core connections, course 2 checkpoint...

Checkpoints © 2013 CPM Educational Program. All rights reserved. 1

Core Connections, Course 2 Checkpoint Materials

Notes to Students (and their Teachers)

Students master different skills at different speeds. No two students learn exactly the same way at the same time. At some point you will be expected to perform certain skills accurately. Most of the Checkpoint problems incorporate skills that you should have been developing in grades 5 and 6. If you have not mastered these skills yet it does not mean that you will not be successful in this class. However, you may 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 and numbered according to the chapter that it is in. After you do each of the Checkpoint problems, check your answers by referring to this section. If your answers are incorrect, you may need some extra practice to develop that skill. The practice sets are keyed to each of the Checkpoint problems in the textbook. Each has the topic clearly labeled, followed by the answers to the corresponding Checkpoint problem and then some completed examples. Next, the complete solution to the Checkpoint problem from the text is given, and there are more problems for you to practice with answers included. Remember, looking is not the same as doing! You will never become good at any sport by just watching it, and in the same way, reading through the worked examples and understanding the steps is not the same as being able to do the problems yourself. How many of the extra practice problems do you need to try? That is really up to you. Remember that your goal is to be able to do similar problems on your own confidently and accurately. This is your responsibility. You should not expect your teacher to spend time in class going over the solutions to the Checkpoint problem sets. If you are not confident after reading the examples and trying the problems, you should get help outside of class time or talk to your teacher about working with a tutor.

Checkpoint Topics

1. Area and Perimeter of Polygons 2. Multiple Representations of Portions 3. Multiplying Fractions and Decimals 5 Order of Operations 6. Writing and Evaluating Algebraic Expressions 7A. Simplifying Expressions 7B. Displays of Data: Histograms and Box Plots 8. Solving Multi-Step Equations 9. Unit Rates and Proportions

2 © 2013 CPM Educational Program. All rights reserved. Core Connections, Course 2

Checkpoint 1 Problem 1-141

Area and Perimeter of Polygons Answers to problem 1-141: a. 96 cm2, 40 cm; b. 22 in.2, 25.05 in.; c. 144 cm2, 52 cm;

d. 696.67 m2, 114.67 m Area is the number of square units in a flat region. The formulas to calculate the area of several kinds of polygons are: RECTANGLE PARALLELOGRAM TRAPEZOID TRIANGLE

h

b

h

b

h

b2

b1

h

b

h

b A = bh A = bh A = 1

2 b1 + b2( ) h A = 12 bh

Perimeter is the distance around a figure on a flat surface. To calculate the perimeter of a polygon, add together the length of each side. Example 1: Example 2: Compute the area and perimeter. Compute the area and perimeter. parallelogram triangle A = bh = 6 !4 = 24 !feet2 A = 1

2 bh =12 !6 !7 = 21!cm

2 P = 6 + 6 + 5 + 5 = 22!feet P = 6 + 8 + 9 = 23!cm Now we can go back and solve the original problem. a. Rectangle: A = bh = 12 !8 = 96!cm2 ; P = 8 + 8 +12 +12 = 40!cm

b. Triangle: A = 12 bh =

12 !11!4 = 22!in.

2 ; P = 11+ 9.05 + 5 = 25.05!in. c. Parallelogram: A = bh = 16 !9 = 144 !cm2 ; P = 16 +16 +10 +10 = 52!cm d. Trapezoid: A = 1

2 b1 + b2( ) h = 12 25 + 44.67( ) !10 = 696.67!m2 ;

P = 21+ 25 + 24 + 44.67 = 114.67!m

6 feet

5 feet

6 feet

5 feet 4 feet 9 cm

8 cm 7 cm

6 cm


Here are some more to try. Find the area and perimeter of each figure.

1. 2.

3. Trapezoid 4. Parallelogram

5. Parallelogram 6.

7. Trapezoid 8.

9. Parallelogram 10.

11 cm

9 cm

10 in.

12 in.

7.2 in. 6 in.

35 feet

17.8 feet 12.5 feet

20 feet feetfeet

12.7 feet

16 cm

10.8 cm

10.8 cm 10 cm

9 fe

et

5 feet

4 feet 9 fe

et

42 fe

et

27 feet

39.2 feet

24.5 feet

29 in.

