ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
QUESTION TYPE
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
CR
MC
MC
MC
MC
MTF
MTF
CR
MC
MC
CR
MC
CR
MC
MC
MC
MC
MC
MC
MC
CR
2015–2016
Clark County School District
CE LEARNING
TARGET
1.1
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.3
1.3
1.3
1.3
1.3
1.4
1.4
1.4
1.4
2.2
2.3
2.5
2.6
2.7
2.7
2.8
2.9
2.9
2.9
3.4
3.5
3.7
4.2
4.2
4.3
4.3
4.3
4.3
4.3
NVACS
S.ID.2
S.ID.1
S.ID.1
S.ID.2
S.ID.2
S.ID.2
S.ID.2
S.ID.8
S.ID.2
S.ID.2
S.ID.2
S.ID.2
S.ID.2
S.ID.5
S.ID.5
S.ID.5
S.ID.5
N.Q.A.3
N.Q.A.1
A.SSE.A.1a
A.SSE.A.Ib
A.REI.A.1
A.REI.A.1
A.CED.A.1, A.SSE.A.1, A.SSE.A.2
A.CED.A.4
A.CED.A.4
N.Q.A.1, N.Q.A.3, A.CED.A.4
A.CED.A.1
A.REI.B.3
A.CED.A.1
F.IF.A.2
F.IF.A.1
A.CED.A.3
F.IF.B.4
F.IF.B.5
F.BF.A.1b
F.IF.B.4, F.IF.B.5, F.BF.A.1a
Page 1 of 7
DOK
LEVEL
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
2
1
2
1
2
1
1
3
1
1
2
1
2
1
1
1
2
2
2
2
2
KEY
D
D
C
B
C
D
D
A
A
A
D
B
D
C
D
A
----C
D
B
C
A
B
----D
A
----B
----D
A
A
B
D
D
B
-----
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
#
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
QUESTION TYPE
CR
MC
MC
MC
MC
MC
CR
MC
MTF
MTF
MC
MC
MC
MC
MC
MTF
MTF
MC
MC
MC
MC
MC
MTF
MTF
MTF
MC
MC
MC
MC
MTF
MTF
MTF
MC
MC
MC
MC
MC
2015–2016
Clark County School District
CE LEARNING
TARGET
4.3
4.6
4.6
4.6
4.6
4.6
4.6
4.7
5.1
5.1
5.2
5.3
5.3
5.3
5.3
5.3
5.3
5.4
5.4
5.4
5.5
5.5
5.5
5.5
5.5
5.5
5.6
5.6
5.7
6.1
6.1
6.1
6.2
6.2
6.3
6.3
6.3
NVACS
F.IF.A.1
F.BF.A.2
F.BF.A.2
F.BF.B.3
F.BF.A.2
F.IF.A.3
F.IF.A.3, F.IF.C.9, F.BF.A.2
F.BF.A.1a
F.LE.A.1b
F.LE.A.1b
F.IF.C.7a
F.IF.B.6
F.IF.C.7a
F.IF.B.6
F.IF.B.6
F.IF.B.6
F.IF.B.6
F.LE.B.5
A.REI.D.10
F.LE.B.5
A.REI.D.12
A.REI.D.12
A.REI.D.12
A.REI.D.12
A.REI.D.12
A.CED.A.2
A.REI.D.11
A.CED.A.2
F.LE.A.2
F.IF.C.9
F.IF.C.9
F.IF.C.9
F.BF.B.3
F.BF.B.3
S.ID.6a
S.ID.6a
S.ID.6a
Page 2 of 7
DOK
LEVEL
2
1
1
1
2
2
2
2
2
2
1
1
1
2
1
1
1
1
1
1
1
1
1
1
1
1
2
2
1
2
2
2
1
1
2
1
2
KEY
----B
C
D
B
D
----C
A
B
C
B
B
A
C
A
B
B
A
C
B
D
B
B
A
D
B
A
B
B
A
B
C
A
C
C
A
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
#
75
76
77
78
79
80
81
82
83
84
85
86
QUESTION TYPE
MC
MC
MC
MC
MC
MTF
MTF
MTF
MTF
MTF
ER
MC
CE LEARNING
TARGET
6.3
6.3
6.3
6.3
6.3
6.3
6.3
6.3
6.3
6.3
6.3
6.5
NVACS
S.ID.6b
S.ID.7
S.ID.7
S.ID.7
S.ID.8
S.ID.8
S.ID.8
S.ID.9
S.ID.8
S.ID.8
S.ID.1, S.ID.2, S.ID.3, S.ID.6
S.ID.6b
DOK
LEVEL
1
1
1
1
1
1
1
1
1
1
3
1
KEY
B
C
B
B
A
A
B
B
A
B
----B
#17
This question assesses the student’s ability to interpret two-way tables of categorical data and
look for relationships between the variables.
Class
Freshmen Sophomores Juniors Total
Yes
56
38
32
126
Favors
No
24
37
58
119
the change
Total
80
75
90
245
(a) 126/245 ≈ 51%
(b) Freshmen: 56/80 = 70% Highest favorability
Sophomores: 38/75 ≈ 51%
Juniors: 32/90 ≈ 36% Lowest favorability
(c) There is a relationship. The older the student, the less chance he favors the change in dress
code.
#24
This question assesses the student’s ability to write mathematical expressions from a context,
where a quantity may be a specific numerical value or a variable, and to see structure in
equivalent mathematical expressions.
2015–2016
Clark County School District
Page 3 of 7
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
8
15
(a) 𝑝 + 0.08𝑝 + 0.15𝑝 or 𝑝 + 100 𝑝 + 100 𝑝 or equivalent
𝑥
𝑔
(b) 𝑝 + 100 𝑝 + 100 𝑝 or equivalent
(c) Yes, this works. Dividing by 10 gives one-tenth or 10% of the price. Dividing that by 2 gives
one-twentieth or 5% of the price. Adding those two numbers gives 15% of the price.
𝑝
𝑝 (10)
𝑝
𝑝
+
=
+
= 0.10𝑝 + 0.05𝑝 = 0.15𝑝
10
2
10 20
#27
This question assesses the student’s ability to determine the units of quantities in a formula,
convert measurements and use unit analysis, and solve literal equations for a given variable.
(a) Unit analysis on the right side of the equation gives
(b) 𝑥0 = 3300 𝑓𝑒𝑒𝑡 ×
1 𝑚𝑖𝑙𝑒
1
𝑚−𝑚
𝑠
= 3300 × 5280 𝑓𝑒𝑒𝑡 ×
5280 𝑓𝑒𝑒𝑡
𝑚
𝑚
= 𝑠 , so 𝑣 is measures in 𝑠 .
𝑚𝑖𝑙𝑒
𝑓𝑒𝑒𝑡
= 0.625 𝑚𝑖𝑙𝑒𝑠
Since the measurement 3,300 ft/s has two significant digits, the conversion should be rounded
to two significant digits, so 𝑥0 ≈ 0.63 miles.
𝑥 −𝑥0
(c)
𝑣 = 𝑓𝑡
𝑣𝑡 = 𝑥𝑓 − 𝑥0
𝑣𝑡 + 𝑥0 = 𝑥𝑓
#29
This question assesses the student’s ability to solve absolute value equations.
(a) 𝑥 = −27, 15
(b) 𝑥 = −5, 7
(c) 𝑦 = ±9
(d) no solution
#37
This question assesses the student’s ability to build a linear function from a context, determine
its domain and range, construct a graph of the function, and identify important points on the
function.
2015–2016
Clark County School District
Page 4 of 7
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
(a) Let 𝐴 represent the amount Steve owes after a payment is made in the month, 𝑚, of the
$150
loan. Then 𝐴 = $4800 − 𝑚𝑜𝑛𝑡ℎ 𝑚.
$150
(b) Steve will pay off the loan when 0 = $4800 − 𝑚𝑜𝑛𝑡ℎ 𝑚, or when 𝑚 = 32 𝑚𝑜𝑛𝑡ℎ𝑠. The
domain of the function is the number of months since the loan was given, namely the integers
from 0 to 32, inclusive. The range of the function is the amounts remaining on the loan, namely
{4800, 4650, 4500, …, 150, 0}.
(c) Important points must include when the loan was taken out (0 months, $4,800), and when
the loan was paid off (32 months, $0). No other points are necessary, but any identified must
be accompanied by a clear and reasonable explanation as to why they are important.
#38
This question assesses the student’s understanding of the concept of a function.
A function is a relation where for each element of the domain, there is only one element in the
range. In 𝑦 = 𝑥 2 , when 𝑥 = −2, 𝑦 = 4. There is only one value of 𝑦 for that value of 𝑥. When
𝑥 = 2, 𝑦 = 4 and again there is only one value of 𝑦 for that value of 𝑥. To be a function, it is
NOT the case that for any of 𝑦 there must be only one 𝑥.
#44
This question assesses the student’s ability to explicitly define an arithmetic sequence, compare
the rates of change of two sequences, and solve a linear equation.
(a) 𝑡(𝑛) = −7 + 3𝑛
(b) These are arithmetic sequences and, thus, linear functions.
The rates of change are their slopes. The slope of 𝑡 is 3 and the
slope of 𝑠 is 2. Sequence 𝑡 grows at a rate 1.5 times that of
sequence 𝑠.
(c) 𝑡(𝑛) = 𝑠(𝑛)
−7 + 3𝑛 = 2 + 2𝑛
𝑛=9
#85
This question assesses the student’s ability create graphical representations of univariate and
bivariate data and describe their characteristics; compare characteristics of multiple data sets;
2015–2016
Clark County School District
Page 5 of 7
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
compute measures of center and spread; and describe the effect of extreme values on
statistical measures.
(a)

