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The Model  distance =3.30+0.235\sqrt { \text { distance } } = 3.30 + 0.235

Question 17

Multiple Choice

The model  distance =3.30+0.235\sqrt { \text { distance } } = 3.30 + 0.235 (speed. can be used to predict the stopping distance (in feet. for a car traveling at a specific speed (in mph.) According to this model, about how much distance will a car going 65mph65 \mathrm { mph } need to stop?


A) 729.0729.0 feet
B) 345.0345.0 feet
C) 27.027.0 feet
D) 4.34.3 feet
E) 18.618.6 feet

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