18 in. 18 in. 16 in.

12 in

.

5 in.

9.9 in.

4 in.

8 cm

9 cm

15 cm

8.1 cm 9.3 cm 6 m

6 m

10 m

6.3 m


11. 12. Trapezoid

13. Trapezoid 14.

15. Parallelogram 16. Trapezoid Answers:

1. 99 cm2, 40 cm 2. 36 in.2, 29.2 in.

3. 343.75 feet2, 85.5 feet 4. 160 cm2, 53.6 cm

5. 36 feet2, 28 feet 6. 24 in.2, 26.9 in.

7. 96 cm2, 41.4 cm 8. 18 m2, 22.3 m

9. 464 in.2, 94 in. 10. 514.5 feet2, 108.2 feet

11. 144 = 3 12 feet

2 , 304 = 7 12 feet 12. 457 in.2, 90.7 in.

13. 400 feet2, 86 feet 14. 7.65 in.2, 13.3 in.

15. 90 cm2, 40 cm 16. 1.78 m2, 9.1 m

2 feet

10 cm

10 cm 10 cm 9 cm

32.7

in.

21 in.

13 in. 20 in.

24 in.

16 feet

32 feet

18 feet

19.8 feet 16.2 feet

0.5 m

4.4 m

2.7 m

0.8 m 1.2 m

5 in.

5.1 in.

3.2 in. 3 in.



Multiple Representations of Portions

Answers to problem 2-120: a. 43%, 43100 , b. 910 , 0.9, 90% , c. 39100 , 0.39 , d. 64%, 0.64 Portions of a whole may be represented in various ways as represented by this web. Percent means “per hundred” and the place value of a decimal will determine its name. Change a fraction in an equivalent fraction with 100 parts to name it as a percent. Example 1: Name the given portion as a fraction and as a percent. 0.3 Solution: The digit 3 is in the tenths place so 0.3= three tenths = 3

10 . On a diagram or a hundreds grid, 3 parts out of 10 is equivalent to 30 parts out of 100 so 310 =

30100 = 30% .

Example 2: Name the given portion as a fraction and as a decimal. 35% Solution: 35% = 35

100 = thirty-five hundredths = 0.35 ; 35100 =720

Now we can go back and solve the original problem. a. 0.43 is forty-three hundredths or 43100 = 43% b. nine tenths is 910 =

910 !

1010 =

90100 = 90% ; 910 = 0.9

c. 39% = 39

100 =!thirty-nine hundredths = 0.39 d. 1625 =

1625 !

44 =

64100 = 0.64 = 64%

words or

pictures

fraction

decimal percent

Representations of a Portion


Here are some more to try. For each portion of a whole, write it as a percent, fraction, and a decimal.

1. 6% 2. 0.35

3. 14 4. 2

5

5. 0.16 6. 87%

7. 1325 8. 21%

9. 750 10. 0.050

11. 65% 12. 3.7%

13. 710 14. 0.66

15. 1920 16. 20%

17. 0.23 18. 1.0

19. 135% 20. 77100

Answers:

1. 6100 =

350 , 0.06 2. 35%, 35100 =

720

3. 25%, 0.25 4. 40%, 0.4

5. 16%, 16100 =425 6. 87

100 , 0.87

7. 52%, 0.52 8. 21100 , 0.21

9. 14%, 0.14 10. 5%, 5100 =120

11. 1320 , 0.65 12. 37

1000 , 0.037

13. 70%, 0.7 14. 66%, 66100 =3350

15. 95%, 0.95 16. 20100 =

15 , 0.2

17. 23%, 23100 18. 100%, 100100 =11

19. 135100 = 1

35100 = 1

720 ,1.35 20. 77%, 0.77



Multiplying Fractions and Decimals

Answers to problem 3-110: a. 920 , b. 15

, c. 4 29 , d. 7 15 , e. 12.308, f. 0.000208 To multiply fractions, multiply the numerators and then multiply the denominators. To multiply mixed numbers, change each mixed number to a fraction greater than one before multiplying. In both cases, simplify by looking for factors than make “one.” To multiply decimals, multiply as with whole numbers. In the product, the number of decimal places is equal to the total number of decimal places in the multiplied numbers. Sometimes zeros need to be added to place the decimal point. Example 1: Multiply

3

8!

