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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.
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