Monday, January 11, 2016 Advanced Algebra Final Exam Study Guide.

Friday, April 21, 2023

Advanced AlgebraFinal Exam Study Guide

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

What is the probability of landing on heads,when flipping a coin?

What is the probability of rolling a die andlanding on 2?

1

2

1

6

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?

1

4

5

Classify the sampling method for each problem.

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.

a.

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.

b.

Convenience Sample

Simple Random Sample

6

Classify the sampling method for each problem.

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.

c.

A school wants to gauge teacher morale. The school decides to survey every 10th person from their list of teachers.

d.

Stratified Sample

Systematic Sample

6

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 representboth groups.

7


Group A 9 12 12 13 14 14 15 16 16 17

Median14

Q1 Q3Min Max

7


Group B 8 9 10 10 10 12 13 13 14 14

Median11

Q1 Q3Min Max

7


Group B

Group A

Do you reject the null hypothesis. Yes. The graphs are very different.

7

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?

Note: Find the probability of the shaded area between 24 and 32.

Find the standard normal values (z) of 24 and 32.

For x = 24: For x = 32:

= mean = 20σ = 6Standard deviation

xz

m-=

s

240.67

6

20xz

m- -= =

s=

32

6

202

xz

m- -= = =

s

8

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?

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)Area isapproximately0.69

z = 2Area = 0.98

AnswerSubtract bothshaded areas0.98 – 0.69

= 0.29

8

Find the zeros and x-intercepts for each function. Write the letter next to each graph that matches the equation.

2( ) ( 1) ( 3)P x x x A.

Zeros

x – 1 = 0

x – 3 = 0

x = 1

x = 3

x-intercepts

(1,0)

(3,0)

9


B.

Zeros

x + 2 = 0

x – 3 = 0

x = –2

x = 3

x-intercepts

(–2,0)

(3,0)

2 21( ) ( 2) ( 3)

12P x x x

9


C.

Zeros

x – 3 = 0

x + 2 = 0

x = 3

x = –2

x-intercepts

(3,0)

(–2,0)

( ) ( 3)( 2)P x x x x

x = 0 (0,0)

9

Subtract

4x(x3 + 2 – 3x) – (–7x2 – x + 9x4)

4x4 + 8x – 12x2 – (–7x2 – x + 9x4)

4x4 + 8x – 12x2 7x2 x 9x4+ + –

–5x4 – 5x2 + 9x

10

2x2

4x

3x

–5

8x3

–4

12x2 –16x

–10x2 –15x 20

8x3 + 2x2 – 31x + 20

(4x – 5)(2x2 + 3x – 4)

(Add matching colors)

11

a3 – b3 = (a – b)(a2 + ab + b2)

Factor 8y3 – 27

Use difference of two cubes formula

a3 – b3

8y3 – 27

a =

b =

2y

3

2y 3 3

322y(2y)2= ( – )( + + )

= (2y – 3)(4y2 + 6y + 9)

12

4x(x2 – x – 6) = 0

Solve the equation

4x(x + 2)(x – 3) = 0x + 2 = 0 x – 3 = 0

x = –2 x = 3

–6

–1

2 –3

4x3 – 4x2 – 24x = 0

4x = 0

x = 0

Use diamond methodto factor trinomial

(Factor GCF)

4 0

4 4

x

13

4 2

5

5 4

24

x z

z x×

4 2

5

4

2

4

5x z

z x= ×

4 2

5

1

6

5x z

z x

5 1

6 1

x

z3

3

3

3

5

6

x

z

4 2

5 1

5 1

6

x z

z x

CircleLargest

Exponents

SubtractExponentsfor x and z

14

1 7 ( 1)

2( 1) 3( 7)

x x x x

x x

2

2

6 7 3 21

2 2

x x x

x x x

- - -¸

- -2 26 7

2 2 3 21

x x x x

x x

1

2 3

x x

( 1)

6

x x

2

6

x x

15

Let denominator equal 0Solve for x.

