Question

(a)
hours: 1, 1.5, 2, 2.5, 3
pay: 18, 23, 28, 33, 38
part a
write a linear function f jack can use to determine his pay.
a) f(x) = 10x + 8
b) f(x) = 15x + 3
c) f(x) = 0.5x + 1
d) f(x) = x + 18
(b) part b
which is a reasonable domain?
a) 0 ≤ x ≤ 6
b) 0 ≤ x ≤ 24
c) 0 ≤ x ≤ 68
d) 0 ≤ x ≤ 248
(c) part c
assume jack works from 1:30 p.m. to 7:30 p.m. how much would he earn?
(a)(b)(c)

Explanation:

Part A

Step1: Recall linear function form

A linear function is in the form \( f(x)=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept.

Step2: Calculate the slope \( m \)

We can use two points from the table. Let's take \( (x_1,y_1)=(1,18) \) and \( (x_2,y_2)=(2,28) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
Substituting the values: \( m=\frac{28 - 18}{2 - 1}=\frac{10}{1}=10 \).

Step3: Find the y - intercept \( b \)

Use the point \( (1,18) \) and \( m = 10 \) in \( f(x)=mx + b \).
\( 18=10\times1 + b \)
\( 18=10 + b \)
Subtract 10 from both sides: \( b=18 - 10 = 8 \).
So the linear function is \( f(x)=10x + 8 \).

The domain represents the number of hours Jack can work. From the linear function \( f(x)=10x + 8 \), we can think about a reasonable number of working hours. If we consider a typical work - day or a reasonable work period, 0 to 6 hours is a reasonable domain as compared to 0 - 24 (which is a full day, but the pay function and the given data suggest a more limited work - time) or 0 - 68/248 (which are too large for a reasonable work - hour context for this pay - per - hour situation).

Step1: Calculate the number of hours worked

From 7:30 a.m. to 2:30 p.m., the number of hours \( x = 7 \) (because 2:30 p.m. - 7:30 a.m. = 7 hours).

Step2: Use the linear function to find pay

We know the linear function is \( f(x)=10x + 8 \). Substitute \( x = 7 \) into the function.
\( f(7)=10\times7+8 \)
\( f(7)=70 + 8=78 \)

Answer:

A. \( f(x)=10x + 8 \)

Part B