The model $ \sqrt{\hat{\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 $ 65 \mathrm{mph} $ need to stop?

A) 4.3 feet
B) 729.0 feet
C) 18.6 feet
D) 27.0 feet
E) 345.0 feet

Question

The model $ \sqrt{\hat{\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 $ 65 \mathrm{mph} $ need to stop?

A) 4.3 feet
B) 729.0 feet
C) 18.6 feet
D) 27.0 feet
E) 345.0 feet

Quizplus · Accepted Answer

First, substitute 65 for speed in the model: $\sqrt{\hat{\text{distance}}} = 3.30 + 0.235(65)$. This gives $\sqrt{\hat{\text{distance}}} = 3.30 + 15.275 = 18.575$. Squaring both sides to solve for $\hat{\text{distance}}$ gives $\hat{\text{distance}} = 18.575^2 \approx 345.0$ feet.

The Model distance ^=3.30+0.235 \sqrt{\hat{\text { distance }}}=3.30+0.235 distance ^​=3.30+0.235 (Speed) Can Be Used to Predict the Stopping Distance (In

The Model $\sqrt{\hat{\text { distance }}}=3.30+0.235$ (Speed) Can Be Used to Predict the Stopping Distance (In