External & Global Environment of Louis Vuitton in Japan Essay

Introduction This case study is on the external and global environment of Louis Vuitton (LV) in Japan. For many years, Japan has been Louis Vuitton’s most profitable market but the economic crisis has had a negative effect on sales; sales have declined in the past few years. According to Tokyo Fashion News in November of 2008, LV has â€Å"seen more than a 5% decline in sales in Japan so far this year† (para. 3). They attribute the decline to one of two things, the decline in the Japanese economy or that LV has fallen out of favor with the Japanese people. Synopsis of the Situation With a decline in the economy, Louis Vuitton has been forced to open stores that offer a lower priced collection. The Japanese economy can no longer support the high-end collector name brands that it used to but Japan is in love with LV. â€Å"Based on sales figures and brand image indicators, we have compiled Asia’s Top Ten. It confirms what every luxe-loving Asian already knows: There is nothing to beat the charms of Louis Vuitton and Rolex† (Chadha, R. & Husband, Paul, 2006, para. 5). Key Issues The number one key issue would be the decline in the economy. The second key issue is that the Japanese are looking for a good deal, good value to product. The third key issue is the number of competitors in the Japanese markets that offer luxury products. Define the Problem In 2008, the economy in Japan and all over the world took a nosedive; people  were worried about making ends meet and severely cut down on buying luxury items. People started looking for products that offer a better value for their yen and became more cost conscious of what they were buying. There is serious competition in Japan for the luxury market. Among the competitors are Rolex, Cartier, Gucci, Chanel, and Prada to name a few (Table 1.1). Alternative Solutions One alternative solution would be to create a less expensive product and market it to a larger segment of the market. A second solution would be to improve the quality of their product and continue to market as a luxury item. A third solution would be to offer a middle of the line product available online that would draw the customers away from the other luxury item companies that would not require a storefront. Selected Solution to the Problem The selected solution to the problem would be to offer a middle of the line product available exclusively online. By offering the products online the company would not be require to have a physical store for customers to come into. This would cut costs dramatically; there would not be the overhead of operating a store, paying employees, or paying the costs of running a store. This would also offer customers the convenience of shopping from home. Implementation Implementation of this plan could be tricky. The company needs to make sure they have the capacity to handle a large volume of internet orders and they need the inventory on hand so they could ship items in a timely manner. Probably a million small details will need to be dealt with so the company needs to do some brainstorming with employees and management as to what problems may arise at implementation. Being prepared for any unforeseen problems would be very important before implementation, do not wait until problems arise to brainstorm solutions. By making sure they are prepared for orders, word-of-mouth advertising could be a huge advantage. Secondly, the company would have to make up a great advertisement campaign targeting the middle-class and the products that would be available. Recommendations Louis Vuitton should start a marketing campaign focusing on the middle class in Japan offering mid-priced products that are available exclusively online. The company should focus on smaller, out of the way cities first that do not have access to the higher priced stores and move out from there. Depending on the success of the marketing campaign in the smaller cities, the company should move to the larger cities where they have stores in order of population starting from smallest to largest. If these campaigns are successful, they could consider moving outside of Japan into China. Conclusion While Louis Vuitton is very popular in Japan, the decline in the economy has affected sales. Because there are so many other luxury stores in Japan, the company should focus on the larger, middle class segment of the population. By offering a lower priced, quality product the company could gain market share by attracting more customers from a larger population of people. Offering these products online opens the door for customers that do not live in an area where their stores are located. References Louis Vuitton Japan Lowering Prices. (2008, November 29). Tokyo Fashion News RSS. Retrieved June 9, 2014, from http://tokyofashion.com/louis-vuitton-japan- lowering-prices/ Chadha, R., & Husband, P. (2006, January 1). The Cult of the Luxury Brand: Inside Asia’s Love Affair With Luxury. . Retrieved June 9, 2014, from http://eds.a.ebscohost.com.lib.kaplan.edu/eds/results?sid=d07a16b5-4279-45c4-925174ed637b11f1%40sessionmgr4004&vid=3&hid=4110&bquery=DE+%22Brand+name+products++Social+aspects+Asia%22&bdata=JmNsaTA9RlQmY2x2MD1ZJnR5cGU9MCZzaXRlPWVkcy1saXZl and image scores.

