anonymous
  • anonymous
i need help solving a x and y table x 30 40 50 60 y 6 10 16 26
Mathematics
katieb
  • katieb
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sgadi
  • sgadi
this is not linear .... do you want a linear fit for this or you want a non-linear relation between them?
anonymous
  • anonymous
non- linear
sgadi
  • sgadi
the equation satisfying the relation is \[y=\frac{x^4}{3000}-\frac{3x^3}{100}+\frac{19x^2}{15}-14\]

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anonymous
  • anonymous
WHAT ABOUT A LINEAR?
sgadi
  • sgadi
the linear is only a approximate fit ... it will not solve anything .... what is your exact problem?
anonymous
  • anonymous
THE PERCENT OF PEOPLE NO WEARING SEAT BELTS WHO ARE INJURED IN AUTOMOBILE ACCIDENTS IS A FUNCTION OF THE CAR'S SPEED SPEED IN MILES PER HOUR, X 30 40 50 60 PERCENT INJURED, Y 6 10 16 26 WHAT IS THE FUNCTION FOR THIS TABLE?
sgadi
  • sgadi
then the equation which I gave before is correct. :)
anonymous
  • anonymous
but when i plug 30 in for x i don't get 6, which is y
sgadi
  • sgadi
ohhh ..... i am soo sorry ... i did a mistake the answer is \[y=\frac{x^3}{3000}-\frac{3x^2}{100}+\frac{9x}{15}-14\] please verify and tell if you are getting :-) sorry again
anonymous
  • anonymous
im still not getting y=6
sgadi
  • sgadi
opsss my mistake again dear ... i m so sorry for this .... \[y=\frac{x^3}{3000}-\frac{3x^2}{100}+\frac{19x}{15}-14\] x=30 (1)...........x^3=27000 (2)...........x^2=900 (3)...........x^3/3000=9 (4)...........3*x^2/100=3*9=27 (5)...........19*x/15=38 then y=(3)-(4)+(5)-14=9-27+38-14=6
anonymous
  • anonymous
thank you sooooo much!!
sgadi
  • sgadi
welcome

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