The interquartile range of the data set is 4.<br>
2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12
Rzqust [24]
Answer:
<em>The IQR or the interquartile range of the following data set would be 4.</em>
Step-by-step explanation:
2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12
Median: 5
Lower quartile: 3
Upper quartile: 7
Interquartile range: 7 - 3 = 4
Answer:
Base of the ladder is 3 meters away from the building.
Step-by-step explanation:
Let's use Pythagoras theorem to solve.
Pythagoras theorem says,

Here let horizontal distance is "a''
Vertical distance of window is 4 m
So, b=4
The Rafi leans 5 m ladder against the wall. So, c=5.

Simplify it

Subtract both sides 16

Take square root on both sides
a=±3
So, base of the ladder is 3 meters away from the building.
Answer:
Step-by-step explanation:
By inscribed angle theorem:
![m\angle R = \frac{1}{2} [360 \degree - (120 \degree + 140 \degree)] \\ \\ m\angle R = \frac{1}{2} [360 \degree -260 \degree] \\ \\ m\angle R = \frac{1}{2} \times 100 \degree \\ \\ m\angle R = 50 \degree \\ \\](https://tex.z-dn.net/?f=m%5Cangle%20R%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5B360%20%5Cdegree%20-%20%28120%20%5Cdegree%20%2B%20140%20%5Cdegree%29%5D%20%20%5C%5C%20%20%5C%5C%20m%5Cangle%20R%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5B360%20%5Cdegree%20-260%20%5Cdegree%5D%20%20%5C%5C%20%20%5C%5C%20%20m%5Cangle%20R%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20100%20%5Cdegree%20%20%5C%5C%20%20%5C%5C%20%20m%5Cangle%20R%20%20%3D%2050%20%5Cdegree%20%20%5C%5C%20%20%5C%5C%20)
Answer:
26
Step-by-step explanation:
5 × 4 = 20 (rectangle)
8-5 = 3 (triangle)
3×4÷2 = 6
20+6 = 26