problem solving data displays lesson 12 8

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Year levels.

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Expected level of development

Australian Curriculum Mathematics V9 : AC9M6ST01

Numeracy Progression : Interpreting and representing data: P5

At this level, students interpret and compare datasets using different data displays and visualisations.

Provide students with the opportunity to investigate different types of questions and related datasets. Make explicit the different types of datasets that students use in the data interpretation and comparison. For example, a summary might include:

• Nominal data: eye colour, ethnicity, favourite food
• Ordinal data: rating scale, ranking items, ordering a list of items
• Discrete: product cost, number of students in a class, days in a week
• Continuous: vehicle speeds, temperature

Students use frequency tables to categorise data. They create column graphs and interpret the frequencies by comparing the heights of the columns. Provide numerical data including continuous data that can be represented as a line graph. Use questioning to prompt students to interpret the graphs and make conclusions.

Demonstrate the link to data and its representation as a percentage, and how these are used to create a pie chart.

Provide spatial data in the form of GPS data as longitude and latitude. Students can visualise the data by plotting on paper – or, more efficiently and effectively, using online mapping software – and can look for patterns or trends.

There is an opportunity to integrate scientific and geographic understanding and skills through relevant contexts to acquire, sort and interpret data.

• In Science, students use their digital literacy skills to access information; analyse and represent data; model and interpret concepts and relationships; and communicate science ideas, processes and information.
• In HASS, students use their digital literacy skills when they locate, select, evaluate, communicate and share geographical information using digital tools, and they learn to use spatial technologies.

Teaching and learning summary:

• Create graphs to display information.
• Choose appropriate graphs and justify the usefulness of each type.
• Make comparisons between different visualisations of the same datasets.

• use a frequency table to sort and categorise data
• represent data using a data display such as a chart or graph
• compare and interpret different data displays.

Some students may:

• have difficulty accurately representing data using a relevant data display. They may use a scale that is inappropriate or inaccurate and does not suitably fit the range of data points. To address this, as a class or in targeted teaching groups, show graphical representations and discuss issues and ask students to identify the issue and how it can be corrected. In terms of the scale being used, ensure that students relate this back to the context, that the choice of scale makes it easy to interpret the graph, and that it is not misleading. Provide guidance to help students make judgements about graphs. Discuss features of a useful and accurate data display and what they should include, for example, the title and subtitles should be clear, data must be understandable, scales should use even and consistent intervals, and in column graphs, the columns should have equal spaces in between and each should be the same width.

The Learning from home activities are designed to be used flexibly by teachers, parents and carers, as well as the students themselves. They can be used in a number of ways including to consolidate and extend learning done at school or for home schooling.

Learning intention

• We are learning to interpret secondary data in order to interpret and compare data.

Why are we learning about this?

• We need to be able to interpret tables and graphs found in digital media to ensure that we are not being misled by data that is presented.
• Look at this frequency table. Decide on what the data might represent.
• Provide a heading for the left-hand column and add labels for each row of data.
• Count the tally and record the totals.

4. Choose a way to represent the data using a relevant data display.

Answer the following questions.

• How many people were surveyed?
• Who would you have chosen to be participants in this survey?
• What graph have you chosen to present this data and why?
• Could you use another graph to represent the same information? Which of the two is clearer?
• What conclusion can you write about your data?

Success criteria

• record data in a frequency table
• represent data using a relevant data display
• justify my choice of graph used to represent the data.

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Teaching strategies.

A collection of evidence-based teaching strategies applicable to this topic. Note we have not included an exhaustive list and acknowledge that some strategies such as differentiation apply to all topics. The selected teaching strategies are suggested as particularly relevant, however you may decide to include other strategies as well. 

Explicit teaching

Explicit teaching is about making the learning intentions and success criteria clear, with the teacher using examples and working though problems, setting relevant learning tasks and checking student understanding and providing feedback.

Questioning

A culture of questioning should be encouraged and students should be comfortable to ask for clarification when they do not understand.

It has been shown that good feedback can make a significant difference to a student’s future performance.

Classroom talks

Classroom talks enable students to develop language, build mathematical thinking skills and create mathematical meaning through collaborative conversations.

Multiple exposures

Providing students with multiple opportunities within different contexts to practise skills and apply concepts allows them to consolidate and deepen their understanding.

Teaching resources

A range of resources to support you to build your student's understanding of these concepts, their skills and procedures. The resources incorporate a variety of teaching strategies.

Osprey migration data

In this lesson, students acquire data related to the migration of the osprey.

Osprey bird dimensions

In this lesson, students investigate measurement data related to the osprey.

Osprey position and location

In this lesson, students describe location and position on a Cartesian plane using paired coordinates.

Osprey ranking threats

In this lesson, students work in groups to rank a list of key threats and collate the data in a class spreadsheet.

Flight of the osprey infographic

In this lesson, students use data acquired during their learning about the osprey and information sourced online to design and create an infographic.

Does using less water make a difference?

