Mktg 301

7) Data from a small bookstore are shown in the accompanying table. The manager wants to predict Sales from Number of Sales People Working. Number of sales people working| Sales (in $1000)| 4| 12| 5| 13| 8| 15| 10| 16| 12| 20| 12| 22| 14| 22| 16| 25| 18| 25| 20| 28| x=11. 9| y=19. 8| SD(x)=5. 30| SD(y)=5. 53| a) Find the slope estimate, b1. Use technology or the formula below to find the slope. b1=rsysx Enter x,y Data in TI-84 under STAT > STAT > CALC > 8: LinReg(a+bx) b1=1. 023 b) What does b1 mean, in this context?The slope tells how the response variable hanges for a one unit step in the predictor Thus, an additional; $1,023 of sales associated with each additional sales person working. c) Find the intercept, b0. b0=y-b1x =19. 8-1. 023(11. 9) For this problem, use technology, rounding to three decimal places. b0=7. 622 d) What does b0 mean in this contet? Is it meanful? The intercept serves as a starting value for the predicitons. It shuld only be interpreted if a 0 value for the predictor variable makes sense for the context of the situation. On average, $7,622 is expected when 0 sales people are working.It is not meaningful because it does not make sense in this context. e) Write the equation that predicts Sales from Number of Sales People Working. Recall that the slope of the equation b1=1. 023 and the intercept is b0=7. 622 Complete the equation. Sales=7. 622+1. 023 *(Number of Sales People Working) f) If 19 people are working, what sales do you predict? Substitute 19 for the number of sales people working in the equation found in the previous step and solve for Sales. Sales=7. 622+1. 023 *(Number of Sales People Working) =7. 622+1. 023*19Substitute. =27. 059Simplify. *Note that each unit of Sales represents $1000. Thus, the predicted sales for 19 people working is 27,059 dollars. g) If sales are actually $26,000, what is the value of the residual? Subtract the predicated value found in the previous step from the actual value. 26,000-27,059=-1059 Thus, t he value of the residual is -1059 dollars. h) Have the sale been overestimated or underestimated The predicted sales are $27,059 and the actual sales are $26,000. Since $27,059 > $26,000, the sales were overestimated. 13) Of the 46 individuals who responded, 25 are concerned, and 21 are not concerned. of those concerned about security are male and 5 of those not concerned are male. If a respondent is selected at random, find each of the fallowing conditional probabilities. | Male| Female| Total| Concerned| 9| 16| 25| Not Concerned| 5| 16| 21| Total| 14| 32| 46| a) The respondent is male, given that the respondent is not concerned about security. P(Male|Not Concerned) = 521 = 0. 238 b) The respondent is not concerned about security, given that is female P(Not Concered|Female) = 1632 = 0. 500 c) The respondent is female, given that the respondent is concerned about security. P(Female|Concerned) = 1625 = 0. 40 14) It was found that 76% of the population were infected with a virus, 21% were without clean water, and 18% were infected and without clean water | Clean Water| | | Yes| No| Total| Infected| 0. 58| 0. 18| 0. 76| Not Infected| 0. 21| 0. 03| 0. 24| Total| 0. 79| 0. 21| 1. 00| a) What’s the probability that a surveyed person had clean water and was not infected? .21 had clean water and was not infected 15) A survey concluded that 54. 4% of the households in a particular country have both a landline and a cell phone, 32. 6% have only cell phone services but no landline, and 4. 6% have no telephone services at all. ) What proportion of households have a landline? Begin by making a contingency table. | Cell Phone| | | Yes| No| Total| Landline| 0. 545| 0. 083| 0. 628| No Landline| 0. 326| 0. 046| 0. 372| Total| 0. 871| 0. 129| 1. 00| The completed contingency tables shows that P(landline) = 0. 628. b) Are having a cell phone and having a landline independent? Explain. Events A and B are independent when P(B|A) = P(B). To determine wheter having a cell pho ne and having a landline are indepented, find P(landline|cell phone) and P(landline). Recall from part a) that P(landline) =0. 628 PBA=P(A and B)P(A)Use the formula to find P(landline|cell phone) Plandlinecell phone=P(landline and cell phone)P(cell phone) Since the contingency table shows that P(landline and cell phone)=0. 545 and P(cell phone)=0. 871, substitute these values into the equation. Divide to find the conditional probability, rounding to three decimal places. Plandlinecell phone=0. 5450. 871=0. 626 Thus, P(landline|cell phone)=0. 626 and P(landline)=0. 628. Because 0. 626 is very close to 0. 628, having a cell phone and having a landline are probably independent. Of the households surveyed, 62. 