Test Bank: Introduction to Probability and Statistics

True/False Questions 1. The standard deviation of any normal random variable is always equal to one. Answer: False Type: Concept Difficulty: Easy 2. For any normal random variable, the probability that the random variable will equal one is always zero. Answer: True Type: Concept Difficulty: Medium 3. The graph of a standard normal random variable is always symmetric. Answer: True Type: Concept Difficulty: Easy 4. The formula will convert any normal distribution into the â€Å"standard normal distribution. † Answer: True Type: Concept Difficulty: Easy 5.Any normal random variable with standard deviation equal to one is a standard normal random variable. Answer: False Type: Concept Difficulty: Medium 6. The notation X – N(4, 32) indicates a normal distribution with mean 2 and standard deviation 3. Answer: False Type: Concept Difficulty: Easy 7. The total area under a normal curve is always equal to one. Answer: True Type: Concept Difficulty: Easy 8. The notation Z  œ N(0, 1) indicates a standard normal distribution. Answer: True Type: Concept Difficulty: Easy 9. The probability that a normal random variable will be within two standard deviations of its mean is approximately 0. 8. Answer: False Type: Concept Difficulty: Easy 10. The normal distribution is a continuous distribution. Answer: True Type: Concept Difficulty: Easy 11. The normal distribution can be used to approximate the binomial distribution when both np and n(1 – p) are at least five. Answer: True Type: Concept Difficulty: Easy 12. The normal distribution approximation to the binomial works best when n is large. Answer: True Type: Concept Difficulty: Easy 13. The formula can be used with both a normal and binominal distribution. Answer: True Type: Concept Difficulty: Easy Multiple Choice Questions 4. Find P(-2 < Z < 2). A)0. 9544 B)0. 4772 C)0. 9772 D)0. 6826 E)none of the above Answer: A Type: Computation Difficulty: Easy 15. Find P(-0. 5 < Z < 0. 5). A)0. 3830 B)0. 1915 C )0. 6515 D)0. 3085 E)none of the above Answer: A Type: Computation Difficulty: Easy 16. What is the probability that a standard normal variable will be between -0. 5 and 1. 00? A)0. 2857 B)0. 5328 C)0. 6687 D)0. 2500 E)none of the above Answer: B Type: Computation Difficulty: Easy 17. Find the probability that a standard normal random variable has a value greater than -1. 56. A)0. 0332 B)0. 0594 C)0. 9406D)0. 9668 E)none of the above Answer: C Type: Computation Difficulty: Easy 18. Let X be a normally distributed random variable with mean 100 and standard deviation 20. Find two values, a and b, symmetric about the mean, such that the probability of the random variable being between them is 0. 99. A)90. 5, 105. 9 B)80. 2, 119. 8 C)22, 78 D)48. 5, 151. 5 E)90. 1, 109. 9 Answer: D Type: Computation Difficulty: Medium 19. A professor grades his students on a normal distribution, with mean at 75 and standard deviation of 15. If there are 39 students in his class, about how many score bet ween 80 and 90? A)5B)21 C)8 D)13 E)none of the above Answer: C Type: Computation Difficulty: Hard 20. A calculator manufacturer performs a test on its calculators and finds their working life to be normally distributed, with a mean of 2,150 hours and a standard deviation of 450 hours. What should the manufacturer advertise as the life of the calculators so that 90% of the calculators are covered? A)2,555 B)1,947 C)1,410 D)1,745 E)1,574 Answer: E Type: Computation Difficulty: Hard 21. You have two stocks: A and B. The price of each stock is normally distributed. Stock A has a mean of 25 and a standard deviation of 3.Stock B also has a mean of 25, but the standard deviation is 5. If I buy stock A at $25 and sell it on a randomly chosen day in the future (without knowing its price then), what is the probability that I will make at least $2 on each share? Answer the same for stock B. A)0. 2514, 0. 1554 B)0. 2514, 0. 3446 C)0. 2486, 0. 1554 D)0. 2486, 0. 3446 E)none of the above Answer: B Type: Computation Difficulty: Hard 22. Find two values symmetric around a mean of 20 such that they include an area equal to 0. 75. (standard deviation = 5). A)16. 65, 23. 35 B)19. 25, 20. 75 C)16. 25, 23. 75 D)14. 25, 25. 75E)none of the above Answer: D Type: Computation Difficulty: Hard 23. A spark plug manufacturer believes that his plug lasts an average of 30,000 miles, with a standard deviation of 2,500 miles. What is the probability that a given spark plug of this type will last 37,500 miles before replacement? A)0. 0228 B)0. 0114 C)0. 0013 D)0. 0714 E)0. 0833 Answer: C Type: Computation Difficulty: Medium 24. Fluctuations in the exchange rate of dollars against the pound sterling over a short time period were approximated by a normal distribution with a mean of 2. 01 and a standard deviation of 0. 13.What is the probability that the rate on a particular day was more than 1. 90? A)0. 8461 B)0. 3023 C)0. 8023 D)0. 