Keystone Exams: Algebra I and...Keystone Exams: Algebra I Assessment Anchors and Eligible Content with Sample Questions and Glossary Pennsylvania Department of Education Pennsylvania

Keystone Exams: Algebra IAssessment Anchors and Eligible Content

with Sample Questions and Glossary

Pennsylvania Department of Education

www.education.state.pa.us

April 2014

Pennsylvania Department of Education—Assessment Anchors and Eligible Content Page 2

PENNSYLVANIA DEPARTMENT OF EDUCATION

General Introduction to the Keystone Exam Assessment Anchors

Introduction

Since the introduction of the Keystone Exams, the Pennsylvania Department of Education (PDE) has been working to create a set of tools designed to help educators improve instructional practices and better understand the Keystone Exams. The Assessment Anchors, as defined by the Eligible Content, are one of the many tools the Department believes will better align curriculum, instruction, and assessment practices throughout the Commonwealth. Without this alignment, it will not be possible to significantly improve student achievement across the Commonwealth.

How were Keystone Exam Assessment Anchors developed?

Prior to the development of the Assessment Anchors, multiple groups of PA educators convened to create a set of standards for each of the Keystone Exams. Enhanced Standards, derived from a review of existing standards, focused on what students need to know and be able to do in order to be college and career ready. (Note: Since that time, PA Core Standards have replaced the Enhanced Standards and reflect the college- and career-ready focus.) Additionally, the Assessment Anchors and Eligible Content statements were created by other groups of educators charged with the task of clarifying the standards assessed on the Keystone Exams. The Assessment Anchors, as defined by the Eligible Content, have been designed to hold together, or anchor, the state assessment system and the curriculum/instructional practices in schools.

Assessment Anchors, as defined by the Eligible Content, were created with the following design parameters: Clear: The Assessment Anchors are easy to read and are user friendly; they clearly detail which

standards are assessed on the Keystone Exams.

Focused: The Assessment Anchors identify a core set of standards that can be reasonably assessed on a large-scale assessment; this will keep educators from having to guess which standards are critical.

Rigorous: The Assessment Anchors support the rigor of the state standards by assessing higher-order and reasoning skills.

Manageable: The Assessment Anchors define the standards in a way that can be easily incorporated into a course to prepare students for success.

How can teachers, administrators, schools, and districts use these Assessment Anchors?

The Assessment Anchors, as defined by the Eligible Content, can help focus teaching and learning because they are clear, manageable, and closely aligned with the Keystone Exams. Teachers and administrators will be better informed about which standards will be assessed. The Assessment Anchors and Eligible Content should be used along with the Standards and the Curriculum Framework of the Standards Aligned System (SAS) to build curriculum, design lessons, and support student achievement.

The Assessment Anchors and Eligible Content are designed to enable educators to determine when they feel students are prepared to be successful in the Keystone Exams. An evaluation of current course offerings, through the lens of what is assessed on those particular Keystone Exams, may provide an opportunity for an alignment to ensure student preparedness.


How are the Assessment Anchors organized?

The Assessment Anchors, as defined by the Eligible Content, are organized into cohesive blueprints, each structured with a common labeling system that can be read like an outline. This framework is organized first by module, then by Assessment Anchor, followed by Anchor Descriptor, and then finally, at the greatest level of detail, by an Eligible Content statement. The common format of this outline is followed across the Keystone Exams.

Here is a description of each level in the labeling system for the Keystone Exams: Module: The Assessment Anchors are organized into two thematic modules for each of the

Keystone Exams. The module title appears at the top of each page. The module level is important because the Keystone Exams are built using a module format, with each of the Keystone Exams divided into two equal-size test modules. Each module is made up of two or more Assessment Anchors.

Assessment Anchor: The Assessment Anchor appears in the shaded bar across the top of each Assessment Anchor table. The Assessment Anchors represent categories of subject matter that anchor the content of the Keystone Exams. Each Assessment Anchor is part of a module and has one or more Anchor Descriptors unified under it.

Anchor Descriptor: Below each Assessment Anchor is a specific Anchor Descriptor. The Anchor Descriptor level provides further details that delineate the scope of content covered by the Assessment Anchor. Each Anchor Descriptor is part of an Assessment Anchor and has one or more Eligible Content statements unified under it.

Eligible Content: The column to the right of the Anchor Descriptor contains the Eligible Content statements. The Eligible Content is the most specific description of the content that is assessed on the Keystone Exams. This level is considered the assessment limit and helps educators identify the range of the content covered on the Keystone Exams.

PA Core Standard: In the column to the right of each Eligible Content statement is a code representing one or more PA Core Standards that correlate to the Eligible Content statement. Some Eligible Content statements include annotations that indicate certain clarifications about the scope of an Eligible Content.

“e.g.” (“for example”)—sample approach, but not a limit to the Eligible Content

“i.e.” (“that is”)—specific limit to the Eligible Content

“Note”—content exclusions or definable range of the Eligible Content

How do the K–12 Pennsylvania Core Standards affect this document?

Assessment Anchor and Eligible Content statements are aligned to the PA Core Standards; thus, the former enhanced standards are no longer necessary. Within this document, all standard references reflect the PA Core Standards.

Standards Aligned System—www.pdesas.org

Pennsylvania Department of Education—www.education.state.pa.us


Keystone Exams: Algebra I

FORMULA SHEET

Formulas that you may need to work questions in this document are found below.You may use calculator π or the number 3.14.

A = lw

l

w

V = lwh

lw

h

Arithmetic Properties

Additive Inverse: a + (ˉa) = 0

Multiplicative Inverse: a · = 1

Commutative Property: a + b = b + a a · b = b · a

Associative Property: (a + b) + c = a + (b + c) (a · b) · c = a · (b · c)

Identity Property: a + 0 = a a · 1 = a

Distributive Property: a · (b + c) = a · b + a · c

Multiplicative Property of Zero: a · 0 = 0

Additive Property of Equality: If a = b, then a + c = b + c

Multiplicative Property of Equality: If a = b, then a · c = b · c

1a

Linear Equations

Slope: m =

Point-Slope Formula: (y – y 1) = m(x – x 1)

Slope-Intercept Formula: y = mx + b

Standard Equation of a Line: Ax + By = C

y 2 – y 1x 2 – x 1



MODULE 1—Operations and Linear Equations & Inequalities

ASSESSMENT ANCHOR

A1.1.1 Operations with Real Numbers and Expressions

Anchor Descriptor Eligible ContentPA Core

Standards

A1.1.1.1 Represent and/or use numbers in equivalent forms (e.g., integers, fractions, decimals, percents, square roots, and exponents).

A1.1.1.1.1 Compare and/or order any real numbers.Note: Rational and irrational may be mixed.

CC.2.1.8.E.1

CC.2.1.8.E.4

CC.2.1.HS.F.1

CC.2.1.HS.F.2

A1.1.1.1.2 Simplify square roots (e.g., Ï}

24 = 2 Ï}

6 ).

Sample Exam Questions

Standard A1.1.1.1.1

Which of the following inequalities is true for all real values of x?

A. x3 ≥ x2

B. 3x2 ≥ 2x3

C. (2x)2 ≥ 3x2

D. 3(x – 2)2 ≥ 3x2 – 2

Standard A1.1.1.1.2

An expression is shown below.

2 Ï}

51x

Which value of x makes the expression equivalent to 10 Ï

}

51 ?

A. 5

B. 25

C. 50

D. 100

An expression is shown below.

Ï}

87x

For which value of x should the expression be further simplified?

A. x = 10

B. x = 13

C. x = 21

D. x = 38




ASSESSMENT ANCHOR



Standards

A1.1.1.2 Apply number theory concepts to show relationships between real numbers in problem-solving settings.

A1.1.1.2.1 Find the Greatest Common Factor (GCF) and/or the Least Common Multiple (LCM) for sets of monomials.

CC.2.1.6.E.3

CC.2.1.HS.F.2

Sample Exam Question

Standard A1.1.1.2.1

Two monomials are shown below.

450x2y5 3,000x4y3

What is the least common multiple (LCM) of these monomials?

A. 2xy

B. 30xy

C. 150x2y3

D. 9,000x4y5




ASSESSMENT ANCHOR



Standards

A1.1.1.3 Use exponents, roots, and/or absolute values to solve problems.

A1.1.1.3.1 Simplify/evaluate expressions involving properties/laws of exponents, roots, and/or absolute values to solve problems. Note: Exponents should be integers from –10 to 10.

CC.2.1.HS.F.1

CC.2.1.HS.F.2

CC.2.2.8.B.1


Standard A1.1.1.3.1

Simplify:

2(2 Ï}

4 ) –2

A. 1 __ 8

B. 1 __ 4

C. 16

D. 32




ASSESSMENT ANCHOR



Standards

A1.1.1.4 Use estimation strategies in problem-solving situations.

A1.1.1.4.1 Use estimation to solve problems. CC.2.2.7.B.3

CC.2.2.HS.D.9


Standard A1.1.1.4.1

A theme park charges $52 for a day pass and $110 for a week pass. Last month, 4,432 day passes were sold and 979 week passes were sold. Which is the closest estimate of the total amount of money paid for the day and week passes for last month?

A. $300,000

B. $400,000

C. $500,000

D. $600,000




ASSESSMENT ANCHOR



Standards

A1.1.1.5 Simplify expressions involving polynomials.

A1.1.1.5.1 Add, subtract, and/or multiply polynomial expressions (express answers in simplest form). Note: Nothing larger than a binomial multiplied by a trinomial.

CC.2.2.HS.D.1

CC.2.2.HS.D.2 CC.2.2.HS.D.3

CC.2.2.HS.D.5

CC.2.2.HS.D.6

A1.1.1.5.2 Factor algebraic expressions, including difference of squares and trinomials.Note: Trinomials are limited to the form ax2 + bx + c where a is equal to 1 after factoring out all monomial factors.

A1.1.1.5.3 Simplify/reduce a rational algebraic expression.


Standard A1.1.1.5.1

A polynomial expression is shown below.

(mx3 + 3 ) (2x2 + 5x + 2) – (8x5 + 20x4 )

The expression is simplified to 8x3 + 6x2 + 15x + 6. What is the value of m?

A. –8

B. –4

C. 4

D. 8

Standard A1.1.1.5.2

When the expression x2 – 3x – 18 is factored completely, which is one of its factors?

A. (x – 2)

B. (x – 3)

C. (x – 6)

D. (x – 9)

Standard A1.1.1.5.3

Simplify:

–3x3 + 9 x2 + 30x }} –3x3 – 18 x2 – 24x ; x ≠ –4, –2, 0

A. – 1 } 2 x2 – 5 }

4 x

B. x3 – 1 } 2

x2 – 5 } 4 x

C. x + 5 } x – 4

D. x – 5 } x + 4




Standard A1.1.1

Keng creates a painting on a rectangular canvas with a width that is four inches longer than the height, as shown in the diagram below.

h

h + 4

A. Write a polynomial expression, in simplifi ed form, that represents the area of the canvas.

Keng adds a 3-inch-wide frame around all sides of his canvas.

B. Write a polynomial expression, in simplified form, that represents the total area of the canvas and the frame.

Continued on next page.

ASSESSMENT ANCHOR






Continued. Please refer to the previous page for task explanation.

Keng is unhappy with his 3-inch-wide frame, so he decides to put a frame with a different width around his canvas. The total area of the canvas and the new frame is given by the polynomial h2 + 8h + 12, where h represents the height of the canvas.

C. Determine the width of the new frame. Show all your work. Explain why you did each step.




Standard A1.1.1

The results of an experiment were listed in several numerical forms as listed below.

5–3 4 } 7 Ï

}

5 3 } 8 0.003

A. Order the numbers listed from least to greatest.

Another experiment required evaluating the expression shown below.

1 } 6 ( Ï

}

36 ÷ 3–2) + 43 ÷ z–8z

B. What is the value of the expression?

value of the expression:






The final experiment required simplifying 7 Ï}

425 . The steps taken are shown below.

7 425

step 1: 7 400 25)

step 2: 7(20 + 5)

step 3: 7(25)

step 4: 175

+

One of the steps shown is incorrect.

C. Rewrite the incorrect step so that it is correct.

correction:

D. Using the corrected step from part C, simplify 7 Ï}

425 .

7 Ï}

425 =




ASSESSMENT ANCHOR

A1.1.2 Linear Equations


Standards

A1.1.2.1 Write, solve, and/or graph linear equations using various methods.

A1.1.2.1.1 Write, solve, and/or apply a linear equation (including problem situations).

CC.2.1.HS.F.3

CC.2.1.HS.F.4

CC.2.1.HS.F.5

CC.2.2.8.B.3

CC.2.2.8.C.1

CC.2.2.8.C.2

CC.2.2.HS.C.3

CC.2.2.HS.D.7

CC.2.2.HS.D.8

CC.2.2.HS.D.9

CC.2.2.HS.D.10

A1.1.2.1.2 Use and/or identify an algebraic property to justify any step in an equation-solving process.Note: Linear equations only.

A1.1.2.1.3 Interpret solutions to problems in the context of the problem situation.Note: Linear equations only.


Standard A1.1.2.1.1

Jenny has a job that pays her $8 per hour plus tips (t). Jenny worked for 4 hours on Monday and made $65 in all. Which equation could be used to find t, the amount Jenny made in tips?

A. 65 = 4t + 8

B. 65 = 8t ÷ 4

C. 65 = 8t + 4

D. 65 = 8(4) + t

Standard A1.1.2.1.2

One of the steps Jamie used to solve an equation is shown below.

