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anonymous

  • 5 years ago

i need help solving a x and y table x 30 40 50 60 y 6 10 16 26

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  1. sgadi
    • 5 years ago
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    this is not linear .... do you want a linear fit for this or you want a non-linear relation between them?

  2. anonymous
    • 5 years ago
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    non- linear

  3. sgadi
    • 5 years ago
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    the equation satisfying the relation is \[y=\frac{x^4}{3000}-\frac{3x^3}{100}+\frac{19x^2}{15}-14\]

  4. anonymous
    • 5 years ago
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    WHAT ABOUT A LINEAR?

  5. sgadi
    • 5 years ago
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    the linear is only a approximate fit ... it will not solve anything .... what is your exact problem?

  6. anonymous
    • 5 years ago
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    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?

  7. sgadi
    • 5 years ago
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    then the equation which I gave before is correct. :)

  8. anonymous
    • 5 years ago
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    but when i plug 30 in for x i don't get 6, which is y

  9. sgadi
    • 5 years ago
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    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
    • 5 years ago
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    im still not getting y=6

  11. sgadi
    • 5 years ago
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    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
    • 5 years ago
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    thank you sooooo much!!

  13. sgadi
    • 5 years ago
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    welcome

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