3461 E)none of the above Answer: C Type: Computation Difficult y: Medium 25. The average time it takes for a letter in the United States to reach from one place in the 48 contiguous states to another is 3. 2 days, with a standard deviation of 0. 85 days. What is the probability of a letter arriving at its destination no more than four days after mailing? Assume a normal distribution. A)0. 1736 B)0. 3264 C)0. 8264 D)0. 6736 E)0. 6528 Answer: C Type: Computation Difficulty: Medium 6. The contents of a particular bottle of shampoo marked as 150 ml are found to be 153 ml at an average, with a standard deviation of 2. 5 ml. What proportion of shampoo bottles contain less than the marked quantity? Assume a normal distribution. A)0. 2192 B)0. 1151 C)0. 4452 D)0. 0548 E)none of the above Answer: B Type: Computation Difficulty: Medium 27. The age of people in a town is normally distributed, with a mean of 34 years and a standard deviation of 11 years. Find two values for age that will give a symmetric 0. 95 probability interval. A)28. 78, 39. 23 B)32. 3 0, 66. 30 C)23. 55, 44. 45D)12. 44, 55. 56 E)15. 91, 51. 10 Answer: D Type: Computation Difficulty: Medium 28. The weight of apples in a farm is normally distributed, with a mean of 110 grams, and a standard deviation of 15 grams. Find the probability that an apple selected at random will weigh between 95 and 105 grams. A)0. 3413 B)0. 4706 C)0. 1293 D)0. 2108 E)0. 5294 Answer: D Type: Computation Difficulty: Medium 29. A grocery store has a mean accounts receivable of $264, with a standard deviation of $55. The accounts receivable are normally distributed. What proportion of all accounts will be greater than $275? A)0. B)0. 1 C)0. 4207 D)0. 0793 E)0. 0228 Answer: C Type: Computation Difficulty: Medium 30. A grocery store has a mean accounts receivable of $264, with a standard deviation of $55. The accounts receivable are approximately normally distributed. Find the value such that 45% of all the accounts exceed this value. That is, find x such that: P(X > x) = 0. 45. A)$257. 13 B)$3 54. 48 C)$270. 91 D)$309. 00 E)none of the above Answer: C Type: Computation Difficulty: Medium 31. The waist measurement of students in a college is normally distributed. The standard deviation is known to be five inches.It is found that 15% of the students have waist sizes less than 28 inches. What proportion of students will have waists between 30 and 35 inches? A)0. 3795 B)0. 2389 C)0. 1406 D)0. 0983 E)none of the above Answer: A Type: Computation Difficulty: Hard 32. The IQs of the employees of a company are normally distributed, with a mean of 127 and a standard deviation of 11. What is the probability that the IQ of an employee selected at random will be between 120 and 130? A)0. 2389 B)0. 3453 C)0. 1064 D)0. 1325 E)0. 4638 Answer: B Type: Computation Difficulty: Medium 33.The mean life of a computer disk drive is 2,000 hours, with a standard deviation of 140 hours. Assuming the life-time of the drives to be normally distributed, find the probability of a disk-drive lasting m ore than 1,800 hours? A)0. 4236 B)0. 9236 C)0. 8472 D)0. 5764 E)0. 2118 Answer: B Type: Computation Difficulty: Medium 34. The average bill for car repairs at a car service center is $196, with a standard deviation of $44. Assuming the bills to be normally distributed, find the probability of a bill exceeding $300. A)0. 4909 B)0. 0182 C)0. 9819 D)0. 1406 E)0. 0090 Answer: E Type: Computation Difficulty: Medium 35.The GMAT scores of students in a college are normally distributed with a mean of 520 and a standard deviation of 41. What proportion of students have a score higher than 600? A)0. 9744 B)0. 2372 C)0. 4774 D)0. 0255 E)none of the above Answer: D Type: Computation Difficulty: Medium 36. The probability that a normal random variable with mean zero and standard deviation one will equal the number 1. 00 is: A)1 B)0. 9 C)0. 3413 D)0. 1587 E)0 Answer: E Type: Concept Difficulty: Medium 37. Suppose that X is a normal random variable with mean 17 and standard deviation 10. The proba bility that the value of X will be between -2. and 36. 6 is: A)0 B)0. 90 C)a number very close to 1 D)0. 95 E)0. 99 Answer: D Type: Computation Difficulty: Medium 38. Suppose that X is a normal random variable with mean 10 and standard deviation 4. Then the probability that X will be greater than 12 is: A)0. 1587 B)0. 3085 C)0. 1915 D)0. 4772 E)none of the above Answer: B Type: Computation Difficulty: Medium 39. A normal random variable has a distribution that is: A)always symmetric B)never symmetric C)sometimes symmetric D)symmetric if the mean is positive E)symmetric if the variance is negative Answer: A Type: Concept Difficulty: Medium 0. The distribution of X, the number of cars sold per day, where X can be 0, 1, 2, 3, 4, or 5 is: A)sometimes normally distributed B)never normal C)always normal D)a uniform distribution E)none of the above Answer: B Type: Concept Difficulty: Medium 41. What is the probability that a normal random variable with mean 15 and standard deviation 5 will have a value of exactly 25? A)0. 