Mlcs handout with math lit lesson 3.12

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Mlcs handout with math lit lesson 3.12

  1. 1. Information for MLCSKathleen Almy, Heather FoesRock Valley CollegeEmails: kathleenalmy@gmail.com heather.foes@gmail.comBlog: http://almydoesmath.blogspot.com Blog contains video, pilot updates, presentations, and more.Mathematical Literacy for College Students (MLCS)3 – 6 semester hoursPrerequisite: Appropriate placement or prealgebra with a grade of “C” or betterThe goal of developmental mathematics education is to provide students with the necessary skills andunderstanding required to be successful in college level mathematics. It is not necessarily intended to be arepeat of high school mathematics. While many programs and initiatives have been developed to improve thestate of developmental education, part of the problem lies in the content and courses taught. MathematicalLiteracy for College Students (MLCS) is a new course being developed at the national level by AMATYC’s NewLife for Developmental Mathematics. Its origins are related to Quantway, funded by the Carnegie Foundation.MLCS is an alternative path to certain college level math courses or further algebra. It integrates numeracy,proportional reasoning, algebraic reasoning, and functions with statistics and geometry as recurring coursethemes. Throughout the course, college success components are integrated with the mathematical topics.The course focuses on developing mathematical maturity through problem solving, critical thinking, writing,and communication of mathematics. Content is developed in an integrated fashion, increasing in depth as thecourse progresses. Upon completion of the course, students will be prepared for a statistics course or ageneral education mathematics course. Students may also take intermediate algebra upon completion if theychoose to pursue STEM courses.MLCS provides an alternative to beginning algebra, creating multiple pathways for the developmentalstudents. However, it is more rigorous than beginning algebra to ensure students are prepared for a collegelevel math course upon successful completion. It allows students to potentially complete their developmentalmath and college level math requirement for an Associate in Arts degree in one year total (one semestereach), working toward the goal of improving college completion rates. It promotes 21st century skills toprepare students for both the workplace and future coursework. Further, it does not diminish requirementsfor non-STEM college level math courses but instead creates appropriate paths to these courses with the samelevel of intensity and complexity as the current path through intermediate algebra. The course has collegelevel expectations and coursework but with a pace and instructional design intended for the adult,developmental learner. This strategy emulates the approach taken by the Common Core Standards and alignswith them as well.1|Page
  2. 2. MLCS Course Description and ObjectivesMathematical Literacy for College Students is a one semester course for non-math and non-science majors integratingnumeracy, proportional reasoning, algebraic reasoning, and functions. Students will develop conceptual and proceduraltools that support the use of key mathematical concepts in a variety of contexts. Throughout the course, college successcontent will be integrated with mathematical topics.Prerequisite: Appropriate placement or prealgebra with a grade of “C” or betterCOURSE OUTCOMES1. Apply the concepts of numeracy in multiple contexts.2. Recognize proportional relationships and use proportional reasoning to solve problems.3. Use the language of algebra to write relationships involving variables, interpret those relationships, and solve problems.4. Interpret and move flexibly between multiple formats including graphs, tables, equations, and words.5. Demonstrate student success skills including perseverance, time management, and appropriate use of resources.6. Develop the ability to think critically and solve problems in a variety of contexts using the tools of mathematics including technology.COURSE OBJECTIVESUpon successful completion of this course, the student will be able to:Numeracy1. Demonstrate operation sense and the effects of common operations on numbers in words and symbols.2. Demonstrate competency in the use of magnitude in the contexts of place values, fractions, and numbers written in scientific notation.3. Use estimation skills.4. Apply quantitative reasoning to solve problems involving quantities or rates.5. Demonstrate measurement sense.6. Demonstrate an understanding of the mathematical properties and uses of different types of mathematical summaries of data.7. Read, interpret, and make decisions based upon data from line graphs, bar graphs, and charts.Proportional reasoning8. Recognize proportional relationships from verbal and numeric representations.9. Compare proportional relationships represented in different ways.10. Apply quantitative reasoning strategies to solve real-world problems with proportional relationships.Algebraic reasoning11. Understand various uses of variables to represent quantities or attributes.12. Describe the effect that changes in variable values have in an algebraic relationship.13. Construct and solve equations or inequalities to represent relationships involving one or more unknown or variable quantities to solve problems.2|Page