4

5 Example 2: Multiply

3

1

3!2

1

2

Solution: Solution: 38 !45 "

3!48!5 "

3! 42! 4 !5 "

310 3 1

3 !212 "

103 ! 5

2 "10!53!2 " 5!2 !5

3!2 " 253 or 8 1

3

Note that we are simplifying using Giant Ones but no longer drawing the Giant One. Example 3: Multiply 12.5 !0.36 Solution: Now we can go back and solve the original problem. a. 23 !

25 "

2!23!5 "

415 b. 710 !

27 "

7 !25!2 ! 7 " 1

5

c. 2 13 !2

12 "

73 !

52 "

7!53!2 " 35

6 or 5 56 d. 1 1

3 !216 "

43 !

136 " 2!2 !13

3!2 !3 " 269 or 2 8

9

e. f.

(one decimal place)

(two decimal places)

(three decimal places)


Here are some more to try. Multiply the fractions and decimals below.

1. 0.08 !4.7 2. 0.21!3.42

3. 4

7!1

2 4. 5

6!3

8

5. 8

9!3

4 6. 7

10!3

4

7. 3.07 !5.4 8. 6.57 !2.8

9. 5

6!3

20 10. 2.9 !0.056

11. 6

7!4

9 12. 3 1

7!1 25

13. 2

3!5

9 14. 3

5!9

13

15. 2.34 !2.7 16. 2 13!4 4

5

17. 4 35!1

2 18. 3

8!5

9

19. 0.235 !0.43 20. 421!0.00005 Answers:

1. 0.376 2. 0.7182

3. 2

7 4. 5

16

5. 2

3 6. 21

40

7. 16.578 8. 18.396

9. 1

8 10. 0.1624

11. 8

21 12. 4 2

5

13. 10

27 14. 27

65

15. 6.318 16. 11 15

17. 2 3

10 18. 5

24

19. 0.10105 20. 0.02105



Order of Operations Answers to problem 5-148: a: 20, b: –4

In general, simplify an expression by using the order of operations: • Evaluate each exponential (for example, 52 = 5 !5 = 25 ). • Multiply and divide each term from left to right. • Combine like terms by adding and subtracting from left to right.

But simplify the expressions in parentheses or any other expressions of grouped numbers first. Numbers above or below a “fraction bar” are considered grouped. A good way to remember is to circle the terms like in the following example. Remember that terms are separated by + and – signs.

Example 1: Simplify 12 ÷ 22 ! 4 + 3(1+ 2)3 Simplify within the circled terms: Be sure to perform the exponent operations before dividing. 12 ÷ 22 = 12 ÷ 2 !2 = 3

Then perform the exponent operation: 33 = 3!3!3= 27 Next, multiply and divide left to right: 3(27) = 81 Finally, add and subtract left to right: 3 – 4 = –1 Example 2: Simplify !32 ! 2+7

3 + 8 ÷ 12( )

Simplify within the circled terms: !32 = !3"3= !9 2+7

3 = 93 = 3!!!!8 ÷

12 = 8 !

21 = 16

Then add and subtract, left to right. Now we can go back and solve the original problem.

a. 16 ! 23 ÷ 8 + 516 ! 2 "2 "2 ÷ 8 + 516 ! 4 "2 ÷ 8 + 516 ! 8 ÷ 8 + 516 !1+ 515 + 520

b. (!2 + 6)2 ! 32( ) "14 +1

(4)2 ! 422 +1

16 ! 21+1!5 +1!4

12 ÷ 22 ! 4 + 3(1+ 2)3

3! 4 + 3(3)3

3! 4 + 3(27)

3! 4 + 81

!1+ 8180

!32 ! 2+73 + 8 ÷ 1

2( )!9 ! 3+16!12 +164


Here are some more to try. 1. 10 ! 12 + ("6)("3) 2.

