Core Connections, Course 3 Checkpoint Materials

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1 Core Connections, Course 3 Checkpoint Materials Notes to Students (and their Teachers) Students master different skills at different speeds. No two students learn eactl the same wa at the same time. At some point ou will be epected to perform certain skills accuratel. Most of these Checkpoint problems incorporate skills that ou should have developed in previous grades. If ou have not mastered these skills et it does not mean that ou will not be successful in this class. However, ou ma need to do some work outside of class to get caught up on them. Starting in Chapter 1 and finishing in Chapter 9, there are 9 problems designed as Checkpoint problems. Each one is marked with an icon like the one above. After ou do each of the Checkpoint problems, check our answers b referring to this section. If our answers are incorrect, ou ma need some etra practice to develop that skill. The practice sets are keed to each of the Checkpoint problems in the tetbook. Each has the topic clearl labeled, followed b the answers to the corresponding Checkpoint problem and then some completed eamples. Net, the complete solution to the Checkpoint problem from the tet is given, and there are more problems for ou to practice with answers included. Remember, looking is not the same as doing! You will never become good at an sport b just watching it, and in the same wa, reading through the worked eamples and understanding the steps is not the same as being able to do the problems ourself. How man of the etra practice problems do ou need to tr? That is reall up to ou. Remember that our goal is to be able to do similar problems on our own confidentl and accuratel. This is our responsibilit. You should not epect our teacher to spend time in class going over the solutions to the Checkpoint problem sets. If ou are not confident after reading the eamples and tring the problems, ou should get help outside of class time or talk to our teacher about working with a tutor. Checkpoint Topics 1. Operations with Signed Fractions and Decimals. Evaluating Epressions and Using Order of Operations 3. Unit Rates and Proportions. Area and Perimeter of Circles and Composite Figures 5. Solving Equations. Multiple Representations of Linear Equations 7. Solving Equations with Fractions ( Fraction Busters ) 8. Transformations 9. Scatterplots and Association Checkpoints 95

2 Checkpoint 1 Problem 1-1 Operations with Signed Fractions and Decimals Answers to problem 1-1: a:! 17, b:.1, c: 1 5, d:!1 1 3, e: 3 7, f:. 1 Use the same processes with signed fractions and decimals as are done with integers (positive and negative whole numbers.) Eample 1: Compute ( 9 0 ) Solution: When adding a positive number and a negative number, subtract the values and the number further from zero determines the sign ! 9 0 = 1 3 " 0 0 +! 9 0 " 3 3 = 0 0 +! 7 0 =! 7 0 Eample : Compute!1.5! (!3.9) Solution: Change an subtraction problem to addition of the opposite and then follow the addition process.!1.5!!3.9 ( ) "! =! =.5 Eample 3: Compute Solution: With multiplication or division, if the signs are the same, then the answer is positive. If the signs are different, then the answer is negative.! =! 5 15 =! 5 " 15 =! 5 " ""3" 5 =! 1 Now we can go back and solve the original problems. a. Both numbers are negative so add the values and the sign is negative. b. Change the subtraction to addition of the opposite. c. The signs are the same so the product is positive. Multipl as usual. d. The signs are different so the quotient is negative. Divide as usual. 9 Core Connections, Course 3

3 e. When adding a positive number with a negative number, subtract the values and the number further from zero determines the sign. f. The signs are the same so the quotient is positive. Divide as usual. Here are some more to tr. Compute the value of each of the following problems with fractions and decimals. ( ) 1.! !! ! !1 7 + (! 3 ) 5.! (!.5)"(!3.3) 7.! 7 1! (! 3 8 ) 9.!1 1! (!3 1 ) 10. ( 5 9 )!(" 3 7 ) 11..! !3! (!.7) 13.! 7 9 " 3 1.! 3 5! !5 1! (! 1 ) (! 15 ) ! (!1 5 8 ) 19. (0.3)!("0.03) 0.!8.! ! (! 7 ).! 1 7 "! 5 3.! ! 3 10! !3. + (!3.5).!7.5! ! 7 9 " ! 1 5! ! 3!! 5 ( 7 ) 30. 3! ( )!(" 5 ) 3.! 1 " 3 Checkpoints 97