6% with cell phones had landlines, and 62. 8% of all households did. 6) A marketing agency has developed three vacation packages to promote a timeshare plan at a new resort. They estimate that 30% of potential customers will choose the Day Plan, which does not include overnight ac commodations; 30% will choose the Overnight Plan, which includes one night at the resort; and 40% will choose the Weekend Plan, which includes two nights. a) Find the expected value of the number of nights that potential customers will need Vacation Package| Nights Included| Probability P(X=x)| | Day Plan| 0| 30100=0. 3| | Overnight Plan| 1| 30100=0. 3| | Weekend Plan| 2| 40100=0. 4| | This, P(X=0)=0. 3, P(X=1) =0. 3, and P(X=2)=0. Use the formula E(X) = ? x †¢ P(x) to detrime the expected value. E(X) = ? x †¢ P(x) = 0(0. 3) +1(0. 3) +2(0. 4) = 1. 1 There, the expected value of the number of night potential customers will need is 1. 1 b) Find the standard deviation of the number of nights potential customers will need. The standard deviation is the square root of the variance. First, Find the Variance: To do so, find the deviation of each value of X from the mean and square each deviation. The variance is the expected value of these squared deviations and is found using th e formula below. = Var(X) = ? (x –  µ)? P(x) Find the deviation for each value of X.Remember that E(x)=1. 1 Vacation Package| Nights Included| Probability P(X=x)| Deviation (x – E(X))| Day Plan| 0| 30100=0. 3| 0 – 1. 1 = -1. 1| Overnight Plan| 1| 30100=0. 3| 1 – 1. 1 = -0. 1| Weekend Plan| 2| 40100=0. 4| 2 – 1. 1 = 0. 9| Now find the variance using the formula =Var(X)=? (x –  µ)? P(x) Var(X) = ? (x –  µ)? P(x) = (-1. 1)? (0. 3) + (-0. 1)? (0. 3) + (0. 9)? (0. 4) = 0. 69 Finally, the standard deviation also known as ? is the square root of the variance. ? = Var(x) = 0. 69 = 0. 83 Therefore, the standard deviation of the number of nights potential customers will need is approximately 0. 83 nights. 7) A grocery supplier believes that in a dozen eggs, the mean number of broken eggs is 0. 2 with a standard deviation of 0. 1 eggs. You buy 3 dozen eggs without checking them. a) How many broken eggs do you get? The expected value of t he sum of random variables is the sum of the expected values of each idividula random variable. Find the sum of the expected values where X is the total number of broken eggs in the three dozen, and X, X, X Represent the three individual dozen eggs. E(X) = E(X1) + EX2+ EX3 = 0. 2 + 0. 2 + 0. 2 = 0. 6 Therefore, the expected value of X is 0. 6 eggs. b) What’s the standard deviation?The variance of the sum of independent variables is the sum of their individual variances. Find the variance for each carton, add the variances, and then take the square root of the sum to find the standard deviation. The variance of each individual dozen is the square of each dozens standard deviation. Var(X1) = Var(X2) = Var(X3) = 0. 12= 0. 01 Find the sum of the variances to find the variance of the sum. Var(X) = VarX1+ VarX2+ VarX3 = 0. 01 + 0. 01 + 0. 01 = 0. 03 Recall that the standard deviation is the square root of the variance. Find the standard deviation. SD(X) = Var(x) = 0. 03 = 0. 17Ther efore, the standard deviation is 0. 17 eggs c) What assumptions did you have to make about the eggs in order to answer this question? The variance for the sum of random variables is only the sun of variances of each random variable in certain cases. Review the assumption that must be made to allow the variance to be the sum of the individual variances. 18) An insurance company estimates that it should make an annual profit of $260 on each homeowner’s policy written, with a standard deviation of $6000. a) Why is the standard deviation so large? Home insurance is used to protect the owner financially in the event of a problem.If a catastrophe occurs, then the insurance company will cover the cost of the damage. If a catastrophe never occurs, then the insurance company pays nothing. Meanwhile, the owner pays the insurance company at regular intervals whether or not a catastrophe occurs. The expected value is the mean annual profit on all of the policies and the standard deviatio n is a measure of how much annual profits can differ from the mean. Use this information with the fact that claims are rare, but very costly, occurrences. b) If the company writes only four of these policies, what are the mean and standard deviation of the annual profit?LetX1,X2, X3,†¦,Xn represent the annual profit on the n policies and let X be the random variable for the total annual profit on n polices written. X=X1+X2+ X3+†¦+Xn