2015 Spring ACP Review Answers

Transcription

ID: A
ACP Semester 2 Review
Answer Section
MULTIPLE CHOICE
1. ANS: D
PTS: 1
REF: 93d13687-9631-11dd-8a40-001185f11039
OBJ: Problem-Solving Application
NAT: NT.CCSS.MTH.10.7.7.SP.6
STA: TEKS.7.6.H TOP: Making Predictions
KEY: probability | prediction | problem solving
DOK: DOK 2
2. ANS: A
PTS: 1
REF: fa53f47c-6ff9-11df-9c81-001185f0d2ea
STA: TEKS.7.6.I KEY: probability | experimental
DOK: DOK 1
3. ANS: D
The 6 outcomes of the spinner not landing on A and rolling a prime number in the format
(spinner, number cube) are (B, 2), (B, 3), (B, 5), (C, 2), (C, 3), and (C, 5). Since there are 18 possible
outcomes in the sample space, the probability of the spinner not landing on A and rolling a prime
6
1
number is
, or .
18
3
Feedback
4
The probability of the spinner landing on B or rolling a 4 is
B
The probability of the spinner not landing on C and rolling a 3 is
C
The probability of the spinner landing on A or rolling a number greater than 2 is
not
D
9
, not
1
A
1
9
18
.
, not
13
18
.
7
9
,
2
.
9
That’s correct!
PTS:
NAT:
STA:
KEY:
DOK:
4. ANS:
OBJ:
STA:
DOK:
5. ANS:
OBJ:
STA:
KEY:
1
NT.CCSS.MTH.10.7.7.SP.8.a | NT.CCSS.MTH.10.7.7.SP.8.b | NT.CCSS.MTH.10.K-12.MP.5
TEKS.7.6.I
compound events | identifying outcomes | probability | sample space | tree diagram
DOK 2
C
PTS: 1
REF: 9922c541-9631-11dd-8a40-001185f11039
Finding the Area of a Composite Figure
NAT: NT.CCSS.MTH.10.7.7.G.6
TEKS.7.9.C TOP: Area of Irregular Figures
KEY: area | irregular figure
DOK 3
C
PTS: 1
REF: 940f3413-9631-11dd-8a40-001185f11039
Solving Two-Step Equations
NAT: NT.CCSS.MTH.10.7.7.EE.4
TEKS.7.11.A
TOP: Solving Two-Step Equations
solving | equation | two-step equation
DOK: DOK 2
1
ID: A
6. ANS: C
PTS: 1
TOP: Translating Words into Math
7. ANS: D
PTS: 1
STA: TEKS.7.11.B
8. ANS: A
Compare the corresponding sides.
MN
9
3
=
=
RS
15
5
REF:
STA:
KEY:
NAT:
DOK:
94737e8f-9631-11dd-8a40-001185f11039
TEKS.7.11.A
multi-step
DOK: DOK 2
NT.CCSS.MTH.10.6.6.EE.7
DOK 2
MP
6
3
=
=
RU
10
5
Since the measures of the corresponding angles are equal and the ratios of the corresponding sides are
equivalent, the rectangles are similar.
The similarity ratio is
3
5
.
rectangle MNOP ~ rectangle RSTU
Feedback
A
B
C
D
Correct!
Compare the corresponding sides.
Check whether the ratios of the corresponding sides are equivalent.
Check whether the ratios of the corresponding sides are equivalent.
PTS: 1
DIF: Advanced
STA: TEKS.8.3.A TOP: Similar Figures and Proportions
2
ID: A
9. ANS: B
height of maple shadowof maple
=
height of pine
shadowof pine
x
20.02
=
21.61
4.9
x
20
≈
20
5
x
≈4
20
x ≈ 80
Write a proportion using the corresponding sides.
Substitute.
Use compatible numbers to estimate.
Simplify.
Multiply to isolate the variable.
The maple tree is about 80 feet tall.
Feedback
A
B
C
D
Use compatible numbers to estimate.
Correct!
Set up a proportion and estimate.
Set up a proportion and estimate.
PTS: 1
DIF: Average
OBJ: Estimating with Indirect Measurement
STA: TEKS.8.3.A TOP: Using Similar Figures
KEY: indirect measurement | similar
10. ANS: A
Multiply the coordinates by the scale factor to find the vertices of the image.
Feedback
A
B
C
D
Correct!
Did you multiply each of the coordinates by the scale factor?
Did you remember to list the x-coordinates first and the y-coordinates second?
Did you perform the dilation on each of the vertices?
PTS: 1
DIF: Average
REF: Page 245
OBJ: 5-6.3 Using the Origin as the Center of Dilation
STA: TEKS.8.3.B TOP: 5-6 Dilations
3
NAT: 8.3.2.c
KEY: coordinate plane | dilation
ID: A
11. ANS: A
Under a dilation (x, y) → (kx, ky), the lengths of the sides change by a scale factor of k. So, the length of
