Question

name date period
algebra 1 unit 3 section b checkpoint

1. a safety inspector records the speed of a car 30 times and how far the car travels until it comes to a complete stop. the results are summarized in the scatter plot.

a. using the linear model ( y = 3.1x - 1 ), what does the slope mean in this situation? does the vertical intercept make sense in this situation? explain your thinking.
b. use the residuals to justify that this linear model is a good fit for the data at slow speeds.

Explanation:

Step1: Analyze the slope

The linear model is \(y = 3.1x-1\), where \(y\) is the stopping distance (in ft) and \(x\) is the speed (in mph). In the linear equation \(y = mx + b\) (slope - intercept form), \(m\) is the slope. Here, \(m = 3.1\). The slope represents the rate of change. So, for every 1 - mph increase in speed, the stopping distance increases by 3.1 feet.

Step2: Analyze the y - intercept

The y - intercept \(b=-1\). In the context of the problem, when \(x = 0\) (speed is 0 mph), \(y=-1\). But the stopping distance cannot be negative. So, the vertical intercept does not make sense in this situation because when the car is not moving (\(x = 0\)), the stopping distance should be 0 (assuming the car is already at rest and not moving forward to have a negative distance).

Answer:

a. The slope \(3.1\) means that for every 1 - mph increase in speed, the stopping distance increases by 3.1 feet. The vertical intercept \(y=-1\) (when \(x = 0\)) does not make sense because when the car's speed \(x = 0\) (car is at rest), the stopping distance \(y\) should be 0 (not - 1).

For part b, since no data points for slow speeds (say \(x\) values like \(x = 2,3,\cdots,10\)) and their corresponding residuals (\(e=y_{actual}-y_{predicted}\), where \(y_{predicted}=3.1x - 1\)) are given in the problem statement, we assume that if the residuals (the differences between the actual stopping distances and the distances predicted by the model \(y = 3.1x-1\)) for slow speeds are small (either close to 0 or with a random pattern around 0), it would justify that the linear model is a good fit for the data at slow speeds. A good fit at slow speeds would mean that the model accurately predicts the stopping distance for lower speed values.