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

Studdy Solution
Since $x$ represents the number of years since2000, and we are looking for the first full year when the pay is above $42,000, we need to round$x2$up to the nearest whole number. This gives us$x =6$.
The first full year when annual pay recovered to be above $42,000 is2000 +6 =2006.