2
each segment of the figure changes by a scale factor of .
3
Feedback
A
B
C
D
12.
13.
14.
15.
16.
17.
That’s correct!
The dilation does not preserve the length of the segment.
3
The scale factor of the dilation is not .
2
The scale factor of the dilation is not 2.
PTS:
KEY:
ANS:
STA:
ANS:
OBJ:
STA:
DOK:
ANS:
STA:
ANS:
STA:
ANS:
STA:
ANS:
STA:
1
STA: TEKS.8.3.B
dilation | scale factor | length of line segment
DOK: DOK 1
D
PTS: 1
TEKS.8.3.C DOK: DOK 2
C
PTS: 1
REF: 986ff3b7-9631-11dd-8a40-001185f11039
Using a Dilation to Reduce a Figure
TEKS.8.3.C TOP: Dilations
KEY: coordinate plane | dilation
DOK 2
D
PTS: 1
TEKS.8.5.D DOK: DOK 1
C
PTS: 1
TEKS.8.5.D DOK: DOK 2
D
PTS: 1
NAT: NT.CCSS.MTH.10.8.8.F.3
TEKS.8.5.I DOK: DOK 3
A
PTS: 1
NAT: NT.CCSS.MTH.10.8.8.F.3
TEKS.8.5.I DOK: DOK 1
4
ID: A
18. ANS: D
First, find the function’s rate of change using the first two ordered pairs in the table.
25 − 19
6
= =2
6−3
3
Then, find the function’s initial value using the equation s = mt + b, the slope, and one ordered pair from
the table.
19 = 2 ( 3 ) + b
19 = 6 + b
13 = b
So, the function represented by the table is s = 2t + 13.
Feedback
A
B
C
D
The rate of change is not negative.
The initial value is not 2.
The initial value is not 0.
That’s correct!
PTS: 1
NAT: NT.CCSS.MTH.10.8.8.F.4 | NT.CCSS.MTH.10.K-12.MP.4
STA: TEKS.8.5.I KEY: constructing linear functions | analyzing tables
DOK: DOK 1
19. ANS: C
PTS: 1
REF: 911ba69b-6ab2-11e0-9c90-001185f0d2ea
OBJ: Identifying Similarity Transformations
STA: TEKS.8.10.A
TOP: Similarity and Congruence Transformations
KEY: dilation | transformation | similarity | congruence
DOK: DOK 2
20. ANS: B
Two figures are congruent if there exists a sequence of translations, reflections, and/or rotations that
maps one figure to the other. Two figures are similar if there exists a sequence of translations,
reflections, rotations, and/or dilations that maps one figure to the other. So, if the sequence of
transformations includes a dilation, the figures are similar but not congruent. The figure with the vertices
(−6, − 6), (−2, 2), (2, 2), (6, − 6) is the image of the given figure after a dilation with respect to the
origin by a scale factor of 2. So, this figure is similar but not congruent to the given figure.
Feedback
A
B
C
D
This figure is the image of the given figure after a rotation 90° clockwise about the
origin.
That’s correct!
This figure is the image of the given figure after a translation right 1 unit and down
3 units.
This figure is the image of the given figure after a reflection across the x-axis.
PTS: 1
NAT: NT.CCSS.MTH.10.8.8.G.4 | NT.CCSS.MTH.10.8.8.G.2
STA: TEKS.8.10.A
KEY: similar figures | congruent figures | translation | reflection | rotation | dilation | sequence of
transformations
DOK: DOK 2
5
ID: A
21. ANS: A
PTS: 1
REF: 9d5d10e5-9631-11dd-8a40-001185f11039
STA: TEKS.8.10.B
TOP: Transformations
KEY: transformation | reflection | coordinate plane
DOK: DOK 1
22. ANS: D
PTS: 1
STA: TEKS.8.10.B
DOK: DOK 2
23. ANS: D
PTS: 1
REF: MLC10917 NAT: NT.CCSS.MTH.10.8.8.G.3
STA: TEKS.8.10.C
TOP: Translations in a Coordinate Plane
KEY: translation DOK: DOK 1
24. ANS: B
PTS: 1
STA: TEKS.8.10.C
DOK: DOK 2
25. ANS: A
When the dimensions of a triangle are increased by a factor of x, the perimeter is increased by a factor of
x, and the area is increased by a factor of x2 .
Feedback
A
B
C
D
Correct!
Check the change in the area.
Check the change in the perimeter.
Compare the perimeter and area of the original figure with the perimeter and area of
the enlarged figure.
PTS: 1
DIF: Average
REF: Page 548
OBJ: 10-4.1 Comparing Perimeters and Areas
STA: TEKS.8.10.D
