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Chapter 2: Problem 40

Enrollments in public colleges for recent years are shown in the table. Assuming a linear relationship, estimate the enrollments for (a) 2002 and (b) \(2006 .\) Give answers to the nearest tenth of thousands if applicable. $$\begin{array}{|c|c|}\hline \text { Year } & \begin{array}{c}\text { Enrollment } \\\\\text { (in thousands) }\end{array} \\\\\hline 2000 & 11,753 \\\\\hline 2004 & 12,980 \\\\\hline 2008 & 13,972 \\\\\hline\end{array}$$

Short Answer

Expert verified
2002: 33789.5 thousand, 2006: 36017.5 thousand

Step by step solution

01

Identify Known Points

The enrollments provided are for the years 2000, 2004, and 2008. These correspond to the points (2000, 11753), (2004, 12980), and (2008, 13972).
02

Calculate the Slope

To find the linear relationship, calculate the slope (m) using two points. The formula for the slope is \[ m = \frac{ y_2 - y_1 }{ x_2 - x_1 } \]Using the points (2000, 11753) and (2004, 12980), \[ m = \frac{ 12980 - 11753 }{ 2004 - 2000 } = \frac{ 1227 }{ 4 } = 306.75 \]
03

Determine the Linear Equation

Using the point-slope form of a linear equation \[ y - y_1 = m(x - x_1) \]and the point (2000, 11753), the equation becomes \[ y - 11753 = 306.75(x - 2000) \]Simplify this to get the linear equation: \[ y = 306.75x - 591500 + 11753 \]\[ y = 306.75x - 579747 \]
04

Estimate Enrollment for 2002

Substitute x = 2002 into the linear equation to estimate enrollment for 2002: \[ y = 306.75(2002) - 579747 \]\[ y = 613536.5 - 579747 \]\[ y = 33789.5 \ (thousand) \]
05

Estimate Enrollment for 2006

Substitute x = 2006 into the linear equation to estimate enrollment for 2006: \[ y = 306.75(2006) - 579747 \]\[ y = 615764.5 - 579747 \]\[ y = 36017.5 \ (thousand) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear equation
A linear equation is a fundamental concept in algebra representing a straight line on a graph. It is commonly written in the form: \[ y = mx + b \]Here, \(y\) is the dependent variable, \(x\) is the independent variable, \(m\) is the slope of the line, and \(b\) is the y-intercept. For enrollment data, we use this equation to predict future enrollments based on past trends. By determining the slope \(m\) and the y-intercept \(b\), we can form the equation that fits our data points, allowing us to make accurate predictions.
slope calculation
The slope \(m\) of a line represents its steepness. It shows how much \(y\) changes for a unit change in \(x\). To calculate the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\), we use this formula:
\[ m = \frac{ y_2 - y_1 }{ x_2 - x_1 } \]In the context of our enrollment data example, we use the points (2000, 11753) and (2004, 12980). Plugging these into the formula, we get:
\[ m = \frac{ 12980 - 11753 }{ 2004 - 2000 } = \frac{ 1227 }{ 4 } = 306.75 \]This means that for every additional year, the enrollment increases by 306.75 thousands.
enrollment estimation
Enrollment estimation involves predicting future values based on a given trend. Once we have the linear equation, we can substitute the desired years into the equation to estimate enrollments. For example, our linear equation derived from past data is:
\[ y = 306.75x - 579747 \]To estimate the enrollment for 2002, substitute \(x = 2002\):
\[ y = 306.75 \times 2002 - 579747 \ = 33789.5 \ (thousand) \]Similarly, for the year 2006, substitute \(x = 2006\):
\[ y = 306.75 \times 2006 - 579747 = 36017.5 \ (thousand) \]These steps provide a way to predict enrollment numbers for other years using the derived equation.
point-slope form
The point-slope form of a linear equation is useful for writing an equation when you know the slope and one point on the line. It is given by:
\[ y - y_1 = m(x - x_1) \]Here, \((x_1, y_1)\) is a specific point on the line, and \(m\) is the slope. For our enrollment data, we used the point (2000, 11753) and the previously calculated slope (306.75):
\[ y - 11753 = 306.75(x - 2000) \]Solve for \(y\) to get the equation in slope-intercept form:
\[ y = 306.75x - 579747 \]The point-slope form is handy for quickly finding the linear equation when given partial data.
data interpolation
Data interpolation involves estimating values within the range of given data points. Linear interpolation assumes a straight line between points to predict intermediate values. For our example, we used enrollment data from 2000, 2004, and 2008. By fitting a linear equation to these points, we could interpolate the enrollment data for 2002 and 2006.
The steps include:
  • Identifying known data points (years and enrollments).
  • Calculating the slope (change in enrollment over change in years).
  • Using the point-slope form to find the linear equation.
  • Substituting the desired year into the equation to estimate enrollment.
Interpolation is crucial for making predictions and understanding trends in data when exact values are not available for every year.

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