In this lesson, students carry out a statistical investigation based on their own inquiry question related to sustainable water use.

Water storage levels across Australia

In this lesson, students interpret and compare real-world data sets and analyse graphs in terms of their range and shape.

Making informed arguments

In this lesson students create an infographic to communicate their findings from their statistical investigation.

Visualize a data set (v2)

This dynamic software tool enables students to explore changes on a column graph and the relationship to the data points.

Rolling two dice

This dynamic software tool enables students to explore the data visualised from the simulation of rolling two dice.

Distance time graphs

This dynamic software tool enables students to explore the data visualised from four modes of transport: a car, a person running, a person walking and a person cycling.

Opinion polls

This lesson focuses on collecting opinion data asking questions where participants select an answer from a 5-point scale.

Turtles: exploring data tracking turtle movements

This lesson uses spatial data recorded as GPS points and free online mapping software to visualise the collected data.

Saltwater crocs: resourceful or a resource?

This lesson follows an inquiry process where students use the dataset to answer relevant questions about the crocodile population.

If the world were a village

In this lesson, students consider how data are presented and interpreted. The same dataset in visualised in different ways.

Paul's basketball challenge

Students interpret a graph that shows the number of baskets Paul scored when he was playing basketball toss with his sister.

Disaster relief

Students develop their understanding of how data can be analysed and represented to make sense of a data set. Refer to page 162.

Relevant assessment tasks and advice related to this topic.

By the end of Year 6, students are comparing distributions of discrete and continuous numerical and ordinal categorical datasets as part of their statistical investigations, using digital tools.

Assessment: statistics – sports data

Use this task to assess how a student uses unorganised data presented in a table to represent the data in graphical form. Teacher assessment guidance is included.

Assessment: line graph

Use this task to assess how a student interprets a line graph.

Assessment: column graph

Use this task to assess how a student analyses and interprets a column graph that has errors.

Interpreting and comparing data

Use this task to assess students’ proficiency in compare distributions of discrete and continuous numerical and ordinal categorical data sets.

Comparing and Contrasting Data Distributions

11.1: Math Talk: Mean (5 minutes)

CCSS Standards

Routines and Materials

Instructional Routines

• MLR8: Discussion Supports

This is the first math talk activity in the course. See the launch for extended instructions for facilitating this activity successfully.

The purpose of this Math Talk is to expand students’ strategies for finding a mean beyond following an algorithm to reasoning that the mean of the values in a symmetric data set is the middle value. The third item is designed to illustrate that this technique only works for symmetric data sets. These understandings help students develop fluency and will be helpful later in this lesson when students will need to use symmetry to match a mean to the distribution.

This Math Talk provides an opportunity for students to notice and make use of the symmetric structure (MP7) of the values to determine the mean. While participating in these activities, students need to be precise in their word choice and use of language (MP6).

Monitor for students who:

• use the standard algorithm for finding mean (sum and divide)
• use the symmetry of the data set

This is the first time students do the  math talk  instructional routine, so it is important to explain how it works before starting.

Explain the math talk routine: one problem is displayed at a time. For each problem, students are given a few minutes to quietly think and give a signal when they have an answer and a strategy. The teacher selects students to share different strategies for each problem, and might ask questions like “Who thought about it a different way?” The teacher records students' explanations for all to see. Students might be asked to provide more details about why they decided to approach a problem a certain way. It may not be possible to share every possible strategy for the given limited time; the teacher may only gather two or three distinctive strategies per problem. 

Consider establishing a small, discreet hand signal that students can display to indicate that they have an answer they can support with reasoning. This signal could be a thumbs-up, a certain number of fingers that tells the number of responses they have, or another subtle signal. This is a quick way to see if the students have had enough time to think about the problem. It also keeps students from being distracted or rushed by hands being raised around the class.

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Student Facing

Evaluate the mean of each data set mentally.

61, 71, 81, 91, 101

0, 100, 100, 100, 100

0, 5, 6, 7, 12

Student Response

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Anticipated Misconceptions

If students struggle to use symmetry as a method for finding the mean, consider asking them to find the mean for the values: 1, 2, 3, 4, 5.

Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

• “Who can restate ___’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to __’s strategy?”
• “Do you agree or disagree? Why?”

Although all correct methods for solving for the mean are valid, highlight the use of symmetry in the data. In previous lessons, students learned that symmetric distributions have a mean in the center of the data. When symmetry is present, it can be used to quickly discover the mean.

11.2: Describing Data Distributions (15 minutes)

Building On

Building Towards

• MLR7: Compare and Connect

Required Materials

• Pre-printed slips, cut from copies of the blackline master

In this activity students take turns with a partner matching data displays with distribution characteristics and determine what measure of center is most appropriate for the data. Students trade roles explaining their thinking and listening, providing opportunities to explain their reasoning and critique the reasoning of others (MP3).

