Chapter6.5slopepointform.pdf: File Size: 2588 kb: File Type: pdf. Analyze the Data. Once you have all the necessary information, analyze it using statistical methods to come up with the best solution, and conclusion for your research project. Lastly, present the results to the target audience. Before conducting a research project, it is essential to learn the use of research frameworks. Chapter 7  Linear Relations and Slope. Powered by Create your own unique website with customizable templates. Welcome Math 10 FP Math 10 A&W Math 11.
Linear functions apply to real world problems that involve a constant rate.
Apply linear equations to solve problems about rates of change
Linear equations often include a rate of change. For example, the rate at which distance changes over time is called velocity. If two points in time and the total distance traveled is known the rate of change, also known as slope, can be determined. From this information, a linear equation can be written and then predictions can be made from the equation of the line.
If the unit or quantity in respect to which something is changing is not specified, usually the rate is per unit of time. The most common type of rate is “per unit of time”, such as speed, heart rate and flux. Ratios that have a nontime denominator include exchange rates, literacy rates, and electric field (in volts/meter).
In describing the units of a rate, the word “per” is used to separate the units of the two measurements used to calculate the rate (for example a heart rate is expressed “beats per minute”).
An athlete begins he normal practice for the next marathon during the evening. At 6:00 pm he starts to run and leaves his home. At 7:30 pm, the athlete finishes the run at home and has run a total of 7.5 miles. How fast was his average speed over the course of the run?
The rate of change is the speed of his run; distance over time. Therefore, the two variables are time [latex](x)[/latex] and distance [latex](y)[/latex]. The first point is at his house, where his watch read 6:00 pm. This is the beginning time so let’s set it to [latex]0[/latex]. So our first point is [latex](0,0)[/latex] because he did not run anywhere yet. Let’s think about our time in hours. Our second point is [latex]1.5[/latex] hours later, and we ran [latex]7.5[/latex] miles. The second point is [latex](1.5,7.5)[/latex]. Our speed (rate of change) is simply the slope of the line connecting the two points. The slope, given by: [latex]m = frac{y_{2}y_{1}}{x_{2}x_{1}}[/latex] becomes [latex]m = frac{7.5}{1.5}=5 [/latex] miles per hour.
To graph this line, we need the [latex]y[/latex]intercept and the slope to write the equation. The slope was [latex]5[/latex] miles per hour and since the starting point was at [latex](0,0)[/latex], the [latex]y[/latex]intercept is [latex]0[/latex]. So our final function is [latex]y=5x[/latex].
Distance and time graph: The graph of [latex]y=5x[/latex]. The two variables are time [latex](x)[/latex] and distance [latex](y)[/latex]. The rate the runner runs is [latex]5[/latex] miles per hour. Using the graph, predictions can be made assuming that his average speed remains the same.
With this new function, we can now answer some more questions.
There are many such applications for linear equations. Anything that involves a constant rate of change can be nicely represented with a line with the slope. Indeed, so long as you have just two points, if you know the function is linear, you can graph it and begin asking questions! Just make sure what you’re asking and graphing makes sense. For instance, in the marathon example, the domain is really only [latex]xgeq0[/latex], since it doesn’t make sense to go into negative time and lose miles!
Linear mathematical models describe real world applications with lines.
Apply linear mathematical models to real world problems
A mathematical model is a description of a system using mathematical concepts and language. Mathematical models are used not only in the natural sciences and engineering disciplines, but also in the social sciences. Linear modeling can include population change, telephone call charges, the cost of renting a bike, weight management, or fundraising. A linear model includes the rate of change [latex](m)[/latex] and the initial amount, the yintercept [latex]b[/latex]. After the model is written and a graph of the line is made, either one can be used to make predictions about behaviors.
Many everyday activities require the use of mathematical models, perhaps unconsciously. One difficulty with mathematical models lies in translating the real world application into an accurate mathematical representation.
A rental company charges a flat fee of [latex]$30[/latex] and an additional [latex]$0.25[/latex] per mile to rent a moving van. Write a linear equation to approximate the cost [latex]y[/latex] (in dollars) in terms of [latex]x[/latex], the number of miles driven. How much would a 75 mile trip cost?
