- What does an input of zero hours mean here? - Does a fitted line have to match a measurement exactly? - Was the candle measured when it was first lit?

1. The y-intercept is the predicted height when \(t=0\). 2. Substituting gives \(\hat h=24\), so the model predicts a height of \(24\,\text{cm}\) when the candle was first lit. 3. No measurement at \(t=0\) was used, so this is a prediction, not the measured starting height.

The line predicts a starting height of \(24\,\text{cm}\). It does not tell us the exact starting height because the measurements began later.

- Interpret the slope as a change for one unit of \(x\). - Scale that change by the requested input interval. - Report a change, not a total predicted distance.

1. The slope predicts an increase of \(0.35\) mile per additional week. 2. Over \(6\) weeks, the predicted change is \(0.35\cdot 6=2.1\) miles.

The predicted distance increases by \(2.1\) miles.

- Rewrite the constant equation with an explicit \(x\)-term. - Identify the coefficient multiplying \(x\). - Evaluate the model at zero.

1. The equation can be written as \(\hat y=0x+37.2\). 2. The slope is \(0\), so the model predicts no change in \(y\) as \(x\) increases. 3. The y-intercept is \(37.2\), which is also the predicted response for every \(x\)-value.

Slope: \(0\), meaning no predicted linear change in \(y\) as \(x\) increases. y-intercept: \(37.2\), the predicted response at \(x=0\).

- Evaluate the equation at zero and at another input. - Separate the intercept from the slope. - Identify exactly which input value the intercept describes.

1. The y-intercept is the prediction at \(x=0\), which is \(0\). 2. For positive \(x\)-values, the term \(4.6x\) is generally positive. 3. The zero intercept describes only the model’s baseline at zero input, not all predictions.

The model predicts \(\hat y=0\) when \(x=0\). It does not predict zero elsewhere because the response changes at a rate of \(4.6\) per unit of \(x\).

- Identify which number multiplies the explanatory variable. - Evaluate the equation at zero. - Separate rate from baseline.

1. The per-unit change is the coefficient of \(x\), which is \(7\). 2. The constant \(42\) is the y-intercept. 3. It represents the predicted response when \(x=0\), not a rate of change.

Slope: \(7\), the predicted change in \(y\) for a one-unit increase in \(x\). y-intercept: \(42\), the predicted \(y\)-value at \(x=0\).

- Put the response variable’s units in the numerator. - Put the explanatory variable’s units in the denominator. - State a predicted change for one additional hour.

1. Slope units are response units divided by explanatory-variable units. 2. The correct units are dollars per hour. 3. The model predicts a \(\$5\) increase in cost for each additional hour of service.

The slope is \(\$5\) per hour, meaning each additional service hour increases predicted cost by \(\$5\).

- Which table entry would determine how a prediction changes when \(x\) increases? - What would the model predict if \(x=0\)? - Be careful to preserve the sign of each reported estimate.

1. The coefficient of \(x\) is the slope, \(-0.4\). 2. The intercept is \(7.5\), so the regression equation is \(\hat y=-0.4x+7.5\).

\(\hat y=-0.4x+7.5\)

- Pick two points on the line whose coordinates can be read exactly. - Compare the change in predicted books with the change in weeks. - Does a fitted line represent exact weekly results or a general trend?

1. The line passes through \((0,3)\) and \((2,7)\). 2. The predicted number of books rises by \(7-3=4\) over \(2\) weeks, so the slope is \(4\div2=2\) books per week. 3. A fitted line predicts a general pattern, not the exact number completed during each week.

The line predicts \(2\) more books per week on average, not necessarily exactly \(2\) in any given week.

- Find the prediction when the input is zero. - Match an input of zero to a calendar day in this situation. - Check which days were actually used to fit the line.

1. The y-intercept is \(12\), which is the model's prediction when \(x=0\), on opening day. 2. The data used to fit the line came from days \(3\) through \(7\), not day \(0\). 3. Predicting at day \(0\) extends the pattern outside the observed range, so the actual number sold could have been different.

The y-intercept predicts \(12\) lunches on opening day. It does not establish actual sales on that day, because opening day was outside the observed range.

- Identify the units attached to one unit of \(x\). - Use the sign to determine increase or decrease. - State the change in the predicted response, not an exact change for every location.

1. The slope is \(-1.7^\circ\text{F}\) per thousand feet. 2. The negative sign indicates that predicted temperature decreases as elevation increases. 3. For each additional \(1000\) feet of elevation, the model predicts an average temperature decrease of about \(1.7^\circ\text{F}\).

