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
As you can see the length of the one with the man is 6 and the length with the lamp post is 18 and we know that if you divide 18 by 6 you get 3 so then you mutiply 5.25 by 3. Then you get 15.75
Therefore the answer is A. 15.75
Answer:
3rd answer : 3n - 8 = n + 2
Step-by-step explanation:
difference means minus or-
more than means plus
is means the equal sign
3n - 8 = 2 + n
The answer is 45 degrees. Hope this helps!