  3. 3. Functions14. Translate problems from a variety of contexts into a mathematical representation and vice versa.15. Describe the behavior of common types of functions using words, algebraic symbols, graphs, and tables.16. Identify the reasonableness of a linear model for given data and consider alternative models.17. Identify important characteristics of functions in various representations.18. Use appropriate terms and units to describe rate of change.19. Understand that abstract mathematical models used to characterize real-world scenarios or physical relationships are not always exact and may be subject to error from many sources.Student success20. Develop written and verbal skills in relation to course content.21. Evaluate personal learning style, strengths, weaknesses, and success strategies that address each.22. Research using print and online resources.23. Apply time management and goal setting techniques.Mathematical success24. Develop the ability to use mathematical skills in diverse scenarios and contexts.25. Use technology appropriately including calculators and computers.26. Demonstrate critical thinking by analyzing ideas, patterns, and principles.27. Demonstrate flexibility with mathematics through various contexts, modes of technology, and presentations of information (tables, graphs, words, equations).28. Demonstrate and explain skills needed in studying for and taking tests.3|Page
  4. 4. Implementation OptionsMLCS is a 3 – 6 credit hour course depending on the depth and breadth desired.1. Replacement Model: Use MLCS to replace beginning algebra.2. Augmented Model: Use MLCS to create a non-STEM alternative to beginning algebra that provides sufficient preparation for statistics or liberal arts math.3. Supplemental Model: Use MLCS lessons for problem solving sessions in an Emporium model (lab-based traditional redesign.), engaging all students and moving beyond skills alone.4. High School Model: Use MLCS lessons for 4th year high school course to develop college readiness and help students place into college level math.4|Page
  5. 5. FP 80–90 min 3.12 Get in Line Explore On a comprehensive final exam, a professor asked each student to list the number of hours spent studying for the test. She chose a few of the students’ results and compiled them into 10 min a table. This helps her see trends that may prove useful for her future classes. Instructor note:  Instruct stu- dents to work on #1–6 in groups. Number of Hours Studied Test Score Out of 100 Go over answers. 0 62 1 66 2 70 3 74 4 78 6 86 9 98 1. Each time the number of hours increases by 1, the test score increases by 4 . Each time the number of hours increases by 2, the test score increases by 8 . Each time the number of hours increases by 6, the test score increases by 24 . Use each of these answers to write a ratio in the form increase in test score 4 = increase in number of hours 1   2. Since the ratios of increase in number of hours studied to increase in test score are ­always 4 , we can say there is a linear relationship b ­ etween the number of hours studied and the test score.  3. Write the slope with units and interpret it. Slope is 4 point/hour studied. For every hour studied, a student will gain 4 points on the final exam.     Instructor note:  We are using 4. Write a formula for the test score, T, based on the number of hours, h, a student studies. patterns and inductive reason- ing here. Formal equation writing T = 4h + 62 comes in Cycle 4. If students need a hint, remind them that the scores start at 62 and increase by 4. This idea is very useful for writing linear 5. Where is the slope in the table? In the equation? equations. The slope is the rate of change in the table. It is the coefficient of the independent variable in the equation. 6. Where is the number 62 in the table? In the equation? It was the score associated with 0 hours studied. It is the constant in the equation.    c
  6. 6. LESSON 3.12 Get in Line 343 Discover Let’s look at this situation and the data using a graph. 20 min 7. Create a graph using the coordinate system provided.Instructor note:  Bring the Final Examclass together to complete #7–10. 100Instructor note:  Discuss the 96setup of the graph and the choices 92for scales on the axes to exposestudents to the idea of domain (rea- 88sonable values for the independentvariable) and range (reasonable val- 84 Test scoreues for the dependent variable). 80It’s reasonable to have fractionsand decimals for the hours studied 76but not negative values. The hourscan continue to increase somewhat, 72but there is a physical limit. Also, 68there is a limit to what the testscore can be. 64Draw a right triangle from (1, 66) 60to (2, 70) on the line. Ask studentsfor the rise and run of the triangleto find the slope of the line. m = 4 1 2 3 4 5 6 7 8 9 10 Hours studied(4 additional points for each hourstudied).Explain that while we graphed the 8. Where do the slope and the number 62 appear on the graph?line by plotting points, we coulduse the slope to find more points The slope is the rise compared to the run when moving between points on the line. Theon the line once a point on it is point (0, 62) is where the line crosses the y-axis.plotted. Let’s look more closely at how the equation of a line and its graph are related. A Closer Look Learn Intercept y look The point at which a graph crosses an axis is 6 called an intercept. If the graph crosses the x-axis, 5 the point of intersection is called an x-intercept. 4 it up If the graph crosses the y-axis, the point of inter- section is called a y-intercept. 3 2 1 For example, this graph has an x-intercept of –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x ( - 2, 0) and a y-intercept of (0, 4). –1 –2 –3 –4 –5 –6 The formula T = 4h + 62 is in a very useful form because it clearly shows the slope and y-intercept of the graph. The equation is in slope-intercept form.