Arrange students in groups of 2. Demonstrate how to set up and and find matches. Choose a student to be your partner. Mix up the cards and place them face-up. Point out that the cards contain either a data display or a written statement. Select one of each style of card and then explain to your partner why you think the cards do or do not match. Demonstrate productive ways to agree or disagree, for example, by explaining your mathematical thinking or asking clarifying questions. Give each group a set of cut-up cards for matching. Ask students to pause after completing the matching for a whole-class discussion. Give students five minutes to work the second question then pause for a whole-class discussion.

Tell students that the appropriate measure of center may not be the one given on the cards.

• For each match that you find, explain to your partner how you know it’s a match.
• For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.
• After matching, determine if the mean or median is more appropriate for describing the center of the data set based on the distribution shape. Discuss your reasoning with your partner. If it is not given, calculate (if possible) or estimate the appropriate measure of center. Be prepared to explain your reasoning.

Much discussion takes place between partners. Once all groups have completed the matching, discuss the following:

• “Which matches were tricky? Explain why.” (The box plot in row 6 was tricky because I had to use process of elimination to figure out that it was the one that was uniform.)
• “Did you need to make adjustments in your matches? What might have caused an error? What adjustments were made?” (Yes. I realized that I thought incorrectly that skewed left meant that most of the data was on the left. However, I learned that skewed left means that there is data to the left of where most of the data is located.)
• “Can you determine the median using only a histogram? Why or why not?” (No, but you can determine the interval that contains the median.)
• “Can you determine if a distribution is uniform from a box plot? Why or why not?” (No. You can determine that the data could possibly be symmetric based on the distribution of the five number summary, but beyond that you would not be able to know that the data is uniform using only a box plot.)

The purpose of the second part of the activity is to discuss the relationship between mean and median based on the shape of the distribution and to make the connection to measures of variability. Ask:

• “If the mean is the appropriate measure of center, should we use the MAD or the IQR to measure variability?” (MAD)
• “If the median is the appropriate measure of center, should we use the MAD or the IQR to measure variability?” (IQR)

11.3: Visual Variability and Statistics (10 minutes)

• MLR2: Collect and Display

This activity prompts students to compare variability in several data sets by analyzing the distributions shown on box plots and dot plots. Some students may reason about variability by observing the shapes and features of the data displays. Others may try to quantify the variability by finding the IQR from each box plot, or by estimating the MAD from each dot plot. Look for students who approach the task quantitatively.

Arrange students in groups of 2. Give students five minutes to work through the questions then pause for a whole-group discussion.

Each box plot summarizes the number of miles driven each day for 30 days in each month. The box plots represent, in order, the months of August, September, October, November, and December.

• The five box plots have the same median. Explain why the median is more appropriate for describing the center of the data set than the mean for these distributions.
• The five dot plots have the same mean. Explain why the mean is more appropriate for describing the center of the data set than the median.

Are you ready for more?

 These two box plots have the same median and the same IQR. How could we compare the variability of the two distributions?

Description: <p>Two box plots on the same number line. From 2 to 20, by 2s.<br> <br> Bottom box plot has whisker from 2 to 8. Box from 8 to 13 with vertical line at 9. Whisker from 13 to 20.<br> <br> Top box plot has whisker from 6 to 8. Box from 8 to 13 with vertical line at 9. Whisker from 13 to 18.</p>

These two dot plots have the same mean and the same MAD. How could we compare the variability of the two distributions?

Students may have forgotten what variability means or which statistic to use to determine the variability in a data set. Refer them to previous work or ask them what measure is useful in determining a data set's tendency to have different values.

The purpose of this discussion is to make the connection between the shape of the distribution and the use of either IQR or MAD to quantify variability. Another goal is to make sure students understand that a greater value from IQR or MAD means greater variability. Display the box plots in order of variability with the IQR included, and then display the dots plots in order of variability with the MAD included.

The IQR for the data in distributions A through E are {40, 60, 50, 40, 20} and the MAD for the data in distributions F through J are approximately {1.56, 1.10, 2.68, 2.22, 0}. Here are some questions for discussion:

• “What are the IQR and MAD measuring?” (They are measuring the spread or variability of the data)
• “Which plots were the most difficult to arrange?” (The dot plots were more difficult because it was easy to find the IQR for the box plots.)
• “Do the orders given by the IQR and MAD match your order?” (Yes, except for the box plots A and D which had the same IQR and I didn’t know how to arrange them.)
• “What do you notice about the values for IQR and MAD?” (The values for the MAD were higher than I thought except for distribution J. I did not know that the MAD could be equal to zero.)
• “What advantages are offered by using IQR and MAD versus visual inspection?” (The IQR and MAD are values that can be easily sorted.)

If some students already arranged the plots using IQR or MAD you should ask them, “Why did you choose to arrange the plots by IQR or MAD?” (I knew that IQR and MAD were measures of variability so I used them.)

Lesson Synthesis

In this lesson, students investigated variability using data displays and summary statistics.

