Linear Models

1. a) Choose any two points, for example $(50, 55)$ and $(100, 80)$:

$m = \frac{80 - 55}{100 - 50} = \frac{25}{50} = 0.50$

The slope is \0.50$per mile. It means the company charges$$0.50$ for each additional mile driven.

b) Using the point-slope form with $(50, 55)$:

$55 = 0.50(50) + b$ $55 = 25 + b$ $b = 30$

The $y$-intercept is \30$. It represents the base rental fee — the cost charged before any miles are driven.

c) The equation of the linear model is:

$y = 0.50x + 30$

2. a) The slope of $-4$ means that for each additional absence, the predicted exam score decreases by $4$ points. More absences are associated with lower scores.

b) The $y$-intercept of $95$ means that a student with $0$ absences is predicted to score $95$ on the exam. This is the predicted score for perfect attendance.

c) Substitute $x = 5$:

$y = -4(5) + 95 = -20 + 95 = 75$

The predicted score for $5$ absences is $75$.

d) Set $y < 60$:

$-4x + 95 < 60$ $-4x < -35$ $x > 8.75$

Since absences are whole numbers, the model predicts a failing score starting at $9$ absences.

3. a) There is a positive linear association between length and weight. As fish length increases, weight increases. The relationship appears to be approximately linear, though larger fish show slightly greater weight gains per cm (suggesting the true relationship may be slightly curved).

b) Substitute $x = 17$:

$y = 7.2(17) - 42 = 122.4 - 42 = 80.4 \text{ g}$

The predicted weight of a $17$ cm fish is $80.4$ g.

c) First, find the predicted weight for $x = 15$:

$\hat{y} = 7.2(15) - 42 = 108 - 42 = 66 \text{ g}$

The residual is actual minus predicted:

$\text{Residual} = 60 - 66 = -6 \text{ g}$

The negative residual means the actual weight ($60$ g) is $6$ g less than what the model predicted ($66$ g). The model overestimated the weight for this fish.

4. a) Students who play both: $45$ out of $200$.

$\frac{45}{200} = 0.225 = 22.5\%$

b) Among students who play an instrument ($80$ total), $45$ play a sport:

$\frac{45}{80} = 0.5625 = 56.25\%$

c) Among students who do NOT play an instrument ($120$ total), $65$ play a sport:

$\frac{65}{120} = 0.5417 = 54.17\%$

d) The percentages are very close: $56.25\%$ of instrument players play a sport compared to $54.17\%$ of non-instrument players. Since these values are nearly the same, there appears to be little to no association between playing an instrument and playing a sport. Playing an instrument does not substantially change the likelihood of also playing a sport.

5. a) Slope ($2.2$): For each $1°$F increase in temperature, the shop is predicted to sell approximately $2.2$ more ice cream cones.

$y$-intercept ($-115$): The model predicts $-115$ cones sold at $0°$F. This is not meaningful in context — you cannot sell a negative number of cones, and $0°$F is far outside the range of the data ($60°$–$95°$F). The intercept is just a mathematical artifact of the linear equation.

b) Substitute $x = 78$:

$y = 2.2(78) - 115 = 171.6 - 115 = 56.6$

The model predicts approximately $57$ cones sold on a $78°$F day.

c) Predicted value for $x = 85$:

$\hat{y} = 2.2(85) - 115 = 187 - 115 = 72$

Residual:

$\text{Residual} = 75 - 72 = 3$

The positive residual means the actual sales ($75$ cones) were $3$ cones more than the model predicted ($72$ cones). The model slightly underestimated sales on this day.

d) No, this model should not be used to predict sales at $30°$F. That temperature is far outside the range of the data ($60°$–$95°$F). Predicting at $x = 30$ gives:

$y = 2.2(30) - 115 = 66 - 115 = -49$

A prediction of $-49$ cones is impossible. This is an example of unreliable extrapolation — the linear relationship observed between $60°$F and $95°$F does not necessarily hold outside that range.