Using the slopeintercept form of a linear equation, with the total cost labeled [latex]y[/latex] (dependent variable) and the miles labeled [latex]x[/latex] (independent variable):
[latex]displaystyle y=mx+b[/latex]
The total cost is equal to the rate per mile times the number of miles driven plus the cost for the flat fee:
[latex]displaystyle y=0.25x+30[/latex]
To calculate the cost of a [latex]75 [/latex] mile trip, substitute [latex]75[/latex] for [latex]x[/latex] into the equation:
[latex]displaystyle begin{align} y&=0.25x+30 &=0.25(75)+30 &=18.75+30 &=48.75 end{align}[/latex]
It’s also possible to model multiple lines and their equations.
Initially, trains A and B are [latex]325[/latex] miles away from each other. Train A is traveling towards B at [latex]50[/latex] miles per hour and train B is traveling towards A at [latex]80[/latex] miles per hour. At what time will the two trains meet? At this time how far did the trains travel?
First, begin with the starting positions of the trains, ([latex]y[/latex]intercepts, [latex]b[/latex]). Train A starts are the origin, [latex](0,0)[/latex]. Since train B is [latex]325[/latex] miles away from train A initially, its position is [latex](0,325)[/latex].
Second, in order to write the equations representing each train’s total distance in terms of time, calculate the rate of change for each train. Since train A is traveling towards train B, which has a greater [latex]y[/latex] value, train A’s rate of change must be positive and equal to its speed of [latex]50[/latex]. Train B is traveling towards A, which has a lesser [latex]y[/latex] value, giving B a negative rate of change: [latex]80[/latex].
The two lines are thus:
[latex]displaystyle y_A=50x [/latex]
And:
[latex]displaystyle y_B=−80x+325[/latex]
The two trains will meet where the two lines intersect. To find where the two lines intersect set the equations equal to each other and solve for [latex]x[/latex]:
[latex]displaystyle y_{A}=y_{B}[/latex]
[latex]displaystyle 50x=80x+325[/latex]
Solving for [latex]x[/latex] gives:
[latex]displaystyle x=2.5[/latex]
The two trains meet after [latex]2.5[/latex] hours. To find where this is, plug [latex]2.5[/latex] into either equation.
Plugging it into the first equation gives us [latex]50(2.5)=125[/latex], which means it meets after A travels [latex]125[/latex] miles.
Here is the distance versus time graphic model of the two trains:
Trains: Train A (red line) is represented by the equation: [latex]y=50x[/latex], and Train B (blue line) is represented by the equation: [latex]y=80x+325[/latex]. The two trains meet at the intersections point [latex](2.5,125)[/latex], which is after [latex]125[/latex] miles in [latex]2.5[/latex] hours.
Curve fitting with a line attempts to draw a line so that it “best fits” all of the data.
Use the least squares regression formula to calculate the line of best fit for a set of points
Curve fitting is the process of constructing a curve, or mathematical function, that has the best fit to a series of data points, possibly subject to constraints. Curve fitting can involve either interpolation, where an exact fit to the data is required, or smoothing, in which a “smooth” function is constructed that approximately fits the data. Fitted curves can be used as an aid for data visualization, to infer values of a function where no data are available, and to summarize the relationships among two or more variables. Extrapolation refers to the use of a fitted curve beyond the range of the observed data, and is subject to a greater degree of uncertainty since it may reflect the method used to construct the curve as much as it reflects the observed data.
In this section, we will only be fitting lines to data points, but it should be noted that one can fit polynomial functions, circles, piecewise functions, and any number of functions to data and it is a heavily used topic in statistics.
Linear regression is an approach to modeling the linear relationship between a dependent variable, [latex]y[/latex] and an independent variable, [latex]x[/latex]. With linear regression, a line in slopeintercept form, [latex]y=mx+b[/latex] is found that “best fits” the data.
The simplest and perhaps most common linear regression model is the ordinary least squares approximation. This approximation attempts to minimize the sums of the squared distance between the line and every point.
[latex]displaystyle m=frac{sum_{i=1}^{n}x_{i}y_{i}frac{1}{n}sum_{i=1}^{n}x_{i}sum_{j=1}^{n}y_{j}}{sum_{i=1}^{n}(x_{i}^{2})frac{1}{n}(sum_{i=1}^{n}x_{i})^{2}}[/latex]
To find the slope of the line of best fit, calculate in the following steps:
[latex]displaystyle begin{align} b&= frac{1}{n} sum_{i=1}^{n} y_{1}  m frac{1}{n} sum_{i=1}^{n} x_{i} &= left (bar{y}  m bar{x} right) end{align}[/latex]
To find the [latex]y[/latex]intercept ([latex]b[/latex]), calculate using the following steps:
Using these values of [latex]m[/latex] and [latex]b[/latex] we now have a line that approximates the points on the graph.