• “One data set’s measure of center is best represented by a median of 7 and another data set by a median of 10. How would you determine which data set has greater variability?” (You calculate the IQR. Whichever one has a larger IQR is more variable.)
• “How do you determine which of two roughly symmetric distributions has less variability?” (You calculate the MAD. Whichever one has a lower MAD has less variability.)
• “What does it mean to say that one data set or distribution has more variability than another?” (The appropriate measure of variability for one data set is greater than the other. Using a data display, one distribution is more spread apart than the other.)

11.4: Cool-down - Which Menu? (5 minutes)

Student lesson summary.

The mean absolute deviation, or MAD, is a measure of variability that is calculated by finding the mean distance from the mean of all the data points. Here are two dot plots, each with a mean of 15 centimeters, displaying the length of sea scallop shells in centimeters.

Notice that both dot plots show a symmetric distribution so the mean and the MAD are appropriate choices for describing center and variability. The data in the first dot plot appear to be more spread apart than the data in the second dot plot, so you can say that the first data set appears to have greater variability than the second data set. This is confirmed by the MAD. The MAD of the first data set is 1.18 centimeters and the MAD of the second data set is approximately 0.94 cm. This means that the values in the first data set are, on average, about 1.18 cm away from the mean and the values in the second data set are, on average, about 0.94 cm away from the mean. The greater the MAD of the data, the greater the variability of the data.

The interquartile range, IQR, is a measure of variability that is calculated by subtracting the value for the first quartile, Q1, from the value for the the third quartile, Q3. These two box plots represent the distributions of the lengths in centimeters of a different group of sea scallop shells, each with a median of 15 centimeters.

Notice that neither of the box plots have a symmetric distribution. The median and the IQR are appropriate choices for describing center and variability for these data sets. The middle half of the data displayed in the first box plot appear to be more spread apart, or show greater variability, than the middle half of the data displayed in the second box plot. The IQR of the first distribution is 14 cm and 10 cm for the second data set. The IQR measures the difference between the median of the second half of the data, Q3, and the median of the first half, Q1, of the data, so it is not impacted by the minimum or the maximum value in the data set. It is a measure of the spread of the middle 50% of the data.

The MAD is calculated using every value in the data while the IQR is calculated using only the values for Q1 and Q3.

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Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays

Students of Grade 8 can get the best solutions with explanations from Big Ideas Math Answers. Our Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays helps to score the highest marks in the exams and also improves your skills. Enhance your Knowledge by referring to BIM Grade 8 Answer Key Chapter 6 Data Analysis and Displays from ccssmathanswers.com

Big Ideas Math Book 8th Grade Answer Key Chapter 6 Data Analysis and Displays

The concepts of maths are applicable in modeling Real life. The Big Ideas Math Textbook 8 Grade Chapter 6 Data Analysis and Displays Answers improves the skills. The solutions are prepared from Big Ideas Math- Modeling Real Life Grade 8 student Edition Set. Learning targets and success criteria help the students to focus on learning the subject. By this teachers and parents can understand the graph of the student’s performance.

Performance Task

Data Analysis and Displays STEAM Video/Performance Task

Data analysis and displays getting ready for chapter 6.

Lesson: 1 Scatter Plots

Lesson 6.1 Scatter Plots

Scatter plots homework & practice 6.1.

Lesson: 2 Lines of Fit

Lesson 6.2 Lines of Fit

Lines of fit homework & practice 6.2.

Lesson: 3 Two-Way Tables

Lesson 6.3 Two-Way Tables

Two-way tables homework & practice 6.3.

Lesson: 4 Choosing a Data Display

Lesson 6.4 Choosing a Data Display

Choosing a data display homework & practice 6.4.

Chapter: 6 – Data Analysis and Displays 

Data Analysis and Displays Connecting Concepts

Data analysis and displays chapter review, data analysis and displays practice test, data analysis and displays cumulative practice.

STEAM Video

Answer: 1.The footprint of a car = 6,466 sq inches.

Explanation: In the above-given question, Tory says that the footprint of a vehicle is the area of the rectangle formed by the wheelbase and the track width. area of rectangle = length  x width Given that the footprint of a car = 106 inches. width with 61 inches. area = 106 x 61 footprint = 6,466 sq inches.

Answer: 2. a.The fuel economy increases when the footprint increases.

Explanation: In the above-shown video, tory says that whenever the footprint increases the fuel economy also increases. whenever the footprint decreases the fuel economy decreases.

Answer: 2.b.The point (50, 40) represents the outlier.

Answer: The relationship between the fuel economy and the purchase price of a vehicle is proportional.

Explanation: In the above-given figure, Given that the city fuel Economy and the purchase price of the cars. for car A (21.8, 24) for car B(22.4, 22) for car C(40.1, 18) if the fuel economy increases the purchase price also increases. whenever the economy decreases the purchase price also decreases.

Answer: a. (0, 95), (3, 88), (2, 90), (5, 83), (7, 79), (9, 70), (4, 85), (1, 94), (10, 65), (8, 75). b. the relationships between the absences and the final grade is decreasing when the absences increases. c. The student’s final grade is 80.

