1. Advanced Algebra Final Exam Study Guide
Wednesday, April 26, 2017

2. 1
The middle value for an ordered set of data. It also represents the 50th percentile of the data set.
B. median 2
The median of the upper half of the data set, representing the 75th percentile of the data.
D. Third quartile

3. 3
The sum of all the values in a data set divided by the number of data values. Also called the average.
A. mean 4
The median of the lower half of the data set, representing the 25th percentile of the data.
C. First quartile

4. 5 What is the probability of landing on heads, when flipping a coin?
What is the probability of rolling a die and
landing on 2?

5. Marcus spins the spinner 50 times and finds that the probability of landing on the letter B
is ¼ . He spins the spinner 8 more times, landing on the letter B the first 7 spins.
What is the probability of the spinner landing
on B on the 8th spin? 5

6. 6 Classify the sampling method for each problem. a.
A startup company wants to do a survey to find out if people would produce its product. Company employees conduct the survey by asking 50 of its friends.
Convenience Sample b.
An apparel company is coming out with a new line of fall clothes. In order to conduct a survey, it uses a computer to randomly select 100 names from its client list.
Simple Random Sample

7. 6 Classify the sampling method for each problem. c.
A couple wants to get a dog for their 6 year old son. They want to find out what dogs are considered friendly for kids. So, they visit a dog shelter, which consists of the Irish Settler, Golden Retriever, and Labrador Retriever breeds.
Stratified Sample d.
A school wants to gauge teacher morale. The school decides to survey every 10th person from their list of teachers.
Systematic Sample

8. 7 Null Hypothesis: The time it takes to complete a task is the
A software company is testing whether a new interface decreases the time it takes to complete a certain task. In a random trial, Group A used the existing interface and
Group B used the new one. The times in seconds are
given for the members of each group.
Null Hypothesis:
The time it takes to complete a task is the
same for Group A and Group B.
Do you reject the null hypothesis?
Use box-and-whisker plots to represent
both groups.

9. 7
A software company is testing whether a new interface decreases the time it takes to complete a certain task. In a random trial, Group A used the existing interface and
Group B used the new one. The times in seconds are
given for the members of each group.
Group A
Min Q1
Median14 Q3
Max

10. 7
A software company is testing whether a new interface decreases the time it takes to complete a certain task. In a random trial, Group A used the existing interface and
Group B used the new one. The times in seconds are
given for the members of each group.
Group B
Min Q1
Median11 Q3
Max

11. 7 Yes. The graphs are very different.
A software company is testing whether a new interface decreases the time it takes to complete a certain task. In a random trial, Group A used the existing interface and
Group B used the new one. The times in seconds are
given for the members of each group. 7
Group A
Group B
Yes. The graphs are very different.
Do you reject the null hypothesis.

12. 8 = mean = 20 σ = 6 Note: Find the probability of the shaded area
The graph is a normal distribution with a standard deviation of 6. What is the best estimate of the probability of the shaded area under the curve? 8
Note: Find the probability
of the shaded area
between 24 and 32.
= mean = 20
σ = 6
Standard deviation
Find the standard normal values (z) of 24 and 32.
For x = 24:
For x = 32:

13. 8 z = 0.67 (close to 0.5) z = 2 Area = 0.98 Area is approximately 0.69
The graph is a normal distribution with a standard deviation of 6. What is the best estimate of the probability of the shaded area under the curve? 8
Find the probability of the shaded area between 24 and 32.
Use the table to find the areas under the curve for all values less than z.
z = 0.67 (close to 0.5)
z = 2
Area = 0.98
Area is
approximately
0.69
Answer
Subtract both
shaded areas
0.98 – 0.69
= 0.29

14. A. Zeros x-intercepts x – 1 = 0 x = 1 (1,0) x – 3 = 0 x = 3 (3,0) 9
Find the zeros and x-intercepts for each function. Write the letter next to each graph that matches the equation. 9 A.
Zeros
x-intercepts
x – 1 = 0
x = 1
(1,0)
x – 3 = 0
x = 3
(3,0)