0228 B)0. 0456 C)0. 9772 D)0 E)1 Answer: D Type: Concept Difficulty: Medium 42. If X is a normal random variable with mean 12 and standard deviation 2, then the probability that X will exceed 16 is? A)0. 4772 B)0. 0228 C)0. 9772 D)0 E)1 Answer: B Type: Computation Difficulty: Medium 43. If X is a normal random variable with mean 15 and standard deviation 10, then the probability that X will have a negative value is: A)0. 0668 B)0. 432 C)0. 9332 D)0. 8664 E)none of the above Answer: A Type: Computation Difficulty: Medium 44. If X is a normally distributed random variable with mean 16 and variance 64, the probability that the random variable will have a value between 0. 32 and 31. 68 is: A)0. 99 B)0. 90 C)0. 85 D)1 E)0. 95 Answer: E Type: Computation Difficulty: Medium 45.For a normally distributed random variable with mean zero and standard deviation five, the probability that its value will be greater than -5 is: A)0. 4772 B)0. 9544 C)0. 3413 D)0. 8 413 E)none of the above Answer: D Type: Computation Difficulty: Medium 46. What is the probability that a standard normal random variable is between -0. 4 and 1. 4? A)0. 3413 B)0. 4254 C)0. 5746 D)0. 2638 E)none of the above Answer: C Type: Computation Difficulty: Easy 47. All of the following are characteristics of the normal distribution, except: A)symmetric about the mean B)bell-shaped curve C)total area under the curve is always oneD)it is a discrete distribution E)probability that x is equal to any specific value is zero Answer: D Type: Computation Difficulty: Medium 48. Find two values symmetric about a mean of 100, standard deviation of 10, such that they include an area equal to 0. 95. A)90, 110 B)80. 4, 119. 6 C)98. 04, 101. 96 D)70, 130 E)none of the above Answer: B Type: Computation Difficulty: Medium 49. A tire manufacturer believes its tires will last an average of 48,000 miles, with standard deviation of 2,000 miles. What is the probability that one of these tires, cho sen at random, will last at least 50,000 miles?A)0. 6587 B)0. 3413 C)0. 1587 D)0. 4772 E)none of the above Answer: C Type: Computation Difficulty: Medium 50. Suppose that an instructor gives an exam. This instructor wants to give those students in the top 2. 5% an A on this exam. What will the cutoff be for an A, if the average score on this exam is 80, with a standard deviation of 5? A)about 80 B)about 90 C)about 85 D)about 86 E)none of the above Answer: B Type: Computation Difficulty: Hard Use the following to answer questions 51-54: LittleAir operates a fleet of regional jets on a contract basis for a major air carrier.LittleAir's jets seat only 50 passengers, but because passengers' travel plans often change, LittleAir books up to 60 reservations for a typical flight. Booked passengers have a â€Å"no-show† probability of 0. 25. 51. Suppose LittleAir loses money if the number of passengers on a flight is less than 40. What is the probability that a randomly selected Littl eAir flight will have fewer than 40 passengers? A)0. 0367 B)0. 0505 C)0. 0681 D)0. 0901 E)0. 1492 Answer: B Type: Computation Difficulty: Medium 52. What is the probability that a randomly selected LittleAir flight will be overbooked (i. . , have more passengers show up than there are seats available)? A)0. 1170 B)0. 0901 C)0. 0681 D)0. 0505 E)0. 0367 Answer: D Type: Computation Difficulty: Medium 53. What is the probability that a randomly selected LittleAir flight will be full? A)0. 1170 B)0. 0901 C)0. 0681 D)0. 0505 E)0. 0367 Answer: B Type: Computation Difficulty: Medium 54. Suppose LittleAir gives compensation vouchers to any passenger who is denied a seat on an overbooked flight. Because these vouchers are valuable (> $200), management would like to keep the number of them on-hand at a minimum.How many vouchers should be held at the gate such that there are enough for at least 99% of all situations? A)1 voucher B)2 vouchers C)3 vouchers D)4 vouchers E)5 vouchers Answer: C Type : Computation Difficulty: Hard Use the following to answer questions 55-57: The ski season at a popular resort destination lasts 120 days. Experience has shown that the probability of snow on any given day is 0. 55 and is independent of whether or not there was snow on the previous day. 55. What is the probability of there being more than 60 days of snow in any given year? A)0. 7967 B)0. 8438 C)0. 8643D)0. 8830 E)0. 8997 Answer: B Type: Computation Difficulty: Medium 56. What is the probability of there being fewer than 55 days of snow in any given year? A)0. 0409 B)0. 0268 C)0. 0217 D)0. 0174 E)0. 0139 Answer: D Type: Computation Difficulty: Medium 57. Suppose a particular hotel at this destination breaks even or makes money so long as there are at least 50 days of snow but no more than 70 days of snow. What is the probability of the hotel's losing money in any given year? A)0. 8438 B)0. 7955 C)0. 7944 D)0. 7664 E)0. 