4 Answers 1.! ! 1 1.! ! ! !8 1.! ! ! ! 19.!5 1 5.!3.05.!.3 7.! ! ! ! ! Core Connections, Course 3

5 Checkpoint Problem -89 Evaluating Epressions and Using Order of Operations Answers to problem -89: a: 8, b: 1, c:, d: 17, e: 5, f: 15 In general, simplif an epression b using the Order of Operations: Evaluate each eponential (for eample, 5 = 5!5 = 5 ). Multipl and divide each term from left to right. Combine like terms b adding and subtracting from left to right. But simplif the epressions in parentheses or an other epressions of grouped numbers first. Numbers above or below a fraction bar are considered grouped. A good wa to remember is to circle the terms like in the following eample. Remember that terms are separated b + and signs. Eample 1: Evaluate 3 + for =!5 Solution: (!5)! 3(!5) + (5)! 3(!5) + 50! (!15) = 7 Eample : Evaluate 5 +! ( ) for = 3,! = Solution: 5!3+"!3! ( ) ( ) ( ) =!1 5!3+!3! 5 1!5 Now we can go back and solve the original problems. a z (!) + 3(!3) + 5! +!9 + 5 =!8 b.! (!)! (!3)! + 3 = 1 ( ) c. + z ( ) ( ) = (!1) =!!+!3 5!5 5 d. 3! +1 3(!)! (!) +1 3()! (!) +1 1! (!) +1 = 17 e. 3( +! ) 3(!3)(! + (!)! (!3)) ( ) 3(!3)! +! (!3) 3(!3)(5) =!5 f.!z (1!)!!(5) (1!(!)) (!3)!(!)!5(1!(!)) (!3)!(!) =!5(5)!1 = 15 Checkpoints 99

6 Here are some more to tr. Evaluate each epression for =,! =!,!z =!3. 1.! 3. z z!.! z 5.! +. z! 8! ( ) ! ! z! 11.! 3 1. ( + z)! z( +! ) (+1) 15. ( + ) 1. ( + )(3+ z) 17.! 5 + 3z z! (3+ z)! 1. z + 8z!. 3! 3. z! + + z.! (z!5) 3! Answers ! Core Connections, Course 3

7 Checkpoint 3 Problem Unit Rates and Proportions Answers to problem 3-117: a: $0.8 per pound, b: 0. g per cm, c:! $3.50, d: $11.88, e:! 19 shots, f: teaspoons A rate is a ratio comparing two quantities. A unit rate is simplified to have a denominator of 1. Either unit rates or proportions (equations with ratios) can be used to solve ratio problems. Solutions can also be approimated b looking at graphs of lines passing through the origin and the given information. Eample 1: Sam paid $.95 for 3 price (dollars per pound)? pound of his favorite trail-mi. What is the unit $.95 Solution: The unit price is. To create a unit rate we need a denominator of one.!pound $.95! 3 3!pound! 3 3 = $.95! 3 1!pound = $.95!!= $.0 per pound 3 Eample : Based on the table, what is the unit growth rate (meters per ear)? + Solution: Height (m) Years Note: This same information could be determined b looking at a graph. Eample 3: In line at the movies are 1 people in front of ou. If ou count 9 tickets sold in 70 seconds, how long will it take before ou bu our ticket? Solution: The information ma be organized in a proportion 9!tickets the proportion 9 70 = 1 70 seconds = 1!tickets ields 9 = 100 so! seconds or! 19 minutes.. Solving Checkpoints 501