–5(3x + 7) = 10–15x + –35 = 10

Which statements describe the procedure Jamie used in this step and identify the property that justifies the procedure?

A. Jamie added –5 and 3x to eliminate the parentheses. This procedure is justifi ed by the associative property.

B. Jamie added –5 and 3x to eliminate the parentheses. This procedure is justifi ed by the distributive property.

C. Jamie multiplied 3x and 7 by –5 to eliminate the parentheses. This procedure is justifi ed by the associative property.

D. Jamie multiplied 3x and 7 by –5 to eliminate the parentheses. This procedure is justifi ed by the distributive property.





Standard A1.1.2.1.3

Francisco purchased x hot dogs and y hamburgers at a baseball game. He spent a total of $10. The equation below describes the relationship between the number of hot dogs and the number of hamburgers purchased.

3x + 4y = 10

The ordered pair (2, 1) is a solution of the equation. What does the solution (2, 1) represent?

A. Hamburgers cost 2 times as much as hot dogs.

B. Francisco purchased 2 hot dogs and 1 hamburger.

C. Hot dogs cost $2 each, and hamburgers cost $1 each.

D. Francisco spent $2 on hot dogs and $1 on hamburgers.




ASSESSMENT ANCHOR



Standards

A1.1.2.2 Write, solve, and/or graph systems of linear equations using various methods.

A1.1.2.2.1 Write and/or solve a system of linear equations (including problem situations) using graphing, substitution, and/or elimination.Note: Limit systems to two linear equations.

CC.2.1.HS.F.5

CC.2.2.8.B.3

CC.2.2.HS.D.7

CC.2.2.HS.D.9

CC.2.2.HS.D.10

A1.1.2.2.2 Interpret solutions to problems in the context of the problem situation.Note: Limit systems to two linear equations.


Standard A1.1.2.2.1

Anna burned 15 calories per minute running for x minutes and 10 calories per minute hiking for y minutes. She spent a total of 60 minutes running and hiking and burned 700 calories. The system of equations shown below can be used to determine how much time Anna spent on each exercise.

15x + 10y = 700

x + y = 60

What is the value of x, the minutes Anna spent running?

A. 10

B. 20

C. 30

D. 40

Standard A1.1.2.2.2

Samantha and Maria purchased flowers. Samantha purchased 5 roses for x dollars each and 4 daisies for y dollars each and spent $32 on the flowers. Maria purchased 1 rose for x dollars and 6 daisies for y dollars each and spent $22. The system of equations shown below represents this situation.

5x + 4y = 32

x + 6y = 22

Which statement is true?

A. A rose costs $1 more than a daisy.

B. Samantha spent $4 on each daisy.

C. Samantha spent more on daisies than she did on roses.

D. Samantha spent over 4 times as much on daisies as she did on roses.




Standard A1.1.2

Nolan has $15.00. He earns $6.00 an hour babysitting. The equation below can be used to determine how much money in dollars (m) Nolan has after any number of hours of babysitting (h).

m = 6h + 15

A. After how many hours of babysitting will Nolan have $51.00?

hours:

Claire has $9.00. She makes $8.00 an hour babysitting.

B. Use the system of linear equations below to find the number of hours of babysitting after which Nolan and Claire will have the same amount of money.

m = 6h + 15

m = 8h + 9

hours:


ASSESSMENT ANCHOR







The graph below displays the amount of money Alex and Pat will each have saved after their hours of babysitting.

0Hours

Mon

ey S

aved

(in

Dol

lars

)

h

m

504540353025201510

5

1 2 3 4 5 6 7 8 9 10

Money Saved

KeyAlex’s money savedPat’s money saved

C. Based on the graph, for what domain (h) will Alex have more money saved than Pat? Explain your reasoning.




Standard A1.1.2

The diagram below shows 5 identical bowls stacked one inside the other.

2 inches

5 inches

Bowls

The height of 1 bowl is 2 inches. The height of a stack of 5 bowls is 5 inches.

A. Write an equation using x and y to find the height of a stack of bowls based on any number of bowls.

equation:






B. Describe what the x and y variables represent.

x-variable:

y-variable:

C. What is the height, in inches, of a stack of 10 bowls?

height: inches




ASSESSMENT ANCHOR

A1.1.3 Linear Inequalities


Standards

A1.1.3.1 Write, solve, and/or graph linear inequalities using various methods.

A1.1.3.1.1 Write or solve compound inequalities and/or graph their solution sets on a number line (may include absolute value inequalities).

CC.2.1.HS.F.5

CC.2.2.HS.D.7

CC.2.2.HS.D.9

CC.2.2.HS.D.10

A1.1.3.1.2 Identify or graph the solution set to a linear inequality on a number line.

A1.1.3.1.3 Interpret solutions to problems in the context of the problem situation.Note: Linear inequalities only.


Standard A1.1.3.1.1

A compound inequality is shown below.

5 < 2 – 3y < 14

What is the solution of the compound inequality?

A. –4 > y > –1

B. –4 < y < –1

C. 1 > y > 4

D. 1 < y < 4

Standard A1.1.3.1.3

A baseball team had $1,000 to spend on supplies. The team spent $185 on a new bat. New baseballs cost $4 each. The inequality 185 + 4b ≤ 1,000 can be used to determine the number of new baseballs (b) that the team can purchase. Which statement about the number of new baseballs that can be purchased is true?

A. The team can purchase 204 new baseballs.

B. The minimum number of new baseballs that can be purchased is 185.

C. The maximum number of new baseballs that can be purchased is 185.

D. The team can purchase 185 new baseballs, but this number is neither the maximum nor the minimum.

Standard A1.1.3.1.2

The solution set of an inequality is graphed on the number line below.

–4–5 –3 –2 –1 0 1 2

The graph shows the solution set of which inequality?

A. 2x + 5 < –1

B. 2x + 5 ≤ –1

C. 2x + 5 > –1

D. 2x + 5 ≥ –1




ASSESSMENT ANCHOR



Standards

A1.1.3.2 Write, solve, and/or graph systems of linear inequalities using various methods.

A1.1.3.2.1 Write and/or solve a system of linear inequalities using graphing.Note: Limit systems to two linear inequalities.

CC.2.1.HS.F.5

CC.2.2.HS.D.7

CC.2.2.HS.D.10

A1.1.3.2.2 Interpret solutions to problems in the context of the problem situation.Note: Limit systems to two linear inequalities.





Standard A1.1.3.2.1

A system of inequalities is shown below.

y < x – 6

y > –2x

Which graph shows the solution set of the system of inequalities?

987654321

–1–2–3–4–5–6–7–8–9

1 2 3 4 5 6 7 8 9–9–8–7–6–5–4–3–2–1

y

x

987654321

–1–2–3–4–5–6–7–8–9

1 2 3 4 5 6 7 8 9–9–8–7–6–5–4–3–2–1

y

x

987654321

–1–2–3–4–5–6–7–8–9

1 2 3 4 5 6 7 8 9–9–8–7–6–5–4–3–2–1

y

x

987654321

–1–2–3–4–5–6–7–8–9

1 2 3 4 5 6 7 8 9–9–8–7–6–5–4–3–2–1

y

x

A. B.

C. D.





Standard A1.1.3.2.2

Tyreke always leaves a tip of between 8% and 20% for the server when he pays for his dinner. This can be represented by the system of inequalities shown below, where y is the amount of tip and x is the cost of dinner.

y > 0.08x

y < 0.2x

Which of the following is a true statement?

A. When the cost of dinner ( x) is $10, the amount of tip ( y) must be between $2 and $8.

B. When the cost of dinner ( x) is $15, the amount of tip ( y) must be between $1.20 and $3.00.

C. When the amount of tip ( y) is $3, the cost of dinner ( x) must be between $11 and $23.

D. When the amount of tip ( y) is $2.40, the cost of dinner ( x) must be between $3 and $6.




Standard A1.1.3

An apple farm owner is deciding how to use each day’s harvest. She can use the harvest to produce apple juice or apple butter. The information she uses to make the decision is listed below.

• A bushel of apples will make 16 quarts of apple juice. • A bushel of apples will make 20 pints of apple butter. • The apple farm can produce no more than 180 pints of apple butter each day. • The apple farm harvests no more than 15 bushels of apples each day.

The information given can be modeled with a system of inequalities. When x is the number of quarts of apple juice and y is the number of pints of apple butter, two of the inequalities that model the situation are x ≥ 0 and y ≥ 0.

A. Write two more inequalities to complete the system of inequalities modeling the information.

inequalities:

B. Graph the solution set of the inequalities from part A below. Shade the area that represents the solution set.

0Quarts of Apple Juice

Pint

s of

App

le B

utte

r

300

240

180

120

60

60 120 180 240 300

Apple Farm Production

x

y


ASSESSMENT ANCHOR







The apple farm makes a profit of $2.25 on each pint of apple butter and $2.50 on each quart of apple juice.

C. Explain how you can be certain the maximum profit will be realized when the apple farm produces 96 quarts of apple juice and 180 pints of apple butter.




Standard A1.1.3

David is solving problems with inequalities.

One of David’s problems is to graph the solution set of an inequality.

A. Graph the solution set to the inequality 4x + 3 < 7x – 9 on the number line below.

–8–10 –9 –7 –6 104–5 7–2 –1 0 1 2 3 5 6 8 9–4 –3

David correctly graphed an inequality as shown below.

104 71 2 3 5 6 8 9–8–10 –9 –7 –6 –5 –2 –1 0–4 –3

The inequality David graphed was written in the form 7 ≤ ? ≤ 9.

B. What is an expression that could be put in place of the question mark so that the inequality would have the same solution set as shown in the graph?

7 ≤ ≤ 9






The solution set to a system of linear inequalities is graphed below.

x

y

1 2 3 4 5–5 –4 –3 –2 –1

54321

–1–2–3–4–5

C. Write a system of two linear inequalities that would have the solution set shown in the graph.

linear inequality 1:

linear inequality 2:



MODULE 2—Linear Functions and Data Organizations

ASSESSMENT ANCHOR

A1.2.1 Functions


Standards

A1.2.1.1 Analyze and/or use patterns or relations.

A1.2.1.1.1 Analyze a set of data for the existence of a pattern and represent the pattern algebraically and/or graphically.

CC.2.2.8.C.1

CC.2.2.8.C.2

CC.2.2.HS.C.1

CC.2.2.HS.C.2

CC.2.2.HS.C.3

CC.2.4.HS.B.2

A1.2.1.1.2 Determine whether a relation is a function, given a set of points or a graph.

A1.2.1.1.3 Identify the domain or range of a relation (may be presented as ordered pairs, a graph, or a table).


Standard A1.2.1.1.1

Tim’s scores the first 5 times he played a video game are listed below.

4,526 4,599 4,672 4,745 4,818

Tim’s scores follow a pattern. Which expression can be used to determine his score after he played the video game n times?

A. 73n + 4,453

B. 73(n + 4,453)

C. 4,453n + 73

D. 4,526n





Standard A1.2.1.1.2

Which graph shows y as a function of x?

x

y

21 3 4–2–3–4

–4–3–2

1234

–1–1

x

y

21 3 4–2–3–4

–4–3–2

1234

–1–1

x

y

21 3 4–2–3–4

–4–3–2

1234

–1–1 x

y

21 3 4–2–3–4

–4–3–2

1234

–1–1

A. B.

C. D.





Standard A1.2.1.1.3

The graph of a function is shown below.

x2 4 6–6 –4 –2

8

6

4

2

–2

y

Which value is not in the range of the function?

A. 0

B. 3

C. 4

D. 5




ASSESSMENT ANCHOR

A1.2.1 Functions


Standards

A1.2.1.2 Interpret and/or use linear functions and their equations, graphs, or tables.

A1.2.1.2.1 Create, interpret, and/or use the equation, graph, or table of a linear function.

CC.2.1.HS.F.3

CC.2.1.HS.F.4

CC.2.2.8.B.2

CC.2.2.8.C.1

CC.2.2.8.C.2

CC.2.2.HS.C.2

CC.2.2.HS.C.3

CC.2.2.HS.C.4

CC.2.2.HS.C.6

CC.2.4.HS.B.2

A1.2.1.2.2 Translate from one representation of a linear function to another (i.e., graph, table, and equation).


Standard A1.2.1.2.1

A pizza restaurant charges for each pizza and adds a delivery fee. The cost (c), in dollars, to have any number of pizzas (p) delivered to a home is described by the function c = 8p + 3. Which statement is true?

A. The cost of 8 pizzas is $11.

B. The cost of 3 pizzas is $14.

C. Each pizza costs $8, and the delivery fee is $3.

D. Each pizza costs $3, and the delivery fee is $8.

Standard A1.2.1.2.2

The table below shows values of y as a function of x.

x y

2 10

6 25

14 55

26 100

34 130

Which linear equation describes the relationship between x and y?

A. y = 2.5x + 5

B. y = 3.75x + 2.5

C. y = 4x + 1

D. y = 5x




Standard A1.2.1

Hector’s family is on a car trip.

When they are 84 miles from home, Hector begins recording the distance they have driven (d ), in miles, after h hours as shown in the table below.

Distance from Home

Time

in Hours

(h)

Distance

in Miles

(d )

0 84

1 146

2 208

3 270

The pattern continues.

A. Write an equation to find the distance driven (d ), in miles, after a given number of hours ( h).


ASSESSMENT ANCHOR

A1.2.1 Functions






B. Hector also kept track of the remaining gasoline. The equation shown below can be used to find the gallons of gasoline remaining ( g) after driving a distance of d miles.

g = 16 – 1 } 20

d

Use the equation to find the missing values for gallons of gasoline remaining.

