+0  
 
0
637
1
avatar+5265 

 

Please answer in something at least somewhat understandable. Thank you! Also, please try to keep it related to the question.

 Oct 25, 2016
 #1
avatar+118587 
+5

Hi rarinstraw,

 

Gino is really good at these he may have a better way to do it.

 

I set up these four equations

\(cost = ax^2+bx+c\\~\\ 550=a*225+b*15+c\\ 700=a*289+b*17+c\\ 1500=a*529+b*23+c\\ 2500=a*900+b*30+c\)

 

Then I solved them simultaneously.

I got 

\(a\approx 3.23\\ b\approx -28.40\\ c\approx 249\\ so\\ cost\approx 3.23x^2-28.40x+249\)

 

https://www.desmos.com/calculator/h94xx9vjun

 

 

 

3.23*42^2-28.40*42+249 = 4753.92

 

The points do not fit on the graph exactly because the these points will not fit any quadratic model exactly.

 

So the estimated cost of a 42cm tele is  $4754

 

BUT since 42cm is an extrapolated point (larger than any of the other sizes) you would need to be very careful about relying on this price.

 Oct 25, 2016

0 Online Users