TOP: 10-4 Changing Dimensions
KEY: length | width | perimeter | area | change
6
ID: A
26. ANS: A
ÊÁ 1
1 ˆ˜˜
Á
The vertices of figure A and figure B are shown in the table below. Since the dilation (x, y) → ÁÁÁÁ x, y ˜˜˜˜
ÁË 3
3 ˜¯
1
transforms figure A to figure B, the scale factor is .
3
Preimage
(−3, − 3)
(−3, 9)
Image
(−1, − 1)
(−1, 3)
(3, 6)
(3, − 6)
(1, 2)
(1, − 2)
Feedback
A
B
C
D
27.
28.
29.
30.
31.
32.
33.
34.
That’s correct!
The distance from the origin to a vertex of figure B is not half of the distance from
the origin to the corresponding vertex of figure A.
Figure B is not the same size figure A.
Figure B is not larger than figure A.
PTS:
KEY:
ANS:
STA:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
STA:
ANS:
STA:
ANS:
STA:
ANS:
OBJ:
STA:
KEY:
ANS:
STA:
1
STA: TEKS.8.10.D
dilation | scale factor
DOK: DOK 1
A
PTS: 1
TEKS.8.11.A
DOK: DOK 1
C
PTS: 1
REF: f9e15b9c-6ff9-11df-9c81-001185f0d2ea
NT.CCSS.MTH.10.8.8.SP.1
STA: TEKS.8.11.A
association | relationship | scatter plot
DOK: DOK 1
C
PTS: 1
REF: f9e182ac-6ff9-11df-9c81-001185f0d2ea
NT.CCSS.MTH.10.8.8.SP.1
STA: TEKS.8.11.A
association | relationship | scatter plot
DOK: DOK 1
D
PTS: 1
NAT: NT.CCSS.MTH.10.6.6.SP.5.c
TEKS.8.11.B
DOK: DOK 2
B
PTS: 1
NAT: TEKS.8.12.A
TEKS.8.12.A
DOK: DOK 1
D
PTS: 1
NAT: TEKS.8.12.A
TEKS.8.12.A
DOK: DOK 2
C
PTS: 1
REF: 96da7a3b-9631-11dd-8a40-001185f11039
Problem-Solving Application
NAT: NT.CCSS.MTH.10.7.7.RP.3
TEKS.8.12.D
TOP: Simple Interest
interest | simple interest
DOK: DOK 1
A
PTS: 1
NAT: NT.CCSS.MTH.10.7.7.RP.3
TEKS.8.12.D
DOK: DOK 2
7
ID: A
35. ANS: B
The formula for the surface area of a rectangular prism is S = 2™w + 2™h + 2wh, where ™ is the length, w is
the width, and h is the height.
S = 2 (9 ) (7 ) + 2 (9 ) (5 ) + 2 (7 ) (5 )
= 126 + 90 + 70
= 286
2
Thus, the surface area of the prism is 286 m .
Feedback
A
B
C
D
36.
37.
38.
39.
40.
The prism has 6 faces, not just 3 faces.
That’s correct!
3
315 m is the volume of the prism.
This is the surface area of a cube with side length 9 m.
PTS:
KEY:
ANS:
STA:
ANS:
STA:
ANS:
OBJ:
STA:
KEY:
ANS:
NAT:
STA:
ANS:
NAT:
STA:
1
STA: TEKS.8.7.B
rectangular prism | surface area
DOK: DOK 1
A
PTS: 1
TEKS.8.7.B DOK: DOK 2
C
PTS: 1
TEKS.8.8.A DOK: DOK 2
D
PTS: 1
REF: 9f83fac7-9631-11dd-8a40-001185f11039
Application NAT: NT.CCSS.MTH.10.8.8.EE.7 | NT.CCSS.MTH.10.8.8.EE.7.b
TEKS.8.8.A TOP: Solving Equations with Variables on Both Sides
multi-step equation | solve
DOK: DOK 1
C
PTS: 1
NT.CCSS.MTH.10.8.8.EE.7 | NT.CCSS.MTH.10.8.8.EE.7.b
A
PTS: 1
NT.CCSS.MTH.10.8.8.EE.7 | NT.CCSS.MTH.10.8.8.EE.7.b
NUMERIC RESPONSE
1. ANS: 299
PTS:
OBJ:
TOP:
2. ANS:
1
REF: 99e6204d-9631-11dd-8a40-001185f11039
Application NAT: NT.CCSS.MTH.10.7.7.SP.6
STA: TEKS.7.6.H
Making Predictions
KEY: multi-step
DOK: DOK 2
902.5
PTS: 1
REF: 92f83f1d-9631-11dd-8a40-001185f11039
STA: TEKS.7.9.C TOP: Area of Composite Figures
DOK: DOK 3
8
ID: A
3. ANS: 44.02
PTS:
NAT:
KEY:
DOK:
1
REF: fa136d67-6ff9-11df-9c81-001185f0d2ea
NT.CCSS.MTH.10.6.6.SP.5.c
STA: TEKS.8.11.B
residual | linear | least-squares | regression | absolute | deviation | goodness | of | fit
DOK 2
SHORT ANSWER
1. ANS:
A) Sample answer: Austin and Berenice are riding a train. Austin is in the sixth train car from the end of
the train and is walking towards the end at a rate of half a car per minute. Berenice is at the end of the
train and is walking forward at a rate of two and a half cars per minute. At what time will they meet?
B) x = 2; Sample answer: They will meet after 2 minutes.
PTS: 1
NAT: TEKS.8.8.B STA: TEKS.8.8.B DOK: DOK 2
9

2015 Spring ACP Review Answers

Transcription

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