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Graphing Linear Inequalities in Two Variables (Part 2) Lesson NarrativeIn a previous lesson, students learned to graphically represent the set of solutions to a linear inequality in two variables. They made a connection between the solutions to a linear inequality and the solutions to a related linear equation. In this lesson, students deepen their understanding of the solutions to linear inequalities by studying them in context. They write inequalities in two variables to represent constraints, and interpret the points on a boundary line and on either side of it in terms of the situation. The work here illustrates that the solution region represents the set of values that satisfy the constraint in a situation (MP2). Interpreting the solutions contextually also engages students in an aspect of mathematical modeling (MP4). It enables students to see that, while some values might make an inequality true, they might not be feasible or appropriate in the situation. The activity Rethinking Landscaping is an opportunity to make a generalization based on repeated reasoning (MP8), since students first find numbers that meet a constraint, and then use variables in place of those numbers to write an equation and an inequality. Because reasoning about the solution region of an inequality is important here, graphing technology should not be used. Students will have opportunities to use graphing technology to solve inequalities in two variables in upcoming lessons. Learning GoalsTeacher Facing  Find the solution to a twovariable inequality by graphing a related twovariable equation and determining the correct region for the solution.
 Interpret, in context, points on the graphs of equations and in the solution region of inequalities in two variables.
 Write inequalities in two variables to represent the constraints in a situation and identify possible solutions by reasoning.
Student Facing Let’s write inequalities in two variables and make sense of the solutions by reasoning and by graphing.
Learning Targets Given a twovariable inequality that represents a situation, I can interpret points in the coordinate plane and decide if they are solutions to the inequality.
 I can find the solutions to a twovariable inequality by using the graph of a related twovariable equation.
 I can write inequalities to describe the constraints in a situation.
CCSS Standards HSACED.A.3
 HSAREI.D.10
 HSAREI.D.12
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Eureka Math Grade 7 Module 5 Lesson 22 Answer KeyEngage ny eureka math 7th grade module 5 lesson 22 answer key, eureka math grade 7 module 5 lesson 22 example answer key. Example 2. Compare the variability in the two data sets using the MAD (mean absolute deviation). Is the variability in each sample about the same? Interpret the MAD in the context of the problem. Answer: The variability in each data set is about the same as measured by the mean absolute deviation (around 4 sec.) For boys and girls, a typical deviation from their respective mean times (35 for boys and 39 for girls) is about 4 sec. Example 3. In the previous lesson, you learned that a difference between two sample means is considered to be meaningful if the difference is more than what you would expect to see just based on sampling variability. The difference in the sample means of the boys’ times and the girls’ times is 4.1 seconds (39.4 seconds – 35.3 seconds). This difference is approximately 1 MAD. a. If4 sec. is used to approximate the value of 1 MAD for both boys and for girls, what is the interval of times that are within 1 MAD of the sample mean for boys? Answer: 35.3 sec.+ 4 sec.=9.3 sec. , and 35.3 sec. 4 sec.=31.4 sec. The interval of times that are within 1 MAD of the boys’ mean time is approximately 31.4 sec. to 39.3 sec. b. Of the 10 sample means for boys, how many of them are within that interval? Answer: Six of the sample means for boys are within the interval. c. Of the 11 sample means for girls, how many of them are within the interval you calculated in part (a)? Answer: Seven of the sample means for girls are within the interval. d. Based on the dot plots, do you think that the difference between the two sample means is a meaningful difference? That is, are you convinced that the mean time for all girls at the school (not just this sample of girls) is different from the mean time for all boys at the school? Explain your choice based on the dot plots. Answer: Answers will vary. Sample answer: I don’t think that the difference is meaningful. The dot plots overlap a lot, and there is a lot of variability in the times for boys and the times for girls. Example 6. Calculate the mean absolute deviation (MAD) for each data set. Based on the MADs, compare the variability in each sample. Is the variability about the same? Interpret the MADs in the context of the problem. Answer: The MAD for the quiet distribution is 2.68 sec. The MAD for the talking distribution is 2.73 sec. The MAD measurements are about the same, indicating that the variability in each data set is similar. In both groups, a typical deviation of students’ minute estimates from their respective means is about 2.7 sec. Example 7. Based on your calculations, is the difference in mean time estimates meaningful? Part of your reasoning should involve the number of MADs that separate the two sample