!5+(!6) 23( )!3

3. (6 ! 8)(9 !10)! (4 + 2)(6 + 3) 4. !8436!12(!5)

5. 12 6 ! 2( )2 ! 4 "3 6. 3 2 1+ 5( ) + 8 ! 32( )

7. 8 +12( ) ÷ 4 ! 6 8. !62 + 4 "8

9. 18 !3÷ 33 10. 10 + 52 ! 25

11. 20 ! (33 ÷ 9) "2 12. 100 ! (23 ! 6)÷ 2

13. 85 ! (4 "2)2 ! 3 14. 22 + (3!2)2 ÷ 2

15. 16 +11(!2)2 ! 25 25( ) 16. 54 ÷ 32 + 4

3( ) 272( )!12

17. 2 3!1( )3 ÷ 8 18. (72 !1)÷ 4 + 2

19. !3"2 ÷ (!2 ! 4) 20. !3"2 ÷ !2 ! 4

21. 12 + 3 8!212!9( )! 2 9!1

19!15( ) 22. 15 + 4 11!29!6( )! 2 12!4

18!10( )

23. 32 ÷16 ! 8 "25 12( )2 24. 36 +16 ! 1

4( )" (50 ÷ 25)2 Answers:

1. 23 2. 3

3. –52 4. ! 78

5. –4 6. 33

7. –1 8. –4

9. 2 10. 10

11. 14 12. 99

13. 18 14. 40

15. 50 16. 23

17. 2 18. 14

19. 1 20. –1

21. 14 22. 25

23. –48 24. 36



Writing and Evaluating Algebraic Expressions Answers to problem 5-133: a. x + 6 , b. y ! 5 , c. 2x + 3 , d. 5y , e. 11; 3; 13; 40 There are some vocabulary words that are frequently used to represent arithmetic operations. Addition is often suggested by: sum, increased, more than, greater than, total

Subtraction is often suggested by: difference, decreased by, less than, smaller than Multiplication is often suggested by: product, times, twice, double

Division is often suggested by: quotient, divided by, shared evenly Examples Five more than m: Five more than a number, increases a number by 5 so it would be m + 5 . Three less than x: Three less than a number makes the number smaller by 3 so it would be x – 3. Triple m: Tripling a number is the same as multiplying the number by 3 so it would be 3m . Five divided by x: Division is usually written as a fraction so it would be 5x . To evaluate an algebraic expression means to calculate the value of the expression when the variable is replaced by a numerical value. Examples Evaluate 2x ! 5 if x = 7 Solution: 2x ! 5" 2 #7 ! 5 = 14 ! 5 = 9 Evaluate 6x + 9 if x = 2 Solution: 6x + 9!

62 + 9 = 3+ 9 = 12

Now we can go back and solve the original problem. Part (e) is included with the writing of the expression. a. Six more suggests adding 6 so it would be x + 6 ; if x = 5 then x + 6! 5 + 6 = 11 . b. Five less suggests subtracting 5 so it would be y ! 5 ; if y = 8 then y ! 5" 8 ! 5 = 3 . c. Twice a number suggests multiplication by 2 and then increasing by 3 means to add 3 to the

previous product so it would be 2x + 3 ; if x = 5 then 2x + 3! 2 "5 + 3= 13 . d. Product suggests multiply so it would be 5y ; if y = 8 then 5y! 5 "8 = 40 .


Here are some more to try. For problems 1 through 16 write an algebraic expression. For problems 17 through 24 evaluate the expression for the given values.

1. 5 greater than m 2. Double s

3. 7 less than t 4. 6 more than y

5. 14 divided by b 6. m subtracted from a

7. x divided among 5 8. The product of 7 and e

9. The sum of d and l 10. 6 times c

11. The product of x, y, and w 12. h times 8, added to j

13. 4 divided by p, increased by 7 14. Half of r

15. Three times q increased by 5 16. Two less than triple n

17. 4x ! 3 if x = 5 18. 7d if d = 10

19. 2xy +1 if x = 3,!y = 4 20. 4 + 8g if g = 6

21. 4b +12 if b = 11 22. 9g27 if g = 3

23. mw + h if m = 5,w = 8, h = 6 24. yz + 7 if y = 4, z = 2

Answers:

1. m + 5 2. 2s

3. t ! 7 4. y + 6

5. 14b 6. a !m

7. x5 8. 7e

9. d + l 10. 6c

11. xyw 12. 8h + j

13. 4p + 7 14. r

2

15. 3q + 5 16. 3n ! 2

17. 17 18. 70

19. 25 20. 52

21. 56 22. 1

23. 46 24. 9


Checkpoint 7A Problem 7-50

Simplifying Expressions

Answers to problem 7-50: a: 2x2 + x !10 , b: x2 ! 9x +11 Like terms are two or more terms that are exactly the same except for their coefficients. That is, they have the same variable(s), with corresponding variable(s) raised to the same power. Like terms can be combined into one quantity by adding and/or subtracting the coefficients of the terms. Terms are usually listed in the order of decreasing powers of the variable. Combining like terms, one way of simplifying expressions, using algebra tiles is shown in the first two examples. Example 1: Simplify (2x2 + 4x + 5)+ (x2 + x + 3) means combine 2x2 + 4x + 5 with x2 + x + 3 .

(2x2 + 4x + 5)+ (x2 + x + 3) = 3x2 + 5x + 8 Example 2:

x2 + 3x ! 4 + 2(2x2 ! x)+ 3 . Simplify Now we can go back and solve the original problem. a. 4x2 + 3x ! 7 + !2x2 ! 2x + (!3)

(4x2 ! 2x2 )+ (3x ! 2x)+ (!7 + (!3))2x2 + x !10

b. !3x2 ! 2x + 5 + 4x2 ! 7x + 6(!3x2 + 4x2 )+ (!2x ! 7x)+ (5 + 6)

x2 ! 9x +11

= +1 = –1

Remember:

0

= x 2 x 2

x 2

x x x

x x 2

x 2 x 2 x

= 5x2 + x !1

x 2 x 2 x

x 2

x 2

x 2 x

x 2

x

x x

x 2

x


Here are some more to try.

1. (x2 + 3x + 4)+ (x2 + 3x + 2) 2. x2 + 4x + 3+ x2 + 2x + 5

3. 2x2 + 2x !1+ x2 ! 4x + 5 4. 3x2 ! x + 7 ! 3x2 + 2x + (!4)

5. 2x2 + 4x + (!3)+ x2 ! 3x + 5 6. 4x2 + 2x ! 8 + (!2x2 )+ 5x +1

7. !4x2 + 2x + 8 ! 3x2 + 5x ! 3 8. 3x2 + 2(4x +1)! 2x2 + 4x + 5

9. 3(5x2 + 4x) – 7 + 2x + 3 10. 3x2 – 4x + 2 + 4(2x2 + 2x)

11. !2(3x2 – x + 2)+ 3x –1 12. 2(x2 – 2x)+ 7 + 5(x2 + 4x) – 3

13. 2x2 – 3(x – 3)+ 5x2 – 4x 14. –3x + 3(x2 + 2)+ (–x) – 4

15. (x2 + 3x + 2)! (x2 ! 3x ! 2) 16. (2x2 ! 5x)! (x2 + 7x)

17. (5x + 6)! (x2 ! 5x + 6) 18. 4a + 2b + 3c ! 6a + 3b ! 6c

19. 3c + 4a ! 7c + 5b ! (!4a)+ 7 20. !5a + 6b ! 7c ! (3c ! 4b ! a) Answers:

1. 2x2 + 6x + 6 2. 2x2 + 6x + 8

3. 3x2 ! 2x + 4 4. x + 3

5. 3x2 + x + 2 6. 2x2 + 7x ! 7

7. !7x2 + 7x + 5 8. x2 +12x + 7

9. 15x2 +14x ! 4 10. 11x2 + 4x + 2

11. !6x2 + 5x ! 5 12. 7x2 +16x + 4

13. 7x2 ! 7x + 9 14. 3x2 ! 4x + 2

15. 6x + 4 16. x2 !12x

17. !x2 +10x 18. !2a + 5b ! 3c

19. 8a + 5b ! 7c + 7 20. !4a +10b !10c


Checkpoint 7B Problem 7-117

Displays of Data: Histograms and Box Plots Answers to problem 7-117: a. b. c. d. c: IQR = 4 d: IQR = 14 Histograms

A histogram is a method of showing data. It uses a bar to show the frequency (the number of times something occurs). The frequency measures something that changes numerically. (In a bar graph the frequency measures something that changes by category.) The intervals (called bins) for the data are shown on the horizontal axis and the frequency is represented by the height of a rectangle above the interval. The labels on the horizontal axis represent the lower end of each interval or bin. Example: Sam and her friends weighed themselves and here is their weight in pounds:

110, 120, 131, 112, 125, 135, 118, 127, 135, and 125. Make a histogram to display the information. Use intervals of 10 pounds.