7657 Answer: B Type: Computation Difficulty: Hard 58.If, for a bi nomially distributed random variable n*p = 5 and n*(1-p) = 5, then a _____________ distribution with a mean equal to _____ and a standard deviation equal to _____ typically can be used. A)Normal; ; B)Normal; ; C)Exponential; ; D)Exponential; ; E)Hypergeometric; ; Answer: A Type: Concept Difficulty: Easy Short Answer Questions Use the following to answer questions 59-67: If x ~ N(40, 32): 59. Find p(X ; 37) Answer: 0. 8413 Type: Computation Difficulty: Medium 60. Find p(X ; 47) Answer: 0. 0099 Type: Computation Difficulty: Medium 61. Find p(42 ; X ; 47) Answer: 0. 415 Type: Computation Difficulty: Medium 62. Find p(X ; 41) Answer: 0. 3707 Type: Computation Difficulty: Medium 63. Find p(36 ; X ; 41) Answer: 0. 5375 Type: Computation Difficulty: Medium 64. Find p(36 ; X ; 39) Answer: 0. 2789 Type: Computation Difficulty: Medium 65. Find x1 such that: p(X ; xl) = 0. 0475 Answer: 45 Type: Computation Difficulty: Medium 66. Find xl such that: p(40 ; X ; xl) = 0. 3770 Answer: 43. 48 Type: Computation Difficulty: Medium 67. Find x1 such that: p(X ; x1) = 0. 0154 Answer: 33. 52 Type: Computation Difficulty: MediumUse the following to answer questions 68-69: There are two shipping routes between a plant and a distribution point. The average time by route A is 220 minutes with a standard deviation of 20 minutes. The average time and standard deviation by route B are 200 and 40, respectively. Assume the distributions of trips can be approximated by normal curves. 68. What proportion of route B trips takes longer than the average route A trip? Answer: 0. 3085 Type: Computation Difficulty: Medium 69. 95% of route A trips take between what two values that are equidistant from the mean? Answer: [180. 8, 259. 2] Type: Computation Difficulty: MediumUse the following to answer questions 70-73: The amount dispensed into bottles by a machine in a ketchup plant is supposed to be normally distributed with a mean of 10 ounces and a standard deviation of 0. 5 ounce. If the machine is working properly, what is the probability that a single bottle chosen at random from the assembly line will have: 70. More than 11 ounces or less than 9. 5 ounces? Answer: 0. 1815 Type: Computation Difficulty: Medium 71. Between 9. 5 and 11 ounces? Answer: 0. 8185 Type: Computation Difficulty: Medium 72. What would you think if a single bottle chosen at random had less than . 5 ounces? Answer: p(x ; 8. 5) = 0. 0013, so we might conclude the machine is operating improperly. Type: Computation Difficulty: Medium 73. Between what two values symmetric about the mean would you expect to find 99% of the bottles filled by the machine, if it is operating properly? Answer: [8. 712, 11. 288] Type: Computation Difficulty: Medium 74. If the contents of bottles coming off a production line are normally distributed with a mean of 16 ounces and a variance of 0. 625, what's the probability of choosing a bottle at random and finding its contents to be less than 15. 1 ounces? Answer: 0. 2676 Type: Com putation Difficulty: Medium 75. A seed packet says that 90% of lettuce seedlings should be between 2 and 2. 5 inches high after 5 weeks. Assuming an average height of 2. 25 inches (and a normal distribution), what's the standard deviation of heights of 5-week-old seedlings? Answer: 0. 1520 Type: Computation Difficulty: Medium Use the following to answer questions 76-78: If the distribution of heights of mature poppy plants is normal, with a mean of 16 inches and a standard deviation of 3 inches, what proportion of the poppies will be: 76.Between 10 and 20 inches? Answer: 0. 8854 Type: Computation Difficulty: Medium 77. Less than 9 inches? Answer: 0. 0099 Type: Computation Difficulty: Medium 78. More than 24 inches? Answer: 0. 0038 Type: Computation Difficulty: Medium 79. You are interested in the incomes of your customers. A random sample of customer incomes yields a mean income of $35,000 with a standard deviation of $4,421. A) Determine what percent of the population would have a salary above $38,000. B) What income range that is symmetric about the mean would include 95% of your customers? Answer:A) 24. 83% have incomes above $35,000. B) 95% of your customers have an income between $26,330 and $43,670. Type: Computation Difficulty: Medium 80. Half of all mutual funds of a particular class charge up-front administration fees. Assuming that a random sample of 60 of these mutual funds is taken, calculate: A) The mean and standard deviation of the normal approximation of the binomial. B) The probability that no more than 40 of the mutual funds sampled charge an up-front administration fee. Answer: A) Mean = 30, standard deviation = 3. 873 B) Prob(# charging fee = 40) = 0. 9966Type: Computation Difficulty: Medium 81. The owner of a 100-room hotel has discovered that his reservations team has booked 110 reservations for an upcoming weekend. Experience has shown that 10% of reservations are â€Å"no shows. † How likely is this hotel to be overbooked (i. e. , have more guests arrive than there are rooms available) for this particular weekend? Answer: 0. 