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QuestionFind when trucker pay PP (in thousands) exceeds \42,using42, using 42=0.449 x^{2}-7.91 x+72.2for for 6 \leq x \leq 14$.

Studdy Solution

STEP 1

Assumptions1. The function given is valid for the range of years 6≤x≤146 \leq x \leq14. . The function accurately predicts the average annual pay for truckers in the country.
3. The year2000 corresponds to x=0x =0.
4. We are looking for the first full year when the pay is above 42,000,soweneedtosolvefor42,000, so we need to solve for xwhen when >42$.

STEP 2

We need to solve the equation for xx when isexactlyequalto42.Theequationisgivenas is exactly equal to42. The equation is given as42=0.449 x^{2}-7.91 x+72.2$$

STEP 3

To find the value of xx, we need to solve this quadratic equation. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c =0. In this case, a=0.449a =0.449, b=−7.91b = -7.91, and c=72.2−42=30.2c =72.2 -42 =30.2.

STEP 4

The quadratic equation is now0.449x2−7.91x+30.2=00.449x^2 -7.91x +30.2 =0

STEP 5

We can solve this quadratic equation using the quadratic formulax=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

STEP 6

Substitute the values of aa, bb, and cc into the quadratic formulax=−(−.91)±(−.91)2−4⋅0.449⋅30.22⋅0.449x = \frac{-(-.91) \pm \sqrt{(-.91)^2 -4 \cdot0.449 \cdot30.2}}{2 \cdot0.449}

STEP 7

implify the equationx=7.91±(7.91)2−4⋅0.449⋅30.22⋅0.449x = \frac{7.91 \pm \sqrt{(7.91)^2 -4 \cdot0.449 \cdot30.2}}{2 \cdot0.449}

STEP 8

Calculate the value under the square rootx=7.91±62.6681−54.17920.898x = \frac{7.91 \pm \sqrt{62.6681 -54.1792}}{0.898}

STEP 9

implify the equationx=7.91±8.4889.898x = \frac{7.91 \pm \sqrt{8.4889}}{.898}

STEP 10

Calculate the square rootx=7.91±2.91450.898x = \frac{7.91 \pm2.9145}{0.898}

STEP 11

Calculate the two possible values for xxx=7.91+.91450.898x = \frac{7.91 +.9145}{0.898}x=7.91−.91450.898x = \frac{7.91 -.9145}{0.898}

STEP 12

Calculate the values for xx and x2x2x=11.42x =11.42x2=5.56x2 =5.56

STEP 13

Since xx represents the number of years since2000, and we are looking for the first full year when the pay is above 42,000,weneedtoround42,000, we need to round x2uptothenearestwholenumber.Thisgivesus up to the nearest whole number. This gives us x =6$.
The first full year when annual pay recovered to be above $42,000 is2000 +6 =2006.

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