The model$\sqrt{\widehat{\text { distance }}}$ = 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) $729.0$ feet
B) $345.0$ feet
C) $27.0$ feet
D) $4.3$ feet
E) $18.6$ feet

Question

The model$\sqrt{\widehat{\text { distance }}}$  = 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) $729.0$ feet
B) $345.0$ feet
C) $27.0$ feet
D) $4.3$ feet
E) $18.6$ feet

Quizplus · Accepted Answer

The Answer of The model$\sqrt{\widehat{\text { distance }}}$  = 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) $729.0$ feet
B) $345.0$ feet
C) $27.0$ feet
D) $4.3$ feet
E) $18.6$ feet

The Model distance ^\sqrt{\widehat{\text { distance }}} distance ​ = 330 + 0235 (Speed 65mph65 \mathrm { mph }65mph

The Model $\sqrt{\widehat{\text { distance }}}$ = 330 + 0235 (Speed $65 \mathrm { mph }$