3156 Type: Computation Difficulty: Medium 82. Harry Highroller likes to bet on the roulette wheel when he is in Las Vegas. Roulette wheels in Las Vegas typically have 38 spaces: 18 of them are red; 18 are black; and 2 are green.Harry's â€Å"strategy† is simple: He bets $2 on every spin, he always bets on red, and he always plays exactly 100 spins. If red comes up, Harry wins $2. If either black or green comes up, Harry loses $2. Suppose Harry has decided to play his usual strategy tonight. A) What are the mean and standard deviation of the normal approximation of the binomial in this instance? B) What is the probability that, after 100 plays, Harry will be ahead (i. e. , have more money than he started with)? Answer: A) Mean = 47. 37 and standard deviation = 4. 993 B) 0. 2643 Type: Computation Difficulty: Medium

The Logistics of Product Recovery (EndofLife) Case Study

The Logistics of Product Recovery (EndofLife) - Case Study Example It is now being realized that producer responsibility needs to be increased in areas of Northern America and to increasingly shift the burden of environmental protection for the government to the producers. This also enables the government to shift the responsibility of economic responsibility from the government to the local taxpayers. The scope for such laws is also being expanded to other non-recyclable wastes such as fluorescent bulbs, paint, mattresses, appliances, mercury thermostats and medical sharps. The use of EPR shall essentially require the formation of a separate and somewhat parallel system of waste management or collection mechanism that is the key to increase the quantity of waste collected. The maximization achieved within the collection system is also responsible for increased industrial as well as consumer participation in management of waste products. The laws help in mandating such convenience in collection methods which is difficult to define (Michelini & Razzoli, 2010). This paper is aimed at analyzing the scope of reverse logistic management and developing of a proper model that would be helpful in EPR management deriving most benefits from reuse and recycle of end-of-life products. The paper suggests the establishment of the OEM takeback methodology for the benefit of companies and the environment because it is the most efficient management technique for wastes. However, it also suggests the use of pooled takeback within the collection mechanism to facilitate convenience and also eliminate the drawbacks of the OEM method by way of using the benefits of pooled takeback in the collection procedure. The components, product, equipment, materials and even the total technical system can go backwards in the supply chain for rework in the manufacturing process so as to facilitate reuse and enhancement of unsatisfactory products on quality and component

A criminial case with relevant, reliable, and competent evidence Essay

A criminial case with relevant, reliable, and competent evidence - Essay Example This was emphasized in the Supreme Court’s ruling in Holbrook v Hymn. The Court held that in accordance with the provisions of the Sixth and Fourteenth Amendments to the Constitution, the guilt of the accused was to be determined exclusively on the basis of the evidence presented during trial. Furthermore, guilt could not be established on the basis of official suspicion, indictment, continued custody or other circumstances (Gardner & Anderson , 2009, p. 31). Evidence that is relevant, reliable and not otherwise inadmissible at trial is deemed to be competent evidence. The defendant in Holmes v South Carolina, endeavored to introduce evidence that the murder had been committed by a third party. In this effort the defendant offered witnesses who exhibited willingness to testify that the third party had committed the crime. This testimony was excluded by the trial court on the grounds that the case against the defendant was strong and that the evidence against the third party merely generated a bare suspicion. The defendant was convicted of murder, and this sentence was upheld by the Supreme Court of South Carolina (Gardner & Anderson , 2009, p. 32). In this case, the Supreme Court of South Carolina affirmed that any evidence of third party guilt had to generate a reasonable implication of innocence. In addition, such evidence had to be restricted to the facts that were not in conformity with the guilt of the defendant. The Court further clarified that the forensic evidence was almost conclusive in indicating the guilt of the defendant, and that the evidence against the third party was effective only to the extent of creating a bare suspicion. In such cases, the evidence against the third party was inadequate for producing a reasonable implication of innocence regarding the defendant (Holmes v. South Carolina). Thereafter, this case was put up for direct review in the US Supreme Court. In its unanimous decision this Court

