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Give the equation and graph for a line with y-intercept and slope given in Exercises. $$y \text { -intercept }=2.5 ; \text { slope }=0$$

Short Answer

Expert verified
Answer: The equation of the line is $$y = 2.5$$, and it represents a horizontal line on the graph passing through the y-intercept at (0, 2.5).

Step by step solution

01

Identify the slope and y-intercept

The problem provides us with the y-intercept and the slope: - y-intercept, b: 2.5 - Slope, m: 0
02

Write the equation of the line

Using the slope-intercept form, $$y = mx + b$$ and plugging in the given values (m=0 and b=2.5), we get the equation of the line as: $$y = 0x + 2.5$$ For simplification, the equation can be written as: $$y = 2.5$$
03

Graph the line

To graph the line $$y = 2.5$$, we can plot two points on the graph that satisfy the equation and connect them to form the line. Since the slope is 0, the line will be a horizontal line that passes through the y-intercept (2.5). 1. Fix x to 0, and find the corresponding y value: $$y = 2.5$$ Point A: (0, 2.5) 2. Fix x to 1, and find the corresponding y value: $$y = 2.5$$ Point B: (1, 2.5) Draw a horizontal line that passes through points A and B. This line represents the equation $$y = 2.5$$. In conclusion, the equation of the line is $$y = 2.5$$, and its graph is a horizontal line passing through the y-intercept at (0, 2.5).

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Key Concepts

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

Y-Intercept
The y-intercept is the point where a line crosses the y-axis of a graph. When you're given an equation of a line in slope-intercept form, which is written as y = mx + b, the y-intercept is represented by b. In practical terms, it tells you the value of y when x is zero.

Let's consider our current example, where the line provided has a y-intercept of 2.5. This means that the line will cross the y-axis at the point (0, 2.5). If you were to draw a vertical line through the graph at x = 0, the spot where it intersects our line is the y-intercept.

Understanding the role of the y-intercept is crucial as it provides a starting point for graphing the line and helps us visually interpret the equation.
Slope
The slope of a line is a measure of its steepness or the degree of its incline and is often represented by the letter m. Mathematically, slope is expressed as the ratio of the change in y (the vertical change) over the change in x (the horizontal change) between two distinct points on the line.

In our example, the slope is given as 0. This indicates that for every step we take to the right or left along the x-axis, there is no change in the y-coordinate. A slope of zero corresponds to a horizontal line – it does not rise or fall but remains constant, parallel to the x-axis.

A firm grasp of the concept of slope is essential, as it can convey the direction and the rate at which a line moves across the graph. Slopes can be positive, negative, zero, or undefined, each reflecting a different direction or type of line.
Slope-Intercept Form
The slope-intercept form is y = mx + b, where m represents the slope, and b represents the y-intercept of the line. This form is one of the most straightforward ways to write the equation of a line. It makes it easy to identify the slope and y-intercept, which are critical when graphing the line or finding points on it.

In our exercise, the slope-intercept form of the line was given as y = 0x + 2.5, which simplifies to y = 2.5, since any number multiplied by zero is zero. This simplified equation still fits within the slope-intercept structure. A slope of zero means the y-value remains constant no matter what x-value we choose. For students looking for clarity in working with linear equations, mastering the slope-intercept form is a valuable skill, as it immediately provides the key components needed to visualize a line.
Graphing Linear Equations
To graph linear equations, one must understand both the slope and the y-intercept, as these two values dictate how the line is drawn on a coordinate plane. The process generally involves plotting the y-intercept on the y-axis and then using the slope to determine another point on the line.

However, in the case of our horizontal line equation y = 2.5, graphing is straightforward. Since the slope is 0, we simply draw a horizontal line through the y-axis at the y-intercept point (0, 2.5). Whether you move left or right from this point along the x-axis, the y-value of the points on the line does not change. Therefore, the line extends left and right indefinitely at the height of y-value 2.5. This graphing method shows that even without a slope, one can still graph a line using just the y-intercept, as long as the slope provided is zero. Teaching students to graph linear equations enhances their ability to visualize algebraic expressions and solve various mathematical problems that involve linear functions.