For each increase of \(1000\) feet in elevation, the predicted afternoon temperature decreases by about \(1.7^\circ\text{F}\).

- Separate a rate of change from a predicted output. - Identify what a one-unit increase in \(x\) represents. - Include the response variable’s units in the corrected statement.

1. The slope measures predicted change in cost for a one-shirt increase, not the total cost. 2. The model predicts that each additional shirt is associated with an increase of \(\$2.75\) in total cost. 3. The separate constant \(18\) represents the predicted cost when \(x=0\).

Each additional shirt increases the predicted total cost by \(\$2.75\). The slope is a rate of change, not the total cost.

- Recall which variable is zero at each type of intercept. - Evaluate the model at \(x=0\). - Solve separately for when the predicted response equals zero.

1. The y-intercept occurs at \(x=0\), so the predicted amount is \(50\) gallons. This is the predicted starting amount. 2. The x-intercept occurs when the predicted response is \(0\): \(50-5x=0\). 3. Solving gives \(x=10\) hours. This is the predicted time when the tank is empty.

x-intercept: \(10\) hours, the predicted time when the tank is empty. y-intercept: \(50\) gallons, the predicted starting amount.

- Use the magnitude of the rate when matching a stated decrease. - Set up a change-in-response equation. - Check that the sign of the slope agrees with the direction described.

1. The model predicts a decrease of \(1.5\) percentage points per additional hour. 2. The required input change satisfies \(1.5\Delta x=12\). 3. Therefore, \(\Delta x=8\) hours.

\(8\) additional hours

- Which number in the fitted equation changes the predicted time when the crate count increases? - What input gives the y-intercept? - Compare that input with the range of crate counts actually recorded.

1. For each extra crate, the line predicts \(4.5\) additional minutes of loading. 2. The y-intercept predicts \(12\) minutes for \(0\) crates. 3. Since the recorded jobs had \(2\) to \(8\) crates, the intercept is an extrapolation, not a measured loading time. Actual jobs may differ from the line.

The slope predicts \(4.5\) more minutes per extra crate. The y-intercept predicts \(12\) minutes for no crates, but that time was not measured and need not be exact.

- Evaluate the model at zero sales. - Translate a negative profit into a loss. - Connect the loss at zero activity with fixed costs.

1. The y-intercept is the predicted profit when \(x=0\) tickets are sold. 2. At zero sales, \(\hat y=-240\), which represents a predicted loss of \(\$240\). 3. In context, this intercept is consistent with a fixed upfront cost that must be recovered through ticket sales.

The y-intercept means the fundraiser has a predicted \(\$240\) loss before any tickets are sold, consistent with fixed upfront costs.

- Translate one \(x\)-unit into the actual number of flyers. - Divide the response change across those individual flyers. - Keep the original per-hundred interpretation as a check.

1. An increase of \(1\) in \(x\) represents \(100\) additional flyers. 2. The model predicts \(6\) additional visits per \(100\) flyers. 3. Per flyer, the rate is \(6\div 100=0.06\) predicted customer visits.

The model predicts \(0.06\) additional customer visits per flyer, equivalent to \(6\) visits per \(100\) flyers.

- Decide whether a fitted line predicts exactly what happens for each student. - Keep the rate's units, points per hour, in your interpretation. - How can observed scores differ from predictions on a regression line?

1. A regression slope describes the predicted response, not a guaranteed change for every individual. 2. A one-hour increase in study time is associated with a \(4.1\)-point increase in predicted quiz score according to this model. 3. Individual quiz scores can be higher or lower than the model's prediction.

The model predicts quiz scores \(4.1\) points higher for students who study one additional hour. This is an average modeled pattern, not a guarantee for every student.

- Is a slope the predicted total time or the change for one extra bike? - Could one shop's line start higher but rise more slowly? - What information fixes the vertical position of each fitted line?

1. Elm's predicted repair time increases by \(12\) minutes for each extra bike; Ridge's increases by \(9\) minutes. 2. Slopes show how predictions change, not the total predicted repair time at four bikes. 3. We also need each line's y-intercept (or one predicted time at a known bike count) to compare the two predictions at four bikes.

Elm's line rises \(12\) minutes per extra bike; Ridge's rises \(9\). That alone does not show which predicts more time for four bikes. We need the y-intercepts or another prediction from each line.

- Compare the coefficients of \(x\) first. - Then compare the constant terms. - Think about the vertical separation of parallel lines.

1. Both models have slope \(1.2\), so they predict the same temperature increase per minute. 2. Model B’s intercept is \(15\), which is \(7\) units above Model A’s intercept of \(8\). 3. Because the lines are parallel, Model B predicts a temperature exactly \(7\) units higher at every common \(x\)-value.

