Problem 90 - College Algebra with Modeling and Visualization

Question

The average hourly wage (adjusted to 1982 dollars) was \(\$ 8.46\) in 1970 and \(\$ 8.18\) in 2005 (Source: Department of Commerce.) (A) Find an equation of a line that passes through the points \((1970,8.46)\) and \((2005,8.18)\) (B) Interpret the slope. (C) Approximate the hourly wage in 2000 . Compare the estimate to the actual value of \(\$ 8.04\)

Step-by-Step Solution

Answer

The equation of the line is \( y = -0.008x + 23.206 \); the slope implies a yearly wage decrease of \$0.008. Estimated wage for 2000 is \$7.206, which underestimates the actual \$8.04 wage.

Step 1: Identify the given points

The problem provides two points: - Point 1: (1970, 8.46) - Point 2: (2005, 8.18) These points represent the year and the average hourly wage in 1982 dollars.

Step 2: Calculate the slope

The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]Substitute the given points into the formula:\[m = \frac{8.18 - 8.46}{2005 - 1970} = \frac{-0.28}{35} = -0.008\]

Step 3: Write the equation of the line

Use the slope-intercept form of a line, \(y = mx + b\). We need to find \(b\). Using the point-slope form can help:\[y - y_1 = m(x - x_1)\]Substitute \(m = -0.008\), \(x_1 = 1970\), and \(y_1 = 8.46\):\[y - 8.46 = -0.008(x - 1970)\]Simplify to find the equation of the line:\[y = -0.008x + 23.206\]

Step 4: Interpret the slope

The slope of the line is \(-0.008\). This means that the average hourly wage decreases by approximately \$0.008\ every year, adjusted to 1982 dollars.

Step 5: Approximate the hourly wage in 2000

Substitute \(x = 2000\) into the equation found in Step 3, \(y = -0.008x + 23.206\):\[y = -0.008(2000) + 23.206 = 7.206\]The estimated wage in 2000 is \$7.206\.

Step 6: Comparison with actual value

The estimated wage for 2000 is \\(7.206\, while the actual value is \\)8.04\. This suggests that the model underestimates the actual wage during this year.

Key Concepts

Slope Calculation

Before diving into equations, it's essential to grasp how to compute a slope, as it describes how steep a line is. The slope is represented by "\(m\)" in equations. Slope calculation uses two points from a line, say

• Point 1: \((x_1, y_1)\)
• Point 2: \((x_2, y_2)\)

The formula for the slope is:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]This formula essentially measures the vertical change (difference in \(y\)) versus the horizontal change (difference in \(x\)) between the two points.
These changes tell us how quickly one variable changes relative to another. In the original exercise, plugging points \((1970, 8.46)\) and \((2005, 8.18)\) into the formula gives:\[ m = \frac{8.18 - 8.46}{2005 - 1970} = -0.008 \]The negative slope of \(-0.008\) suggests a slight decline, hinting at a gradual decrease in wages over the years in the context provided.

Point-Slope Form

Having a slope is one piece of the puzzle. Next, we use it to find the equation of the line, starting with the point-slope form. This form is great when you have the slope and just one point. The point-slope form can be written as:\[ y - y_1 = m(x - x_1) \]Here, \(m\) is the slope found already, and \((x_1, y_1)\) is any of the two given points you have. Applying the point-slope formula in the problem with point \((1970, 8.46)\) and slope \(-0.008\), we get:\[ y - 8.46 = -0.008(x - 1970) \]This is an intermediate step that shows the relation between \(x\) and \(y\) with respect to the given point and slope.
Expanding or simplifying this equation can transition into other forms, like the slope-intercept form, which students often deal with in further problems.

Equation of a Line

Once the point-slope form is set up, transforming it into the equation of the line can provide a clearer insight, especially when predicting other values.To convert from the point-slope form to the slope-intercept form \(y = mx + b\), you simply need to solve for \(y\). This represents the line where \(m\) is the slope and \(b\) the y-intercept. Taking the point-slope equation:\[ y - 8.46 = -0.008(x - 1970) \]You simplify it to:\[ y = -0.008x + 23.206 \]This is now in the slope-intercept form, allowing us to easily substitute a year (as \(x\)) to predict a wage (as \(y\)).
In the exercise, inserting the year 2000 gives us a predicted wage of \$7.206, showing practical use of linear equations in estimating and comparing variations over time.
This approach is not only vital in mathematics but also widely applicable in economics, sciences, and everyday decision-making.