1 / 28

Math 140 Quiz 6 - Summer 2004 Solution Review

Math 140 Quiz 6 - Summer 2004 Solution Review. (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.). Problem 1 (12). Test for arithmetic sequence:. No common difference => not arithmetic sequence Test for geometric sequence:.

amal-riggs
Download Presentation

Math 140 Quiz 6 - Summer 2004 Solution Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Math 140Quiz 6 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

  2. Problem 1 (12) Test for arithmetic sequence: No common difference => not arithmetic sequence Test for geometric sequence: Find a general term, an: 4 , 16 , 64 , 256 , 1024. Common ratio r = 4 => an = a1r(n-1) = 4(4)(n-1) = 4n

  3. Problem 2 (19) Approach when choices are given - Test each choice: Find a general term, an: 0, 2, 6, 12, 20.

  4. Problem 2 cont’d (19) Approach if no choices - Test for arithmetic sequence: No common difference => not arithmetic sequence Test for geometric sequence: Find a general term, an: 0, 2, 6, 12, 20. No common ratio => not geometric sequence

  5. Problem 2 cont’d (19) Note pattern in arithmetic sequence test: an+1 - an = 2n. Find a general term, an: 0, 2, 6, 12, 20. Hence, an = an -1+ 2(n - 1) =an -2+ 2(n - 1) + 2(n - 2) Thus, an = a1 + n(n - 1) =n2- n.

  6. Problem 3 (31) Put n = 1, 2, 3, 4, 5 in the expression for an and evaluate to get: a1 = (-1)(1- 1) (1 + 1)/(2•1 - 1) = 2 a2 = (-1)(2- 1) (2 + 1)/(2•2 - 1) = -3/3 = -1 a3 = (-1)(3- 1) (3 + 1)/(2•3 - 1) = 4/5 = 0.80 a4 = (-1)(4- 1) (4 + 1)/(2•4 - 1) = -5/7 = -0.71 a5 = (-1)(5- 1) (5 + 1)/(2•5 - 1) = 6/9 = 0.67 Write out the first five terms of the sequence to two decimal places. an = (-1)(n - 1) (n + 1)/(2n - 1)

  7. Problem 4 (37) Note choices are of arithmetic sequence: (a1 = a) _____________an = a + d(n - 1). a10 = a + d(10 - 1) = 16 (a) a15 = a + d(15 - 1) = -29 (b) Compute: (b) minus (a) & substitute in (a) => d(14 - 9) = -29 –16 => d = -45/5 = -9 a - 9(9) = 16 => a = 97 an = a + d(n - 1) = 97 - 9(n - 1) = 106 - 9n Find the first term and give a formula for the sequence: 10th term is 16; 15th term is –29.

  8. NO YES NO NO Problem 5 (12) Since choices are given, test each choice: Express the sum using summation notation, S = 3 + 12 + 27 + . . . + 108

  9. Problem 6 (12) Find the common difference, an: 3.15, 4.82, 6.49, 8.16, .... Common difference d = 1.67

  10. Problem 7 (31) Test for geometric sequence: Find the common ratio for the geometric sequence. If a sequence is not geometric, say so. an: 1, -3, 9, -27, 81 Common ratio r = -3 => an = a1r(n-1) = (-3)(n-1)

  11. Problem 8 (81) Use the formula for the sum of first n integers: Find the requested sum of the arithmetic sequence. Put n = 1368 and find: S1368 = 1368(1368 + 1)/2 = 936,396

  12. Problem 9 (81) Factor one power of (1/5) to put into standard infinite geometric series form and use formula for the sum of an infinite geometric series with a = 2/5 & r = 1/5 < 1. Find the sum, if it exists, for the infinite geometric sequence.

  13. Problem 10 (25) Because n is squared, {an} is not an arithmetic sequence. Because n is not as power of a constant, rn, {an} is not a geometric sequence. Numerical tests for arithmetic & geometric sequences are on next slide. Determine whether the given sequence is arithmetic, geometric, or neither. If arithmetic, find the common difference. If geometric, find the common ratio. {an}= {5n2 - 4}

  14. Problem 10 cont’d (25) Arithmetic sequence test: an+1 - an=5[(n+1)2 - n2]= 10n + 5 No common difference => not arithmetic sequence Geometric sequence test: an+1 /an=[5(n+1)2 –4]/(5n2-4) {an}= {5n2 - 4} No common ratio => not geometric sequence

  15. Problem 11 (44) Note: in the formula for the binomial expansion, Write the indicated term of the binomial expansion. (3x + 2)5; 5th term the mth term has i = m – 1. Thus, the mth term is: Here we have n = 5, m = 5, a = 2, & y = 3x. Thus, Next slide has alternate method.