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Most popular questions from this chapter

Is there any relationship between these two variables? To find out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate. \({ }^{19}\) $$\begin{array}{lllllll}\hline \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline \text { Temperature } & 96.3 & 97.4 & 98.9 & 99.0 & 99.0 & 96.8 \\ \text { (degrees) } & & & & & & \\\\\text { Heart Rate } & 70 & 68 & 80 & 75 & 79 & 75 \\\\\text { (beats per } & & & & & & \\\\\text { minute) } & & & & & & \\\ \hline\end{array}$$ $$\begin{array}{lllllll}\hline \text { Person } & 7 & 8 & 9 & 10 & 11 & 12 \\\\\hline \text { Temperature } & 98.4 & 98.4 & 98.8 & 98.8 & 99.2 & 99.3 \\\\\multicolumn{2}{l} {\text { (degrees) }} & & & & & & \\\\\text { Heart Rate } & 74 & 84 & 73 & 84 & 66 & 68 \\\\\text { (beats per } & & & & & & \\\\\text { minute) } & & & & & & & \\\& & & & \\\\\hline\end{array}$$ a. Find the correlation coefficient \(r\), relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the 5\% level of significance.

Of two personnel evaluation methods, the first requires a two-hour test interview while the second can be completed in less than an hour. The scores for each of the 15 individuals who took both tests are given in the next table. $$\begin{array}{ccc}\hline \text { Applicant } & \text { Test } 1(x) & \text { Test } 2(y) \\\\\hline 1 & 75 & 38 \\\2 & 89 & 56 \\\3 & 60 & 35 \\\4 & 71 & 45 \\\5 & 92 & 59 \\\6 & 105 & 70 \\\7 & 55 & 31 \\\8 & 87 & 52 \\\9 & 73 & 48 \\\10 & 77 & 41\end{array}$$ $$\begin{array}{ccc}\hline \text { Applicant } & \text { Test } 1(x) & \text { Test } 2(y) \\\\\hline 11 & 84 & 51 \\\12 & 91 & 58 \\\13 & 75 & 45 \\\14 & 82 & 49 \\\15 & 76 & 47 \\\\\hline\end{array}$$ a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. Find the least-squares line for the data. c. Use the regression line to predict the score on the second test for an applicant who scored 85 on Test 1 . d. Construct the ANOVA table for the linear regression relating \(y\) to \(x\).

What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used?

Subjects in a sleep deprivation experiment were asked to solve a set of simple addition problems after having been deprived of sleep for a specified number of hours. The number of errors was recorded along with the number of hours without sleep. The results, along with the MINITAB output for a simple linear regression, are shown below. $$ \begin{aligned} &\begin{array}{l|l|l|l} \text { Number of Errors, } y & 8,6 & 6,10 & 8,14 \\ \hline \text { Number of Hours without Sleep, } x & 8 & 12 & 16 \end{array}\\\ &\begin{array}{l|l|l} \text { Number of Errors, } y & 14,12 & 16,12 \\ \hline \text { Number of Hours without Sleep, } x & 20 & 24 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Analysis of Variance }\\\ &\begin{array}{lcrrrr} \text { Source } & \text { DF } & \text { Adj SS } & \text { Adj MS } & \text { F-Value } & \text { P-Value } \\ \hline \text { Regression } & 1 & 72.20 & 72.200 & 14.37 & 0.005 \\ \text { Error } & 8 & 40.20 & 5.025 & & \\ \text { Total } & 9 & 112.40 & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Model Summary }\\\ &\begin{array}{rrr} \mathrm{S} & \text { R-sq } & \text { R-sq(adj) } \\ \hline 2.24165 & 64.23 \% & 59.76 \% \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Coefficients }\\\ &\begin{array}{lrrrr} \text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \hline \text { Constant } & 3.00 & 2.13 & 1.41 & 0.196 \\ \mathrm{x} & 0.475 & 0.125 & 3.79 & 0.005 \end{array} \end{aligned} $$ Regression Equation $$ y=3.00+0.475 x $$ a. Do the data present sufficient evidence to indicate that the number of errors is linearly related to the number of hours without sleep? Identify the two test statistics in the printout that can be used to answer this question. b. Would you expect the relationship between \(y\) and \(x\) to be linear if \(x\) varied over a wider range \((\) say \(, x=4\) to \(x=48\) )? c. How do you describe the strength of the relationship between \(y\) and \(x ?\) d. What is the best estimate of the common population variance \(\sigma^{2} ?\) e. Find a \(95 \%\) confidence interval for the slope of the line.

Use the data set and the MINITAB output (Exercise I8, Section 12.1) below to answer the questions. $$ \begin{array}{l|llllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7 \end{array} $$ Find a \(95 \%\) prediction interval for some value of \(y\) to be observed in the future when \(x=2\).

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