The data given as a whole would be called ungrouped data. Now to get the variance, you will need the formula:
s^2= <u>Σ(x-mean)^2</u>
n
x = raw data
mean = average of all data
n = no. of observations
s^2 = variance
Now we do not have the mean yet, so you have to solve for it. All you need to do is add up all the data and divide it by the number of observations.
Data: <span>90, 75, 72, 88, 85 n= 5
</span>Mean=<u>Σx</u>
n
Mean = <u>90+75+72+88+85 </u> = <u>410</u> = 82
5 5
The mean is 82. Now we can make a table using this.
The firs column will be your raw data or x, the second column will be your mean and the third will be the difference between the raw data and the mean and the fourth column will be the difference raised to two.
90-82 = 8
8^2 =64
75-82 = -7
-7^2 =49
72-82 = -10
-10^2=100
88-82=6
6^2 = 12
85-82=3
3^2=9
Now you have your results, you can now tabulate the data:
x mean x-mean (x-mean)^2
90 82 8 64
75 82 -7 49
72 82 -10 100
88 82 6 36
85 82 3 9
Now that you have a table, you will need the sum of (x-mean)^2 because the sigma sign Σ in statistics, means "the sum of."
64+49+100+36+9 = 258
This will be the answer to your question. The value of the numerator of the calculation will be 258.
<u>
</u>
Answer:
b = - 5
Step-by-step explanation:
(k + a )(k + x) + 1 = k^2 + kx + ak + ax + 1
I think the way to solve this is to worry about the 36
k^s + 1 + ak should equal 36
We know that a = 2
k^2 + 1 + 2k = 36
k^2 + 2k + 1 - 36 = 0
k^2 + 2k - 35 = 0
(k + 7)(k - 5) = 0
k = -7 is the only acceptable answer. It is given that K < 0.
bx = kx + ax
b = k + a
b = - 7 + 2
b = - 5
Answer:
it's b 59° because it's at the side
Answer:
<em>i: </em>x=-2, x=1
<em>ii: </em>x=-1/2
Step-by-step explanation:
Quadratic form:
You solve <em>i </em>by using FOIL (First, Outside, Inside, Last) because it is a multiplication problem.
![(x+2)(x-1)=0](https://tex.z-dn.net/?f=%28x%2B2%29%28x-1%29%3D0)
<em>"first"</em> would be
, which would equal ![x^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D)
<em>"outside"</em> would be
, which would equal ![-1x, or -x](https://tex.z-dn.net/?f=-1x%2C%20or%20-x)
<em>"inside"</em> would be
, which would equal ![2x](https://tex.z-dn.net/?f=2x)
<em>"last" </em>would be
, which would equal ![-2](https://tex.z-dn.net/?f=-2)
Now you need to combine the terms so that they are one after the other
![+2x-2](https://tex.z-dn.net/?f=%2B2x-2)
Combine like terms, and you should get:
![x^{2} +x-2](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Bx-2)
i Solution
<em>You need to get the variable by itself.</em>
<em>Subtract two from both sides</em>
![x+2=0\\x=-2](https://tex.z-dn.net/?f=x%2B2%3D0%5C%5Cx%3D-2)
<em>Add one to both sides.</em>
![x-1=0\\x=1](https://tex.z-dn.net/?f=x-1%3D0%5C%5Cx%3D1)
ii Solution
<em>Add all the terms.</em>
![x+2+x-1=0\\2x+1=0\\2x=-1\\x=-\frac{1}{2}](https://tex.z-dn.net/?f=x%2B2%2Bx-1%3D0%5C%5C2x%2B1%3D0%5C%5C2x%3D-1%5C%5Cx%3D-%5Cfrac%7B1%7D%7B2%7D)
Answer:
(m³/3 + 5m/2 + 3)pi
Step-by-step explanation:
pi integral [(f(x))² - (g(x))²]
Limits 0 to 1
pi × integral [(2+mx)² - (1-mx)²]
pi × integral[4 + 4mx + m²x² - 1 + 2mx - m²x²]
pi × integral [m²x² + 5mx + 3]
pi × [m²x³/3 + 5mx²/2 + 3x]
Upper limit - lower limit
pi × [m²/3 + 5m/2 + 3]
Verification:
m = 0
[pi × 2² × 1] - [pi × 1² × 1] = 3pi
[m³/3 + 5m/2 + 3]pi
m = 0
3pi