The expected value of the sum is the sum of the expected values. Find the expected value of the annual profit on each policy. EX1=EX2=EX3=EX4=$260 Now find the sum of the expected values. EX=EX1+EX2+EX3+EX4 =260+260+260+260 = $1040 Therefore, the mean annual profit is $1040 To find the standard deviation of the annual profit, use the fact that te variances of the sum of independent variables is the sum of their individual variances. First find the variance for each policy.The variance for the policy is the square of the standard deviation. VarX 1=VarX2=VarX3=VarX4=60002=36,000,000 VarX=VarX1+VarX2+VarX3+VarX4 = 4(36,000,000) = 144,000,000 Evaluate the square root of the variance to find the standard deviation. SDX=VarX =144,000,000 =$12,000 Therefore, the standard deviation is $12,000 c) If the company writes 10,000 of these policies, what are the mean and standard deviation of annual profit? The expected value of the sum is the sum of the expected values. The expected value of each policy was found earlier. EX1=EX2=EX3=†¦ =EX10,000=$260 Now find the sum of expected values. EX=EX1+EX2+EX3+†¦ +EX10,000 10,000(260) =$2,600,000 Therefore, the mean annual profit is $2,600,000 To find the standard deviation of the annual profit, use the fact that the variance of the sum of independent variables is the sum of their individual variances. First find the variance for each policy. The variance for the policy is the square of the standard deviation and was found earlier. VarX1=VarX2=VarX3= †¦ =VarX10,000=36,000,000 Now sum the variances to find the variances of the sum. VarX=VarX1+VarX2+VarX3+†¦ +VarX10,000 =10,000(36,000,000) =360,000,000,000 Evaluate the square root of the variance to find the standard deviation. SDX=Var(X) =360,000,000,000 $600,000 Therefore, the standard deviation is $600,000. d) Do you think the company is likely to be profitable? Recall that the mean annual profit for 10,000 policies is $2,600,000. While this number seems quite large, it is necessary to determine how likely a profit is to ensure that this company will be profitable. Find the distance in standard deviation of $0 from the mean to determine how rare an occurrence of no profit would be. z=x- =0-2,600,000600,000 =-4. 3 Thus, $0 is 4. 3 standard deviation below the mean. **Note that approximately 95% of the annual profits should lie within two standard deviations of the mean.Evaluate whether the distance of $0 from the mean is convincing enough to determine whether or not the company will be profitable. e) W hat assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? The variance of the sum of random variables is only the sum of the variances of each random variables in certain cases. Review the assumption that must be made to allow the variance to be the sum of the individual variances. Then chose the situation that would create an association among policy losses. 19) A farmer has 130 lbs. of apples and 60 lbs. f potatoes for sale. The market price for apples (per pound) each day is a random variable with a mean of 0. 8 dollars and a standard deviation of 0. 4 dollars. Similarly, for a pound of potatoes, the mean price is 0. 4 dollars and the standard deviation is 0. 2 dollars. It also costs him 5 dollars to bring all the apples and potato’s to the market. The market is busy with shoppers, so assume that he’ll be able to sell all of each type of produce at the day’s price. a) Define your random variables, and use them to express the farmer’s net income. A random variable’s outcome is bases on a random event.Therefore let the random variables represent the factors that will be randomly determined each day. The random variables should represent the market prices of the two items. A = price per pound of apples P = price per pound of potatoes The profit is equal to the total income minus the total cost. The income is found by multiplying the market price for apples by the total number of pounds sold and adding it to the product of the market price for potatoes and the number of pounds of potatoes sold. The total cost is the transparent cost. Profit = 130A + 60P – 5 b) Find the mean. The mean of the net income is the expected value of the profit.Profit = 130A + 60P – 5 E(Proft) = E(130A + 60P – 5) Use the property E(X + Y) = E(X) + E(Y) to express the expected value of the profit as the sum of two separate expected values E=(Profit) = E(130A +60P -5) = E [(130A) + (60P – 5)] = E(130A) + E(60P – 5) Now use the property EX ±c= E(X) ±c E(Profit) = E(130A) + E(60P – 5) = E(130A) + E(60P) – 5 Finally, use the property E(aX) = aE(X) to remove the coefficient from the expected values. E(Profit) = E(130A) + E(60P) – 5 = 130E(A) + 60E(P) – 5 Substitute the known expected values of the prices of apples and potatoes in the equation. E(Profit) = E(130A) + E(60P) – 5 E(0. 