15. B. Zeros x-intercepts x + 2 = 0 x = –2 (–2,0) x – 3 = 0 x = 3 (3,0) 9
Find the zeros and x-intercepts for each function. Write the letter next to each graph that matches the equation. 9 B.
Zeros
x-intercepts
x + 2 = 0
x = –2
(–2,0)
x – 3 = 0
x = 3
(3,0)

16. C. Zeros x-intercepts x = 0 (0,0) x – 3 = 0 x = 3 (3,0) x + 2 = 0
Find the zeros and x-intercepts for each function. Write the letter next to each graph that matches the equation. 9 C.
Zeros
x-intercepts
x = 0
(0,0)
x – 3 = 0
x = 3
(3,0)
x + 2 = 0
x = –2
(–2,0)

17. 4x(x3 + 2 – 3x) – (–7x2 – x + 9x4) 4x4 + 8x – 12x2 – (–7x2 – x + 9x4)10
Subtract
4x(x3 + 2 – 3x) – (–7x2 – x + 9x4)
4x4 + 8x – 12x2 – (–7x2 – x + 9x4)
4x4 + 8x – 12x2 +
7x2 x 9x4 + –
–5x4 – 5x x

18. 2x2 3x –4 4x 8x3 12x2 –16x –5 –10x2 –15x 20 (4x – 5)(2x2 + 3x – 4)11
(4x – 5)(2x2 + 3x – 4)
2x2 3x –4 4x
8x3
12x2
–16x –5
–10x2
–15x 20
8x3 + 2x2 – 31x + 20
(Add matching colors)

19. Use difference of two cubes formula12
a = 2y
Factor
8y3 – 27
a3 – b3
b = 3
Use difference of two cubes formula
a3 – b3 = (a – b)(a2 + ab + b2)
8y3 – 27
= ( – )( ) 2y 3
(2y)2 2y 3 32
= (2y – 3)(4y2 + 6y + 9)

20. 4x3 – 4x2 – 24x = 0 4x(x2 – x – 6) = 0 4x(x + 2)(x – 3) = 0
Solve the equation 13
4x3 – 4x2 – 24x = 0
4x(x2 – x – 6) = 0
(Factor GCF) –6
Use diamond method
to factor trinomial 2 –3 –1
4x(x + 2)(x – 3) = 0
4x = 0
x + 2 = 0
x – 3 = 0
x = –2
x = 3
x = 0

21. 14
Circle
Largest
Exponents
Subtract
Exponents
for x and z

22. 15

23. x2 – 1 = 0 (x + 1)(x – 1) = 0 x + 1 = 0 x – 1 = 0 x = –1 x = 1 16
Vertical Asymptote
x2 – 1 = 0
Let denominator equal 0
Solve for x.
(x + 1)(x – 1) = 0
x + 1 = 0
x – 1 = 0
x = –1
x = 1
Horizontal Asymptote
y = 0

24. 16 Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 None 1
Vertical Asymptote
Let denominator equal 0
x – 1 = 0
Solve for x.
x = 1
Horizontal Asymptote
None

25. x2 – 9 = 0 (x + 3)(x – 3) = 0 x + 3 = 0 x – 3 = 0 x = –3 x = 3 16
Vertical Asymptote
Let denominator equal 0
x2 – 9 = 0
Solve for x.
(x + 3)(x – 3) = 0
x + 3 = 0
x – 3 = 0
x = –3
x = 3
Horizontal Asymptote
y = 3

26. 17
Subtract the
numerators

27. x – 5 = 0 x = 5 3(x – 5) = (12)(1) 3x – 15 = 12 x = 5 3x = 27 x = 918
Solve
Step 1A
Step 1B
Let denominator = 0
x – 5 = 0
x = 5
3(x – 5) = (12)(1)
Endpoints
3x – 15 = 12
x = 5
3x = 27
x = 9
x = 9

28. 5 < x < 9 2 –4 > 3 F 6 12 > 3 T 11 2 > 3 F 18 Solve A B
Step 2 A B C 5 9
Step 3
Interval
Test
Number
Test of
Inequality
True / False A B C 2
–4 > 3 F 6
12 > 3 T 11
2 > 3 F
Step 4
5 < x < 9

