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Question Number 118612 by rexfordattacudjoe last updated on 18/Oct/20

Answered by 1549442205PVT last updated on 18/Oct/20

$$\mathrm{y}=0.{4\mathrm{x}}^2-36\mathrm{x}+1000=$$$$\frac{4}{10}({\mathrm{x}}^2-90\mathrm{x}+2500)=\frac{2}{5}[{(\mathrm{x}-45)}^2+475]$$$$⩾\frac{2.475}{5}=190$$$$\Rightarrow {\mathrm{y}}_{\min }=190\mathrm{when}\mathrm{x}=45(\mathrm{years}-\mathrm{old})$$$$\mathrm{It}\mathrm{means}\mathrm{that}\mathrm{numbers}\mathrm{of}\mathrm{car}\mathrm{accidents}$$$$\mathrm{are}\mathrm{lowest}\mathrm{at}\mathrm{age}45.\mathrm{That}\mathrm{number}\mathrm{is}$$$$190\mathrm{cases}\mathrm{per}50\mathrm{million}\mathrm{km}\mathrm{driven}$$

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