A study of relationship between service advertising strategies and Dissertation

A study of relationship between service advertising strategies and consumers responses within hospitality industry - Dissertation Example Results The following mean ranges have been used for the substantive interpretations of the means: 1.00-1.49 – strongly disagree; 1.50-2.49 – slightly disagree; 2.50 – 3.49 – neutral; 3.50-4.49 – slightly agree; and 4.50 – 5.00 – strongly agree. Table 1a. Descriptive statistics: The advertisement caught my attention. Mean Std. Deviation Advert 1 3.94 0.80 Advert 2 3.76 0.86 Advert 3 4.17 0.66 Total 3.96 0.79 On the capacity of the advertisement to catch their attention, the means for all three advertisements all suggest agreement. Table 1b. One-way ANOVA: The advertisement caught my attention. Sum of Squares df Mean Square F Sig. Between Groups 12.40 2.00 6.20 10.30 0.00 Within Groups 261.77 435.00 0.60 Â   Â   Total 274.18 437.00 Â   Â   Â   The one-way ANOVA suggests that there is a significant difference among the three advertisements’ ratings on attention getting capacity (F=10.30, p=.00). The post hoc tests in Table A1 (Appendix A) indicates that Ad 3 received significantly higher means than the other two advertisements. Ad 1 is likewise more effective than Ad 2 on this aspect. Table 2a. Descriptive statistics: The advertisement elicited my interest in the hotel being advertised. Mean Std. Deviation Advert 1 3.84 0.63 Advert 2 3.79 0.72 Advert 3 3.99 0.58 On the capability of the advertisement to elicit interest in the hotel, all means suggest agreement by the respondents. Table 2b. One-way ANOVA: The advertisement elicited my interest in the hotel being advertised Sum of Squares df Mean Square F Sig. Between Groups 3.31 2.00 1.65 3.98 0.02 Within Groups 180.79 435.00 0.42 Â   Â   Total 184.09 437.00 Â   Â   Â   On the capability of the advertisement to elicit interest in the hotel being advertised, the F-value indicates that there is indeed a significant difference among the ratings given to the 3 advertisements (F=3.98, p=.02). Table A2 (Appendix A) indicates that Ad 3 garnered significant ly higher ratings than the other two advertisements on this facet. Table 3a. Descriptive statistics: I felt the conviction that what is shown in this advertisement must be true, and developed a positive disposition on my part. Mean Std. Deviation Advert 1 3.33 0.82 Advert 2 4.10 0.78 Advert 3 4.26 0.62 On the conviction that what is shown in this advertisement must be true and developing a positive disposition on the respondents, Ad 2 and 3 garnered agreement. However, Ad 1 received a neutral rating. Table 3b. One-way ANOVA: I felt the conviction that what is shown in this advertisement must be true, and developed a positive disposition on my part. Sum of Squares df Mean Square F Sig. Between Groups 72.18 2.00 36.09 64.61 0.00 Within Groups 242.99 435.00 0.56 Â   Â   Total 315.17 437.00 Â   Â   Â   The one-way ANOVA for statement 3 suggests that there are significant differences yielded for the ratings given to the 3 groups (F=64.61, p=.00). The results in Table A3 (Appendix A) indicates that Ad 3 got higher ratings than the other two ads on this facet, whereas Ad 2 received a higher rating compared to Ad 1. Table 4a. Descriptive statistics: I would like to know more information about this hotel by looking it up in the telephone directory and calling this hote

Planning Science Lessons Essay Example | Topics and Well Written Essays - 1500 words

Planning Science Lessons - Essay Example Students must also do the following: ".demonstrate an awareness of how scientific evidence is collected and are aware that scientific knowledge and theories can be changed by new evidence" "describe how and why decisions about uses of science are made in some familiar contexts" "demonstrate good understanding of the benefits and risks of scientific advances and identify ethical issues related to these." Students should be able to address these issues better if they are able to argue and support their points. While learning how to do this, students must be aware that their arguments must have a clear goal. Thus, the teachers need to make sure the students are taught how to approach this when doing argument lessons and combining them with science lessons. Using argument to teach science should also help to effectively stimulate the students since most students enjoy debating, and this enjoyment usually makes the learning much easier. Learning how to argue scientific concepts is an impo rtant concept for students to learn because science is based on facts; therefore, students must use facts to back up their arguments and prove their points. ... In order to stimulate thought-provoking questions that have to do with science, teachers need to make sure that they use open-ended questions or statements while allowing the students to interact in groups. Therefore, Shakespeare's book demonstrates fantastic concepts and lesson ideas that will help get