means. Note that if the MADs differ, use the larger one in determining how many MADs separate the two means. Answer: The number of MADs that separate the two sample means is \(\frac{6}{2.73}\), or 2.2. There is a meaningful difference between the means. Eureka Math Grade 7 Module 5 Lesson 22 Problem Set Answer KeyQuestion 1. A school is trying to decide which reading program to purchase. a. How many MADs separate the mean reading comprehension score for a standard program (mean = 67.8, MAD = 4.6, n = 24) and an activitybased program (mean = 70.3, MAD = 4.5, n = 27)? Answer: The number of MADs that separate the sample mean reading comprehension score for a standard program and an activitybased program is \(\frac{70.367.8}{4.6}\), or 0.54, about half a MAD. b. What recommendation would you make based on this result? Answer: The number of MADs that separate the programs is not large enough to indicate that one program is better than the other program based on mean scores. There is no noticeable difference in the two programs. a. Calculate the difference between the sample mean distance for the football filled with air and for the one filled with helium. Answer: The 17 airfilled balls had a mean of 27 yd. compared to 23.8 yd. for the 17 heliumfilled balls, a difference of 3.2 yd. c. Calculate the MAD for each distribution. Based on the MADs, compare the variability in each distribution. Is the variability about the same? Interpret the MADs in the context of the problem. Answer: The MAD is 2.59 yd. for the airfilled balls and 2.07 yd. for the heliumfilled balls. The typical deviation from the mean of 27.0 is about 2.59 yd. for the airfilled balls. The typical deviation from the mean of 23.8 is about 2.07 yd. for the heliumfilled balls. There is a slight difference in variability. d. Based on your calculations, is the difference in mean distance meaningful? Part of your reasoning should involve the number of MADs that separate the sample means. Note that if the MADs differ, use the larger one in determining how many MADs separate the two means. Answer: \(\frac{3.2}{2.59}\) = 1.2 There is a separation of 1.2 MADs. There is no meaningful distance between the means. c. Calculate the MAD for each distribution. Based on the MADs, compare the variability in each distribution. Is the variability about the same? Interpret the MADs in the context of the problem. Answer: The MAD is 5.15 for college graduates and 5.17 for high school graduates. The typical deviation from the mean of 52.4 is about 5.15 (or $5,150) for college graduates. The typical deviation from the mean of 32.8 is about 5.17 ($5,170) for high school graduates. The variability in the two distributions is nearly the same. d. Based on your calculations, is going to college worth the effort? Part of your reasoning should involve the number of MADs that separate the sample means. Answer: \(\frac{19.6}{5.17}\) = 3.79 There is a separation of 3.79 MADs. There is a meaningful difference between the population means. Going to college is worth the effort. Eureka Math Grade 7 Module 5 Lesson 22 Exit Ticket Answer KeyQuestion 2. Compare the variability in the two data sets using the MAD. Interpret the result in the context of the problem. Answer: The MAD for the boys’ number of minutes spent texting is 7.9 min., which is higher than that for the girls, which is 5.3 min. This is not surprising, as seen in the dot plots. The typical deviation from the mean of 70.9 is about 7.9 min. for boys. The typical deviation from the mean of 97.3 is about 5.3 min. for girls. Question 3. From 1 and 2, does the difference in the two means appear to be meaningful? Answer: 97.3 – 70.9 = 26.4 The difference in means is 26.4 min. \(\frac{26.4}{7.9}\) = 3.3 Using the larger MAD of 7.9 min., the means are separated by 3.3 MADs. Looking at the dot plots, it certainly seems as though a separation of more than 3 MADs is meaningful. 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Lesson 22 : Add a fraction less than 1 to, or subtract a fraction less than 1 from, a whole number using decomposition and visual models. 22 Homework 4•Lesson 5 Name Date 1. Draw a tape diagram to match each number sentence. Then, complete the number sentence. a. 2 + 1 4 = _____ b. 3 + 2 3 = _____ c. 2 F1 5
Answer Key  Chapter 22 (31.0K) Answer Key  Chapter 23 (32.0K) Answer Key  Chapter 24 (27.0K) Answer Key  Chapter 25 (31.0K) Answer Key  Chapter 26 (36.0K) To learn more about the book this website supports, please visit its Information Center. ...
YS. COMMON CORE MATHEMATICS CURRICULUMLesson 22 Homework 42. Find all factors. r the following numbers and classify as prime or composite. F. ctor Pairs for 19Factor Pairs for 21Factor Pai. s for 243. Bryan says that only even numbers are composite. ist all of the odd numbers less than 20 in numerical. rd.
Eureka Math Grade 3 Module 5 Lesson 22 Homework Answer Key. Question 1. Write the shaded fraction of each figure on the blank. Then, draw a line to match the equivalent fractions. The fraction of shaded parts = Number of shaded parts ÷ Total Number of Parts . All the fractions are written and the respective fractions are matched . Question 2.
It's Homework Time! Help for fourth graders with Eureka Math Module 5 Lesson 22.
Divide three and four digit dividends by two digit divisors resulting in two and three digit quotients, reasoning about the decomposition of successive remai...