B. whenever the final grade is decreasing the absences also decrease. whenever the final grade increases the absence also increases. c. Given that the student has been absent for 6 days. The student’s final grade is 80.

Vocabulary The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts. scatter plot two-way table line of fit joint frequency

Answer: Scatter plot = A scatter plot uses dots to represent values for two different numeric variables. The position of each dot on the horizontal and vertical axis indicates values for an individual data point. Two-way table = A two-way table is a way to display frequencies or relative frequencies for two categorical variables. Line of fit = Line of fit refers to a line through a scatter plot of data points that best expresses the relationship between those points. Joint frequency = Joint frequency is joining one variable from the row and one variable from the column.

Explanation: Scatter plot = A scatter plot uses dots to represent values for two different numeric variables. The position of each dot on the horizontal and vertical axis indicates values for an individual data point. Two-way table = A two-way table is a way to display frequencies or relative frequencies for two categorical variables. Line of fit = Line of fit refers to a line through a scatter plot of data points that best expresses the relationship between those points. Joint frequency = Joint frequency is joining one variable from the row and one variable from the column.

EXPLORATION 1

Answer: b. The weight is measured in inches and size is measured in ounces.

Explanation: In the above-given figure, the size and the weight of the balls are given. size and weight of basketball = (21, 30). size and weight of baseball = (5, 9). size and weight of golfball = (1.6, 5.3). size and weight of soccerball = (16, 28). size and weight of tennis = (2, 8). size and weight of racquetball = (1.4, 7). size and weight of softball = (7, 12). size and weight of volleyball = (10, 26)

Answer: c. No, it is not reasonable to use the graph.

Answer: outliers = (120, 70) gaps =(10, 62) to (45, 85) clusters =(80, 95), (90, 97), (80, 91)

Explanation: outliers =(120, 70) gaps = (10, 62) to (45, 85) clusters = (80, 95), (90, 97), (80, 91)

Question 2. Describe the relationship between the data in Example 1.

Answer: Linear relationship.

Explanation: In the above-given graph, the relationship used is a linear relationship.

Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal.

Explanation: outliers = (3,24) clusters = 22 to 36 gaps = (4, 27), (8, 36)

Answer: The point (3.5, 3) does not belong with the other three.

Explanation: In the above-given figure The points (1,8),  (3, 6.5), and (8, 2) lies in the coordinate plane. the point (3.5, 3) does not belong with the other three. the point (3.5, 3) is an outlier. Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal.

Answer: The college GPA I expect for a high school student with a GPA of 2.7 is 2.45.

Explanation: In the above-given points, given that the college GPA for high school students. college GPA for 2.4 = high school students of 2.6 so I am expecting the 2.45 for 2.7.

Answer: outliers = (40, 6) clusters = (20, 2) to (70, 1) gaps = (0, 30), (1, 35), (2, 50) and so on.

Explanation: Given that, the person’s age (years) in the x-axis. a number of pets owned in the y-axis. outliers = (40, 6) clusters = (20, 2) to (70, 1) gaps = (0, 30), (1, 35), (2, 50) and so on.

Review & Refresh

Solve the system. Check your solution. Question 1. y = – 5x + 1 y = – 5x – 2

Answer: There is no solution for the given equation.

Explanation: Given that y = – 5x + 1 y = – 5x – 2 so there is no solution for the given equation.

Question 2. 2x + 2y = 9 x = 4.5 – y

Answer: 9 = 9

Explanation: Given that, 2x + 2y = 9 x = 4.5 – y 2(4.5 – y) + 2y = 9 9 – 2y + 2y = 9 -2y and + 2y get cancelled on both sides. 9 = 9

Question 3. y = – x 6x + y = 4

Answer: x = (4/5 , -4/5)

Explanation: Given that y = -x 6x + y = 4 6x + (-x) = 4 6x – x = 4 5x = 4 x = (4/5)

Question 4. When graphing a proportional relationship represented by y = mx, which point is not on the graph? A. (0, 0) B. (0, m) C. (1, m) D. (2, 2m)

Answer: Point A is not on the graph.

Explanation: In the above question, given that the points are: (0, 0) (0, m) (1, m) (2, 2m) the point (0, 0) is not in the graph.

Concepts, Skills, &Problem Solving

USING A SCATTER PLOT The table shows the average prices (in dollars) of jeans sold at different stores and the numbers of pairs of jeans sold at each store in one month. (See Exploration 1, p. 237.)

Answer: The points are (22, 152), (40, 94), (28, 134), (35, 110), and (46, 81)

Question 6. Is there a relationship between the average price and the number sold? Explain your reasoning.

Answer: The linear relationship.

Explanation: In the above-given figure, the relationship given is linear relationship.

