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- Guesstimation: Solving the World's Problems on the Back of a Cocktail Napkin
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- Guesstimation: Solving the World's Problems on the Back of a Cocktail Napkin

Guesstimationis a book that unlocks the power of approximation--it's popular mathematics rounded to the nearest power of ten! The ability to estimate is an. Guesstimation reveals the simple and effective techniques needed to estimate virtually Try searching on JSTOR for other items related to this book. That book is Guesstimation. —Solving the World's Problems on the Back of a Cocktail Napkin, by Lawrence Weinstein and John A. Adams, both.

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In this notebook (article) I have shown the solutions of some problems from the book “Guesstimation: Solving the World's Problems on the Back of a Cocktail. Read here calivekospa.ml?book= Read [PDF] Download Guesstimation: Solving the World's Problems on. eBook (PDF): Course Book: Publication Date: September ; Copyright year: Guesstimation reveals the simple and effective techniques needed to estimate A stimulating follow-up to Guesstimation, this is the must-have book for.

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You would immediately estimate that it should take about four or five hours, based on an average speed of 50—60 mph. This is enough information to decide whether or not you will drive to Boston for the weekend. If you do decide to drive, you will look at maps or the Internet and figure out the exact route and the exact expected driving time. Similarly, before you go into a store, you usually know how much you are willing to spend.

We will apply the same reasoning here. Why a factor of ten? Because that is good enough to make most decisions. Once you have estimated the answer to a problem, the answer will fall into one of the three Goldilocks categories:. If the answer is too big or too small, then you know what to do e.

Only if the answer is just right will you need to put more work into solving the problem and refining the answer. We just aim to help you estimate the answer to within a factor of ten.

Many problems are too complicated for you to come up with an immediate correct answer. These problems will need to be broken down into smaller and smaller pieces. Eventually, the pieces will be small enough and simple enough that you can estimate an answer for each one.

And so we come to. STEP 2: You only need to estimate each answer to within a factor of ten. How hard can that be?

It is often easier to establish lower and upper bounds for a quantity than to estimate it directly. If we are trying to estimate, for example, how many circus clowns can fit into a Volkswagen Beetle, we know the answer must be more than one and less than We could average the upper and lower bounds and use 50 for our estimate. This is not the best choice because it is a factor of 50 greater than our lower bound and only a factor of two lower than our upper bound.

Since we want our estimate to be the same factor away from our upper and lower bounds, we will use the geometric mean. To take the approximate geometric mean of any two numbers, just average their coefficients and average their exponents.

Then you should decrease the exponent sum by one so it is even, and multiply the final answer by three.

Your chance of winning the MongaMillions lottery is one in million. Which distance is this closest to: Imagine trying to pick the single winning ticket from a stack this high. To solve this problem, we need two pieces of information: One ream of copier or printer paper sheets is about 1. A pack of 52 playing cards is also about 1 cm. This means that the thickness of one ticket is.

If stacked vertically, it would be twice as high as Mt Everest 30, ft or 10 km and twice as high as jumbo jets fly. Now perhaps you used the thickness of regular paper so your stack is a few times shorter. Perhaps you used 1mm per ticket so your stack is a few times taller.

Does it really matter whether the stack is 10 km or 50 km? Either way, your chance of pulling the single winning ticket from that stack is pretty darn small. These problems are great fun because, first, we are not looking for an exact answer, and second, there are many different ways of estimating the answer.

Here is a slightly harder question with multiple solutions. We can estimate this from the top down or from the bottom up. We can start with the number of airports or with the number of Americans. Solution 1: Start with the number of Americans and estimate how many plane flights each of us take per year. Therefore, the total number of flights per year is. Solution 2: Start with the number of airports and then estimate the flights per airport and the passengers per flight.

There are several reasonable size airports in a medium-sized state e. If each of the fifty states has three airports then there are airports in the US.

Each airport can handle at most one flight every two minutes, which is 30 flights per hour or flights per hour day. Most airports will have many fewer flights than the maximum. Each airplane can hold between 50 and passengers. This means that we have about. The actual number of US domestic airline passengers in was 6. This is the classic example originated by Enrico Fermi [5] and used at the beginning of many physics courses because it requires employing the methods and reasoning used to attack these problems but does not need any physics concepts.

This is a sufficiently complicated problem that we cannot just estimate the answer. To solve this, we need to break down the problem. We need to estimate 1 how many pianos there are in Los Angeles and 2 how many pianos each tuner can care for. To estimate the number of pianos, we need 1 the population of the city, 2 the proportion of people that own a piano, and 3 the number of schools, churches, etc.

To estimate the number of pianos each tuner can care for, we need to estimate 1 how often each piano is tuned, 2 how much time it takes to tune a piano, and 3 how much time a piano tuner spends tuning pianos. This action might not be possible to undo.

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Save For Later. Create a List. Summary Guesstimation is a book that unlocks the power of approximation--it's popular mathematics rounded to the nearest power of ten! Read on the Scribd mobile app Download the free Scribd mobile app to read anytime, anywhere. Princeton University Press Released: Feb 9, ISBN: Prices are subject to change without notice.

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Overview Aims and Scope Guesstimation 2. English Readership: He is the coauthor of Guesstimation: Patricia Edwards is senior lecturer in art at Old Dominion University.

The breadth of scope of the problems is truly impressive. Weinstein's arguments are always convincing and, in many cases, very clever. His sense of humor provides a pain-free tutorial on how analysts can make real progress in understanding vaguely defined problems. Nahin, author of Number-Crunching: Taming Unruly Computational Problems from Mathematical Physics to Science Fiction Certainly a good read for any teacher who enjoys numbers and the world around us.

Whether or not a fan of numbers, it's always cool to appear smart, therefore Guesstimation 2. I can easily see having this book close by and returning to it again and again. And 76 Other Physical Paradoxes and Puzzles Readers who enjoyed Weinstein's first volume will be pleased with this instalment. A delightful volume. I hope to be able to use many of the tricks I learned in the future. I also hope to teach some of them to students. This would make a great secondary textbook in many classes, ranging from quantitative literacy to a science methods class for future educators.

A careful study of this book would certainly improve a student's ability to take a complicated question, break it down into solvable parts, and assemble the parts to find an answer.

Because this is quite close to what I want my students to do when faced with a difficult problem in pure mathematics as well, I consider this to be a very valuable book indeed.