For [latex]n=8[/latex] points: [latex](1,0),(0,0),(1,1),(2,2),(3,1),(4,2.5),(5,3) [/latex] and [latex](6,4)[/latex].
Example Points: The points are graphed in a scatterplot fashion.
First, find the slope [latex](m)[/latex] and [latex]y[/latex]intercept [latex](b)[/latex] that best approximate this data, using the equations from the prior section:
To find the slope, calculate:
[latex]displaystyle begin{align} sum_{i=1}^{n}x_{i}y_{i}&=0+0+1+4+3+10+15+24&=57 end{align} [/latex][latex]displaystyle begin{align} sum_{i=1}^{n}x_{i}&=1+0+1+2+3+4+5+6&=20 end{align}[/latex][latex]displaystyle begin{align} sum_{i=1}^{n}y_{i}&=0+0+1+2+1+2.5+3+4&=13.5 end{align}[/latex]
[latex]displaystyle m=frac{sum_{i=1}^{n}x_{i}y_{i}frac{1}{n}sum_{i=1}^{n}x_{i}sum_{j=1}^{n}y_{j}}{sum_{i=1}^{n}(x_{i}^{2})frac{1}{n}(sum_{i=1}^{n}x_{i})^{2}}[/latex]
4. Calculate the numerator: The product of the [latex]x[/latex]
and [latex]y[/latex]coordinates
minus oneeighth the product of the sum of the [latex]x[/latex]coordinates and the sum of the [latex]y[/latex]coordinates:
[latex]displaystyle sum_{i=1}^{n}x_{i}y_{i}frac{1}{n}sum_{i=1}^{n}x_{i}sum_{j=1}^{n}y_{j}[/latex]
The numerator in the slope equation is:
[latex]displaystyle 57frac{1}{8}(20)(13.5)=23.25[/latex]
5. Calculate the denominator: The
sum of the squares of the [latex]x[/latex]coordinates minus oneeighth the sum of the [latex]x[/latex]coordinates squared:
[latex]displaystyle sum_{i=1}^{n}(x_{i}^{2})frac{1}{n}(sum_{i=1}^{n}x_{i})^{2}[/latex]
[latex]displaystyle begin{align} sum_{i=1}^{n}(x_{i}^{2})&=1+0+1+4+9+16+25+36&=92 end{align}[/latex]
The denominator is [latex]92frac{1}{8}(20)^{2}=9250=42[/latex] and the slope is the quotient of the numerator and denominator: [latex]frac{23.25}{42}approx0.554.[/latex]
Now for the [latex]y[/latex]intercept, ([latex]b[/latex]) oneeighth times the average of the [latex][/latex][latex]x[/latex]coordinates: [latex]bar{x}=frac{20}{8}=2.5[/latex] and oneeighth times the average of the [latex]y[/latex]coordinates: [latex]bar{y}=frac{13.5}{8}=1.6875[/latex].
Therefore [latex]b=frac{1}{n} sum_{i=1}^{n} y_{1}  m frac{1}{n} sum_{i=1}^{n} x_{i} [/latex]:
[latex]displaystyle bapprox1.68750.554(2.5)=0.3025.[/latex]
Our final equation is therefore [latex]y=0.554x+0.3025[/latex], and this line is graphed along with the points.
Least Squares Fit Line: The line found by the least squares approximation, [latex]y = 0.554x+0.3025[/latex]. Notice 4 points are above the line, and 4 points are below the line.
If we have a point that is far away from the approximating line, then it will skew the results and make the line much worse. For instance, let’s say in our original example, instead of the point [latex](1,0)[/latex] we have [latex](1,6)[/latex].
Using the same calculations as above with the new point, the results are:[latex]mapprox0.0536[/latex] and [latex]bapprox2.3035[/latex], to get the new equation [latex]y=0.0536x+2.3035[/latex].
