Review for Final Exam
Transcription
Review for Final Exam
ANSWER SHEET Advanced Algebra Review for Final Exam 1. Joey is a pretty good student. He estimates that his probability of making an A in Algebra is 94% and he has a 82% probability for making an A in Chemistry. What is the probability he makes: a) An A in both? P(A in Algebra AND A in Chemistry) = P(A in Algebra) ∙ P(A in Chemistry) = 77.08% b) An A in either class? P(A in Algebra OR A in Chemistry) = P(A in Algebra) + P(A in Chemistry) − P(A in Algebra AND A in Chemistry) = 98.92% c) No A in Chemistry? P(No A in Chemistry) = 1 − P(A in Chemistry) = 18% d) No A in either class? P(No A in either class) = 1 − P(A in Algebra OR A in Chemistry) = 1.08% Another way to do problem 6 d): P(No A in either class) = P(No A in Algebra AND No A in Chemistry) = P(No A in Algebra) ∙ P(No A in Chemistry) = 1.08% 2. From a standard deck of 52 cards, 4 cards are randomly chosen. Find each probability: P(all 4 cards are aces) P(all 4 cards are black) P(all 4 cards are face cards) 3. Without Replacement With Replacement 4 3 2 1 0.000004 52 51 50 49 26 25 24 23 0.0552 52 51 50 49 12 11 10 9 0.0018 52 51 50 49 4 4 4 4 0.000035 52 52 52 52 26 26 26 26 0.0625 52 52 52 52 12 12 12 12 0.0028 52 52 52 52 Thirteen boys and sixteen girls are in first period geometry class. Six will be chosen at random to represent the class in a contest.. a) In how many different ways can six students be selected? 29 𝐶6 = 475,020 b) How many six-student arrangements will be made of all girls? 16 𝐶6 = 8,008 c) How many six-student arrangements will be made of all boys? 13 𝐶6 = 1,716 d) How many six-student arrangements will include exactly four girls and two boys? 4 girls AND 2 boys: 16 𝐶4 ∙ 13𝐶2 = 1,820 ∙ 78 = 141,960 e) What is the probability that there are four girls and two boys? 141,960 26 16 𝐶4 ∙ 13𝐶2 = = ≈ 0.2989 475,020 87 29 𝐶5 4. A multiple choice quiz has 6 questions, each of them with four different possible answers. Each question has only one correct answer. The probability of correctly guessing a multiple-choice question is 0.25. a) What is the probability that the student misses all 6 questions? P(all incorrect) = P(1st incorrect) ∙ P(2nd incorrect) ∙ … ∙ P(6th incorrect) = 0.756 .1779 b) What is the probability that a student correctly guesses all 6 questions? P(all correct) = P(1st correct) ∙ P(2nd correct) ∙ … ∙ P(6th correct) = 0.256 .00024 5. Liberty High School is hosting a track meet. For the 400m event there are 6 seniors signed up, 3 juniors, and only 2 sophomore. What is the probability that the first place winner is a senior and the second one a junior. 𝐏(𝟏𝐬𝐭 𝐬𝐞𝐧𝐢𝐨𝐫 𝐀𝐍𝐃 𝟐𝐧𝐝 𝐣𝐮𝐧𝐢𝐨𝐫) = 𝐏(𝟏𝐬𝐭 𝐬𝐞𝐧𝐢𝐨𝐫) ∙ 𝐏(𝟐𝐧𝐝 𝐣𝐮𝐧𝐢𝐨𝐫) 𝟔 𝟑 𝟗 ∙ = 𝟏𝟏 𝟏𝟎 𝟓𝟓 6. A survey designed to determine family size among teenagers from Bakersfield yielded the results shown in the table below. Based on the data, find the probability of randomly selecting one teenager from Bakersfield with the given characteristics. Male Female Only child 25 21 2 siblings 41 51 3 siblings 6 9 Other 2 3 a) probability that the teenager is a male, given that he lives in a 2-sibling family 41 𝑃(male|2 siblings) = 92 b) probability that the teenager is an only child, given that he is a male 25 𝑃(only child|male) = 74 c) probability that the teenager lives in a 3-sibling family, given that she is a female 9 3 𝑃(3 siblings|female) = = 84 28 d) probability that the teenager is a female, given that she lives in a 3-sibling