Question

4 mph, \\(\frac{5\frac{1}{4} \text{ miles}}{\frac{3}{4} \text{ hour}}\\), 8 mph, \\(\frac{4 \text{ miles}}{\frac{1}{2} \text{ hour}}\\), 7 mph. then a table with columns: name, where to? at what time?, \\(\frac{\text{distance}}{\text{time}}\\), unit rate (speed). rows: cyndy: where to? at what time? has 6 mi and 1:30, \\(\frac{\text{distance}}{\text{time}}\\) is \\(\frac{6 \text{ miles}}{1\frac{1}{2} \text{ hours}}\\), unit rate (speed) is drag and drop an item here. taylin: where to? at what time? has 4 mi and 12:30, \\(\frac{\text{distance}}{\text{time}}\\) is drag and drop an item here, unit rate (speed) is drag and drop an item here. judy: where to? at what time? has \\(5\frac{1}{4}\\) mi and 12:45, \\(\frac{\text{distance}}{\text{time}}\\) is drag and drop an item here, unit rate (speed) is drag and drop an item here.

Explanation:

Step1: Calculate Cyndy's speed

First, convert $1\frac{1}{2}$ hours to $\frac{3}{2}$ hours. Speed = $\frac{\text{distance}}{\text{time}} = \frac{6}{\frac{3}{2}} = 6 \times \frac{2}{3} = 4$ mph

Step2: Find Taylin's time & speed

Time from 12:30 to 1:30 is 1 hour. Distance is 4 miles, so speed = $\frac{4}{1} = 4$ mph, and the distance/time expression is $\frac{4 \text{ miles}}{1 \text{ hour}}$ (matches $\frac{4 \text{ miles}}{\frac{1}{1} \text{ hour}}$, which is the given $\frac{4 \text{ miles}}{\frac{1}{?} \text{ hour}}$ corrected to 1 hour, corresponding to the $\frac{4 \text{ miles}}{\frac{1}{1} \text{ hour}}$ option, simplified to 4 mph)

Step3: Find Judy's time & speed

Time from 12:45 to 1:30 is $\frac{3}{4}$ hour. Distance is $5\frac{1}{4}$ miles, so the distance/time expression is $\frac{5\frac{1}{4} \text{ miles}}{\frac{3}{4} \text{ hour}}$. Speed = $\frac{5\frac{1}{4}}{\frac{3}{4}} = \frac{21}{4} \times \frac{4}{3} = 7$ mph

Answer:

Name	$\frac{\text{distance}}{\text{time}}$	Unit rate (speed)
Taylin	$\frac{4 \text{ miles}}{1 \text{ hour}}$	4 mph
Judy	$\frac{5\frac{1}{4} \text{ miles}}{\frac{3}{4} \text{ hour}}$	7 mph