29. x – 8 = 0 x = 8 3(x – 8) = (6)(1) 3x – 24 = 6 x = 8 3x = 30 x = 1019
Solve
Step 1A
Step 1B
Let denominator = 0
x – 8 = 0
x = 8
3(x – 8) = (6)(1)
Endpoints
3x – 24 = 6
x = 8
3x = 30
x = 10
x = 10

30. x < 8 or x > 10 7 –6 < 3 T 9 6 < 3 F 11 2 < 3 T 19
Solve
Step 2 A B C 8 10
Step 3
Interval
Test
Number
Test of
Inequality
True / False A B C 7
–6 < 3 T 9
6 < 3 F 11
2 < 3 T
Step 4
x < 8 or
x > 10

31. Simplify each expression20
Simplify each expression
ln e0.15t
ln e0.15t = 0.15t
3eln(x+1)
3(x + 1) = 3x+3

32. log42 + 2log43 – log46 log42 + log432 – log46 log42 + log49 – log4620
log42 + 2log43 – log46
log42 + log432 – log46
log42 + log49 – log46
log4(2·9) – log46
log4(18) – log46
log4(18÷6)
log4(3)

33. 21
Evaluate log9489.
Use a calculator.

34. 3x = 25 22 Divide both sides by 2 Exponentiate each side

35. 23
P(–3, 6) is a point on the terminal side of θ in standard position. Find the exact value of the six trigonometric functions for θ.
Step 1 Plot point P, and use it to sketch a right triangle and angle θ in standard position. Find r.
x2 + y2 = r2
45 = r2
(–3)2 + (6)2 = r2
= r2

36. 23
Step 2 Find sin θ, cos θ, and tan θ.

37. 23
Step 3 Use reciprocals to find csc θ, sec θ, and cot θ.

38. Convert radians from radians to degrees.24
Convert radians from radians to degrees.
= 100°

39. 25
Verify the identity.

40. Verify the identity. 26

41. Amplitude Period          27A 1 π 2π –1
Wednesday, April 26, 2017

42. |2| = 2 = 6π Amplitude Period          27B 2 12π 6π –2
Wednesday, April 26, 2017

43. 28 B This problem involves daily profits for an amusement park.
Identify the graph to represent the situation.
Ticket sales were good until a massive power outage happened on Saturday that was not repaired until late Sunday. B
The graph will show decreased sales until Sunday.

44. 28 C This problem involves daily profits for an amusement park.
Identify the graph to represent the situation.
The weather was beautiful on Friday and Saturday, but it rained all day on Sunday and Monday. C
The graph will show decreased sales on Sunday and Monday.

45. 28 A This problem involves daily profits for an amusement park.
Identify the graph to represent the situation.
Only of the rides were running on Friday and Sunday. A
The graph will show decreased sales on Friday and Sunday.

46. g(f(4)) = –9 f(x) = 2x and g(x) = 7 – x Find g(f(4)) Step 1: Find f(4)29
f(x) = 2x and g(x) = 7 – x
Find g(f(4))
Step 1: Find f(4)
Step 2: Find g(16)
Use f(x) = 2x
Use g(x) = 7 – x
f(4) = 24
g(16) = 7 – 16
= 16
= –9
g(f(4)) = –9

47. Given the quadratic function h(t) = 2t2 – 8t + 5 ,30
Given the quadratic function h(t) = 2t2 – 8t + 5 ,
verify that the points (0, 5), (3, –1), and (5,15)
are on the graph of the function.
(0,5)
5 = 2(0)2 – 8(0) + 5
5 = 2(0) – 8(0) + 5
5 = –
5 = 5

48. Given the quadratic function h(t) = 2t2 – 8t + 5 ,30
Given the quadratic function h(t) = 2t2 – 8t + 5 ,
verify that the points (0, 5), (3, –1), and (5,15)
are on the graph of the function.
(3,-1)
-1 = 2(3)2 – 8(3) + 5
-1 = 2(9) – 8(3) + 5
-1 = 18 –
-1 = -1

49. Given the quadratic function h(t) = 2t2 – 8t + 5 ,30
Given the quadratic function h(t) = 2t2 – 8t + 5 ,
verify that the points (0, 5), (3, –1), and (5,15)
are on the graph of the function.
(5,15)
15 = 2(5)2 – 8(5) + 5
15 = 2(25) – 8(5) + 5
15 = –
15 = 15

50. 31
Swimming
Biking
Running
The piecewise function represents the distance that Jennifer traveled when competing in a 15.5 mile
triathlon in hours. The equations representing each activity are listed above.