Gasoline Remaining by Mile

Distance

in Miles

(d )

Gallons of Gasoline

Remaining

( g)

100

200

300

C. Draw the graph of the line formed by the points in the table from part B.

g

d25 50 75 100 125 150 175 200 225 250 275 300

20

18

16

14

12

10

8

6

4

2

0

Gasoline Remaining by Mile

Ga

llo

ns o

f G

aso

lin

e R

em

ain

ing

Distance in Miles






D. Explain why the slope of the line drawn in part C must be negative.




Standard A1.2.1

Last summer Ben purchased materials to build model airplanes and then sold the finished models. He sold each model for the same amount of money. The table below shows the relationship between the number of model airplanes sold and the running total of Ben’s profit.

Ben’s Model-Airplane Sales

Model

Airplanes

Sold

Total Profi t

12 $68

15 $140

20 $260

22 $308

A. Write a linear equation, in slope-intercept form, to represent the amount of Ben’s total profit (y) based on the number of model airplanes (x) he sold.

y =

B. Determine a value of y that represents a situation where Ben did not make a profit from building model airplanes.

y-value:






C. How much did Ben spend on the materials he needed to build his models?

$

D. What is the least number of model airplanes Ben needed to sell in order to make a profit?

least number:




ASSESSMENT ANCHOR

A1.2.2 Coordinate Geometry


Standards

A1.2.2.1 Describe, compute, and/or use the rate of change (slope) of a line.

A1.2.2.1.1 Identify, describe, and/or use constant rates of change.

CC.2.2.8.C.2

CC.2.2.HS.C.1

CC.2.2.HS.C.2

CC.2.2.HS.C.3

CC.2.2.HS.C.5

CC.2.2.HS.C.6

CC.2.4.HS.B.1

A1.2.2.1.2 Apply the concept of linear rate of change (slope) to solve problems.

A1.2.2.1.3 Write or identify a linear equation when given • the graph of the line, • two points on the line, or • the slope and a point on the line.

Note: Linear equation may be in point-slope, standard, and/or slope-intercept form.

A1.2.2.1.4 Determine the slope and/or y-intercept represented by a linear equation or graph.


Standard A1.2.2.1.1

Jeff’s restaurant sells hamburgers. The amount charged for a hamburger ( h) is based on the cost for a plain hamburger plus an additional charge for each topping (t ) as shown in the equation below.

h = 0.60t + 5

What does the number 0.60 represent in the equation?

A. the number of toppings

B. the cost of a plain hamburger

C. the additional cost for each topping

D. the cost of a hamburger with 1 topping





Standard A1.2.2.1.2

A ball rolls down a ramp with a slope of 2 } 3 . At one

point the ball is 10 feet high, and at another point

the ball is 4 feet high, as shown in the diagram

below.

x ft

4 ft

10 ft

What is the horizontal distance (x), in feet, the ball travels as it rolls down the ramp from 10 feet high to 4 feet high?

A. 6

B. 9

C. 14

D. 15

Standard A1.2.2.1.3

A graph of a linear equation is shown below.

x

y

8642

–2–4–6–8

2 4 6 8–8 –6 –4 –2

Which equation describes the graph?

A. y = 0.5x – 1.5

B. y = 0.5x + 3

C. y = 2x – 1.5

D. y = 2x + 3

Standard A1.2.2.1.4

A juice machine dispenses the same amount of juice into a cup each time the machine is used. The equation below describes the relationship between the number of cups (x) into which juice is dispensed and the gallons of juice (y) remaining in the machine.

x + 12y = 180

How many gallons of juice are in the machine when it is full?

A. 12

B. 15

C. 168

D. 180




ASSESSMENT ANCHOR



Standards

A1.2.2.2 Analyze and/or interpret data on a scatter plot.

A1.2.2.2.1 Draw, identify, fi nd, and/or write an equation for a line of best fi t for a scatter plot.

CC.2.2.HS.C.6

CC.2.4.8.B.1

CC.2.4.HS.B.2

CC.2.4.HS.B.3


Standard A1.2.2.2.1

The scatter plot below shows the cost ( y) of ground shipping packages from Harrisburg, Pennsylvania, to Minneapolis, Minnesota, based on the package weight ( x).

x

y

Weight (in pounds)

Cos

t (in

dol

lars

)

10 20 30 40

Ground Shipping

0

30

20

10

Which equation best describes the line of best fit?

A. y = 0.37x + 1.57

B. y = 0.37x + 10.11

C. y = 0.68 x + 2.32

D. y = 0.68 x + 6.61




Standard A1.2.2

Georgia is purchasing treats for her classmates. Georgia can spend exactly $10.00 to purchase 25 fruit bars, each equal in price. Georgia can also spend exactly $10.00 to purchase 40 granola bars, each equal in price.

A. Write an equation that can be used to find all combinations of fruit bars ( x) and granola bars ( y) that will cost exactly $10.00.

equation:

B. Graph the equation from part A below.

Fruit Bars0

Gra

nola

Bar

s

50

40

30

20

10

10 20 30 40 50

Purchasing Treats

x

y


ASSESSMENT ANCHOR







C. What is the slope of the line graphed in part B?

slope:

D. Explain what the slope from part C means in the context of Georgia purchasing treats.




Standard A1.2.2

Ahava is traveling on a train.

The train is going at a constant speed of 80 miles per hour.

A. How many hours will it take for the train to travel 1,120 miles?

hours:

Ahava also considered taking an airplane. The airplane can travel the same 1,120 miles in 12 hours less time than it takes the train.

B. What is the speed of the airplane in miles per hour (mph)?

speed of the airplane: mph






Ahava is very concerned about the environment. The graph below displays the carbon dioxide (CO2), in metric tons, for each traveler on an airplane and each traveler on a train.

0Miles Traveled

Met

ric T

ons

of C

O2

x

y

0.40

0.32

0.24

0.16

0.08

200 400 600 800 1000

Carbon Footprint

KeyTraveler on anairplaneTraveler on atrain

C. What equation could be used to find the metric tons of CO2 produced (y) by a traveler on an airplane for x miles traveled?

equation:






On another trip, Ahava traveled to her destination on a train and returned home on an airplane. Her total carbon footprint for the trip was 0.42 metric tons of CO2 produced.

D. How far, in miles, is Ahava’s destination from her home?

miles:




ASSESSMENT ANCHOR

A1.2.3 Data Analysis


Standards

A1.2.3.1 Use measures of dispersion to describe a set of data.

A1.2.3.1.1 Calculate and/or interpret the range, quartiles, and interquartile range of data.

CC.2.4.HS.B.1

CC.2.4.HS.B.3


Standard A1.2.3.1.1

The daily high temperatures, in degrees Fahrenheit (°F), of a town are recorded for one year. The median high temperature is 62°F. The interquartile range of high temperatures is 32. Which statement is most likely true?

A. Approximately 25% of the days had a high temperature less than 30°F.

B. Approximately 25% of the days had a high temperature greater than 62°F.

C. Approximately 50% of the days had a high temperature greater than 62°F.

D. Approximately 75% of the days had a high temperature less than 94°F.




ASSESSMENT ANCHOR



Standards

A1.2.3.2 Use data displays in problem-solving settings and/or to make predictions.

A1.2.3.2.1 Estimate or calculate to make predictions based on a circle, line, bar graph, measure of central tendency, or other representation.

CC.2.4.HS.B.1

CC.2.4.HS.B.3

CC.2.4.HS.B.5

A1.2.3.2.2 Analyze data, make predictions, and/or answer questions based on displayed data (box-and-whisker plots, stem-and-leaf plots, scatter plots, measures of central tendency, or other representations).

A1.2.3.2.3 Make predictions using the equations or graphs of best-fi t lines of scatter plots.


Standard A1.2.3.2.1

Vy asked 200 students to select their favorite sport and then recorded the results in the bar graph below.

Favorite Sport

Sports

Vote

s

8070605040302010

0football basketball baseball hockey volleyball track

and field

Vy will ask another 80 students to select their favorite sport. Based on the information in the bar graph, how many more students of the next 80 asked are likely to select basketball rather than football as their favorite sport?

A. 10

B. 20

C. 25

D. 30





Standard A1.2.3.2.2

The points scored by a football team are shown in the stem-and-leaf plot below.

Football-Team Points

0123

6 2 3 4 7 0 3 4 4 7 8 8 80 7 8

Key1 | 3 = 13 points

What was the median number of points scored by the football team?

A. 24

B. 27

C. 28

D. 32





Standard A1.2.3.2.3

John recorded the weight of his dog Spot at different ages as shown in the scatter plot below.

Spot’s Weight

Age (in months)

Wei

ght (

in p

ound

s)

504540353025201510

5

20 4 6 8 1210 14 16

Based on the line of best fit, what will be Spot’s weight after 18 months?

A. 27 pounds

B. 32 pounds

C. 36 pounds

D. 50 pounds




ASSESSMENT ANCHOR



Standards

A1.2.3.3 Apply probability to practical situations.

A1.2.3.3.1 Find probabilities for compound events (e.g., fi nd probability of red and blue, fi nd probability of red or blue) and represent as a fraction, decimal, or percent.

CC.2.4.7.B.3

CC.2.4.HS.B.4

CC.2.4.HS.B.7


Standard A1.2.3.3.1

A number cube with sides labeled 1 through 6 is rolled two times, and the sum of the numbers that end face up is calculated. What is the probability that the sum of the numbers is 3?

A. 1 } 18

B. 1 } 12

C. 1 } 9

D. 1 } 2




Standard A1.2.3

The box-and-whisker plot shown below represents students’ test scores on Mr. Ali’s history test.

52 60 68 76 84 92 100

History-Test Scores

A. What is the range of scores for the history test?

range:

B. What is the best estimate for the percent of students scoring greater than 92 on the test?

percent: %


ASSESSMENT ANCHOR







Mr. Ali wanted more than half of the students to score 75 or greater on the test.

C. Explain how you know that more than half of the students did not score greater than 75.

Michael is a student in Mr. Ali’s class. The scatter plot below shows Michael’s test scores for each test given by Mr. Ali.

0Test Number

Te

st

Sc

ore

100908070605040302010

1 2 3 4 5 6 7 8

Michael’s Test Scores

D. Draw a line of best fit on the scatter plot above.




Standard A1.2.3

The weight, in pounds, of each wrestler on the high school wrestling team at the beginning of the season is listed below.

178 142 112 150 206 130

A. What is the median weight of the wrestlers?

median: pounds

B. What is the mean weight of the wrestlers?

mean: pounds






Two more wrestlers join the team during the season. The addition of these wrestlers has no effect on the mean weight of the wrestlers, but the median weight of the wrestlers increases 3 pounds.

C. Determine the weights of the two new wrestlers.

new wrestlers: pounds and pounds


KEYSTONE ALGEBRA I ASSESSMENT ANCHORS

KEY TO SAMPLE MULTIPLE-CHOICE ITEMS

Algebra I

Eligible Content Key

A1.1.1.1.1 C

A1.1.1.1.2 (top) B

A1.1.1.1.2 (bottom) C

A1.1.1.2.1 D

A1.1.1.3.1 A

A1.1.1.4.1 A

A1.1.1.5.1 C

A1.1.1.5.2 C

A1.1.1.5.3 D


A1.1.2.1.1 D

A1.1.2.1.2 D

A1.1.2.1.3 B

A1.1.2.2.1 B

A1.1.2.2.2 A


A1.1.3.1.1 B

A1.1.3.1.2 D

A1.1.3.1.3 D

A1.1.3.2.1 A

A1.1.3.2.2 B


A1.2.1.1.1 A

A1.2.1.1.2 B

A1.2.1.1.3 B

A1.2.1.2.1 C

A1.2.1.2.2 B


A1.2.2.1.1 C

A1.2.2.1.2 B

A1.2.2.1.3 D

A1.2.2.1.4 B

A1.2.2.2.1 D


A1.2.3.1.1 C

A1.2.3.2.1 A

A1.2.3.2.2 A

A1.2.3.2.3 B

A1.2.3.3.1 A

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va

ria

ble

in a

te

rm (

e.g

., 3

5 is t

he

co

eff

icie

nt of

35

x2y);

the

ab

sen

ce

of a

co

eff

icie

nt is

th

e s

am

e a

s a

1 b

ein

g p

rese

nt (e

.g.,

x is t

he s

am

e a

s 1

x).

Co

mb

ina

tio

n

An

un

ord

ere

d a

rra

nge

me

nt,

lis

tin

g o

r sele

ction

of

ob

jects

(e.g

., t

wo

-le

tte

r com

bin

ation

s o

f th

e th

ree

le

tte

rs X

, Y

, a

nd Z

wo

uld

be

XY

, X

Z,

an

d Y

Z; X

Y is t

he

sa

me

as Y

X a

nd

is n

ot co

un

ted

as a

diffe

ren

t co

mb

ina

tio

n).

A c

om

bin

ation

is s

imila

r to

, b

ut n

ot

the

sa

me

as,

a p

erm

uta

tio

n.

Co

mm

on

Lo

ga

rith

m

A loga

rith

m w

ith

ba

se

10

. It is w

ritt

en

log x

. T

he

co

mm

on

lo

ga

rith

m is t

he

po

we

r o

f 10

ne

cessa

ry t

o

equ

al a

giv

en

nu

mb

er

(i.e

., log x

= y

is e

qu

ivale

nt

to 1

0y =

x).

