Confused on how to do this picture attached
Did you find the slope?
No how would I?
I know that much but where do I get x and y from
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.
I graphed it and drew a best fit line now what?
y 2 - y 1 / x 2 - x 1
I get that but what points do I input
consider two points.. just take any 2 points.. and use that y2-y1 equation
for example use 1990 and 1950
somethings not right here, isn't time supposed to be on x-axis?
Lol yes it is sorry
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 \[\frac{ 95000-35000 }{ 1990-1950 }\]
1500?
looks right.. :)
I'm still confuse on how I get the answer after finding the slope
i'll see to it, sorry
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" \[y = 1488.33*100 + 23800\]
Thanks so much!
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