College Gift

Somerville College - Gift aid and deed of covenant for UK Tax Payers Somerville College Skip to Content The College | Prospective Students | Current Students | Former Students | Conferences | Notices | Contacts Home | Alumni and Friends Events | Degree Day | The Association of Senior Members | Missing Members | Development | Contact Us | Site Map Gift aid and deed of covenant for UK Tax Payers Deeds of Covenant have been superseded by the new Gift Aid scheme. However, if you have a covenanted gift to the College which was set up before 6 April 2000, you need take no action. Your donation will continue as before and will be tax effective under the Gift Aid scheme until the deed expires. If you set up a deed of covenant on or after the 6 April 2000, you will need to make a Gift Aid declaration to ensure that your gifts remain tax effective. Gift Aid is a scheme by which you can give a sum of money to charity and the charity can reclaim from the Inland Revenue the basic rate tax (currently 22%) on your gift, at no extra cost to you. That increases the value of the gift you make to the charity. Under current regulations for example, if you give £10 using Gift Aid, your gift would be worth £12.82 to Somerville. If you are a higher rate tax payer, you can claim relief on the difference between the basic rate and higher rate of tax on your annual tax return. Please note:You must pay at least as much tax as Somerville will reclaim on your gift(s) in the tax year in which you make them (tax credits on dividend income will count towards the tax paid). The tax year runs from 6 April in one year to 5 April in the next. If you have any queries about Gift Aid, please speak to the Development Office. You can make payments by cash, cheque, postal order, direct debit, standing order, debit or credit card or even in a foreign currency (including the euro). Subject to a few rules, you can give any amount, large or small, regular or one-off, and Somerville can reclaim the tax. In order for Somerville to claim Gift Aid on your gift, you need to make a declaration confirming that you want your gifts to be treated as Gift Aid donation. Please click here to access a Gift Aid declaration form or contact the Development Office. Accessibility Statement © Somerville College 2003. Last Modified: 2:27pm on the 28th of January 2005

Birthday Present

Math Forum: Ask Dr. Math FAQ: The Birthday Problem -- Ask Dr. Math: FAQ The Birthday Problem Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time. What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet? This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised? How large must a class be to make the probability of finding two people with the same birthday at least 50%? Let's forget about leap year when we solve this problem (no February 29 birthdays!) This way, we can assume that a year is always 365 days long. We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%. Now, what about three people ? The chance of the first and second person sharing a birthday is still 1/365. The first and third person might share a birthday instead. The probability of that is 1/365 as well. But what if the second and third person shared a birthday? And what if all three of them had the same birthday? Things are getting complicated fast. Four or five people would be even messier. Is there a simpler way? To solve the birthday problem, we need to use one of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won't happen is always 1. (In other words, the chance that anything might or might nothappen is always 100%.) If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn't happen) = 1 P(two people share birthday) + P(no two people share birthday) = 1 P(two people share birthday) = 1 - P(no two people share birthday). So, what is the probability that no two people will share a birthday? Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person. To find the probability that both the second person and the third person will have different birthdays, we have to multiply: (365/365) * (364/365) * (363/365) = 132 132/133 225, which is about 99.18%. If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125, or about 98.36%. We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365). If you know permutation notation, you can write this formula as (365_P_n)/(365^n). That's the same as 365! / ((365-n)! * 365^n). We've made some progress, but we still haven't answered the original question: how large must a class be to make the probability of finding two people with the same birthday at least 50%? We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday, and we know how to find the probability that everybody has a different birthday for any number of people. The easiest way to find the right class size is to use a calculator to try different numbers in the formula. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is... a class of 23 people . (The probability is about 50.73%.) From the Dr. Math archives: Probability Theory: Coincidental Birthday Probability of the Same Birthday within a Group Birthday Probabilities Three Share a Birthday The Birthday Problem; Queuing at a Bank Birthday Probability, Class of 25 One Person of Seven Born on Monday Odds of Left-Handedness in a Group From the Web: The Birthday Problem: A short lesson in probability , George Reese A Java applet that you can use to test different class sizes (it works better with small classes) and graphs of the probability for different numbers of people. The Law of Small Errors , Keith Devlin The birthday problem, and related questions - what's the probability that someone will have your birthday? Birthday Surprises, Ivars Peterson Birthday Problem, Eric Weisstein's World of Mathematics Coincidence, Alexander Bogomolny How to Read Mathematics, Shai Simonson and Fernando Gouveau This article uses an explanation of the birthday problem as an example. An Introduction to Mathematica and the "Birthday Problem," Louie Beuschlein For a general review of probability: Probability, Dr. Math FAQ Probability in the Real World, Dr. Math FAQ - Ursula Whitcher, for the Math Forum Submit your ownquestion to Dr. Math [ Privacy Policy ] [ Terms of Use ] Math Forum Home || Math Library || Quick Reference || Math Forum Search Ask Dr. Math ® © 1994-2005 The Math Forum http://mathforum.org/dr.math/

Creative Gifts, Valentine's Day

Compare Prices and Read Reviews on Creative Gifts Valentine's Day Jewelry at Epinions.com Join Epinions | Help | Sign In All Categories Beauty Books Business & Technology Cars & Motorsports Computer Hardware Education Electronics Games Gourmet Health Home & Garden Hotels & Travel Kids & Family Magazines & Newspapers Movies Music Musical Equipment Online Stores & Services Outdoor Gear Personal Finance Pets Restaurants & Gourmet Software Sporting Goods ---------------------------------- Related Deals Home > Gifts > Jewelry We found 1 result for , Creative Gifts, Valentine's Day Search Results Showing 1 item Hide photos Sort by Name or Rating Sort by Price See specs Creative Gifts Beaded Double Jewelry Box Model 22040 22040 Jewelry Boxes · Creative Gifts · Women's · Valentine's Day · Silver Showing 1 item Featured Resources Diamond Jewelry Forbes Favorite Online Jeweler Free FedEx & 30-Day Returns. www.bluenile.com Diamond Necklace Sale 40-80% off Necklaces & Pendants. Quality selection, $2.95 shipping! Overstock.com Diamond Jewelry Shop for fine earrings, pendants, rings & more. Save with us today. www.Diamond.com Netaya Fine Jewelry Fine Diamond, Gemstone, Gold & Silver Jewelry - Up to 80% Off www.netaya.com Designer Engagement Rings 3000 Designs. 60000 GIA Diamonds Non-Designer Prices From $1500 www.UnitedDiamonds.com Help | Member Center | Message Boards | Privacy Statement | Site Index About Epinions | Careers | Contact Epinions | Merchant Center | New Merchants | Advertising Epinions | DealTime USA | DealTime UK | PriceTool | Shopping.com © 1999-2005 Epinions, Inc. Trademark Notice Epinions.com periodically updates pricing and product information from third-party sources, so some information may be slightly out-of-date. You should confirm all information before relying on it.

Birthday Present

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