Solution: See histogram at right. Note that the person weighing 120 pounds is counted in the next higher bin. Box Plots

A box plot displays a summary of data using the median, quartiles, and extremes of the data. The box contains the “middle half” of the data. The right segment represents the top 25% of the data and the left segment represent the bottom 25% of the data.

Weight (pounds)

Num

ber o

f peo

ple

1 2 3 4 5 0

2 4 6 8 10

Hours of Homework

Freq

uenc

y

5 10 15 20 25 0

2

4 6

8

Score

Freq

uenc

y

45 50 55

46

46.5 50.5

48 52

65 75 85 95

70

73.5

80

87.5

93


Example: Create a box plot for the set of data given in the previous example. Solution: Place the data in order

to find the median (middle number) and the quartiles (middle numbers of the upper half and the lower half.)

Based on the extremes,

first quartile, third quartile, and median, the box plot is drawn.

The interquartile range

IQR = 131–118 = 13. Now we can go back to the original problem. a. The 0–1 bin contains the six students who do less than one hour of homework. The 1–2 bin

contains the 10 students who do at least one hour but less than two hours. The 2–3 bin contains the seven students who do at least two hours but less than three hours. There are no students who do at least three hours and less than four. Two students did four hours and less than five. See the histogram above.

b. The 0–5 bin contains two scores less than 5 points. The 5–10 bin contains the two scores

of a least five but less than 10. The 10–15 bin contains the eight scores at least 10 but less than 15. The 15–20 bin contains the seven scores at least 15 but less than 20. See the histogram above.

c. Place the ages in order: 46, 46, 47, 47, 48, 49, 50, 51, 52. The median is the middle age: 48. The first quartile is the median of the lower half of the

ages. Since there are four lower-half ages, the median is the average of the middle two: 46+47

2= 46.5 . The third quartile is the median of the upper half ages. Again, there are four

upper-half ages, so average the two middle ages: 50+512

= 50.5 . The interquartile range is the difference between the third quartile and the first quartile: 50.5–46.5 = 4. See the box plot above.

d. Place the scores in order: 70, 72, 75, 76, 80, 82, 85, 90, 93. The median is the middle score: 80. The lower quartile is the median of the lower half of

the scores. Since there are four lower-half scores, the median is the average of the middle two: 72+75

2= 73.5 . The third quartile is the median of the upper half of the scores. Again,

there are four upper-half scores, so average the two middle ages: 85+902

= 87.5 . The interquartile range is the difference between the third quartile and the first quartile: 87.5–73.5 = 14. See the box plot above.

110 112 118 120 125

first quartile =118 (median of lower half)

third quartile= 131 (median of upper half)

median = 125

125 127 131 135 135

110 120 130 140

110 118 125 131 135


Here are some more to try. For problems 1 through 6, create a histogram. For problems 7 through 12, create a box plot. State the quartiles and the interquartile range.

1. Number of heads showing in 20 tosses of three coins: 2, 2, 1, 3, 1, 0, 2, 1, 2, 1, 1, 2, 0, 1, 3, 2, 1, 3, 1, 2

2. Number of even numbers in 5 rolls of a dice done 14 times: 4, 2, 2, 3, 1, 2, 1, 1, 3, 3, 2, 2, 4, 5 3. Number of fish caught by 7 fishermen: 2, 3, 0, 3, 3, 1, 5

