Bias In Printmedia :: essays research papers

Print media provides its readers with information, but what the reader very often does not recognize is the bias within the articles. Bias is not so easily recognized. Writers have the gift to blend the bias in with their work. It is so well done, that in order to see the bias, one must thoroughly analyze the article. A person must also know what the types of bias are and how they are used. There are many different types of bias that are used in health related articles such as statistics and crowd counts, word choice and tone, and through omission. Print media demonstrates these types of bias in many articles.   Ã‚  Ã‚  Ã‚  Ã‚  One method of bias being used is print media is through statistics and crowd counts. A writer can manipulate the reader into thinking that the results are very high or very low in some cases. In The Toronto Star on October 23, 1999, the article â€Å" Pregnancy biggest threat to women, V.N. says† uses statistics to give an estimation, â€Å" an estimated 585,000 women do every year†. This article explains how pregnancy affects many women. By using this statistic, it gives the reader an approximation, but not an exact number. This is used to make the reader think that the statistic is very high. Another article in The Toronto Star, â€Å"Tamil health crisis probed†, on October 29, 1999,demostrates bias by saying, â€Å" At least 70,000 people†. The article is talking about the Tamil community and how 70,000 people have been affected, but it does not give the amount of people in that community. This type of bias is often used in print med ia to make an article more important than it is.   Ã‚  Ã‚  Ã‚  Ã‚  Bias through word choice and tone is often used in print media sources. By choosing specific words, the writer can easily influence the reader’s opinion about the article. Certain words give the reader a different meaning. In an article, â€Å"Health care to receive $3.8 billion injection†, in The Toronto Star, on October 22, 1999, the Governor, Hilary Weston, is reading a passage from a â€Å"throne speech† given by the government to introduce a Patient’s Bill of Rights. By using the word, throne, the writer suggests that the speech given is very important. During this speech, there is a â€Å"caucus chuckle† from a Liberal at the meeting. The writer uses caucus to indicate that it is more than a normal chuckle during an important speech.

Martial Arts :: essays research papers

Martial Arts To follow is my report on martial arts in Asia. This a very interesting subject, and a very good report. It will describe martial arts and some types of martial arts. It will also say where they originated from. The term "martial arts" is a general term used to describe general types of fighting. Most martial arts practised today came from China, Japan, and Korea. There are hundreds of types of martial arts, each divided into specific styles or systems. Technically, martial arts fall into two categories: percussive, and non-percussive. In percussive martial arts such as Karate, Tae Kwon Do, and Kung-Fu, people strike with their hands, feed, elbows, and head. This type of martial arts is very aggressive. On the other hand, in non-percussive martial arts, such as Judo, involve throwing, locking, and neutralising the opponent. They are far less aggressive. Kung-Fu is an interesting type of martial art, the origins of which are unknown. Some historians believe it started as early as 1500 BC There are two major types of Japanese martial arts. They are Bujitsu, and Budo. The bujitsu martial art is a relatively new one. It emphasises combat and willingness to face death as a matter of honour. Budo, which was started during the late 1800's, focuses on developing moral and aesthetic developments. Karate-do and Judo are forms of Budo. People who learn budo learn it to use it only as a last resort. Another martial art that developed in Japan is ninjitsu, which means "the art of stealing in!" People who practice ninjitsu are called ninjas. Ninjitsu was developed in the late 1200's.