Answer: Outliers = (102, 63) gaps = x from 40 to 44 clusters = 82 to 89

Answer: Outliers = (0, 5.5) gaps = x from 4.5 to 5.5 clusters = 1.5 to 2.5

Answer: Outliers = (15, 10) gaps = from x = 15 to x = 25 clusters = 0 Negative linear relationship.

Explanation: Outliers = (15, 10) gaps = from x = 15 to x = 25 clusters = 0 There are no clusters.

Answer: There are no clusters. gaps = from x = 4 to x = 36 outliers.

Explanation: In the above-given figure, there are no clusters. gaps = from x = 4 to x = 36 no outliers.

Answer: There is no relationship. there are no clusters. no gaps. no outliers.

Explanation: In the above-given graph, there are no clusters. no gaps. no clusters. there is no relationship.

Answer: The relationship is a positive linear relationship.

Explanation: In the above-figure, given points are: (2014, $4.65), (2015, $5.90), (2016, $6.50), and (2017, $7.70) so the above given is a positive linear relationship.

Explanation: In  the above-given figure, outliers = (49, 80) clusters = from x = 190 to 220.

Question 14. OPEN-ENDED Describe a set of real-life data that has a negative linear relationship. Answer:

Answer: a. 3.5 h b. 85 $ c. positive linear relationship.

Explanation: In the above-given graph, given that, a. the hours must server work to earn $70 = 3.5 h b. The server earns for 5 hours of work = $ 85. c. the relationship is shown by the data = positive linear relationship.

Answer: Outliers =(16, 50) gaps = 128 on x. clusters = 64, 32, 64

Explanation: Outliers =(16, 50) gaps =128 on x. clusters = 64, 32, 64.

Answer: a. 2014 b. about 950 scooters. c. negative linear relationship. d. 2019.

Explanation: In the above-given figure, Given that the number of vehicles sold in the year. a. 2014 b. about 950 scooters. c. negative linear relationship. d. 2019

Question 18. DIG DEEPER! Sales of sunglasses and beach towels at a store show a positive linear relationship in the summer. Does this mean that the sales of one item cause the sales of the other item to increase? Explain.

Answer: Yes.

Explanation: In the above-figure, given that the sales of the sunglasses and beach towels at a store show a positive linear relationship. yes the sales of one item cause the sales of the other item to increase.

Answer: a. The relation is a linear relationship.

Answer: The order pairs (1, 420), (2, 500), (3, 650), (4, 900), (5, 1100), (6, 1500), (7, 1750), (8, 2400)

Question 2. Find an equation of the line of best fit for the data in Example 1. Identify and interpret the correlation coefficient. Answer:

Question 4. IDENTIFYING RELATIONSHIPS Find an equation of the line of best fit for the data at the left. Identify and interpret the correlation coefficient Answer:

Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5. The ordered pairs show amounts y (in inches) of rainfall equivalent x to inches of snow. About how many inches of rainfall are equivalent to 6 inches of snow? Justify your answer. (16, 1.5) (12, 1.3) (18, 1.8) (15, 1.5) (20, 2.1) (23, 2.4) Answer:

Answer: Negative linear relationship. outliers = (6, 10) clusters = 0 gaps = 0

Explanation: In the above-given figure, The relationship is negative linear relationship. outliers = (6, 10) cluster = 0 gaps = 0 there are no clusters and no gaps.

positive linear relationships. outliers = 0 gaps = 0 clusters = x = 11 to x = 15

Explanation: In the above-given figure, given that positive linear relationship. outliers = 0 gaps = 0 clusters = x = 11 to x = 15

Write the fraction as a decimal and a percent. Question 4. \(\frac{29}{100}\)

Answer: Decimal = 0.29 percent = 29 %

Explanation: Given that (29/100) 0.29 percent = 29% decimal = 0.29

Question 5. \(\frac{7}{25}\)

Answer: Decimal = 0.28 percent = 28%

Explanation: Given that (7/25) = 0.28 decimal = 0.28 percent = 28

Question 6. \(\frac{35}{50}\)

Answer: Decimal = 0.7 percent = 0.007

Explanation: Given that (35/50) = 0.7 decimal = 0.7 percent = 0.007

Answer: The points are (0,0), (1, 0.8), (2, 1.50), (3, 2.20), (4, 3.0), (5, 3.75)

Answer: The given points are (0,91), (2, 82), (4, 74), (6, 65), (8, 55), (10, 43).

Answer: a.The given points are (30, 45), (36, 43), (44, 36), (51, 35), (60, 30), (68, 27), (75, 23), (82, 17). b. y = -0.5x + 60 c. you could expect that 60 hot chocolates are sold when the temperature is 0 degree f, and the sales decrease by 1 hot chocolate for every 2 degrees f increase in temperature.

Explanation: a.The given points are (30, 45), (36, 43), (44, 36), (51, 35), (60, 30), (68, 27), (75, 23), (82, 17). b. y = -0.5x + 60 c. you could expect that 60 hot chocolates are sold when the temperature is 0 degree f, and the sales decrease by 1 hot chocolate for every 2 degrees f increase in temperature.