  7. 7. 344 Cycle 3Instructornote:  Although it Slope-intercept formis conventional to look Linear graphs have equations that can be written in the form y = mx + b, called slope-use y = mx + b, it intercept form. x represents the independent variable, and y is the dependent variable.is often easier for When an equation is in slope-intercept form, the coefficient of x is the slope. The con-students to thinkof linear equationsas y = b + mx. it up stant b represents the y-coordinate of the point where the line intersects the y-axis. The ordered pair (0, b) is the y-intercept.This order is usedwhen graphing dependent variable independent variablewith the slope and graphed on vertical axis graphed on horizontal axisy-intercept and it’s y = 50x + 200how most contex-tual problems aredescribed with an slope y-coordinate of y-interceptequation: a starting (change) (starting value)point and changefrom it. For example, imagine you have $200 in a savings account and you deposit $50 each month. This equation has m = 50 and b = 200. So the line has a slope of 50 and y-intercept of (0, 200). We can read these values only from an equation in slope-intercept form. Practice We have now learned that linearity can be seen from a table when there is a constant rate of change, on a graph when the points fall in a line, and in an equation of the form y = mx + b.Instructor note:  Remind Example 1  For each linear equation, list the slope and y-intercept. Remember,students that every number has a the y-intercept is an ordered pair, not a number.sign. Don’t separate the sign froma number. a. y = -3x - 2  m = -3, y-intercept is (0, -2) x 1 b. y = - 8  m = , y-intercept is (0, -8) 2 2 c. y = 4 - x  m = -1, y-intercept is (0, 4) practice 1  For each linear equation, list the slope and y-intercept. Remember, the y-intercept is an ordered pair, not a number. 5 min a. y = -x + 1  m = -1, y-intercept is (0, 1)Instructor note:  Instructstudents to work in groups. Check b. y = -x  m = -1, y-intercept is (0, 0).answers and clear up issues. Signsand implied coefficients usually c. y = 4  m = 0, y-intercept is (0, 4).present a challenge. d. T = 4h + 62  m = 4, y-intercept is (0, 62). Let’s graph a linear equation in two ways. The first method requires finding points that appear on the line. Organizing the points into a table is helpful. The second method uses the slope and y-intercept. Example 2  Complete this table for the following linear equation: y = -5x - 6. 10–15 min x y Instructor note:  After going -1 -1 over answers to the practice problems, work with the class on 0 -6­example 2. 1 -11 2 -16
  8. 8. LESSON 3.12 Get in Line 345 We can choose any values for x. The numbers in the table are whole numbers and are small, which make calculations faster. Notice that we are picking x, plugging it into the equation, and finding y. The value of y depends on whatever we choose forInstructor note:  Remind stu- x. We can connect the points with a line because the rate of change is constant, mak-dents that although only 2 pointsare needed to draw a line, a third ing the pattern linear.point provides a check. Plot the points and graph the line. y 20 18 16 14 12 10 8 6 4 2 x –20 –16 –12 –8 –4 0 2 4 6 8 10 12 14 16 18 20 –2 –4 –6 –8 –10 –12 –14 –16 –18Instructor note:  The slope is –20- 5, which is easier to work with as -5the fraction . We can move down If you recognize the slope and y-intercept of a line, you can often graph it very 15 and to the right 1 from the quickly. To graph a line, we need to know a point to get a rough location on the gridy-intercept. Or we can think of the and also how the line is tilted. The y-intercept gives us a location, and the slope tells 5 the steepness.slope as and move up 5 and left -11. Do this again to get a third point. Find the slope from the equation y = -5x - 6 and verify it by finding the rateStudents should notice that this of change in the table.  m = -5graph is the same as the first graph. Find the y-intercept from the equation and verify it in the table.  (0, -6)Instruct students to work on prac-tice 2. Check answers. Ask which Plot the y-intercept. Use the rise and run of the slope to move to another pointmethod they prefer. on the line. y 20 18 16 14 12 10 8 6 4 2 x –20 –16 –12 –8 –4 0 2 4 6 8 10 12 14 16 18 20 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20
  9. 9. 346 Cycle 3 3 practice 2  Suppose y = - x + 5. 4 5–10 min a. Graph this linear equation by making a table of points. To avoid fractional y-values, choose x-values that are multiples of the denominator of the slope. y 10 x y 9 8 -4 8 7 6 0 5 5 4 4 2 3 2 1 x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 b. Graph this linear equation by using the slope and y-intercept. -3 m = , y-intercept is (0, 5) 4   y 10 9 8 7 6 5 4 3 2 1 x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 –1 –2 –3 –4 –5Instructor note:  Remind –6students that we have used the –7method of plotting points (as in –8part a) throughout the book so far. –9The method used in part b is a –10shortcut but it only works for lines.
