<span>•<span>If
the raw score is transformed into a z-score, however, the value of the z-score
tells exactly where the score is located relative to all the other scores in
the distribution. </span></span>
Answer:
125.6 cm²
Step-by-step explanation:
Area of the shaded region = area of larger circle - area of the smaller circle
Area of the smaller circle = πr²
π = 3.14, r = 3 cm
Area of smaller circle = 3.14*3² = 3.14*9 = 28.26 cm²
Area of larger circle = πr²
π = 3.14, r = 7
Area of larger circle = 3.14*7² = 3.14*49 = 153.86 cm²
Area of the shaded region = 153.86 - 28.26 = 125.6 cm²
Answer:
Answer:
Distance is 13.6 units.
Which means it is 13
Step-by-step explanation:
Given a grid shows the positions of subway stop and house. The subway stop is located at (-7, 8) and house is located at (6, 4).
we have to find the distance between house and the subway stop.
Points are (-7, 8) and (6, 4).
Using distance formula,
D= /(x2-x1)^2+(y2-y1)^2} = (6-(-7))^2+(4-8)^2=(13)^2+16}={169+16}={185}=13.60147\sim13.6units.
Hence, distance between house and the subway stop is 13.6 units.
Answer:
Step-by-step explanation:
y= 1/2x + −7/2
Answer:
![A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B1%7D%7B9%7D%20%26%20%5Cfrac%7B4%7D%7B27%7D%20%26%20-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%20%5Cfrac%7B8%7D%7B9%7D%20%26%20%5Cfrac%7B5%7D%7B27%7D%20%26%20%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%20-%20%5Cfrac%7B4%7D%7B9%7D%20%26%20%5Cfrac%7B2%7D%7B27%7D%20%26%20-%20%5Cfrac%7B1%7D%7B27%7D%20%5Cend%7Barray%7D%20%5Cright%5D)
Step-by-step explanation:
We want to find the inverse of ![A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%20-2%20%5C%5C%5C%5C%204%20%26%201%20%26%203%20%5C%5C%5C%5C%20-4%20%26%202%20%26%203%20%5Cend%7Barray%7D%20%5Cright%5D)
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
So, augment the matrix with identity matrix:
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%204%261%263%260%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 4 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 1 multiplied by 4 to row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%262%26-5%264%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 2 from row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%26-27%2612%26-2%261%5Cend%7Barray%7D%5Cright%5D)
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 2 to row 1
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 11 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%260%26%5Cfrac%7B8%7D%7B9%7D%26%5Cfrac%7B5%7D%7B27%7D%26%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
As can be seen, we have obtained the identity matrix to the left. So, we are done.
