The line of best fit went through the points (80,48.3) and (90,54.3). To find the slope of the line we used the slope formula: M=(Y2 -- Y1)/(X2 -- X1). If (80,48.3) is X1 and Y1 then (90,54.3) is X2 and Y2. After this you take X2 and subtract it by X1 and the same with Y1 and Y2. So 90-80 is 10 and 54.3 48.3 is 6. …show more content…
For the least square regression line all you have to do is plug 300cm into the X spot in the equation. Our equation Y=.6X+3.3 will now look like Y=.6(300)+3.3. We first multiplied .6 and 300 because of order of operations. Then we added 3.3 and the bounce height was 183.3. Using the line of best fit we first used the formula Y=MX+B and plugged in the M and B we found which looked like this y=0.6X+0.3. Then for X we put in 300, Y=0.6(300)+0.3. Just like the last equation we multiply .6 and 300, then added .3 so we got 180.3 as the bounce height for slope intercept. Both the predictions are so close that you could choose either equations because they are both extremely close to each …show more content…
We chose to use the least square regression line because it is more accurate than the slope intercept line, because the calculator takes our data and gives us an exact number whereas the slope intercept form is an educated guess we make with our eyes. We use the same formula as before Y=.6X+3.3 and instead of plugging in X we plugged in Y. We did this because we need to use the exact formula as before in order to find our prediction. Our equation now looks like 200=.6X+3.3. We need to get X alone so we subtracted 3.3 from 200 which is 196.7. Then we divided .6 from 196.7 and we got 327.8. To check our math we put 327.8 back into our last equation Y=.6(327.8)+3.3. and we got 199.98 which rounded up is