Vertical Asymptote

Horizontal Asymptote

y = 0

x2 – 1 = 0 (x + 1)(x – 1) = 0

x + 1 = 0 x – 1 = 0x = –1 x = 1

2

2( )

1

xf x

x

Numerator Degree smaller

Denominator Degree larger

116


Vertical Asymptote


x – 1 = 0+1 +1x = 1

None

2 2 15( )

1

x xf x

x

Numerator Degree larger

Denominator Degree smaller

1

16


Vertical Asymptote


y = 3

x2 – 9 = 0 (x + 3)(x – 3) = 0

x + 3 = 0 x – 3 = 0x = –3 x = 3

Numerator Degree equal

Denominator Degree equal

3

1y

2

2

3( )

9

x xf x

x

16

Subtract thenumerators

206

232

206

114

x

x

x

x

4 11 (2 23)

6 20

x x

x

4 11 2 23

6 20

x x

x

2 12

6 20

x

x

2( 6)

2(3 10)

x

x

6

3 10

x

x

17

123

5x

Solve

123

5x

12 3

5 1x

3(x – 5) = (12)(1)

3x – 15 = 123x = 27x = 9

Step 1A

Let denominator = 0

x – 5 = 0

x = 5

Step 1B

Endpointsx = 5x = 9

18

Interval TestNumber

Test of Inequality

True / False

A

B

C

5 9

123

5x

Solve

Step 2 CA B

Step 3

2

611

–4 > 3 F

12 > 3 T2 > 3 F

Step 4 5 < x < 9

18

Solve

3(x – 8) = (6)(1)

3x – 24 = 63x = 30x = 10

Step 1A

Let denominator = 0

x – 8 = 0

x = 8

Step 1B

Endpointsx = 8x = 10

63

8x

63

8x

6 3

8 1x

19

Interval TestNumber

Test of Inequality

True / False

A

B

C

8 10

Step 2 CA B

Step 3

7

911

–6 < 3 T

6 < 3 F2 < 3 T

Step 4 x < 8

Solve6

38x

x > 10or

19

Simplify each expression

ln e0.15t

ln e0.15t = 0.15t

3eln(x+1)

3(x + 1) = 3x+3

20

log42 + 2log43 – log46

log4(2·9) – log46

log4(18) – log46log4(18÷6)

log4(3)

20

log42 + log432 – log46log42 + log49 – log46

21Evaluate log

9489.

Use a calculator.

9log 489 =log489log9

9542.0

6893.2

82.2

52log 3 4x

5log 3 2x Divide both sides by 2

Exponentiate each side23 5x

25

3x

Divide both sides by 3

22

3x = 25

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

(–3)2 + (6)2 = r2

9 + 36 = r2

45 = r2

245 = r

45 = r

9 5× = r

3 5 = r

23

Step 2 Find sin θ, cos θ, and tan θ.

3, 6, 3 5=- = =x y r

2

5=

52

5 5= ×

2 5

25= 2 5

5=

1

5

-=

5

5

1

5

-= ×

1 5

25

-=

6

3=

-

23

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

3, 6, 3 5=- = =x y r

1csc

sinq= =

qr

y

1sec

cosq= =

qr

x

1cot

tanq= =

qx

y

3 5

6=

5

2=

3 5

3=

-5=-

3

6

-=

1

2=-

23

Convert radians from radians to degrees.5

9

= 100°

1805 rad

raians

dian9 s

æ ö÷ç ÷ç ÷ç ÷çp

æ öp ÷ç ÷ç ÷ç øè øè

o

185

9

0p= ´

p

o 5 180

9 1= ´

o 900

9=

o

24

Verify the identity.25

cscseccot

csc

cos

1

sin

cos

cscsin

1

csccsc

Verify the identity.26

1csccoscsc 222

1sin

1

1

coscsc

2

22

1sin

coscsc

2

22

1cotcsc 22 22 csccsc


Amplitude

1( ) sin(2 )

2g x x

1 12 2

Period2 2

2 2

1

–1

π

2π

27A


Amplitude

Period2

–26π

12π

1( ) 2cos

3h x x

|2| = 2

2 211

33

= 6π

2 1 2 3 6

1 3 1 1 1

27B

This problem involves daily profits for an amusement park.

