Algebra 2 B
Semester Exam Review Answers
Algebra 2 B
Semester Exam Review
2015-2016
Answers
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Algebra 2 B
Semester Exam Review Answers
Unit 2, Topic 3
1.
a.
2.
a.
3.
a.
y  1 
1
x2
y
2
x
b.
LCM =  x  3 x  2 
Solution is x  
b.
b.
29
2
LCM = x  x  2 
A student who solves by cross multiplication will get x  0, x  12 , with 0 being
extraneous. Solution is x  12 . A student who multiplies by the LCM will not get
the extraneous solution.
4.
Function f  x  
1
x
Function g  x  
1
x2
Even, odd, or neither
Domain
Range
End behavior as x  
Odd
All real numbers but 0
All real numbers but 0
f  x  0
Even
All real numbers but 0
All positive real numbers
g  x  0
End behavior as x  
f  x  0
g  x  0
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Algebra 2 B
5.
Semester Exam Review Answers
a.
15 hours
b.
inversely
c.
H
d.
3.75 hours
6.
H
7.
a.
15
n
200
s
Method 1
1
x3
2  x  3
or
2
x  3 2x  7
2x  6
2

x3
2x  6 1
x3
2x  7
x3
8.
Method 2
1
x3
1
2
1
x3
b.
Vertical asymptote is x  3 . Horizontal asymptote is y  2 .
a.
Method 1
1
x4
5  x  4

x4
5 x  20  1
x4
5 x  21
x4
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Method 2
5
x  4 5 x  21
5
b.
or
5 x  20
1
1
x4
5
1
x4
Vertical asymptote is x  4 . Horizontal asymptote is y  5 .
Algebra 2 B
9.
a.
Semester Exam Review Answers
250  6 10 
 $31
10
b.
250  6 x
 11
x
250  6 x  11x
5 x  250
x  50 cakes
10.
a.
500  8  22  676

 $33.80
22  2
20
b.
500  8 x
 18.75
x2
500  8 x  18.75  x  2 
500  8 x  18.75 x  37.50
537.50  10.75 x
x  50 fruitcakes
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Algebra 2 B
Semester Exam Review Answers
Unit 3
11.
Values of x and y
x is positive, y is positive
x is positive, y is negative
x is negative, y is positive
x is negative, y is negative
Quadrant of the Angle 
1
4
2
3
12. a.
Sign of sin 
Positive
Negative
Positive
Negative
Sign of cos 
Positive
Positive
Negative
Negative
b.
1
17
sin  
5
3
sin  
cos  
2
3
cos   
tan  
5
2
tan   
4
17
1
4
  166.0o
  48.2o
12
13
13.

14.
All statements should be selected.
15.
a.
60
120 
or or 1.047 units
 2  or
360
360
3
b.
RF
c.
OF
d.
RF
OF
e.
120o or 420o (or any integer multiples of 360o added to or subtracted from these)
f.
300o or 420o (or any integer multiples of 360o added to or subtracted from these)
a.
Amplitude: 20, Period: 8, Equation of midline: y  30
b.
Amplitude: 8, Period:
16.
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
2
, Equation of midline: y  10
Algebra 2 B
17.
18.
Semester Exam Review Answers
a.
29.6 feet
b.
33.4 feet
c.
8.5 feet
d.
20  15sin  40o   29.6
a.
22.6 feet
b.
7.0 feet
c.
8.5 feet
d.
Since one complete circle takes 36 seconds, the pony rotates 10 degrees per
second. Since we want to know where the pony is was 31 seconds, there is a 310
degree rotation, so 20  15sin  310o   8.5 .
19.
d
t
20.
Amplitude: 15, Period: 36, Equation of midline: y  20
21.
 
d  20  15sin 10o t  or d  20  15sin   t 
 18 
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Algebra 2 B
22.
Semester Exam Review Answers
a.
This is not a good idea. There is no explanation as to what was done, and there are
only 5 values in the table.
b.
d  2 
c.
All positive real numbers
d.
If the table were used, I’d have to continue the table until I got to 1.6 lux, or keep
measuring using my device until I got 1.6 lux, then measure the distance.
320 320

 80;
22
4
d  4 
320 320

 20
42
16
If the formula were used, I would solve the equation 1.6 
320
. It looks like a
d2
simple equation to solve. I would probably do this.
If the graph were used, I would go up the y-axis to 1.6, then go across until I
found the corresponding x-value on the graph. It would be hard to do on this
graph due to its scale.
23.
24.
25.
a.
sine (or cosine) and exponential
b.
sine (or cosine) and quadratic
a.
Some kind of trig function (sine or cosine). The data appear to be periodic.
b.
This is not a good model. When I tried to evaluate it at x  0 , I got 40, not 0. If I
replace sin with cos it will work. (There are other possibilities.)
a.
The domain is 0  x  10 . Since x is a length it must be positive. Its value must
be less than 10. Otherwise, the volume will be negative (The 10  x term in the
formula will be negative.
b.
It looks like the side of the square should be a little less than 6 inches, with a
corresponding volume of approximately 96 or 97 cubic inches.
c.
A table would probably not be a good model, since it appears that the maximum
volume does not occur at a whole number value.
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Algebra 2 B
Semester Exam Review Answers
Unit 5
26.
12
27.
a.
R
B
R
B
b.
c.
R
B
R
B
d.
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Algebra 2 B
28.
29.
Semester Exam Review Answers
a.
80 4
  0.8  80%
100 5
b.
30
3
  0.3  30%
100 10
c.
14
7

 0.14  14%
100 50
d.
24
6

 0.24  24%
100 25
e.
30  20  6 44 11


 0.44  44%
100
100 25
f.
70  80  56 94 47


 0.94  94%
100
100 50
g.
14 2

 0.2  20%
70 10
h.
24 3

 0.3  30%
80 10
i.
Yes they are independent. The probability that a person likes hockey is the same
whether the person is male or female. P  H | M   0.7  P  H | F 
a.
50%
b.
60%
c.
8%
d.
82%
e.
8%
8

 0.16  16%
50% 50
f.
8%
8

 0.2  20%
40% 40
g.
18%
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Algebra 2 B
Semester Exam Review Answers
30.
 0.4  0.4  0.4   0.064  6.4%
31.
 0.7  0.6   0.42  42%
32.
a.
68%
b.
2.5%
c.
49.85%
d.
16%
e.
33.
24  14
 5 , meaning that this insects
2
life span is 5 standard deviations above the mean. This would be quite unusual.
If I compute a z-score for 24 days, I get z 
a.
greater
b.
greater
34.
All should be selected.
35.
z-scores should be computed for each store and compared.
Goldie’s Store: z 
25.9  22.6
 1.5
2.2
Ahmed’s Store:
17.9  15.1
2
1.4
Ahmed’s sales were two standard deviations above the mean sales for New York, while
Goldie’s sales were 1.5 standard deviations above the mean sales for Milwaukee.
Ahmed’s store should get the reward.
36.
z-scores should be computed for each store and compared.
Jose’s Store: z 
31.7  52.9
 3.365
6.3
Jamal’s Store:
14.4  30.2
 3.95
4
Jose’s sales were about 3.365 standard deviations below the mean for New York stores,
while Jamal’s sales were 3.95 standard deviations below the mean for Seattle stores.
The manager of Jamal’s store should be fired.
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