Question

constructing a two-way table using driving data
distribution of mileage efficiencies across different weight classes
what is the missing value for heavyweight cars in the medium-mileage category?
(table with weight, high mileage, medium mileage, low mileage, total columns; light: 2, 1, 0, 3; medium: ?, 2, 1, 3; heavy: 0, ?, 2, 2; total: 2, 3, 3, 8; answer options: 0, 2, 3, 1)

Explanation:

Step1: Recall total for Heavy

The total number of heavy - weight cars is 2. We know the number of heavy - weight cars with high mileage is 0 and with low mileage is 2.

Step2: Use total formula

Let the number of heavy - weight cars with medium mileage be \(x\). The formula for the total of a row is \( \text{High Mileage}+\text{Medium Mileage}+\text{Low Mileage}=\text{Total}\). So for the "Heavy" row, we have \(0 + x+2 = 2\).

Step3: Solve for \(x\)

Simplify the equation \(x + 2=2\). Subtract 2 from both sides: \(x=2 - 2=0\)? Wait, no, wait. Wait, the total for the "Medium Mileage" column: the total of medium mileage is 3. The number of light - weight cars with medium mileage is 1, medium - weight cars with medium mileage is 2. Let the heavy - weight cars with medium mileage be \(y\). Then \(1 + 2+y=3\). So \(y = 3-(1 + 2)=0\)? Wait, no, the row total for heavy is 2. High (0)+Medium (\(y\))+Low (2) = 2. So \(y=2-(0 + 2)=0\)? But also, column total for medium mileage: light (1)+medium (2)+heavy (\(y\)) = 3. So \(1+2 + y=3\), so \(y = 0\). Wait, but let's check again. Wait the row for heavy: total is 2. High: 0, Low: 2. So Medium must be \(2-(0 + 2)=0\). But also, column medium mileage: 1 (light)+2 (medium)+y (heavy)=3. So \(y = 0\). So the missing value is 0.

Wait, maybe I made a mistake. Wait the total for the "Medium Mileage" column: light has 1, medium has 2, so heavy should have \(3-(1 + 2)=0\). And the row for heavy: total is 2, high is 0, low is 2, so medium is \(2-(0 + 2)=0\). So the missing value is 0.

Answer:

0