Question 10. NUMBER SENSE Which correlation coefficient indicates a stronger relationship: – 0.98 or 0.91? Explain.

Answer: 0.91 indicates a stronger correlation coefficient.

Explanation: In the above-given question, -0.98 is a negative value and 0.91 is a positive value. So 0.91 indicates a stronger correlation coefficient.

Answer: The equation for the line of best fit is Y = -4.9x + 1042 about -0.969. strong negative correlation.

Explanation: In the above-given figure, The given points are (20, 940), (21, 935), (22, 940), (24, 925), (25, 920), (27, 905), (28, 910), and (30, 890) The equation for the line of best fit is y = -4.9x + 1042. about -0.969. strong negative correlation.

Answer: a. y = 1.3 x + 2; about 0.9995; strong positive correlation. b. The number of electoral votes increases by 1.3 for every increase of 1  million people in the state. c. A state with a population of 0 has 2 electoral votes. d. The number of electoral votes a state has is based on the number of members that the state has in congress. Each state has 2 senators, plus a number of members of the House of Representatives based on its population. so, the y-intercept is 2 because a hypothetical state with no population would still have 2 senators.

Explanation: a. y = 1.3 x + 2; about 0.9995; strong positive correlation. b. The number of electoral votes increases by 1.3 for every increase of 1  million people in the state. c. A state with a population of 0 has 2 electoral votes. d. The number of electoral votes a state has is based on the number of members that the state has in congress. Each state has 2 senators, plus a number of members of the House of Representatives based on its population. so, the y-intercept is 2 because a hypothetical state with no population would still have 2 senators.

Answer: a. 251 ft. b. The height of the baseball is not linear.

Explanation: a. The height after 5 seconds is 251 feet. Given that the seconds on the x-axis and height on the y-axis. the points are (0, 3), (0.5, 39), (1, 67), (1.5, 87), and (2, 99). b. The actual height after 5 seconds is about 3 feet.

Question 1. How many students in the survey above studied for the test and failed? Answer:

Question 7. You randomly survey 40 students about whether they play an instrument. You find that8 males play an instrument and 13 females do not play an instrument. A total of 17 students in the survey play an instrument. Make a two-way table that includes the marginal frequencies. Answer:

Question 8. Collect data from each student in your math class about whether they like math and whether they like science. Is there a relationship between liking math and liking science? Justify your answer. Answer:

Answer: The line y = 12.6x + 75.8 best fit for the data.

Explanation: In the above-given figure, Given that the points are (0,75), (1, 91), (2, 101), (3, 109) and (4, 129). The line y = 12.6x + 75.8 is the best fit for the data.

The vertices of a triangle are A (1, 2), B (3, 1), and C (1, – 1). Draw the figure and its image after the translation. Question 3. 4 units left Answer:

Question 4. 2 units down Answer:

Question 5. (x – 2, y + 3) Answer:

ANALYZING DATA In Exploration 1, determine how many of the indicated T-shirt are in stock at the end of the soccer season. (See Exploration 1, p. 249.) Question 6. black-and-white M

Answer: 4 T-shirts are in stock at the end of the soccer season.

Explanation: In the above-given Exploration 1, Given that The T-shirts are in stock. 4 T-shirts are in stock at the end of the soccer season.

Question 7. blue-and-gold XXL

Answer: 0 shirts.

Explanation: In the above-given Exploration 1, Given that The T-shirts are in stock. 0 T-shirts are in stock at the end of the soccer season.

Question 8. blue-and-white L

Answer: 1 T-shirt.

Explanation: In the above-given Exploration 1, Given that the T-shirts are in stock. 1 T-shirt is in stock at the end of the soccer season.

Answer: 51 students participate.

Explanation: In the above-given table, Given that male and female students are participated in the fundraiser. so 51 female students participate.

Question 10. How many male students do not participate in the fundraiser?

Answer: 30 male students do not participate.

Explanation: In the above-given table, Given that male and female students are participated in the fundraiser. so 30 male students do not participate.

Answer: 71 students are juniors. 75 students are seniors. 93 students will attend the school play. 53 students will not attend the school play. 146 students were surveyed.

Explanation: In the above-given table, Given that students of the class participate in the school play. 71 students are juniors. 75 students are seniors. 93 students will attend the school play. 53 students will not attend the school play. 146 students were surveyed.

Answer: The data plan of 78 people is limited for the cell phone company A. The data plan of  94 people is limited for the cell phone company B. The data plan of 175 people is unlimited for the cell phone company A. The data plan of 135 people is unlimited for the cell phone company B. 482 people were surveyed.

Explanation: In the above-given table, The data plan of the cell phone company are given. The data plan of 78 people is limited for the cell phone company A. The data plan of  94 people is limited for the cell phone company B. The data plan of 175 people is unlimited for the cell phone company A. The data plan of 135 people is unlimited for the cell phone company B. 482 people were surveyed.

