Question

5-2 at this rate...
draw a line to connect each rate to the equivalent unit rate.

1. $\frac{$22.80}{8 \text{lb}}$
2. $\frac{288 \text{ft}}{16 \text{s}}$
3. $\frac{$16.50}{6 \text{lb}}$
4. $\frac{792 \text{ft}}{45 \text{s}}$
5. $\frac{$3.74}{68 \text{min}}$
6. $\frac{$4.05}{75 \text{min}}$

$\frac{18 \text{ft}}{1 \text{s}}$, $\frac{$2.75}{1 \text{lb}}$, $\frac{17.6 \text{ft}}{1 \text{s}}$, $\frac{$0.05}{1 \text{min}}$, $\frac{$2.85}{1 \text{lb}}$, $\frac{$0.06}{1 \text{min}}$

Explanation:

Step1: Calculate unit rate for 1

$\frac{\$22.80}{8\ \text{lb}} = \$2.85\ \text{per lb}$

Step2: Calculate unit rate for 2

$\frac{288\ \text{ft}}{16\ \text{s}} = 18\ \text{ft per s}$

Step3: Calculate unit rate for 3

$\frac{\$16.50}{6\ \text{lb}} = \$2.75\ \text{per lb}$

Step4: Calculate unit rate for 4

$\frac{792\ \text{ft}}{45\ \text{s}} = 17.6\ \text{ft per s}$

Step5: Calculate unit rate for 5

$\frac{\$3.74}{68\ \text{min}} = \$0.055 \approx \$0.06\ \text{per min}$

Step6: Calculate unit rate for 6

$\frac{\$4.05}{75\ \text{min}} = \$0.054 \approx \$0.05\ \text{per min}$

Answer:

1. $\frac{\$22.80}{8\ \text{lb}}$ connects to $\frac{\$2.85}{1\ \text{lb}}$ (K)
2. $\frac{288\ \text{ft}}{16\ \text{s}}$ connects to $\frac{18\ \text{ft}}{1\ \text{s}}$ (O)
3. $\frac{\$16.50}{6\ \text{lb}}$ connects to $\frac{\$2.75}{1\ \text{lb}}$ (I)
4. $\frac{792\ \text{ft}}{45\ \text{s}}$ connects to $\frac{17.6\ \text{ft}}{1\ \text{s}}$ (M)
5. $\frac{\$3.74}{68\ \text{min}}$ connects to $\frac{\$0.06}{1\ \text{min}}$ (A)
6. $\frac{\$4.05}{75\ \text{min}}$ connects to $\frac{\$0.05}{1\ \text{min}}$ (N)