Two-variable data models and scatterplots: worksheet with answers

Use the two points on the line of best fit to determine the predicted score. The line passes through (2, 70) and (8, 88).

The slope is \(\frac{88-70}{8-2}=\frac{18}{6}=3\), so the score increases by 3 points for each additional practice session.

From 2 practice sessions to 7 practice sessions is 5 more sessions, so add \(5\times 3=15\) points to 70.

\(70+15=85\)

Therefore, the predicted average performance score is 85.

As the x-values increase from \(-2\) to \(2\), the y-values decrease from \(16\) to \(1\), so the function is decreasing.

The graph is not a straight line, and the outputs are multiplied by the same factor each time x increases by 1: \(16\to 8\to 4\to 2\to 1\). Because the function changes by a constant ratio rather than a constant difference, it is exponential.

Therefore, the function is a decreasing exponential function.

To answer the question, identify the largest value in the Tickets Sold column. The greatest number of tickets sold is 176, which occurred in August. Therefore, the correct answer is August.

The line of best fit rises from left to right, so the slope is positive. The line crosses the y-axis below 0, so the y-intercept is negative. Among the choices, only \(y=2x-6\) has a positive slope and a negative y-intercept.

The data show a clear negative linear trend, so the slope of the model should be negative. The points near \(x=0\) have \(y\)-values around 11, so the \(y\)-intercept should be close to 11.

The only choice with a negative slope and a \(y\)-intercept near 11 is \(y=11-1.1x\).

Choice A has a positive slope and a negative intercept. Choice B has the correct negative slope but the intercept is about 1 instead of 11. Choice D has the correct intercept but the slope is positive.

The line of best fit decreases from left to right, so the slope is negative. Using the two points on the line, the slope is approximately

\(\frac{5-10}{8-2}=\frac{-5}{6}\approx -0.83\).

The choice closest to \(-0.83\) is \(-0.8\).

Choose two convenient points that lie on the line of best fit, such as approximately \((2,5)\) and \((8,14)\). The slope is

\(\frac{14-5}{8-2}=\frac{9}{6}=1.5\).

Because the line rises from left to right, the slope is positive. Therefore, the slope closest to the line of best fit is \(1.5\).

Use two approximate points on the line of best fit, such as (2, 21) and (14, 7).

The slope is

\(\frac{7-21}{14-2}=\frac{-14}{12}\approx -1.17\).

The choice closest to \(-1.17\) is \(-1.2\).

Because \(145\% = 1.45\), the relationship can be written as \(g(t)=1.45t\). A function of the form \(y=ax\), where \(a\) is a constant, is linear. Since the coefficient \(1.45\) is greater than 1 and positive, the function increases as \(t\) increases. Therefore, the function is increasing and linear.

On the scatterplot, the point for 35 miles after leaving the warehouse is \((35, 22)\), meaning 22 liters of fuel remained. The point for 40 miles after leaving the warehouse is \((40, 16)\), meaning 16 liters remained.

The average rate of change is

\(\frac{16-22}{40-35}=\frac{-6}{5}=-1.2\).

So the average rate of change is \(-\frac{6}{5}\) liters per mile, or \(-1.2\) liters per mile.

The amount of chemical decreases over time, so the model represents exponential decay. In an exponential model \(y=a(b)^x\), exponential decay occurs when \(0<b<1\).

Among the choices, only \(0.74\) is between 0 and 1, so it is the only reasonable value for \(b\). The data also support this because each hour the amount is multiplied by about the same factor: \(\frac{178}{240}\approx0.74\) and \(\frac{132}{178}\approx0.74\).

Therefore, the value closest to \(b\) is \(0.74\).

The original line of best fit can be estimated from the two points. The slope is \(\frac{24-6}{10-0}=\frac{18}{10}=1.8\), and the y-intercept is about \(6\). So an approximate equation for the original line is \(y=1.8x+6\).

Multiplying every \(y\)-value by \(2.5\) multiplies both the slope and the y-intercept by \(2.5\). The new slope is \(1.8(2.5)=4.5\), and the new intercept is \(6(2.5)=15\).

Therefore, a possible line of best fit for the transformed data set is \(y=4.5x+15\).