July
Frequency























Rainfall (inches)



December
Frequency







(b) July’s distribution is skewed right. December’s distribution is symmetric. (Students may
choose to make the histogram using bin widths of 0.5 inches or 1.0 inches.)
(c) Since the distribution of July rainfall is skewed right, the mean will be greater than the
median.
(d) The median rainfall for December is 5.15 inches. The median rainfall for July is between 0.5
inches and 1.0 inch. The median rainfall for December is at least 4 inches greater than July.
(e) 1) Compute the mean rainfall for the 20 observations.
2) Subtract the mean from each of the 20 observations.
3) Square those values.
4) Add up those squares.
5) Divide by 19.
6) Take the square root of that value. The standard deviation will have units of inches.
2015–2016
Clark County School District
Page 6 of 7
Revised September 2015
ALGEBRA I
2015-2016 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 1
(f) December’s rainfall has the larger standard deviation because its distribution has a much
wider spread than July’s.
(g) The median is between 0.5 and 1.0 inches. Changing 2.4 to 1.4 will not affect the median
because the value will remain greater than the median. Changing 2.4 to 1.4 will decrease the
mean by
1.0
 0.05
20
inches.
(h)
December
Rainfall (inches)










Year



(i) There is no relationship between rainfall and year. That is, the rainfall amounts over time
seem to vary randomly with no trend of increase or decrease.
2015–2016
Clark County School District
Page 7 of 7
Revised September 2015