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

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 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

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?

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 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

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

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.

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 9 12 12 13 14 14 15 16 16 17 Min Q1 Median14 Q3 Max

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 8 9 10 10 10 12 13 13 14 14 Min Q1 Median11 Q3 Max

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.

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:

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

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)

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)

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)

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 – 5x2 + 9x

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)

Use difference of two cubes formula 12 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)

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

14 Circle Largest Exponents Subtract Exponents for x and z

15

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

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 +1 +1 Solve for x. x = 1 Horizontal Asymptote None

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

17 Subtract the numerators

x – 5 = 0 x = 5 3(x – 5) = (12)(1) 3x – 15 = 12 x = 5 3x = 27 x = 9 18 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

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

x – 8 = 0 x = 8 3(x – 8) = (6)(1) 3x – 24 = 6 x = 8 3x = 30 x = 10 19 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

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

Simplify each expression 20 Simplify each expression ln e0.15t ln e0.15t = 0.15t 3eln(x+1) 3(x + 1) = 3x+3

log42 + 2log43 – log46 log42 + log432 – log46 log42 + log49 – log46 20 log42 + 2log43 – log46 log42 + log432 – log46 log42 + log49 – log46 log4(2·9) – log46 log4(18) – log46 log4(18÷6) log4(3)

21 Evaluate log9489. Use a calculator.

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

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 9 + 36 = r2

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

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

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

25 Verify the identity.

Verify the identity. 26

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

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

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.

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.

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.

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

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 = 0 – 0 + 5 5 = 5

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 – 24 + 5 -1 = -1

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 = 50 – 40 + 5 15 = 15

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.

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)

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) + 3.5 = 15.5 (2 , 15.5)

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

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

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

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

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

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

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

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

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

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

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