bx%7D%7Bx%5E2-1%7D)
This one is odd because
![f(-x) = \dfrac{-x}{(-x)^2-1} = -\dfrac{x}{x^2-1} = -f(x)](https://tex.z-dn.net/?f=f%28-x%29%20%3D%20%5Cdfrac%7B-x%7D%7B%28-x%29%5E2-1%7D%20%3D%20-%5Cdfrac%7Bx%7D%7Bx%5E2-1%7D%20%3D%20-f%28x%29)
![g(x)=x+x^2](https://tex.z-dn.net/?f=g%28x%29%3Dx%2Bx%5E2)
g(1)=1+1=2
g(-1)=-1+1=0
neither odd nor even
Average rate of change in Section A:
![\dfrac{h(1)-h(0)}{1-0}=\dfrac{4-0}{1-0}=4](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%281%29-h%280%29%7D%7B1-0%7D%3D%5Cdfrac%7B4-0%7D%7B1-0%7D%3D4)
Average rate of change in Section B:
![\dfrac{h(3)-h(2)}{3-2}=\dfrac{12-8}{3-2}=4](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%283%29-h%282%29%7D%7B3-2%7D%3D%5Cdfrac%7B12-8%7D%7B3-2%7D%3D4)
As you can see, the average rates of change are the same, as expected.
![h(x)=4x](https://tex.z-dn.net/?f=h%28x%29%3D4x)
is linear, which means it has a constant rate of change over any interval in its domain.
Answer:
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
![\sigma = 5 .7](https://tex.z-dn.net/?f=%5Csigma%20%3D%205%20.7)
The standard deviation will remain unchanged.
Step-by-step explanation:
Given
![Data: 136, 129, 141, 139, 138, 127](https://tex.z-dn.net/?f=Data%3A%20136%2C%20129%2C%20141%2C%20139%2C%20138%2C%20127)
Solving (a): The range
This is calculated as:
![Range = Highest - Least](https://tex.z-dn.net/?f=Range%20%3D%20Highest%20-%20Least)
Where:
![Highest = 141; Least = 127](https://tex.z-dn.net/?f=Highest%20%3D%20141%3B%20Least%20%3D%20127)
So:
![Range=141-127](https://tex.z-dn.net/?f=Range%3D141-127)
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
Solving (b): The variance
First, we calculate the mean
![\bar x = \frac{1}{n} \sum x](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%5Csum%20x)
![\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%28136%2B%20129%2B%20141%2B%20139%2B%20138%2B%20127%29)
![\bar x = \frac{1}{6} *810](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%2A810)
![\bar x = 135](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20135)
The variance is calculated as:
![\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%28x%20-%20%5Cbar%20x%29%5E2)
So, we have:
![\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B6-1%7D%2A%5B%28136%20-%20135%29%5E2%20%2B%28129%20-%20135%29%5E2%20%2B%28141%20-%20135%29%5E2%20%2B%28139%20-%20135%29%5E2%20%2B%28138%20-%20135%29%5E2%20%2B%28127%20-%20135%29%5E2%5D)
![\sigma^2 =\frac{1}{5}*[162]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B5%7D%2A%5B162%5D)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
Solving (c): The standard deviation
This is calculated as:
![\sigma = \sqrt {\sigma^2 }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B%5Csigma%5E2%20%7D)
![\sigma = \sqrt {32.4}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B32.4%7D)
--- approximately
Solving (d): With the stated condition, the standard deviation will remain unchanged.
Answer:
b
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given:
Games:
0
0
1
1
2
2
3
5
5
5
6
8
8
A.
Median is the middle value after the ste of values are arranged in ascending order, M which is the 7th value = 3
B.
Median increased by 2, M + 2
Therefore, new Median = 5
This means that the 2 runs are less than the previous median, 3 or greater than the new median, 5.
Possible pairs:
1, 2 or 6, 8
C.
Median can be 2.5 when the total number of runs scored are even numbers, therefore if the middle numbers are 2 and 3. In this case a number greater than or equal to 2;
The median = (2 + 3)/2
= 2.5
D.
Range is the difference between the highest value of runs scored to the lowest value of runs. In the data given above, range = 8 - 0
= 8.
It is possible to play 2 more games and no change in the range, 8 if the value of the 2 games (or 1 game) is/are greater than 0, the lowest value of runs scored and/or less than 8, which is the greatest value of runs scored.