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QuestionFind the first full year after 2000 when truckers' pay PP exceeds \41,given41, given P=0.393 x^{2}-7.82 x+76.4for for 7 \leq x \leq 14$.

Studdy Solution

STEP 1

Assumptions1. The average annual pay for truckers $$ is given by the function $=0.393 x^{}-7.82 x+76.4$ where $x$ represents the number of years since2000. . The value of $x$ is between7 and14, inclusive.
3. We are looking for the first full year when the annual pay is above $\$41$ thousand.

STEP 2

We need to find the value of xx when >41 >41. To do this, we set up the inequality 0.393x2−7.82x+76.4>410.393 x^{2}-7.82 x+76.4 >41.

STEP 3

Subtract41 from both sides of the inequality to simplify it.
0.393x2−7.82x+76.−41>00.393 x^{2}-7.82 x+76. -41 >0

STEP 4

implify the inequality.
0.393x2−7.82x+35.4>00.393 x^{2}-7.82 x+35.4 >0

STEP 5

This is a quadratic inequality. We can solve it by first finding the roots of the corresponding quadratic equation 0.393x2−7.82x+35.4=00.393 x^{2}-7.82 x+35.4 =0.

STEP 6

To find the roots of the quadratic equation, we can use the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}, where a=0.393a =0.393, b=−.82b = -.82, and c=35.4c =35.4.

STEP 7

Substitute the values of aa, bb, and cc into the quadratic formula to find the roots.
x=−(−7.82)±(−7.82)2−4∗0.393∗35.42∗0.393x = \frac{-(-7.82) \pm \sqrt{(-7.82)^2 -4*0.393*35.4}}{2*0.393}

STEP 8

implify the equation.
x=7.82±(−7.82)2−4∗0.393∗35.42∗0.393x = \frac{7.82 \pm \sqrt{(-7.82)^2 -4*0.393*35.4}}{2*0.393}

STEP 9

Calculate the value under the square root.
x=7.82±(7.82)2−4∗.393∗35.42∗.393x = \frac{7.82 \pm \sqrt{(7.82)^2 -4*.393*35.4}}{2*.393}

STEP 10

Calculate the values of xx.
x=7.82±61.1524−55.632240.786x = \frac{7.82 \pm \sqrt{61.1524 -55.63224}}{0.786}

STEP 11

implify the equation.
x=7.82±5.520160.786x = \frac{7.82 \pm \sqrt{5.52016}}{0.786}

STEP 12

Calculate the square root.
x=7.82±2.350.786x = \frac{7.82 \pm2.35}{0.786}

STEP 13

Calculate the two possible values of xx.
x=7.82+2.350.786x = \frac{7.82 +2.35}{0.786}x2=7.82−2.350.786x2 = \frac{7.82 -2.35}{0.786}

STEP 14

Calculate the values of xx and x2x2.
x=12.94x =12.94x2=6.95x2 =6.95

STEP 15

Since xx represents the number of years since2000, it must be a whole number. Also, according to the problem, 7≤x≤147 \leq x \leq14. So, we round xx and x2x2 to the nearest whole numbers within this range.

STEP 16

Round xx and x2x2 to the nearest whole numbers.
x=13x =13x2=x2 =

STEP 17

Since we are looking for the first full year when the annual pay is above $41\$41 thousand, we choose the smaller value of xx, which is x2=7x2 =7.

STEP 18

Add7 to2000 to find the first full year when the annual pay is above $41\$41 thousand.
Year=2000+7Year =2000 +7

STEP 19

Calculate the year.
Year=2007Year =2007So, the first full year when the annual pay recovered to be above $41\$41 thousand is2007.

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