Answer :
To solve the problem and determine the appropriate piecewise equation for the engineering technician's weekly pay, let's break down the situation step by step:
1. Understanding the Pay Structure:
- For the first 40 hours worked in a week, the technician earns [tex]$\$[/tex]25[tex]$ per hour. - For any hours worked over 40 in a week, the technician earns $[/tex]\[tex]$32$[/tex] per hour for those extra hours.
2. Determining the Equations for Different Scenarios:
- Scenario 1: If the technician works between 0 and 40 hours (inclusive), the weekly pay [tex]\( y \)[/tex] is simply the number of hours [tex]\( x \)[/tex] multiplied by the standard hourly rate of [tex]$\$[/tex]25[tex]$. - Mathematically, this can be expressed as: \[ y = 25x \quad \text{for} \quad 0 \leq x \leq 40 \] - Scenario 2: If the technician works more than 40 hours, the weekly pay \( y \) includes two components: - Payment for the first 40 hours at the standard rate of $[/tex]\[tex]$25$[/tex] per hour.
- Payment for any additional hours (over 40) at the overtime rate of [tex]$\$[/tex]32$ per hour.
Mathematically, this can be expressed as:
- Payment for the first 40 hours: [tex]\( 25 \times 40 \)[/tex]
- Payment for additional hours (overtime [tex]\( x - 40 \)[/tex]): [tex]\( 32 \times (x - 40) \)[/tex]
Combining these components, the equation for the pay when [tex]\( x > 40 \)[/tex] is:
[tex]\[ y = 32(x - 40) + 25 \times 40 \][/tex]
Simplifying [tex]\( 25 \times 40 \)[/tex] gives us 1000, so:
[tex]\[ y = 32(x - 40) + 1000 \quad \text{for} \quad x > 40 \][/tex]
3. Matching the Equations with the Given Options:
The piecewise function needs to cover both scenarios accurately.
Comparing our derived equations with the given options:
- Option A: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32(x - 40) \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because it does not include the base pay of 1000 for hours over 40.
- Option B: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32x \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because the second part doesn't correctly account for the first 40 hours of standard pay.
- Option C: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32x + 1000 \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because the term [tex]\(32x + 1000\)[/tex] overestimates the pay for hours beyond 40.
- Option D: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32(x - 40) + 1000 \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Correct as it matches exactly with our derived equations. It accurately represents the weekly pay calculation for both scenarios.
Hence, the correct piecewise equation is:
[tex]\[ \boxed{D} \][/tex]
1. Understanding the Pay Structure:
- For the first 40 hours worked in a week, the technician earns [tex]$\$[/tex]25[tex]$ per hour. - For any hours worked over 40 in a week, the technician earns $[/tex]\[tex]$32$[/tex] per hour for those extra hours.
2. Determining the Equations for Different Scenarios:
- Scenario 1: If the technician works between 0 and 40 hours (inclusive), the weekly pay [tex]\( y \)[/tex] is simply the number of hours [tex]\( x \)[/tex] multiplied by the standard hourly rate of [tex]$\$[/tex]25[tex]$. - Mathematically, this can be expressed as: \[ y = 25x \quad \text{for} \quad 0 \leq x \leq 40 \] - Scenario 2: If the technician works more than 40 hours, the weekly pay \( y \) includes two components: - Payment for the first 40 hours at the standard rate of $[/tex]\[tex]$25$[/tex] per hour.
- Payment for any additional hours (over 40) at the overtime rate of [tex]$\$[/tex]32$ per hour.
Mathematically, this can be expressed as:
- Payment for the first 40 hours: [tex]\( 25 \times 40 \)[/tex]
- Payment for additional hours (overtime [tex]\( x - 40 \)[/tex]): [tex]\( 32 \times (x - 40) \)[/tex]
Combining these components, the equation for the pay when [tex]\( x > 40 \)[/tex] is:
[tex]\[ y = 32(x - 40) + 25 \times 40 \][/tex]
Simplifying [tex]\( 25 \times 40 \)[/tex] gives us 1000, so:
[tex]\[ y = 32(x - 40) + 1000 \quad \text{for} \quad x > 40 \][/tex]
3. Matching the Equations with the Given Options:
The piecewise function needs to cover both scenarios accurately.
Comparing our derived equations with the given options:
- Option A: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32(x - 40) \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because it does not include the base pay of 1000 for hours over 40.
- Option B: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32x \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because the second part doesn't correctly account for the first 40 hours of standard pay.
- Option C: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32x + 1000 \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Incorrect because the term [tex]\(32x + 1000\)[/tex] overestimates the pay for hours beyond 40.
- Option D: [tex]\( y = \left\{ \begin{array}{l} 25x \\ 32(x - 40) + 1000 \end{array} \right. \quad 0 \leq x \leq 40 \\ x > 40 \)[/tex]
- Correct as it matches exactly with our derived equations. It accurately represents the weekly pay calculation for both scenarios.
Hence, the correct piecewise equation is:
[tex]\[ \boxed{D} \][/tex]