Looking at the points and line in the new figure below, this new line does not fit the data well, due to the outlier [latex](1,6)[/latex]. Indeed, trying to fit linear models to data that is quadratic, cubic, or anything nonlinear, or data with many outliers or errors can result in bad approximations.
Outlier Approximated Line: Here is the approximated line given the new outlier point at (1, 6).

Goals  


Textbooks 

Supplies  



Course Content:
Descriptive Statistics and Probability
Chapter1:Data and Statistics
Chapter2:Descriptive Statistics: Tabular andGraphical Presentation
Chapter3:Descriptive Statistics: NumericalMeasures
Chapter4:Introduction to Probability
Probability Distributions
Chapter5:Discrete Probability Distributions
Chapter6:Continuous Probability Distributions
Chapter7:Sampling and Sampling Distributions
Inferential Statistics
Chapter8:Interval Estimation
Chapter9:HypothesisTests
Chapter 11:Tests ofGoodness of Fit and Independence
Chapter 12:LinearRegression and Correlation
I.P R E R E Q U I S I T E S
A good quality Algebra II duringhigh school or a college course at the level of Intermediate Algebra (Math 114)will provide you with the needed problem solving skills.
II.G R A D E DM A T E R I A LA N DG R A D E S
Expectto Think and Evaluate, not just regurgitate!!!
Each chapter section will have a problem assignmentto be completed:Practice problemassignments and HandIn problem assignments.The appendices of the textbook contains answers to thepractice problems.Student groupswho work together on practice assignments may find that team study isbeneficial to all students of the team.Practice problems will NOT be graded.
When assigned the handin problems of asection assignment will be graded. The average of all graded homework during thesemester will be equivalent to one test grade in determining your final coursegrade. A due date will beannounced for each homework assignment and the assignment must be handed in atbeginning of class on the given date.Late homework will be penalizedper the Homework Requirements handout.
The handin assignment will not provide enoughpractice to be able to remember a concept and be able to demonstrate thatconcept accurately on a test.GradedHomework should be an individual project rather than a team effort.Students who cover their lack ofunderstanding by depending too much on others may fail tests and eventuallyfail the course.
Graded homework provides a student with feedback regarding how well theyare understanding and performing the concepts of a chapter plus utilization ofproper notation and work shown.
There will be computer lab projectsassigned.We will be using the Excelspreadsheet for Windows software.Each project will be evaluated and collectively they will be equivalentto 1 test.Due dates for computerprojects will generally precede the test which covers the concepts in theproject.These computer activitieswill give you some experience in working with computer software, which is apowerful tool to handle the number crunching and data analysis for applicationproblems.
Tests will be announced in advance and will beabout 50 minutes in length.Eachtest will be normalized to 100 pts.These tests should be taken during the scheduled period unless prior arrangements aremade.As soon as you know that youwill be absent on a test date, please notify the instructor.
There isno provision for makeup tests or retests.
Graded test will usually be available by the nextclass session, thus if you miss thescheduled test you must contact the instructor prior to the next scheduledclass session else receive a zero on the exam.
Formulasheets are provided with each test.
Test 1:Chapters1 & 2 100points
Test 2: Chapters3 & 4100points
Test 3: Chapters5 & 6100points
Test 4: Chapters7, 8100points
Test 5:Chapters9, 11, 12100points
Homework 100points
Computer Project(s)100points
TOTAL POINTSPOSSIBLE  700
Your final letter grade is determined by dividing yourtotal points earned by the 700 total possible points. Letter grade isdetermined by scale listed earlier.
V.P H I L O S O P H YO NP U R P O S EOFG R A D E S
Gradesare determined by the level of mastery of the subject that you have demonstrated.Grades do not measure your abilityto learn.Grades do notmeasure how much effort has been applied.Grades do not reflect information about exterior reasons fora student's performance.But keepin mind that mastery is hardest to achieve when attendance is poor, study isnot regular, practice exercises are not done regularly, nor completely.
Manythings impact a grade:personalhealth, family health, work, family obligations, social life, math background,effort, priorities, college activities, etc.
Thisis college, I expect you to be responsible for your actions, good or not sogood.
Math/Science CenterRoom 411 ( faculty Pod )
Math faculty schedule office hours so that they areavailable to help students.Student tutors in math are employed by the college in an effort toexpand upon the hours per week during which help with mathematics isavailable.