  10. 10. LESSON 3.12 Get in Line 347 When working with real data, we need to be able to interpret the slope and y-intercept, not just find them. Keep in mind that slope compares the rise and run or the dependent and independent variables. We can use this idea to write a r ­ atio to interpret the slope. Example 3  If gas costs $3.65 per gallon with a $5.50 car wash, a formula for the cost (C) of g gallons of gas and the car wash is given by C = 3.65g + 5.50. 5 min This equation, or model of the situation, is also a function since there is only one cost ­ ssociated with each value of g. The equation is linear. If it helps, think of the a ­equation as y = 3.65x + 5.50. List the slope and y-intercept. m = 3.65, y-intercept is (0, 5.50)     The values of the slope and y-intercept matter, but their meaning is also important. change in y change in cost 3.65 Slope = = = change in x change in gallons 1 Interpret the slope in words:  For each gallon of gas, we will pay an additional $3.65. Interpret the y-intercept in words:  The car wash, without any gas, would cost $5.50.Instructor note:  Explain the StickySticky Note. Instruct students to notework in groups for practice 3. Goover answers. Keep in mind that slope-intercept form does not have to use the letters y and x. If you see an equ ation in this form, it is in slope-intercept form: Dependent = Number  ~  Independ ent + Number     Variable        Variable          y =     mx       +   b The numbers, which represent the slope and y-intercept, can be positive, negative, or zero, and are not necessarily identical. Practice 3  For each problem, write an equation to model the situation. Interpret the slope and y-intercept in words. Also list which variable would be on the hori- 5–10 min zontal axis and which would be on the vertical axis. a. Model the cost if you buy an iPod for $200 and download songs for $1.29 each. C = 1.29s + 200 would give the cost, C, of the iPod and s songs. The slope gives the price per song of $1.29. The y-intercept would be (0, 200), indi- cating the cost of the iPod with 0 songs downloaded. The number of songs, s, would be on the horizontal axis. The cost, C, would be on the vertical axis.          
  11. 11. 348 Cycle 3 b. Model the weekly salary a salesman earns if the salary is a combination of his base salary and commissions on his sales. His base weekly salary is $500 and he makes $50 for each sale. S = 500 + 50n, where n is the number of sales he makes and S is his weekly salary. The slope gives the commission per sale of $50. The y-intercept would be (0, 500), indicating his salary if he made 0 sales. The number of sales, n, would be on the hori- zontal axis. S would be on the vertical axis.           Connect In this cycle, we have defined linear equations as equations of the form ax + b = c, where a is not zero. But in this lesson, we have also referred to equations in y = mx + b form as 10 min linear. Let’s connect these ideas.Instructor note:  Complete #9 9. Suppose y = 2x + 4. yas a class. This ensures studentswill make the necessary connec- a. Graph the equation using the slope 6tions between solving and graphing and y-intercept. 5linear equations. 4 3 2 1 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 b. Pick an ordered pair from the graph that appears to be on the line and confirm that the ordered pair is on the graph in part a. Answers will vary.    c. Pick another ordered pair from the graph that appears to be on the line and con- firm that the ordered pair is on the graph in part a. Answers will vary. d. How are the points on the line related to the equation y = 2x + 4? The coordinates of the points on the line are the solutions of the equation.    e. How do you know for sure that the point (-100, -196) is on the graph of the line y = 2x + 4 even though you cannot see it in the graph in part a? Since the ordered pair is a solution to the equation, the point must lie on the line.   