Co

mp

lex

Nu

mb

er

The

su

m o

r diffe

ren

ce

of

a r

ea

l nu

mb

er

an

d a

n im

agin

ary

nu

mb

er.

It

is w

ritt

en

in th

e fo

rm a

+ b

i,

wh

ere

a a

nd

b a

re r

ea

l n

um

be

rs a

nd i

is th

e im

agin

ary

un

it (

i.e

., i

=

1

). T

he

a is c

alle

d t

he r

ea

l p

art

,

an

d th

e b

i is

ca

lled

th

e im

agin

ary

pa

rt.

Co

mp

osit

e N

um

be

r A

ny n

atu

ral n

um

be

r w

ith

mo

re t

ha

n t

wo

fa

cto

rs (

e.g

., 6

is a

co

mp

osite

nu

mb

er

sin

ce

it h

as fo

ur

facto

rs: 1

, 2,

3,

and

6).

A c

om

po

site

nu

mb

er

is n

ot a

prim

e n

um

be

r.

Co

mp

ou

nd

(o

r C

om

bin

ed

) E

ve

nt

An

eve

nt th

at

is m

ad

e u

p o

f tw

o o

r m

ore

sim

ple

even

ts,

su

ch

as t

he

flip

pin

g o

f tw

o o

r m

ore

co

ins.

Co

mp

ou

nd

In

eq

ua

lity

W

hen t

wo

or

mo

re ine

qu

alit

ies a

re ta

ke

n t

oge

the

r a

nd w

ritt

en

with

th

e ine

qu

alit

ies c

on

ne

cte

d b

y t

he

w

ord

s a

nd

or

or

(e.g

., x

> 6

and

x <

12

, w

hic

h c

an a

lso b

e w

ritt

en

as 6

< x

< 1

2).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 6

A

pri

l 20

14

Co

ns

tan

t A

te

rm o

r exp

ressio

n w

ith

no v

aria

ble

in it. It

ha

s t

he

sa

me

va

lue a

ll th

e t

ime

.

Co

ord

ina

te P

lan

e

A p

lane

fo

rme

d b

y p

erp

end

icula

r nu

mb

er

line

s.

The

ho

rizo

nta

l nu

mb

er

line

is t

he

x-a

xis

, a

nd

the

ve

rtic

al n

um

be

r lin

e is th

e y

-axis

. T

he

po

int

wh

ere

th

e a

xe

s m

ee

t is

ca

lled

th

e o

rigin

. E

xa

mp

le o

f a

co

ord

inate

pla

ne

:

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 7

A

pri

l 20

14

Cu

be

Ro

ot

On

e o

f th

ree

equ

al fa

cto

rs (

roo

ts)

of

a n

um

be

r or

exp

ressio

n; a

rad

ica

l e

xp

ressio

n w

ith

a d

egre

e o

f 3

(e.g

.,

3a

). T

he

cu

be r

oo

t of

a n

um

be

r or

exp

ressio

n h

as t

he

sa

me

sig

n a

s t

he

nu

mb

er

or

exp

ressio

n

un

de

r th

e r

ad

ical (e

.g.,

3

_6

343

x =

–(7

x2)

an

d

36

343

x =

7x

2).

Cu

rve

of

Bes

t F

it (

for

a

Sc

att

er

Plo

t)

Se

e lin

e o

r curv

e o

f b

est fit

(fo

r a s

ca

tte

r plo

t).

De

gre

e (

of

a P

oly

no

mia

l)

The

va

lue

of

the

gre

ate

st

exp

one

nt in

a p

oly

no

mia

l.

De

pe

nd

en

t E

ve

nts

T

wo

or

mo

re e

ve

nts

in

wh

ich t

he

outc

om

e o

f o

ne

eve

nt

aff

ects

or

influ

en

ce

s t

he o

utc

om

e o

f th

e o

the

r e

ve

nt(

s).

De

pe

nd

en

t V

ari

ab

le

The

ou

tpu

t nu

mb

er

or

va

ria

ble

in a

rela

tio

n o

r fu

nctio

n th

at d

ep

end

s u

pon

ano

the

r variab

le, ca

lled

th

e

ind

epe

nde

nt va

ria

ble

, o

r in

pu

t n

um

be

r (e

.g.,

in

the

equ

atio

n y

= 2

x +

4,

y is th

e d

ep

end

ent va

ria

ble

sin

ce

its

va

lue

de

pe

nds o

n th

e v

alu

e o

f x).

It

is t

he

va

ria

ble

fo

r w

hic

h a

n e

qu

atio

n is s

olv

ed

. It

s v

alu

es

ma

ke

up

the

ran

ge

of

the

rela

tio

n o

r fu

nctio

n.

Do

main

(o

f a

Rela

tio

n o

r F

un

cti

on

) T

he

se

t of

all

po

ssib

le v

alu

es o

f th

e ind

epe

nde

nt

va

ria

ble

on

wh

ich a

fu

nctio

n o

r re

lation is a

llow

ed

to

o

pe

rate

. A

lso,

the

first n

um

be

rs in

th

e o

rde

red

pa

irs o

f a

re

lation

; th

e v

alu

es o

f th

e x

-co

ord

ina

tes in

(x

, y).

Eli

min

ati

on

Me

tho

d

Se

e lin

ea

r com

bin

ation

.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 8

A

pri

l 20

14

Eq

ua

tio

n

A m

ath

em

atica

l sta

tem

en

t o

r sen

ten

ce

tha

t says o

ne

ma

the

ma

tica

l e

xp

ressio

n o

r qu

an

tity

is e

qu

al to

a

no

the

r (e

.g.,

x +

5 =

y –

7).

An

equ

atio

n w

ill a

lwa

ys c

on

tain

an

equ

al sig

n (

=).

Es

tim

ati

on

Str

ate

gy

An

ap

pro

xim

atio

n b

ased

on a

ju

dgm

en

t; m

ay in

clu

de d

ete

rmin

ing a

pp

roxim

ate

va

lue

s, e

sta

blis

hin

g

the

rea

so

na

ble

ne

ss o

f a

nsw

ers

, a

sse

ssin

g t

he

am

ou

nt of

err

or

resu

ltin

g f

rom

estim

atio

n,

an

d/o

r d

ete

rmin

ing if

an

err

or

is w

ith

in a

cce

pta

ble

lim

its.

Ex

po

nen

t T

he

po

we

r to

wh

ich a

nu

mb

er

or

exp

ressio

n is r

ais

ed

. W

hen

th

e e

xpo

nen

t is

a f

raction

, th

e n

um

be

r or

exp

ressio

n c

an b

e r

ew

ritt

en

with

a r

ad

ical sig

n (

e.g

., x

3/4

=

43

x).

Se

e a

lso

po

sitiv

e e

xpo

ne

nt a

nd

ne

ga

tive

exp

on

ent.

Ex

po

nen

tia

l E

qu

ati

on

A

n e

qu

atio

n w

ith

va

ria

ble

s in its

exp

on

en

ts (

e.g

., 4

x =

50

). It

can

be

so

lved

by t

akin

g loga

rith

ms o

f b

oth

sid

es.

Ex

po

nen

tia

l E

xp

res

sio

n

An

exp

ressio

n in

wh

ich

th

e v

aria

ble

occu

rs in

th

e e

xp

on

ent

(such a

s 4

x r

ath

er

than

x4).

Oft

en

it

occu

rs

wh

en

a q

ua

ntity

ch

an

ge

s b

y t

he

sa

me

fa

cto

r fo

r ea

ch u

nit o

f tim

e (

e.g

., “

do

ub

les e

ve

ry y

ea

r” o

r “d

ecre

ase

s 2

% e

ach m

on

th”)

.

Ex

po

nen

tia

l F

un

cti

on

(o

r M

od

el)

A

fu

nctio

n w

ho

se

ge

nera

l e

qu

atio

n is y

= a

• b

x w

he

re a

an

d b

are

con

sta

nts

.

Ex

po

nen

tia

l G

row

th/D

ec

ay

A s

itu

atio

n w

he

re a

qu

an

tity

in

cre

ase

s o

r d

ecre

ase

s e

xp

on

entia

lly b

y t

he

sa

me

fa

cto

r o

ve

r tim

e;

it is

use

d fo

r such

ph

eno

me

na

as inflatio

n, p

op

ula

tio

n g

row

th,

rad

ioa

ctivity o

r de

pre

cia

tion

.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 9

A

pri

l 20

14

Ex

pre

ss

ion

A

ma

the

ma

tica

l p

hra

se t

ha

t in

clu

de

s o

pe

ration

s,

nu

mb

ers

, a

nd

/or

varia

ble

s (

e.g

., 2

x +

3y is a

n

alg

eb

raic

exp

ressio

n,

13

.4 –

4.7

is a

nu

me

ric e

xp

ressio

n).

An e

xp

ressio

n d

oe

s n

ot

co

nta

in a

n e

qu

al

sig

n (

=)

or

an

y t

yp

e o

f in

equ

alit

y s

ign

.

Fa

cto

r (n

ou

n)

The

nu

mb

er

or

exp

ressio

n t

ha

t is

mu

ltip

lied

by a

no

the

r to

ge

t a

pro

du

ct

(e.g

., 6

is a

fa

cto

r of

30

, a

nd

6x is a

fa

cto

r of

42

x2).

Fa

cto

r (v

erb

) T

o e

xp

ress o

r w

rite

a n

um

be

r, m

on

om

ial, o

r po

lyn

om

ial a

s a

pro

du

ct of

two

or

mo

re f

acto

rs.

Fa

cto

r a

Mo

no

mia

l T

o e

xp

ress a

mo

no

mia

l a

s t

he

pro

du

ct of

two

or

mo

re m

on

om

ials

.

Fa

cto

r a

Po

lyn

om

ial

To e

xp

ress a

po

lyn

om

ial a

s th

e p

rod

uct

of

mo

no

mia

ls a

nd

/or

po

lyn

om

ials

(e.g

., f

acto

rin

g t

he

po

lyn

om

ial x

2 +

x –

12

resu

lts in

th

e p

rod

uct

(x –

3)(

x +

4))

.

Fre

qu

en

cy

Ho

w o

ften

so

me

thin

g o

ccu

rs (

i.e

., th

e n

um

be

r of

tim

es a

n ite

m,

nu

mb

er,

or

eve

nt h

ap

pen

s in a

se

t of

da

ta).

Fu

nc

tio

n

A r

ela

tion

in w

hic

h e

ach

va

lue

of

an ind

ep

en

de

nt

va

ria

ble

is a

sso

cia

ted

with

a u

niq

ue

va

lue

of

a

de

pe

nde

nt

va

ria

ble

(e

.g.,

on

e e

lem

en

t of

the

do

ma

in is p

aire

d w

ith

on

e a

nd

on

ly o

ne

ele

me

nt

of

the

ra

nge

). I

t is

a m

ap

pin

g w

hic

h invo

lves e

ithe

r a o

ne

-to

-on

e c

orr

espo

nd

en

ce

or

a m

an

y-t

o-o

ne

co

rre

sp

ond

en

ce

, b

ut no

t a

one

-to

-ma

ny c

orr

esp

ond

en

ce

.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

0

Ap

ril 2

01

4

Fu

nd

am

en

tal

Co

un

tin

g

Pri

nc

iple

A

wa

y t

o c

alc

ula

te a

ll of

the p

ossib

le c

om

bin

ation

s o

f a

giv

en

nu

mb

er

of

eve

nts

. It

sta

tes t

hat

if t

he

re

are

x d

iffe

rent

wa

ys o

f d

oin

g o

ne

th

ing a

nd y

diffe

rent

wa

ys o

f do

ing a

noth

er

thin

g,

the

n the

re a

re

xy d

iffe

ren

t w

ays o

f d

oin

g b

oth

th

ings.

It u

se

s th

e m

ultip

lication

rule

.

Ge

om

etr

ic S

eq

uen

ce

A

n o

rde

red

lis

t of

nu

mb

ers

tha

t ha

s th

e s

am

e r

atio

be

twe

en

co

nse

cu

tive

te

rms (

e.g

., 1

, 7

, 4

9, 3

43, …

is

a g

eo

me

tric

se

qu

ence

tha

t ha

s a

ratio o

f 7/1

betw

ee

n c

on

se

cu

tive t

erm

s;

ea

ch

te

rm a

fte

r th

e f

irst

term

ca

n b

e fo

un

d b

y m

ultip

lyin

g t

he

pre

vio

us t

erm

by a

co

nsta

nt,

in

th

is c

ase

th

e n

um

be

r 7 o

r 7/1

).

Gre

ate

st

Co

mm

on

Fac

tor

(GC

F)

The

la

rge

st fa

cto

r th

at tw

o o

r m

ore

nu

mb

ers

or

alg

eb

raic

te

rms h

ave

in

co

mm

on

. In

so

me

ca

se

s th

e

GC

F m

ay b

e 1

or

on

e o

f th

e a

ctu

al n

um

be

rs (

e.g

., t

he

GC

F o

f 1

8x

3 a

nd

24

x5 is 6

x3).

Ima

gin

ary

Nu

mb

er

The

squ

are

roo

t of

a n

ega

tive

nu

mb

er,

or

the o

ppo

site

of

the

squ

are

roo

t of

a n

ega

tive n

um

be

r. It is

wri

tte

n in

th

e fo

rm b

i, w

he

re b

is a

rea

l n

um

be

r a

nd i

is th

e im

agin

ary

roo

t (i.e

., i

=

1

or

i2 =

–1

).