51. 31    Swimming Points d = t t = 0 t = 0.5 d = 0 d = 0.5 (0,0)
(0.5 , 0.5)
Biking Points
d = 12t – 5.5  
t = 0.5
d =12(0.5) – 5.50 = 0.5
(0.5 , 0.5)
t = 1.5
d =12(1.5) – 5.50 = 12.5
(1.5 , 12.5)

52. 31     Running Points d = 6t + 3.5 t = 1.5 d = 6(1.5) + 3.5 = 12.5
(1.5 , 12.5) 
t = 2 
d = 6(2) = 15.5
(2 , 15.5)

53. 31
Swimming
Time = 0.5 – 0 = 0.5
0.5
0.5
Distance
= 0.5 – 0 = 0.5 1 12
0.5 3
Biking
Time = 1.5 – 0.5 = 1
Distance = 12.5 – 0.5 = 12
Running
Time = 2 – 1.5 = 0.5
Distance = 15.5 – 12.5 = 3

54. 32A g(x) = 2x+4 – 3 f(x) = 2x Parent Function:________ y = –3
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = 2x+4 – 3
f(x) = 2x
Parent Function:________
y = –3
Asymptote:_______
Transformation:
Shift left 4 units
Shift down 3 units

55. 32A g(x) = 2x+4 – 3 f(x) = 2x Parent Function:________ y = –3
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = 2x+4 – 3
f(x) = 2x
Parent Function:________
y = –3
Asymptote:_______
Transformation:
Shift left 4 units
Shift down 3 units

56. 32B g(x) = –5x f(x) = 5x Parent Function:________ y = 0
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = –5x
f(x) = 5x
Parent Function:________
y = 0
Asymptote:_______
Transformation:
Reflect across x-axis

57. 32B g(x) = –5x f(x) = 5x Parent Function:________ y = 0
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = –5x
f(x) = 5x
Parent Function:________
y = 0
Asymptote:_______
Transformation:
Reflect across x-axis

58. 32C g(x) = e–x – 6 f(x) = ex Parent Function:________ y = –6
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = e–x – 6
f(x) = ex
Parent Function:________
y = –6
Asymptote:_______
Transformation:
Reflect across y-axis
Shift down 6 units

59. 32C g(x) = e–x – 6 f(x) = ex Parent Function:________ y = –6
Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = e–x – 6
f(x) = ex
Parent Function:________
y = –6
Asymptote:_______
Transformation:
Reflect across y-axis
Shift down 6 units

60. 33A g(x) = log (x+4) – 2 f(x) = log x Parent Function:________ x = –4
Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = log (x+4) – 2
f(x) = log x
Parent Function:________
x = –4
Asymptote:_______
Transformation:
Shift left 4 units
Shift down 2 units

61. 33A g(x) = log (x+4) – 2 f(x) = log x Parent Function:________ x = –4
Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = log (x+4) – 2
f(x) = log x
Parent Function:________
x = –4
Asymptote:_______
Transformation:
Shift left 4 units
Shift down 2 units

62. 33B g(x) = log (–x) f(x) = log x Parent Function:________ x = 0
Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = log (–x)
f(x) = log x
Parent Function:________
x = 0
Asymptote:_______
Transformation:
Reflect across y-axis

63. 33B g(x) = log (–x) f(x) = log x Parent Function:________ x = 0
Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?
g(x) = log (–x)
f(x) = log x
Parent Function:________
x = 0
Asymptote:_______
Transformation:
Reflect across y-axis