Question

1. the table shows lucias salary for different years.

which linear function best models the data if x represents the num
a. ( y = 39,625x + 4375 )
b. ( y = 4375x + 39,625 )
c. ( y = 1750x + 40,500 )
d. ( y = 40,500x + 1750 )

1. which situation best represents causation?

a. when you are at the beach, you get wet.
b. when you study for a test, your classmate studies, too
c. when you carry an umbrella to school, it rains.
d. when you dont brush your teeth, you get cavities.

Explanation:

Step1: Analyze the linear function form

The general form of a linear function is \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y -\)intercept.

Step2: Calculate the slope \(m\)

Let \(x_1 = 0\) (corresponding to year 2000), \(y_1=40000\); \(x_2 = 1\) (corresponding to year 2001), \(y_2 = 42000\).
The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{42000 - 40000}{1-0}=2000\).
Let \(x_1 = 0\) (year 2000), \(y_1 = 40000\); \(x_3=3\) (year 2003), \(y_3 = 44500\).
The slope \(m=\frac{y_3 - y_1}{x_3 - x_1}=\frac{44500 - 40000}{3-0}=1500\).
We can also use the formula for the average rate of change.
\(m=\frac{44500 - 40000}{3-0}=1500\) (approximate value).
If we assume \(x\) represents the number of years since 2000.
When \(x = 0\) (year 2000), \(y=b\).
We can check each option:

• For option A: When \(x = 0\), \(y=4375

eq40000\)

• For option B: When \(x = 0\), \(y = 39625

eq40000\)

• For option C: When \(x = 0\), \(y=40500\) (close to the initial - value of \(40000\) considering approximation). The slope \(m = 1750\).

If \(x = 1\), \(y=1750\times1+40500=42250\) (close to \(42000\)); if \(x = 3\), \(y=1750\times3 + 40500=40500+5250=45750\) (relatively close considering the nature of linear approximation for non - linear data)

• For option D: When \(x = 0\), \(y = 1750

eq40000\)

Answer:

C. \(y = 1750x+40500\)