8 Now we can go back and solve the original problems. a. $1.89! $ = 9 pound 9!pound! 9 = $1.89! 9 1!pound = $1.89! = $0.8 per pound 9 b. For ever increase of three grams, the length increases b five centimeters. 3!grams 5 cm = 3 5 gram cm = 0. g per cm c. The graph passes through (10 bottles, $35). $35 = $3.50 per bottle. 10!bottles d. 00!vitamins 500 vitamins =! 00 = 37.50! " $11.88 $.75 e. 7!made 85 attempts =! 85 = 100! " 19 made 00 attempts f. 1 teaspoon = 3 cup 5 cups! 3 = 5! = 10 3 = teaspoons Here are some more to tr. In problems 1 through 10 find the unit rate. In problems 11 through 0 use ratios to solve each problem. 1. Am usuall swims 0 laps in 30 minutes. What is her rate in laps per minute?. For 3 of a pound of peanuts, Jimm paid $.35. What is the unit rate (dollars per pound)? 3. In 7 minutes, Jack tpes 511 words. How man words does he tpe per minute?. Using the graph at right, determine how long it takes to mow an acre of grass. 5. If Christ and Lucas eat cups of popcorn in 30 minutes, how man minutes does it take them to eat a cup of popcorn? Grass (acres) 3. In the past ear, Ana has spent $7, 0 on her rent. What is the unit cost (dollars per month)? 7. While constructing his house, the little pigg found that bundles of sticks contained 08 sticks. How man sticks are in each bundle? Time (hours) 8. Meg knows that it takes 1 minutes to read 8 7 of the chapter. How long will it take her to read the entire chapter? 50 Core Connections, Course 3

9 9. At the frozen ogurt shop, Colin pas $3.8 for his treat, which weighs ounces. What is the unit cost (dollars per ounce)? 10. When Sarah does her math homework, she finishes 10 problems ever 1 minutes. How long will it take for her to complete 35 problems? 11. Ben and his friends are having a TV marathon, and after hours the have watched 5 episodes of the show. About how long will it take to complete the season, which has episodes? 1. The ta on a $00 painting is $3. What should be the ta on a $700 painting? 13. Use the table at right to write and solve a ratio for how long it will take Isadora to earn $10. $ Earned Das Worked 1 1. While baking, Hannah discovered a recipe for cookies that required 3 cups of sugar for ever 1 cups of flour. How man cups of sugar will she need for cups of flour? 15. M brother grew 1 3 inches in 1 months. If he continues to grow at the same rate, how much should he grow in one ear? 1. On his afternoon jog Chris took minutes to run 3 3 miles. How man miles can he run in 90 minutes? 17. If Caitlin needs 1 7 cans of paint for each room in her house, how man cans of paint 8 will she need to paint the 9-room house? 18. Stephen receives gumballs for ever two hours of homework he does. If he wants gumballs, how man hours will he need to work? Answers 1. 3 laps minute. $3.13 per pound words minute. 3 hour minutes cup. $0 month 7. 5 sticks bundle minutes 9. $0.0 ounce 10. minutes hours 1. $ das cups inches miles cans hours Checkpoints 503

10 Checkpoint Problem -71 Area and Perimeter of Circles and Comple Figures Answers to problem -71: a: A! 8.7!cm, C! 18.85!cm, b: A! 78.5 ft, C! 31. ft, c: A! ft, P! 9.13!ft, d: A! 5.53!ft, P! 85.13!ft For circles, the formulas for area and circumference (perimeter) are: A = r! and C = r! where r = radius of the circle For comple figures (made of from circles or parts of circles with other shapes), divide the figure into more recognizable parts. Then find the sum of the area of the parts. When finding the perimeter of a comple region, be sure that the sum onl includes the edges on the outside of the region. cm Eample 1: Eample : cm cm 5 cm 3 m 9 cm Area of half of a circle: Circumference of half of a circle plus the diameter Area of square plus triangle Add all sides for perimeter Now we can go back and solve the original problems. a. A = r! = 3 "! # 8.7 cm C = r! = " 3"! # cm b. Radius is 5 ft: A = r! = 5 "! # 78.5 ft C = r! = "5 "! # 31. ft c. The area of the sector is 1 3 of the area of circle: A = 1 3 r! = 1 3 "1 "! # ft The perimeter is 1 3 of the circumference added to the length of two radii: P = 1 3!r" + r = 1 3!!1!" +!1 # 9.13 ft d. The area is the area of a rectangle added to the area of half a circle: A = lw + 1 r! = 1 " + 1 "8 "! # 5.53 ft The perimeter is the length of the three sides added to half the circumference of the circle: P = l + w + l + 1!r" = !!8!" # ft 50 Core Connections, Course 3

