Grayson: 35h + 35 = y
Ian: 45h + 15 = y
(Substituting first hour in h)
35h + 35 = y
35(1) + 35
y = 70$ in the first hour for Grayson
45h + 15 = y
45(1) + 15
y = 60$ in the first hour for Ian
Grayson charges > Ian charges
(Substituting second hour in h)
35h + 35 = h
35(2) + 35
105$ in the second hour for Grayson
45h + 15 = h
45(2) + 15
135$ in the second hour for Ian
Grayson charges < Ian charges
Answer:
In the first hour Grayson charges more than Ian.
Answer:
Try c am sure it is correct
Answer:
5.36
Step-by-step explanation:
Given that:
<BAD = <CAE, therefore, BD = EC
Let's take x to be the length of BD = EC
BD + DE + EC = BC
BC = 20,
BD = EC = x
DE ≈ 9.28
Thus,
x + 9.28 + x = 20
x + x + 9.28 = 20
2x + 9.28 = 20
Subtract 9.28 from both sides
2x + 9.28 - 9.28 = 20 - 9.28
2x = 10.72
Divided both sides by 2 to solve for x
BD ≈ 5.36
The correct answer is x= -1. Hope this helps!!
<h3>
Answer: Choice B) 0.5</h3>
=============================================================
Explanation:
For now, focus solely on the orange boxplot. The first quartile Q1 is the left edge of the box, which is at 3.5; while the value of Q3 is 7.5 (right edge of the box). The interquarile range (IQR) is ...
IQR = Q3 - Q1
IQR = 7.5 - 3.5
IQR = 4
This is basically the width of the box. Ignore the whiskers when it comes to the IQR.
Let A = 4 as we'll use it later.
---------------------------------
Find the IQR for the blue box plot
Q1 = 6 = left edge of the blue box
Q3 = 9.5 = right edge of the blue box
IQR = Q3 - Q1
IQR = 9.5 - 6
IQR = 3.5
Let B = 3.5 as we'll use it later
---------------------------------
Subtract the values of A and B to find the difference of the IQR values
A - B = 4 - 3.5 = 0.5