8) + E(0. 4) – 5 Evaluate the expected profit. E(Profit) = 130(0. 8) + 60(0. 4) – 5 = 123 Therefore, the mean is 123 dollars. c) Find the standard deviation of the net income. To find the standard deviation, first find the variance and then take the square root, since the properties useful in this case are in terms of variance and not standard deviation SD(Profit) = Var(Profit) = Var(130A+60P-5) First use the property Var(X + Y) = Var(X) + Var(Y) to express the variance of the profit as the sum of two separate v ariance Var(Profit) = Var(130A + 60P – 5) = Var[(130A) + (60P – 5)] =Var(130A) + Var(60P – 5)Now use the property Var(X  ± c) = Var(X) to simplify the second variance Vr(Profit) = Var(130A) + Var(60P – 5) = Var(130A) + Var(60P) Finally, use the property VaraX=a2VarX to restate each variance. Var(Profit) = Var(130A) + Var(60P) = 1302VarA+ 602VarP = 16,900Var(A) + 3600Var(P) Evaluate the variance of the profit. Var(Profit) = 16,900(0. 16) + 3600(0. 04) = 2848 Lastly, find the standard deviation, rounding to two decimal place. SD(Profit) = VarProfit = 2848 = 53. 37 Therefore, the standard deviation of the net income is 53. 37 dollars. d) Do you need to make any assumptions in calculating the mean?Recall that the mean of the sum of two or more random variables is the sum of the means. Determine what, if any, assumptions are made to use this property. Do you need to make any assumptions in calculating the standard deviation? Recall that the variance of the s um of two random variables is only the sum of their individual variances in certain cases, Determine what, if any, assumptions are made to use this property. 20) A salesman normally makes a sale (closes) on 65% of his presentations. Assuming the presentations are independent, find the probability of the following. ) He fails to close for the first time on his sixth attempt. Use the formula below to determine the probability, where p is the probability success, q=1 – p and X is the number of trails until the first success occurs. P(X=x) = qx-1p Find the values for p and q. **Note that in this case that a success is defined as failed to close p = 0. 35 q = 0. 65 Substitute and solve to find P(X=6). Rounding to four decimal places P(X=6) = qx-1p = 0. 656-1(0. 35) = 0. 0406 Therefore, the probability he fails to close for the first tie on his sixth attempt is 0. 0406 b) He closes his first presentation on his fifth attempt.Find the values for p and q. **Note that in this case tha t a success is defined as making a sale p = 0. 65 q = 0. 35 Substitute and solve to find P(X = 5), rounding to four decimal places P(X=5) = qx-1p = 0. 355-1(0. 65) = 0. 0098 Therefore, the probability he closes his first presentation on his fifth attempt is 0. 0098 c) The first presentation he closes will be on his second attempt. Find the values for p and q. Note that in this case that a success is defined as making a sale. p = 0. 65 q = 0. 35 Substitute and solve to find P(X=2) P(X=2) = qx-1p = 0. 352-1(0. 65) = 0. 2275Therefore, the probability the first presentation he closes will be on his second attempt is 0. 2275 d) The first presentation he closes will be on one of his first three attempts. Use the fact that the compliment of an even is equal to 1 – P(X=x) to find the probability. The compliment event is that he will not close a sale on any of his first three attempts. Find the probability that he does not close on his first three attempts, rounding to four decimal pl aces. 0. 353=0. 0429 Subtract from 1 to find the probability the first presentation he closes will be on one of his first three attempts 1 – 0. 429 = 0. 9571 Therefore, the probability the first presentation he closes will be on one of his first three attempts is 0. 9571 21) College students are a major target for advertisements for credit cards. At a university, 73% of students surveyed said that they had opened a new credit card account within the past year. If that percentage is accurate, how many students would you expect to survey before finding one who had not opened a new account in the past year? First check to see that the cells are Bernoulli trials. Trials are Bernoulli if the following three conditions are satisfied. 1.There are only two possible outcomes (called success and failure) for each trial. 