students interested in learning about science, and to help them adequately learn science so that they can meet and pass expectations in the subject. The lesson plans and ideas offered help to stimulate student thought and critical thinking. In the researcher's opinion, this is a very effective book, and can greatly assist teachers with getting their students interested. Science has always rated as one of the more difficulty subjects for teachers to teach, simply because of lack of resources, and many times, because of lack of student interest. However, by using Shakespeare's approach of asking questions and getting the students ready to participate in a discussion by argument, it should be easier for the teacher to both capture and retain the attention of the students. This book is well formed and would help the UK science student gain a better understanding of the concepts of science t hrough critical thinking and argument. These lessons could truly help many UK secondary students prepare for the debates and other arguments that may come to light when they are in college. Thus, this practice will not only help to develop their understanding of science, but it will also help to develop their critical thinking skills. Rosalind Driver Rosalind Driver's book Making Sense of Secondary Science was inspired by the fact that she understood many students had a lack of understanding regarding the

Network Societies and the Implications for their Privacies Essay

Network Societies and the Implications for their Privacies - Essay Example The popularity of the SNSs is quite evident when we find in a 2009 report, which stated that globally almost 38% internet users are a member of one or many of the SNSs, and maintain regular profiles in the social networking sites (Wray, Social Networking Booming with Doubling of Online Profiles, 2009). Facebook, at present is the most popular SNS, with a rise of nearly 86.1% in user percentage (ibid). One major characteristic of these SNSs is that the users can upload their personal data on these sites on a daily basis. As per the recent study made by OfCom in 2010, â€Å"Social networking accounts for nearly a quarter of all time spent on the internet (23 per cent compared to 9 per cent in 2007).   This has been driven by the rapid growth of Facebook, which grew by 31 per cent. The average Facebook user spent 6 and 30 minutes on the site during May 2010,† (OfCom, Consumers spend almost half of their waking hours using media and communications, 2010. The 2008 OfCom report no ted that an adult user, on an average, maintained his/her profile on around 1.6 SNSs, while enter their profiles at least once, every two days (OfCom, Social networking: a quantitative and qualitative research report into attitudes, behaviours & use, 2008). This expeditious rise in the usage of social networking sites in the past decade, has created new problems, where there are increased instances of user personal data being misused through identity theft and cyber stalking, for various commercial activities related to unauthorised searching for employees, or fishing for prospective clients (Brown, Edwards, and Marsden, Staking 2.0: privacy protection in a leading social networking site, nd). The internet and SNSs being open to all, the uploaded user information (even personal information) becomes accessible to a much wider user spectrum, besides the intended user group. Often user inexperience and a general unawareness coupled with inappropriate SNS website designing, unintentiona lly help in the misuse of private information by various commercial organisations. These misuses and the future potentiality of fraudulent activities using the obtained personal information have raised questions and concerns over the issue of creating a stronger security system that would assure SNS user privacy, and the inaccessibility of the uploaded information outside the targeted viewer group. As for example, a member of the medical SNS PatientsLikeMe, may opt to discuss his/her condition only with a specific group of people (like those sharing similar medical problems), thus making it imperative that the site gives the user his/her right to privacy. In this context, we will discuss network societies and the implications for their privacies, focussing on Facebook, as it is the leading SNS now. Discussion What is a SNS? Boyd and Ellison, defined SNSs as services provided that are internet-based and allow its users to: Create user profiles which can be kept partially public or co mpletely public, within the provide domain of the site; Create a ‘friends’ list and a group where they can upload and share private information; Have an access to the friends’ profiles, and to these friends’ ‘friend list,’ where the user can view all the connections made by their friends and often by ‘other users’ (who are not direct friends, but may have common friends or common interests) within the domain of the same SNS (Boyd, and Ellison, 2007, 210-211). The ‘