The slopes are equal, so the predicted rates of change match. Model B’s intercept is \(7\) higher, so it predicts values \(7\) units above Model A throughout.

- Compare the constants separately from the \(x\)-coefficients. - Find the difference between the two rates. - Consider how repeated rate differences accumulate.

1. Both models have y-intercept \(20\), so they predict the same score at \(x=0\). 2. Model B’s slope, \(1.1\), is larger than Model A’s slope, \(0.5\). 3. Model B gains \(0.6\) more predicted points per day, so the difference between predictions grows as \(x\) increases.

The models share the same intercept, but Model B has the larger slope. Its prediction pulls farther above Model A by \(0.6\) points for each additional day.

- Evaluate the model at zero occupants. - Think about systems that operate without people present. - Separate baseline use from per-person use.

1. The y-intercept predicts \(180\,\text{kWh/day}\) when \(x=0\) occupants. 2. Empty buildings can still use electricity for lighting, security, ventilation, refrigeration, and other systems. 3. The intercept represents predicted baseline electricity use at zero occupancy.

The y-intercept predicts baseline use of \(180\,\text{kWh/day}\) even with no occupants.

- Locate where the line crosses the y-axis. - Choose two grid points on the line and compute change in \(y\) divided by change in \(x\). - Interpret the negative sign as a decrease in predicted completion time.

1. The line crosses the y-axis at \(20\), so the y-intercept is \(20\) minutes. 2. Using \((0,20)\) and \((4,14)\), the slope is \(\frac{14-20}{4-0}=-1.5\) minutes per practice session. 3. The intercept predicts a completion time of \(20\) minutes before any practice sessions. 4. The slope means each additional practice session is associated with a decrease of \(1.5\) minutes in predicted completion time.

a) \(20\) minutes b) \(-1.5\) minutes per practice session c) The model predicts \(20\) minutes before practice and a decrease of \(1.5\) predicted minutes for each additional practice session.

- What change in battery percentage does the negative slope represent? - What does an input of zero hours represent? - Was the phone's battery level measured at that time?

1. Each additional hour of streaming is associated with a decrease of \(3.5\) percentage points in the predicted battery level. 2. The y-intercept predicts \(80\%\) battery at the start, when \(t=0\). 3. The measurements began after the first hour, so the starting value is a prediction rather than an observed result.

The line predicts \(3.5\) percentage points less battery per extra hour. It predicts \(80\%\) at the start but does not prove that was the exact starting level.

- Read the equation as prediction of \(y\) from \(x\). - Keep the explanatory and response variables in their stated roles. - Use the sign to state the direction of change.

1. Regression slope describes predicted change in \(y\) for a one-unit increase in \(x\). 2. The slope is \(-4\), so \(y\) is predicted to decrease by \(4\) units when \(x\) increases by \(1\). 3. Reversing the variables creates a different rate and is not the interpretation of this equation.

For each one-unit increase in \(x\), the predicted \(y\)-value decreases by \(4\) units.

- Read the two coefficients directly from the equation. - Substitute the observed input to find the prediction. - Keep the residual separate from both coefficients.

1. The slope is \(4\), and the y-intercept is \(9\). 2. At \(x=5\), the predicted response is \(\hat y=4\cdot 5+9=29\). 3. The residual is \(32-29=3\).

Slope: \(4\) y-intercept: \(9\) Predicted response: \(29\) Residual: \(3\)

- Use change in predicted cost divided by change in distance. - Substitute one table row into slope-intercept form. - Interpret the slope as a per-mile rate and the intercept as the zero-mile prediction without assigning the intercept a billing mechanism that is not stated.

1. From \(x=2\) to \(x=4\), the predicted cost increases by \(13-8=5\) dollars while distance increases by \(2\) miles. 2. The slope is \(\frac{5}{2}=2.5\) dollars per mile. 3. Substitute \((2, 8)\) into \(\hat y=2.5x+b\): \(8=2.5\cdot2+b\), so \(b=3\). 4. The model is \(\hat y=2.5x+3\). 5. The slope means the predicted delivery cost increases by \(\$2.50\) per additional mile. The intercept means the model predicts a cost of \(\$3\) at a distance of \(0\) miles; the table alone does not establish that this amount is an actual fixed charge.

a) \(2.5\) dollars per mile b) y-intercept: \(\$3\); model: \(\hat y=2.5x+3\) c) The predicted cost increases by \(\$2.50\) per mile. At \(0\) miles, the model predicts a cost of \(\$3\).