Dr. Martin Luther King Jr. :: essays research papers

Dr. Martin Luther King Jr. was a very important leader of the American Civil Rights movement as well as a Nobel Prize winner. He proved that Civil Disobedience was an effective weapon against depression. King’s challenges to segregation and racial discrimination in the 1950’s and 1960’s helped convince many white Americans to support the cause of Civil Rights in the United States.   Ã‚  Ã‚  Ã‚  Ã‚  Dr. King was born into the American Civil Rights movement in Atlanta, Georgia on January 15, 1929. His grandfather was the founder of the Atlanta Chapters of the NAACP, and his father was the Pastor of the Eboniza Baptist Church where he worked as a Civil Rights Leader. Dr. King attended Morehouse College and graduated with a bachelor’s degree in sociology in 1948. Dr. King married Coretta Scott King in 1953. After graduating with honors from Crozer Theological Seminary in Pennsylvania in 1951, he went to Boston University where he earned a PHD in Divinity in 1955. After graduating from Boston University, Dr. King became the Pastor of the Dexter Avenue Baptist Church in Montgomery, Alabama where he began the activities that would make him an American Civil Rights Leader.   Ã‚  Ã‚  Ã‚  Ã‚  In many states, African Americans were denied voting rights and access to schools, buses, and other public facilities that were segregated. They were also denied accommodations in hotels that were for whites only. Discrimination was openly practiced and in some places sanctioned by law. Dr. King’s goal was to protest segregation until it was declared unconstitutional. In 1955 Rosa Parks was ordered by a bus driver to give up her seat to a white passenger. When she refused, she was arrested and taken to jail. King started the Boycott of the Montgomery Bus System. In 1956 the Supreme Court declared Segregation Laws unconstitutional which ended Bus Segregation. King learned Civil Disobedience from Gandhi and proved that peaceful non-violent protests, such as marches, boycotts, and sit-ins, were an effective weapon against depression. In 1957 Dr. King and other ministers formed the SCLC (Southern Christian Leadership Conference) which fought for the Civil Right s of all Americans. In 1959 King returned to Atlanta, Georgia, the headquarters of the SCLC, to assist his dad and work for Civil Rights. In August of 1963 Dr. Martin Luther King gave his â€Å"I Have A Dream† speech during the March on Washington. Later in the spring of 1963 President JFK introduced the Civil Rights Act (the single most important piece of Civil Rights Legislature) which was passed by President Johnson.

The Women Rag Pickers of Mumbai

As the term ‘rag-picker’ sounds very low so does their job. My study includes several visits to the Govandi Dumping Ground, Mumbai, interviewing the women rag-pickers working there, visiting their slums, meeting with the social workers of Street Mukti Sangathan, studying the problems faced by these women and also keeping a track of their progress. I used personal interviewing of women rag pickers, their boss (the person to whom they sold their collected rags), the social workers, who had devoted their entire time working for them and some printed facts, as my research and analytical methodology. I interviewed 20 women rag pickers, from a diverse age group and varied religions. The following are the main outcomes of my study. Starting with the history of the plights of the women rag pickers, these women came to Mumbai due to droughts in their villages or they were married to men residing in Mumbai. What started of as additional income for the family eventually became the only source of income because 90 % of men stopped working or got into alcohol consumption or left their wives for other younger women. Their day started from as early as 5 am in the morning and went on until it was evening. Their breakfast and lunch were light, comprising of tea and bread or roti and onion, so as to enable them to work efficiently through out the day without feeling sleepy. Infant girls, aged 10 to 15 and older women aged 50 to 65, earned less in comparison to the younger women, due to their higher efficiency levels to work for longer hours. They earned on daily basis and the money depended on the weight of the rag collected by them. So the day they fell ill or couldn’t collect a good amount of scrap meant a day without food. Even their bosses exploited them to the core, by paying them very nominal prices for the scraps that they collected, their bosses in turn earned much more by selling it to companies who recycled these scraps. The government never played any role in their lives. The slum where they resided was rite next to the dumping yard, making it almost impossible for me to breath due to the abundance of flies and stinky smell. Even though majority of the city’s waste was dumped here, no precautionary measures were taken by the govt. to ensure that people residing near by were safe. A lot of rag pickers and their family members were hurt due to the sudden blast that happened when two poisonous particles came into contact with each other. The waste was just dumped as it was; they were not bifurcated on the terms of harmful and unharmful particles. Around 15 people had lost their lives over the years, due to this ignorance of the Govt. Even the middle men who employed these rag pickers didn’t care for their safety. I was horrified to see them collect waste with bare hands and with uncovered mouth. Severely wounded hands, lack of affordable medical facility, unhygienic environment, acute back problems, where just the physical pains they suffered. The mental pressure to progress, the tension of repaying the loans that they took from their bosses on high interest at the times of diseases or marriage, the feeling of being helpless and vulnerable all the time were more killing than the physical pains they suffered. The worse part that I discovered during my study was that, even their children remained uneducated and joined their mothers to support their families. Their lives were surrounded by such dark shadows that even a ray of better future seemed unimaginable. It was in these dark times that Stree Mukti Sangathan came to their rescue. They formed a group of women rag pickers and helped them open a bank account, convincing them to save a small amount of their earnings every month to avoid taking high interest loans at difficult times. By organizing them into a group, the Sangathan saw to it that these women were no longer exploited and worked under much safer conditions. The social workers opened primary schools for the children of the rag pickers and also provided them with proper medical facilities at affordable rates. The rag pickers finally marched towards their independence from the vicious dominance of its exploiters. Once the Sangathan had a decent number of rag picker women members, it then trained them and their children to paint, make hand bags, stitch fancy kurtis, crafting, wall hangings, etc. he ones with higher potential to learn and adapt were taught to produce bio – gas from waste materials and they now work at work stations where machines are operated on bio-gas. After investing 2 decades for the upliftment of these women rag pickers, Sangathan’s effort finally showed drastically positive result; in the form of some of women rag pickers now working as social workers in the Sangathan and helping the other women rag pickers to earn a better living, some have acquired professional education and now work as nurses, teachers, etc. Although the situation hasn’t changed for all the women rag pickers in Mumbai and vulnerability and sufferings still prevails in their lives; it was quiet a relief to learn that a number of women rag pickers whom I interviewed were now proud mothers of engineer sons, owned houses and were spending a decent living. The once very vulnerable and invisible group of my society now has a solid foundation. They conduct annual exhibitions of their hand made products all over India. It wasn’t surprising to learn that the women behind this Sangathan and social revolution of these rag pickers, Ms. Jyoti Mhapsekar was the first Indian Women who won an award at the Clinton Global Initiatives for her committed work towards women and environment. This project was chosen as the best project of that year in my college.

