Question

i need help solving a x and y table

x 30 40 50 60
y 6 10 16 26

Answer 1

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

Answer 2

non- linear

Answer 3

the equation satisfying the relation is

\[y=\frac{x^4}{3000}-\frac{3x^3}{100}+\frac{19x^2}{15}-14\]

Answer 4

WHAT ABOUT A LINEAR?

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

im still not getting y=6

Answer 11

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

Answer 12

thank you sooooo much!!

Answer 13

welcome