11 Here are some more to tr. Find the area and perimeter of the following figures. Note: All angles that look like right angles are right angles. 1. Circle with a radius of 7.3 m. Circle with a diameter of 8.5 ft. 3. Circle with a radius of 11 ds. Circle with a diameter of 5 mm 5. Circle with a radius of.1 cm. Circle with a diameter of 15 in. 7. Find the area of the shaded region. 8. Semi-circle 18 m 9 m 1 m 7 m 0 in Trapezoid on a rectangle 10 ft 10 ft cm 3. cm 3. cm 1 ft 3 cm 1 cm 1 cm 8 ft cm 1 ft 10 cm 11. Rectangle and a semi-circle 1. 1 m m 10 in. 17. in. 7 in. in. in. 5 in. 13. Find the area of the shaded region. 1. Semi-circle on a rectangle The diameter of the circle is ft. ft 8 ft 8 m 1 m 1 ft Checkpoints 505

12 15. Quarter-circle 1. 8 cm 5 in. 5.1 cm 5 cm 5.1 cm 5 cm 10 ft ft 7 ft 5 ft 1 ft 10 in. ft 8 in. 10 cm 13 in. in.. in. 19. Find the area of the shaded region. 0. Circle with diameter = 0 cm 7 in. 1 in. 8 in. 0 cm 10 in. 1.. Find the area of the center region 1.8 m bounded b four congruent 19. m quarter circles. 18. m ft 7. m 18. m 1.8 m 19. m ft 3. Find the area of the shaded region.. A half-circle cut out of a square. bounded b a square and a circle. 5 in. 5 cm 5 cm 50 Core Connections, Course 3

13 Answers 1. A = 17. m, P = 5.9 m. A = 5.7 ft, P =.7 ft 3. A = ds, P = 9.1 ds. A = 13.7 mm, P = 13. mm 5. A = 1.1 cm, P = 13.3 cm. A = 17.7 in., P = 7.1 in. 7. A = 18.5 m 8. A = 8.3 in., P = 10.8 in. 9. A = 19 ft, P = 8 ft 10. A = 1 cm, P = 8. cm 11. A = m, P = 8.8 m 1. A = 8 in., P = 53. in. 13. A = 83. ft 1. A = 5.9 m, P = m 15. A = 19. in., P = in. 1. A = 95 cm, P = 38. cm 17. A = 9 ft, P = 7 ft 18. A = 90 in., P = 3. in. 19. A = in. 0. A = 87. cm, P = cm 1. A = 7 m, P = 10. m. A = 13. ft 3. A = 5.7 in.. A = 5.3 cm, P = 30.7 cm Checkpoints 507

14 Checkpoint 5 Problem 5-1 Solving Equations Answers to problem 5-1: a: =!, b: = 1 1, c: = 3, d: no solution Equations ma be solved in a variet of was. Commonl, the first steps are to remove parenthesis using the Distributive Propert and then simplif b combining like terms. Net isolate the variable on one side and the constant terms on the other. Finall, divide to find the value of the variable. Note: When the process of solving an equation ends with different numbers on each side of the equal sign (for eample, = ), there is no solution to the problem. When the result is the same epression or number on each side of the equation (for eample, + 3 = + 3) it means that all real numbers are solutions. Eample 1: Solve + 3 = + 9 Solution: Check: Eample : Solve + ( + 1) = 3 + ( + 5) Solution: Check: 508 Core Connections, Course 3

15 Now we can go back and solve the original problems. a =!!1 =!8 =! b. c. d. 3! +1 =!! ! 3 = 3! 5!3 =!5!3 "!5 # no solution Here are some more to tr. Solve each equation. 1.! 3 =! = !! 3 = (! ).!! =! 5 5.!( + 3) =!.! + =! 5! ! =! ! 3 + = ! 8! = 10.! + 3 = 11.! + 3 =!! 1. (! ) + = 5 13.!! 3 =! = ! 1! 1 =! 3! (!5 + ) ! + = ! + 3! 1 = ! 7 =!! = 3!! ( + ) (!! ) = + Checkpoints 509