The Fencing Problem

A farmer has exactly 1000 metres of fencing; with it she wishes to fence off a plot of land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. So it could be or anything else with a perimeter (or circumference) of 1000m What she does wish to do is fence off the plot of land which contains the maximum area. Investigate the shape, or shapes of the plot of land which have a maximum area. Throughout this investigation I will check that the perimeter is 1000 meters by finding the total of all the outer sides. Also I will use refining as a way of finding the maximum area. When I talk about using the maximum area of the previous table the maximum area of each table will be highlighted. Rectangles The first shape I will test will be a rectangle. Having been told that the perimeter must be 1000 meters I will find the areas of three rectangles, each with different lengths of sides, making sure that the perimeter is kept the same. To calculate the area I will use the formula LENGTH x WIDTH = AREA or Area = lw. Rectangle A: l = 450m w = 10m Area = 450 x 10 Area = 4500m2 Rectangle B: l = 300m w = 200m Area = 300 x 200 Area = 60000m2 Rectangle C: l = 100m w = 400m Area = 100 x 400 Area = 40000m2 Having carried out the above calculations I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of Length, Width, Perimeter and Area. Under length there will be a variable number (less than 500 and greater than 0). The first formula will be put under the width heading. The width will be calculated by taking the length away from 500. This will guarantee the perimeter to be 1000m. The formula will be =500-B2 where B2 is the cell in which the length is. To double check that the perimeter is 1000m under the perimeter heading there will be another formula. This will be =(B2+C2)*2 where B2 is the length and C2 is the perimeter. It will be multiplied by 2 because the answer in the brackets would be just the total of two sides and not all four. Finally under the area heading there will be a formula. This will be =B2*C2 where B2 is the length and C2 is the width. This formula is the same as the one used previously to calculate the area of a rectangle. The formulas and headings will be entered in as shown in the table below. Length (m) Width (m) Perimeter (m) Area (square m) 490 =500-B2 =(B2+C2)*2 =B2*C2 Having entered the correct information I will be able to calculate the areas of many different sizes of rectangles with a perimeter of 1000m. I can do this in Microsoft Excel by dragging the formula boxes down, thus duplicating them but allowing them to refer to different lengths. (Please see tables and graphs [Fencing Problem for Rectangles]) To start with I used my spreadsheet to find the area of a few rectangles within the range of 1m and 499m.I then plotted a graph showing length against area. It showed a perfect curve. I decided that the line of symmetry of this curve would help to find the length that would give me the maximum area. I found the line of symmetry to be along the 250m mark on the x axis of the graph. Hypothesis I predict that the length of a rectangle that will give me the maximum area will be 250m. I have decided this having found the line of symmetry on the graph. Poof (Please see tables and graphs [Fencing Problem for Rectangles]) To prove my hypothesis I refined my search around the maximum area of the first table and then the second table, followed by the third table and so on. Eventually I found that, even to 1 decimal place above or below 250m that, the maximum area was given by rectangle of sides 250m by 250m. This shows that a square gives the maximum area for a rectangle. Isosceles Triangles The second shape that I will test will be an isosceles triangle. Having carried out tests for a rectangle I am going to see whether the maximum area will be bigger, smaller or the same as that of a rectangle. I am also going to find out whether the number of sides affects the results and whether there are any similarities in results to a triangle. This will help me find the shape that gives the maximum area. As previously for rectangles I will test some different sized isosceles triangles that have an area of 1000m. The formula for the area of a triangle is BASE x HEIGHT divided by 2 or bh/2. I cannot find the area without knowing what the height of the triangle is. To find the height of the triangle I must use Pythagoras. This states that for a right-angled triangle a2+b2=c2 or the square hypotenuse is equal to the sum of square of the other two sides. Therefore to find the height I must split the triangle in half and then use half of the base to help me find the height. The square height will therefore be equal to the square of the hypotenuse minus the square of half the base. In the below examples b = base, s = one equal side of the triangle and h = height. Triangle A: b = 500m s = 250m b/2 = 250m h = 2502-2502 h = 0m Area = 250 x 0 / 2 Area = 0m2 Triangle B: b = 400m s = 300m b/2 = 200m h = 3002-1002 h = ?50000m h = 223.6068m Area = 400 x 223.6068 / 2 Area = 44721.35955m2 Triangle C: b = 200m s = 400m b/2 = 100m h = 4002-1002 h = ?150000m h = 387.29833m Area = 200 x 387.29833 / 2 Area = 38729.38466m2 After completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of Base, 1 equal side, Perimeter, Height and Area. Under the base heading there will be a variable number between 1 and 500. The first formula will be used to calculate the length of one equal side of the isosceles triangle. The formula will be =(1000-B2)/2 where B2 is the base. It will be divided by 2 because 1000-B2 would give the sum of the two equal sides together. As previously , for the rectangles, there will be a formula to check that the perimeter is 1000m. This will be the base plus, one equal side multiplied by two or =B2+(C2*2). The main formula in this spreadsheet will be the one used to find the height. In a spreadsheet there are codes that represent calculations carried out. These are put at the front of the formula and the substitute for square root is SQRT. So my formula will be the square root of 1 equal side squared, minus half the base squared. However before entering my formula I found out that using the power sign (^) doesn't give accurate results and in order to square numbers I must multiply the number by itself instead of using such a sign. Therefore the formula entered into the spreadsheet will be =SQRT((C2*C2)-((B2/2)*(B2/2))) Finally under the area heading there will be a formula. This will be =(B2*E2)/2 where B2 is the base and E2 is the height. This formula is the same as the one used previously to calculate the area of a triangle. The formulas and headings will be entered in as shown in the table below. Base (m) 1 Equal Side (m) Perimeter (m) Height (m) Area (square m) 200 =(1000-B2)/2 =B2+(C2*2) =SQRT((C2*C2)-((B2/2)*(B2/2))) =(B2*E2)/2 Having entered the correct information I will be able to calculate the areas of many different sizes of isosceles triangles with a perimeter of 1000m. I can do this in Microsoft Excel by dragging the formula boxes down, thus duplicating them but allowing them to refer to a different base. (Please see tables and graphs [Fencing Problem for Isosceles Triangles}) As before I entered a range bases between 1m and 499m. I then plotted a graph of base against area and found that unlike the results for a rectangle there wasn't a perfect curve in order to find the line of symmetry, to aid my search. However I could tell that the maximum area would be given by a triangle with a base between 300m and 400m Hypothesis I predict that the maximum area will be given by a triangle with equal sides. I have decided this because the maximum area for a rectangle was given by a square and that my graph shows that the base must be between 300m and 400m. For a triangle with equal sides and a perimeter of 1000m the base would be 333.33†¦meters. Poof (Please see tables [Fencing Problem for Isosceles Triangles}) To prove my hypothesis I refined my search around the maximum area of the first table and then the second table, followed by the third table and so on. Eventually I found that, to 2 decimal places, the maximum area was given by a triangle of equal sides which is 333.33m to every side. This shows that an equilateral triangle gives the maximum area for a triangle and this proves my hypothesis right. Regular Polygons Having tested isosceles triangles and rectangles I found that regular sided shapes give the maximum area. I know this because the maximum area of an isosceles triangle is given when the sides are each 333.33m. The maximum area given by a rectangle is give by a square with 250m sides. I have also that as you increase the number of sides the area increases because the maximum area for a rectangle is 62500m2, and the maximum area for an isosceles triangle is 48112.52243m2. As a result of these findings I am going to test regular sided polygons. Having split the pentagon into isosceles triangles and then into right angled triangles I can now find the area. I know that the base of the triangle is 100m however I do not know the height. Before finding the height I must work out what the internal angle is. To find this I will divide 360 by the number of right-angled triangles (in this case 10). I can now tell the following about the triangle: – I can now use Trigonometry to find the height of the triangle. SOH CAH TOA I know what the opposite is and the angle, and I want to know what the adjacent is. I will therefore use the formula TAN=Opposite/Adjacent. Therefore Adjacent=Opposite/TAN. So the height in metres will be: Height = 100/TAN36 Height = 137.638192m Area of 1 Isosceles Triangle = (200*137.638192)/2 Area of 1 Isosceles Triangle = 13763.819205m2 Area