I would say that the median is the best measure for the center as there are two major outliers (16 and 18)
Answer:
![\displaystyle Range: Set-Builder\:Notation → [y|-2 ≤ y] \\ Interval\:Notation → [-2, ∞) \\ \\ Domain: Set-Builder\:Notation → [x|-4 ≤ x] \\ Interval\:Notation → [-4, ∞)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Range%3A%20Set-Builder%5C%3ANotation%20%E2%86%92%20%5By%7C-2%20%E2%89%A4%20y%5D%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%5B-2%2C%20%E2%88%9E%29%20%5C%5C%20%5C%5C%20Domain%3A%20Set-Builder%5C%3ANotation%20%E2%86%92%20%5Bx%7C-4%20%E2%89%A4%20x%5D%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%5B-4%2C%20%E2%88%9E%29)
Explanation:
<em>See above graph</em>
I am joyous to assist you anytime.
Answer: 2 4/5
Step-by-step explanation:
14/3 - 2 2/5
14/3 - 12/5
= 34/15
= 2 4/15
Answer:
w=67°
Step-by-step explanation:
First you solve for x. 12x+11=x. X would be 1. Then you plug in 1 for x and add the w to the equation. 12(1)+11+w=90. after solving that you will get the total for w. which should be 67°
Note: Consider we need to find the vertices of the triangle A'B'C'
Given:
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).
To find:
The vertices of the triangle A'B'C'.
Solution:
If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

Using this rule, we get



Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).