2. The probability of success, denoted p, is the same on every trial. (The probability of failure, 1 – p is often denoted q. ) 3. The trials are independent Ther e are only two possible outcomes for each trial because a student either opened a credit card account in the past year or they did not. The probability of success is the same on every trial, based on the percent given in the problem statement. The trails are independent because each student’s response is not dependent on any other student’s response.Thus, the trials of surveying the students are Bernoulli trials. A geometric probability model models how long it will take to achieve the first success in a series of Bernoulli trials. Let X be the number of students that will have to be surveyed before finding the first student who did not open a credit card in the past year. The two outcomes are a student who did not open a credit card account in the past year *success) and a student who opened a credit card account in the past year (failure). The probability of a failure is given in the problem statement as q = 73% = 0. 73.Find the probability of success by subtracting this from 1. P = 1 – 0. 73 = 0. 27 Find the expected value of X. In a geometric model, the expected value is EX= 1p , where p is the probability of success. Round up to the nearest integer. EX=10. 27=4 Therefore, on average, you would expect to survey 4 students before finding one who had not opened a new account in the past year. 22) A certain tennis player makes a successful first serve 82% of the time/ Assume that each serve is independent of the others. If she serves 7 times, what’s the probability she gets a) all 7 serves in? b) exactly 5 serves in? ) at least 5 serves in? d) no more than 5 serves in? The first step is to check to see that these are Bernoulli trails. The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0. 82. Finally, it is assumed that each serve is independent of the others. Next define the random variable. Each questi on deals with the number of serves, so let X be the number of successful serves in n = 7 first serves. Now determine which probability model is appropriate for these problems.Recall that geometric probability models deal with how long it will take to achieve a success. A binomial probability model describes the number of successes in a specific number of trails. All the question deal with the number of successful serves so the binomial probability model Binom(7,0. 82 is appropriate here. a) all 7 serves in? The probability that she ges all 7 serves in is P(X=7). To use the binomial probability model Binom(n,p), use the fallowing formula, where n is the number of trials, p is the probability of success, q is the probability of failure (q = 1 – p), and X is the number of successes in n trials.PX=x= nxpxqn – x, where nx= n! x! n-x! First substitute the correct values into the formula PX=7= 770. 8270. 187- 7 Now simplify. P(X = 7) ? 0. 249 Therefore, the probability that s he gets all 7 serves in is approximately 0. 249 binomPDF(7, . 82, 7) = b) exactly 5 serves in? The probability she gets exactly 5 serves in is P(X = 5). As in part a, use the formula PX=x= nxpxqn – x to find this probability PX=5= 750. 8250. 187 – 5 ?0. 252 Therefore, the probability she gets exactly 5 serves in is approximately 0. 252 binomPDF(7, . 82, 5) = c) at least 5 serves in?To find P(at least 5 serves in), first determine and an expression that is equal to this probability. Note that the wording â€Å"at least 5†, means 5 or more, meaning that there can 5, 6, or 7 serves in. Thus, the probability equals P(X=5) + P(X=6) + P(X=7). So to find the probability that she got at least 5 serves in, evaluate. P(X=5) + P(X=6) + P(X=7) = 75(0. 82)50. 187-5+76(0. 82)6(0. 18)7-6+77(0. 82)7(0. 18)7-7 ?0. 885 Therefore, the probability she gets at least 5 serves in is approximately 0. 885 binomPDF(7, . 82, 5) + binomPDF(7, . 82, 6) + binomPDF(7, . 82, 7) = d) no more th an 5 serves in?To find P(no more than 5), first determine an expression that is equal to this probability. Note that the wondering â€Å"no more than 5† means 5 or less, meaning that there can be 0 thru 5 successful serves. Thus, the probability equals P(X? 5). So to find the probability that there are no more than 5 serves in, evaluate P(X? 5), which is equal to P(X=0) + P(X=1) + †¦ + P(X=5), using the formula PX=x= nxpxqn – x P(X=0) + P(X=1) + †¦ + P(X=5) = 70(0. 82)00. 187-0+71(0. 82)1(0. 18)7-1 + †¦ + 75(0. 82)5(0. 18)7-5 ? 0. 368 Therefore, the probability that there are no more than 5 serves in is approximately 0. 368 binomCDF(7, . 82, 5)

Comparison and Analysis of the use of Camerawork in Friends and The Office