  12. 12. LESSON 3.12 Get in Line 349 Reflect wrap-up lesson 5 minInstructor note:  Explain the What’s the point?contents of the Wrap-Up. Have stu- Linear situations can be seen in tables, graphs, and equations.dents write an answer to the cyclequestion, “Why does it matter?” What did you learn?A prompt is provided to help stu-dents. Discuss homework. How to find and interpret the slope and y-intercept from a linear equation How to graph lines using tables and slope and y-intercept Cycle 3 Question: When is it worth it? When is it worth drawing a graph to see and understand the slope? When are you able to get the information you need from the equation alone? Answers will vary. 3.12  Homework Skills  MyMathLab • Find and interpret the slope and y-intercept from a linear equation. • Graph lines using tables and slope and y-intercept. 1. List the slope and y-intercept:  y = 3 - 2x. m = -2, y-intercept is (0, 3)     2. Use the slope and y-intercept from #1 to form the graph. y 6 5 4 3 2 1 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6    
  13. 13. 350 Cycle 3 Concepts and Applications • Find and interpret the slope and y-intercept from a linear equation. 3. For each situation, write a linear model. Find and interpret the slope and y-intercept. a. A student has $2,000 in debt from college loans, along with $500 payment for every month she has left to pay off her car. Model her total debt. D = 2,000 + 500m, where m = number of months and D = total debt The slope is 500, meaning she has an additional $500 of debt for each month that passes in which she has not paid off her car. The y-intercept is (0, 2000). This means that without the car debt, she would have $2,000 of debt.     b. A man has a starting weight of 205 pounds and is losing 10 pounds each month. Model his weight. W = 205 - 10m, where m = number of months and W = the man’s weight The slope is -10, meaning he is losing 10 pounds per month. The y-intercept is (0, 205). This means before he starts losing weight, when no months have passed, his weight is 205 pounds.     4. Imagine a company is offering newly hired employees three salary-pricing structures. One option computes raises by adding $1,000 and then a 3% increase. Another option computes raises by adding a 3% increase and then $1,000. A final option option com- putes raises by adding a 5% increase. a. Write a formula for computing the new salary under each plan, where NS = New Salary and S = Old Salary. Add $1,000; increase by 3%  NS1 = 1.03(S + 1,000) Increase by 3%; add $1,000   NS2 = 1.03S + 1,000 Increase by 5% NS3 = 1.05S b. In each formula, what is the independent variable?  S, the original salary in dollars What are the dependent variables? NS1, NS2, NS3 c. Using the formulas you created, complete the table below. All amounts are in dollars. Old Salary, S NS1 = 1.03(S + 1,000) NS2 = 1.03S + 1,000 NS3 = 1.05S 10,000 11,330 11,300 10,500 20,000 21,630 21,600 21,000 30,000 31,930 31,900 31,500 40,000 42,230 42,200 42,000 50,000 52,530 52,500 52,500
  14. 14. LESSON 3.12 Get in Line 351 d. Simplify the first equation and list the three equations below. What do you notice about their slopes and y-intercepts? NS1 = 1.03S + 1,030 NS2 = 1.03S + 1,000 NS3 = 1.05S The slopes are the same for Options 1 and 2. The first equation’s value of b is exactly $30 more than the value of the second equation. The third equation has a different slope and y-intercept than either of the first two.        e. Of Options 1 and 2, which will always be better and why? Simplified, we can see that the rate of change is the same, but Option 1 will always be $30 more than Option 2.       f. Since the three lines are very close to each other, a specific graph is not helpful in this case. Instead, use only the y-intercepts and slopes to make a rough sketch of all three lines on the same graph. Use the fact that lines with the same slope are parallel. New Salary vs. Old Salary New salary Option 1 Option 2 Option 3 Old salary g. Solve an equation to determine for which original salary amount the first two o ­ ptions produce equal new salaries. 1.03(S + 1,000) = 1.03S + 1,000; no solution   There are no salaries for which the first two plans produce equal new salaries.     h. Solve an equation to determine for which original salary amount the second and third options produce equal new salaries. 1.03S + 1,000 = 1.05S; S = $50,000   For an original salary of $50,000, the second and third plans both produce the same new salary.    
  15. 15. 352 Cycle 3 5. In the Explore, you looked at how the number of study hours is related to a student’s exam score. Imagine the professor’s data had instead been the following: Number of Hours Studied Test Score Out of 100 0 62 1 66 2 48 2 70 3 74 4 78 6 86 6 78 9 98 9 74 a. Does this data represent a function? Did the original data in the Explore represent a function? Explain. The original data represented a function, but this data does not since there are values of the independent variable that are matched with more than one value of the dependent variable.       b. What could the instructor do to get an equation to model the data?  Draw a scatterplot and find a trendline. 6. a. Do all lines have a y-intercept? Why or why not? No, vertical lines do not have a y-intercept (unless the vertical line is the y-axis).     b. What kinds of lines are not functions? Vertical     7. Decide whether or not each equation is linear. If necessary, complete a table to help you decide. If the equation is linear, identify the slope and the y-intercept. a. y = x - 4 linear slope = 1, y@intercept = (0, -4) b. y = 4 - x linear slope = -1, y@intercept = (0, 4) 1 c. y = - 4  not linear x
  16. 16. LESSON 3.13 Gas Up and Go 353 1 d. y =   not linear (x - 4) 1 1 e. y = x  linear slope = , y@intercept = (0, 0) 4 4 , y@intercept = a0, bU      (x + 1) 1 1 f. y =   linear slope = 4 4 4

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