family 9 3 𝑃(female|3 siblings) = = 15 5 7. A geneticists studies the probabilities of children born with certain physical traits. See the chart below. New born 55% 45% Girl Boy 12% Left Handed 88% 9% Right Handed Left Handed 91% Right Handed Find the probability that a new born is a lefty. Clearly set up and show all your work for this problem. P((Left Handed AND Boy) OR (Left Handed AND Girl)) = P(Left Handed AND Boy) + P(Left Handed AND Girl) = P(Boy) ∙ P(Left Handed|Boy) + P(Girl) ∙ P(Left Handed|Girl) = 0.55 ∙ 0.12 + 0.45 ∙ 0.09 = 0.1065 8. 9. As a promotion, Dr. Pepper advertises that one in eight bottle caps will offer a free 20 oz bottle of Dr. Pepper. Billy Bob randomly selected 3 bottles of Dr. Pepper at a store. What is the probability that all of them will have winning bottle caps? 𝐏(𝐚𝐥𝐥 𝐰𝐢𝐧𝐧𝐢𝐧𝐠 𝐜𝐚𝐩𝐬) = 𝐏(𝟏𝐬𝐭 𝐰𝐢𝐧𝐧𝐢𝐧𝐠 𝐜𝐚𝐩 𝐀𝐍𝐃 𝟐𝐧𝐝 𝐰𝐢𝐧𝐧𝐢𝐧𝐠 𝐜𝐚𝐩 𝐀𝐍𝐃 𝟑𝐫𝐝 𝐰𝐢𝐧𝐧𝐢𝐧𝐠 𝐜𝐚𝐩) 𝟏 𝟏 𝟏 𝟏 ∙ ∙ = = 𝟎. 𝟎𝟎𝟏𝟗𝟓 𝟖 𝟖 𝟖 𝟓𝟏𝟐 How many distinguishable arrangements of the 10. letters in the word HORROR are there? A B 6! 6! 3! C 6! 2! 3! D 6! 2!3! (C) (B) The weight in pounds of each starting player for Liberty’s frosh/soph-boys basketball team are shown below. 162, 164, 175, 180, 189 a) Find the mean, range, and standard deviation for the data. Mean: 174; Range: 27; Standard Deviation: 10.05 b) During the preseason, each member of the team lost 5 pounds. What are the new mean, range, and standard deviation? Mean: 169; Range: 27; Standard Deviation: 10.05 c) By their senior year, each member of the team is increase their weight by 10%. What will the mean, range, and standard deviation be in their senior year? Mean: 191.4; Range: 29.7; Standard Deviation: 11.055 11. 12. In 2012 the mean for the Mathematics section of the SAT was 514 points with a standard deviation of 117. Assume the scores for the Mathematics section of the SAT are normally distributed. a) What percent of the students who took the test scored below 631 points? Justify your answer. 631 = 514 + 117, so 631 is one standard deviation above the mean. Therefore 𝑷(𝐬𝐜𝐨𝐫𝐞 < 𝟔𝟑𝟏) = 𝟖𝟒% b) What percent of the students who took the test scored above 280 points? Justify your answer. 280 = 514 − 2×117, so 280 is two standard deviations below the mean. Therefore 𝑷(𝐬𝐜𝐨𝐫𝐞 > 𝟐𝟖𝟎) = 𝟗𝟕. 𝟓% c) What percent of the students who took the test scored between 397 and 748 points? Justify your answer. 397 = 514 – 117 and 748 = 514+ 2×117, so the data is contained between one standard deviation below the mean and two standard deviations above it Therefore 𝑷(𝟑𝟗𝟕 < 𝐬𝐜𝐨𝐫𝐞 < 𝟕𝟒𝟖) = 𝟖𝟏. 𝟓% d) In 2012, 1,600,000 people took the SAT. How many of them scored below 631 points? Show all your work. 𝟖𝟒% of 1,600,000 is 1,344,000 13. θ = −250° y x 14. θ = 190° 𝛉 = −𝟏𝟕𝟎° & 𝛉 = 𝟓𝟓𝟎° 15. 16. b 60° a 8 in b 18 m 45° a 𝒂 = 𝟔√𝟑 meters; 𝒃 = 𝟏𝟐√𝟑 meters 𝒂 = 𝒃 = 𝟒√𝟐 inches 17. 𝟏𝟖. 260° to radians 𝟏𝟑 𝟐𝟔𝟎° = 𝛑 𝐫𝐚𝐝𝐢𝐚𝐧𝐬 𝟗 19. 𝐬𝐢𝐧 𝟑𝟎° = 𝟏 𝟐 22. 𝐭𝐚𝐧 𝟑𝟑𝟎° = √𝟑 − 𝟑 𝟐𝟓. (𝟏𝟐, − 𝟓) 𝛑 𝐭𝐨 𝐝𝐞𝐠𝐫𝐞𝐞𝐬 𝟏𝟓 𝛑 𝐫𝐚𝐝𝐢𝐚𝐧𝐬 = 𝟏𝟐° 𝟏𝟓 20. 𝐜𝐨𝐭 𝟏𝟐𝟎° = − 21. 𝐬𝐞𝐜 𝟑𝟏𝟓° = √𝟐 √𝟑 𝟑 23. 𝐜𝐨𝐬 𝟐𝟏𝟎° = 𝟓 𝟏𝟑 𝐬𝐢𝐧 𝛉 = − → 𝐜𝐬𝐜 𝛉 = − 𝟏𝟑 𝟓 𝟏𝟐 𝟏𝟑 𝐜𝐨𝐬 𝛉 = → 𝐬𝐞𝐜 𝛉 = 𝟏𝟑 𝟏𝟐 𝟓 𝟏𝟐 𝐭𝐚𝐧 𝛉 = − → 𝐜𝐨𝐭 𝛉 = − 𝟏𝟐 𝟓 − 24. 𝐜𝐬𝐜 𝟐𝟐𝟓° = −√𝟐 √𝟑 𝟐 𝟐𝟔. (−𝟒, 𝟒) √𝟐 → 𝐜𝐬𝐜 𝛉 = √𝟐 𝟐 √𝟐 𝐜𝐨𝐬 𝛉 = − → 𝐬𝐞𝐜 𝛉 = −√𝟏 𝟐 𝐭𝐚𝐧 𝛉 = −𝟏 → 𝐜𝐨𝐭 𝛉 = −𝟏 𝐬𝐢𝐧 𝛉 = 27. 𝑦 = 5 sin 2𝑥 Amplitude: 5 Period: 180 Midline: 𝒚 = 𝟎 𝑥 28. 𝑦 = −4 cos ( ) + 2 3 Amplitude: 4 Period: 1,080 Midline: 𝒚 = 𝟐 Amplitude: 1 Period: 90 Midline: 𝑦 = −3 29. 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧: 𝒚 = 𝐜𝐨𝐬(𝟒𝒙) − 𝟑