The Office has a â€Å"docusoap† format, this is supported by ingenious camerawork, shots are often shaky, and frames are crowded. The raw documentary style is engrossing, the camera often moves with uncertainty as if it is not known what will happen next. The camera often pans quickly and unsteadily in order to obtain reaction shots. There is little consideration to position of elements within the frame, this all adds to the raw docusoap style. Zooming is often done shakily in order to catch reactions. The Office is made up of many two shots and the camerawork is unsteady, also characters often slightly slip out of frame momentarily and the camera must quickly adjust when this character becomes active. Friends is found in stark contrast to the office, it optimises exceptional quality camerawork. Many medium close shots are to be found, with all characters held neatly within the frame. The group shots are framed perfectly with all elements positioned appropriately. Everything is more calm and neat hich demonstrates the high quality of the programme. Editing Firstly the title sequence of The Office has several cross-fades to set the location of Slough. The edit technique used most frequently is the cut, often to catch reactions it also maintains a pace to the episode. Commonly an establishing pan-shot around the office is used before turning focus to a particular interaction between characters. Friends however optimises the cross-fade for transition between establishing hots of city buildings and actual shots of characters. All edit points are dealt with smoothly, little snippets of music often signify a change of scene and add to the general flow of the episode. Application of the Above When Gareth is in the â€Å"Invetigation Office† talking to Glen, the camera shot is really unsteady and is through a set of blinds, this gives the feel that you shouldn't be watching this happening, it zooms in beyond the blinds to see reactions but always zooms back out to remind you of the blinds being a barrier.

Free Essays on Ligeia

A short story written by Edgar Allan Poe which fuses the themes of transcendence and lost love is "Ligeia,†. The narrator of this story meets and marries a woman of exquisite beautya woman named Ligeia. To the narrator, she is the perfect woman. She possesses classical beauty, expanded intellect, and spiritual purity. The narrator describes at length the strange attributes of this woman Ligeiaher raven-black hair; her low, musical voice; her ivory skin, lofty forehead; her delicate nose and radiant smile. However, Ligeia's most striking feature was the presence of her dark, large eyes, which the narrator is obsessed with. He seems to exert quite a bit of emotion when describing her eyes. By reading this story, you would realize that her eyes were the things that fascinated the narrator the most about Ligeia. The narrator describes his beautiful spouse almost like a ghost: "She came and departed as a shadow." He also thinks her beauty, more specifically her eyes, as a "! strange mystery." Her eyes make her seem unreal or superhuman because of her large "expressive" eyes that the narrator cannot explain except that they are "far larger than the ordinary eyes of our own race." Ligeia's unusual beauty represents a reoccurring theme throughout the story. The text portrays a rejection of the features that society would call "ordinary" beauty. One example of this is how Poe repeatedly points out how flaws in the classical appearance of Rowena, "the fair-haired, the blue-eyed," by comparing her to Ligeia whose "features were not of that regular mould which we have been falsely taught to worship in the classical labors of the heathen." Poe explains through the narrator how more exalted and meaningful Ligeia's beauty is specifically because she exhibits more natural features instead of the classical features. The rejection of the classical beauty and the welcoming of the unusual, mysterious beauty indicates Poe's bias towards Romanti... Free Essays on Ligeia Free Essays on Ligeia A short story written by Edgar Allan Poe which fuses the themes of transcendence and lost love is "Ligeia,†. The narrator of this story meets and marries a woman of exquisite beautya woman named Ligeia. To the narrator, she is the perfect woman. She possesses classical beauty, expanded intellect, and spiritual purity. The narrator describes at length the strange attributes of this woman Ligeiaher raven-black hair; her low, musical voice; her ivory skin, lofty forehead; her delicate nose and radiant smile. However, Ligeia's most striking feature was the presence of her dark, large eyes, which the narrator is obsessed with. He seems to exert quite a bit of emotion when describing her eyes. By reading this story, you would realize that her eyes were the things that fascinated the narrator the most about Ligeia. The