sertain map shows two roads. Road \( A \) is \( 1 \frac{1}{4} \) miles long but is \( 1 \frac{1}{2} \) inches long on the map. What is the unit r mile on this map? If road B is 15 miles long, how long is road \( B \) on the map?

To find the scale of the map, we can create a ratio of the actual distance to the distance on the map for Road \( A \). Road \( A \) is \( 1 \frac{1}{4} \) miles, which is equal to \( 1.25 \) miles, and it is represented by \( 1 \frac{1}{2} \) inches on the map, or \( 1.5 \) inches. The ratio of actual miles to inches on the map is: \[ \frac{1.25 \text{ miles}}{1.5 \text{ inches}} \] To simplify this, we find the number of miles per inch: \[ \frac{1.25}{1.5} = \frac{25}{30} = \frac{5}{6} \] Thus, the scale of the map is \( \frac{5}{6} \) miles per inch. Now, to find out how long Road \( B \) (which is 15 miles long) is on the map, we can set up a proportion using the scale we just found. If \( x \) is the length of Road \( B \) on the map in inches: \[ \frac{5}{6} \text{ miles} \equiv 1 \text{ inch} \] This gives us: \[ \frac{15 \text{ miles}}{x \text{ inches}} = \frac{5}{6} \] Cross-multiplying gives: \[ 5x = 15 \times 6 \] \[ 5x = 90 \] Now, divide both sides by 5 to solve for \( x \): \[ x = \frac{90}{5} = 18 \] So, Road \( B \) is \( 18 \) inches long on the map.