lesson 22 homework answer key

Graphing Linear Inequalities in Two Variables (Part 2)

Lesson Narrative

In 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 Goals

Teacher Facing

• Find the solution to a two-variable inequality by graphing a related two-variable 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 two-variable 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 two-variable inequality by using the graph of a related two-variable equation.
• I can write inequalities to describe the constraints in a situation.

CCSS Standards

• HSA-CED.A.3
• HSA-REI.D.10
• HSA-REI.D.12

Print Formatted Materials

Teachers with a valid work email address can  click here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials.

Additional Resources

• 888-309-8227
• 732-384-0146

New User Registration

Forgot Password

Textbook Resources

textbook resources

• Call us toll-free
• FAQs – Frequently Asked Questions
• Contact Lumos Learning – Proven Study Programs by Expert Teachers

Follow us: Lumos Learning -->

• 2024 © Lumos Learning
• Privacy Policy - Terms of Service - Disclaimers

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc... Read More

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc., the Partnership for the Assessment of Readiness for College and Careers, nor any state of the Union. Neither PARCC, Inc., nor The Partnership for the Assessment of Readiness for College and Careers, nor any member state has endorsed this product. No portion of any fees or charges paid for any products or services Lumos Learning offers will be paid or inure to the benefit of PARCC, Inc., or any state of the Union

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not aff... Read More

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not affiliated to Lumos Learning. The Regents of the University of California – Smarter Balanced Assessment Consortium, was not involved in the production of, and does not endorse these products or this site.

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC, was not... Read More

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC,was not involved in the production of, and does not endorse these products or this site.

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the... Read More

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the production of, and does not endorse these products or this site.

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the... Read More

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the production of, and does not endorse these products or this site.

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved... Read More

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved in the production of, and does not endorse these products or this site.

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the prod... Read More

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the production of, and does not endorse these products or this site.

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved... Read More

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved in the production of, and does not endorse these products or this site.

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved... Read More

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved in the production of, and does not endorse these products or this site.

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved... Read More

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved in the production of, and does not endorse these products or this site.

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved... Read More

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved in the production of, and does not endorse these products or this site.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Eureka Math Grade 7 Module 5 Lesson 22 Answer Key

Engage 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 Key

Question 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 activity-based 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 activity-based program is \(\frac{70.3-67.8}{4.6}\), 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 air-filled balls had a mean of 27 yd. compared to 23.8 yd. for the 17 helium-filled 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 air-filled balls and 2.07 yd. for the helium-filled balls. The typical deviation from the mean of 27.0 is about 2.59 yd. for the air-filled balls. The typical deviation from the mean of 23.8 is about 2.07 yd. for the helium-filled 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 Key

Question 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.

Leave a Comment Cancel Reply

You must be logged in to post a comment.