of Pentagon = 13763.819205*5 Area of Pentagon = 68819.09602 m2 After completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of Number of Sides, 1 Equal Side, Perimeter, Internal Angle of 1 Triangle, Half Angle, Height (of internal isosceles triangle), Area of 1 Triangle and Total Area. Under the first heading (Number of Sides) there will be a variable, whole, number between 3 and as higher number as desired (e.g. 30). Under the second heading there will be a formula to calculate the length of one equal side. The formula will be =1000/A3 where A3 is the number of sides. As in all the other tests there will be a formula to check that the perimeter is 1000m. This will tell me if I have made an error in any of the previous cells. So far so good, however before I continue I must point out that a computer spreadsheet doesn't work in degrees to measure angles. It measures in radians where a complete rotation is 2?. Also ? is represented by PI() in a spreadsheet. So instead of using 360 in my formula under the Internal Angle of 1 Triangle heading I will use 2*PI()/A3 where A3 is the number of sides. Under the Half Angle heading there will be a formula that will be =D3/2 where D3 is the internal angle of one triangle. This gives the internal angle of 1 right-angled triangle. My main formula will go under the Height heading and it will use Tan which is substituted by TAN in a spreadsheet. It will be =(B3/2)/TAN(E3) where B3 is 1 equal side and E3 is the angle inside a right-angled triangle. The area of one isosceles triangle will be calculated using the formula =(B3*F3)/2 where B3 is one equal side and F3 is the height. Finally the total area will be calculated by multiplying the area of one isosceles triangle by the number of sides. The formula entered will be =G3*A3 where G3 is the area of one triangle and A3 is the number of sides. The formulas and headings will be entered in as shown in the table below. Number 1 Equal Side Perimeter Internal Angle Half Angle Height Area of 1 Triangle Total Area of Sides (m) (m) of 1 Triangle (rad.) (rad.) (m) (square m) (square m) 5 =1000/A3 =B3*A3 =2*PI()/A3 =D3/2 =(B3/2)/TAN(E3) =(B3*F3)/2 =G3*A3 Having entered the correct information I will be able to calculate the areas of many regular polygons with different numbers of sides and with a perimeter of 1000m. I can do this in Microsoft Excel by dragging the formula boxes down, thus duplicating them but allowing them to refer to a different number of sides. Hypothesis I predict that as you increase the number of sides the area increases because the maximum area for a rectangle is 62500m2, and the maximum area for an isosceles triangle is 48112.52243m2. Proof (Please see graph and table [Fencing Problem for Regular Polygons]) Used my spreadsheet to calculate the areas of polygons with sides ranging from 3 to 30. The polygons with 3 and 4 sides were used to test that my formula worked correctly. I plotted a graph showing the number of sides against the area and found that, as predicted, as the number of sides increased so too did the area. Circle After my findings from carrying out tests on regular polygons I have decided to test circle. I have decided this because as the number of sides of a regular polygon increase so too does the area and a circle is an infinitely sided regular polygon. Hypothesis I predict that a circle will give the largest area because of my tests on regular polygons. I also predict that the maximum area given will be pretty close to that of a regular polygon with 30 sides (79286.37045m2) because of the curve on the graph plotted for the regular polygon section. To find the area of a circle I will be required to use the formulae 2?r and ?r2. The circumference must be 1000m and before finding the area I need to find the radius. Radius = (1000/2)/? r = 500/? r = 159.1549431m Area = ?*159.15494312 Area = 79577.47155m2 To complete this in a spreadsheet under the circumference heading I would enter 1000. Under the radius heading I would use the formula =(C2/2)/PI() where C2 is the circumference. Finally under the Area heading I would enter the formula =PI()*(D2*D2) where D2 is the radius. The headings and formulas will be entered as shown in the table below. Number of Sides Circumference (m) Radius (m) Area (square m) Infinite 1000 =(C2/2)/PI() =PI()*(D2*D2) Formula – 2?r (Circumference/2)/? ?r2 Proof Number of Sides Circumference (m) Radius (m) Area (square m) Infinite 1000 159.1549431 79577.47155 The table above clearly proves my hypothesis correct. The working out also proves my hypothesis correct. Conclusion Having completed the spreadsheet table I can conclude that a circle gives the maximum area and that the result was close to that given by a 30 sided regular polygon. A circle provides the maximum area possible for fencing of length 1000m. The maximum area possible is: – 79577.47155m2