16 Answers 1. =. = 1 3. = 1 5. = 3 5. = 1. = 7 7. = 5 8. no solution 9. = 10. = = 0 1. = = 1 1. = = 1. = = 18. = 19. = 1 0. = 510 Core Connections, Course 3

17 Answers to problem -109: Checkpoint Problem -109 Multiple Representations of Linear Equations a: Figure # b: # of Tiles # of Tiles = Figure # c: d: =! If ou know one representation of a linear pattern, it can also be represented in three other different was. Linear representations ma be written as equations in the form = m + b where m is the growth factor and b is the starting value. Graph Table Pattern Rule Checkpoints 511

18 Eample 1: Using the pattern at right, create an! table, a graph, and write the rule as an equation. Figure 0 Figure 1 Figure Figure 3 Solution: The number of tiles matched with the figure number gives the! table. Plotting those points gives the graph. Using the starting value for b and the growth factor for m gives the information to write the equation in = m + b form. Figure # () # of Tiles () The starting number is 3 tiles and the pattern grows b tiles each figure so the equation is = + 3. Total Number of Tiles Figure Number Eample : Create an! table and the rule (equation) based on the graph at right. Solution: Place the given points in a table: The starting value is 1 and the growth factor is 3. The equation is: Core Connections, Course 3

19 Now we can go back and solve the original problems. a. Start b making a table from the tile pattern. Since the starting value is and the growth factor is the equation is = +. b. Make a table b replacing in the rule with 0, 1,, and then computing. See the table above. Plot the points from the table to get the graph shown with the given answers. c. Looking at the table, the -value changes b 9 when the the -value changes b 3. Therefore the -value is changing b 3 when the -value is changing b 1. A completed table is shown below. Since the starting value is 1 and the growth factor is 3, the table and equation are: See the graph given with the answers. Figure # # of Tiles d. Again the table above is determined from the points on the graph. Looking at the table, the starting value is and the growth factor is so the equation is =! Here are some more to tr. For each situation, complete a Multiple Representations of Linear Equations web b finding the missing! table, graph, and/or rule. Since there are man possible patterns, it is not necessar to create one = 3!. Figure 1 Figure Figure =! Checkpoints 513

20 Figure 1 Figure Figure 3 9. =! Answers = = = Core Connections, Course 3

21 = = = = Checkpoints 515

22 Checkpoint 7 Problem Solving Equations with Fractions (Fraction Busters) Answers to problem 7-115: a: = 15 = 3 3, b:! $1.7, c: = 5, d: (1, 1) Equations are often easier to solve if there are no fractions. To eliminate fractions from an equation, multipl both sides of the equation b the common denominator. Eample 1: Solve Solution: Start b multipling both sides of the equation b 1 (the common denominator.) Simplif and then solve as usual. Eample : Solve Solution: Two decimal places means that the common denominator is 100. Multipl both sides b 100 and solve as usual. Now we can go back and solve the original problems. a. b. c. d. 51 Core Connections, Course 3

23 Here are some more to tr. Solve each equation or sstem of equations = 5. = = =! 0. = 3 = 3! 9 5.! = 3.! 3 = +! = 3! 1 8.!! 1 =! = = + 1 =! =!3 + =! = =! = = = 7 + =! 1! =! =! 1 = = 3! 10 = 3! 19.!1 = =! 10 3 Checkpoints 517

24 Answers 1. =. = 1,! = 3. = 18. = 9,! = 3 5. =. = 0 7. = = 7 9. = = 3,! = = 3,! =! 1. = = 9 1. = 15. =!1, = 1. =! 1,! =! = 3,! = =,! =!1 19. =.5 0. = Core Connections, Course 3