Ind

ep

en

de

nt

Eve

nt(

s)

Tw

o o

r m

ore

eve

nts

in

wh

ich t

he

outc

om

e o

f o

ne

eve

nt

do

es n

ot aff

ect

the

ou

tco

me

of

the o

the

r e

ve

nt(

s)

(e.g

., to

ssin

g a

co

in a

nd r

olli

ng a

num

be

r cub

e a

re in

dep

en

den

t e

ve

nts

). T

he

pro

ba

bili

ty o

f tw

o in

de

pen

den

t e

ve

nts

(A

and

B)

occu

rrin

g is w

ritte

n P

(A a

nd

B)

or

P(A

B

) an

d e

qu

als

P(A

) •

P(B

)

(i.e

., t

he

pro

du

ct of

the p

rob

ab

ilities o

f th

e t

wo

in

div

idu

al e

ven

ts).

Ind

ep

en

de

nt

Va

ria

ble

T

he

in

pu

t n

um

be

r or

va

ria

ble

in a

rela

tio

n o

r fu

nction

wh

ose

va

lue

is s

ub

ject

to c

ho

ice.

It is n

ot

de

pe

nde

nt u

po

n a

ny o

the

r valu

es.

It is u

sua

lly t

he

x-v

alu

e o

r th

e x

in f(x

). I

t is

gra

ph

ed

on

the

x-a

xis

. It

s v

alu

es m

ake

up

th

e d

om

ain

of

the

re

lation o

r fu

nctio

n.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

1

Ap

ril 2

01

4

Ine

qu

ali

ty

A m

ath

em

atica

l se

nte

nce

tha

t con

tain

s a

n in

equ

alit

y s

ym

bo

l (i.e

., >

, <

, ≥, ≤,

or

≠).

It

co

mp

are

s t

wo

qu

an

titie

s.

Th

e s

ym

bo

l >

me

an

s g

rea

ter

than

, th

e s

ym

bo

l <

me

an

s le

ss t

ha

n, th

e s

ym

bo

l

me

an

s

gre

ate

r th

an

or

equ

al to

, th

e s

ym

bo

l m

ea

ns le

ss t

ha

n o

r equ

al to

, a

nd

th

e s

ym

bo

l m

ea

ns n

ot

equ

al to

.

Inte

ge

r A

na

tura

l nu

mb

er,

th

e a

dd

itiv

e in

ve

rse

of

a n

atu

ral nu

mb

er,

or

ze

ro.

An

y n

um

be

r fr

om

th

e s

et

of

nu

mb

ers

re

pre

se

nte

d b

y {

…,

–3

, –2

, –1,

0, 1

, 2

, 3

, …

}.

Inte

rqu

art

ile

Ra

ng

e (

of

Da

ta)

The

diffe

ren

ce

be

twe

en

the

first

(lo

we

r) a

nd

th

ird

(up

pe

r) q

ua

rtile

. It

rep

rese

nts

the

sp

rea

d o

f th

e

mid

dle

50%

of

a s

et

of d

ata

.

Inve

rse

(o

f a

Re

lati

on

) A

rela

tion

in w

hic

h t

he

co

ord

inate

s in e

ach

ord

ere

d p

air

are

sw

itch

ed

fro

m a

giv

en

rela

tion

. T

he

po

int

(x,

y)

be

co

me

s (

y,

x),

so

(3

, 8

) w

ou

ld b

eco

me

(8,

3).

Irra

tio

nal

Nu

mb

er

A r

ea

l nu

mb

er

tha

t ca

nn

ot b

e w

ritte

n a

s a

sim

ple

fra

ction

(i.e

., t

he

ratio

of

two

in

tege

rs).

It

is a

no

n-

term

ina

tin

g (

infinite

) and

non

-rep

ea

tin

g d

ecim

al. T

he

squa

re r

oot

of a

ny p

rim

e n

um

be

r is

irr

atio

na

l, a

s

are

π a

nd

e.

Le

as

t (o

r L

ow

es

t) C

om

mo

n

Mu

ltip

le (

LC

M)

The

sm

alle

st

nu

mb

er

or

exp

ressio

n th

at

is a

co

mm

on

mu

ltip

le o

f tw

o o

r m

ore

nu

mb

ers

or

alg

eb

raic

te

rms,

oth

er

tha

n z

ero

.

Lik

e T

erm

s

Mo

no

mia

ls t

ha

t con

tain

the

sa

me

va

ria

ble

s a

nd

co

rre

sp

ond

ing p

ow

ers

an

d/o

r ro

ots

. O

nly

th

e

co

eff

icie

nts

can

be

diffe

ren

t (e

.g.,

4x

3 a

nd

12x

3).

Lik

e t

erm

s c

an

be

ad

de

d o

r su

btr

acte

d.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

2

Ap

ril 2

01

4

Lin

e G

rap

h

A g

rap

h th

at u

se

s a

lin

e o

r lin

e s

egm

en

ts t

o c

on

ne

ct

da

ta p

oin

ts,

plo

tte

d o

n a

coo

rdin

ate

pla

ne

, u

su

ally

to

sh

ow

tre

nd

s o

r cha

nge

s in

da

ta o

ver

tim

e.

Mo

re b

roa

dly

, a

gra

ph

to

re

pre

sen

t th

e

rela

tion

sh

ip b

etw

ee

n t

wo

co

ntin

uou

s v

aria

ble

s.

Lin

e o

r C

urv

e o

f B

es

t F

it (

for

a S

ca

tte

r P

lot)

A

lin

e o

r cu

rve d

raw

n o

n a

sca

tte

r p

lot

to b

est e

stim

ate

th

e r

ela

tio

nship

be

twe

en

tw

o s

ets

of d

ata

. It

d

escrib

es th

e tre

nd o

f th

e d

ata

. D

iffe

ren

t m

ea

su

res a

re p

ossib

le t

o d

escrib

e t

he

be

st fit. T

he

mo

st

co

mm

on

is a

lin

e o

r curv

e t

ha

t m

inim

ize

s t

he

su

m o

f th

e s

qu

are

s o

f th

e e

rro

rs (

ve

rtic

al d

ista

nce

s)

from

th

e d

ata

po

ints

to

th

e lin

e.

The

lin

e o

f b

est fit is

a s

ub

se

t of

the

cu

rve o

f be

st fit. E

xa

mp

les o

f a

lin

e o

f b

est fit a

nd

a c

urv

e o

f b

est fit:

Lin

ea

r C

om

bin

ati

on

A

me

tho

d b

y w

hic

h a

syste

m o

f lin

ea

r e

qu

ation

s c

an b

e s

olv

ed.

It u

se

s a

dd

itio

n o

r sub

tractio

n in

co

mb

ina

tio

n w

ith

mu

ltip

lica

tio

n o

r div

isio

n to

elim

ina

te o

ne

of

the

varia

ble

s in o

rde

r to

so

lve

fo

r th

e

oth

er

va

ria

ble

.

Lin

ea

r E

qu

ati

on

A

n e

qu

atio

n fo

r w

hic

h t

he

gra

ph

is a

str

aig

ht

line

(i.e., a

po

lyn

om

ial e

qu

atio

n o

f th

e f

irst d

egre

e o

f th

e

form

Ax +

By =

C,

wh

ere

A,

B,

an

d C

are

re

al n

um

be

rs a

nd

wh

ere

A a

nd

B a

re n

ot

both

ze

ro;

an

e

qu

atio

n in

wh

ich t

he

va

ria

ble

s a

re n

ot

mu

ltip

lied

by o

ne

ano

the

r or

rais

ed

to

an

y p

ow

er

oth

er

than

1).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

3

Ap

ril 2

01

4

Lin

ea

r F

un

cti

on

A

fu

nctio

n fo

r w

hic

h t

he

gra

ph

is a

non

-ve

rtic

al str

aig

ht

line

. It

is a

first

de

gre

e p

oly

no

mia

l o

f th

e

co

mm

on

fo

rm f

(x)

= m

x +

b,

wh

ere

m a

nd

b a

re c

on

sta

nts

and

x is a

rea

l va

ria

ble

. T

he

consta

nt m

is

ca

lled

th

e s

lop

e a

nd b

is c

alle

d th

e y

-inte

rcep

t. I

t ha

s a

co

nsta

nt

rate

of

cha

nge

.

Lin

ea

r In

eq

uali

ty

The

rela

tio

n o

f tw

o e

xp

ressio

ns u

sin

g th

e s

ym

bo

ls <

, >

, ≤,

≥,

or

≠ a

nd

wh

ose

bou

nd

ary

is a

str

aig

ht

line

. T

he lin

e d

ivid

es the

co

ord

ina

te p

lan

e into

tw

o p

art

s.

If t

he

ine

qua

lity is e

ith

er

≤ o

r ≥, th

en

th

e

bo

un

da

ry is s

olid

. If

the

in

equ

alit

y is e

ith

er

< o

r >

, th

en

the

bo

un

da

ry is d

ash

ed

. If

th

e in

equ

alit

y is ≠

, th

en t

he

so

lutio

n c

on

tain

s e

ve

ryth

ing e

xce

pt fo

r th

e b

ou

nda

ry.

Lo

ga

rith

m

Th

e e

xp

one

nt

requ

ire

d t

o p

rod

uce

a g

iven n

um

be

r (e

.g.,

sin

ce

2 r

ais

ed

to

a p

ow

er

of

5 is 3

2,

the

loga

rith

m b

ase

2 o

f 3

2 is 5

; th

is is w

ritt

en

as log

2 3

2 =

5).

Tw

o f

reque

ntly u

se

d b

ase

s a

re 1

0 (

co

mm

on

loga

rith

m)

an

d e

(n

atu

ral lo

ga

rith

m).

Wh

en

a lo

ga

rith

m is w

ritte

n w

ith

ou

t a

ba

se

, it is u

nd

ers

too

d t

o b

e

ba

se

10

.

Lo

ga

rith

mic

Eq

uati

on

A

n e

qu

atio

n w

hic

h c

onta

ins a

loga

rith

m o

f a

va

ria

ble

or

nu

mb

er.

Som

etim

es it

is s

olv

ed

by r

ew

ritin

g

the

equ

atio

n in

exp

on

en

tia

l fo

rm a

nd

so

lvin

g f

or

the

va

ria

ble

(e

.g.,

log

2 3

2 =

5 is the

sa

me

as 2

5 =

32

).

It is a

n in

ve

rse fu

nctio

n o

f th

e e

xp

on

entia

l fu

nctio

n.

Ma

pp

ing

T

he

ma

tch

ing o

r pa

irin

g o

f o

ne

se

t of

nu

mb

ers

to

an

oth

er

by u

se

of

a r

ule

. A

nu

mb

er

in t

he d

om

ain

is

ma

tch

ed

or

pa

ire

d w

ith

a n

um

be

r in

th

e r

an

ge (

or

a r

ela

tion

or

fun

ctio

n).

It m

ay b

e a

on

e-t

o-o

ne

co

rre

sp

ond

en

ce

, a

on

e-t

o-m

an

y c

orr

esp

on

den

ce

, o

r a m

an

y-t

o-o

ne

co

rre

sp

ond

en

ce

.

Ma

xim

um

Va

lue

(o

f a

Gra

ph

) T

he

va

lue

of

the

de

pe

nd

en

t va

ria

ble

fo

r th

e h

igh

est

po

int

on

the

gra

ph

of

a c

urv

e.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

4

Ap

ril 2

01

4

Me

an

A

me

asu

re o

f ce

ntr

al te

nde

ncy t

ha

t is

ca

lcula

ted

by a

dd

ing a

ll th

e v

alu

es o

f a s

et of

da

ta a

nd

div

idin

g

tha

t su

m b

y t

he

to

tal nu

mb

er

of

va

lue

s.

Unlik

e m

ed

ian

, th

e m

ea

n is s

en

sitiv

e to

ou

tlie

r va

lue

s.

It is

als

o c

alle

d “

arith

me

tic m

ea

n”

or

“ave

rage

”.

Me

as

ure

of

Cen

tral

Te

nd

en

cy

A m

ea

su

re o

f lo

ca

tion

of

the

mid

dle

(cen

ter)

of

a d

istr

ibu

tion

of

a s

et

of

da

ta (

i.e., h

ow

da

ta c

luste

rs).

T

he

th

ree

mo

st

co

mm

on

me

asu

res o

f cen

tral te

nd

en

cy a

re m

ea

n,

me

dia

n,

and

mo

de

.

Me

as

ure

of

Dis

pers

ion

A

me

asu

re o

f th

e w

ay in

wh

ich t

he

dis

trib

ution o

f a s

et of

da

ta is s

pre

ad

ou

t. In

ge

ne

ral th

e m

ore

sp

rea

d o

ut a

dis

trib

ution

is,

the

la

rge

r th

e m

ea

su

re o

f d

ispe

rsio

n.

Ran

ge

an

d inte

rqu

art

ile r

an

ge

are

tw

o m

ea

su

res o

f d

ispers

ion

.

Me

dia

n

A m

ea

su

re o

f ce

ntr

al te

nde

ncy t

ha

t is

th

e m

idd

le v

alu

e in a

n o

rde

red

se

t of

da

ta o

r th

e a

vera

ge

of

the

tw

o m

idd

le v

alu

es w

he

n t

he

se

t h

as t

wo

mid

dle

va

lue

s (

occu

rs w

he

n t

he

se

t of

da

ta h

as a

n e

ve

n

nu

mb

er

of

da

ta p

oin

ts).

It

is th

e v

alu

e h

alfw

ay t

hro

ugh

th

e o

rde

red

se

t of

data

, b

elo

w a

nd

ab

ove

wh

ich

the

re is a

n e

qu

al nu

mb

er

of

da

ta v

alu

es.

