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Mathematics 8 Online
OpenStudy (dobby1):

2. Which set of ordered pairs and equation for a best-fit line represents data from the table? (x1, y1) = (64, 135), (x2, y2) = (74, 166) y = 3.14x – 66.42 (x1, y1) = (62, 131), (x2, y2) = (70, 152) y = 2.625x – 31.75 (x1, y1) = (62, 131), (x2, y2) = (72, 158.5) y = 2.75x – 12 (x1, y1) = (64, 130), (x2, y2) = (72, 167) y = 3.5625x – 98

OpenStudy (dobby1):

@phi

OpenStudy (praxer):

do you know the 2 point form representation of a straight line.

OpenStudy (dobby1):

OpenStudy (dobby1):

@mathstudent55

OpenStudy (mathstudent55):

Are you told which method to use?

OpenStudy (dobby1):

no

OpenStudy (dobby1):

that is all the info

OpenStudy (mathstudent55):

You must have a textbook, or some other source of math info. What methods have been shown to you?

OpenStudy (dobby1):

I left it at school I was staying for a club and left it in my locker

OpenStudy (dobby1):

so i dont remember

OpenStudy (dobby1):

all I know is it is pre calc

OpenStudy (dobby1):

and right now I am in a panic

OpenStudy (mathstudent55):

One method is to plot all the points. Then you find which line best averages the points. Then you pick two points on the line and find its slope. Then you use a point on the line and the slope to find the equation of the line.

OpenStudy (mathstudent55):

Another method is to use a calculator that can do linear regression for you.

OpenStudy (mathstudent55):

There is also a way of doing all the work manually and figuring it out manually without plotting.

OpenStudy (dobby1):

Ok I will plot the points and see how it goes

OpenStudy (mathstudent55):

Also, notice that for each age, you need to plot the age vs the average cholesterol count, so first, you need to find all the average cholesterol counts for all ages.

OpenStudy (dobby1):

I just got an A on the practice thx for the help

OpenStudy (mathstudent55):

62, 131 63,133 64, 135 65, 137.5 66, 140 67, 143 68, 146 etc.

OpenStudy (mathstudent55):

Ok, great!

OpenStudy (dobby1):

thx so much

OpenStudy (mathstudent55):

You're welcome.

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