Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To solve this problem, we need to determine the set fee and hourly rate for each shop, and then determine the number of hours at which the costs from both shops are equal.
First, let's define the cost function for each shop:
- For Shop A: [tex]\( \text{Cost}_A = \text{set\_fee}_A + \text{hourly\_rate}_A \times \text{hours} \)[/tex]
- For Shop B: [tex]\( \text{Cost}_B = \text{set\_fee}_B + \text{hourly\_rate}_B \times \text{hours} \)[/tex]
We know from the table:
- For Shop A, 1 hour costs [tex]$9 and 5 hours costs $[/tex]25.
- For Shop B, 1 hour costs [tex]$7 and 5 hours costs $[/tex]27.
From these points, we can set up two equations for each shop:
Shop A:
1. [tex]\( 9 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 1 \)[/tex]
2. [tex]\( 25 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 5 \)[/tex]
Subtract the first equation from the second:
[tex]\[ 25 - 9 = (\text{set\_fee}_A + 5 \times \text{hourly\_rate}_A) - (\text{set\_fee}_A + 1 \times \text{hourly\_rate}_A) \][/tex]
[tex]\[ 16 = 4 \times \text{hourly\_rate}_A \][/tex]
[tex]\[ \text{hourly\_rate}_A = 4 \][/tex]
Substitute back into the first equation:
[tex]\[ 9 = \text{set\_fee}_A + 4 \][/tex]
[tex]\[ \text{set\_fee}_A = 5 \][/tex]
Shop B:
1. [tex]\( 7 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 1 \)[/tex]
2. [tex]\( 27 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 5 \)[/tex]
Subtract the first equation from the second:
[tex]\[ 27 - 7 = (\text{set\_fee}_B + 5 \times \text{hourly\_rate}_B) - (\text{set\_fee}_B + 1 \times \text{hourly\_rate}_B) \][/tex]
[tex]\[ 20 = 4 \times \text{hourly\_rate}_B \][/tex]
[tex]\[ \text{hourly\_rate}_B = 5 \][/tex]
Substitute back into the first equation:
[tex]\[ 7 = \text{set\_fee}_B + 5 \][/tex]
[tex]\[ \text{set\_fee}_B = 2 \][/tex]
Now, we have:
- [tex]\( \text{Cost}_A = 5 + 4 \times \text{hours} \)[/tex]
- [tex]\( \text{Cost}_B = 2 + 5 \times \text{hours} \)[/tex]
We need to find the number of hours where the costs are equal:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]
Solving for hours:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]
[tex]\[ 5 - 2 = 5 \times \text{hours} - 4 \times \text{hours} \][/tex]
[tex]\[ 3 = \text{hours} \][/tex]
Hence, Mike wants to rent a bike for [tex]\( \boxed{3} \)[/tex] hours.
First, let's define the cost function for each shop:
- For Shop A: [tex]\( \text{Cost}_A = \text{set\_fee}_A + \text{hourly\_rate}_A \times \text{hours} \)[/tex]
- For Shop B: [tex]\( \text{Cost}_B = \text{set\_fee}_B + \text{hourly\_rate}_B \times \text{hours} \)[/tex]
We know from the table:
- For Shop A, 1 hour costs [tex]$9 and 5 hours costs $[/tex]25.
- For Shop B, 1 hour costs [tex]$7 and 5 hours costs $[/tex]27.
From these points, we can set up two equations for each shop:
Shop A:
1. [tex]\( 9 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 1 \)[/tex]
2. [tex]\( 25 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 5 \)[/tex]
Subtract the first equation from the second:
[tex]\[ 25 - 9 = (\text{set\_fee}_A + 5 \times \text{hourly\_rate}_A) - (\text{set\_fee}_A + 1 \times \text{hourly\_rate}_A) \][/tex]
[tex]\[ 16 = 4 \times \text{hourly\_rate}_A \][/tex]
[tex]\[ \text{hourly\_rate}_A = 4 \][/tex]
Substitute back into the first equation:
[tex]\[ 9 = \text{set\_fee}_A + 4 \][/tex]
[tex]\[ \text{set\_fee}_A = 5 \][/tex]
Shop B:
1. [tex]\( 7 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 1 \)[/tex]
2. [tex]\( 27 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 5 \)[/tex]
Subtract the first equation from the second:
[tex]\[ 27 - 7 = (\text{set\_fee}_B + 5 \times \text{hourly\_rate}_B) - (\text{set\_fee}_B + 1 \times \text{hourly\_rate}_B) \][/tex]
[tex]\[ 20 = 4 \times \text{hourly\_rate}_B \][/tex]
[tex]\[ \text{hourly\_rate}_B = 5 \][/tex]
Substitute back into the first equation:
[tex]\[ 7 = \text{set\_fee}_B + 5 \][/tex]
[tex]\[ \text{set\_fee}_B = 2 \][/tex]
Now, we have:
- [tex]\( \text{Cost}_A = 5 + 4 \times \text{hours} \)[/tex]
- [tex]\( \text{Cost}_B = 2 + 5 \times \text{hours} \)[/tex]
We need to find the number of hours where the costs are equal:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]
Solving for hours:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]
[tex]\[ 5 - 2 = 5 \times \text{hours} - 4 \times \text{hours} \][/tex]
[tex]\[ 3 = \text{hours} \][/tex]
Hence, Mike wants to rent a bike for [tex]\( \boxed{3} \)[/tex] hours.
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.