To determine which option is cheaper per day, we can evaluate the cost of each option for different numbers of rides (x) using the equations provided.
Option A: \( y = 4.25x \)
Now, let's calculate the cost for Option A for several values of x (number of rides):
- For \( x = 0 \): \( y = 4.25(0) = 0 \)
- For \( x = 5 \): \( y = 4.25(5) = 21.25 \)
- For \( x = 10 \): \( y = 4.25(10) = 42.50 \)
- For \( x = 15 \): \( y = 4.25(15) = 63.75 \)
Now, let's summarize the costs for Option B from the graph:
- For \( x = 0 \): Cost = 0
- For \( x = 5 \): Cost = 20
- For \( x = 10 \): Cost = 40
- For \( x = 15 \): Cost = 60
Now, we can compare the costs for Option A and Option B:
| Rides (x) | Option A Cost | Option B Cost | |-----------|----------------|----------------| | 0 | $0 | $0 | | 5 | $21.25 | $20 | | 10 | $42.50 | $40 | | 15 | $63.75 | $60 |
From the table, we can see:
- For 5 rides, Option B ($20) is cheaper than Option A ($21.25).
- For 10 rides, Option B ($40) is cheaper than Option A ($42.50).
- For 15 rides, Option B ($60) is cheaper than Option A ($63.75).
Therefore, Option B is cheaper per day for all ride counts (except for 0 rides, where both options cost nothing).
So, the answer is:
2 is cheaper per day.