Eureka Math Grade 5 Module 4 Lesson 22 Problem Set Answer Key. Question 1. Solve for the unknown. Rewrite each phrase as a multiplication sentence. Circle the scaling factor and put a box around the number of meters. a. 12 as long as 8 meters = ______ meter (s) Answer: 4 meters. Explanation:
Eureka Essentials: Grade 4. An outline of learning goals, key ideas, pacing suggestions, and more! Fluency Games. Teach Eureka Lesson Breakdown. Downloadable Resources. Teacher editions, student materials, application problems, sprints, etc. Application Problems. Files for printing or for projecting on the screen.
LESSON 22 GRADE 8 LESSON 22 Page 2 of 2 Adding and Subtracting with Scientific Notation continued 7 (4 3 102) 1 120.5 8 (2.75 3 103) 2 100 9 (9.5 3 102) 2 (4.3 3 101) 10 18 2 (2 3 1021) 11 0.071 1 (6 3 1022) 12 2,000 1 (8 3 103) 13 When adding or subtracting with scientific notation, why is it important to have the same power of 10?
Lesson 4 Answer Key 2• 8 Lesson 4 Problem Set 1. 2 parallel lines of different lengths drawn 2. 2 parallel lines of the same length drawn 3. a. Both pairs of sides highlighted b. 1 pair of sides highlighted c. Both pairs of sides highlighted d. Both pairs of sides highlighted and boxes drawn around all 4 angles e. 1 pair of sides highlighted f.
In this lesson, students deepen their understanding of the solutions to linear inequalities by studying them in context. They write inequalities in two variables to represent constraints, and interpret the points on a boundary line and on either side of it in terms of the situation. The work here illustrates that the solution region represents ...
Engage NY Eureka Math 4th Grade Module 5 Lesson 22 Answer Key Eureka Math Grade 4 Module 5 Lesson 22 Sprint Answer Key A Add Fractions Answer: 1 + 1 = 2, 1/5 + 1/5. Skip to content. Above Header. Go ... Eureka Math Grade 4 Module 5 Lesson 22 Homework Answer Key. Question 1. Draw a tape diagram to match each number sentence. Then, complete the ...
You can find the source for the homework pages at the link below. click on the "full module" PDF:https://www.engageny.org/resource/grade2mathematicsmodule4
Find stepbystep solutions and answers to Algebra 1 Common Core  9780133185485, as well as thousands of textbooks so you can move forward with confidence. ... Exercise 22. Exercise 23. Exercise 24. Exercise 25. Exercise 26. Exercise 27. Exercise 28. Exercise 29. Exercise 30. Exercise 31. Exercise 32. Exercise 33. Exercise 34. Exercise 35 ...
Eureka Math Grade 2 Module 7 Lesson 22 Problem Set Answer Key. Question 1. Each unit length on both number lines is 10 centimeters. (Note: Number lines are not drawn to scale.) a. Show 30 centimeters more than 65 centimeters on the number line. b. Show 20 centimeters more than 75 centimeters on the number line.
Georgia Milestones Assessment System Test Prep: Grade 3 English Language Arts Literacy (ELA) Practice Workbook and Fulllength Online Assessments: GMAS Study Guide. Lumos Learning. 3. 2017. ACT Aspire Test Prep: 3rd Grade Math Practice Workbook and Fulllength Online Assessments: ACT Aspire Study Guide. Lumos Learning.
EngageNY/Eureka Math Grade 4 Module 5 Lesson 22For more videos, please visit http://bit.ly/eurekapusdPLEASE leave a message if a video has a technical diffic...
Engage NY Eureka Math 5th Grade Module 2 Lesson 22 Answer Key Eureka Math Grade 5 Module 2 Lesson 22 Problem Set Answer Key. Question 1. Divide. Then, check using multiplication. The first one is done for you. a. 580 ÷ 17. Answer: 580/17 = 2. Explanation: In the abovegiven question, given that, Divide, and then check. 17 x 3 = 51. 58  51 ...
Assessments, Answer Keys, and Sample Questions. Follow. This section includes helpful information about assessments, including where to locate answer keys for module problem and homework sets as well as the location of sample questions for state assessments that are released to the public. When will the Grades 38 ELA and mathematics Common ...
EngageNY/Eureka Math Grade 3 Module 5 Lesson 22For more videos, please visit http://EMBARC.onlinePLEASE leave a message if a video has a technical difficulty...
Eureka Math Grade 4 Module 3 Lesson 22 Problem Set Answer Key. Question 1. Record the factors of the given numbers as multiplication sentences and as a list in order from least to greatest. Classify each as prime (P) or composite (C). The first problem is done for you.
Answer: The mean of the quiet estimates is 58.8 sec. The mean of the talking estimates is 64.8 sec. 64.8  58.8 = 6. The difference between the two means is 6 sec. Example 5. On the same scale, draw dot plots of the two data distributions, and discuss the similarities and differences in the two distributions. Answer: