Answer:
C
Step-by-step explanation:
A patio=(x+15)*(x+10)
A bench=5*1=5
A painted=A patio-A bench
A painted=(x+15)*(x+10)-5
thus, the answer is C
Answer:
B. The coefficient of determination is 54.76%. Therefore, 54.76% of the variation in weight can be explained by the regression line.
Step-by-step explanation:
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
The coefficient of determination is a measure to quantify how a model explains an dependent variable.
The formula for the correlation coeffcient is given by:
![r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bn%28%5Csum%20xy%29-%28%5Csum%20x%29%28%5Csum%20y%29%7D%7B%5Csqrt%7B%5Bn%5Csum%20x%5E2%20-%28%5Csum%20x%29%5E2%5D%5Bn%5Csum%20y%5E2%20-%28%5Csum%20y%29%5E2%5D%7D%7D)
The formula for the coefficient of determination is 
In our case the correlation coefficient obtained was 0.74
And the determination coefficient is
, and if we convert this into % we got 54.76%
Assume that height is the predictor (X) and weight is the response (Y)
And the best answer for this case is:
B. The coefficient of determination is 54.76%. Therefore, 54.76% of the variation in weight can be explained by the regression line.
Answer:
Step-by-step explanation:
you would multiply by 3 on both sides to get the division to go away.
then subtract 8 from both sides.
then you have your answer
Answer:
0.2036
Step-by-step explanation:
u = arcsin(0.391) ≈ 23.016737°
tan(u/2) = tan(11.508368°)
tan(u/2) ≈ 0.2036
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You can also use the trig identity ...
tan(α/2) = sin(α)/(1+cos(α))
and you can find cos(u) as cos(arcsin(0.391)) ≈ 0.920391
or using the trig identity ...
cos(α) = √(1 -sin²(α)) = √(1 -.152881) = √.847119
Then ...
tan(u/2) = 0.391/(1 +√0.847119)
tan(u/2) ≈ 0.2036
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<em>Comment on the solution</em>
These problems are probably intended to have you think about and use the trig half-angle and double-angle formulas. Since you need a calculator anyway for the roots and the division, it makes a certain amount of sense to use it for inverse trig functions. Finding the angle and the appropriate function of it is a lot easier than messing with trig identities, IMO.