<span>The first step you need to do is to rewrite
x+y=3
as y=3-x
</span><span>Now we can replace the y in the equation with our newfound value for y, which is
y=3-x.
So lets write it now like this
Q=x^2+2(3-y)^2
=x^2+18-12x+2x^2
=3x^2+18-12x
</span>Hope this helps
We can find the side length of square 3 by dividing by 4, which is 9.
Then, we find the side length of square 2 by dividing it by 4, which is 12.
To find the AREA of square 1, we do a^2+b^2=c^2. This is basically adding up area of square 1 and square 2 to get square 3.
a^2+b^2=c^2
9^2+12^2=c^2
81+144=225
So the area is 225 units.
The product of a scalar and a matrix is found by multiplying each element of the matrix by the scalar. Multiply each element by -4.
... [4 -12 -24 12]
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given set of values
STEP 2: Write the formula for calculating the Standard deviation of a set of numbers
![\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20S%5Ctan%20dard%5Ctext%7B%20deviation%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B%5Csum%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%20%5C%5C%20where%5Ctext%7B%20%7Dx_i%5Ctext%7B%20are%20data%20points%2C%7D%20%5C%5C%20%5Cbar%7Bx%7D%5Ctext%7B%20is%20the%20mean%7D%20%5C%5C%20%5Ctext%7Bn%20is%20the%20number%20of%20values%20in%20the%20data%20set%7D%20%5Cend%7Bgathered%7D)
STEP 3: Calculate the mean
STEP 4: Calculate the Standard deviation
![\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20S%5Ctan%20dard%5Ctext%7B%20deviation%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B%5Csum%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%20%5C%5C%20%5Csum%20%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%5CRightarrow%5Ctext%7BSum%20of%20squares%20of%20differences%7D%20%5C%5C%20%5CRightarrow10332.7225%2B657.9225%2B18591.3225%2B982.8225%2B2740.52251%2B9731.8225%2B3522.4225%2B18319.6225%2B2878.3225%20%5C%5C%20%2B8163.1225%2B1417.5225%2B3925.0225%2B1321.3225%2B386.1225%2B5677.6225%2B2953.9225%2B3800.7225%20%5C%5C%20%2B3209.2225%2B2565.4225%2B10537.0225%20%5C%5C%20%5Ctext%7BSum%7D%5CRightarrow108974.0275%20%5C%5C%20%20%5C%5C%20S%5Ctan%20dard%5Ctext%7B%20deviation%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B111714.55%7D%7B20-1%7D%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B111714.55%7D%7B19%7D%7D%20%5C%5C%20%5CRightarrow%5Csqrt%5B%5D%7B5879.713158%7D%3D76.67928767%20%5C%5C%20%20%5C%5C%20S%5Ctan%20dard%5Ctext%7B%20deviation%7D%5Capprox76.68%20%5Cend%7Bgathered%7D)
Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.
STEP 5: Calculate the First and third quartile
STEP 6: Find the Interquartile Range
Hence, the interquartile range of the data is 116
3x-x+2=4
<span>P = t + r + s
P - r - s = t + r - r + s - s
P - r - s = t</span>