narrator describes his beautiful spouse almost like a ghost: "She came and departed as a shadow." He also thinks her beauty, more specifically her eyes, as a "! strange mystery." Her eyes make her seem unreal or superhuman because of her large "expressive" eyes that the narrator cannot explain except that they are "far larger than the ordinary eyes of our own race." Ligeia's unusual beauty represents a reoccurring theme throughout the story. The text portrays a rejection of the features that society would call "ordinary" beauty. One example of this is how Poe repeatedly points out how flaws in the classical appearance of Rowena, "the fair-haired, the blue-eyed," by comparing her to Ligeia whose "features were not of that regular mould which we have been falsely taught to worship in the classical labors of the heathen." Poe explains through the narrator how more exalted and meaningful Ligeia's beauty is specifically because she exhibits more natural features instead of the classical features. The rejection of the classical beauty and the welcoming of the unusual, mysterious beauty indicates Poe's bias towards Romanti... Free Essays on Ligeia Two Portraits of women in Poe's tale Ligeia. Not only does Ligeia’s unusual beauty represents the main theme throughout the story, but the text reflects Poe’s rejecting the â€Å"ordinary†, a common theme in literature. The writter rejects classical values and welcomes supernatural through unusual, mysterious beauty. Ligeia was extremely beautiful, she knew a lot. A relationship with the narrator was a deep affection. He describes her very precisely as being afraid to omit any perfect feature: â€Å"I examined the contour of the lofty and pale forehead – it was faultless (†¦), the skin rivalling the purest ivory (†¦), the gentle prominence of the region above the temples†Ã¢â‚¬ ¦ The speaker portrays his perfect spouse almost like a ghost : â€Å"She came and departed as a shadow. I was never made aware of her entrance into my closed study†. He also thinks her beauty, more specifically her eyes, as a â€Å"strange mystery†. The narrator sees a secret in his beloved eyes. Her eyes make her seem unreal because they are so â€Å"expresive† and the narrator cannot explain except that they â€Å"far larger than the ordinary eyes of our own race†. He even compares Ligeia’s eyes to the stars: â€Å"Those eyes! Those large, those shining, th! ose divine orbs! They became to me twin stars of Leda†¦Ã¢â‚¬ . The narrator is impressed by her voice also: â€Å"which at once so delighted and appalled me – by the almost magical melody, modulation, distinctness and placidity of her low voice.† At first it seems that the narrator is interested in Ligeia’s body only, because the largest part of her portrait consists of her physical appearance. But later it appears that the narrator is also impressed by her knowledge, she was the first woman met by him who was so intelligent. And in some spheres she knew even much more than her husband: â€Å"In the classical tongues was she deeply proficient and as far as my own acquin...

How an Ice Cream Soda or Float Works

How an Ice Cream Soda or Float Works An ice cream soda or ice cream float (called a spider in Australia and New Zealand) is made by adding soda pop or seltzer to ice cream. Some people add flavoring, like chocolate syrup, or a little milk. However you make it, as soon as the soda hits the ice cream you get fizzy, frothy, tasty bubbles.Do you know how it works? Its basically the same as what is going on with the Mentos and Soda Fountain, except not as messy. You are knocking the carbon dioxide in the soda out of solution. Bubbles of air in the ice cream provide nucleation sites around which carbon dioxide bubbles can form and grow. Some ingredients in the ice cream lower the surface tension of the soda so the gas bubbles can expand, while other ingredients trap the bubbles in much the same way as small amounts of protein in seawater trap air to form seafoam.You can make all types of floats, including black cows (coke floats with cola and vanilla ice cream), brown cows (root beer float with root beer and vanilla ice cream ), and purple cows (grape soda and vanilla ice cream), but you can use other ingredients. Heres a recipe for a coffee cola Float, which is bubbly and caffeinated and therefore a double-win: 2-1/2 cups coffee (room temperature or chilled)2/3 cup light cream or milkcoffee, chocolate or vanilla ice creamcola Mix the coffee and cream or milk, pour it into glasses, add scoops of ice cream, and top it off with cola. You can garnish it with whipped cream, chocolate covered coffee beans, or a little coffee powder or cocoa.