It is g

en

era

lly a

go

od

de

scriptive m

ea

su

re f

or

skew

ed

da

ta o

r d

ata

with

ou

tlie

rs.

Min

imu

m V

alu

e (

of

a G

rap

h)

The

va

lue

of

the

de

pe

nd

en

t va

ria

ble

fo

r th

e low

est

po

int

on

th

e g

raph

of

a c

urv

e.

Mo

de

A

me

asu

re o

f ce

ntr

al te

nde

ncy t

ha

t is

th

e v

alu

e o

r va

lue

s th

at

occu

r(s)

mo

st

oft

en

in

a s

et of

da

ta.

A

se

t of

data

ca

n h

ave o

ne

mo

de

, m

ore

tha

n o

ne

mo

de

, o

r no m

od

e.

Mo

no

mia

l A

po

lyn

om

ial w

ith

on

ly o

ne

te

rm;

it c

on

tain

s n

o a

dd

itio

n o

r sub

traction

. It

ca

n b

e a

nu

mb

er,

a v

aria

ble

,

or

a p

rod

uct

of

nu

mb

ers

an

d/o

r m

ore

va

ria

ble

s (

e.g

., 2

• 5

or

x3y

4 o

r 2

4 3r

).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

5

Ap

ril 2

01

4

Mu

ltip

lic

ati

ve

In

ve

rse

T

he

recip

roca

l of

a n

um

be

r (i.e

., f

or

an

y n

on

-ze

ro n

um

be

r a

, th

e m

ultip

licative inve

rse is 1 a

; fo

r an

y

ratio

na

l n

um

be

r b c

, w

he

re b

≠ 0

an

d c

≠ 0

, th

e m

ultip

licative inve

rse is c b

). A

ny n

um

be

r an

d its

mu

ltip

licative inve

rse h

ave

a p

rodu

ct of

1 (

e.g

.,

1 4 is th

e m

ultip

licative in

ve

rse o

f 4

sin

ce

4 •

1 4

= 1

;

like

wis

e,

the

mu

ltip

licative inve

rse o

f 1 4

is 4

sin

ce

1 4

• 4

= 1

).

Mu

tuall

y E

xc

lus

ive

Eve

nts

T

wo

eve

nts

tha

t ca

nnot

occu

r at

the

sa

me

tim

e (

i.e.,

eve

nts

tha

t h

ave

no o

utc

om

es in

co

mm

on

). If

two

e

ve

nts

A a

nd

B a

re m

utu

ally

exclu

siv

e,

the

n th

e p

rob

ab

ility

of

A o

r B

occu

rrin

g is t

he

su

m o

f th

eir

ind

ivid

ua

l p

rob

ab

ilities:

P(A

B

) =

P(A

) +

P(B

). A

lso d

efined

as w

he

n th

e inte

rsection

of

two

se

ts is

em

pty

, w

ritte

n a

s A

B

= Ø

.

Na

tura

l L

og

ari

thm

A

loga

rith

m w

ith

ba

se

e.

It is w

ritt

en

ln

x.

The

na

tura

l lo

ga

rith

m is th

e p

ow

er

of

e n

ece

ssa

ry t

o e

qu

al a

giv

en

nu

mb

er

(i.e

., ln

x =

y is e

qu

ivale

nt to

e y =

x).

The

con

sta

nt

e is

an

irr

atio

na

l n

um

be

r w

ho

se

va

lue

is

ap

pro

xim

ate

ly 2

.71

82

8…

.

Na

tura

l N

um

be

r A

co

un

tin

g n

um

be

r. A

nu

mb

er

rep

rese

ntin

g a

po

sitiv

e,

wh

ole

am

ou

nt.

An

y n

um

be

r fr

om

th

e s

et

of

nu

mb

ers

re

pre

se

nte

d b

y {

1,

2,

3, …

}. S

om

etim

es,

it is r

efe

rred

to a

s a

“p

ositiv

e inte

ge

r”.

Ne

ga

tive

Ex

po

nen

t A

n e

xp

one

nt th

at

ind

icate

s a

recip

roca

l th

at

ha

s t

o b

e ta

ken

befo

re th

e e

xp

one

nt ca

n b

e a

pp

lied

(e.g

.,

2

215

5

or

1x

xa

a

). It

is u

se

d in

scie

ntific n

ota

tio

n f

or

nu

mb

ers

be

twe

en

–1

and

1.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

6

Ap

ril 2

01

4

Nu

mb

er

Lin

e

A g

rad

ua

ted

str

aig

ht

line

th

at

rep

resen

ts th

e s

et

of

all

rea

l n

um

be

rs in o

rde

r. T

yp

ically

, it is m

ark

ed

sh

ow

ing inte

ge

r va

lue

s.

Od

ds

A

co

mp

ariso

n,

in r

atio

fo

rm (

as a

fra

ctio

n o

r w

ith

a c

olo

n),

of

ou

tco

me

s.

“Odd

s in f

avo

r” (

or

sim

ply

“o

dd

s”)

is th

e r

atio o

f fa

vo

rab

le o

utc

om

es t

o u

nfa

vo

rab

le o

utc

om

es (

e.g

., t

he o

dd

s in

fa

vo

r of

pic

kin

g a

re

d h

at

wh

en

th

ere

are

3 r

ed

ha

ts a

nd

5 n

on

-red

ha

ts is 3

:5).

“O

dd

s a

ga

inst”

is th

e r

atio o

f u

nfa

vo

rab

le

ou

tco

me

s t

o fa

vo

rab

le o

utc

om

es (

e.g

., th

e o

dd

s a

ga

inst p

ickin

g a

red

hat

wh

en

th

ere

are

3 r

ed

ha

ts

an

d 5

no

n-r

ed

ha

ts is 5

:3).

Ord

er

of

Op

era

tio

ns

Rule

s d

escrib

ing w

ha

t o

rde

r to

use in

eva

lua

tin

g e

xp

ressio

ns:

(1)

Pe

rfo

rm o

pe

ration

s in

gro

up

ing s

ym

bo

ls (

pa

ren

the

se

s a

nd b

racke

ts),

(2

) E

va

lua

te e

xp

on

ential exp

ressio

ns a

nd r

ad

ical e

xp

ressio

ns f

rom

le

ft to

rig

ht,

(3

) M

ultip

ly o

r d

ivid

e f

rom

left

to

rig

ht,

(4

) A

dd

or

su

btr

act fr

om

le

ft t

o r

igh

t.

Ord

ere

d P

air

A

pa

ir o

f nu

mb

ers

used

to

lo

ca

te a

po

int o

n a

co

ord

ina

te p

lane

, o

r th

e s

olu

tio

n o

f an

equ

atio

n in t

wo

va

ria

ble

s.

The

first n

um

be

r te

lls h

ow

fa

r to

move

ho

rizo

nta

lly,

an

d th

e s

eco

nd

nu

mb

er

tells

ho

w f

ar

to

mo

ve

ve

rtic

ally

; w

ritte

n in

the

fo

rm (

x-c

oo

rdin

ate

, y-c

oo

rdin

ate

). O

rde

r m

att

ers

: th

e p

oin

t (x

, y)

is n

ot

the

sa

me

as (

y,

x).

Ori

gin

T

he

po

int

(0,

0)

on

a c

oo

rdin

ate

pla

ne

. It is t

he

po

int

of

inte

rsection fo

r th

e x

-axis

and

the

y-a

xis

.

Ou

tlie

r A

va

lue t

ha

t is

mu

ch

gre

ate

r o

r m

uch

le

ss t

han

the

rest of

the

da

ta.

It is d

iffe

ren

t in

so

me

wa

y f

rom

th

e

ge

ne

ral p

att

ern

of

da

ta.

It d

ire

ctly s

tan

ds o

ut fr

om

th

e r

est

of

the d

ata

. S

om

etim

es it

is r

efe

rre

d t

o a

s

an

y d

ata

po

int

mo

re t

ha

n 1

.5 inte

rqu

art

ile r

ange

s g

rea

ter

than

the

up

pe

r (t

hird

) qu

art

ile o

r le

ss t

ha

n

the

lo

we

r (f

irst)

qu

art

ile.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

7

Ap

ril 2

01

4

Pa

tte

rn (

or

Se

qu

en

ce)

A s

et of

nu

mb

ers

arr

ange

d in

ord

er

(or

in a

sequ

en

ce

). T

he

nu

mb

ers

an

d t

he

ir a

rra

nge

me

nt

are

d

ete

rmin

ed

by a

rule

, in

clu

din

g r

epe

titio

n a

nd g

row

th/d

eca

y r

ule

s.

Se

e a

rith

me

tic s

equ

en

ce

and

ge

om

etr

ic s

equ

en

ce

.

Pe

rfec

t S

qu

are

A

nu

mb

er

wh

ose

squ

are

ro

ot

is a

wh

ole

nu

mb

er

(e.g

., 2

5 is a

pe

rfect

squ

are

sin

ce

25

= 5

). A

pe

rfe

ct

squa

re c

an

be

fo

und

by r

ais

ing a

wh

ole

nu

mb

er

to t

he

se

cond

po

we

r (e

.g.,

52 =

25

).

Pe

rmu

tati

on

A

n o

rde

red

arr

an

ge

me

nt

of

ob

jects

fro

m a

giv

en

set

in w

hic

h t

he

ord

er

of

the

ob

jects

is s

ign

ific

an

t (e

.g.,

tw

o-lett

er

pe

rmu

tatio

ns o

f th

e th

ree le

tters

X,

Y,

an

d Z

wo

uld

be

XY

, Y

X,

XZ

, Z

X,

YZ

, a

nd

ZY

). A

p

erm

uta

tio

n is s

imila

r to

, bu

t no

t th

e s

am

e a

s, a

co

mb

ina

tion

.

Po

int-

Slo

pe

Fo

rm (

of

a

Lin

ea

r E

qu

ati

on

) A

n e

qu

atio

n o

f a

str

aig

ht,

no

n-v

ert

ical lin

e w

ritt

en

in t

he

fo

rm y

– y

1 =

m(x

– x

1),

wh

ere

m is th

e s

lop

e

of

the

lin

e a

nd

(x

1,

y1)

is a

giv

en

po

int o

n t

he

lin

e.

Po

lyn

om

ial

A

n a

lge

bra

ic e

xp

ressio

n t

ha

t is

a m

on

om

ial o

r th

e s

um

or

diffe

ren

ce

of

two

or

mo

re m

on

om

ials

(e.g

.,

6a

or

5a

2 +

3a

– 1

3 w

he

re t

he

exp

one

nts

are

na

tura

l n

um

be

rs).

Po

lyn

om

ial

Fu

nc

tio

n

A f

un

ctio

n o

f th

e fo

rm f

(x)

= a

nx

n +

an–1x

n–1 +

… +

a1x +

a0,

wh

ere

an ≠

0 a

nd

na

tura

l nu

mb

er

n is the

d

egre

e o

f th

e p

oly

no

mia

l.

Po

sit

ive

Ex

po

nen

t In

dic

ate

s h

ow

ma

ny t

ime

s a

ba

se

nu

mb

er

is m

ultip

lied

by its

elf.

In th

e e

xp

ressio

n x

n,

n is th

e p

ositiv

e

exp

on

en

t, a

nd

x is th

e b

ase

nu

mb

er

(e.g

., 2

3 =

2 •

2 •

2).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

8

Ap

ril 2

01

4

Po

we

r T

he

va

lue

of

the

exp

one

nt

in a

te

rm.

The

exp

ressio

n a

n is r

ea

d “

a to

th

e p

ow

er

of

n.”

To r

ais

e a

nu

mb

er,

a,

to th

e p

ow

er

of

ano

the

r w

ho

le n

um

be

r, n

, is

to

mu

ltip

ly a

by its

elf n

tim

es (

e.g

., th

e n

um

be

r

43 is r

ea

d “

fou

r to

th

e th

ird

po

we

r” a

nd

rep

rese

nts

4 •

4 •

4).

Po

we

r o

f a P

ow

er

An

exp

ressio

n o

f th

e form

(a

m)n

. It

ca

n b

e f

ou

nd

by m

ultip

lyin

g t

he

exp

one

nts

(e.g

.,

(23)4

= 2

3•4

= 2

12 =

4,0

96

).

Po

we

rs o

f P

rod

ucts

A

n e

xp

ressio

n o

f th

e form

am •

an. It

ca

n b

e fou

nd

by a

dd

ing t

he

expo

nen

ts w

he

n m

ultip

lyin

g p

ow

ers

tha

t h

ave

the

sa

me

base

(e.g

., 2

3 •

24 =

23+

4 =

27 =

12

8).

Pri

me

Nu

mb

er

An

y n

atu

ral n

um

be

r w

ith

exa

ctly t

wo

fa

cto

rs, 1 a

nd its

elf (

e.g

., 3

is a

prim

e n

um

be

r sin

ce

it

ha

s o

nly

tw

o f

acto

rs: 1

and

3).

[N

ote

: S

ince

1 h

as o

nly

on

e f

acto

r, its

elf, it is n

ot

a p

rim

e n

um

be

r.]

A p

rim

e

nu

mb

er

is n

ot a

co

mp

osite

nu

mb

er.