- Find two points on the part of the line that is drawn. - How can a point on a line help you find its value when the input is zero? - Which day is \(x=0\), and is it in the range of days used for fitting?

1. The line passes through \((2,5)\) and \((4,9)\). 2. Its slope is \(m=(9-5)\div(4-2)=2\). 3. Substitute a point: \(5=2\cdot2+b\), giving \(b=1\). 4. The fitted equation is \(\hat y=2x+1\). At \(x=0\), the day the shop opened, it predicts \(1\) bike rental. That day was not included in the data used to fit the line.

a) Slope \(m=2\), equation \(\hat y=2x+1\). b) The line predicts \(1\) bike rental on opening day; that day's count was not used to fit the line.

- Compare the two line heights at the first requested x-value. - Repeat at the second x-value rather than assuming the same order. - How do fitted predictions differ from actual results for individual students?

1. At \(x=2\), line a predicts \(6\), while line b predicts \(5\), so a is higher. 2. At \(x=5\), line a predicts \(9\), while line b predicts \(11\), so b is higher. 3. Line b starts lower but rises faster. The lines cross at \(x=3\), changing which prediction is greater. 4. A fitted line predicts a typical pattern; individuals can differ from that prediction.

a) Line a. b) Line b. c) Line b has a greater slope and overtakes line a where they cross. Neither line guarantees individual results.

- Identify which \(x\)-value makes the parenthetical expression zero. - Rewrite the equation in slope-intercept form. - Evaluate the model at \(x=0\) to confirm the intercept.

1. The value \(40\) is the predicted response when \(x=12\), because the centered term is zero there. 2. Expanding gives \(\hat y=40+2.5x-30=2.5x+10\). 3. The actual y-intercept is \(10\).

The value \(40\) is the predicted response at \(x=12\). The y-intercept is \(10\).

- Translate zero in each variable into a calendar year. - Use the common slope to connect the two baseline predictions. - Distinguish the underlying trend from the coordinate system used to describe it.

1. In Model A, \(x=0\) means \(2020\), so the intercept predicts \(50\) in \(2020\). 2. In Model B, \(z=0\) means \(2022\), so the intercept predicts \(58\) in \(2022\). 3. Two years of growth at \(4\) per year changes \(50\) to \(58\), so the models are consistent. 4. Intercepts depend on the chosen zero point for the explanatory variable.

The intercepts refer to different calendar years. Model A’s \(50\) is the prediction for \(2020\), while Model B’s \(58\) is the prediction for \(2022\).

- Solve the new-variable definition for the old input. - Substitute before simplifying. - Translate zero in the new scale back to the original scale.

1. From \(z=x-5\), write \(x=z+5\). 2. Substitute: \(\hat y=3(z+5)+11=3z+26\). 3. The new intercept \(26\) is the predicted response at \(z=0\), which corresponds to \(x=5\).

\(\hat y=3z+26\) The new intercept \(26\) is the predicted response at the original input \(x=5\).

- Evaluate the model at zero customers. - Compare zero with the observed \(x\)-range. - Judge the model where data exist rather than only at the intercept.

1. The y-intercept predicts \(-4\) minutes when \(x=0\). 2. A negative wait time is impossible, and zero customers lies outside the observed range. 3. The intercept therefore has no practical meaning, but the line may still describe the relationship reasonably within the observed range of \(5\) to \(20\) customers.

The intercept is the model’s unsupported prediction of \(-4\) minutes at zero customers. It is not meaningful in context, but the model can still be useful over the data range.

- Substitute the given input for the forward prediction. - For the reverse question, set the model equal to the target time and solve. - Attach “per dozen” to the slope, and interpret the intercept only as the model’s prediction at zero dozens unless the context gives a more specific meaning.

1. For \(x=7\), \(\hat y=18+2.4\cdot7=34.8\) minutes. 2. Set the prediction equal to \(42\): \(18+2.4x=42\). 3. Solving gives \(2.4x=24\), so \(x=10\) dozens. 4. The slope \(2.4\) means each additional dozen cookies is associated with an increase of \(2.4\) minutes in predicted packaging time. 5. The y-intercept \(18\) means the model predicts \(18\) minutes of packaging time when \(0\) dozen cookies are packaged. The model alone does not identify those \(18\) minutes as a particular setup process.

a) \(34.8\) minutes b) \(10\) dozen cookies c) Slope: \(2.4\) additional predicted minutes per dozen cookies. y-intercept: \(18\) predicted minutes when \(0\) dozen cookies are packaged.