Ticket sales were good until a massive power outage happened on Saturday that was not repaired until late Sunday.

The graph will show decreased sales until Sunday.

Identify the graph to represent the situation.

B

28


The weather was beautiful on Friday and Saturday, but it rained all day on Sunday and Monday.

The graph will show decreased sales on Sunday and Monday.


C

28


The graph will show decreased sales on Friday and Sunday.

Only of the rides were running on Friday and Sunday.


A

28

f(x) = 2x and g(x) = 7 – x

Find g(f(4))

Step 1: Find f(4)

g(16) = 7 – 16

= 16

Step 2: Find g(16)

f(4) = 24

= –9

g(f(4)) = –9

Use g(x) = 7 – xUse f(x) = 2x

29

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) + 55 = 2(0) – 8(0) + 55 = 0 – 0 + 55 = 5

30

(3,-1) -1 = 2(3)2 – 8(3) + 5-1 = 2(9) – 8(3) + 5-1 = 18 – 24 + 5-1 = -1

30


(5,15) 15 = 2(5)2 – 8(5) + 515 = 2(25) – 8(5) + 515 = 50 – 40 + 515 = 15

30


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.

if 0 0.5

12 5.5 if 0.5 1.5

6 3.5 if 1.5 2

t t

d t t t

t t

Swimming

Biking

Running

31

Swimming Points d = tt = 0 t = 0.5d = 0 d = 0.5

(0,0) (0.5 , 0.5)

Biking Points d = 12t – 5.5t = 0.5

t = 1.5d =12(0.5) – 5.50 = 0.5 (0.5 , 0.5)

(1.5 , 12.5)

if 0 0.5

12 5.5 if 0.5 1.5

6 3.5 if 1.5 2

t t

d t t t

t t

d =12(1.5) – 5.50 = 12.5

31

Running Points d = 6t + 3.5t = 1.5

t = 2

d = 6(1.5) + 3.5 = 12.5(1.5 , 12.5)

(2 , 15.5)

if 0 0.5

12 5.5 if 0.5 1.5

6 3.5 if 1.5 2

t t

d t t t

t t

d = 6(2) + 3.5 = 15.5

31

Swimming

Time = 0.5 – 0 = 0.5

Biking Time = 1.5 – 0.5 = 1

0.51

Running Time = 2 – 1.5 = 0.5

0.5

Distance = 0.5 – 0 = 0.5

0.5

Distance = 12.5 – 0.5 = 12

12

Distance = 15.5 – 12.5 = 3

3

31

Graph the exponential function. Find the asymptote. How is the graph transformed from the graph of its parent function?

Parent Function:________

g(x) = 2x+4 – 3

Asymptote:_______

Transformation:

f(x) = 2x

y = –3

Shift down 3 unitsShift left 4 units

32A



g(x) = –5x

Asymptote:_______

Transformation:

f(x) = 5x

y = 0

Reflect across x-axis

32B



g(x) = e–x – 6

Asymptote:_______

Transformation:

f(x) = ex

y = –6

Shift down 6 unitsReflect across y-axis

32C

Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?


Asymptote:_______

Transformation:

f(x) = log x

x = –4

Shift down 2 unitsShift left 4 units

g(x) = log (x+4) – 2

33A

Graph the logarithmic function. Find the asymptote. How is the graph transformed from the graph of its parent function?


Asymptote:_______

Transformation:

f(x) = log x

x = 0

Reflect across y-axis

g(x) = log (–x)

33B

Monday, January 11, 2016 Advanced Algebra Final Exam Study Guide.

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