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An engineering technician makes \[tex]$25 an hour for the first 40 hours she works during a week and \$[/tex]32 an hour for each hour over 40 hours. Which piecewise equation models her total weekly pay [tex]\(y\)[/tex] in dollars as it relates to the number of hours [tex]\(x\)[/tex] that she has worked during the week?

A. [tex]\(y=\left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32x + 1000 & x \ \textgreater \ 40\end{array}\right.\)[/tex]

B. [tex]\(y=\left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32(x-40) + 1000 & x \ \textgreater \ 40\end{array}\right.\)[/tex]

C. [tex]\(y=\left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32x & x \ \textgreater \ 40\end{array}\right.\)[/tex]

D. [tex]\(y=\left\{\begin{array}{l}25x \\ 32(x-48)\end{array}\right.\)[/tex]
[tex]\[
\begin{array}{l}
0 \leq x \leq 40 \\
x \ \textgreater \ 40
\end{array}
\][/tex]


Sagot :

To determine the correct piecewise equation that models the total weekly pay [tex]\( y \)[/tex] for an engineering technician, let's break the problem into the two distinct scenarios provided:

1. For [tex]\( 0 \leq x \leq 40 \)[/tex]:

- The technician is paid [tex]$25 per hour for the first 40 hours. - Therefore, the pay for \( x \) hours worked is simply \( y = 25x \). 2. For \( x > 40 \): - For the first 40 hours, the technician earns \( 25 \times 40 = 1000 \) dollars. - For any hours beyond 40, the technician is paid $[/tex]32 per hour.
- If the total hours [tex]\( x \)[/tex] exceeds 40, then the pay for the hours beyond 40 hours (i.e., for [tex]\( x - 40 \)[/tex] hours) is [tex]\( 32 \times (x - 40) \)[/tex].
- Therefore, for [tex]\( x > 40 \)[/tex], the total pay [tex]\( y \)[/tex] is the sum of these two components:
- [tex]\( 1000 \)[/tex] dollars for the first 40 hours,
- Plus [tex]\( 32 \times (x - 40) \)[/tex] dollars for the hours exceeding 40.
- This results in the equation [tex]\( y = 1000 + 32(x - 40) \)[/tex].

Given these analyses, we can now match the piecewise functions with the choices provided:

A. [tex]\( y = \left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32x + 1000 & x > 40 \end{array}\right. \)[/tex] – Incorrect, as the component for [tex]\( x > 40 \)[/tex] does not consider the reduction of 40 hours from [tex]\( x \)[/tex] before multiplying by 32.

B. [tex]\( y = \left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32(x - 40) + 1000 & x > 40 \end{array}\right. \)[/tex] – Correct, as it correctly calculates the pay for [tex]\( x > 40 \)[/tex] hours using [tex]\( y = 1000 + 32(x - 40) \)[/tex].

C. [tex]\( y = \left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32x & x > 40 \end{array}\right. \)[/tex] – Incorrect, as it wrongly applies the rate of [tex]$32 to all hours worked beyond 40 without considering the first 40 hours at $[/tex]25 per hour.

D. [tex]\( y = \left\{\begin{array}{l}25x \\ 32(x - 48) \end{array}\right. \)[/tex] for [tex]\( 0 \leq x \leq 40 \)[/tex] and [tex]\( x > 40 \)[/tex] – Incorrect, as the formulation for [tex]\( x > 40 \)[/tex] does not align with the given pay rates and hours.

Therefore, the correct piecewise equation that models her total weekly pay [tex]\( y \)[/tex] in dollars as it relates to the number of hours [tex]\( x \)[/tex] worked during the week is:

B. [tex]\( y = \left\{\begin{array}{ll}25x & 0 \leq x \leq 40 \\ 32(x - 40) + 1000 & x > 40 \end{array}\right. \)[/tex]