1.5m = 150cm
150cm ∠ 145cm
Jayleen ordered more
First, multiply each side by 5
6(x+1) = 25x-15
distribute the 6
6x+6=25x-15
put the x's on the same side and the other numbers on the opposite side
21=19x
Divide by 19 to isolate the x
21/19=x
Make the length x+4 and the width x and then make a formula x(x+4)=21
then u ll get x^2+4x-21=0 and then solve for x ..... the width is 3
Expand the factored expression (multiply all the terms back together), and see if it matches up with the original expression.
Assume that the data for both movies and basketball games are normally distributed.
Therefore, the median and the mean are assumed equal.
The standard deviation, σ, is related to the interquartile range by
IQR = 1.35
From the data, we can say the following:
Movies:
Range = 150 - 60 = 90 (approx)
Q1 = 62 (approx), first quartile
Q3 = 120 (approx), third quartlie
Q2 (median) = 90 (approx)
IQR = Q3 - Q1 = 58
σ = IQR/1.35 = 58/1.35 = 43
Basketball:
Range = 150 - 90 = 60 approx
Q1 = 95 (approx)
Q3 = 145 (approx)
Q2 = 125 (approx)
IQR = 145 - 95 = 50
σ = 50/1.35 = 37
Test the given answers.
A. The IQRs are approximately equal, so they are not good measures of spread. This is not a good answer.
B. The std. deviation is a better measure of spread for basketball. This is not a good answer.
C. IQR is not a better measure of spread for basketball games. This is not a good answer.
D. The standard deviation is a good measure of spread for both movies and basketball. This is the best answer.
Answer: D