Defender of the Faith/ Philip Roth.

â€Å"l refuse, I can't stop being me, that's all there is to it†. Tears came to his eyes. â€Å"It's a hard thing to be a Jew. But now I understand what Mackey says- it's a harder thing to stay one†. He raised a hand sadly toward me. â€Å"Look at you. † Defender of the Faith/ Philip Roth. The quote above is the part in the Defender of Faith which is a conversation between Crossbars and Marx. Crossbars asks Marx to go out of the army to attend thePassover dinner. Throughout the story we come to understand that Crossbars is taking advantage of the fact that both he and Marx are Jewish for his own benefit. Crossbars tries to â€Å"bond† with Marx on the basis of their common religion, but soon we realize that he uses this common fact to try to get benefits and other privileges. At some point Marx understands that Crossbars is being selfish and is using the religion for his own needs, and gets tired from Crossbar's requests.Crossbars insists that he shouldn' t be treated like everyone else; because he claims that he is better. In addition he throws at Marx accusations saying that he is denting his roots and his family. Crossbars says this only because he is unhappy and doesn't get from Marx what he wants, he calls him a go. â€Å"You even talk like a go. † â€Å"It's a hard thing to be a Jew. But now I understand what Mackey says- It's a harder thing to stay one†.

Ownership and Control Essay

The first business decision that an entrepreneur faces when he is opening his private business is the type of enterprise he/they will create. Basically there are three main forms of business enterprises that can be opened, a sole trader, a partnership and a limited liability company. Such enterprises comprise different legal status and ownership, plus different control measures. In the following paragraphs we will explain such issues of these forms of organizations. A sole trader is the simplest form of business enterprise. It usually consists of one individual who opened a business operation. Sole traders are usually small firms due to the limited amount of capital that is invested by a single owner. The main benefit of a sole trader is that the owner has full control over the assets of the business and all the critical decisions are usually taken solely by him. The main limitation of such a form of business, apart from limited capital is the unlimited liability on the organization’s debts. Partnership arises whenever two or more people invest money in a business to commence trading. A partnership also possesses unlimited liability like a sole trader. However more capital is available since money is invested by more people. In a partnership not all the partners exercise control on the firm especially if it is a limited partnership. In case of companies, which are limited liability due to separate legal entity, give room to shareholders and directors being formed. The shareholders are the owners, which can be a significant number, while the directors are the individuals managing the company which are not necessarily the shareholders particularly in public companies. Companies are frequently formed due to higher availability of finance.