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Mathematics 10 Online
OpenStudy (anonymous):

Confused on how to do this picture attached

OpenStudy (anonymous):

OpenStudy (mertsj):

Did you find the slope?

OpenStudy (anonymous):

No how would I?

OpenStudy (anonymous):

I know that much but where do I get x and y from

OpenStudy (mertsj):

From the table. The top row contains x values. The bottom row contains y values. You should graph the ordered pairs and draw the line of best fit.

OpenStudy (anonymous):

I graphed it and drew a best fit line now what?

OpenStudy (anonymous):

y 2 - y 1 / x 2 - x 1

OpenStudy (anonymous):

I get that but what points do I input

OpenStudy (anonymous):

consider two points.. just take any 2 points.. and use that y2-y1 equation

OpenStudy (anonymous):

for example use 1990 and 1950

OpenStudy (anonymous):

somethings not right here, isn't time supposed to be on x-axis?

OpenStudy (anonymous):

Lol yes it is sorry

OpenStudy (anonymous):

to find the slope, for example, draw a line from 1990, to a point on the line looks about 95000 to me.. Then draw a line from 1950 to a point on the line as well..looks about 35000 to me then divide \[\frac{ 95000-35000 }{ 1990-1950 }\]

OpenStudy (anonymous):

1500?

OpenStudy (anonymous):

looks right.. :)

OpenStudy (anonymous):

I'm still confuse on how I get the answer after finding the slope

OpenStudy (anonymous):

i'll see to it, sorry

OpenStudy (dumbcow):

for these trend line problems ,its best to use the end points to estimate slope \[slope = \frac{113,100-23,800}{2000-1940} = 1488.33\] for part b) plug in (2040 -1940 = 100) for "x" \[y = 1488.33*100 + 23800\]

OpenStudy (anonymous):

Thanks so much!

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