Question

which situation represents a proportional relationship?
a. a gym charges a $20 membership fee plus $5 for each class.
b. a recipe uses 3 cups of flour for every 2 cups of sugar.
c.

x	y
1	5.5
2	11
7	23.5

d.

x	y
6	-1
12	-5
18	-9

Explanation:

Step1: Recall Proportional Relationship

A proportional relationship has a constant ratio (or unit rate) and passes through the origin (for linear cases). For two quantities \( x \) and \( y \), it can be written as \( y = kx \) where \( k \) is the constant of proportionality.

Step2: Analyze Option A

The gym's cost: Let \( x \) be the number of classes, \( y \) be total cost. Then \( y = 20 + 5x \). This is a linear equation but has a y - intercept (\( 20 \)), so it's not proportional (since proportional relationships have \( y - \)intercept \( 0 \)).

Step3: Analyze Option B

Let \( f \) be flour and \( s \) be sugar. The ratio \( \frac{f}{s}=\frac{3}{2}= 1.5 \) (constant). So \( f=\frac{3}{2}s \), which is a proportional relationship (passes through origin when \( s = 0,f = 0 \)).

Step4: Analyze Option C

Check the ratio \( \frac{y}{x} \): For \( x = 1,y = 5.5 \), ratio is \( 5.5 \); for \( x = 2,y = 11 \), ratio is \( \frac{11}{2}=5.5 \); for \( x = 7,y = 23.5 \), ratio is \( \frac{23.5}{7}\approx3.36
eq5.5 \). So the ratio is not constant, not proportional.

Step5: Analyze Option D (the last table)

Check the ratio \( \frac{y}{x} \): For \( x = 6,y=- 1 \), ratio is \( \frac{-1}{6}\approx - 0.167 \); for \( x = 12,y=-5 \), ratio is \( \frac{-5}{12}\approx - 0.417
eq - 0.167 \). Ratio not constant, not proportional.

Answer:

B. A recipe uses 3 cups of flour for every 2 cups of sugar.