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Daniel [21]
3 years ago
5

Two quantities of data have a strong positive correlation. Can the line of best fit for the data set have a correlation coeffici

ent of 0.25? Explain.
Choose the correct answer below.
O A. Yes. The sign of the correlation coefficient indicates the type of correlation data have. Because 0.25 is positive, it is possible for data with a strong
positive correlation to have it as the correlation coefficient
OB. No. By definition, data that have a strong positive correlation have a strong linear relationship. The correlation coefficient would be closer to 1 than 0.25 is
OC. No. By definition, data that have a strong positive correlation have a strong linear relationship. The correlation coefficient would be closer to - 1 than 0.25
OD. No. By definition, data that have a strong positive correlation have a strong linear relationship. The correlation coefficient would be closer to than 0.25 is.
O E. Yes. By definition, data that have a strong positive correlation have a strong linear relationship. A correlation coefficient of 0.25 indicates a fairly strong
is.
correlation
Mathematics
1 answer:
zhuklara [117]3 years ago
4 0
I think it’s c for this answer
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kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

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