Pro

ba

bilit

y

A n

um

be

r fr

om

0 t

o 1

(o

r 0%

to

10

0%

) th

at

ind

icate

s h

ow

lik

ely

an

eve

nt

is to

ha

pp

en

. A

very

un

like

ly

eve

nt h

as a

pro

bab

ility

ne

ar

0 (

or

0%

) w

hile

a v

ery

lik

ely

eve

nt

ha

s a

pro

bab

ility

ne

ar

1 (

or

10

0%

). It

is

wri

tte

n a

s a

ratio (

fra

ctio

n, d

ecim

al, o

r e

qu

ivale

nt p

erc

ent)

. T

he n

um

be

r of

wa

ys a

n e

ve

nt

co

uld

h

app

en

(fa

vo

rab

le o

utc

om

es)

is p

lace

d o

ve

r th

e to

tal n

um

be

r of

even

ts (

tota

l p

ossib

le o

utc

om

es)

tha

t co

uld

ha

pp

en.

A p

robab

ility

of

0 m

ea

ns it

is im

po

ssib

le,

an

d a

pro

ba

bili

ty o

f 1

me

an

s it

is c

ert

ain

.

Pro

ba

bilit

y o

f a

Co

mp

ou

nd

(o

r C

om

bin

ed

) E

ve

nt

The

re a

re t

wo

typ

es:

1.

Th

e u

nio

n o

f tw

o e

ve

nts

A a

nd

B,

wh

ich is th

e p

rob

ab

ility

of

A o

r B

occu

rrin

g.

This

is

rep

rese

nte

d a

s P

(A

B)

= P

(A)

+ P

(B)

– P

(A)

• P

(B).

2.

Th

e in

ters

ection o

f tw

o e

ve

nts

A a

nd B

, w

hic

h is th

e p

roba

bili

ty o

f A

an

d B

occu

rrin

g.

This

is

rep

rese

nte

d a

s P

(A

B)

= P

(A)

• P

(B).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 1

9

Ap

ril 2

01

4

Qu

ad

ran

ts

The

fou

r re

gio

ns o

f a

co

ord

inate

pla

ne

tha

t a

re s

epa

rate

d b

y t

he

x-a

xis

an

d th

e y

-axis

, a

s s

ho

wn

b

elo

w.

(1

) T

he

first

qu

ad

ran

t (Q

ua

dra

nt I)

con

tain

s a

ll th

e p

oin

ts w

ith

po

sitiv

e x

an

d p

ositiv

e y

coo

rdin

ate

s

(e.g

., (

3,

4))

. (2

) T

he

se

co

nd q

ua

dra

nt

(Qu

ad

ran

t II)

co

nta

ins a

ll th

e p

oin

ts w

ith

ne

ga

tive

x a

nd

po

sitiv

e y

co

ord

inate

s (

e.g

., (

–3, 4

)).

(3)

Th

e t

hird q

ua

dra

nt

(Qua

dra

nt II

I) c

on

tain

s a

ll th

e p

oin

ts w

ith

ne

ga

tive x

and

ne

ga

tive

y

co

ord

inate

s (

e.g

., (

–3,

–4

)).

(4)

Th

e f

ou

rth q

ua

dra

nt

(Qu

ad

ran

t IV

) con

tain

s a

ll th

e p

oin

ts w

ith

po

sitiv

e x

an

d n

ega

tive

y

co

ord

inate

s (

e.g

., (

3,

–4

)).

Qu

ad

rati

c E

qu

ati

on

A

n e

qu

atio

n th

at

can

be

writt

en

in

the

sta

nda

rd f

orm

ax

2 +

bx +

c =

0,

wh

ere

a,

b,

an

d c

are

rea

l

nu

mb

ers

an

d a

do

es n

ot

equ

al ze

ro.

Th

e h

ighe

st

po

we

r o

f th

e v

aria

ble

is 2

. It h

as,

at

mo

st,

tw

o

so

lution

s.

The g

rap

h is a

pa

rab

ola

.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

0

Ap

ril 2

01

4

Qu

ad

rati

c F

orm

ula

T

he

so

lutio

ns o

r ro

ots

of

a q

ua

dra

tic e

qu

atio

n in t

he

fo

rm a

x2 +

bx +

c =

0,

wh

ere

a ≠

0, a

re g

iven

by

the

fo

rmu

la

24

2

bb

ac

xa

.

Qu

ad

rati

c F

un

cti

on

A

fu

nctio

n th

at

can

be

exp

ressed

in

th

e fo

rm f

(x)

= a

x2 +

bx +

c,

wh

ere

a ≠

0 a

nd

th

e h

igh

est

po

we

r o

f

the

va

ria

ble

is 2

. T

he

gra

ph

is a

pa

rab

ola

.

Qu

art

ile

On

e o

f th

ree

va

lue

s tha

t d

ivid

es a

se

t of

da

ta in

to fo

ur

equa

l p

art

s:

1.

Me

dia

n d

ivid

es a

se

t of

da

ta into

tw

o e

qu

al p

art

s.

2.

Lo

we

r qu

art

ile (

25

th p

erc

en

tile

) is

th

e m

ed

ian

of

the

lo

we

r ha

lf o

f th

e d

ata

. 3

. U

pp

er

qu

art

ile (

75

th p

erc

en

tile

) is

th

e m

ed

ian

of

the

up

pe

r ha

lf o

f th

e d

ata

.

Ra

dic

al E

xp

res

sio

n

An

exp

ressio

n c

on

tain

ing a

rad

ical sym

bo

l (

na

). T

he

exp

ressio

n o

r nu

mb

er

insid

e t

he

rad

ical (a

) is

ca

lled

th

e r

ad

ican

d,

and

the

nu

mb

er

ap

pea

ring a

bo

ve

th

e r

ad

ical (n

) is

th

e d

egre

e.

Th

e d

egre

e is

alw

ays a

po

sitiv

e inte

ge

r. W

hen

a r

ad

ical is

writt

en

with

ou

t a

de

gre

e, it is u

nd

ers

too

d to

be

a d

egre

e

of

2 a

nd

is r

ead

as “

the s

qu

are

ro

ot of

a.”

When

the

de

gre

e is 3

, it is r

ea

d a

s “

the

cub

e r

oo

t of

a.”

Fo

r

an

y o

the

r de

gre

e,

the

exp

ressio

n

na

is r

ea

d a

s “

the

nth

roo

t of

a.”

Whe

n th

e d

egre

e is a

n e

ve

n

nu

mb

er,

th

e r

ad

ical e

xp

ressio

n is a

ssu

me

d t

o b

e th

e p

rin

cip

al (p

ositiv

e)

roo

t (e

.g.,

altho

ugh (

–7

)2 =

49

,

49

= 7

).

Ra

ng

e (

of

a R

ela

tio

n o

r F

un

cti

on

) T

he

se

t of

all

po

ssib

le v

alu

es fo

r th

e o

utp

ut

(de

pen

den

t va

riab

le)

of

a f

un

ctio

n o

r re

lation;

the

se

t of

se

co

nd

nu

mb

ers

in

th

e o

rde

red

pa

irs o

f a f

unctio

n o

r re

latio

n; th

e v

alu

es o

f th

e y

-co

ord

ina

tes in (

x,

y).

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

1

Ap

ril 2

01

4

Ra

ng

e (

of

Da

ta)

In s

tatistics,

a m

ea

su

re o

f d

isp

ers

ion

th

at

is the

diffe

ren

ce b

etw

ee

n th

e g

rea

test

va

lue

(m

axim

um

va

lue

) an

d th

e le

ast

valu

e (

min

imu

m v

alu

e)

in a

se

t of

da

ta.

Ra

te

A r

atio

th

at co

mp

are

s tw

o q

ua

ntitie

s h

avin

g d

iffe

ren

t u

nits (

e.g

., 1

68 m

iles

3.5

hours

or

122

.5 c

alo

rie

s

5 c

ups

). W

hen

the

rate

is s

imp

lifie

d s

o t

ha

t th

e s

eco

nd (

ind

epe

nde

nt)

qu

an

tity

is 1

, it is c

alle

d a

un

it r

ate

(e.g

.,

48

mile

s p

er

hou

r o

r 24

.5 c

alo

rie

s p

er

cup

).

Ra

te (

of

Ch

an

ge

) T

he

am

ou

nt

a q

uan

tity

ch

an

ge

s o

ve

r tim

e (

e.g

., 3

.2 c

m p

er

ye

ar)

. A

lso t

he

am

ou

nt a

fu

nction

’s o

utp

ut

ch

an

ge

s (

incre

ase

s o

r d

ecre

ase

s)

for

ea

ch

unit o

f cha

nge

in t

he in

put.

See

slo

pe

.

Ra

te (

of

Inte

res

t)

The

pe

rcen

t b

y w

hic

h a

mo

ne

tary

acco

un

t a

ccru

es in

tere

st.

It

is m

ost

co

mm

on

fo

r th

e r

ate

of

inte

rest

to b

e m

ea

su

red

on

an

an

nu

al ba

sis

(e.g

., 4

.5%

pe

r ye

ar)

, e

ve

n if

the

in

tere

st

is c

om

po

unde

d

pe

rio

dic

ally

(i.e

., m

ore

fre

qu

en

tly t

ha

n o

nce

pe

r ye

ar)

.

Ra

tio

A

co

mp

ariso

n o

f tw

o n

um

be

rs, qu

an

titie

s o

r exp

ressio

ns b

y d

ivis

ion

. It

is o

fte

n w

ritt

en

as a

fra

ctio

n,

bu

t n

ot a

lwa

ys (

e.g

., 2 3

, 2

:3,

2 to

3,

2 ÷

3 a

re a

ll th

e s

am

e r

atios).

Ra

tio

na

l E

xp

ress

ion

A

n e

xp

ressio

n th

at

can b

e w

ritte

n a

s a

po

lyn

om

ial d

ivid

ed

by a

po

lyn

om

ial, d

efin

ed o

nly

wh

en

th

e

latt

er

is n

ot

equa

l to

zero

.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

2

Ap

ril 2

01

4

Ra

tio

na

l N

um

be

r A

ny n

um

be

r th

at

ca

n b

e w

ritt

en

in

the

fo

rm a b

wh

ere

a is a

ny inte

ge

r a

nd

b is a

ny inte

ge

r e

xce

pt

ze

ro.

All

rep

eatin

g d

ecim

al a

nd

te

rmin

atin

g d

ecim

al n

um

be

rs a

re r

ation

al n

um

be

rs.

Re

al

Nu

mb

er

The

co

mb

ine

d s

et of

ratio

na

l a

nd

irr

ation

al n

um

be

rs. A

ll n

um

be

rs o

n t

he

nu

mb

er

line

. N

ot

an

im

agin

ary

nu

mb

er.

Re

gre

ssio

n C

urv

e

The

lin

e o

r cu

rve o

f b

est fit

tha

t re

pre

sen

ts the

le

ast d

evia

tio

n f

rom

th

e p

oin

ts in a

sca

tte

r plo

t of

da

ta.

Mo

st

co

mm

on

ly it

is lin

ea

r a

nd

use

s a

“le

ast

squ

are

s”

me

tho

d.

Exa

mp

les o

f re

gre

ssio

n c

urv

es:

Re

lati

on

A

se

t of

pa

irs o

f va

lue

s (

e.g

., {

(1,

2),

(2, 3

) (3

, 2

)}).

The

first

va

lue

in

ea

ch

pa

ir is t

he

inp

ut

(ind

ep

en

de

nt

va

lue

), a

nd

th

e s

eco

nd

va

lue

in t

he

pa

ir is t

he

ou

tpu

t (d

epe

nde

nt

va

lue

). I

n a

rela

tion

, n

eithe

r th

e in

pu

t va

lues n

or

the

ou

tpu

t va

lue

s n

eed

to

be

un

iqu

e.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

3

Ap

ril 2

01

4

Re

pe

ati

ng

De

cim

al

A d

ecim

al w

ith

on

e o

r m

ore

dig

its t

ha

t re

pea

ts e

nd

lessly

(e.g

., 0

.666

…,

0.7

27

272…

, 0.0

83

33…

). T

o

ind

icate

th

e r

epe

titio

n, a

ba

r m

ay b

e w

ritt

en

ab

ove

th

e r

ep

ea

ted

dig

its (

e.g

., 0

.66

6…

= 0

.6,

0.7

27

272…

= 0

.72

, 0

.08

333

… =

0.0

83

). A

de

cim

al th

at h

as e

ith

er

a 0

or

a 9

rep

ea

tin

g e

nd

lessly

is

equ

ivale

nt

to a

te

rmin

atin

g d

ecim

al (e

.g.,

0.3

75

000…

= 0

.37

5, 0

.19

99

… =

0.2

). A

ll re

pea

ting d

ecim

als

a

re r

ation

al nu

mb

ers

.

Ris

e

The

ve

rtic

al (u

p a

nd

dow

n)

cha

nge

or

diffe

rence

betw

ee

n a

ny t

wo

po

ints

on a

lin

e o

n a

co

ord

ina

te

pla

ne (

i.e

., fo

r po

ints

(x

1,

y1)

an

d (

x2,

y2),

the

ris

e is y

2 –

y1).

Se

e s

lope

.

Ru

n

The

ho

rizo

nta

l (left

an

d r

igh

t) c

ha

nge

or

diffe

ren

ce

be

twe

en

an

y t

wo

po

ints

on

a lin

e o

n a

co

ord

inate

p

lan

e (

i.e

., fo

r po

ints

(x

1,

y1)

an

d (

x2,

y2),

the

ru

n is x

2 –

x1).

See

slo

pe

.

