Question

I need help with this (don't answer if you don't know)

Answer 1

Answer:

Step-by-step explanation:

Here's a step by step tutorial on your calculator on how to do this.  

Hit "stat" then "1: Edit".  If there are numbers there, arrow up to highlight L1, or L2, or wherever there are numbers.  Hit "clear" then "enter" and the numbers will be gone.  In L1, enter the Practice throws values.  Press 3 then enter, then 12 then enter, then 6 then enter, etc. til all of them are in L1's column.  Then arrow over to L2 and do the same with entering all the Free Throw values.

When you're done, hit "stat" again, then arrow over once to "calc" and #4 should say LinReg.  That's a linear regression equation.  If you have a TI 83, just hit enter and you'll get the equation in the form y = mx + b.  If you have a TI 84 or 84+, you'll need to arrow down to the word "calculate" and then you'll see the equation.

One thing...if you have not turned your diagnostics on, you wont be able to see the coefficient of determination (the r-squared value).  To make sure it's on:

Hit 2nd, then 0.  You have opened up the catalog which lists every single thing your calculator can do in alphabetical order.  The hit the button UNDER the MATH button (x to the negative 1) and scroll down until you see "diagnosticsON" and hit enter twice.

If you need to recall the linear regression equation, hit "stat", then "calc" then either enter or calculate and you'll get the linear equation again and an r value and an r-squared value.  The closer that number is to 1, the better the data fits that model.  I got that the equation is

y = 2.352x - .852 with an r-squared value of .844

To get the quadratic regression equation, hit "stat" then "calc" then choose #5, QuadReg.  Repeat the process to calculate your equation.  Mine was

$y=.073x^2+1.205x+2.555; r^2=.855$

And for the exponential regression, choose ExpReg (mine is under 0.  Yours may be some other number.  Just arrow down til you find it, it's there!) My ExpReg equation was

$y=4.229(1.170)^x; r^2=.833$

It appears that the r-squared value is the highest in the quadratic regression equation, so that's the best fit.