## anonymous 5 years ago i need help solving a x and y table x 30 40 50 60 y 6 10 16 26

this is not linear .... do you want a linear fit for this or you want a non-linear relation between them?

2. anonymous

non- linear

the equation satisfying the relation is $y=\frac{x^4}{3000}-\frac{3x^3}{100}+\frac{19x^2}{15}-14$

4. anonymous

the linear is only a approximate fit ... it will not solve anything .... what is your exact problem?

6. 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?

then the equation which I gave before is correct. :)

8. anonymous

but when i plug 30 in for x i don't get 6, which is y

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

10. anonymous

im still not getting y=6

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

12. anonymous

thank you sooooo much!!