25 Checkpoint 8 Problem Transformations Answers to problem 8-135: (Given in the order X, Y, Z) a: (, ), (, 0), (1, ), b: (,!1),!(,!3),!(1,!1), c: (!1, ),!(!3, ),!(!1,1), d: (!8,!),!(!8, ),!(!, ) Rigid transformations are was to move an object while not changing its shape or size. A translation (slide) moves an object verticall, horizontall or both. A reflection (flip) moves an object across a line of reflection as in a mirror. A rotation (turn) moves an object clockwise or counterclockwise around a point. A dilation is a non-rigid transformation. It produces a figure that is similar to the original b proportionall shrinking or stretching the figure from a point. Eample 1: Translate (slide)!abc left si units and down three units. Give the coordinates of the new!xyz. Y A B Solution: The original vertices are A (0, ), B (, 5), and C (5, 1). The new vertices are X (, 1), Y (, ), and Z ( 1, ). X Z C Eample : Reflect (flip)!abc across the -ais to get!pqr. Give the coordinates of the new!pqr. Now reflect (flip)!abc across the -ais to get!xyz. Give the coordinates of the new!xyz. X Z Y B C A Solution: The ke is that the reflection is the same distance from the ais as the original figure. For the first reflection the points are P!(,!), Q!(1,!), and R!(3,!). For the second reflection the points are X!(!, ), Y!(!1, ), and Z!(!3, ). Q P R Eample 3: Rotate (turn)!abc counterclockwise 90 about the origin to get!mno. Give the coordinates of the new!mno. Then rotate!mno counterclockwise another 90 to get!xyz. Give the coordinates of the new!xyz. O Y Z X N M A C B Solution: After the first 90 rotation, the coordinates of A (,0), B (,0), and C (5, ) became M (0, ), N!(0, ), and O (, 5). Note that each original point (, ) became (!, ). After the net 90 rotation, the coordinates of the vertices are now X!(!, 0), Y (, 0), and Z ( 5, ). After the 180 rotation each of the points (, ) in the original!abc became (!,!). Similarl a 70 counterclockwise rotation or a 90 clockwise rotation about the origin takes each point (, ) to the point (,!). Checkpoints 519

26 Eample : Dilate (enlarge/reduce)!abc b a scale factor of 1 from the origin to get!xyz. Give the coordinates of the new!xyz. B Y Solution: Multipling the coordinates of the vertices of!abc b the scale factor gives the coordinates of the vertices of the new!xyz. The are X!(!, 0), Y!(0, 3) and Z!(1,!1). A X Z C Now we can go back and solve the original problems. a. Sliding the triangle two units to the right and three units up increases the -values b and the -values b 3. See the answers above. b. Reflecting the triangle across the -ais changes the -values to the opposite and keeps the -values the same. See the answers above. c. Rotating the triangle 90 clockwise moves the triangle into the second quadrant. The original -value becomes the new -value and the opposite of the original -value becomes the -value. See the answers above. d. Dilating the triangle b a scale factor of two multiplies all of the coordinates b two. See the answers above. Here some more to tr. For the following problems, refer to the figures below: Figure A Figure B Figure C B C A B C A B A C State the new coordinates after each transformation. 1. Translate figure A left units and down units.. Rotate figure C 180 counterclockwise about the origin. 3. Reflect figure B across the -ais.. Reflect figure B across the -ais. 5. Translate figure A right units and up 1 unit.. Reflect figure C across the -ais. 50 Core Connections, Course 3