Individual critical review report Essay Example | Topics and Well Written Essays - 3000 words

Individual critical review report - Essay Example Secondly, change must not be ambiguous to the employees within the organization. For that reason, the change agents should communicate the intended change and give such people time to think it over. If change is to be adopted and effectively impact on the organization, all the affected parties must be given an opportunity to decision making. Change is a continuous process that an organization seeks to undertake. It is determined by the demands within the environment. For instance, the current economic depression has made several organizations focus on reduction of cost of operation. To achieve this, the management teams in these organizations are carrying out retrenchments. On the other hand, an organization may require all the employees to have technological skills. This is especially to enable them handle the technological advancement within the organization. Recent change currently experienced In the recent past, I have been able to appreciate the need for change within an organiz ation. In the past few months, I had volunteered myself to work in a given nongovernmental organization. It deals with the services to the community including and not limited to the response and mitigation of drug use amongst the young. As one of the volunteers, my main work was to collect and file data for future reference. However, most of the work within the organization was done manually and every individual seemed comfortable about that. With the dream of expanding the organization and ensuring efficiency, the management announced that it was going to avail new technology in the organization. All the employees were therefore expected to be ready for the new organization culture. One amazing thing is that the news was received with mixed reactions by most of the employees. There are those who felt as if the new technology was going to take away their jobs. Others like me focused on the positive aspect of the new development. I knew that with the new technological advancement, th e quality of work will improve and documents will be much safer. The main aim of the proposed change apart from efficiency was to cut down the cost of operation. The organization was experiencing an inflated cost of operation owing to the increased cost of purchasing materials. Papers and books were mostly used for keeping data. There was need for downsizing to enable the organization run effectively and sustainably. It is this thought that created tension amongst the employees. The employer had to release some of them to remain with a more manageable number of employees. However bad it sounds, it is part of organization change and therefore must be carried out. The main purpose should be to ensure that organization could sustain itself. Theories employed There are various theories of change that can be used in explaining the alterations witnessed in many organizations. However, for the purpose of the organization I was working at, two main theories of change were in use. These incl uded Lewin’s 3 state model theory of change and 8-step change model. In the first category, Lewin states that change goes through three main stages. These include the unfreezing, change and the refreezing stages (Burnes 2004, p.89). This represents exactly what was happening in the organization where I was working. The first state according to this theory is to alter the culture that people have been used at. This is the

Why are you passionate about leadership Essay Example | Topics and Well Written Essays - 250 words

Why are you passionate about leadership - Essay Example Hence, for me leadership is about the ability to interpret events and concepts in a way that allows for individual views, while making way for a sense of commonality and a shared sense of purpose. Doing so, a leader not only gives meaning to a collective vision, but also helps bring about the much needed alignment on strategies and goals. To me a leader is a person who ushers in an atmosphere of cooperation and trust by the dint of one’s personal integrity and charisma. Enterprises are always in the need of individuals who facilitate the quintessential backbone to the overall organizational culture, thereby ushering in a professional environment imbued with trust and integrity. The biggest source of satisfaction for a true leader is one’s ability to empower and enable people and groups. Leadership is about what happens when the leader is not present at the scene of action. So I hold that the biggest responsibility of a leader is to create an environment that motivates others to assume leadership and be willing to face

Analysis of the UK Anti-Money Laundering legislation in the light of Coursework

Analysis of the UK Anti-Money Laundering legislation in the light of implementation of the EU Directives - Coursework Example The UK, being home to the global financial hub of London, has necessarily had to enact a number of measures traceable to the 1993 Money Laundering Regulations. The paper is an analysis of the UK Anti-Money Laundering legislation in the light of implementation of the EU Directives Money laundering is a criminal process of accruing supposedly legitimate assets or moneys, through diverse means and procedures, which are able to ‘clean’ illegal criminal proceeds. Fundamentally, due to its diverse impacts upon the economy, it has been vital for not only localized and regional attention but also global implementation of watchdog policy initiatives. The dynamic nature of criminal enterprise, has necessarily dictated the need for constant updating of prevailing legislation in line with the new methods, processes and avenues of laundering ‘dirty’ money. As a result, the term ‘Money laundering’, has come to be categorized and connected to other diverse forms of financial crime in relation to the general misuse of existing financial systems. Accordingly, under this categorization, diverse forms of financial system misuse i.e. of traditional and digital currencies, credit cards and securities amongst others, have been interlinked to the criminal act1. Furthermore, this has led to the association of terrorism funding, as well as diverse evasion measures of international sanctions with the criminal act of money laundering. Initially, pertinent legislation dealing with this form of criminal activity was mainly focused on local and regional settings, as various nations experienced the positive impacts of industrial growth. Accordingly, the diverse legislations enacted were principled along these contexts, mainly dealing on a local and national scale. However, due to the influence of gradual globalization process, there was need for a re-think of the prevailing preventative measures. Of fundamental importance, because of the link