4. Number of girls in grades K-8 at local schools: 12, 13, 15, 10, 11, 12, 15, 11, 12

5. Number of birthdays in each March in various 2nd grade classes: 5, 1, 0, 0, 2, 4, 4, 1, 3, 1, 0, 4

6. Laps jogged by 15 students: 10, 15, 10, 13, 20, 14, 17, 10, 15, 20, 8, 7, 13, 15, 12

7. Number of days of rain: 6, 8, 10, 9, 7, 7, 11, 12, 6, 12, 14, 10

8. Number of times a frog croaked per minute: 38, 23, 40, 12, 35, 27, 51, 26, 24, 14, 38, 41, 23, 17

9. Speed in mph of 15 different cars: 30, 35, 40, 23, 33, 32, 28, 37, 30, 31, 29, 33, 39, 22, 30

10. Typing speed of 12 students in words per minute: 28, 30, 60, 26, 47, 53, 39, 42, 48, 27, 23, 86

11. Number of face cards pulled when 13 cards are drawn 15 times: 1, 4, 2, 1, 1, 0, 0, 2, 1, 3, 3, 0, 0, 2, 1

12. Height of 15 students in inches: 48, 55, 56, 65, 67, 60, 60, 57, 50, 59, 62, 65, 58, 70, 68


Answers:

1 2 3 4 0

2 4 6 8

10

Number of Heads Showing

Freq

uenc

y

1.

1 2 3 4 5 6 0

1 2 3 4 5

Number of Even Rolls Fr

eque

ncy

2.

10 11 12 13 14 15 16 0

1 2 3 4 5

Number of Girls

Freq

uenc

y

4.

1 2 3 4 5 6 0

1 2 3 4 5

Number of Birthdays

Freq

uenc

y

5.

5 10 15 20 25 0

2 4 6 8

10

Number of Laps Jogged

Freq

uenc

y

6.

10 35 60

8.

Q1 = 23; Med = 26.5; Q3 = 38; IQR = 15

0 25 50

9.

Q1 = 29; Med = 31; Q3 = 35; IQR = 6

0 50 100

10.

Q1 = 27.5; Med = 40.5Q3 = 50.5; IQR = 23

0 5

11.

Q1 = 0; Med = 1; Q3 = 2; IQR = 2

40 55 70

12.

Q1 = 56; Med = 60; Q3 = 65; IQR = 9

1 2 3 4 5 6 0

1 2 3 4 5

Number of Fish Caught

Freq

uenc

y

3.

5 10 15

7.

Q1 = 7; Med = 9.5; Q3 = 11.5; IQR = 4.5



Solving Multi-Step Equations Answers to problem 8-109: a: x = 7, b: x = –2, c: x = 0, d: x = 23 A general strategy for solving equations is to first simplify each side of the equation. Next isolate the variable on one side and the constants on the other by adding equal values on both sides of the equation or removing balanced sets or zeros. Finally determine the value of the variable–usually by division. Note: When the process of solving an equation ends with different numbers on each side of the equal sign (for example, 2 = 4), there is no solution to the problem. When the result is the same expression or number on each side of the equation (for example, x + 2 = x + 2 ) it means that all numbers are solutions. Example 1: Solve 3x + 3x – 1 = 4x + 9

3x + 3x !1= 4x + 96x !1= 4x + 92x = 10x = 5

Example 2: Solve –2x + 1 + 3(x – 1) = –4 + – x – 2

!2x +1+ 3(x !1) = !4 + !x ! 2!2x +1+ 3x ! 3= !x ! 6

x ! 2 = !x ! 62x = !4x = !2

Now we can go back and solve the original problem. a. 24 = 3x + 3

21= 3x7 = x

b. 6x +12 = !x ! 26x = !x !147x = !14x = !2

c. 3x + 3! x + 2 = x + 52x + 5 = x + 52x = xx = 0

d. 5(x !1) = 5(4x ! 3)5x ! 5 = 20x !155x = 20x !10

!15x = !10x = !10

!15 =23

Solution

Solution


Here are some more to try.

1. 2x + 3= !7 2. !3x ! 2 = !1

3. 3x + 2 + x = x + 5 4. 4x ! 2 ! 2x = x ! 5

5. 2x ! 3= !x + 3 6. 1+ 3x ! x = x ! 4 + 2x

7. 4 ! 3x = 2x ! 6 8. 3+ 3x ! x + 2 = 3x + 4

9. !x ! 3= 2x ! 6 10. !4 + 3x !1= 2x +1+ 2x

11. !x + 3= 6 12. 5x ! 3+ 2x = x + 2 + x

13. 2x ! 7 = !x !1 14. !2 + 3x = x ! 2 ! 4x

15. !3x + 7 = x !1 16. 1+ 2x ! 4 = !3+ x

17. 3(x + 2) = x + 2 18. 2(x ! 2)+ x = 5

19. 10 = x + 5 + x 20. !x + 2 = x ! 5 ! 3x

21. 3x + 2 ! x = x ! 2 + x 22. 4x +12 = 2x ! 8

23. 3(4 + x) = x + 6 24. 6 ! x ! 3= 4(x ! 2) Answers:

1. x = –5 2. x = ! 13

3. x = 1 4. x = –3

5. x = 2 6. x = 5

7. x = 2 8. x = 1

9. x = 1 10. x = –6

11. x = –3 12. x = 1

13. x = 2 14. x = 0

15. x = 2 16. x = 0

17. x = –2 18. x = 3

19. x = 2 12 20. x = –7

21. no solution 22. x = –10

23. x = –3 24. x = 2 15



Unit Rates and Proportions Answers to problem 9-92: a: 24 mpg; b: $0.19; c: x = 8 ; d: m = 32.5 A rate is a ratio comparing two quantities and a unit rate has a denominator of one after simplifying. Unit rates or proportions may be used to solve ratio problems. Solutions may also be approximated by looking at graphs of lines passing through the origin and the given information. Example 1: Judy’s grape vine grew 15 inches in 6 weeks. What is the unit growth rate

(inches per week)? Solution: The growth rate is 15 !inches

6 !weeks. To create a unit rate we need a denominator of “one.”

15 !inches

6!weeks=x !inches

1!week! . Solve by using a Giant One: 15 !inches

6!weeks=6

6!x !inches

1!week!" 2.5 inches

week

Example 2: Bob’s favorite oatmeal raisin cookie recipe use 3 cups of raisins for 5 dozen

cookies. How many cups are needed for 40 dozen cookies? Solution: The rate is 3!cups

5!dozen so the problem may be written as this proportion: 3

5=

c

40.

One method of solving the proportion is to use a Giant One:

3

5=

c

40!

3

5"8

8=24

40! c = 24

Another method is to think about unit rates. Since the unit rate is 3

5cup per dozen, one could

also take the unit rate and multiply by the number of units needed: 35!40 = 24 .

Using either method the answer is 24 cups of raisins. Now we can go back and solve the original problem. a. 108 !miles

4.5!gallons=4.54.5

!x !miles1!gallon

" 24 milesgallon

b. $3.2317 !oranges

=1717!

x

1!orange" 0.19 $

orange

c. Using a Giant One: d. Using the Giant One

x12 !

2.52.5 =

2.5x30 = 20

30 " 20 = 2.5x" x = 8 1340 !

2.52.5 =

32.5100 " m = 32.5


Here are some more to try. For problems 1 through 8 find the unit rate. For problems 17 through 24 solve the proportion.

1. Typing 544 words in 17 minutes ( words per minute )

2. Taking 92 minutes to run 10 miles ( minutes per mile )

3. Reading 258 pages in 86 minutes ( pages per minute )

4. Falling 385 feet in 35 seconds ( feet per second )

5. Buying 15 boxes of cereal for $39.75 ( $ per box)

6. Drinking 28 bottles of water in 8 days ( bottles per day )

7. Scoring 98 points in a 40 minute game (points per minute)

8. Planting 76 flowers in 4 hours (flowers per hour)

9. 3

8=

x

50 10. 2

5=

x

75 11. 7

9=14

x 12. 24

25=96

x

13. 15

9=12

x 14. 45

60=

x

4 15. 4

7=18

x 16. 8

9=72

x

17. 3

5=

x

17 18. 17

30=51

x 19. 5

8=16

x 20. 3

22=15

x

21. 1

5=

x

27 22. x

11=8

15 23. 14

17=

x

34 24. 12

15=36

x

Answers:

1. 32 words

minute 2. 9.2 minutes

mile 3. 3pages

minute 4. 11 feet

second

5. 2.65 dollarsbox

6. 3.5 bottlesday 7. 2.45

points

minute 8. 19 flowers

hour

9. x = 18.75 10. x = 30 11. x = 18 12. x = 100 13. x = 7.2 14. x = 3 15. x = 31.5 16. x = 81 17. x = 10.2 18. x = 90 19. x = 25.6 20. x = 110

21. x = 5.4 22. x = 5 1315 23. x = 28 24. x = 45

core connections, course 2 checkpoint...

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