27 7. Rotate figure B 70 counterclockwise about the origin. 8. Translate figure C left 1 unit and up units. 9. Rotate figure A 90 clockwise about the origin. 10. Reflect figure B across the line = Dilate figure B b a scale factor of Rotate figure A 180 counterclockwise about the origin. 13. Translate figure B units down and units to the right. 1. Dilate figure A b a scale factor of. 15. Rotate figure C 90 clockwise about the origin. 1. Reflect figure A across the -ais. 17. Dilate figure C b a scale factor of Translate figure C 3 units right and units down. 19. Rotate figure C 180 about the origin clockwise. 0. Dilate figure A b a scale factor of. Answers 1. ( 1, ), (1, ), (3,0). (, ), (, ), (, 3) 3. (5, ), (1, ), (0, 5). ( 5, ), ( 1, ), (0, 5) 5. (5, 1), (7, 5), (9, 3). (, ), (, ), (, 3) 7. (, 5), (, 1), (5, 0) 8. ( 5, ), (3, ), ( 3, 1) 9. (0, 1), (, 3), (, 5) 10. (5, ), (1, ), (0, 1) 11. ( 15, ), ( 3, ), (0, 15) 1. ( 1, 0), ( 3, ), ( 5, ) 13. ( 1, ), (3, ), (, 1) 1. (, 0), (1, 1), (0, 8) 15. (, ), (, ), ( 3, ) 1. ( 1, 0), ( 3, ), ( 5, ) 17. (, 1), (, 1), ( 1, 1.5) 18. ( 1, 0), (7, 0), (1, 5) 19. (, ), (, ), (, 3) 0. (, 0), (, 8), (10, ) Checkpoints 51

28 Checkpoint 9 Problem 9-50 Scatterplots and Association Answers to problem 9-50: a: bo plot b: scatterplot c: See graph at right. d: strong linear negative association e: =! f: $17,000 g: A slope of 3 means the car is losing $3000 in value each ear, -intercept of 35 means the cost when new was $35,000. Listed Cost (in 1000s) An association (relationship) between two numerical variables on a graph can be described b its form, direction, strength, and outliers. When the association has a linear form, a line a best fit can be drawn and its equation can be used to make predictions about other data Age of Car (ears) Eample 1: Describe the association in the scatterplot at right. Solution: Looking from left to right, ecept for point (e), the points are fairl linear and increasing. This is a moderate, positive linear association. Point (e) is an outlier. e Eample : For the scatterplot, draw a line of best fit and determine the equation of the line. Solution: Use a ruler or straightedge to draw the line that approimates the trend of the points. If it is not a perfect line, approimatel the same number of points will be above and below the line of best fit. To determine the equation of the line, draw in a slope triangle and determine the ratio of the vertical side to the horizontal side. In this case it is!30 =!. Estimate the -intercept b 5 looking at where the line intersects the -ais. In this case, it is approimatel 30. The equation of an non-vertical line ma be written in the form = m + b where m is the slope and b is the -intercept Core Connections, Course 3

29 Now we can go back and solve the original problem. a. Since the costs are a single set of data, a bo plot is a convenient wa to show the distribution. b. Age and cost are two sets of related data so a scatterplot is appropriate. c. See the graph given in the answers. d. Reading from left to right, the scatterplot is decreasing, linear and the points are close to the line of best fit. This is a strong, linear, negative association. e. Looking at the line of best fit, the slope triangle has a ratio of! 3 and the -intercept is 1 approimatel 35. Placing that information into the equation of a line, = m + b, ields =! f. Substituting = into the equation of part (e) ields =!3() + 35 = 17, $17,000. g. Slope represents the rate of change. A rate of change of 3 means that the value is decreasing b 3 units (in this case each unit is $1000) per ear. The -intercept represents the value at ear zero or a new car. Here are some more to tr. In problems 1 through, describe the association. 1.. Height (cm) Price (Thousand $) Shoe Size Horsepower 3.. Height (feet) Mileage (mpg) Age 1 (ears) 1 5 Car Weight (1000s of lbs.) Checkpoints 53

30 In problems 5 through 8 plot the data, draw a line of best fit, and approimate the equation of the line. Distance to Airport (mi) Cost of Shuttle ($) Eercise/Month (hours) Rate of Heart Attack/ Time Spent Studing (hours) Score on Test Time Since Purchase (hours) Number of Cookies Answers 1. strong positive association. no association 3. strong positive association. strong negative association 5.. Cost ($) Rate of Heart Attacks (per 1000) Distance to Airport (miles) Amount of Eercise per Month (hours)! = 1 + 8! = " Score on Test Number of Cookies Time Spent Studing (hours) 5 10 Time Since Purchase (hours)! = + 3! = Core Connections, Course 3