Sc

att

er

Plo

t A

gra

ph

th

at

sh

ow

s t

he

“ge

ne

ral” r

ela

tio

nsh

ip b

etw

ee

n t

wo

se

ts o

f d

ata

. F

or

ea

ch p

oin

t th

at

is b

ein

g

plo

tted

th

ere

are

tw

o s

ep

ara

te p

iece

s o

f da

ta. It

sho

ws h

ow

on

e v

aria

ble

is a

ffe

cte

d b

y a

no

the

r.

Exa

mp

le o

f a s

ca

tte

r plo

t:

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

4

Ap

ril 2

01

4

Sim

ple

Eve

nt

When a

n e

ve

nt co

nsis

ts o

f a s

ingle

outc

om

e (

e.g

., r

olli

ng a

nu

mb

er

cu

be

).

Sim

ple

st

Fo

rm (

of

an

E

xp

ress

ion

) W

hen a

ll lik

e te

rms a

re c

om

bin

ed

(e.g

., 8

x +

2(6

x –

22

) b

eco

me

s 2

0x –

44

wh

en

in

sim

ple

st

form

).

The

fo

rm w

hic

h n

o lo

nge

r co

nta

ins a

ny lik

e t

erm

s,

pa

ren

the

se

s,

or

red

ucib

le f

raction

s.

Sim

plify

T

o w

rite

an

exp

ressio

n in

its

sim

ple

st

form

(i.e

., r

em

ove

an

y u

nn

ece

ssa

ry t

erm

s,

usu

ally

by c

om

bin

ing

se

ve

ral o

r m

an

y t

erm

s into

fe

we

r te

rms o

r b

y c

an

ce

lling t

erm

s).

Slo

pe

(o

f a

Lin

e)

A r

ate

of

ch

an

ge

. T

he

me

asu

rem

en

t o

f th

e s

tee

pn

ess,

inclin

e, o

r gra

de

of

a lin

e f

rom

le

ft to

rig

ht.

It

is

the

ratio

of

ve

rtic

al chan

ge

to

ho

rizo

nta

l cha

nge

. M

ore

sp

ecific

ally

, it is t

he

ratio

of

the

cha

nge

in

th

e

yc

oo

rdin

ate

s (

rise

) to

th

e c

orr

esp

ond

ing c

hange

in

th

e x

- co

ord

ina

tes (

run

) w

he

n m

ovin

g f

rom

on

e

po

int to

an

oth

er

alo

ng a

lin

e. It

als

o in

dic

ate

s w

he

the

r a lin

e is t

ilte

d u

pw

ard

(p

ositiv

e s

lop

e)

or

do

wn

wa

rd (

ne

ga

tive

slo

pe

) an

d is w

ritt

en

as th

e le

tte

r m

wh

ere

m =

rise

run

=

21

21

yy

xx

.

Exa

mp

le o

f slo

pe:

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

5

Ap

ril 2

01

4

Slo

pe

-In

terc

ep

t F

orm

A

n e

qu

atio

n o

f a

str

aig

ht,

no

n-v

ert

ical lin

e w

ritt

en

in t

he

fo

rm y

= m

x +

b,

wh

ere

m is th

e s

lop

e a

nd

b is

the

yinte

rcep

t.

Sq

ua

re R

oo

t O

ne

of

two

equ

al fa

cto

rs (

roo

ts)

of

a n

um

be

r o

r e

xp

ressio

n;

a r

ad

ical e

xp

ressio

n (

a)

with

an

un

de

rsto

od

de

gre

e o

f 2

. T

he s

qu

are

roo

t of

a n

um

be

r or

exp

ressio

n is a

ssu

me

d t

o b

e th

e p

rin

cip

al

(po

sitiv

e)

roo

t (e

.g.,

4

49

x =

7x

2).

Th

e s

qu

are

roo

t of

a n

ega

tive

nu

mb

er

resu

lts in

an

im

agin

ary

nu

mb

er

(e.g

.,

_49

= 7

i).

Sta

nd

ard

Fo

rm (

of

a L

ine

ar

Eq

ua

tio

n)

An

equ

atio

n o

f a

str

aig

ht

line

writt

en

in

th

e f

orm

Ax +

By =

C,

wh

ere

A,

B, a

nd

C a

re r

ea

l num

be

rs a

nd

w

he

re A

an

d B

are

no

t b

oth

ze

ro.

It in

clu

de

s v

aria

ble

s o

n o

ne

sid

e o

f th

e e

qu

ation

an

d a

con

sta

nt

on

th

e o

the

r sid

e.

Ste

m-a

nd

-Lea

f P

lot

A v

isua

l w

ay t

o d

ispla

y t

he

sh

ap

e o

f a d

istr

ibutio

n th

at

sh

ow

s g

rou

ps o

f da

ta a

rra

nge

d b

y p

lace

va

lue

; a

wa

y t

o s

ho

w t

he

fre

qu

en

cy w

ith

wh

ich c

ert

ain

cla

sse

s o

f da

ta o

ccur.

Th

e s

tem

co

nsis

ts o

f a

co

lum

n

of

the

la

rge

r pla

ce v

alu

e(s

); t

he

se

nu

mb

ers

are

not

repe

ate

d.

Th

e lea

ve

s c

on

sis

t of

the

sm

alle

st

pla

ce

va

lue

(usu

ally

th

e o

ne

s p

lace

) of

eve

ry p

iece

of

da

ta; th

ese

nu

mb

ers

are

arr

an

ged

in

nu

me

rica

l o

rde

r in

th

e r

ow

of

the a

pp

rop

ria

te s

tem

(e.g

., t

he

nu

mb

er

36

wo

uld

be in

dic

ate

d b

y a

lea

f of

6 a

pp

ea

rin

g in

th

e s

am

e r

ow

as t

he

ste

m o

f 3

). E

xa

mp

le o

f a s

tem

-an

d-leaf

plo

t:

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

6

Ap

ril 2

01

4

Su

bs

titu

tio

n

The

rep

lace

me

nt

of

a t

erm

or

va

ria

ble

in a

n e

xp

ressio

n o

r e

qu

ation

by a

no

the

r th

at

ha

s th

e s

am

e

va

lue

in o

rde

r to

sim

plif

y o

r eva

lua

te t

he

exp

ressio

n o

r equ

ation

.

Sys

tem

of

Lin

ea

r E

qu

ati

on

s

A s

et of

two

or

mo

re lin

ea

r equ

atio

ns w

ith

th

e s

am

e v

aria

ble

s.

The

so

lution t

o a

syste

m o

f lin

ea

r e

qu

atio

ns m

ay b

e f

ou

nd

by lin

ea

r com

bin

atio

n,

sub

stitu

tio

n, o

r gra

ph

ing

. A

syste

m o

f tw

o lin

ea

r e

qu

atio

ns w

ill e

ith

er

have

one

so

lutio

n, in

finitely

ma

ny s

olu

tion

s,

or

no

so

lutio

ns.

Sys

tem

of

Lin

ea

r In

eq

uali

ties

Tw

o o

r m

ore

lin

ea

r in

equ

alit

ies w

ith

th

e s

am

e v

aria

ble

s. S

om

e s

yste

ms o

f in

equ

alit

ies m

ay inclu

de

e

qu

atio

ns a

s w

ell

as in

equ

alit

ies.

The

so

lutio

n r

egio

n m

ay b

e c

lose

d o

r bo

un

de

d b

ecau

se

the

re a

re

line

s o

n a

ll sid

es,

wh

ile o

the

r solu

tion

s m

ay b

e o

pe

n o

r u

nb

ou

nde

d.

Sys

tem

s o

f E

qu

ati

on

s

A s

et of

two

or

mo

re e

qu

atio

ns c

on

tain

ing a

se

t of

co

mm

on

va

ria

ble

s.

Te

rm

A p

art

of

an

alg

eb

raic

exp

ressio

n.

Term

s a

re s

ep

ara

ted

by e

ith

er

an

ad

ditio

n s

ym

bo

l (+

) or

a

su

btr

action

sym

bo

l (–

). I

t can

be

a n

um

be

r, a

va

ria

ble

, o

r a p

rod

uct

of

a n

um

be

r an

d o

ne

or

mo

re

va

ria

ble

s (

e.g

., in t

he

exp

ressio

n 4

x2+

6y,

4x

2 a

nd

6y a

re b

oth

te

rms).

Te

rmin

ati

ng

De

cim

al

A d

ecim

al w

ith

a f

inite

nu

mb

er

of

dig

its.

A d

ecim

al fo

r w

hic

h t

he

div

isio

n o

pe

ration

resu

lts in e

ith

er

rep

ea

tin

g z

ero

es o

r re

pe

atin

g n

ine

s (

e.g

., 0

.37

500

0…

= 0

.375

, 0.1

99

9…

= 0

.2).

It is

ge

ne

rally

wri

tte

n

to t

he

la

st n

on

-ze

ro p

lace

va

lue

, bu

t ca

n a

lso

be

writt

en

with

ad

ditio

na

l ze

roe

s in

sm

alle

r pla

ce

va

lue

s

as n

ee

ded

(e.g

., 0

.25

ca

n a

lso

be w

ritt

en

as 0

.25

00

). A

ll te

rmin

ating d

ecim

als

are

ratio

na

l nu

mb

ers

.

Tri

no

mia

l A

po

lyn

om

ial w

ith

th

ree

un

like t

erm

s (

e.g

., 7

a +

4b

+ 9

c).

Ea

ch t

erm

is a

mo

no

mia

l, a

nd

the

mo

no

mia

ls a

re jo

ine

d b

y a

n a

dd

itio

n s

ym

bo

l (+

) or

a s

ub

tractio

n s

ym

bo

l (–

). I

t is

co

nsid

ere

d a

n

alg

eb

raic

exp

ressio

n.

K

ey

sto

ne

Ex

am

s: A

lge

bra

Ass

ess

me

nt

An

cho

r &

Eli

gib

le C

on

ten

t G

loss

ary

A

pri

l 2

01

4

Pen

nsy

lva

nia

Dep

art

men

t o

f E

du

cati

on

Pag

e 2

7

Ap

ril 2

01

4

Un

it R

ate

A

rate

in

wh

ich t

he

se

co

nd

(in

de

pe

nde

nt)

quan

tity

of

the

ratio

is 1

(e.g

., 6

0 w

ord

s p

er

min

ute

, $4

.50

p

er

pou

nd

, 21

stu

den

ts p

er

cla

ss).

Va

ria

ble

A

le

tte

r o

r sym

bo

l u

se

d t

o r

ep

rese

nt

an

y o

ne

of

a g

iven s

et of

nu

mb

ers

or

oth

er

ob

jects

(e

.g.,

in t

he

equ

atio

n y

= x

+ 5

, th

e y

an

d x

are

va

ria

ble

s).

Sin

ce

it

ca

n ta

ke

on

diffe

ren

t va

lue

s,

it is th

e o

ppo

site

of

a c

on

sta

nt.

Wh

ole

Nu

mb

er

A n

atu

ral nu

mb

er

or

ze

ro.

An

y n

um

be

r fr

om

th

e s

et of

nu

mb

ers

re

pre

se

nte

d b

y {

0,

1,

2, 3

, …

}.

So

me

tim

es it

is r

efe

rred

to

as a

“no

n-n

ega

tive inte

ge

r”.

x-A

xis

T

he

ho

rizo

nta

l n

um

be

r lin

e o

n a

co

ord

ina

te p

lan

e th

at

inte

rsects

with

a v

ert

ical n

um

be

r lin

e, th

e

ya

xis

; th

e lin

e w

ho

se

equ

atio

n is y

= 0

. T

he

x-a

xis

co

nta

ins a

ll th

e p

oin

ts w

ith

a z

ero

y-c

oo

rdin

ate

(e

.g.,

(5,

0))

.

x-I

nte

rce

pt(

s)

The

x-c

oo

rdin

ate

(s)

of

the

po

int(

s)

at

wh

ich t

he

gra

ph

of

an

equ

atio

n c

rosse

s t

he

x-a

xis

(i.e

., t

he

va

lue

(s)

of

the x

-co

ord

ina

te w

he

n y

= 0

). T

he

so

lution

(s)

or

roo

t(s)

of

an

equ

atio

n t

ha

t is

set e

qu

al

to 0

.

y-A

xis

T

he

ve

rtic

al n

um

be

r lin

e o

n a

co

ord

ina

te p

lane

tha

t in

ters

ects

with

a h

orizo

nta

l n

um

be

r lin

e, th

e

xa

xis

; th

e lin

e w

ho

se

equ

atio

n is x

= 0

. T

he

y-a

xis

co

nta

ins a

ll th

e p

oin

ts w

ith

a z

ero

x-c

oo

rdin

ate

(e

.g.,

(0,

7))

.

y-I

nte

rce

pt(

s)

The

y-c

oo

rdin

ate

(s)

of

the

po

int(

s)

at

wh

ich t

he

gra

ph

of

an

equ

atio

n c

rosse

s t

he

y-

axis

(i.e.,

the

va

lue

(s)

of

the y

-co

ord

ina

te w

he

n x

= 0

). F

or

a lin

ea

r e

qu

atio

n in s

lop

e-inte

rcep

t fo

rm (

y =

mx +

b),

it is

in

dic

ate

d b

y b

.

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Copyright © 2014 by the Pennsylvania Department of Education. The materials contained in this publication may be

duplicated by Pennsylvania educators for local classroom use. This permission does not extend to the duplication

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Assessment Anchors and Eligible Contentwith Sample Questions and Glossary

April 2014

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Keystone Exams: Algebra I and...Keystone Exams: Algebra I Assessment Anchors and Eligible Content with Sample Questions and Glossary Pennsylvania Department of Education Pennsylvania