  16. Problem 11 Cont’d (62) Here is an alternate method. Consider Pascal’s Triangle where each entry is sum of 2 above. 5th power => 5th line Write the indicated term of the binomial expansion. (3x + 2)5; 5th term 5th term => next to last entry Thus, term in (y + a)5 is 5a4y. Here a = 2& y = 3x => 5a4y = 5•24•3x = 240x

  17. Problem 12 (44) P(n, r) = n!/(n – r)! P(8, 7) = 8!/(8 – 7)! = 8•7•6•5•4•3•2•1!/1! = 40,320 Evaluate the expression: P(8, 7).

  18. Problem 13 (56) P(n, r) = n!/(n – r)! P(11, 11) = 11!/(11 – 11)! = 11•10•9•8•7•6•5•4•3•2•1•0!/0! = 39,916,800 Evaluate the expression: P(11, 11).

  19. Problem 14 (44) C(n, r) = n!/[r!(n – r)!] C(9, 3) = 9!/[3!(9 – 3)!] = 9•8•7•6!/[3•2•1•6!] = 84 Evaluate the expression: C(9, 3).

  20. Problem 15 (62) Quick solution: In a survey of 48 hospital patients, 15 said they were satisfied with the nursing care, 23 said they were satisfied with the medical treatment, and 5 said they were satisfied with both. a) How many patients were satisfied with neither? b) How many were satisfied with only the medical treatment? B: medical satisfied A: nursing satisfied = 15 + 23 - 5 = 33 48 - 33 = 15 satisfied with neither

  21. Problem 15 cont’d (62) Universe is 48 people. Satisfied by in A = 15 nursing and in B = 23 medical. In AB = 5 satisfied by both nursing & medical. In AB = 15+23-5 = 33 ok with nursing or medical. In (AB) = 48-33 = 15 satisfied by neither. In B - AB= 23-5 = 18 only medical satisfied. A= 15 B= 23 AB=5 (AB)=15

  22. Problem 16 (69) The first code position can be filled in 6 ways, the second in 5, the third in 4, & the fourth in 3. Thus, by the Multiplication Rule there are: 6•5•4•3 = 360 ways. Or, by the Permutation Rule for 6 objects taken 4 at a time, there are: P(n, r) = n!/(n – r)! = 6!/(6–4)! = 6•5•4•3•2! /2! = 6•5•4•3 = 360 ways. How many 4-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once.

  23. Problem 17 (56) The first booth can be staffed in 8 ways, the second in 7, the third in 6,.... By the Multiplication Rule there are: 8•7•6•...1 = 8! ways. That is, 8•7•6•5•4•3•2•1 = 40,320 ways. In how many ways can 8 volunteers be assigned to 8 booths for a charity bazaar?

  24. Problem 18 (37) The probabilities in Model I all are in [0, 1] and add to 1. So YES, it is a probability model. The probabilities in Model II all are in [0, 1] but add to 1.27 > 1. So NO, it is not a probability model. Tell whether or not the given model is a probability model.

  25. Problem 19 (31) There are 7 + 3 + 1 = 11 marbles in the bag. Picking a “not blue” one means picking a red or green one of which there are 8. The probability is, thus, 8/11. A bag contains 7 red marbles, 3 blue marbles, and 1 green marble. What is the probability of choosing a marble that is not blue when one marble is drawn from the bag?

  26. Problem 20 (25) There are 6•6 = 36 ways that the dice may fall. To get a sum > 10 one must have a sum of 11 or 12. There are only 3 ways to do this: 5+6, 6+5, or 6+6. The probability is, thus, 3/36 = 1/12. Two 6-sided dice are rolled. What is the probability that the sum of the two numbers on the dice will be greater than 10?

  27. Problem 21 (75) Since order is unimportant, we seek the number of combinations of 7 dressings taken 3 at a time. By the Combination Rule for 7 objects taken 3 at a time, there are: C(n, r) = n!/[r!(n – r)!] = 7!/[3!(7–3)!] = 7•6•5•4! /(3•2•1•4!) = 35 ways. A hamburger shop sells hamburgers with cheese, relish, lettuce, tomato, onion, mustard, or ketchup. How many different hamburgers can be concocted using any 3 of the extras?

More Related