chapter 9 review

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Pre-Calculus 1/31/2007 Chapter 9 Review a n = a 1 + (n – 1)d a n = a 1 • r (n – 1) 1 a S 1 r 1 n a a n 2 1 n 2a (n 1)d 2 n k k1 a n 1 a(1 r) 1 r n k k1 a

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a n = a 1 + (n – 1)d. a n = a 1 • r (n – 1). Chapter 9 Review. Probability Review (Sections 9.1 – 9.3). Fundamental Principle of Counting (Multiplication Principle). n! = n(n – 1)(n – 2)(n – 3)… (2)(1). How many ways can you arrange:. (5)(4)(3)(2)(1) = 120. A, B, C, D, E. n! - PowerPoint PPT Presentation

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Page 1: Chapter 9 Review

Pre-Calculus

1/31/2007

Chapter 9 ReviewChapter 9 Review

an = a1 + (n – 1)d an = a1 • r (n – 1)

1aS

1 r

1 na an

2 1

n2a (n 1)d

2

n

kk 1

a

n1a (1 r )

1 r

n

kk 1

a

Page 2: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability Review(Sections 9.1 – 9.3)

Probability Review(Sections 9.1 – 9.3)

Page 3: Chapter 9 Review

Pre-Calculus

1/31/2007

Fundamental Principle of Counting(Multiplication Principle)

Fundamental Principle of Counting(Multiplication Principle)

n! = n(n – 1)(n – 2)(n – 3)… (2)(1)

A, B, C, D, E (5)(4)(3)(2)(1) = 120

How many ways can you arrange:

Page 4: Chapter 9 Review

Pre-Calculus

1/31/2007

PermutationsOrder is important!!!

PermutationsOrder is important!!!

permutations of “n” objects taken “r” at a time:

using “n” objects to fill “r” blanks in order.

How many ways can 6 runners finish 1 – 2 – 3?

nPr =n!

(n – r)!

6P3 =6!

(6 – 3)!720

6120= =

Page 5: Chapter 9 Review

Pre-Calculus

1/31/2007

only interested in the ways to select the “r” objects regardless of the order in which we arrange them.

How many ways can 2 cards be picked from a deck of 10:

nCr =n!

(n – r)! • r!

10C2 = = 4510!

(8)! • 2!= 3,628,800

40320 • 2

CombinationsOrder is NOT important!!!

CombinationsOrder is NOT important!!!

n

r

Page 6: Chapter 9 Review

Pre-Calculus

1/31/2007

Subsets of an “n” – setSubsets of an “n” – set

There are ________ subsets of a set with n objects (including the empty set and the entire set).

2n

DiMaggio’s Pizzeria offers patrons any combination of up to 10 different pizza toppings. How many different pizzas can be ordered if we can choose any number of toppings (0 through 10)?We could add up all the numbers of the form for r = 0, 1, …, 10 but there is an easier way. In considering each option of a topping, we have 2 choices: __________ or __________.

Therefore the number of different possible pizzas is:

Example:

Yes No

2n = 210 = 1024

Page 7: Chapter 9 Review

Pre-Calculus

1/31/2007

Binomial TheoremBinomial Theorem

n n 1 n r r nn n n n

(a b) a a b ... a b ... b0 1 r n

n r

nC

r Don’t forget:

Page 8: Chapter 9 Review

Pre-Calculus

1/31/2007

4 4 8 2

or 0.15452 52 52 13

Also, recall that in mathematics, the word or signifies addition; the word and signifies multiplication .

Also, recall that in mathematics, the word or signifies addition; the word and signifies multiplication .

Find the probability of selecting an ace or a king from a draw of one card from a standard deck of cards.

Find the probability of selecting an ace and a king from a draw of one card from a standard deck of cards.

4 4 16 1

or 0.005952 52 2704 169

Page 9: Chapter 9 Review

Pre-Calculus

1/31/2007

Venn Diagrams

sample space (all students)

subsets to represent “girls” and “sports”

1

(54%) 0.183

“boys”

“no sports”

decimals

1

students

girls sports

0.36 0.18 0.23

0.23

.54 .18 .36

1 .36 .18 .23 .23

Page 10: Chapter 9 Review

Pre-Calculus

1/31/2007

0.5

conditional probability

0.25 1

0.125 + 0.125 + 0.5 = 0.75

dependent

of the event A, given that event B occurs

2/4 or 0.5 1

along the branches that come out of the two jars

the probability P(A B)

Page 11: Chapter 9 Review

Pre-Calculus

1/31/2007

P(A andB)

P(A)

the ends of the branches

conditional probability formula

P(jar A and )

P(chocolate chip)

chocolate chip

P(A) • P(B A)