Answer: The people who improved with treatment = 34. The people who did not improve with treatment = 10 The people who improved with no treatment = 12. The people who did not improve with no treatment = 29 Totally are about 85 people.

Explanation: The people who improved with treatment = 34. The people who did not improve with treatment = 10 The people who improved with no treatment = 12. The people who did not improve with no treatment = 29 Totally are about 85 people.

Question 16. CRITICAL THINKING What percent of students in the survey in Exercise 14 are either female or have green eyes? What percent of students in the survey are males who do not have green eyes? Find and explain the sum of these two percents. Answer:

Displaying Data Work with a partner. Analyze and display each data set in a way that best describes the data. Explain your choice of display.

Choose an appropriate data display for the situation. Explain your reasoning. Question 1. the population of the United States divided into age groups Answer:

Question 2. the number of students in your school who play basketball, football, soccer, or lacrosse Answer:

Tell whether the data display is appropriate for representing the data in Example 2. Explain your reasoning. Question 3. dot plot Answer:

Question 4. circle graph Answer:

Question 5. stem-and-leaf plot Answer:

CHOOSING A DATA DISPLAY Choose an appropriate data display for the situation. Explain your reasoning. Question 7. the percent of band students playing each instrument Answer:

Question 8. a comparison of the amount of time spent using a tablet computer and the remaining battery life Answer:

Question 2. Find and interpret the marginal frequencies. Answer:

Find the slope and the y-intercept of the graph of the linear equation. Question 3. y = 4x + 10 Answer:

Question 4. y = – 3.5x – 2 Answer:

Question 5. y – 8 = – x Answer:

Concepts, Skills, & Problem Solving

CHOOSING A DATA DISPLAY Choose an appropriate data display for the situation. Explain your reasoning. Question 7. a student’s test scores and how the scores are spread out

Answer: stem and leaf plot shows how data is distributed.

Question 8. the prices of different televisions and the numbers of televisions sold Answer:

Question 9. the outcome of rolling a number cube Answer:

Question 10. the distance a person drives each month Answer:

Question 13. WRITING When should you use a histogram instead of a bar graph to display data? Use an example to support your answer. Answer:

Question 16. REASONING What type of data display is appropriate for showing the mode of a data set? Answer:

Make a plan. Find the marginal frequencies for the data. Then use the marginal frequencies to find the probability that a randomly selected middle school student prefers action movies.

Solve and check. Use the plan to solve the problem. Then check your solution. Answer:

Question 2. An equation of the line of best fit for a data set is y = – 0.68x + 2.35. Describe what happens to the slope and the y-intercept of the line when each y-value in the data set increases by 7. Answer:

Review Vocabulary

Graphic Organizers

Chapter Self-Assessment

6.1 Scatter Plots   (pp. 237–242) Learning Target: Use scatter plots to describe patterns and relationships between two quantities.

Question 6. Describe a set of real-life data that has each relationship. a. positive linear relationship b. no relationship Answer:

6.2 Lines of Fit   (pp. 243–248) Learning Target: Use lines of fit to model data.

Question 9. Find an equation of the line of best fit for the data in Exercise 8. Identify and interpret the correlation coefficient. Answer:

6.3 TwoWay Tables   (pp. 249–254) Learning Target: Use two-way tables to represent data. You randomly survey students about participating in the science fair. The two-way table shows the results.

Question 12. How many female students do not participate in the science fair? Answer:

Question 15. For each group, what percent of the people surveyed like the food court? dislike the food court? Organize your results in a two-way table. Answer:

Question 16. Does your table in Exercise 15 show a relationship between age and whether people like the food court? Answer:

6.4 Choosing a Data Display (pp. 255–262)

Learning Target: Use appropriate data displays to represent situations.

Choose an appropriate data display for the situation. Explain your reasoning. Question 17. the numbers of pairs of shoes sold by a store each week Answer:

Question 18. the percent of votes that each candidate received in an election. Answer:

Question 20. Give an example of a bar graph that is misleading. Explain your reasoning. Answer:

Question 21. Give an example of a situation where a dot plot is an appropriate data display. Explain your reasoning. Answer:

Choose an appropriate data display for the situation. Explain your reasoning. Question 4. magazine sales grouped by price range Answer:

Question 5. the distance a person hikes each week Answer:

Question 5. A system of two linear equations has no solution. What can you conclude about the graphs of the two equations? F. The lines have the same slope and the same y-intercept. G. The lines have the same slope and different y-intercepts. H. The lines have different slopes and the same y-intercept. I. The lines have different slopes and different y-intercepts. Answer:

Question 6. What is the solution of the equation? 0.22(x + 6) = 0.2x + 1.8 A. x = 2.4 B. x = 15.6 C. x = 24 D. x = 156 Answer:

Question 8. A store records total sales (in dollars) each month for three years. Which type of graph can best show how sales increase over this time period? F. circle graph G. line graph H. histogram I. stem-and-leaf plot Answer:

Conclusion:

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