I think it’s C but I’m not so sure it’s been a while
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
Step-by-step explanation:
I4(-3)+5| +8(-3)
|-12+5| -24
7-24 = -17
Answer:
23w
Step-by-step explanation:
For 6days, She skates 3 * 6 = 18 miles
on weekend she skates 5 miles
Thus, In one week, she skates 23miles
In w week, she skates: 23w
For f(x), when x=1, y=1, and when x=2, y=4. Since the graphs are both quadratic, you can assume f(x)=x².
For g(x), when x=2, y=1, and when x=4, y=4. So g(x)=(x²)÷4.
Now it's a given that g(x) is equivalent to f(k*x), so we can say
(x²)/4=(kx)²
4kx²=x²
k=1/4