1 20.25 12 4

0.75 0.75 3

Page 12: Chapter 9 Review

Pre-Calculus

1/31/2007

1

6

binomial distribution

binomial Theorem

41 1 1 1 1

0.000776 6 6 6 65

6

45

0.482256

2 21 5

0.019296 6

2 21 5

6 6

2 2

4 2

1 5C 0.11574

6 6

4C2 = 6 6

Page 13: Chapter 9 Review

Pre-Calculus

1/31/2007

Homework AnswersHomework Answers# 54, 56, 61, 65, 68, 70, 76, 77 , 17, 23

(p. 748 – 749)

5

5 4 3 2 2 3 4 5

5 4 3 2 2 3 4 5

5 4 3 2 2 3 4 5

(2x y)

5 5 5 5 5 5(2x) (2x) (y) (2x) (y) (2x) (y) (2x)(y) (y)

0 1 2 3 4 5

(1)32x (5)16x y (10)8x y (10)4x y (5)2xy (1)y

32x 80x y 80x y 40x y 10xy y

11

8 3 8 8

(x 2)

11x ( 2) 165 ( 8)x 1320x

3

Page 14: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ReviewQuestions

Probability ReviewQuestions

Page 15: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

Suppose there is a 70% chance of rain tomorrow. If it rains, there is a 10% chance that all of the rides at an amusement park will be operating. If it doesn’t rain, there is a 95% chance all of the rides will be operating. What is the probability that all of the rides will be operating tomorrow?

.7(.1) .3(.95) .355

Page 16: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

Binomial Theorem:

Bubba rolls a fair die 6 times. What is the probability that he will roll exactly two 2’s?

2 4

6 2

1 5C 0.2009

6 6

Page 17: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

There are 20 runners on a track team.

How many groups of 4 can be selected to run the 4 x 100 relay?

How many ways can 4 runners be selected to run 1st – 2nd – 3rd – 4th?

20 4C 4845

20 4P 116280

Page 18: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

A fair coin is tossed 10 times.

Find the probability of tossing HHHHHTTTTT.

Find the probability of tossing exactly 5 tails in those 10 tosses

10

10 5

1 252C .2461

2 1024

101 1

2 1024

Page 19: Chapter 9 Review

Pre-Calculus

1/31/2007

Review QuestionReview Question

# 43

(p. 748)

Page 20: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

Find the x4 term in the expansion of:

5(4x y)

4 1 45 1C (4x) (y) 1280x y

Page 21: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

License plates are created using 3 letters of the alphabet for the first 3 characters and 4 numbers for the last 4 characters.

How many possible different license plates are there if the letters and numbers are NOT allowed to repeat?

26 25 24 10 9 8 7 78,624,000

Page 22: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

A spinner, numbered 1 through 10, is spun twice.

What is the probability of spinning a 1 and a 10 in any order?

What is the probability of not spinning the same number twice?

2 1 1

10 10 50

Page 23: Chapter 9 Review

Pre-Calculus

1/31/2007

Probability ExerciseProbability Exercise

Expand: 4(3x 2y)

4 0 3 1 2 24 0 4 1 4 2

1 3 0 44 3 4 4

C (3x) (2y) C (3x) (2y) C (3x) (2y)

C (3x) (2y) C (3x) (2y)

4 3 2 2 3 4(3x) 4(3x) (2y) 6(3x) (2y) 4(3x)(2y) (2y)

4 3 2 2 3 481x 216x y 216x y 96xy 16y

Page 24: Chapter 9 Review

Pre-Calculus

1/31/2007

an = 12 – 2.5(n – 1)an = an-1 – 2.5

Review QuestionReview Question

Is the series arithmetic or geometric?Find the explicit formulaFind the recursive formulaFind the 100th termFind the sum for a1 through a100

12, 9.5, 7, 4.5, …

10, 12, 14.4, 17.28, …

12 – 2.5(100 – 1) = – 235.5

12 235.5

100 1117.52

Page 25: Chapter 9 Review

Pre-Calculus

1/31/2007

an = 10(1.2)(n – 1)

Review QuestionReview Question

Is the series arithmetic or geometric?Find the explicit formulaFind the recursive formulaFind the 100th termFind the sum for a1 through a100

12, 9.5, 7, 4.5, …

10, 12, 14.4, 17.28, …

an = an-1(2.5)

10(1.2)99 = 690,149,787.7

10010(1 1.2 )4,140,898,676

1 1.2

Page 26: Chapter 9 Review

Pre-Calculus

1/31/2007

Review QuestionReview Question

Evaluate:

6

212

k 3

( k )

– ½(3)2 – ½(4)2 – ½(5)2 – ½(6)2

– ½(9) – ½(16) – ½(25) – ½(36)

– 43

Page 27: Chapter 9 Review

Pre-Calculus

1/31/2007

Review QuestionReview Question

Is this sequence arithmetic or geometric?

9, 18, … 144