I think the statement given above is false. The variable used to predict changes in the values of another value is not called the response variable. The variable used to predict another variable is called the independent, <span>predictor or </span><span>explanatory variable. Hope this answers the question.</span>
Answer:
side c = 10.2
A = 52.6°
B = 37.4°
Step-by-step explanation:
c² = 8.1² + 6.2²
c² = 104.05
c = 10.20049018
Using cosine law:
8.1² = 10.2² + 6.2² - 2(10.2)(6.2)cos
cosA = 7687/11648
A = 52.57199409
A = 52.6°
B = 180 - 90 - 52.6 = 37.4°
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
Answer:
i'm not sure what is it ?
Step-by-step explanation:
