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https://www.mathworks.com/matlabcentral/cody/problems/1702-maximum-value-in-a-matrix/solutions/2996713 | 1,606,407,014,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188800.15/warc/CC-MAIN-20201126142720-20201126172720-00533.warc.gz | 770,467,600 | 18,634 | Cody
# Problem 1702. Maximum value in a matrix
Solution 2996713
Submitted on 27 Sep 2020 by Rodrigo García
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3; 4 5 6; 7 8 9]; y_correct = 9; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = -10:0; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 17; y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = magic(6); y_correct = 36; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
x = [5 23 6 2 9 0 -1]'; y_correct = 23; assert(isequal(your_fcn_name(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 270 | 827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.654408 |
https://hwhshop.com/qa/quick-answer-how-often-does-week-53-occur.html | 1,618,499,718,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00166.warc.gz | 375,950,365 | 6,747 | # Quick Answer: How Often Does Week 53 Occur?
## Is there a Week 53 in 2019 payroll?
Pay period 53 takes place when there are 53 weekly pay days in the year – usually when it’s in a leap year.
For the 2019/2020 tax year, pay period 53 is Saturday 4 April and Sunday 5 April 2020..
## Why is 52 weeks in a year?
The year has 365 days, leaving one extra day. A leap year has 366 days, adding a second extra day. This makes 52 1/7 weeks in a normal year and 52 2/7 weeks in a leap year..
## Which week is now?
The current Week Number is WN 01.
## What years have 53 Fridays?
The year 2021 has exactly 365 days. Now if the year starts on a Friday in a non-leap year, you end up with 53 Fridays. Or if either of the first two days lands on a Friday during a leap year, then you can also get 53 Fridays.
## How often are there 53 Thursdays in a year?
0.14 or 1/7 is probability for 53 Thursdays in a non-leap year.
## How often do you get a 53 week year?
There are 52 weeks in a normal year, meaning 52 weekly charges. However, every five to six years, there are 53 Mondays in a year instead.
## Are there 52 or 53 weeks in a year?
The weeks of the year in a Gregorian calendar are numbered from week 1 to week 52 or 53, depending on several varying factors. Most years have 52 weeks, but if the year starts on a Thursday or is a leap year that starts on a Wednesday, that particular year will have 53 numbered weeks.
## Can there ever be 53 weeks in a year?
An ISO week-numbering year (also called ISO year informally) has 52 or 53 full weeks. That is 364 or 371 days instead of the usual 365 or 366 days. The extra week is sometimes referred to as a leap week, although ISO 8601 does not use this term. … Each week’s year is the Gregorian year in which the Thursday falls.
## Are there 53 weeks in 2021?
The year 2021 has 52 calendar weeks.
## How many pay weeks are there in 2020?
“That same employer would have 52 paydays in 2020, which is a leap year.” For those years when an employer finds itself with 53 or 27 paydays, there are two general options, Trabold pointed out: Do nothing and pay the same amount for each payday, recognizing one extra paycheck in the year.
## Is 2020 a 53 week tax year?
“Week 53” occurs when there are fifty-three weekly pay days in the year: In 2020, employers who pay wages on either a Wednesday or Thursday will have 53 pay days, thus a Week 53.
## Is there an extra week in 2020?
Of course next year is a leap year, and when you receive your 2020 calendars you’ll notice that 1st January falls on a Wednesday, meaning that the year will have 53 weeks. That’s a whole extra week to enjoy your award winning Rose calendar! | 706 | 2,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-17 | latest | en | 0.962143 |
https://fr.mathworks.com/matlabcentral/cody/problems/135-inner-product-of-two-vectors/solutions/104570 | 1,569,231,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576345.90/warc/CC-MAIN-20190923084859-20190923110859-00237.warc.gz | 478,264,151 | 15,451 | Cody
# Problem 135. Inner product of two vectors
Solution 104570
Submitted on 28 Jun 2012 by Søren
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1:3; y= 3:-1:1; z_correct = 10; assert(isequal(your_fcn_name(x,y),z_correct))
2 Pass
%% x = 1:6; y= ones(1,6); z_correct = 21; assert(isequal(your_fcn_name(x,y),z_correct)) | 150 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-39 | latest | en | 0.710008 |
https://www.mathsmoves.co.uk/multiplication-and-division | 1,627,448,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00499.warc.gz | 939,682,841 | 143,504 | ## Multiplication and Division Ideas
As with addition and subtraction, pupils need a sound understanding of multiplication and division concepts, and the links between them, in order to progress within many areas of maths.
## Multiplication Facts
There has to be an emphasis on pupils being able to efficiently use multiplication facts (or times tables) since they are essential for so many aspects of maths - pupils will not be able to divide or work with fractions accurately without having quick access to times table facts. However, we don't have to rely entirely on rote learning, which will not work for some pupils and for others it does not expose pupils to the underlying patterns within and between the multiple groups.
Make sure to emphasise the multiples of six, seven, eight and nine since pupils can struggle with these facts compared to the earlier multiples so provide lots of opportunities to work with these. Also, don't forget to include multiplying by zero and one which can sometimes catch pupils out because as teachers, we can assume they are obvious and therefore spend little time on them.
100 Grid Patterns
Exploring multiple patterns will help pupils to assimilate multiplication and division facts (which can be a struggle for some pupils) and this can be started on the hundred square, where multiples of numbers can be shaded and any patterns noticed can be gathered together. Allowing time to discuss the patterns and why they occur is time well spent. This can be as simple as looking at a set of multiples and looking for patterns in the digits within the multiples, or shading several multiple sets such as threes, sixes and nines on the same grid. It is often surprising how many older pupils - including some who are proficient at recalling multiplication facts - are surprised about insights such as the connections between the multiples of three, six and nine! When developing divisibility tests, the 100 grid provides a useful investigation tool.
Multiplication Grid
Make sure that pupils are familiar with the multiplication grid and allow time to discover the symmetry within it (thanks to the commutative rule where 3 x 8 = 8 x 3) which cuts the number of facts to learn by half. Connections between different sets of multiples can also be explored such as doubling twos to get fours and doubling fours to get eights. It is also useful for exploring the links between multiplication and division, (for many pupils, 24 ÷ 3 is better as 3 x ? = 24) and for showing the shape of numbers. The array model of multiplication is also emphasised, which can be linked to area.
## Written Methods
In its latest incarnation, the English National Curriculum has become more prescriptive regarding the written methods that pupils should have mastered by the end of their primary schooling. These are specified as: columnar addition and subtraction, short and long multiplication and short and long division.
Previously, pupils were allowed to use whatever method suited them best and were able to utilise a flexible approach which made sense to them. However, these approaches are still useful for helping pupils to make sense of multiplication and division calculations.
Grid Method
This method connects to arrays and by partitioning numbers before multiplying, it helps pupils to understand what is happening when we multiply larger numbers. If this is explained alongside the long multiplication algorithm it can help pupils to see why the algorithm works since some pupils struggle to remember the process for calculating using the algorithm. It is also an excellent approach for multiplying brackets when pupils meet algebra later on in their maths journey, rather than relying on remembering the process involved.
Chunking
Based on repeated subtraction, this became a popular method for division with some pupils partly because they had control over the size of the chunks to choose each time and partly because they understood the process. In fact, when I showed this to a group of fourteen year olds, who had always struggled with division calculations, they were thrilled that they could now use a process that they understood. Of course, it does rely on the ability to subtract accurately and familiarity with the number line can help pupils to decide on the size of their chunks; multiples of ten can be used initially to reduce the likelihood of errors when subtracting.
Short and Long Division
These have been put together because I suggest that they are taught together. The reason for this is that many pupils (and adults) will attest that they can apply one method but are confused by the other, so they are not seeing the links between the two methods where we are simply more explicit in our writing when undertaking long division. I would also suggest that either method can be employed for any size divisor.
Of course, it is always worth writing the calculation as a fraction (if the dividend and divisor have common factors), returning to the method once the equivalent fraction with the smallest divisor has been found in order to work with an easier calculation.
See Multiplication Facts in Resources which will help pupils to explore
multiplication facts by focusing on patterns and connections.
Visuals
An array is a useful image for multiplication and pupils can be provided with 2cm square paper, together with counters, to see how many rectangles they can create for a given number of counters.
1cm squared paper is also useful when introducing two-digit multiplication since it provides a visual grid where the side length and width of the original rectangle can be partitioned into multiples of ten and the remaining units, so that pupils can see how many squares there are in each section using multiplication. Make sure to point out the link between this and area since they are in fact finding the area of each section. Using base 10 materials on squared paper can also help pupils to understand the grid method and can illustrate the link between arrays and division.
It is worth displaying a 100 square and a multiplication grid in the classroom, as well as having individual smaller grids available for pupils to use as needed, alongside laminated L shapes to assist in reading the grid, as needed.
Fluency
Pupils sometimes claim that they like multiplication but dislike division, even when working with numbers less than one hundred, which implies that they are not connecting multiplication and division. I have some quick starters where numbers appear and then the required operation and often faces drop when they see say, 72 ÷ 8, but when I ask the question, '8 x what results in 72?' then those who are secure with their multiplication facts can answer easily.
When dividing with larger numbers, using factors or writing the division as a fraction and cancelling, can result in a much easier calculation depending on the numbers involved. It is therefore worth exploring different approaches to calculations so that pupils can choose the most appropriate method or use an alternative to check a formal written calculation if the required method has been specified.
See Multiplication Facts in Resources, which contains images of multiplication grids that can be printed out and laminated. | 1,397 | 7,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-31 | latest | en | 0.958832 |
https://www.chess.com/blog/Win_Like_McEntee/chess-notation-and-fool-s-mate | 1,545,069,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00391.warc.gz | 843,157,005 | 12,147 | # Chess Notation and Fool's Mate
Apr 5, 2017, 9:09 PM |
0
Chess notation (aka scorekeeping) is required of players during chess tournament play. The immediate practical purpose of this is to allow refs to reference the players' moves in the case a dispute arises over the board, allow a player to prove three-fold repetition in the case of a draw (and 50 move rule), and lets players to review their games later.
Algebraic notation is the form of chess notation used. Chess.com uses this method. Checkmate is #, check is +, castling kingside is o-o and castling queenside is o-o-o. You should understand from examples on chess.com how the rest works.
Example game for you to see notation in action for certain moves.
Find the winning move in what is known as Fool's Mate.
Here is a game that shows how the position on the board may be reached. You'll note that this checkmate got its name from the fact that your opponent will need to practically walk into it.
Here are some other variants of it where you use the Fool's Mate motif. Find the winning moves.
Your neighbor said you get \$4,000 if you beat him in under 3 moves.
You know you only have one chance, go for it.
Blogs | 275 | 1,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-51 | latest | en | 0.961581 |
https://calculator-online.org/mathlogic | 1,721,599,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00667.warc.gz | 142,081,055 | 8,659 | Mister Exam
# Mathematical logic step by step
The teacher will be very surprised to see your correct solution 😉
• Use symbolic logic and logic algebra
• Place brackets in expressions, given the priority of operations
• Simplify logical expressions
• Build a truth table for the formulas entered
• Find Normal Forms of Boolean Expression:
• Conjunctive normal form (CNF), including perfect
• Disjunctive normal form (DNF), including perfect
### Supported symbols in logical expressions
Here are the symbols that should be specified when entering a logical formula into the calculator
¬a
negation
a⇒b
material implication
a∧b
logical conjunction
a∨b
logical disjunction
a⇔b
logical equality
a⊕b
– exclusive or ( Exclusive disjunction)
a|b
– Nand (not and) (Sheffer stroke)
a↓b
– Not-Or (logical NOR)
a⊙b
– XNOR gate ( Exclusive AND)
In the calculator, you can simplify expressions with the following operations: NOT, XOR, AND, OR, NAND, NOR, NOT, XNOR | 241 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.787467 |
https://blog.ielliott.io/a-practical-introduction-to-monad-transformers | 1,701,615,060,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00408.warc.gz | 171,203,491 | 4,773 | # A Practical Introduction to Monad Transformers
1 February, 2015
Monad transformers combine the functionality of two monads into one. They are often used as a way to “flatten” nested monads, but are also used to enable interactions between monads that, when used seperately, would be incredibly difficult to implement.
You input a number and want to manipulate it while printing the result each time. If there were no intermediate IO operations we could use the state monad with the following state changes:
``````add :: Int -> State Int ()
add n = state \$ \s -> ((),s + n)
subtract :: Int -> State Int ()
subtract n = state \$ \s -> ((),s - n)``````
chain them together:
``````manyOperations :: State Int ()
manyOperations = do
subtract 3
subtract 22``````
then get the result:
``(_,result) = runState manyOperations 5 :: ((),Int)``
Now let’s consider how to print the state. If we want to preserve the above chaining syntax, we need a monad where:
1. We can do stateful computations
2. Our IO functions have access variables inside the computation
## The Solution
The state monad transformer is defined as:
``newtype StateT s m a = StateT { runStateT :: s -> m (a,s) }``
1Meaning that given an initial state `s` and a state transformer `st`, we can call `runStateT st s` to get a monad containing the state tuple.
The real beauty (or magic, as some would say) of this monad comes from the bind function. Let’s take a look at its definition:
``````(>>=) :: Monad n => StateT s n a -> (a -> StateT s n a) -> StateT s n a
m >>= k = StateT \$ \ s -> do
~(a, s') <- runStateT m s
runStateT (k a) s'``````
Time to break it down. `m` is a state transformer. `k` is a function that takes a result of type `a`, and returns a state transformer. The final state transformer, when run with an initial state, does the following:
1. Gets the result and state of running the computation in `m` with the initial state `s`
2. Passes the result of (1) to the function `k`, returning a different state transformer
3. Runs the computation created in (2) using the state returned in (1)
4. Wraps the result
This means that we will be able to keep using a simple chained sequence of monads.
How does this relate to the problem at hand?
The monad component of the state transformer allows us to execute IO operations which have access to the state during the computation. Here is how the `add` and `subtract` functions can be written using the state transformer monad:
``````add :: Int -> StateT Int IO ()
add n = StateT \$ \s -> do
print (s+n)
return ((),s+n)
subtract :: Int -> StateT Int IO ()
subtract n = StateT \$ \s -> do
print (s-n)
return ((),s-n)``````
We can still chain them using the same syntax as before:
``````manyOperations :: StateT Int IO ()
manyOperations = do
subtract 3
``````main = runStateT manyOperations 5 | 739 | 2,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | longest | en | 0.858652 |
https://fcmorkovice.cz/site/2lza5r.php?id=how-to-calculate-revenue-percentage-8d67e2 | 1,679,848,062,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00653.warc.gz | 289,293,620 | 25,028 | # how to calculate revenue percentage
Company A's revenue growth might be high, but its churn is so high that their MRR is going down. Prior to ProfitWell Patrick led Strategic Initiatives for Boston-based Gemvara and was an Economist at Google and the US Intelligence community. Example. Units are selling at \$20 per unit and 400 sell. But if you only have the final amount know the percentage added, you need to work in reverse to find the original amount. To calculate the percentage increase in sales, you simply compare the sales figures for one period with the sales figures for a comparable period. Or are customers leaving your service before they've even used it? What Is Revenue Growth and How to Calculate It. Is your pricing generally correct? Look at the company's balance sheet for total revenues and its breakdown for each department. Selling more product from one period to the next shows that you're attracting more customers and thus have a better chance of sustaining your business. On the other hand based on the engineer's survey the revenue recognized should be \$80 million (40% multiplied by \$200 million). Growth rate is one of the most important metrics for subscription businesses. If the company has a balance sheet, you can usually find department revenue listed there. Are you striking the right balance with your onboarding?
in International Law from the University of East London. It helps you estimate how much revenue you'll take in this year. That's 0.5, which times 100 gives us 50 percent. For example, if you had the final cost and the percentage of sales … The current accounting year versus the previous accounting year. A positive number indicates that your sales revenue is growing. Or are customers leaving your service before they've even used it? We explain how to calculate it. That's fine if you're selling bath products to consumers.
The amount before the sales tax was added is \$200.
Her articles have appeared on numerous business sites including Typefinder, Women in Business, Startwire and Indeed.com. Business Literacy Institute: Revenue Growth. UX: Is your product performing at the right level? In this example, your revenue dropped by almost 18 percent (17.6 actual percent) in 2019 from 2018's revenues. Once you’ve established those core competencies, craft ideas for new revenue drivers to complement them. Freemium Offerings: Offering the basic version of your product for free is an excellent way of raising awareness of its value and gaining early traction.
For example, if a company generated \$50 million in revenue during one business year and \$75 million in revenue the next, it saw a 50% revenue growth. The basic EBITDA formula is: EBITDA = Net income + interest expenses + tax + depreciation + amortization Are you updating regularly? Any revenue growth strategies you do implement will only be successful if you can figure out what exactly is undermining your revenue growth in the first place. Next, gather revenue figures for the two periods you're comparing. In addition to nailing your expansion revenue, revenue growth is about making the best of what you already have as much it is about the next big idea. The 2018 \$450,000 minus the 2017 \$300,000 is \$150,000 in actual sales revenue growth.
Once you pass even relatively low rates of churn, adding new customers won’t be enough to see stable revenue growth. Total Revenue = \$20 x 400 = \$8,000. For example, if your total revenue in 2019 was \$100,000 and your total revenue in 2018 was \$85,000, subtract \$85,000 from \$100,000 for a difference of \$15,000.
To calculate revenue growth as a percentage, you subtract the previous period’s revenue from the current period’s revenue, and then divide that number by the previous period’s revenue. The formula calculates both positive and negative changes in revenue growth. In math terms, it looks like this: (Current period's revenue - prior period's revenue) ÷ by prior period's revenue x 100 = revenue percentage change. Divide this difference in revenue by the later year's revenue.
## Další aktuality:
### Tréninky se opět rozběhly
Vzhledem k vývoji korona viru v ČR a opatření vlády, bylo za přísných opatření opět povoleno konání tréninků.Výbor FC Morkovice
### První ročník sportovního plesu v Morkovicích
Datum konání: 29.2.2020 od 20:00 hodin Místo konání: Sokolovna Morkovice Pořadatel: FC Morkovice Vážení sportovní přátelé, děkujeme za hojnou účast a skvělou atmosféru na prvním
### FC Morkovice – příprava podzim 2020
Přátelské zápasy našich mužstev: Muži A So. 30.5. 16:00 Morkovice – Kroměříž U19 6:2 (Berky 3x, Oplt M. 2x, Knap) So. 13.6. 15:00 Lutín – Morkovice | 1,068 | 4,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.961716 |
https://www.numwords.com/words-to-number/en/125000 | 1,606,579,931,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00392.warc.gz | 770,237,368 | 2,485 | NumWords.com
# How to write One hundred twenty-five thousand in numbers in English?
We can write One hundred twenty-five thousand equal to 125000 in numbers in English
< One hundred twenty-four thousand nine hundred ninety-nine :||: One hundred twenty-five thousand one >
Two hundred fifty thousand = 250000 = 125000 × 2
Three hundred seventy-five thousand = 375000 = 125000 × 3
Five hundred thousand = 500000 = 125000 × 4
Six hundred twenty-five thousand = 625000 = 125000 × 5
Seven hundred fifty thousand = 750000 = 125000 × 6
Eight hundred seventy-five thousand = 875000 = 125000 × 7
One million = 1000000 = 125000 × 8
One million one hundred twenty-five thousand = 1125000 = 125000 × 9
One million two hundred fifty thousand = 1250000 = 125000 × 10
One million three hundred seventy-five thousand = 1375000 = 125000 × 11
One million five hundred thousand = 1500000 = 125000 × 12
One million six hundred twenty-five thousand = 1625000 = 125000 × 13
One million seven hundred fifty thousand = 1750000 = 125000 × 14
One million eight hundred seventy-five thousand = 1875000 = 125000 × 15
Two million = 2000000 = 125000 × 16
Two million one hundred twenty-five thousand = 2125000 = 125000 × 17
Two million two hundred fifty thousand = 2250000 = 125000 × 18
Two million three hundred seventy-five thousand = 2375000 = 125000 × 19
Two million five hundred thousand = 2500000 = 125000 × 20
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(Difference between revisions)
Revision as of 02:22, 23 August 2012 (view source) (→Acceleration Factor)← Older edit Revision as of 02:25, 23 August 2012 (view source) (→T-NT Exponential)Newer edit → Line 70: Line 70: =T-NT Exponential= =T-NT Exponential= - By setting $m=L(U,V)$, the exponential $pdf$ becomes: + By setting $m=L(U,V)\,\!$, the exponential $pdf$ becomes: ::$f(t,U,V)=\frac{{{U}^{n}}}{C}{{e}^{-\tfrac{B}{V}}}\cdot {{e}^{-\tfrac{{{U}^{n}}}{C}\left( {{e}^{-\tfrac{B}{V}}} \right)t}}$ ::$f(t,U,V)=\frac{{{U}^{n}}}{C}{{e}^{-\tfrac{B}{V}}}\cdot {{e}^{-\tfrac{{{U}^{n}}}{C}\left( {{e}^{-\tfrac{B}{V}}} \right)t}}$ Line 76: Line 76: ==T-NT Exponential Statistical Properties Summary== ==T-NT Exponential Statistical Properties Summary== ===Mean or MTTF=== ===Mean or MTTF=== - The mean, $\overline{T},$ or Mean Time To Failure (MTTF) for the T-NT exponential model is given by: + The mean, $\overline{T},\,\!$ or Mean Time To Failure (MTTF) for the T-NT exponential model is given by: ::\begin{align} ::[itex]\begin{align} Line 83: Line 83: ===Median=== ===Median=== - The median, [itex]\breve{T}, + The median, $\breve{T},\,\!$ for the T-NT exponential model is given by: - for the T-NT exponential model is given by: + ::$\breve{T}=\frac{1}{\lambda }0.693=0.693\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$ ::$\breve{T}=\frac{1}{\lambda }0.693=0.693\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$ ===Mode=== ===Mode=== - The mode, $\tilde{T},$ + The mode, $\tilde{T},\,\!$ for the T-NT exponential model is given by: - for the T-NT exponential model is given by: + ::$\tilde{T}=0$ ::$\tilde{T}=0$ ===Standard Deviation=== ===Standard Deviation=== - The standard deviation, ${{\sigma }_{T}}$ , for the T-NT exponential model is given by: + The standard deviation, ${{\sigma }_{T}}\,\!$ , for the T-NT exponential model is given by: ::${{\sigma }_{T}}=\frac{1}{\lambda }=m=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$ ::${{\sigma }_{T}}=\frac{1}{\lambda }=m=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$ Line 119: Line 117: ====Reliable Life==== ====Reliable Life==== - For the T-NT exponential model, the reliable life, or the mission duration for a desired reliability goal, ${{t}_{R}}$ , is given by: + For the T-NT exponential model, the reliable life, or the mission duration for a desired reliability goal, ${{t}_{R}}\,\!$ , is given by: ::$R({{t}_{R}},U,V)={{e}^{-\tfrac{{{t}_{R}}\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$ ::$R({{t}_{R}},U,V)={{e}^{-\tfrac{{{t}_{R}}\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$ Line 147: Line 145: and: and: - *${{F}_{e}}$ is the number of groups of exact times-to-failure data points. + *${{F}_{e}}\,\!$ is the number of groups of exact times-to-failure data points. - *${{N}_{i}}$ is the number of times-to-failure data points in the ${{i}^{th}}$ time-to-failure data group. + *${{N}_{i}}\,\!$ is the number of times-to-failure data points in the ${{i}^{th}}\,\!$ time-to-failure data group. - *$B$ is the T-NT parameter (unknown, the first of three parameters to be estimated). + *$B\,\!$ is the T-NT parameter (unknown, the first of three parameters to be estimated). - *$C$ is the second T-NT parameter (unknown, the second of three parameters to be estimated). + *$C\,\!$ is the second T-NT parameter (unknown, the second of three parameters to be estimated). - *$n$ is the third T-NT parameter (unknown, the third of three parameters to be estimated). + *$n\,\!$ is the third T-NT parameter (unknown, the third of three parameters to be estimated). - *${{V}_{i}}$ is the temperature level of the ${{i}^{th}}$ group. + *${{V}_{i}}\,\!$ is the temperature level of the ${{i}^{th}}\,\!$ group. - *${{U}_{i}}$ is the non-thermal stress level of the ${{i}^{th}}$ group. + *${{U}_{i}}\,\!$ is the non-thermal stress level of the ${{i}^{th}}\,\!$ group. - *${{T}_{i}}$ is the exact failure time of the ${{i}^{th}}$ group. + *${{T}_{i}}\,\!$ is the exact failure time of the ${{i}^{th}}\,\!$ group. - *$S$ is the number of groups of suspension data points. + *$S\,\!$ is the number of groups of suspension data points. - *$N_{i}^{\prime }$ is the number of suspensions in the ${{i}^{th}}$ group of suspension data points. + *$N_{i}^{\prime }\,\!$ is the number of suspensions in the ${{i}^{th}}\,\!$ group of suspension data points. - *$T_{i}^{\prime }$ is the running time of the ${{i}^{th}}$ suspension data group. + *$T_{i}^{\prime }\,\!$ is the running time of the ${{i}^{th}}\,\!$ suspension data group. - *$FI$ is the number of interval data groups. + *$FI\,\!$ is the number of interval data groups. - *$N_{i}^{\prime \prime }$ is the number of intervals in the ${{i}^{th}}$ group of data intervals. + *$N_{i}^{\prime \prime }\,\!$ is the number of intervals in the ${{i}^{th}}\,\!$ group of data intervals. - *$T_{Li}^{\prime \prime }$ is the beginning of the ${{i}^{th}}$ interval. + *$T_{Li}^{\prime \prime }\,\!$ is the beginning of the ${{i}^{th}}\,\!$ interval. - *$T_{Ri}^{\prime \prime }$ is the ending of the ${{i}^{th}}$ interval. + *$T_{Ri}^{\prime \prime }\,\!$ is the ending of the ${{i}^{th}}\,\!$ interval. - The solution (parameter estimates) will be found by solving for the parameters $B,$ $C$ and $n$ so that $\tfrac{\partial \Lambda }{\partial B}=0,$ $\tfrac{\partial \Lambda }{\partial C}=0$ and $\tfrac{\partial \Lambda }{\partial n}=0$ . + The solution (parameter estimates) will be found by solving for the parameters $B,\,\!$ $C\,\!$ and $n\,\!$ so that $\tfrac{\partial \Lambda }{\partial B}=0,\,\!$ $\tfrac{\partial \Lambda }{\partial C}=0\,\!$ and $\tfrac{\partial \Lambda }{\partial n}=0\,\!$ . =T-NT Weibull= =T-NT Weibull=
## Revision as of 02:25, 23 August 2012
Chapter 8: Temperature-NonThermal Relationship
Chapter 8 Temperature-NonThermal Relationship
## Contents
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When temperature and a second non-thermal stress (e.g., voltage) are the accelerated stresses of a test, then the Arrhenius and the inverse power law relationships can be combined to yield the Temperature-NonThermal (T-NT) relationship. This relationship is given by:
$L(U,V)=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$
where:
• $U\,\!$ is the non-thermal stress (i.e., voltage, vibration, etc.)
• $V\,\!$ is the temperature (in absolute units).
• $B\,\!$, $C\,\!$, $n\,\!$ are the parameters to be determined.
The T-NT relationship can be linearized and plotted on a Life vs. Stress plot. The relationship is linearized by taking the natural logarithm of both sides in the T-NT relationship or:
$\ln (L(V,U))=\ln (C)-n\ln (U)+\frac{B}{V}$
Since life is now a function of two stresses, a Life vs. Stress plot can only be obtained by keeping one of the two stresses constant and varying the other one. Doing so will yield the straight line described by the above equation, where the term for the stress which is kept at a fixed value becomes another constant (in addition to the $\ln (C)\,\!$ constant). When the non-thermal stress is kept constant, then the linearized T-NT relationship becomes:
$\ln (L(V))=const.+\frac{B}{V}$
This is the Arrhenius equation and it is plotted on a log-reciprocal scale. When the thermal stress is kept constant, then the linearized T-NT relationship becomes:
\begin{align} \ln (L(U))=const.-n\ln (U) \end{align}
This is the inverse power law equation and it is plotted on a log-log scale. In the next two figures, data obtained from a temperature and voltage test were analyzed and plotted on a log-reciprocal scale. In the first figure, life is plotted versus temperature, with voltage held at a fixed value. In the second figure, life is plotted versus voltage, with temperature held at a fixed value.
### A look at the Parameters B and n
Depending on which stress type is kept constant, it can be seen from the linearized T-NT relationship that either the parameter $B\,\!$ or the parameter $n\,\!$ is the slope of the resulting line. If, for example, the non-thermal stress is kept constant then $B\,\!$ is the slope of the life line in a Life vs. Temperature plot. The steeper the slope, the greater the dependency of the product's life to the temperature. In other words, $B\,\!$ is a measure of the effect that temperature has on the life and $n\,\!$ is a measure of the effect that the non-thermal stress has on the life. The larger the value of $B,\,\!$ the higher the dependency of the life on the temperature. Similarly, the larger the value of $n,\,\!$ the higher the dependency of the life on the non-thermal stress.
### Acceleration Factor
The acceleration factor for the T-NT relationship is given by:
${{A}_{F}}=\frac{{{L}_{USE}}}{{{L}_{Accelerated}}}=\frac{\tfrac{C}{U_{u}^{n}}{{e}^{\tfrac{B}{{{V}_{u}}}}}}{\tfrac{C}{U_{A}^{n}}{{e}^{\tfrac{B}{{{V}_{A}}}}}}={{\left( \frac{{{U}_{A}}}{{{U}_{u}}} \right)}^{n}}{{e}^{B\left( \tfrac{1}{{{V}_{u}}}-\tfrac{1}{{{V}_{A}}} \right)}}$
where:
• ${{L}_{USE}}\,\!$ is the life at use stress level.
• ${{L}_{Accelerated}}\,\!$ is the life at the accelerated stress level.
• ${{V}_{u}}\,\!$ is the use temperature level.
• ${{V}_{A}}\,\!$ is the accelerated temperature level.
• ${{U}_{A}}\,\!$ is the accelerated non-thermal level.
• ${{U}_{u}}\,\!$ is the use non-thermal level.
The acceleration factor is plotted versus stress in the same manner used to create the Life vs. Stress plots. That is, one stress type is kept constant and the other is varied.
\begin{align} & \overline{T}= & \int\limits_{0}^{\infty }t\cdot f(t,U,V)dt = & \int\limits_{0}^{\infty }t\cdot \frac{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}{{e}^{-\tfrac{t\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}dt = & \frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}} \end{align}
# T-NT Exponential
By setting $m=L(U,V)\,\!$, the exponential pdf becomes:
$f(t,U,V)=\frac{{{U}^{n}}}{C}{{e}^{-\tfrac{B}{V}}}\cdot {{e}^{-\tfrac{{{U}^{n}}}{C}\left( {{e}^{-\tfrac{B}{V}}} \right)t}}$
## T-NT Exponential Statistical Properties Summary
### Mean or MTTF
The mean, $\overline{T},\,\!$ or Mean Time To Failure (MTTF) for the T-NT exponential model is given by:
\begin{align} & \overline{T}= & \int\limits_{0}^{\infty }t\cdot f(t,U,V)dt = & \int\limits_{0}^{\infty }t\cdot \frac{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}{{e}^{-\tfrac{t\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}dt = & \frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}} \end{align}
### Median
The median, $\breve{T},\,\!$ for the T-NT exponential model is given by:
$\breve{T}=\frac{1}{\lambda }0.693=0.693\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$
### Mode
The mode, $\tilde{T},\,\!$ for the T-NT exponential model is given by:
$\tilde{T}=0$
### Standard Deviation
The standard deviation, ${{\sigma }_{T}}\,\!$ , for the T-NT exponential model is given by:
${{\sigma }_{T}}=\frac{1}{\lambda }=m=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}$
### T-NT Exponential Reliability Function
The T-NT exponential reliability function is given by:
$R(T,U,V)={{e}^{-\tfrac{T\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$
This function is the complement of the T-NT exponential cumulative distribution function or:
$R(T,U,V)=1-Q(T,U,V)=1-\int_{0}^{T}f(T)dT$
and,
$R(T,U,V)=1-\int_{0}^{T}\frac{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}{{e}^{-\tfrac{T\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}dT={{e}^{-\tfrac{T\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$
### Conditional Reliability
The conditional reliability function for the T-NT exponential model is given by,
$R(T,t,U,V)=\frac{R(T+t,U,V)}{R(T,U,V)}=\frac{{{e}^{-\lambda (T+t)}}}{{{e}^{-\lambda T}}}={{e}^{-\tfrac{t\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$
#### Reliable Life
For the T-NT exponential model, the reliable life, or the mission duration for a desired reliability goal, ${{t}_{R}}\,\!$ , is given by:
$R({{t}_{R}},U,V)={{e}^{-\tfrac{{{t}_{R}}\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$
$\ln [R({{t}_{R}},U,V)]{{=}^{-\tfrac{{{t}_{R}}\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}}}$
or:
${{t}_{R}}=-\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}\ln [R({{t}_{R}},U,V)]$
## Parameter Estimation
### Maximum Likelihood Estimation Method
Substituting the T-NT relationship into the exponential log-likelihood equation yields:
\begin{align} & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{U_{i}^{n}}{C}{{e}^{-\tfrac{B}{{{V}_{i}}}}}\cdot {{e}^{-\tfrac{U_{i}^{n}}{C}\left( {{e}^{-\tfrac{B}{{{V}_{i}}}}} \right){{T}_{i}}}} \right] -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{U_{i}^{n}}{C}\left( {{e}^{-\tfrac{B}{{{V}_{i}}}}} \right)T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }] \end{align}
where:
$R_{Li}^{\prime \prime }={{e}^{-\tfrac{T_{Li}^{\prime \prime }}{C}U_{i}^{\prime \prime n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}}$
$R_{Ri}^{\prime \prime }={{e}^{-\tfrac{T_{Ri}^{\prime \prime }}{C}U_{i}^{\prime \prime n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}}$
and:
• ${{F}_{e}}\,\!$ is the number of groups of exact times-to-failure data points.
• ${{N}_{i}}\,\!$ is the number of times-to-failure data points in the ${{i}^{th}}\,\!$ time-to-failure data group.
• $B\,\!$ is the T-NT parameter (unknown, the first of three parameters to be estimated).
• $C\,\!$ is the second T-NT parameter (unknown, the second of three parameters to be estimated).
• $n\,\!$ is the third T-NT parameter (unknown, the third of three parameters to be estimated).
• ${{V}_{i}}\,\!$ is the temperature level of the ${{i}^{th}}\,\!$ group.
• ${{U}_{i}}\,\!$ is the non-thermal stress level of the ${{i}^{th}}\,\!$ group.
• ${{T}_{i}}\,\!$ is the exact failure time of the ${{i}^{th}}\,\!$ group.
• $S\,\!$ is the number of groups of suspension data points.
• $N_{i}^{\prime }\,\!$ is the number of suspensions in the ${{i}^{th}}\,\!$ group of suspension data points.
• $T_{i}^{\prime }\,\!$ is the running time of the ${{i}^{th}}\,\!$ suspension data group.
• $FI\,\!$ is the number of interval data groups.
• $N_{i}^{\prime \prime }\,\!$ is the number of intervals in the ${{i}^{th}}\,\!$ group of data intervals.
• $T_{Li}^{\prime \prime }\,\!$ is the beginning of the ${{i}^{th}}\,\!$ interval.
• $T_{Ri}^{\prime \prime }\,\!$ is the ending of the ${{i}^{th}}\,\!$ interval.
The solution (parameter estimates) will be found by solving for the parameters $B,\,\!$ $C\,\!$ and $n\,\!$ so that $\tfrac{\partial \Lambda }{\partial B}=0,\,\!$ $\tfrac{\partial \Lambda }{\partial C}=0\,\!$ and $\tfrac{\partial \Lambda }{\partial n}=0\,\!$ .
# T-NT Weibull
By setting η = L(U,V), the T-NT Weibull model is given by:
$f(t,U,V)=\frac{\beta {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}{{\left( \frac{t\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{t\cdot {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }}}}$
## T-NT Weibull Statistical Properties Summary
### Mean or MTTF
The mean, $\overline{T}$ , for the T-NT Weibull model is given by:
$\overline{T}=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}\cdot \Gamma \left( \frac{1}{\beta }+1 \right)$
where $\Gamma \left( \tfrac{1}{\beta }+1 \right)$ is the gamma function evaluated at the value of $\left( \tfrac{1}{\beta }+1 \right)$ .
### Median
The median, $\breve{T},$ for the T-NT Weibull model is given by:
$\breve{T}=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{{\left( \ln 2 \right)}^{\tfrac{1}{\beta }}}$
### Mode
The mode, $\tilde{T},$ for the T-NT Weibull model is given by:
$\tilde{T}=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{{\left( 1-\frac{1}{\beta } \right)}^{\tfrac{1}{\beta }}}$
### Standard Deviation
The standard deviation, σT, for the T-NT Weibull model is given by:
${{\sigma }_{T}}=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}\cdot \sqrt{\Gamma \left( \frac{2}{\beta }+1 \right)-{{\left( \Gamma \left( \frac{1}{\beta }+1 \right) \right)}^{2}}}$
### T-NT Weibull Reliability Function
The T-NT Weibull reliability function is given by:
$R(T,U,V)={{e}^{-{{\left( \tfrac{T{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }}}}$
### Conditional Reliability Function
The T-NT Weibull conditional reliability function at a specified stress level is given by:
$R(T,t,U,V)=\frac{R(T+t,U,V)}{R(T,U,V)}=\frac{{{e}^{-{{\left( \tfrac{\left( T+t \right){{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }}}}}{{{e}^{-{{\left( \tfrac{T{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }}}}}$
or:
$R(T,t,U,V)={{e}^{-\left[ {{\left( \tfrac{\left( T+t \right){{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }}-{{\left( \tfrac{T{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta }} \right]}}$
### Reliable Life
For the T-NT Weibull model, the reliable life, TR , of a unit for a specified reliability and starting the mission at age zero is given by:
${{T}_{R}}=\frac{C}{{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{{\left\{ -\ln \left[ R\left( {{T}_{R}},U,V \right) \right] \right\}}^{\tfrac{1}{\beta }}}$
### T-NT Weibull Failure Rate Function
The T-NT Weibull failure rate function, λ(T) , is given by:
$\lambda \left( T,U,V \right)=\frac{f\left( T,U,V \right)}{R\left( T,U,V \right)}=\frac{\beta {{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C}{{\left( \frac{T{{U}^{n}}{{e}^{-\tfrac{B}{V}}}}{C} \right)}^{\beta -1}}$
## Parameter Estimation
### Maximum Likelihood Estimation Method
Substituting the T-NT relationship into the Weibull log-likelihood function yields:
\begin{align} & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{\beta U_{i}^{n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}{C}{{\left( \frac{U_{i}^{n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}{C}{{T}_{i}} \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{U_{i}^{n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}{C}{{T}_{i}} \right)}^{\beta }}}} \right] -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }{{\left( \frac{U_{i}^{n}{{e}^{-\tfrac{B}{{{V}_{i}}}}}}{C}T_{i}^{\prime } \right)}^{\beta }}+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }] \end{align}
where:
$R_{Li}^{\prime \prime }={{e}^{-{{\left( \tfrac{T_{Li}^{\prime \prime }}{C}U_{i}^{\prime \prime n}{{e}^{-\tfrac{B}{{{V}_{i}}}}} \right)}^{\beta }}}}$
$R_{Ri}^{\prime \prime }={{e}^{-{{\left( \tfrac{T_{Ri}^{\prime \prime }}{C}U_{i}^{\prime \prime n}{{e}^{-\tfrac{B}{{{V}_{i}}}}} \right)}^{\beta }}}}$
and:
• Fe is the number of groups of exact times-to-failure data points.
• Ni is the number of times-to-failure data points in the ith time-to-failure data group.
• β is the Weibull shape parameter (unknown, the first of four parameters to be estimated).
• B is the first T-NT parameter (unknown, the second of four parameters to be estimated).
• C is the second T-NT parameter (unknown, the third of four parameters to be estimated).
• n is the third T-NT parameter (unknown, the fourth of four parameters to be estimated).
• Vi is the temperature level of the ith group.
• Ui is the non-thermal stress level of the ith group.
• Ti is the exact failure time of the ith group.
• S is the number of groups of suspension data points.
• $N_{i}^{\prime }$ is the number of suspensions in the ith group of suspension data points.
• $T_{i}^{\prime }$ is the running time of the ith suspension data group.
• FI is the number of interval data groups.
• $N_{i}^{\prime \prime }$ is the number of intervals in the ith group of data intervals.
• $T_{Li}^{\prime \prime }$ is the beginning of the ith interval.
• $T_{Ri}^{\prime \prime }$ is the ending of the ith interval.
The solution (parameter estimates) will be found by solving for the parameters B, C, n and β so that $\tfrac{\partial \Lambda }{\partial B}=0,$ $\tfrac{\partial \Lambda }{\partial C}=0,$ $\tfrac{\partial \Lambda }{\partial n}=0$ and $\tfrac{\partial \Lambda }{\partial \beta }=0$ .
# T-NT Lognormal
The pdf of the lognormal distribution is given by:
$f(T)=\frac{1}{T\text{ }{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\overline{{{T}'}}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}$
where:
\begin{align} {T}'=\ln (T) \end{align}
and:
• T = times-to-failure.
• $\overline{{{T}'}}=$ mean of the natural logarithms of the times-to-failure.
• σT' = standard deviation of the natural logarithms of the times-to-failure.
The median of the lognormal distribution is given by:
$\breve{T}={{e}^{{{\overline{T}}^{\prime }}}}$
The T-NT lognormal model pdf can be obtained by setting $\breve{T}=L(V)$. Therefore:
$\breve{T}=L(V)=\frac{C}{{{U}^{n}}}{{e}^{\tfrac{B}{V}}}$
or:
${{e}^{{{\overline{T}}^{\prime }}}}=\frac{C}{{{U}^{n}}}{{e}^{\tfrac{B}{V}}}$
Thus:
${{\overline{T}}^{\prime }}=\ln (C)-n\ln (U)+\frac{B}{V}$
Substituting the above equation into the lognormal pdf yields the T-NT lognormal model pdf or:
$f(T,U,V)=\frac{1}{T\text{ }{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\ln (C)+n\ln (U)-\tfrac{B}{V}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}$
## T-N-T Lognormal Statistical Properties Summary
### The Mean
The mean life of the T-NT lognormal model (mean of the times-to-failure), $\bar{T}$ , is given by:
\begin{align} & \bar{T}= & {{e}^{\bar{{T}'}+\tfrac{1}{2}\sigma _{{{T}'}}^{2}}} = & {{e}^{\ln (C)-n\ln (U)+\tfrac{B}{V}+\tfrac{1}{2}\sigma _{{{T}'}}^{2}}} \end{align}
The mean of the natural logarithms of the times-to-failure, ${{\bar{T}}^{^{\prime }}}$ , in terms of $\bar{T}$ and σT is given by:
${{\bar{T}}^{\prime }}=\ln \left( {\bar{T}} \right)-\frac{1}{2}\ln \left( \frac{\sigma _{T}^{2}}{{{{\bar{T}}}^{2}}}+1 \right)$
### The Standard Deviation
The standard deviation of the T-NT lognormal model (standard deviation of the times-to-failure), σT , is given by:
\begin{align} & {{\sigma }_{T}}= & \sqrt{\left( {{e}^{2\bar{{T}'}+\sigma _{{{T}'}}^{2}}} \right)\left( {{e}^{\sigma _{{{T}'}}^{2}}}-1 \right)} = & \sqrt{\left( {{e}^{2\left( \ln (C)-n\ln (U)+\tfrac{B}{V} \right)+\sigma _{{{T}'}}^{2}}} \right)\left( {{e}^{\sigma _{{{T}'}}^{2}}}-1 \right)} \end{align}
The standard deviation of the natural logarithms of the times-to-failure, σT' , in terms of $\bar{T}$ and σT is given by:
${{\sigma }_{{{T}'}}}=\sqrt{\ln \left( \frac{\sigma _{T}^{2}}{{{{\bar{T}}}^{2}}}+1 \right)}$
### The Mode
The mode of the T-NT lognormal model is given by:
\begin{align} & \tilde{T}= & {{e}^{{{\overline{T}}^{\prime }}-\sigma _{{{T}'}}^{2}}} = & {{e}^{\ln (C)-n\ln (U)+\tfrac{B}{V}-\sigma _{{{T}'}}^{2}}} \end{align}
### T-NT Lognormal Reliability
For the T-NT lognormal model, the reliability for a mission of time T , starting at age 0, for the T-NT lognormal model is determined by:
$R(T,U,V)=\int_{T}^{\infty }f(t,U,V)dt$
or:
$R(T,U,V)=\int_{{{T}^{^{\prime }}}}^{\infty }\frac{1}{{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\ln (C)+n\ln (U)-\tfrac{B}{V}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}dt$
### Reliable Life
For the T-NT lognormal model, the reliable life, or the mission duration for a desired reliability goal, tR, is estimated by first solving the reliability equation with respect to time, as follows:
$T_{R}^{\prime }=\ln (C)-n\ln (U)+\frac{B}{V}+z\cdot {{\sigma }_{{{T}'}}}$
where:
$z={{\Phi }^{-1}}\left[ F\left( T_{R}^{\prime },U,V \right) \right]$
and:
$\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z({T}',U,V)}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt$
Since T' = ln(T) the reliable life, tR , is given by:
${{t}_{R}}={{e}^{T_{R}^{\prime }}}$
### Lognormal Failure Rate
The T-NT lognormal failure rate is given by:
$\lambda (T,U,V)=\frac{f(T,U,V)}{R(T,U,V)}=\frac{\tfrac{1}{T\text{ }{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\ln (C)+n\ln (U)-\tfrac{B}{V}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}}{\int_{{{T}'}}^{\infty }\tfrac{1}{{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\ln (C)+n\ln (U)-\tfrac{B}{V}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}dt}$
## Parameter Estimation
### Maximum Likelihood Estimation Method
The complete T-NT lognormal log-likelihood function is:
\begin{align} & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{{{\sigma }_{{{T}'}}}{{T}_{i}}}{{\phi }_{pdf}}\left( \frac{\ln \left( {{T}_{i}} \right)-\ln (C)+n\ln ({{U}_{i}})-\tfrac{B}{{{V}_{i}}}}{{{\sigma }_{{{T}'}}}} \right) \right] \text{ }+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ln \left[ 1-\Phi \left( \frac{\ln \left( T_{i}^{\prime } \right)-\ln (C)+n\ln ({{U}_{i}})-\tfrac{B}{{{V}_{i}}}}{{{\sigma }_{{{T}'}}}} \right) \right] +\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime })] \end{align}
where:
$z_{Ri}^{\prime \prime }=\frac{\ln T_{Ri}^{\prime \prime }-\ln C+n\ln U_{i}^{\prime \prime }-\tfrac{B}{{{V}_{i}}}}{\sigma _{T}^{\prime }}$
$z_{Li}^{\prime \prime }=\frac{\ln T_{Li}^{\prime \prime }-\ln C+n\ln U_{i}^{\prime \prime }-\tfrac{B}{{{V}_{i}}}}{\sigma _{T}^{\prime }}$
$\phi \left( x \right)=\frac{1}{\sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( x \right)}^{2}}}}$
$\Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt$
and:
• Fe is the number of groups of exact times-to-failure data points.
• Ni is the number of times-to-failure data points in the ith time-to-failure data group.
• σT' is the standard deviation of the natural logarithm of the times-to-failure (unknown, the first of four parameters to be estimated).
• B is the first T-NT parameter (unknown, the second of four parameters to be estimated).
• C is the second T-NT parameter (unknown, the third of four parameters to be estimated).
• n is the third T-NT parameter (unknown, the fourth of four parameters to be estimated).
• Vi is the stress level for the first stress type (i.e., temperature) of the ith group.
• Ui is the stress level for the second stress type (i.e., non-thermal) of the ith group.
• Ti is the exact failure time of the ith group.
• S is the number of groups of suspension data points.
• $N_{i}^{\prime }$ is the number of suspensions in the ith group of suspension data points.
• $T_{i}^{\prime }$ is the running time of the ith suspension data group.
• FI is the number of interval data groups.
• $N_{i}^{\prime \prime }$ is the number of intervals in the ith group of data intervals.
• $T_{Li}^{\prime \prime }$ is the beginning of the ith interval.
• $T_{Ri}^{\prime \prime }$ is the ending of the ith interval.
The solution (parameter estimates) will be found by solving for ${{\widehat{\sigma }}_{{{T}'}}},$ $\widehat{B},$ $\widehat{C},$ $\widehat{n}$ so that $\tfrac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}=0,$ $\tfrac{\partial \Lambda }{\partial B}=0,$ $\tfrac{\partial \Lambda }{\partial C}=0$ and $\tfrac{\partial \Lambda }{\partial n}=0$ .
### T-NT Lognormal Example
12 electronic devices were put into a continuous accelerated life test and the following data were collected.
Using ALTA and the T-NT lognormal model, the following parameters were obtained:
\begin{align} \widehat{Std}=\ & 0.182558 \\ \widehat{B}=\ & 3729.650303 \\ \widehat{C}=\ & 0.035292 \\ \widehat{n}=\ & 0.776797 \end{align}\,\!
A probability plot, with the 2-sided 90% confidence bounds for the use stress levels of 323K and 2V, is shown next.
An acceleration factor plot, in which one of the stresses must be kept constant, can also be obtained. For example, in the following plot, the acceleration factor is plotted versus temperature given a constant voltage of 2V.
# T-NT Confidence Bounds
## Approximate Confidence Bounds for the T-NT Exponential
### Confidence Bounds on the Mean Life
The mean life for the T-NT model is given by setting m = L(V) . The upper (mU) and lower (mL) bounds on the mean life (ML estimate of the mean life) are estimated by:
${{m}_{U}}=\widehat{m}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{m})}}{\widehat{m}}}}$
${{m}_{L}}=\widehat{m}\cdot {{e}^{-\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{m})}}{\widehat{m}}}}$
where Kα is defined by:
$\alpha =\frac{1}{\sqrt{2\pi }}\int_{{{K}_{\alpha }}}^{\infty }{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt=1-\Phi ({{K}_{\alpha }})$
If δ is the confidence level, then $\alpha =\tfrac{1-\delta }{2}$ for the two-sided bounds, and α = 1 − δ for the one-sided bounds. The variance of $\widehat{m}$ is given by:
\begin{align} Var(\widehat{m})= & {{\left( \frac{\partial m}{\partial B} \right)}^{2}}Var(\widehat{B})+{{\left( \frac{\partial m}{\partial C} \right)}^{2}}Var(\widehat{C}) +{{\left( \frac{\partial m}{\partial n} \right)}^{2}}Var(\widehat{b}) +2\left( \frac{\partial m}{\partial B} \right)\left( \frac{\partial m}{\partial C} \right)Cov(\widehat{B},\widehat{C}) \\ & +2\left( \frac{\partial m}{\partial B} \right)\left( \frac{\partial m}{\partial n} \right)Cov(\widehat{B},\widehat{n}) +2\left( \frac{\partial m}{\partial C} \right)\left( \frac{\partial m}{\partial n} \right)Cov(\widehat{C},\widehat{n}) \end{align}
or:
\begin{align} Var(\widehat{m})= & \frac{1}{{{U}^{2\widehat{n}}}}{{e}^{2\tfrac{\widehat{B}}{V}}}[\frac{{{\widehat{C}}^{2}}}{{{V}^{2}}}Var(\widehat{B})+Var(\widehat{C}) +{{\widehat{C}}^{2}}{{\left( \ln (U) \right)}^{2}}Var(\widehat{n}) +\frac{2\widehat{C}}{V}Cov(\widehat{B},\widehat{C}) \\ & -\frac{2{{\widehat{C}}^{2}}\ln (U)}{V}Cov(\widehat{B},\widehat{n}) -2\widehat{C}\ln (U)Cov(\widehat{C},\widehat{n})] \end{align}
The variances and covariance of B, C and n are estimated from the local Fisher matrix (evaluated at $\widehat{B},$ $\widehat{C},$ $\widehat{n})$ as follows:
$\left[ \begin{matrix} Var(\widehat{B}) & Cov(\widehat{B},\widehat{C}) & Cov(\widehat{B},\widehat{n}) \\ Cov(\widehat{C},\widehat{B}) & Var(\widehat{C}) & Cov(\widehat{C},\widehat{n}) \\ Cov(\widehat{n},\widehat{B}) & Cov(\widehat{n},\widehat{C}) & Var(\widehat{n}) \\ \end{matrix} \right]={{\left[ F \right]}^{-1}}$
where:
$F=\left[ \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{B}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{C}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{n}^{2}}} \\ \end{matrix} \right].$
### Confidence Bounds on Reliability
The bounds on reliability at a given time, T , are estimated by:
\begin{align} & {{R}_{U}}= & {{e}^{-\tfrac{T}{{{m}_{U}}}}} \\ & {{R}_{L}}= & {{e}^{-\tfrac{T}{{{m}_{L}}}}} \end{align}
### Confidence Bounds on Time
The bounds on time for a given reliability (ML estimate of time) are estimated by first solving the reliability function with respect to time:
$\widehat{T}=-\widehat{m}\cdot \ln (R)$
The corresponding confidence bounds are estimated from:
\begin{align} & {{T}_{U}}= & -{{m}_{U}}\cdot \ln (R) \\ & {{T}_{L}}= & -{{m}_{L}}\cdot \ln (R) \end{align}
## Approximate Confidence Bounds for the T-NT Weibull
### Bounds on the Parameters
Using the same approach as previously discussed ( $\widehat{\beta }$ and $\widehat{C}$ positive parameters):
\begin{align} & {{\beta }_{U}}= & \widehat{\beta }\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{\beta })}}{\widehat{\beta }}}} \\ & {{\beta }_{L}}= & \widehat{\beta }\cdot {{e}^{-\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{\beta })}}{\widehat{\beta }}}} \end{align}
\begin{align} & {{B}_{U}}= & \widehat{B}+{{K}_{\alpha }}\sqrt{Var(\widehat{B})} \\ & {{B}_{L}}= & \widehat{B}-{{K}_{\alpha }}\sqrt{Var(\widehat{A})} \end{align}
\begin{align} & {{C}_{U}}= & \widehat{C}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{C})}}{\widehat{C}}}} \\ & {{C}_{L}}= & \widehat{C}\cdot {{e}^{-\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{C})}}{\widehat{C}}}} \end{align}
and:
\begin{align} & {{n}_{U}}= & \widehat{n}+{{K}_{\alpha }}\sqrt{Var(\widehat{n})} \\ & {{n}_{L}}= & \widehat{n}-{{K}_{\alpha }}\sqrt{Var(\widehat{n})} \end{align}
The variances and covariances of β, B, C, and n are estimated from the Fisher matrix (evaluated at $\widehat{\beta },$ $\widehat{B},$ $\widehat{C},$ $\widehat{n})$ as follows:
$\left[ \begin{matrix} Var(\widehat{\beta }) & Cov(\widehat{\beta },\widehat{B}) & Cov(\widehat{\beta },\widehat{C}) & Cov(\widehat{\beta },\widehat{n}) \\ Cov(\widehat{B},\widehat{\beta }) & Var(\widehat{B}) & Cov(\widehat{B},\widehat{C}) & Cov(\widehat{B},\widehat{n}) \\ Cov(\widehat{C},\widehat{\beta }) & Cov(\widehat{C},\widehat{B}) & Var(\widehat{C}) & Cov(\widehat{C},\widehat{n}) \\ Cov(\widehat{n},\widehat{\beta }) & Cov(\widehat{n},\widehat{B}) & Cov(\widehat{n},\widehat{C}) & Var(\widehat{n}) \\ \end{matrix} \right]={{\left[ F \right]}^{-1}}$
where:
$F=\left[ \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \beta \partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \beta \partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \beta \partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial \beta } & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{B}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial \beta } & -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{C}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial \beta } & -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{n}^{2}}} \\ \end{matrix} \right]$
### Confidence Bounds on Reliability
The reliability function (ML estimate) for the T-NT Weibull model is given by:
$\widehat{R}(T,U,V)={{e}^{-{{\left( \tfrac{{{U}^{\widehat{n}}}{{e}^{-\tfrac{\widehat{B}}{V}}}}{\widehat{C}}T \right)}^{\widehat{\beta }}}}}$
or:
$\widehat{R}(T,U,V)={{e}^{-{{e}^{\ln \left[ {{\left( \tfrac{{{U}^{\widehat{n}}}{{e}^{-\tfrac{\widehat{B}}{V}}}}{\widehat{C}}T \right)}^{\widehat{\beta }}} \right]}}}}$
Setting:
$\widehat{u}=\ln \left[ {{\left( \frac{{{U}^{\widehat{n}}}{{e}^{-\tfrac{\widehat{B}}{V}}}}{\widehat{C}}T \right)}^{\widehat{\beta }}} \right]$
or:
$\widehat{u}=\widehat{\beta }\left[ \ln (T)-\frac{\widehat{B}}{V}-\ln (\widehat{C})+\widehat{n}\ln (U) \right]$
The reliability function now becomes:
$\widehat{R}(T,U,V)={{e}^{-e\widehat{^{u}}}}$
The next step is to find the upper and lower bounds on u :
${{u}_{U}}=\widehat{u}+{{K}_{\alpha }}\sqrt{Var(\widehat{u})}$
${{u}_{L}}=\widehat{u}-{{K}_{\alpha }}\sqrt{Var(\widehat{u})}$
where:
\begin{align} Var(\widehat{u})= & {{\left( \frac{\partial \widehat{u}}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial \widehat{u}}{\partial B} \right)}^{2}}Var(\widehat{B}) +{{\left( \frac{\partial \widehat{u}}{\partial C} \right)}^{2}}Var(\widehat{C})+{{\left( \frac{\partial \widehat{u}}{\partial n} \right)}^{2}}Var(\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{\beta },\widehat{B}) +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{\beta },\widehat{C}) \\ & +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{\beta },\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial B} \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{B},\widehat{C}) +2\left( \frac{\partial \widehat{u}}{\partial B} \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{B},\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial C} \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{C},\widehat{n}) \end{align}
or:
\begin{align} Var(\widehat{u})= & {{\left( \frac{\widehat{u}}{\widehat{\beta }} \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\widehat{\beta }}{V} \right)}^{2}}Var(\widehat{B}) +{{\left( \frac{\widehat{\beta }}{\widehat{C}} \right)}^{2}}Var(\widehat{C})+{{\left( \widehat{\beta }\ln (U) \right)}^{2}}Var(\widehat{n}) -\frac{2\widehat{u}}{V}Cov(\widehat{\beta },\widehat{B})-\frac{2\widehat{u}}{\widehat{C}}Cov(\widehat{\beta },\widehat{C}) \\ & +2\widehat{u}\ln (U)Cov(\widehat{\beta },\widehat{n}) +\frac{2{{\widehat{\beta }}^{2}}}{\widehat{C}V}Cov(\widehat{B},\widehat{C})-\frac{2{{\widehat{\beta }}^{2}}\ln (U)}{V}Cov(\widehat{B},\widehat{n}) -\frac{2{{\widehat{\beta }}^{2}}\ln (U)}{\widehat{C}}Cov(\widehat{C},\widehat{n}) \end{align}
The upper and lower bounds on reliability are:
\begin{align} & {{R}_{U}}= & {{e}^{-{{e}^{\left( {{u}_{L}} \right)}}}} \\ & {{R}_{L}}= & {{e}^{-{{e}^{\left( {{u}_{U}} \right)}}}} \end{align}
### Confidence Bounds on Time
The bounds on time (ML estimate of time) for a given reliability are estimated by first solving the reliability function with respect to time as follows:
\begin{align} \ln (R)=\ & -{{\left( \frac{{{U}^{\widehat{n}}}{{e}^{-\tfrac{\widehat{B}}{V}}}}{\widehat{C}}\widehat{T} \right)}^{\widehat{\beta }}} \\ \ln (-\ln (R))=\ & \widehat{\beta }\left( \ln (\widehat{T})-\frac{\widehat{B}}{V}-\ln (\widehat{C})+\widehat{n}\ln (U) \right) \end{align}
or:
$\widehat{u}=\frac{1}{\widehat{\beta }}\ln (-\ln (R))+\frac{\widehat{B}}{V}+\ln (\widehat{C})-\widehat{n}\ln (U)$
where $\widehat{u}=\ln \widehat{T}.$
The upper and lower bounds on u are estimated from:
${{u}_{U}}=\widehat{u}+{{K}_{\alpha }}\sqrt{Var(\widehat{u})}$
${{u}_{L}}=\widehat{u}-{{K}_{\alpha }}\sqrt{Var(\widehat{u})}$
where:
\begin{align} Var(\widehat{u})= & {{\left( \frac{\partial \widehat{u}}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial \widehat{u}}{\partial B} \right)}^{2}}Var(\widehat{B}) +{{\left( \frac{\partial \widehat{u}}{\partial C} \right)}^{2}}Var(\widehat{C})+{{\left( \frac{\partial \widehat{u}}{\partial n} \right)}^{2}}Var(\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{\beta },\widehat{B}) +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{\beta },\widehat{C}) \\ & +2\left( \frac{\partial \widehat{u}}{\partial \beta } \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{\beta },\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial B} \right)\left( \frac{\partial \widehat{u}}{\partial C} \right)Cov(\widehat{B},\widehat{C}) +2\left( \frac{\partial \widehat{u}}{\partial B} \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{B},\widehat{n}) +2\left( \frac{\partial \widehat{u}}{\partial C} \right)\left( \frac{\partial \widehat{u}}{\partial n} \right)Cov(\widehat{C},\widehat{n}) \end{align}
or:
\begin{align} Var(\widehat{u})= & \frac{1}{{{\widehat{\beta }}^{4}}}{{\left[ \ln (-\ln (R)) \right]}^{2}}Var(\widehat{\beta }) +\frac{1}{{{V}^{2}}}Var(\widehat{B})+\frac{1}{{{\widehat{C}}^{2}}}Var(\widehat{C})+{{\left[ \ln (U) \right]}^{2}}Var(\widehat{n}) -\frac{2\ln (-\ln (R))}{{{\widehat{\beta }}^{2}}V}Cov(\widehat{\beta },\widehat{B}) -\frac{2\ln (-\ln (R))}{{{\widehat{\beta }}^{2}}\widehat{C}}Cov(\widehat{\beta },\widehat{C}) \\ & +\frac{2\ln (-\ln (R))\ln (U)}{{{\widehat{\beta }}^{2}}}Cov(\widehat{\beta },\widehat{n}) +\frac{2}{\widehat{C}V}Cov(\widehat{B},\widehat{C}) -\frac{2\ln (U)}{V}Cov(\widehat{B},\widehat{n})-\frac{2\ln (U)}{\widehat{C}}Cov(\widehat{C},\widehat{n}) \end{align}
The upper and lower bounds on time are then found by:
\begin{align} & {{T}_{U}}= & {{e}^{{{u}_{U}}}} \\ & {{T}_{L}}= & {{e}^{{{u}_{L}}}} \end{align}
## Approximate Confidence Bounds for the T-NT Lognormal
### Bounds on the Parameters
Since the standard deviation, ${{\widehat{\sigma }}_{{{T}'}}}$ , and $\widehat{C}$ are positive parameters, $\ln ({{\widehat{\sigma }}_{{{T}'}}})$ and $\ln (\widehat{C})$ are treated as normally distributed and the bounds are estimated from:
\begin{align} {{\sigma }_{U}}=\ & {{\widehat{\sigma }}_{{{T}'}}}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{{{T}'}}})}}{{{\widehat{\sigma }}_{{{T}'}}}}}} &\text{ (Upper bound)} \\ & & \\ {{\sigma }_{L}}=\ & \frac{{{\widehat{\sigma }}_{{{T}'}}}}{{{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{{{T}'}}})}}{{{\widehat{\sigma }}_{{{T}'}}}}}}} &\text{ (Lower bound)} \end{align}
and:
\begin{align} {{C}_{U}}= & \widehat{C}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{C})}}{\widehat{C}}}} &\text{ (Upper bound)} \\ & & \\ {{C}_{L}}= & \frac{\widehat{A}}{{{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var(\widehat{C})}}{\widehat{C}}}}} &\text{ (Lower bound)} \end{align}
The lower and upper bounds on B and n are estimated from:
\begin{align} & {{B}_{U}}= & \widehat{B}+{{K}_{\alpha }}\sqrt{Var(\widehat{B})}\text{ (Upper bound)} \\ & & \\ & {{B}_{L}}= & \widehat{B}-{{K}_{\alpha }}\sqrt{Var(\widehat{B})}\text{ (Lower bound)} \end{align}
and:
\begin{align} & {{n}_{U}}= & \widehat{n}+{{K}_{\alpha }}\sqrt{Var(\widehat{n})}\text{ (Upper bound)} \\ & & \\ & {{n}_{L}}= & \widehat{n}-{{K}_{\alpha }}\sqrt{Var(\widehat{n})}\text{ (Lower bound)} \end{align}
The variances and covariances of B , C, n, and σT' are estimated from the local Fisher matrix (evaluated at $\widehat{B},$ $\widehat{C},$ $\widehat{n}$ , ${{\widehat{\sigma }}_{{{T}'}}})$ as follows:
$\left( \begin{matrix} Var\left( {{\widehat{\sigma }}_{{{T}'}}} \right) & Cov\left( \widehat{B},{{\widehat{\sigma }}_{{{T}'}}} \right) & Cov\left( \widehat{C},{{\widehat{\sigma }}_{{{T}'}}} \right) & Cov\left( \widehat{n},{{\widehat{\sigma }}_{{{T}'}}} \right) \\ Cov\left( {{\widehat{\sigma }}_{{{T}'}}},\widehat{B} \right) & Var\left( \widehat{B} \right) & Cov\left( \widehat{B},\widehat{C} \right) & Cov\left( \widehat{B},\widehat{n} \right) \\ Cov\left( {{\widehat{\sigma }}_{{{T}'}}},\widehat{C} \right) & Cov\left( \widehat{C},\widehat{B} \right) & Var\left( \widehat{C} \right) & Cov\left( \widehat{C},\widehat{n} \right) \\ Cov\left( \widehat{n},{{\widehat{\sigma }}_{{{T}'}}} \right) & Cov\left( \widehat{n},\widehat{B} \right) & Cov\left( \widehat{n},\widehat{C} \right) & Var\left( \widehat{n} \right) \\ \end{matrix} \right)={{\left[ F \right]}^{-1}}$
where:
$F=\left( \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial \sigma _{{{T}'}}^{2}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\sigma }_{{{T}'}}}\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\sigma }_{{{T}'}}}\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\sigma }_{{{T}'}}}\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial {{\sigma }_{{{T}'}}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{B}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial B\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial {{\sigma }_{{{T}'}}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{C}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial C\partial n} \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial {{\sigma }_{{{T}'}}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial B} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial n\partial C} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{n}^{2}}} \\ \end{matrix} \right)$
### Bounds on Reliability
The reliability of the lognormal distribution is given by:
$R({T}',U,V;B,C,n,{{\sigma }_{{{T}'}}})=\int_{{{T}'}}^{\infty }\frac{1}{{{\widehat{\sigma }}_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\ln (\widehat{C})+\widehat{n}\ln ({{U}_{i}})-\tfrac{\widehat{B}}{{{V}_{i}}}}{{{\widehat{\sigma }}_{{{T}'}}}} \right)}^{2}}}}dt$
Let $\widehat{z}(t,U,V;B,C,n,{{\sigma }_{T}})=\tfrac{t-\ln (\widehat{C})+\widehat{n}\ln (U)-\tfrac{\widehat{B}}{V}}{{{\widehat{\sigma }}_{{{T}'}}}},$ then $\tfrac{d\widehat{z}}{dt}=\tfrac{1}{{{\widehat{\sigma }}_{{{T}'}}}}.$ For t = T' , $\widehat{z}=\tfrac{{T}'-\ln (\widehat{C})+\widehat{n}\ln (U)-\tfrac{\widehat{B}}{V}}{{{\widehat{\sigma }}_{{{T}'}}}}$ , and for $t=\infty ,$ $\widehat{z}=\infty .$
The above equation then becomes:
$R(\widehat{z})=\int_{\widehat{z}({T}',U,V)}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz$
The bounds on z are estimated from:
\begin{align} & {{z}_{U}}= & \widehat{z}+{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \\ & {{z}_{L}}= & \widehat{z}-{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \end{align}
where:
\begin{align} Var(\widehat{z})= & \left( \frac{\partial \widehat{z}}{\partial B} \right)_{\widehat{B}}^{2}Var(\widehat{B})+\left( \frac{\partial \widehat{z}}{\partial C} \right)_{\widehat{C}}^{2}Var(\widehat{C}) +\left( \frac{\partial \widehat{z}}{\partial n} \right)_{\widehat{b}}^{2}Var(\widehat{n})+\left( \frac{\partial \widehat{z}}{\partial {{\sigma }_{{{T}'}}}} \right)_{{{\widehat{\sigma }}_{{{T}'}}}}^{2}Var({{\widehat{\sigma }}_{{{T}'}}}) +2{{\left( \frac{\partial \widehat{z}}{\partial B} \right)}_{\widehat{B}}}{{\left( \frac{\partial \widehat{z}}{\partial C} \right)}_{\widehat{C}}}Cov\left( \widehat{B},\widehat{C} \right) \\ & +2{{\left( \frac{\partial \widehat{z}}{\partial B} \right)}_{\widehat{B}}}{{\left( \frac{\partial \widehat{z}}{\partial b} \right)}_{\widehat{n}}}Cov\left( \widehat{B},\widehat{n} \right) +2{{\left( \frac{\partial \widehat{z}}{\partial C} \right)}_{\widehat{C}}}{{\left( \frac{\partial \widehat{z}}{\partial n} \right)}_{\widehat{n}}}Cov\left( \widehat{C},\widehat{n} \right) +2{{\left( \frac{\partial \widehat{z}}{\partial B} \right)}_{\widehat{B}}}{{\left( \frac{\partial \widehat{z}}{\partial {{\sigma }_{{{T}'}}}} \right)}_{{{\widehat{\sigma }}_{{{T}'}}}}}Cov\left( \widehat{B},{{\widehat{\sigma }}_{{{T}'}}} \right) \\ & +2{{\left( \frac{\partial \widehat{z}}{\partial C} \right)}_{\widehat{C}}}{{\left( \frac{\partial \widehat{z}}{\partial {{\sigma }_{{{T}'}}}} \right)}_{{{\widehat{\sigma }}_{{{T}'}}}}}Cov\left( \widehat{C},{{\widehat{\sigma }}_{{{T}'}}} \right) +2{{\left( \frac{\partial \widehat{z}}{\partial n} \right)}_{\widehat{n}}}{{\left( \frac{\partial \widehat{z}}{\partial {{\sigma }_{{{T}'}}}} \right)}_{{{\widehat{\sigma }}_{{{T}'}}}}}Cov\left( \widehat{n},{{\widehat{\sigma }}_{{{T}'}}} \right) \end{align}
or:
\begin{align} Var(\widehat{z})= & \frac{1}{\widehat{\sigma }_{{{T}'}}^{2}}[\frac{1}{{{V}^{2}}}Var(\widehat{B})+\frac{1}{{{C}^{2}}}Var(\widehat{C})+\ln {{(U)}^{2}}Var(\widehat{n})+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{{{T}'}}}) +\frac{2}{C\cdot V}Cov\left( \widehat{B},\widehat{C} \right)-\frac{2\ln (U)}{V}Cov\left( \widehat{B},\widehat{n} \right) \\ & -\frac{2\ln (U)}{C}Cov\left( \widehat{C},\widehat{n} \right)+\frac{2\widehat{z}}{V}Cov\left( \widehat{B},{{\widehat{\sigma }}_{{{T}'}}} \right) +\frac{2\widehat{z}}{C}Cov\left( \widehat{C},{{\widehat{\sigma }}_{{{T}'}}} \right)-2\widehat{z}\ln (U)Cov\left( \widehat{n},{{\widehat{\sigma }}_{{{T}'}}} \right)] \end{align}
The upper and lower bounds on reliability are:
\begin{align} & {{R}_{U}}= & \int_{{{z}_{L}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (Upper bound)} \\ & {{R}_{L}}= & \int_{{{z}_{U}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (Lower bound)} \end{align}
### Confidence Bounds on Time
The bounds around time for a given lognormal percentile (unreliability) are estimated by first solving the reliability equation with respect to time, as follows:
${T}'(U,V;\widehat{B},\widehat{C},\widehat{n},{{\widehat{\sigma }}_{{{T}'}}})=\ln (\widehat{C})+\widehat{n}\ln (U)-\frac{\widehat{B}}{V}+z\cdot {{\widehat{\sigma }}_{{{T}'}}}$
where:
\begin{align} {T}'(U,V;\widehat{A},\widehat{\phi },\widehat{b},{{\widehat{\sigma }}_{{{T}'}}})=\ & \ln (T) \\ z=\ & {{\Phi }^{-1}}\left[ F({T}') \right] \end{align}
and:
$\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z({T}',U,V)}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz$
The next step is to calculate the variance of ${T}'(U,V;\widehat{B},\widehat{C},\widehat{n},{{\widehat{\sigma }}_{{{T}'}}})$ :
\begin{align} Var({T}')= & {{\left( \frac{\partial {T}'}{\partial B} \right)}^{2}}Var(\widehat{B})+{{\left( \frac{\partial {T}'}{\partial C} \right)}^{2}}Var(\widehat{C}) +{{\left( \frac{\partial {T}'}{\partial n} \right)}^{2}}Var(\widehat{n})+{{\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)}^{2}}Var({{\widehat{\sigma }}_{{{T}'}}}) +2\left( \frac{\partial {T}'}{\partial B} \right)\left( \frac{\partial {T}'}{\partial C} \right)Cov\left( \widehat{B},\widehat{C} \right) \\ & +2\left( \frac{\partial {T}'}{\partial B} \right)\left( \frac{\partial {T}'}{\partial n} \right)Cov\left( \widehat{B},\widehat{n} \right) +2\left( \frac{\partial {T}'}{\partial C} \right)\left( \frac{\partial {T}'}{\partial n} \right)Cov\left( \widehat{C},\widehat{n} \right) +2\left( \frac{\partial {T}'}{\partial B} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)Cov\left( \widehat{B},{{\widehat{\sigma }}_{{{T}'}}} \right) \\ & +2\left( \frac{\partial {T}'}{\partial C} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)Cov\left( \widehat{C},{{\widehat{\sigma }}_{{{T}'}}} \right) +2\left( \frac{\partial {T}'}{\partial n} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)Cov\left( \widehat{n},{{\widehat{\sigma }}_{{{T}'}}} \right) \end{align}
or:
The upper and lower bounds are then found by:
\begin{align} & T_{U}^{\prime }= & \ln {{T}_{U}}={T}'+{{K}_{\alpha }}\sqrt{Var({T}')} \\ & T_{L}^{\prime }= & \ln {{T}_{L}}={T}'-{{K}_{\alpha }}\sqrt{Var({T}')} \end{align}
Solving for ${{T}_{U}}\,\!$ and ${{T}_{L}}\,\!$ yields:
\begin{align} & {{T}_{U}}= & {{e}^{T_{U}^{\prime }}}\text{ (Upper bound)} \\ & {{T}_{L}}= & {{e}^{T_{L}^{\prime }}}\text{ (Lower bound)} \end{align} | 18,955 | 50,160 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 226, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-51 | longest | en | 0.511491 |
https://lesson.moneyinstructor.com/tag/6th-grade/page/5 | 1,695,638,930,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00782.warc.gz | 408,621,617 | 9,414 | ### Money Math: Lessons for Life
Students learn money math. Table of Contents: Foreword Correlations to NCTM Principles and Standards of Mathematics Lesson 1 The Secret to Becoming a Millionaire Lesson 2 Wallpaper Woes Lesson 3 Math and Taxes: A Pair to Count On Lesson 4 Spreading the Budget Suggested Grade Level 4th – 8th Grade Lesson Excerpt: Let’s face it—kids … Continue reading Money Math: Lessons for Life
#### LESSON CATEGORIES
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Categories Money Math,
Students learn about credit scores. Teaching Objectives: Describe the purpose of a credit report. Define the role of credit scores. Explain the importance of a good credit score. Standard 7: The student will identify the procedures and analyze the responsibilities of borrowing money. Suggested Grade Level 6th – 12th Grade Lesson Excerpt: When people apply … Continue reading Your Credit Score
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### Sources of Credit: It Is In Your Interest?
Students learn about sources of credit and calculating interest rates. Teaching Objectives: Identify potential sources of credit. Compare credit sources. Evaluate credit practices. Calculate credit costs. Demonstrate the ability to make good credit choices. Standard 7: The student will identify the procedures and analyze the responsibilities of borrowing money. Suggested Grade Level 6th – 12th … Continue reading Sources of Credit: It Is In Your Interest?
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### Cooking for a Living: Becoming a Chef
Learn what chefs do and how a person can become a chef. Teaching Objectives: Students will be able to list several places where a chef might work. Students will be able to describe the main responsibilities of a head chef. Students will be able to name other staff members that might work in a kitchen … Continue reading Cooking for a Living: Becoming a Chef
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### Protecting the Environment: Becoming an Environmental Engineer
Learn how environmental engineers protect the environment and how a person can become one. Teaching Objectives: Students will be able to list and describe various capacities in which environmental engineers work. Students will understand that environmental engineers design ways to protect the environment. Students will understand that environmental engineers can work for companies or for … Continue reading Protecting the Environment: Becoming an Environmental Engineer
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### Figuring Costs Using Fractions
Students will learn to use fractions to figure out costs. Teaching Objectives: Students will apply fractions to real- life situations involving costs. Students will review multiplying fractions. Students will learn to take a fraction of a cost. Students will apply these skills when problem solving. Suggested Grade Level 5th – 8th Grade Lesson Excerpt: There … Continue reading Figuring Costs Using Fractions
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Categories Money Math, | 661 | 3,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-40 | latest | en | 0.853321 |
https://www.physicsforums.com/threads/gravitational-force-speed-and-period-of-a-satellite.310872/ | 1,674,925,187,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00191.warc.gz | 986,250,733 | 13,752 | # Gravitational force, Speed, and period of a Satellite.
Imuell1
## Homework Statement
A satellite that has a mass of 300 kg moves in a circular orbit 5.00 X 107 m above Earth's surface. (a) What is the gravitational force on the satellite? (b) What is the speed of the satellite? (c) What is the period of the satellite?
## Homework Equations
Fg= GMEm/(r+rE
T2= [4pi2/G(ME+m)]*(r+rE)3
## The Attempt at a Solution
Fg= 38N
T=37 hours
V=9572517 m/h
I think the gravitational force might be right but the speed and period seem to be ridiculously too large. | 166 | 562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-06 | latest | en | 0.88287 |
https://recipes4day.com/how-many-tablespoon-in-4-oz/ | 1,659,956,145,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00443.warc.gz | 459,018,802 | 27,185 | # How many tablespoon in 4 oz?
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How many dry tablespoons is 4 oz?
Dry Measures
3 teaspoons 1 tablespoon 1/2 ounce
5 1/3 tablespoons 1/3 cup 2.6 fluid ounces
8 tablespoons 1/2 cup 4 ounces
12 tablespoons 3/4 cup 6 ounces
32 tablespoons 2 cups
How much is 4 oz in cups or tablespoons?
Dry Measures
3 teaspoons 1 tablespoon 1/2 ounce
5 1/3 tablespoons 1/3 cup 2.6 fluid ounces
8 tablespoons 1/2 cup 4 ounces
12 tablespoons 3/4 cup 6 ounces
32 tablespoons 2 cups
How many dry teaspoons is 4 oz?
How Many Teaspoons are in a Ounce?
Weight in Ounces: Volume in Teaspoons of:
Water All Purpose Flour
2 oz 11.5 tsp 21.75 tsp
3 oz 17.26 tsp 32.62 tsp
4 oz 23.01 tsp 43.49 tsp
Does 2 tablespoons equal 1 oz? There are 2 tablespoons in a fluid ounce, which is why we use this value in the formula above. Fluid ounces and tablespoons are both units used to measure volume.
## How many tablespoon in 4 oz? – Related Asked Question
### How do you convert dry ounces to tablespoons?
To convert an ounce measurement to a tablespoon measurement, divide the weight multiplied by 1.917223 by the density of the ingredient or material. Thus, the weight in tablespoons is equal to the ounces times 1.917223, divided by the density of the ingredient or material.
### How many dry ounces equal a tablespoon?
In this example, 1 oz. corresponds to 28.35 / 10.67 = 2.6 tbsp.
### How do I measure 4 oz?
By referring to the chart, you will see that 4 fluid ounces equals 1/2 cup. How can I measure 3 ounces?
How many cups does 4 oz equal?
Spoon &, Cups Fluid Oz. Metric
1/2 cup 4 ounces 120 mL
3/4 cup 6 ounces 180 mL
1 cup 8 ounces or 1/2 pint 240 mL
1 1/2 cups 12 ounces 350 mL
### How do I measure 4 oz in cups?
How many ounces in a cup
1. 1 cup = 8 fluid ounces, 16 tablespoons.
2. ¾ cup = 6 fluid ounces, 12 tablespoons.
3. ½ cup = 4 fluid ounces, 8 tablespoons.
4. ¼ cup = 2 fluid ounces, 4 tablespoons.
5. 1 cup = 4.5 dry weight ounces (dry weight is measuring flour and dry ingredients)
### How many cups is 4 oz dry?
Dry measures
3 teaspoons 1 tablespoon 1/2 ounce
8 tablespoons 1/2 cup 4 ounces
12 tablespoons 3/4 cup 6 ounces
32 tablespoons 2 cups 16 ounces
64 tablespoons 4 cups 32 ounces
### How much is 1 oz in teaspoons or tablespoons?
There are 6 teaspoons in a fluid ounce, which is why we use this value in the formula above.
### How do you measure dry ounces?
The convention in the US is this: If a dry ingredient is listed in ounces, it’s a unit of weight and should be measured on a scale. If a wet ingredient is listed in ounces, it’s fluid ounces and should be measured in a wet measuring cup.
### How many TSP is an oz?
Teaspoon to Ounce Conversion Table
Volume in Teaspoons: Weight in Ounces of:
Water Cooking Oil
1/4 tsp 0.043466 oz 0.03825 oz
1/3 tsp 0.057954 oz 0.051 oz
1/2 tsp 0.086931 oz 0.0765 oz
### Is a cup 8 oz?
Liquid measuring cups indicate that 1 cup = 8 ounces. But what they really mean is 1 cup of liquid = 8 fluid ounces. For dry measurements, the rules change. Because dry ingredients vary greatly in weight, you can’t rely on the same conversion.
### What’s 1 oz in cups?
1 fluid ounce is equal to 0.12500004 cups, which is the conversion factor from ounces to cups.
### What’s 6 oz in cups?
6 oz = 0.75 cups
You may also be interested to know that 1 oz is 1/8 of a cup.
### Is a dry cup 8 oz?
On average, one dry cup is equal to 6.8 US dry ounces. One cup equals 16 tablespoons equals 8 ounces equals. 5 pounds equals 221.23 grams.
### What number is a 1 oz scoop?
Tiger Chef Stainless steel construction 1 oz. capacity Black Ergo handle Scoop size: #30 Lever Disher (1, Black 1 oz.)
### How do I measure 4 ounces of ground beef?
3 oz portion is similar in size to a deck of cards ▪ 1 oz of cooked meat is similar in size to 3 dice. A 1-inch meatball is about one ounce. 4 oz of raw, lean meat is about 3 ounces after cooking. 3 oz of grilled fish is the size of a checkbook.
### How many cups is 4 oz shredded cheese?
When measuring semi-hard cheeses, such as cheddar, Swiss or mozzarella, by weight, it is generally accepted that 4 ounces yields 1 cup shredded cheese, or in answer your question, yes, 8 ounce of shredded cheese will fit into a 2-cup volume measuring cup.
### How many cups of chocolate chips is 4 oz?
Weight Conversions for Common Baking Ingredients
Ingredient Ounces Grams
1 cup powdered sugar 4 oz. 110 g
1 cup chocolate chips 6 oz. 170 g
1 cup chocolate chunks 5 oz. 140 g
1 cup chopped nuts 4 oz. 110 g
### How many ounces Makes 2 cups?
US cups to ounces chart
Cups (US) Ounces (fluid) Tablespoons (US)
1 cup 8 fl oz 16 tbsp
3/4 cup 6 fl oz 12 tbsp
2/3 cup 5.3 fl oz 10.6 tbsp
1/2 cup 4 fl oz 8 tbsp
### How do you convert fluid ounces to dry ounces?
The answer is: The change of 1 fl oz ( fluid ounce US ) unit for a volume and capacity measure equals = into 0.027 qt dry ( quart dry US ) as per its equivalent volume and capacity unit type measure often used.
### What does 0.5 cups mean?
1/2 (0.5) cup. 113 g. 4 oz. 5/8 (0.625) cup.
### How many tablespoons is 1 cup dry ingredients?
The short answer for how many tablespoons are in a cup is 16 tablespoons. So when you need 4 tablespoons, you can use ¼ cup.
Basic Tablespoon Conversions.
Tablespoons (TBSP) Dry Measurement (Cups) Liquid Measurement (fl. oz.)
16 tablespoons 1 cup 8 fluid ounces
### How many oz of dry pasta is 4 cups?
Long Pastas
Pastas Uncooked Cooked
Fettuccine 8 oz. 4 cups
Linguine 8 oz. 3 3/4 cups
Pappardelle 8 oz. 4 cups
Spaghetti 8 oz. 3 1/2 cups
### Are dry ounces the same as liquid ounces?
In its simplest possible explanation, a fluid ounce (abbreviated as fl. oz.) is used to measure fluids while an ounce (abbreviated as oz.) is for dry measurements. This gives us the idea that a fluid ounce is a measurement of volume while the other is a measurement of weight.
### How do you measure 3 oz?
3 oz = 0.375 cups
You may also be interested to know that 1 oz is 1/8 of a cup.
### How many ounce are in a mile?
How many milliliters in an ounce? 1 fluid ounce is equal to 29.57353193 milliliter, which is the conversion factor from ounces to milliliter.
### How can I measure a tablespoon without a measuring spoon?
Use a measuring cup.
One tablespoon is equal to about one-sixteenth of a dry measuring cup. If the set comes with a one-eighth dry measuring cup, use half of that to approximate one tablespoon.
### Which unit of measure is equivalent to 1 oz?
Differences between the U.S. and Imperial Systems
Unit of Measurement Imperial System Metric Equivalent
1 ounce 1 (fluid) oz. 28.41 mL
1 gill 5 (fluid) oz. 142.07 mL
1 cup Not commonly used
1 pint 20 (fluid) oz. 568.26 mL
### Does 10 tablespoons equal 1 cup?
In ⅔ of a cup, there are 10 tablespoons and 2 teaspoons. There are also 4 and ⅔ fluid ounces, and 151.4 grams.
### How many cups is 1pound?
16 ounces equals one pound or two cups. Another way to look at the equivalent is that one cup weighs eight ounces and therefore two cups equal 16 ounces and this is the same weight of one pound–16 ounces.
### Is 1 cup dry the same as 1 cup liquid?
Technically, yes. They both measure the same amount of volume. There is no liquid to dry measuring cup conversion. 1 cup in a dry measuring cup is the same as 1 cup in a liquid measuring cup.
### How do I measure 1 oz?
To measure ounces, select either a liquid or dry measuring cup. If you’re measuring a liquid, place the measuring cup on a flat surface and pour the liquid into the cup. Be sure to bend down to view the measurements on the side of the cup. Make sure the liquid reaches the right measurement.
### Is 5 oz a cup?
5 oz = 0.625 cups
You may also be interested to know that 1 oz is 1/8 of a cup.
### How much of a cup is 8 oz?
Based on the measurement chart, eight ounces is equal to one cup. Typically you might want to ask if 8 ounces always equals a cup?
### How many cups equals 2 ounces?
2 oz = 0.25 cups
You may also be interested to know that 1 oz is 1/8 of a cup.
### How many cups is 6 oz shredded cheese?
FAQs: Measuring Cheese
For soft or crumbly cheeses, 1 cup is equivalent to 6 ounces. For semi-hard cheeses like cheddar, 1 cup is equivalent to 4 ounces. Finally, for un-grated hard cheeses like parmesan, 1 cup is equivalent to 3 ounces. For smaller amounts use ½ a cup and divide the corresponding ounces by half.
Sharing is caring! | 2,367 | 8,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-33 | latest | en | 0.80965 |
http://mathhelpforum.com/statistics/122185-show-variance-binomial-distribution.html | 1,524,355,469,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00220.warc.gz | 197,500,045 | 11,052 | # Thread: Show that the variance of a binomial distribution
1. ## Show that the variance of a binomial distribution
Show that the variance of a binomial distribution with parameters n and p cannot exceed n/4.
2. 1. What is the variance of a binomial distribution (n,p) ?
2. What is the maximum value of the function $\displaystyle f(x)=x(1-x)$, when $\displaystyle 0\leq x\leq 1$ ?
3. Originally Posted by Moo
1. What is the variance of a binomial distribution (n,p) ?
2. What is the maximum value of the function $\displaystyle f(x)=x(1-x)$, when $\displaystyle 0\leq x\leq 1$ ?
$\displaystyle \sigma^2 = n \theta(1- \theta)$
and max value of $\displaystyle f(x) = x(1-x) = 0$ when we substitute the value of x= 1
$\displaystyle \sigma^2 = n \theta(1- \theta)$
and max value of $\displaystyle f(x) = x(1-x) = 0$ when we substitute the value of x= 1
Maximum value! NOT minimum value. Draw the graph of $\displaystyle y = x(1 - x)$ over the domain $\displaystyle 0 \leq x \leq 1$. (Did you bother to do that before just plucking a random [wrong] answer out of the air?). The maximum value occurs at the turning point. Drawing parabolas and finding turning points is something you should have learned how to do long ago. Do you realise that work previously studied is assumed to be understood?
5. ## Re: Show that the variance of a binomial distribution
I believe that the maximum variance for a binomial distribution occurs when the probability is 1/2 and the minimum variance occurs when probably is 0 or 1. The function of variance is in fact an inverted parabola however the maximal value occurs at 1/2 not 0. A quick check is to plot the variance over different value of p in the range of 0 and 1. A better way is to calculate first derivative = 0 and solve the for p which shows the apex of the parabola --- in this case p is also 1/2.
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# the variance of a binomial distribution cannot exceed n/4
Click on a term to search for related topics. | 538 | 1,988 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-17 | latest | en | 0.818584 |
https://leetcode.ca/2022-11-13-2425-Bitwise-XOR-of-All-Pairings/ | 1,713,440,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00727.warc.gz | 320,750,260 | 7,103 | ##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/2425.html
# 2425. Bitwise XOR of All Pairings
• Difficulty: Medium.
• Related Topics: .
• Similar Questions: .
## Problem
You are given two 0-indexed arrays, nums1 and nums2, consisting of non-negative integers. There exists another array, nums3, which contains the bitwise XOR of all pairings of integers between nums1 and nums2 (every integer in nums1 is paired with every integer in nums2 exactly once).
Return** the bitwise XOR of all integers in **nums3.
Example 1:
Input: nums1 = [2,1,3], nums2 = [10,2,5,0]
Output: 13
Explanation:
A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3].
The bitwise XOR of all these numbers is 13, so we return 13.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 0
Explanation:
All possible pairs of bitwise XORs are nums1[0] ^ nums2[0], nums1[0] ^ nums2[1], nums1[1] ^ nums2[0],
and nums1[1] ^ nums2[1].
Thus, one possible nums3 array is [2,5,1,6].
2 ^ 5 ^ 1 ^ 6 = 0, so we return 0.
Constraints:
• 1 <= nums1.length, nums2.length <= 10^5
• 0 <= nums1[i], nums2[j] <= 10^9
## Solution (Java, C++, Python)
• class Solution {
public int xorAllNums(int[] nums1, int[] nums2) {
int ans = 0;
if (nums2.length % 2 == 1) {
for (int v : nums1) {
ans ^= v;
}
}
if (nums1.length % 2 == 1) {
for (int v : nums2) {
ans ^= v;
}
}
return ans;
}
}
• class Solution {
public:
int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
int ans = 0;
if (nums2.size() % 2 == 1) {
for (int v : nums1) {
ans ^= v;
}
}
if (nums1.size() % 2 == 1) {
for (int v : nums2) {
ans ^= v;
}
}
return ans;
}
};
• class Solution:
def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
ans = 0
if len(nums2) & 1:
for v in nums1:
ans ^= v
if len(nums1) & 1:
for v in nums2:
ans ^= v
return ans
• func xorAllNums(nums1 []int, nums2 []int) int {
ans := 0
if len(nums2)%2 == 1 {
for _, v := range nums1 {
ans ^= v
}
}
if len(nums1)%2 == 1 {
for _, v := range nums2 {
ans ^= v
}
}
return ans
}
• function xorAllNums(nums1: number[], nums2: number[]): number {
let ans = 0;
if (nums2.length % 2 != 0) {
ans ^= nums1.reduce((a, c) => a ^ c, 0);
}
if (nums1.length % 2 != 0) {
ans ^= nums2.reduce((a, c) => a ^ c, 0);
}
return ans;
}
Explain:
nope.
Complexity:
• Time complexity : O(n).
• Space complexity : O(n). | 821 | 2,347 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-18 | latest | en | 0.265624 |
https://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=ri&To=width | 1,686,347,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00524.warc.gz | 214,781,420 | 3,885 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Japanese ri = 3926.79936 length (length) ``` Related Measurements: Try converting from "ri" to agate (typography agate), archin (Russian archin), cable length, caliber (gun barrel caliber), cloth quarter, engineers chain, finger, foot, furlong (surveyors furlong), Greek cubit, Greek fathom, Greek palm, inch, nail (cloth nail), parasang, rod (surveyors rod), Roman mile, sazhen (Russian sazhen), spindle, sun (Japanese sun), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: ri = 39,267,993,600,000 angstrom, 8,588.8 Biblical cubit, 6,183,936 bottom measure, 15,459,840 caliber (gun barrel caliber), 34,355.2 cloth finger, 17,177.6 cloth quarter, 128.83 engineers chain, 2,147.2 fathom, 42.94 football field, 6.09 li (Chinese li), 1,855,180.8 line, 154,598,400 mil, 2.44 mile, 68,710.4 nail (cloth nail), .70676734 nautical league, 5,153.28 pace, 51,532.8 palm, 42.94 soccer field, .29822222 spindle, 21.26 stadium (Roman stadium).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 472 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-23 | latest | en | 0.738701 |
https://gmatclub.com/forum/since-chromosome-damage-may-be-caused-by-viral-infections-84266.html?fl=similar | 1,506,366,108,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693240.90/warc/CC-MAIN-20170925182814-20170925202814-00517.warc.gz | 647,070,143 | 48,049 | It is currently 25 Sep 2017, 12:01
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# Since chromosome damage may be caused by viral infections,
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Since chromosome damage may be caused by viral infections, [#permalink]
### Show Tags
23 Sep 2009, 12:56
1
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Since chromosome damage may be caused by viral infections, medical x-rays, and exposure to sunlight, it is important that the chrosomes of a population to be tested for chemically induced damage be compared with those of a control population.
(A) to be tested for chemically induced damage be compared with
(B) being tested for damage induced chemically are compared with
(C) being tested for chemically induced damage should~be compared to
(D) being tested for chemfEally induced damage are to be compared to
(E) that is to be tested for chemically induced damage are to be comparable with
please some 1 explain this
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### Show Tags
23 Sep 2009, 13:43
sagarsabnis wrote:
Since chromosome damage may be caused by viral infections, medical x-rays, and exposure to sunlight, it is important that the chrosomes of a population to be tested for chemically induced damage be compared with those of a control population.
(A) to be tested for chemically induced damage be compared with
(B) being tested for damage induced chemically are compared with
(C) being tested for chemically induced damage should~be compared to
(D) being tested for chemfEally induced damage are to be compared to
(E) that is to be tested for chemically induced damage are to be comparable with
please some 1 explain this
compare to Vs compare with - compare with wins
C&D are out.
B - being, out
between A & E - I dont see glaring issues (expect for "are", but I'm not sure), however, E looks more wordy...so, I'll go with A
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### Show Tags
24 Sep 2009, 02:08
1
KUDOS
The subjunctive mood introduced by "that the chromosomes..." needs
a verb form "be compared". Also, we have to use "compared with"
here since the objects being compared are similar.
B, C, D and E use incorrect "are compared", "to be compared" etc.
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### Show Tags
25 Sep 2009, 08:50
B,C, D go with "being" which renders them incorrect. E is wordy, so it's A.
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Re: Since chromosome damage may be caused by viral infections, [#permalink]
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Re: Since chromosome damage may be caused by viral infections, [#permalink]
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Re: Since chromosome damage may be caused by viral infections, [#permalink] 03 Jul 2016, 11:54
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# Since chromosome damage may be caused by viral infections,
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,400 | 5,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-39 | latest | en | 0.917496 |
https://www.ibsurvival.com/profile/146054-watermelonprincess/ | 1,561,418,325,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999779.19/warc/CC-MAIN-20190624231501-20190625013501-00525.warc.gz | 801,463,045 | 23,316 | # watermelonprincess
Members
16
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## Profile Information
• Gender
Female
• Exams
May 2015
• Country
United Arab Emirates
1. ## Design and Technology paper's 1 and 2
yeah haha i figured yeah guess what about that stupid sohcahtoa thing... i did like a pass paper question once A LOOONG time ago and some question like that popped up too and I remmbered it was sin so i put like some sin question but ive never been told by my teaacher about questions like that.. my dt teacher aint really good either.. but paper 3 wow cant wait to talk about it because that one.. WOW like it was a one where i just sat and i waited for time to pass. you know for that diving question the first one where it asked about the limitations to the wildlife using the shark thing. i put that it can only go 30m deep. and you get to see like only a few wildlife since u can't go further.. idk it was like some made up answers.. DT has always been like that for me haha
2. ## Design and Technology paper's 1 and 2
I found Paper 2 quite okay... Paper 1 has always been tough for me - (the batch production one was right but thank God I got it right, it was like set of number). What did everyone else put for that tan or sin question with the house thing? It was like tan = x/10 or sin = 10/x there were various answers so tell me what you guys put! Also, what did everyone choose for Section B? I chose the Sharkdiving thing, well that diving thing for inexperienced divers.. if anyone else chose the same as me what did you put for the question where it was like regarding ONE limitation with the sea wild life? Also the maintenance one?
3. ## Math Studies Paper 1
i got 14 bikes for profit. Agree with the rest... Btw for normal distribution did you use the 5% they give you and just do 1-5/100=0.95 and thats the Area you enter into Inv.norm I used the 0.95 because it's the inverse whenever they give you a percentage
4. ## Math Studies Paper 1
Yeah I got like 33% or something like that!
5. ## Math Studies Paper 1
Eh that was okay, I got like a 2 'parabolas' (they weren't perfect) in my graph. I don't know if you got the same.. haha
6. ## Math Studies Paper 1
I'm the opposite and yeah that profit thing was Paper 2 but thank God it's already been 24 hours haha. I couldn't do that one. But the mango, kiwi thing I was able to do it except for like e) and f) I think. Paper 1 was meh.. wish I had more time.
7. ## Math Studies Paper 1
If it was the question about that profit thing, I couldn't do it either haha
8. ## Math Studies Paper 1
How did everyone find paper 1 of Math Studies TZ2? Or TZ1 either ways
9. ## May15 English A: language and literature: May 2015 exams
I said the exactly same thing as you (Britain being isolated). I also added the theme of 'isolation'.
10. ## May15 English A: language and literature: May 2015 exams
HEY! WHICH TEXT DID YOU DO?! TEXT 1 OR 2?! HEY! I'm sorry to hear that! I also did text 1 and I found it great! Found so many stylistic devices, and could say so much the topic. But this was quite possibly because I wrote about this in my EE as well. That's so good! I'm happy for you! I found it really hard, I had to read the text multiple times to understand what he was saying - but I did mention some personification and symbolism (from the image - Britain's the left fishbowl and the other countries were the right). But aah.. I did the Big 5 though so hopefully it'll be fine Thank you I'm sure it wasn't too bad! The image was quite good to talk about. You'll be fine, what other subjects do you have? I have Maths, Design Technology, French B and Geography left. What about you? Fun! I have Physics, Spanish A, Maths and German B left. Done with English and Eco woooo I wish you the best of luck!
11. ## May15 English A: language and literature: May 2015 exams
HEY! WHICH TEXT DID YOU DO?! TEXT 1 OR 2?! HEY! I'm sorry to hear that! I also did text 1 and I found it great! Found so many stylistic devices, and could say so much the topic. But this was quite possibly because I wrote about this in my EE as well. That's so good! I'm happy for you! I found it really hard, I had to read the text multiple times to understand what he was saying - but I did mention some personification and symbolism (from the image - Britain's the left fishbowl and the other countries were the right). But aah.. I did the Big 5 though so hopefully it'll be fine Thank you I'm sure it wasn't too bad! The image was quite good to talk about. You'll be fine, what other subjects do you have? I have Maths, Design Technology, French B and Geography left. What about you?
12. ## May15 English A: language and literature: May 2015 exams
HEY! WHICH TEXT DID YOU DO?! TEXT 1 OR 2?! HEY! I'm sorry to hear that! I also did text 1 and I found it great! Found so many stylistic devices, and could say so much the topic. But this was quite possibly because I wrote about this in my EE as well. That's so good! I'm happy for you! I found it really hard, I had to read the text multiple times to understand what he was saying - but I did mention some personification and symbolism (from the image - Britain's the left fishbowl and the other countries were the right). But aah.. I did the Big 5 though so hopefully it'll be fine
13. ## May15 English A: language and literature: May 2015 exams
Is anyone in English A: Language & Literature SL? I found it extremely hard to analyze text 1 due to the language (the British slang).. if anyone is in SL please reply back so I don't freak out even more! HEY! WHICH TEXT DID YOU DO?! TEXT 1 OR 2?!
14. ## How many studio works are required for Visual Arts HL?
It's definitely 8-12 pieces of work but I have a feeling your teacher is trying to get you on a safe path (it also depends on your quality of work perhaps). My teacher said the exact same thing, HL would have a minimum of 12. I had 17 pieces.
15. ## May15 May 2015 - Exam DISCUSSION Schedule
Really confused, can we discuss it here? It's already 1:50pm and nobody has written anything!
× | 1,529 | 6,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-26 | latest | en | 0.971314 |
https://metanumbers.com/15016 | 1,623,785,610,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00330.warc.gz | 348,999,620 | 7,511 | ## 15016
15,016 (fifteen thousand sixteen) is an even five-digits composite number following 15015 and preceding 15017. In scientific notation, it is written as 1.5016 × 104. The sum of its digits is 13. It has a total of 4 prime factors and 8 positive divisors. There are 7,504 positive integers (up to 15016) that are relatively prime to 15016.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 15 thousand 16 fifteen thousand sixteen
## Notation
Scientific notation 1.5016 × 104 15.016 × 103
## Prime Factorization of 15016
Prime Factorization 23 × 1877
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 3754 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 15,016 is 23 × 1877. Since it has a total of 4 prime factors, 15,016 is a composite number.
## Divisors of 15016
1, 2, 4, 8, 1877, 3754, 7508, 15016
8 divisors
Even divisors 6 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 28170 Sum of all the positive divisors of n s(n) 13154 Sum of the proper positive divisors of n A(n) 3521.25 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 122.54 Returns the nth root of the product of n divisors H(n) 4.26439 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 15,016 can be divided by 8 positive divisors (out of which 6 are even, and 2 are odd). The sum of these divisors (counting 15,016) is 28,170, the average is 3,521,.25.
## Other Arithmetic Functions (n = 15016)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7504 Total number of positive integers not greater than n that are coprime to n λ(n) 1876 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1759 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 7,504 positive integers (less than 15,016) that are coprime with 15,016. And there are approximately 1,759 prime numbers less than or equal to 15,016.
## Divisibility of 15016
m n mod m 2 3 4 5 6 7 8 9 0 1 0 1 4 1 0 4
The number 15,016 is divisible by 2, 4 and 8.
## Classification of 15016
• Refactorable
• Deficient
• Polite
### By Shape (2D, centered)
• Centered Pentagonal
• Centered Heptagonal
## Base conversion (15016)
Base System Value
2 Binary 11101010101000
3 Ternary 202121011
4 Quaternary 3222220
5 Quinary 440031
6 Senary 153304
8 Octal 35250
10 Decimal 15016
12 Duodecimal 8834
20 Vigesimal 1hag
36 Base36 bl4
## Basic calculations (n = 15016)
### Multiplication
n×i
n×2 30032 45048 60064 75080
### Division
ni
n⁄2 7508 5005.33 3754 3003.2
### Exponentiation
ni
n2 225480256 3385811524096 50841345845825536 763433649220916248576
### Nth Root
i√n
2√n 122.54 24.6709 11.0698 6.84401
## 15016 as geometric shapes
### Circle
Diameter 30032 94348.3 7.08367e+08
### Sphere
Volume 1.41825e+13 2.83347e+09 94348.3
### Square
Length = n
Perimeter 60064 2.2548e+08 21235.8
### Cube
Length = n
Surface area 1.35288e+09 3.38581e+12 26008.5
### Equilateral Triangle
Length = n
Perimeter 45048 9.76358e+07 13004.2
### Triangular Pyramid
Length = n
Surface area 3.90543e+08 3.99022e+11 12260.5
## Cryptographic Hash Functions
md5 7f808f3226c4a95302489cfc3778fde0 3a4e69866cbb28c81d17fe637ce2d4b10cc6e9f5 88a743788a4ac320e338e25db4f91407cd46dc13cfe221df10f985f48df1f596 020baf123858996538f333949ef8a5aa0e96ca9e4e512cdf7417dea8900d062d88df57b8281f563aea2930331bf778c9f01f4589fc2911a95aee06e86a231d42 ddee17207ece46b2fc6b043b11bd8d025af6028c | 1,464 | 4,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-25 | latest | en | 0.80349 |
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# Fukushima Now 72,000 Times Hiroshima Radiation (I hope this is alarmist reporting)
page: 2
10
share:
posted on Mar, 24 2011 @ 03:00 AM
Lord, our schools need overhauling.
Half-life is how long it takes for a radioactive isotope to lose HALF of its radioactivity. You can calculate how long it will take to lose its radioactivity, at least to an extremely small amount (because it will always be a little radioactive, just so low you can't measure it). If a half-life is 8 days, after 8 days it will be half as radioactive. If it started at 100 (totally arbitrary measurement), in 8 days it will be 50, 8 more days it will be 25, 8 ,more and it's 12.5, etc.
The isotope does not "become another substance" based on half-life decay.
posted on Mar, 24 2011 @ 04:20 AM
I hope these articles will explain some more to the radiation spill and the cover-up.
What They're Covering Up at Fukushima
this second article is regrettable in german language.
German scientists
Update 23/03/2011 at 18.10 clock: + + + Committee on Radiation Protection warns that meltdown in Fukushima 1 + + + IAEA measurements carried out up to 200 km from Fukushima I: High levels of beta-gamma contamination 16-58 km from the NPP found + + + Values are comparable to Chernobyl + + + IAEA can not rule out that such high values also occur at greater distances + + + More evacuation measures urgently required + + + There is warned against further trivialization of the contamination of the sea + + + We have now a worst-case scenario to do + + +
posted on Mar, 24 2011 @ 11:58 AM
Originally posted by 00nunya00
Lord, our schools need overhauling.
Half-life is how long it takes for a radioactive isotope to lose HALF of its radioactivity. You can calculate how long it will take to lose its radioactivity, at least to an extremely small amount (because it will always be a little radioactive, just so low you can't measure it). If a half-life is 8 days, after 8 days it will be half as radioactive. If it started at 100 (totally arbitrary measurement), in 8 days it will be 50, 8 more days it will be 25, 8 ,more and it's 12.5, etc.
The isotope does not "become another substance" based on half-life decay.
- You've just proved yourself a rude, ignorant fool. -
The substance loses it's radioactivity because it's unstable and breaks down to a less radioactive substance. This breakdown is what causes radioactivity. Radioactivity the energy released from this breakdown.
The parent can break down to a daughter which is a different isotope, or to a daughter which is a completely different element. Either way, a new substance has been formed.
The decay, or loss of energy, results when an atom with one type of nucleus, called the parent radionuclide, transforms to an atom with a nucleus in a different state, or a different nucleus, either of which is named the daughter nuclide. Often the parent and daughter are different chemical elements, and in such cases the decay process results in nuclear transmutation. In an example of this, a carbon-14 atom (the "parent") emits radiation (a beta particle, antineutrino, and a gamma ray) and transforms to a nitrogen-14 atom (the "daughter"). By contrast, there exist two types of radioactive decay processes (gamma decay and internal conversion decay) that do not result in transmutation, but only decrease the energy of an excited nucleus. This results in an atom of the same element as before but with a nucleus in a lower energy state. An example is the nuclear isomer technetium-99m decaying, by the emission of a gamma ray, to an atom of technetium-99.
posted on Mar, 24 2011 @ 12:38 PM
Did you read what you just quoted?
You're wrong. End of story. You were asked "what is half-life?" And you answered "it's apples becoming oranges." FAIL. (And you had to U2U me to repeat your little insults? Reported).
And since you're dependent on Wikipedia, here
Half-life is the period of time it takes for a substance undergoing decay to decrease by half. The name was originally used to describe a characteristic of unstable atoms (radioactive decay), but may apply to any quantity which follows a set-rate decay. The original term, dating to 1907, was "half-life period", which was later shortened to "half-life" in the early 1950s.[1] Half-lives are very often used to describe quantities undergoing exponential decay—for example radioactive decay—where the half-life is constant over the whole life of the decay, and is a characteristic unit (a natural unit of scale) for the exponential decay equation. However, a half-life can also be defined for non-exponential decay processes, although in these cases the half-life varies throughout the decay process. For a general introduction and description of exponential decay, see the article exponential decay. For a general introduction and description of non-exponential decay, see the article rate law.
edit on 24-3-2011 by 00nunya00 because: (no reason given)
posted on Mar, 24 2011 @ 01:10 PM
post removed because the user has no concept of manners
posted on Mar, 24 2011 @ 01:43 PM
What you're talking about is called "transmutation". Not half-life.
I'm not engaging you any more; you're not worth my time. Have a nice day, sweetheart.
posted on Mar, 24 2011 @ 04:00 PM
Originally posted by 00nunya00
What you're talking about is called "transmutation". Not half-life.
I'm not engaging you any more; you're not worth my time. Have a nice day, sweetheart.
Wow, more rude and stupid ignorance.
You still think half-life involves a substance disappearing???
And you're querying other people's educational background?
Any child with a basic understanding of science knows matter and energy do not just disappear.
The half-life of any substance is the time in which half of that substance transmutes into another substance.
This may be a different element, a different isotope of the same element, or the ground form of the element.
The energy given off by the decay into a daughter substance is the energy that is radiation.
Half-life is the time it takes for such transmutation to occur.
Transmutation describes a process by which the nucleus of a radioactive atom undergoes decay into an atom with a different number of protons, until such time as a stable nucleus is produced.
An alpha particle (i.e., a helium nucleus) is released during alpha decay of a radioactive substance. An element with a lower mass is formed. Mass is not conserved. Atomic mass number (or nucleon number, or baryon number) is conserved.
Beta decay (beta negative decay) occurs when a beta (negative) particle is released from the nucleus (i.e., electron). Mass is also not conserved in beta decay. Nucleon number is conserved. In beta decay, the beta particle released originated in the nucleus of the atom, not in the electron orbital. A neutron is lost, and in its place a proton and an electron are formed.
Gamma decay is the release of excess stored energy from the nucleus. No transmutation occurs. However, gamma decay often accompanies alpha and beta negative decay in a decay series. (The series of steps in the transmutations occurring until a stable nucleus results, is called a decay series.) Gamma decay occurs when an excited nucleus (excited by photon or particle bombardment, or it may be a decay product in an excited state) returns to the ground state. An excited nucleus is heavier than the ground state, by a mass equal to the mass/energy equivalent of the energy of the emitted gamma ray.
Each radioactive nuclide emits radioactivity at its characteristic rate, different from that of other nuclides. The rate of radioactive decay is related to the energy change that accompanies the transformation, but it is not a direct relationship. The rate of radioactive emissions of a radioactive nuclide is directly proportional to the amount of radioactive material present. The rate of decay of a radioactive nuclide is measured by its half-life. Half-life is the time required for one half of the atoms in any starting sample of a radioisotope to decay.
Get that yet?
"Half-life is the time required for one half of the atoms in any starting sample of a radioisotope to decay."
Transformation describes that decay.
edit on 24/3/11 by Kailassa because: (no reason given)
posted on Mar, 25 2011 @ 05:22 AM
Sooooo moving on.... has anyone found out if the claim that 'it' is 72,000 times Hiroshima true or rubbish? I saw something pop up on RT as I walked past the TV on their scroller that said China was reporting severely high radiation.. haven't had a chance to look it up yet though.
posted on Mar, 25 2011 @ 01:15 PM
Originally posted by yzzyUK
Sooooo moving on.... has anyone found out if the claim that 'it' is 72,000 times Hiroshima true or rubbish? I saw something pop up on RT as I walked past the TV on their scroller that said China was reporting severely high radiation.. haven't had a chance to look it up yet though.
Any simple quantification of a comparison of complex events is rubbish. There are many different factors to consider.
On the one hand, we don't have ~100.000 people being suddenly blasted to death from Fukushima.
On the other hand, the Hiroshima bomb was an air blast which supposedly, (I'm sceptical,) did not drop nuclear fall-out over Hiroshima.
The Fukushima reactors each have a storage pool above them, right there in the same building, containing hundreds of spent fuel assemblies. Spent fuel from a reactor is even more deadly than fresh fuel as the fission process creates plutonium from uranium, so each spent uranium rod contains ~1% plutonium. Plutonium is thousands times more dangerous than uranium. To make things worse, at least one plant, No. 3, had MOX rods, made with 5-6% weapons-grade plutonium from decommissioned weapons, mixed in with the uranium during manufacture. There was no plutonium in the Hiroshima bomb.
With the external containment structures blown open, the spent rods are open to the atmosphere, and there have been fires burning over them, releasing huge amounts of radioactive particles into the environment. There also appear to be containment leaks in the cores. Prevailing winds from Fukushima have tended include Tokyo in their curved sweep out over the ocean, thus Tokyo may be experiencing fall-out on a scale Hiroshima did not.
Statistically, this is going to cause an increased cancer/death rate, the only question is how much. I'm still hoping the long-term effects in Tokyo won't be too serious.
The core of Little Boy contained 64 kg of uranium,
According to the latest TEPCO report, on the Dai-Ichi site there was 577 tons of fuel in reactors
+ 1840 tons in storage =
= 2417 tons
Therefore there is 37765 times more radioactive fuel, by weight, on the Fukushima Dai-Ichi plant as there was in the Hiroshima bomb. It would not be a stretch to say the addition of plutonium makes this at least twice as deadly as uranium. So that gets you close to the OP's figure of 72,000, except for the fact that much of this fuel will not be released into the environment.
The dangers from this disaster are great, and terrible harm has already been done to the surrounding area. However a general comparison to a tragedy like Hiroshima is not helpful, as they are completely different situations.
For a good article on the dangers of the spent fuel: Greater Danger Lies in Spent Fuel Than in Reactors
posted on Mar, 27 2011 @ 12:18 PM
Originally posted by 00nunya00
And since you're dependent on Wikipedia, here
Half-life is the period of time it takes for a substance undergoing decay to decrease by half.
A kind Mod helped me get back the important part of my deleted post:
As your quote says, Half-life is the period of time it takes for a substance undergoing decay to decrease by half.
The substance decreases by half because half of it changes to another substance!
Did you think matter or energy simply disappear?
Nothing just disappears in this world. Things can move, things can change, but matter+energy is a constant, as any basic physics text will tell you. Half life is not about substances disappearing. It's about entropy causing a substance in a higher energy state to change into a substance in a lower energy state. This is done by releasing energy, and that energy is radiation. After the energy is released, what's left of the substance the energy was released from is a new substance. It can be a new element, a new isotope or the ground state of the original isotope, depending on what the initial substance was.
For example, carbon-14 atom (the "parent") emits radiation (a beta particle, antineutrino, and a gamma ray) and transforms to a nitrogen-14 atom (the "daughter"). Technetium-99m decays by the emission of a gamma ray, to an atom of technetium-99.
Just as I illustrated by talking of the half-life of oranges leaving half, over that period of time, turning into lemons, the half lives of the various plutonium isotopes mean that in that time half turns into uranium or americium isotopes, depending on the initial isotope of plutonium.
Decay modes of plutonium isotopes
Strontium 90, a product of nuclear fission, has a half life of 28.8 years. 90Sr undergoes β− decay with decay energy of 0.546 MeV to an electron and the yttrium isotope 90Y, which in turn undergoes β− decay with half life of 64 hours and decay energy 2.28 MeV for beta particles to an electron and 90Zr (zirconium), which is stable.
If you can't understand that carbon-14 is a different substance to nitrogen-14, that plutonium is a different substance to uranium, or that strontium 90 is a different substance to zirconium, I guess no physics book is going to help you.
edit on 27/3/11 by Kailassa because: formatting
new topics
top topics
10 | 3,168 | 13,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-43 | longest | en | 0.915474 |
http://mathhelpforum.com/advanced-algebra/202107-kernal-image-complex-numbers.html | 1,527,289,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00382.warc.gz | 180,123,888 | 12,743 | # Thread: kernal and image with complex numbers
1. ## kernal and image with complex numbers
C* is the group of non sero complex numbers under multiplication, and C is the group of all complex numbers under addition.
iii) C* ---> C*
z --> z + 3iz
iv) C* --> C*
z --> z / (the complex conjugate) z
I have already proved that these are homomorphism. Just stuck on the next bit of finding the image and kernel of the groups.
2. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
C* is the group of non sero complex numbers under multiplication, and C is the group of all complex numbers under addition.
iii) C* ---> C*
z --> z + 3iz
Do you mean C ---> C? It is not a homomorphism C* ---> C*.
Originally Posted by fireychariot
Just stuck on the next bit of finding the image and kernel of the groups.
For the first one, can you solve the equation z + 3iz = 0? For the second one, represent z as $\displaystyle re^{i\varphi}$ and solve $\displaystyle z/\bar{z}=1$.
3. ## Re: kernal and image with complex numbers
Originally Posted by emakarov
Do you mean C ---> C? It is not a homomorphism C* ---> C*.
For the first one, can you solve the equation z + 3iz = 0? For the second one, represent z as $\displaystyle re^{i\varphi}$ and solve $\displaystyle z/\bar{z}=1$.
Yes sorry C --> C for the first one
why do I represent z as $\displaystyle re^{i\varphi}$ ?
4. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
why do I represent z as $\displaystyle re^{i\varphi}$ ?
Because every complex number can be represented in polar form and it may simplify the solution.
5. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
C* is the group of non sero complex numbers under multiplication
iv) C* --> C*
z --> z / (the complex conjugate) z
Just stuck on the next bit of finding the image and kernel of the groups.
Recall that $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{{{\left| z \right|}^2}}}$. Thus $\displaystyle \frac{z}{{\overline z }} = \frac{{{z^2}}}{{{{\left| z \right|}^2}}}$
6. ## Re: kernal and image with complex numbers
I am sorry but I still don't know how to work each out. Start with the image for the first one
iii) C ---> C
z --> z + 3iz
So from what I am trying to understand is that the Image is what maps z to z +3iz? Am I thinking in the right terms there? The kernel is what maps z to the identity, am I right there? How do you know what the identity maybe?
7. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
So from what I am trying to understand is that the Image is what maps z to z +3iz?
No, what maps z to z +3iz is the map, the homomorphism. Why don't you look up the definition of image in your textbook or Wikipedia?
Originally Posted by fireychariot
The kernel is what maps z to the identity, am I right there?
No, what maps is the map. The kernel is what is mapped to identity, and the identity is 0 here.
Originally Posted by fireychariot
How do you know what the identity maybe?
Sorry, "what the identity maybe" is an incomplete sentence; I don't know what it means.
8. ## Re: kernal and image with complex numbers
Sorry it was meant to read how do we work out the identity? You say it is 0 but how did you know that?
9. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
how do we work out the identity? You say it is 0 but how did you know that?
Because 0 is the identity in the group of all complex numbers under addition. It satisfies the group axiom about the identity, and group identity is unique.
10. ## Re: kernal and image with complex numbers
Oh right thanks good point to know. So for the image of
iii) C ---> C
z --> z + 3iz
Im = { w in C : w = z - 4iz, some z in C}
= { w in C : w/(1-4i) in C}
= C
Have I written that out correct? If so I will have a go at the kernel next.
11. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
Have I written that out correct?
Yes, except I am not sure how z + 3iz became z - 4iz.
12. ## Re: kernal and image with complex numbers
ooops wrong question I was looking at. But same applies I take it?
is the Ker = {z in C : z+3iz = 0} = {0}?
13. ## Re: kernal and image with complex numbers
Originally Posted by fireychariot
ooops wrong question I was looking at. But same applies I take it?
is the Ker = {z in C : z+3iz = 0} = {0}?
Yes.
14. ## Re: kernal and image with complex numbers
ok back to the next question then,
iv) C* --> C*
z --> z / (the complex conjugate) z
Im = { w in C* : w = z / (the complex conjugate) z, for some z in C*}
However stuck on how to maniplulate this for the next line??
15. ## Re: kernal and image with complex numbers
What do you think about suggestions in posts #2 and #5? Also, it may be a little easier to find the kernel.
Page 1 of 2 12 Last | 1,356 | 4,880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-22 | latest | en | 0.854691 |
https://www.alexejgossmann.com/auc/ | 1,606,647,200,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00358.warc.gz | 538,172,236 | 7,954 | Unfortunately this was not taught in any of my statistics or data analysis classes at university (wtf it so needs to be ). So it took me some time until I learned that the AUC has a nice probabilistic meaning.
## What’s AUC anyway?
AUC is the area under the ROC curve. The ROC curve is the receiver operating characteristic curve. AUC is simply the area between that curve and the x-axis. So, to understand AUC we need to look at the concept of an ROC curve.
Consider:
1. A dataset $S$ : $(\mathbf{x}_1, y_1), \ldots, (\mathbf{x}_n, y_n) \in \mathbb{R}^p \times \{0, 1\}$, where
• $\mathbf{x}_i$ is a vector of $p$ features collected for the $i$th subject,
• $y_i$ is the $i$th subject’s label (binary outcome variable of interest, like a disease status, class membership, or whatever binary label).
2. A classification algorithm (such as logistic regression, SVM, deep neural net, or whatever you like), trained on $S$, that assigns a score (or a “probability”) $\hat{p}(\mathbf{x}_{\ast})$ to any new observation $\mathbf{x}_{\ast} \in \mathbb{R}^p$ signifying the algorithm’s confidence that the label (or class) of $\mathbf{x}_{\ast}$ is $y_{\ast} = 1$.
Then:
1. A decision threshold (or operating point) can be chosen to assign a class label ($y_{\ast} = 0$ or $1$) to $\mathbf{x}_{\ast}$ based on the value of $\hat{p}(\mathbf{x}_{\ast})$. The chosen threshold determines the balance between how many false positives and false negatives will result from this classification.
2. Plotting the true positive rate (TPR) against the false positive rate (FPR) as the operating point changes from its minimum to its maximum value yields the receiver operating characteristic (ROC) curve. Check the confusion matrix if you are not sure what TPR and FPR refer to.
3. The area under the ROC curve, or AUC, is used as a measure of classifier performance.
Here is some R code for clarification:
# load some data, fit a logistic regression classifier
data(iris)
versicolor_virginica <- iris[iris$Species != "setosa", ] logistic_reg_fit <- glm(Species ~ Sepal.Width + Sepal.Length, data = versicolor_virginica, family = "binomial") y <- ifelse(versicolor_virginica$Species == "versicolor", 0, 1)
y_pred <- logistic_reg_fit$fitted.values # get TPR and FPR at different values of the decision threshold threshold <- seq(0, 1, length = 100) FPR <- sapply(threshold, function(thresh) { sum(y_pred >= thresh & y != 1) / sum(y != 1) }) TPR <- sapply(threshold, function(thresh) { sum(y_pred >= thresh & y == 1) / sum(y == 1) }) # plot an ROC curve plot(FPR, TPR) lines(FPR, TPR) A rather ugly ROC curve emerges: The area under the ROC curve, or AUC, seems like a nice heuristic to evaluate and compare the overall performance of classification models independent of the exact decision threshold chosen. $\mathrm{AUC} = 1.0$ signifies perfect classification accuracy, and $\mathrm{AUC} = 0.5$ is the accuracy of making classification decisions via coin toss (or rather a continuous coin that outputs values in $[0,1]$…). Most classification algorithms will result in an AUC in that range. But there’s more to it. ## Probabilistic interpretation As above, assume that we are looking at a dataset where we want to distinguish data points of type 0 from those of type 1. Consider a classification algorithm that assigns to a random observation $\mathbf{x}\in\mathbb{R}^p$ a score (or probability) $\hat{p}(\mathbf{x}) \in [0,1]$ signifying membership in class 1. If the final classification between class 1 and class 0 is determined by a decision threshold $t\in[0, 1]$, then the true positive rate (a.k.a. sensitivity or recall) can be written as a conditional probability and the false positive rate (or 1 - specificity) can be written as For brevity of notation let’s say $y(\mathbf{x}) = 1$ instead of “$\mathbf{x}$ belongs to class 1”, and $y(\mathbf{x})=0$ instead of “$\mathbf{x}$ doesn’t belong to class 1”. The ROC curve simply plots $T(t)$ against $F(t)$ while varying $t$ from 0 to 1. Thus, if we view $T$ as a function of $F$, the AUC can be rewritten as follows. where we used the fact that the probability density function is the derivative with respect to $t$ of the cumulative distribution function So, given a randomly chosen observation $\mathbf{x}$ belonging to class 1, and a randomly chosen observation $\mathbf{x^{\prime}}$ belonging to class 0, the AUC is the probability that the evaluated classification algorithm will assign a higher score to $\mathbf{x}$ than to $\mathbf{x^{\prime}}$, i.e., the conditional probability of $\hat{p}(\mathbf{x}) > \hat{p}(\mathbf{x^{\prime}})$. An alternative purely geometric proof can be found in the Scatterplot Smoothers blog. In other words, if the classification algorithm distinguishes “positive” and “negative” examples (e.g., disease status), then AUC is the probability of correct ranking of a random “positive”-“negative” pair. ## Computing AUC The above probabilistic interpretation suggest a simple formula to compute AUC on a finite sample: Among all “positive”-“negative” pairs in the dataset compute the proportion of those which are ranked correctly by the evaluated classification algorithm. Here is an inefficient implementation using results from the above logistic regression example: s <- 0 for (i in which(y == 1)) { for (j in which(y == 0)) { if (y_pred[i] > y_pred[j]) { s <- s + 1 } else if (y_pred[i] == y_pred[j]) { s <- s + 0.5 } } } s <- s / (sum(y == 1) * sum(y == 0)) s # [1] 0.7918 The proportion of correctly ranked “positive”-“negative” pairs yields estimated $\mathrm{AUC} = 0.7918$. We can compare this value to the area under the ROC curve computed with the trapezoidal rule. s <- 0 for (i in 1:(length(FPR) - 1)) { dFPR <- abs(FPR[i+1] - FPR[i]) s <- s + 0.5 * dFPR * (TPR[i+1] + TPR[i]) } s # [1] 0.7922 Trapezoidal rule yields estimated $\mathrm{AUC} = 0.7922$. The difference of $0.0004$ can be explained by the fact that we evaluated the ROC curve at only 100 points. Since there is a minor disagreement, let’s use some standard R package to compute AUC. library(ROCR) pred <- prediction(y_pred, y) auc <- as.numeric(performance(pred, measure = "auc")@y.values) auc # [1] 0.7918 Same as the proportion of correctly ranked pairs! #### Wilcoxon-Mann-Whitney test By analysing the probabilistic meaning of AUC, we not only got a practically relevant interpretation of this classification performance metric, but we also obtained a simple formula to estimate the AUC of a trained classification algorithm. Well, it turns out that taking the proportion of correctly ranked “positive”-“negative” pairs as a formula to estimate the AUC is equivalent to the Wilcoxon-Mann-Whitney statistical test. This fact can also be easily demonstrated in a couple lines of R code. y_is_1 <- which(y == 1) y_is_0 <- which(y == 0) n_pairs <- length(y_is_1) * length(y_is_0) WMW_test <- wilcox.test(y_pred[y_is_1], y_pred[y_is_0]) WMW_test$statistic / n_pairs
# W
# 0.7918 | 1,883 | 6,950 | {"found_math": true, "script_math_tex": 40, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-50 | longest | en | 0.871736 |
https://math.stackexchange.com/questions/39161/centralizer-of-a-p-element-modulo-the-p-core-and-conjugacy-class-sizes-in-q | 1,620,250,419,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00449.warc.gz | 413,012,871 | 38,595 | # Centralizer of a $p$-element modulo the $p'$-core and conjugacy class sizes in quotient groups
Does $[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ]$ hold for all finite groups $G$ and $p$-elements $x$, where $K = O_{p'}(G)$ is the largest normal subgroup of $G$ with order coprime to $p$?
A related statement is true: Let $G = H \ltimes K$ be a semidirect product of $H$ with the (arbitrary) normal subgroup $K$, and let $x$ in $H$. Suppose $hk$ centralizes $x$. Then considering the group mod $K$, one gets that $h$ centralizes $x$. Hence $k$ centralizes $x$, and $C_G(x) = C_H(x) \cdot C_K(x)$. Importantly, we have the equality:
$$[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$
If $G$ splits over $K = O_{p'}(G)$ so that $G = H \ltimes K$, then every $p$-element is conjugate to some element of $H$, and the previous implies the conjugacy class sizes behave as I expect. However, $G$ often does not split over $K$ and I am not sure what happens then.
So suppose $G$ is a finite group with normal subgroup $K$. One still has that every $h$ centralizing $x$ gives rise to an $hK$ in $C_{G/K}(x)$ and of course $C_G(x)$ is a union of cosets of $C_K(x)$. However, it is not clear that all elements $hK$ of $C_{G/K}(x)$ give rise to elements $h$ in $C_G(x)$. In other words, I can only prove that:
$$[ G : C_G(x) ] \geq [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$
I checked for counterexamples (small groups up to order $1000$ except $768$, primitive groups up to degree $500$ except $343$, and perfect groups up to order $10^6$ available in GAP), but found no counterexamples to equality. I am not sure if the coprime hypothesis is truly relevant, but it is the case of interest and I could not think of any other way to say "$x$ in $H$" when $G$ did not split. With absolutely no hypotheses on $x$ or $K$, then $G$ dihedral of order 8 with $K = Z(G)$ and $x$ a non-central involution gives a counterexample, but this is not at all similar to the case I am interested in.
If $H$ is a p-subgroup of $G$, then it is true that: $$[G:C_G(H)]=[G/K:C_{G/K}(H)]\ [K:C_K(H)].$$
$C_K(H) = C_G(H)\cap K$; this is clear.
$C_{G/K}(H)=C_G(H)K/K$. To prove this, note first that $N_{G/K}(H)=N_G(H)K/K$. This is equivalent to $N_G(HK)=N_G(H)K$, and this follows from the Frattini argument applied to $HK$ in $N_G(HK)$. Now the centralizer case follows because $N_G(H)$ is mapped to $N_{G/K}(H)$, $H$ is mapped to an isomorphic copy of itself, and the quotient map respects the action of $N_G(H)$ on $H$ (so, in particular, respects the kernel of this action, $C_G(H)$). | 846 | 2,575 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.889152 |
https://www.coursehero.com/file/6006594/cs682-canny-edges/ | 1,521,332,082,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00106.warc.gz | 775,541,286 | 220,089 | {[ promptMessage ]}
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cs682-canny-edges
# cs682-canny-edges - Image Features Edges Formal Design of...
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1 Image Features Edges 2/11/2004 Octavia I. Camps 2 Formal Design of an Optimal Edge Detector Edge detection involves 3 steps: – Noise smoothing – Edge enhancement – Edge localization J. Canny formalized these steps to design an optimal edge detector
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2 2/11/2004 Octavia I. Camps 3 Canny Edge Detector Experiments consistently show that it performs very well Probably, the most used by C.V. practitioners 2/11/2004 Octavia I. Camps 4 Canny Edge Detector Uses a mathematical model of the edge and the noise Formalizes a performance criteria Synthesizes the best filter
3 2/11/2004 Octavia I. Camps 5 Edge Model (1D) An ideal edge can be modeled as an step A < = 0 if 0 if 0 ) ( x A x x G 2/11/2004 Octavia I. Camps 6 Performance Criteria (1) Good detection – The filter must have a stronger response at the edge location (x=0) than to noise
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4 2/11/2004 Octavia I. Camps 7 Performance Criteria (2) Good Localization – The filter response must be maximum very close to x=0 X=0 X=0 2/11/2004 Octavia I. Camps 8 Performance Criteria (3) Low False Positives – There should be only one maximum in a reasonable neighborhood of x=0 large
5 2/11/2004 Octavia I. Camps 9 Canny Edge Detector Canny found a linear, continuous filter that maximized the three given criteria. There is no close-form solution for the optimal filter. However, it looks VERY SIMILAR to the derivative of a Gaussian. 2/11/2004 Octavia I. Camps 10 Algorithm CANNY_ENHANCER The input is image I; G is a zero mean Gaussian filter (std = σ ) 1. J = I * G (smoothing) 2. For each pixel (i,j): (edge enhancement) Compute the image gradient » J(i,j) = (J x (i,j),J y (i,j))’
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Ask a homework question - tutors are online | 658 | 2,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-13 | latest | en | 0.747814 |
https://www.coursehero.com/tutors-problems/Economics/11881338-I-am-having-hard-time-to-solve-this-I-need-help-with-explanation-Tha/ | 1,544,432,176,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823320.11/warc/CC-MAIN-20181210080704-20181210102204-00610.warc.gz | 856,384,910 | 21,374 | View the step-by-step solution to:
I am having hard time to solve this. Thank you! Use the function below (whose parameters qualify it as a STC function) to answer the questions....
I am having hard time to solve this. I need help with explanation. Thank you!
Use the function below (whose parameters qualify it as a STC function) to answer the questions.
STC=30+18Q-2.7Q2+ 0.15Q3
1)Total fixed cost is \$________
2)Obtain the AFC function from (a) and write it here _____________
3)Obtain the AVC function contained in the STC function and write the AVC function here __________________________
4) Write here the MC function __________________________
5) Find the value which AFC approaches as Q gets very large.
What happens fixed costs per unit (AFC) as production quantities get ever larger.
6) Find the value of Q at which AVC is a minimum.
7) Is productive efficiency at the value in (f) greatest or least?
8) Demonstrate that the value of SMC equals the value of AVC at the value of Q where AVC is a minimum. Hint: The level of Q you found in (f) is where AVC is a minimum. If you plug this level of Q into AVC (see c) and also into MC (see d) the two outcomes should be the same if SMC crosses AVC at this level of Q.
9) Find the value of Q where increasing returns ceases and diminishing returns begins. Hint: Diminishing returns begins at the level of Q where MC is a minimum.
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Browse Documents | 421 | 1,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-51 | latest | en | 0.877447 |
https://medium.com/trick-the-interviwer/regular-expression-matching-9972eb74c03?source=post_internal_links---------7---------------------------- | 1,656,992,715,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00229.warc.gz | 424,344,167 | 42,530 | # Regular Expression Matching
Hi! Today we are going to talk about a very useful concept known as regular expressions also called regex.
## What is regex?
Regex is basically a pattern that can represent a variety of strings. In regex vocabulary, few characters carry a special meaning. Two of those that are relevant to the current post are:
. in regex -> can match any single character
.* in regex -> can match zero or more characters preceding *
Regex is used everywhere. For example, regex is used in search engines. In websites also, where users are required to enter their email address, you can validate the field using a regular expression that matches the structure of an email address. Regex has many use cases in a string validation and string searching.
Suppose our pattern is, p= aa.c and the search string is, s=aabc
We can search this string with our pattern, or we can say that this pattern can represent s. Here . can represent a single character, so b can be represented using dot character. This pattern can also represent other strings like, aacc, aayc, and so on. Similarly, for the patterns like a.*b, they can effectively represent strings like “ab”, “acb”, “accb”, “actb”, “ahhkb”, and so on.
## Let’s understand how we can implement an algorithm for achieving the target of regular expression matching. Here is our problem statement.
Problem Statement: Given an input string (`s`) and a pattern (`p`), implement regular expression matching with support for `'.'` and `'*'`.
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Before we begin, please make sure you read the below assumptions we are making to go about solving this problem.
There are a few cases to consider. For all the cases, we will be iterating our strings. Here, p will represent our pattern string and s will represent our input string. The character i will denote the index of the input string and j will denote the index of our pattern string. Initially, i and j represent the last characters of both the strings.
There are basically three cases that we need to consider. Without further delay, let’s see them all.
# CASE 1: Last character of both the strings matches.
s[i] == p[j] or p[j] == ‘.’
Here, s can be represented by p, if, the last character of both s and p matches and the substring of s and p, before the last character, also matches. Basically, the prefix of s and p are matching.
Here, s[i] == p[j], we need to check whether, s[0..i-1] can be represented by p[0..j-1]. In this case, ab can be formed using ab. That means s can be formed using p. Therefore, we checked for smaller subproblems. We can break a bigger problem into smaller subproblems. Let’s consider another case.
Here, s[i] != p[j], but p[j] == ‘.’ and by the definition of regex, ‘.’ can represent any single character. Therefore, we can say, ‘d’ can be represented using ‘.’. Now we need to check for smaller subproblem. Check whether s[0..i-1] can be represented by p[0..j-1]. In this case, ab can be formed using ab. That means s can be formed using p.
Let’s see another case. Here, s[i] == p[j]. Now we need to check for smaller subproblem. Check whether s[0..i-1] can be represented by p[0..j-1]. In this case, ab can not be formed using string cb. That means s can not be formed using p.
This suggests that when the last characters of s and p matches, the result of smaller subproblems will be our result. That is, we are recursively checking for the smaller subproblem.
# CASE 2: Last character pattern string is *
It has two cases. Either the character preceding * is occurring once or more than once or it is not occurring at all.
## Character preceding * occurring once or more than once
If the last character in the pattern is *. It means there can be either zero or more occurrence of character before * in the input string. We will check if the character before * matches with the current character, i.e. whether s[i] == p[j-1] or p[j-1]==’.’. The characters match, as b matches with b, check for the result of the smaller subproblem. That is, by removing the last character from the input string. Check for s[0..i-1] and p[0..j]. As we can see ab matches with ab*, therefore s can be generated using p. Let’s see for the smaller problem just generated.
We are left with this input string s after breaking it into smaller sub-problem. Here again, we can see that the character preceding the last character of * matches with the current character of the input string, i.e. b matches with b (s[i] == p[j-1] || or p[j-1]==’.’). We again need to break the problem into smaller subproblems. Remove the last character of the input string and again try to match.
## Character preceding * not present in the input string
We are left with this input string s after breaking it into smaller sub-problem. By the definition of regex, we know that character before * may not occur in the input string even once as well. Therefore match the current character of the input string with the character before the preceding character of * in the pattern string. That is, match s[i] with p[j-2] or if p[j-2]=’.’ . We can see that s[i]=a and p[j-2]=a. We have just broken down the problem as matching of string s[0..i] and p[0..j-2]. These smaller subproblems match. Therefore, the final result is, s can be formed using p.
# CASE 3: Last character pattern string does not match the last character of the input string and it is not * or .
In this case, we can’t break the problem into smaller subproblems as the last characters don’t match. Therefore, simply return that the strings can not match.
These are the basic cases involved in this problem. As it is a recursive algorithm there must be a base case as well to stop the recursion at some point.
You can actually reach the base condition yourself. Try to figure out yourself first and then continue here.
## BASE CONDITION
1. When both the input string and pattern becomes null, then that actually means pattern can represent input string, return true in that case. (i<0 and j<0)
2. When the pattern is an empty string and input string is non-empty, then we can not generate input string using the pattern string, return false in this case. (i>0 and j==0)
3. This one is a tricky base case now. When the input string is empty and the pattern is non-empty. We can still generate the input string using the pattern. Let’s see how.
Suppose pattern = a*b*
and input = “”
As you can see, pattern means, zero or more times a followed by zero or more times b. Therefore, strings that can be generated by the pattern are: “”, “a”, “ab”, “aab”, “aa”, “abb”, “b”, “bb”, and so on.
Suppose in our recursive call, the input string is empty, therefore i=-1. And in the pattern string j=0, we see that we can’t generate empty string using “a”. But at j=1, we see that we can generate empty string using “a*”. Similarly, at j=3 we can generate empty string using “a*b*”. So our third base case will be,
if(i<0 && p[j]==’*’) regex(s,p, i, j-2);
if(i<0 && p[j]!=’*’) return false;
So, these are the three base cases for recursion. I hope the solution is clear to you. Try to create a recursive function using the concepts explained above.
## The recursive solution, Top-down approach
`boolean regex(s, p, i, j){ //base conditions if(i<0 && j<0) return true; if(j<0 || (i<0 && p[j]!='*')) return false; if(i<0 && p[j]=='*') return regex(s,p,i,j-2); if(s[i]==p[j] || p[j]=='.') return regex(s,p,i-1,j-1); if(p[j]=='*'){ boolean result = false; if(s[i]==p[j-1] || p[j]=='.'){ result = regex(s,p,i-1,j) } if(!result){ return (result || regex(s,p,i,j-2)); } } return false;}`
As in a recursive function, there are many instances where a subproblem is calculated multiple times. In the recursive call, time complexity will be exponential. We need to memoize the solution and reduce the time complexity. For that, we can use the concept of dynamic programming.
## Memoization and Bottom-up approach
Our DP array size will be (s.length()+1) x (p.length()+1)
This is a boolean array, where s[i][j] will store whether it’s possible to generate string s[0..i] using pattern p[0..j]. I am filling this DP array in a bottom-up manner.
The first row and first column of the DP will correspond to the empty input string and empty pattern string respectively. You can fill the first row and first column of the boolean DP array using the base conditions explained above.
Considering the size of the input string as m and size of pattern string as n.
This solution has a space complexity of O((m+1) x (n+1)). Along with that it has a time complexity of O((m+1) x (n+1)).
Hurrah! We are done! Thanks for devoting your time to this article. I hope you find it helpful and it gives you a clear insight into the topic.
If you have any doubts, feel free to ask in the comment section below. If you like this blog, please share it with your friends. Your claps motivate me for writing more such blogs.
Stay tuned for the upcoming explanatory posts by following me on medium.
Have a good day and enjoy coding!! :D
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## More from TRICK THE INTERVIEWER
Programming interview is the one where the interviewer will make all possible attempts to trick you. Trick the interviewer this time! :)
## Jhanak Didwania
Software Engineer | Programmer | 2,272 | 9,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.89449 |
https://matheducators.stackexchange.com/questions/12308/is-estimating-still-considered-a-valuable-skill/20669 | 1,723,676,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00739.warc.gz | 300,575,257 | 56,264 | Is 'estimating' still considered a valuable skill?
I was with a 2nd year high school class, preparing for our (US) state's standardized test. I asked the class how they would solve this, and they flipped through the sheets to find
$$V=\frac{1}{3}\pi r^2h$$
(To be clear, there is a 'formula' page that's given with the exam.) As they fumbled with their calculators, I asked them what the radius was. 3.5. and I wrote the numbers on the board.
$$V=\frac{1}{3}\pi (3.5)^2(9)$$
I attempted to show them how 3.5 squared is close enough to 12. Divide by 3 to get 4, and we are left with 36pi. Since 36x3 is 108, the only answer that can make sense is 115.
In my lecture, I offered 3 benefits of this process. (1) The speedy approach can save precious time on this type of question so more time remains for those requiring more thought and time. (2) If you still use the calculator, this is a good double check to be sure the answer makes sense, that you didn't hit the wrong key. (3) If you forget your calculator, the proctor won't always have an extra, and the whole exam can be done by hand.
Out of the 12 questions we did as a group, a full half lent themselves to this method, as number were easy to round, and the answers were different enough so cumulative rounding errors didn't inch results too far away from the true answer. After my second attempt, on the next appropriate question, to explain this approach, I realized it wasn't helpful to this class and I stopped.
I recall that as a student, estimating was a skill that we were taught as part of the regular math class. Is this no longer considered to be valuable?
• You might be interested in Fermi estimates en.wikipedia.org/wiki/Fermi_problem also see lesswrong.com/lw/h5e/fermi_estimates . Used rather a lot in computational physics (to determine is a result is reasonable), probably other disciplines too. Commented May 13, 2017 at 18:36
• @TheoreticalPerson Post an answer about that! That's probably the motherlode of insights into this topic. Commented May 13, 2017 at 18:50
• I'm a quant. I spend all day eyeballing, estimating, and mentally confirming the numbers on my screens. I would say the skill is absolutely invaluable.
– Kaz
Commented May 13, 2017 at 22:31
• @Kaz Post an answer about that! A few real-life examples reported first-person would likely shed a lot of light on why estimation is valuable. Commented May 14, 2017 at 9:13
• I agree 100% that estimation is useful, but I would be careful with this particular problem, because 132 is dangerously close to 115. If I was estimating, I would have complete confidence in eliminating A and D, not so much C. Commented May 14, 2017 at 22:54
Yes, of course it is. You've identified many of the important reasons in your question. In practice in quantitative fields, we are all estimating as a first-pass on whether problems are soluble all the time. Estimation is incorporated throughout Common Core standards (e.g., Grade 3, Grade 4, Grade 7; there are more). And it was part of official state standards long before that (link).
That said, I am concerned that in some locations it may not in fact be taught. When I teach remedial arithmetic courses at my community college, the entire classroom of students will claim to have never heard the term before (either "estimate" or "approximate", etc.). It is, in fact, utterly bewildering to them. Therefore, in that course I make it the topmost priority and try to work estimation skills into the exercises every single day. Nevertheless, at the end of a semester, many students will say something like, "I never got this estimation thing".
Often when I share this with other people I get great skepticism ("but my child is learning that in the third grade right now"). We might theorize how it came to be: Maybe my students come from specially broken-down school systems. Or maybe many of the students have intellectual disabilities and were taught to rely on the calculator in the absence of basic numeracy. Or maybe they've just never built anything practical such that they saw the value in double-checking for errors. At any rate, it's that kind of gap for which Common Core is trying to give guidance, I think.
I can't speak for a broad consensus, but I personally certainly think that estimation is still a valuable skill.
First, here's my rough, loose, quick way of solving the problem, using only easy mental arithmetic, no pencil, paper, or calculator. Seeing this will make my reasons below clearer.
1. This problem would be a whole lot easier if it were a rectangular prism. The base would be 7 ⨉ 7 = 49 in2, which is almost 50. Multiplying that by the height of 9 would yield a volume of 450 in^2. Now we just have to "shave off" everything that isn't part of the cone. Notice that this already shows that answer D is hopeless.
2. OK, let's shave it down to a cylinder. That's pretty easy. The base is a circle, which has an area of a little more than 3/4 of the square with the same "diameter" (i.e. a circumscribed square). 3/4 of 49…ecch, let's make this easier and say 3/4 of 48…that's 36. So the cylinder has a volume of a little more than 9⨉ that…10⨉36 = 360, subtract 36, that's 324.
3. The idea of the volume formula for the cone is just that it's 1/3 of the cylinder. So, to shave away the part of the cylinder that's not the cone, just divide by 3. 324/3 = 300/3 + 24/3 = 100 + 8. So the volume of the cone is a little more than 108 in3. Looks like answer B.
Now here are some reasons why being able to estimate is a valuable skill:
1. As you said, the estimate is a double-check on the precise calculation. Really, you should never do a precise calculation without also doing an estimate. Precise calculations, especially if a calculator is involved, can easily be off by orders of magnitude if you make a tiny mistake, like a wrong keypress. You should always know the "ball park" before you do the calculation, or your common sense can't function. Notice how considering a rectangular prism set an upper bound on reasonable answers. Just from looking at the picture, you should be thinking that the right answer is a little less than a third of that. Estimates can be wrong, too, but the point is: when the calculation disagrees with your estimate, you know something is wrong. Without an estimate, you can't tell.
2. When students are taught math as a bunch of formulas and procedures to memorize, they often can't think fluidly or flexibly about it. When they meet a problem for which they don't know a formula or procedure, they're instantly stumped and give up. Estimating gives the student easy practice in thinking mathematically. Most of mathematical reasoning consists of seeing the same thing as it is involved in different relationships, each of which sheds some light on it or constrains it in some way. Estimating gives you practice in feeling around for helpful relationships, and in distinguishing between important and unimportant. It's easier than usual because you don't need to be exact. It gives you practice in the ubiquitous mathematical technique of solving a much easier problem than you were given and then figuring out how to correct for the difference. And it shows you that mathematics makes sense.
3. Estimating teaches through direct experience that there is not just one right way to solve a problem. Notice that your way of reaching the same estimate was more efficient than mine. That's to be expected. I don't mess with geometry much. I just fumbled around a bit, but I was still able to find the answer completely in my head. I don't need to be boned up on solid geometry to find a passable way to solve the problem. I can cook something up that's good enough, without being an expert. There are many, many paths, and they all lead to the same answer.
4. Consequently, estimating builds confidence. The more you estimate, the more you see math as a system of relationships that tend to guide you to the thing you're looking for. If you miss one relationship, no big deal, math is abundant in relationships. And if you're off by a little bit, no big deal. You see that math is best approached loosely, lightly, playfully. This is the opposite of the feeling that students usually get: that math is a minefield—make one false step and it's over.
5. In real life, you almost never need an exact answer. Estimating is the main practical skill of mathematics beyond arithmetic. In real life, outside of fields like accounting, usually you can't even get very precise measurements. The first couple digits and the order of magnitude are usually all you need to know—often, all you can know. Also in real life, you seldom encounter problems fully spelled out. Usually you have to figure out what available information would enable you to make a good estimate. You can't really do that effectively without experience estimating.
For inspiration, read about Enrico Fermi. He is said to have been able to solve pretty much any physics problem in his head to one significant digit and the order of magnitude. And yes, he taught this skill to his students.
I have found that estimating is a very valuable skill, particularly for finding orders of magnitude of solutions without having to use a calculator. It comes in very useful for helping students work out problems on the fly.
Unfortunately, I have found that most students now have no feel for this skill, opting instead to use a calculator for any calculation, even those that could easily be worked out exactly in one's head. When we get into calculations involving quantities with uncertainties, this lack of understanding of really how precisely one can express an answer becomes even more pronounced. Students seem to always think the correctness of the answer corresponds to the number of decimal places that are written down, and they seemingly have no feel for whether a possible answer is feasible or not without working it out with a calculator.
I think we should sometimes require students to write exams without a calculator. That's how many of us learned not to require one for every little calculation. Estimating the answer in their head beforehand would also help students to realize when an answer from their calculator might be questionable.
In general, yes, estimation is a great way to get a quick sanity check.
In this particular example it looks like they chose values such that using $22/7$ as an approximation for $\pi$ lets you quickly cancel stuff out,
$$\begin{eqnarray*} V = {\frac{1}{3}}\times{\frac{22}{7}}\times{\frac{7}{2}}\times{\frac{7}{2}}\times9 = {\frac{11\times7\times3}{2}} = 115.5 \end{eqnarray*}$$
• Please don't write mixed fraction. Pretty please. Commented May 14, 2017 at 7:55
• @JessicaB What is the trouble with the mixed fraction? Commented May 14, 2017 at 9:09
• "Mixed fractions" are terrible notation — they look like multiplication, but are actually addition, in a weird, unnecessary exception to the usual rules of mathematical notation. It only takes one extra symbol to write $115 + \frac{1}{2}$, and that avoids all the ambiguity and confusion of mixed fractions. Commented May 14, 2017 at 15:09
• @DanielHast Weird: to me, a mixed fraction doesn't look like multiplication, it looks like a mixed fraction—that is, like a number (so you're done) rather than a calculation (so you're not done). But math is positively overflowing with ambiguous notations that require context to sort out—and that the uninitiated can't sort out. Do you object to function notation, like $f(x)$, for the same reason: it looks like multiplication? I could sympathize with that: that one had me very confused for a long time when I was 13. Commented May 14, 2017 at 19:02
• @BenKovitz I can tell $f(x)$ is not multiplication because $f$ and $x$ are different types of object that cannot be multiplied. If it worked out they did need to be multiplied, you could use $f\cdot x$ or $xf$. With mixed fractions, there is no way of telling that a different convention is being used. I read this answer as 'half of 115' until I scrolled to another answer and saw 115.5. Commented May 15, 2017 at 6:41
First, an even easier estimate: $$V = \frac{1}{3} \pi r^2 h \approx \frac{1}{3} \frac{22}{7} \left( \frac{7}{2} \right)^2 9 = \frac{2 * 11 * 7^2 * 3^2}{3 * 7 * 2^2} = \frac{1}{2}*11*7*3=115.5.$$ Never underestimate the power of "engineer's $$\pi$$" ! Admittedly, some of the cancellations make this problem even easier.
I would whole-heartedly agree with the other answers that YES estimation is vitally important. I'll try to flush out a few more reasons:
Daily Life
While the old teacher reply that "you might not have a calculator with you" has been somewhat invalidated by most people carrying around a computer in their pocket, many daily problems just don't require precise answers or have vaguely defined inputs. For example,
• I'm throwing a party. How many pizzas to order / cases of ...soda...to buy / pounds of ice? Especially since I have a few flaky friends that might not show up or might bring along dates.
• I'm driving on the interstate and have a quarter tank left. Do I need to pull off at the next exit or can I make it to the next town that's 100 miles down the road? What if the gas at the next exit is overpriced?
• I'm painting my house. Do I need 5 gallons of paint or 10? I got a quote from a professional painter...is their price reasonable, considering that I would spend the entire weekend doing it myself?
• I'm splitting a meal at a restaurant with some friends. How much do I need to throw in (including tip) if I don't want to spend 20 minutes breaking down everyone's exact costs?
• Is my water bill lower if I take a 10-minute shower every day or a bath every other day? If I buy the more energy-efficient (but expensive) LED light bulbs with redtooth interweb connectivity, how long will it take to recoup that investment?
The list goes on and on. Most of these sorts of problems are related to basic numeracy, of which estimation is a huge component.
Preparation for Calculus
At its core, I would argue that calculus is the study of making a sequence of approximations that become "infinitely accurate."
• How can I derive that $$V = \frac{1}{3} \pi r^2 h$$ if I don't know how to estimate the volume of a cone to begin with? Will increasing the radius or height of this cone increase the volume quicker? This problem states that the height is exactly $$9in$$ and the diameter is exactly $$7in$$. If you actually measured a similar cone with a ruler, would the answer you got out of your calculator likely be higher or lower than the true volume? What if you spent a lot of time measuring really carefully?
• The speedometer in my car is broken (or rather, was until I replaced it). How did I manage to not get pulled over for speeding on the highway? What if the odometer was broken instead of the speedometer? Speaking of which, how does your smartphone give you a current speed?
• How can I find the volume of water in a lake or volume of water moving down a river, when they aren't convenient geometric shapes? If you live close to said lake and it rains for forty days, do you need sandbags or an ark?
• Do I need to pedal fast, really fast, or really really fast to clear this jump on my bike? If I ski off this cliff, is it going to hurt or hurt a lot?
Science & Engineering
Never underestimate the power of a good estimation or toy problem to sanity check a result. Even with "hardcore applied math," one of the first steps is to define your assumptions for modeling a physical system---which assumptions are reasonable and which are more likely to introduce errors? Is it the identical spherical cows or the flat and infinite grazing pasture that's the problem?
Since other answers have already touched on Fermi estimations and such, I'd like to give one of them Real World Examples™. A calculator (or computer) does a fine job of returning a Number™, but doesn't tell you if that Number™ is the Answer™.
My uncle works as a lead airframe engineer on fighter jets. He once tasked a junior engineer with figuring out what the fuel economy of their current design was, i.e., how many gallons of fuel the jet would use in an hour while flying at cruising speed. The junior engineer went off and did a week of computation and computer modeling before proudly returning with a Number™. Said number seemed a bit high to my uncle, but he was busy at the time and couldn't look over the report until the end of the day. A second look showed said number to be...really big. A bit of estimation later, he found that the junior engineer had proudly reported that the jet would use almost 2 Pentagons of fuel per hour of flight (yes, as in the building used as a unit of volume). The junior engineer had apparently flubbed a unit conversion at the outset and had absolutely no idea how absurd the number they got out of the program was, at least until they were called into my uncle's office the next day.
My calculator will return 13 digits. Not nearly enough to approximate my release point to slingshot around Saturn on my way to Betelgeuse. But way more than I need to slice an apple pie for the kids. Estimation, iteration, approximation. All tools in your toolbox for answering questions and solving problems. Don't handicap yourself or your students.
Estimating is absolutely crucial for high level success in mathematics, not because of a specific application (though there are plenty of uses for mental estimation in daily life), but because constantly making estimates is helpful exercise to make the brain more adept at processing numerical calculations and connecting them to context.
Why does a soccer (i.e. football, if you're not American) player run sprints during practice? Because they may want to race a teammate later on? Because they will be sprinting in an exact straight line for a set distance during their upcoming matches? No, but they will be executing more complicated runs that involve additional skills such as cutting, sliding and kicking all while trying to track the motion of the ball and/or key players on the pitch. But in order to do that effectively, they need to be able to run very quickly, and that can only be developed over time through extensive practice. Sprinting isn't the end goal during practice; it's a way to develop an important skill that will be used to accomplish a greater goal.
If you want to solve challenging mathematical problems, you have to be willing to run through many different ideas and a first start is trying to do easy calculations, which often will require you to do some estimation in order to keep the calculations simple. If you don't develop a good number sense through practicing exercises that cover, among other topics, estimation, then you will be missing a key tool needed to deal with more advanced problems. Estimation likewise is not an end goal; it's a way to develop an important skill that will be used to accomplish a greater goal.
It's a skill that we have to encourage physics postgrads to practice and use.
My physics degree included a course on estimation/approximate methods (the only time I've considered a spherical mammal on an infinite plane in a vacuum). I wish it was taught more. But I suggest it's more like a practical skill than say finding the roots of a quadratic. Certainly I've done this sort of estimation more in practical problems (How many tins of paint to paint a room? How much cable should I buy? Have I got enough petrol for the whole journey?) than in any classroom
IT is common in many technical fields, in business consulting and in engineering in the field. I've also seen it in the Navy in nuclear power ("radcon math"), in TMA (maneuvering boards), and in submarine TMA (where you lack range information and have only bearing, assuming passive sonar).
I'm not sure it needs to be stressed in math class if it is addressed in physics, chemistry, etc.
It is absolutely a valuable skill. HOWEVER, students should also learn to be aware of the difference between finding the true answer and approximations. Approximations involve errors no matter how small it is unless it is zero. | 4,605 | 20,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.969839 |
http://www.thefreedictionary.com/permute | 1,558,411,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00373.warc.gz | 340,098,395 | 14,064 | # permute
Also found in: Thesaurus, Medical, Legal, Encyclopedia, Wikipedia.
## per·mute
(pər-myo͞ot′)
tr.v. per·mut·ed, per·mut·ing, per·mutes
1. To change the order of.
2. Mathematics To subject to permutation.
[Middle English permuten, from Old French permuter, from Latin permūtāre : per-, per- + mūtāre, to change; see mei- in Indo-European roots.]
per·mut′a·bil′i·ty n.
## permute
(pəˈmjuːt)
vb (tr)
1. to change the sequence of
2. (Mathematics) maths to subject to permutation
[C14: from Latin permūtāre, from per- + mūtāre to change, alter]
perˌmutaˈbility, perˈmutableness n
## per•mute
(pərˈmyut)
v.t. -mut•ed, -mut•ing.
1. to alter; change.
2. Math. to subject to permutation.
[1350–1400; Middle English < Latin permūtāre to exchange, transform. See per-, mutate]
per•mut`a•bil′i•ty, per•mut′a•ble•ness, n.
## permute
Past participle: permuted
Gerund: permuting
Imperative
permute
permute
Present
I permute
you permute
he/she/it permutes
we permute
you permute
they permute
Preterite
I permuted
you permuted
he/she/it permuted
we permuted
you permuted
they permuted
Present Continuous
I am permuting
you are permuting
he/she/it is permuting
we are permuting
you are permuting
they are permuting
Present Perfect
I have permuted
you have permuted
he/she/it has permuted
we have permuted
you have permuted
they have permuted
Past Continuous
I was permuting
you were permuting
he/she/it was permuting
we were permuting
you were permuting
they were permuting
Past Perfect
Future
I will permute
you will permute
he/she/it will permute
we will permute
you will permute
they will permute
Future Perfect
I will have permuted
you will have permuted
he/she/it will have permuted
we will have permuted
you will have permuted
they will have permuted
Future Continuous
I will be permuting
you will be permuting
he/she/it will be permuting
we will be permuting
you will be permuting
they will be permuting
Present Perfect Continuous
I have been permuting
you have been permuting
he/she/it has been permuting
we have been permuting
you have been permuting
they have been permuting
Future Perfect Continuous
I will have been permuting
you will have been permuting
he/she/it will have been permuting
we will have been permuting
you will have been permuting
they will have been permuting
Past Perfect Continuous
I had been permuting
you had been permuting
he/she/it had been permuting
we had been permuting
you had been permuting
they had been permuting
Conditional
I would permute
you would permute
he/she/it would permute
we would permute
you would permute
they would permute
Past Conditional
I would have permuted
you would have permuted
he/she/it would have permuted
we would have permuted
you would have permuted
they would have permuted
ThesaurusAntonymsRelated WordsSynonymsLegend:
Verb 1 permute - change the order or arrangement of; "Dyslexics often transpose letters in a word"change by reversal, reverse, turn - change to the contrary; "The trend was reversed"; "the tides turned against him"; "public opinion turned when it was revealed that the president had an affair with a White House intern"map, represent - to establish a mapping (of mathematical elements or sets)
Translations
## permute
[pəˈmjuːt] VTpermutar
## permute
vtpermutieren
References in periodicals archive ?
trump, tempter, tempt, temper, rupee, rump, repute, putter, puttee, putt, purr, purer, puree, pure, petter, peter, pert, permute, perm, peer, erupt, trumpet.
In order to reformulate relation (1) as a pseudo-band correlation form we introduce some auxiliary sequences and use the proprieties of the Galois Field of indexes to appropriate permute the input and output sequences.
Fake bowlers: England does not permute good bowlers after Swann's departure.
where rand is a random number in [0,1] range, permute 1 and permute 2 are different rows of permutation functions i is the number of water molecule, j is the number of dimensions of the problem.
Its core technology is called a mixnet: a series of servers that permute the order of traffic received, before passing it through.
i] are fabricated by permute the coefficients of polynomials [k.
4) Use the same seed number to permute the triangles and assign the faces to them.
The value rand() is a random number generator with an interval between 0 and 1 and permute 1 and permute 2 are different rows permutation functions applied to the nests matrix [28].
Treize coursiers de race pur-sang ont pris rendez-vous aujourd'hui a l'hippodrome du Parc des Loisirs de Laghouat qui a permute avec celui du Caroubier, en raison de la commemoration des festivites du 1er Novembre, ou une magnifique fresque sera organisee en la circonstance.
Before matricizing, we permute the indices of X where the modes are reordered from 1, 2, 3,4 to 1,3,2,4.
2,n]] Choose two crossover points a and b such that 1 [less than or equal to] a [less than or equal to] b [less than or equal to] n; Repeat Permute ([x.
Since most of the exercises are repeated in semitone intervals, there is adequate opportunity to permute through the basic vowel set /a/, /e/, /i/, /o/, /u/.
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Open / Close | 1,473 | 5,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | latest | en | 0.912548 |
http://www.top-law-schools.com/forums/viewtopic.php?f=3&t=119289 | 1,524,789,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948738.65/warc/CC-MAIN-20180427002118-20180427022118-00621.warc.gz | 517,313,036 | 10,791 | ## Math-wiz help with curves
(Study Tips, Dealing With Stress, Maintaining a Social Life, Financial Aid, Internships, Bar Exam, Careers in Law . . . )
stinger35
Posts: 614
Joined: Thu Jan 07, 2010 2:37 pm
### Math-wiz help with curves
I was wondering if anyone could tell me what these curves work out to be (i.e. 2.9, 3.0, whatever). I am sure it's not hard but I am absolutely terrible at Math (I thought that's why we went to law school)
A 10%-15%
B+ 20%-25%
B 25%-30%
C+ 20%-25%
C or below 15%-20%
and
A 12%-20%
B+ 21%-27%
B 25%-31%
C+ 16%-22%
C or below 10%-16%
Thanks dudes.
d34d9823
Posts: 1879
Joined: Wed Apr 14, 2010 2:52 pm
### Re: Math-wiz help with curves
Not that hard, but if you want a precise answer, you need more precise distributions. I've commented in the calc.
stinger35 wrote:I was wondering if anyone could tell me what these curves work out to be (i.e. 2.9, 3.0, whatever). I am sure it's not hard but I am absolutely terrible at Math (I thought that's why we went to law school)
A 10%-15% 4*.125
B+ 20%-25% 3.3*.225
B 25%-30% 3*.275
C+ 20%-25% 2.3*.225
C or below 15%-20% unsure what to use here, call it 1.5*.175
Sum them and we get 2.84
and
A 12%-20%
B+ 21%-27%
B 25%-31%
C+ 16%-22%
C or below 10%-16%
do same thing here and get 2.88
Thanks dudes.
pinkzeppelin
Posts: 231
Joined: Mon Dec 07, 2009 11:51 pm
### Re: Math-wiz help with curves
d34dluk3 wrote:Not that hard, but if you want a precise answer, you need more precise distributions. I've commented in the calc.
stinger35 wrote:I was wondering if anyone could tell me what these curves work out to be (i.e. 2.9, 3.0, whatever). I am sure it's not hard but I am absolutely terrible at Math (I thought that's why we went to law school)
A 10%-15% 4*.125
B+ 20%-25% 3.3*.225
B 25%-30% 3*.275
C+ 20%-25% 2.3*.225
C or below 15%-20% unsure what to use here, call it 1.5*.175
Sum them and we get 2.84
and
A 12%-20%
B+ 21%-27%
B 25%-31%
C+ 16%-22%
C or below 10%-16%
do same thing here and get 2.88
Thanks dudes.
This is good, but since the summation of the assumed percentiles don't sum to 1 for the first one, you are over estimating. Since the percentages sum to 1.025... multiply each by 1/1.025 and then run the calculation again. You get 2.78 for the first one. The second calculation is fine.
roofles
Posts: 97
Joined: Mon Feb 22, 2010 10:05 pm
### Re: Math-wiz help with curves
In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
stinger35
Posts: 614
Joined: Thu Jan 07, 2010 2:37 pm
### Re: Math-wiz help with curves
roofles wrote:In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
Yea, that actually is true. B+ is 3.5, C+ is 2.5, etc.
So I guess it needs to be recalculated. (I don't understand why they don't have minus grades. I would have much preferred A-'s over the two B+'s I received, if only for transfer purposes.
d34d9823
Posts: 1879
Joined: Wed Apr 14, 2010 2:52 pm
### Re: Math-wiz help with curves
stinger35 wrote:
roofles wrote:In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
Yea, that actually is true. B+ is 3.5, C+ is 2.5, etc.
So I guess it needs to be recalculated. (I don't understand why they don't have minus grades. I would have much preferred A-'s over the two B+'s I received, if only for transfer purposes.
Yes for school GPA, no for LSAC GPA. From how I read their chart LSAC counts +/- grades as += +.3, -= -.3
stinger35
Posts: 614
Joined: Thu Jan 07, 2010 2:37 pm
### Re: Math-wiz help with curves
d34dluk3 wrote:
stinger35 wrote:
roofles wrote:In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
Yea, that actually is true. B+ is 3.5, C+ is 2.5, etc.
So I guess it needs to be recalculated. (I don't understand why they don't have minus grades. I would have much preferred A-'s over the two B+'s I received, if only for transfer purposes.
Yes for school GPA, no for LSAC GPA. From how I read their chart LSAC counts +/- grades as += +.3, -= -.3
True, but that doesn't matter for transferring. Any help figuring it out with the different numbers? I think it should be around a 2.8 for the first 2.9 for the second?
d34d9823
Posts: 1879
Joined: Wed Apr 14, 2010 2:52 pm
### Re: Math-wiz help with curves
stinger35 wrote:
d34dluk3 wrote:
stinger35 wrote:
roofles wrote:In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
Yea, that actually is true. B+ is 3.5, C+ is 2.5, etc.
So I guess it needs to be recalculated. (I don't understand why they don't have minus grades. I would have much preferred A-'s over the two B+'s I received, if only for transfer purposes.
Yes for school GPA, no for LSAC GPA. From how I read their chart LSAC counts +/- grades as += +.3, -= -.3
True, but that doesn't matter for transferring. Any help figuring it out with the different numbers? I think it should be around a 2.8 for the first 2.9 for the second?
What pinkzep said.
Tautology
Posts: 434
Joined: Thu Mar 18, 2010 12:40 pm
### Re: Math-wiz help with curves
stinger35 wrote:
d34dluk3 wrote:
stinger35 wrote:
roofles wrote:In the absence of minus grades, wouldn't the plus grades be 3.5 for a B+, 2.5 for a C+, etc.? My undergrad didn't issue minus grades and that is how they calculated gpa.
Yea, that actually is true. B+ is 3.5, C+ is 2.5, etc.
So I guess it needs to be recalculated. (I don't understand why they don't have minus grades. I would have much preferred A-'s over the two B+'s I received, if only for transfer purposes.
Yes for school GPA, no for LSAC GPA. From how I read their chart LSAC counts +/- grades as += +.3, -= -.3
True, but that doesn't matter for transferring. Any help figuring it out with the different numbers? I think it should be around a 2.8 for the first 2.9 for the second?
I'll do the first one for you, and maybe you can do the second; the steps are pretty straightforward.
First, you'll want to decide on a precise percentage who gets each grade. Even if you don't know exactly what they are, you want to pick one for the calculation. Probably the best thing to do (as was done above) is to just go right down the middle (take the average) of the two ends of the range. Thus if we know that 10%-15% get A's, we'll make that number 12.5%. Doing this for all of them in the first example yields
A 12.5%
B+ 22.5%
B 27.5%
C+ 22.5%
F-C 17.5%
A problem arises here though, because those numbers don't add up to 100%, but 102.5%. I will fix this at the end of my calculation because it's easier there. Since this problem doesn't exist with the second set of numbers you provided, you won't have to worry about that step.
Now, remember that 12.5% can be expressed in decimal form as .125. Similarly, 22.5% can be expressed as .225. We need to convert the percentages we have above to decimals so that we can multiply them by the value given for each letter grade. That looks like this.
A .125
B+ .225
B .275
C+ .225
F-C .175
Now all we have to do is multiply each of those numbers by the value assigned to each letter grade, and add those together. Thus we get:
A .125 x 4.0 = .5
B+ .225 x 3.5 = .7875
B .275 x 3.0 = .825
C+ .225 x 2.5 = .5625
F-C .175 x 1.5 = .2625
That last category, everything below a C+, doesn't have a specific number attached to it like the rest, so we sort of have to guess. I used 1.5, the value used above, but in reality it may be higher than 1.5 if Fs and Ds are very rare, and lower than 1.5 if Fs and Ds are common.
Finally, we just add all those numbers up and get 2.94 (rounded). That would be the average GPA given that curve, and is the end of the steps for the second set of numbers, but since we had a problem (mentioned above) with our numbers not adding up to 100%, we have to divide it by 1.025 (102.5%). This leaves us with 2.87 (rounded) for our average GPA. | 2,561 | 8,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-17 | latest | en | 0.917076 |
https://eomys.com/e-nvh/technical-notes-on-electromagnetically-excited-noise-and-vibrations/article/what-is-a-frf-frequency-response-function | 1,657,059,780,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00192.warc.gz | 282,086,956 | 6,431 | # What is a FRF (Frequency Response Function)?
### Definition
A Frequency Response Function (FRF) is a function used to quantify the response of a system to an excitation, normalized by the magnitude of this excitation, in the frequency domain.
For instance, impacting a structure with an impact hammer and measuring its structural response with an accelerometer normalized by the injected force, the structural FRF is obtained. Then the information of the modal response of the structure is included in the FRF, in particular modal damping and natural frequencies.
### Application of the vibration FRF to e-NVH
The structural response of electrical machines under electromagnetic excitations is generally quantified using the normal complex displacement [m] of the envelopped nodes (stator or rotor), while the excitation of the structure is quantified using Maxwell stress waves [N/m^2] or equivalent magnetic forces per tooth [N]. The vibration FRF unit is thefore in [m/N/m^2] or [m/N]. The FRF can be expressed at each point of the outer motor surface, or as a RMS value over the vibrating surface responsible for acoustic noise radiation. This average vibration velocity can be used as an indicator of the sound power level radiated by the electrical machine assuming that the modal radiation efficiency is close to one (ERP model).
The response of the outer rotor or stator under magnetic forces can be quantified using different approaches: the stator teeth can be excited by radial & tangential elementary tooth forces, giving as many FRF as the stator teeth number, or they can be excited by the Maxwell stress waves, giving as many FRF as the number of Maxwell stress wavenumbers coming from its Fourier decomposition along the airgap.
### Application of the vibration FRF to Manatee
These two excitation methods are used in the Electromagnetic Vibration Synthesis algorithm of Manatee software.
An example of FRF is given in the following figure. One can see that the FRF is maximum when there is a resonance, for instance when the FRF of Maxwell stress wavenumber 2 meets the natural frequencies of the elliptical model (2,0) of the stator stack.
### Application of the acoustic FRF to e-NVH
The acoustic response of electrical machines under electromagnetic excitations is generally quantified using the Sound Power Level radiated by the outer yoke (stator or rotor), while the excitation of the structure is quantified using Maxwell stress waves [N/m^2] or equivalent magnetic forces per tooth [N]. The acoustic FRF unit is thefore in [W/(N/m^2)^2] or [W/N^2] as the acoustic power evolves with the vibration to the power 2. The acoustic FRF can also be locally expressed using Sound Pressure Level on a pre-defined acoustic mesh simulating virtual microphones around the outer yoke.
The response of the outer rotor or stator under magnetic forces can be quantified using Maxwell stress wave excitations, giving as many FRF as the number of Maxwell stress wavenumbers coming from its Fourier decomposition along the airgap. | 640 | 3,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.904309 |
http://openstudy.com/updates/521b7a4ee4b06211a67d2d41 | 1,519,035,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812579.21/warc/CC-MAIN-20180219091902-20180219111902-00262.warc.gz | 269,393,168 | 8,357 | • anonymous
MEDAL FOR THE CORRECT ANSWER! As part of an Earth Science project, Erin tracked the amount of rain fall for 4 weeks. She started tracking after finding some initial data on the year's rainfall to date. Week Rainfall 1 5.5 2 8 3 10.5 4 13 If the linear function representing the amount of rainfall each week is f(w) = 2.5w + 3. What does the 3 represent in this function? The amount of rainfall in week 0. The week that Erin collected data. The amount of total rainfall. The amount the rainfall increased each week.
Mathematics
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Not the answer you are looking for? Search for more explanations. | 380 | 1,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-09 | latest | en | 0.498543 |
https://www.examrace.com/CAT/CAT-Mock-Practice-Questions/CAT-Model-Paper-5-Questions-and-Answers-with-Explanation-Part-7.html | 1,600,752,356,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00681.warc.gz | 867,493,949 | 9,290 | # CAT Model Paper 5 Questions and Answers with Explanation Part 7
Get unlimited access to the best preparation resource for CTET/Paper-2 : get questions, notes, tests, video lectures and more- for all subjects of CTET/Paper-2.
Start Passage
Directions for Questions 31 to 33: Solve the questions given below on the basis on the following information:
There are six males A, B, C, D, E and F with professions Professor, Lawyer, Doctor, Engineer, Teacher and Businessman, not necessarily in that order. They sit around a circular table with six females L, M, N, P, Q and R who occupy Six chairs around the table, not necessarily in that order. Each male stands behind his female partner. The following information is also known:
A is two places away from F, who is two places to the right of the lawyer.
The Doctor is to the right of B and to the left of P.
E stands behind M and they are adjacent to R.
The Doctor stands behind L.
The Professor stands between A and E.
C is a Teacher and is standing adjacent to F, who is a Businessman.
Q: 31. What is the profession of partner?
(A) Teacher
(B) Lawyer
(C) Professor
(D) Engineer
Ans: (C)
Solution
In the diagram, the position of Q and N can be interchanged.
As can be seen from the diagram, P’s partner is D, who is a professor.
Hence, option C
Q: 32. Who is partner?
(A)
(B)
(C) R
(D) L
Ans: (R)
Solution
In the diagram, the position of Q and N can be interchanged.
As can be seen from the diagram, C’s partner is R.
Hence, option C
Q: 33. If C is not adjacent to N, then who is opposite Q?
(A) Teacher
(B) Professor
(C) Doctor
(D) Lawyer
Ans: (B)
Solution:
In the diagram, the position of Q and N can be interchanged.
If C is not adjacent to N, then D, who is a Professor, will be opposite to Q.
End Passage
Q: 34. The sentence below has two blanks followed by four pairs of words as choices. From the choices, select the pair of words that can best complete the given sentence.
_________ of the armed forces could yet be ________ if the military refuses to accept into service locally developed weaponry on the grounds that better equipment can be bought elsewhere.
(A) Indigenization, scuttled
(B) Globalisation, stopped
(C) Modernization, encouraged
(D) Transformation, stymied
Ans: (A)
Solution:
The sentence says that if the military refuses to accept locally made weaponry, then what?? Local goes with indigenization which means developing equipment at home rather than importing it from abroad. If the military does not accept this, indigenization plans with be stopped or scuttled, which means cannot be carried out. So Indigenization, scuttled. Stymied also means scuttled, but transformation is not as relevant as indigenization to locally developed weaponry.
Developed by: | 670 | 2,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-40 | longest | en | 0.936524 |
https://sourcetable.com/formula/intercept | 1,718,908,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00380.warc.gz | 475,251,814 | 12,843 | # INTERCEPT
Formulas / INTERCEPT
Calculate the intercept of a regression line.
`INTERCEPT(known_y's,known_x's)`
• known_x - required, independent set of data
• known_y - required, dependent set of data
## Examples
• `=INTERCEPT({2;0},{-1;1})`
The function is used to calculate the point of intersection between two linear functions. For example, the function returns 1 because the first argument is {2;0} which is the line 2x+0 and the second argument is {-1;1} which is the line -1x+1. The result is 1, which is the point of intersection between the two linear functions.
• `=INTERCEPT({2;3},{-1;2})`
The function can be used to calculate the slope of a line. For example, the function returns -1. This is because the first argument is {2;3} which is the line 2x+3 and the second argument is {-1;2} which is the line -1x+2. The result is -1, which is the slope of the line.
• `=INTERCEPT({3;-2},{1;3})`
The function can also be used to calculate the y-intercept of a line. For example, the function returns -2. This is because the first argument is {3;-2} which is the line 3x-2 and the second argument is {1;3} which is the line 1x+3. The result is -2, which is the y-intercept of the line.
## Summary
The INTERCEPT function allows users to calculate the intercept point of a line by providing existing x-values and y-values. It is a simple and convenient way to calculate this point without having to manually calculate it.
• The INTERCEPT function calculates the intercept of a regression line, using an array or range of known_ys and known_xs values as arguments.
• The INTERCEPT function can be used to calculate the value of a dependent variable when the independent variable is zero.
What is the INTERCEPT function?
The INTERCEPT function is available in Excel 2007 and later versions. It returns the intercept for a linear regression. The INTERCEPT function is improved in later versions of Excel.
How do I use the INTERCEPT function?
To use the INTERCEPT function, enter the known x-values in one column, the known y-values in a second column, and the INTERCEPT function in the third column. The function will return the intercept of the linear regression based on the data.
What versions of Excel support the INTERCEPT function?
The INTERCEPT function is available in Excel 2007 and later versions.
What are the benefits of using the INTERCEPT function?
• It quickly returns the intercept of a linear regression.
• The function is improved in later versions of Excel.
• It is simple to use. | 605 | 2,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.855476 |
https://brainly.ph/question/251730 | 1,485,259,958,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284405.58/warc/CC-MAIN-20170116095124-00361-ip-10-171-10-70.ec2.internal.warc.gz | 790,507,344 | 10,339 | # What is the image distance and image height if a 7.00 cm tall object is place 30.0 cm from a concave mirror having a focal length of 10.0 cm?
1
by llantojoyceann
2015-11-01T19:28:39+08:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Given:
ho = 7 cm
do = 30 cm
f = 10 cm
Find:
di = ?
hi = ?
Solution:
di = 1.499 cm
m = -0.0499
hi = m x ho
hi = (-0.0499)(7)
hi = -0.3493 cm | 196 | 644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-04 | latest | en | 0.900174 |
http://mathhelpforum.com/algebra/40202-algebra-print.html | 1,503,341,514,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00482.warc.gz | 277,218,154 | 4,898 | # algebra
• May 31st 2008, 02:58 PM
Rakaa
algebra
I'm helping my little sister with her HW and its been years since i've seen this stuff...
How would I solve
z-6 - 5z
2z+3 2z+3
• May 31st 2008, 03:02 PM
Plato
The two fractions have the same denominator.
• May 31st 2008, 03:06 PM
angel.white
Quote:
Originally Posted by Rakaa
I'm helping my little sister with her HW and its been years since i've seen this stuff...
How would I solve
z-6 - 5z
2z+3 2z+3
You are wanting to simplify? (solve implies that you want to find the value of Z, but you can't do that if you don't know what it is equal to)
• May 31st 2008, 03:25 PM
Rakaa
Quote:
Originally Posted by angel.white
You are wanting to simplify? (solve implies that you want to find the value of Z, but you can't do that if you don't know what it is equal to)
Yes I want to simplify
• May 31st 2008, 04:00 PM
angel.white
Quote:
Originally Posted by Rakaa
I'm helping my little sister with her HW and its been years since i've seen this stuff...
How would I solve
z-6 - 5z
2z+3 2z+3
$\frac {z-6}{2z+3} - \frac {5z}{2z+3}$
They both have the same denominator, so you can combine them:
$=\frac {z-6-5z}{2z+3}$
z and -5z have the same variable so you could look at it like this: z -z -z -z -z -z
it should be clear then that they combine to equal -4z
$=\frac {-4z-6}{2z+3}$
both terms in the numerator are multiples of -2, so factor out a -2
$=\frac {-2(2z+3)}{2z+3}$
And since -2 is multiplied and divided by 2z+3, the 2z+3 cancels out to equal one
$=-2*1$
Anything times 1 is itself
$=-2$
• May 31st 2008, 09:05 PM
Rakaa
How would I do
y-3 - 2y-7
y+5 y+5
It would become
y-3 - 2y-7
y+5
Then
-1y-(-4)
y+5
And then I become totally lost...
• May 31st 2008, 09:09 PM
o_O
$\frac{y-3}{y+5} - \frac{2y - 7}{y+5}$
$= \frac{y - 3 {\color{red} \: - \:} (2y - 7)}{y + 5}$
$= \frac{y - 3 {\color{red} \: - \:}2y {\color{red} \: - \:} (-7)}{y+5}$
Negative of a negative = Positive
• May 31st 2008, 09:13 PM
angel.white
Quote:
Originally Posted by Rakaa
How would I do
y-3 - 2y-7
y+5 y+5
It would become
y-3 - 2y-7
y+5
Then
-1y-(-4)
y+5
And then I become totally lost...
The minus sign applies to both terms look at it like this:
$\frac {y-3}{y+5} - \frac {2y-7}{y+5}$
combine
$\frac {y-3-(2y-7)}{y+5}$
distribute the minus sign
$\frac {y-3-2y+7}{y+5}$
combine like terms
$\frac {-y+4}{y+5}$
There are other ways to simplify from here, but "simplify" is kind of ambiguous, this answer is probably what her book is looking for.
Anyway, with that negative sign, it applies to the whole term. You could think of it like this also:
$\frac {y-3}{y+5} +(-1)* \frac {2y-7}{y+5}$
Or it may help to work backwards:
$\frac {-2y+7}{y+5}$
= $\frac {(-1)2y+(-1)(-1)7}{y+5}$
= $\frac {(-1)(2y+(-1)7)}{y+5}$
= $\frac {(-1)(2y-7)}{y+5}$
= $(-1)\frac {(2y-7)}{y+5}$
= $-\frac {2y-7}{y+5}$
However you want to look at it, just understand the sign out front needs to get distributed.
• May 31st 2008, 09:30 PM
Rakaa
Thanks a million.. :)
I really wish I wouldve kept my mind sharp with all of this...
z+6 + 3z2+19z+19
z+5 z+5
next
z+6 + 3z2+19z+19
z+5
29z+26
z+5
Whats the next step? I'm still semi confused about that..
• May 31st 2008, 09:44 PM
angel.white
Quote:
Originally Posted by Rakaa
Thanks a million.. :)
I really wish I wouldve kept my mind sharp with all of this...
z+6 + 3z2+19z+19
z+5 z+5
next
z+6 + 3z2+19z+19
z+5
29z+26
z+5
Whats the next step? I'm still semi confused about that..
with this one, you cannot combine $z^2$ and z. Another way of looking at it which might help is to factor out the Z.
so z+19z = z(1+19) = 20z
but $3z^2 + 19z = z(3z+19)$
3z and 19 are not like terms. So $z^2$ and z are not like terms.
You would do it like this:
$\frac {z+6+3z^2+19z+19}{z+5}$
$= \frac {3z^2+20z+25}{z+5}$
Split up the numerator
$= \frac {3z^2+15z+5z+25}{z+5}$
group them (you don't have to do all these steps I'm doing, but I want to hit every step along the way to make it easier to see)
$= \frac {(3z^2+15z)+(5z+25)}{z+5}$
Factor out a 3z
$= \frac {3z(z+5)+(5z+25)}{z+5}$
Factor out a 5
$= \frac {3z(z+5)+5(z+5)}{z+5}$
Factor out a (z+5)
$= \frac {(z+5)(3z+5)}{z+5}$
z+5 is in the numerator and the denominator, so it cancels out
$= 3z+5$
• Jun 1st 2008, 11:41 AM
Rakaa
One more.. I'm starting to get it..
4y2+zy-3 - 3y2-2y-4
5y+1 z+5
• Jun 1st 2008, 01:06 PM
Reckoner
Quote:
Originally Posted by Rakaa
One more.. I'm starting to get it..
4y2+zy-3 - 3y2-2y-4
5y+1 z+5
The only difference here is that you don't have a common denominator, so you will have to make one before you can combine the fractions.
In simple arithmetic, you combine fractions without a common denominator by first finding the least common multiple of both denominators, and then multiplying the numerator and denominator of both fractions by the necessary amount to change the denominator to that multiple. For example:
$\frac12 + \frac23$
The least common multiple of 2 and 3 is 6, so we do
$\frac12 + \frac23$
$=\frac{1\color{red}\cdot3}{2\color{red}\cdot3} + \frac{2\color{red}\cdot2}{3\color{red}\cdot2}$
$=\frac36 + \frac46 = \frac{3 + 4}6 = \frac76$
This basic process doesn't change with the introduction of variable expressions. We have:
$\frac{4y^2+zy-3}{5y+1} - \frac{3y^2-2y-4}{z+5}$
The least common multiple of $5y + 1$ and $z + 5$ is $(5y + 1)(z + 5)$, so we simplify thus:
$\frac{4y^2+zy-3}{5y+1} - \frac{3y^2-2y-4}{z+5}$
$=\frac{\left(4y^2+zy-3\right){\color{red}(z + 5)}}{(5y+1){\color{red}(z + 5)}} - \frac{\left(3y^2-2y-4\right){\color{red}(5y + 1)}}{{\color{red}(5y + 1)}(z+5)}$
$=\frac{\left(4y^2+zy-3\right)(z + 5) - \left(3y^2-2y-4\right)(5y + 1)}{(5y + 1)(z+5)}$
You should be able to take it from here. | 2,261 | 5,731 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 42, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-34 | longest | en | 0.942278 |
https://help.xlstat.com/customer/es/portal/articles/2752199-skewness-and-kurtosis-in-excel?b_id=9283 | 1,566,488,992,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00126.warc.gz | 491,319,911 | 12,823 | Su solución de análisis de datos
# Skewness and Kurtosis in Excel
20/10/2017
This tutorial shows how to compute and interpret skewness and kurtosis in Excel using the XLSTAT software.
## Dataset for computing skewness and kurtosis
An Excel sheet with both the data and the results can be downloaded by clicking on the button below:
The data represent the time needed to complete two online assessments, one in maths and another one in logical reasoning, by pupils of three different schools. Time is measured in minutes. Rows correspond to pupils and columns to the time spent for each of the two assessments as well as the school they belong to.
Our goal here is to study two specific characteristics of a given distribution:
1. The skewness, that reflects the asymmetry of a distribution
2. The kurtosis, that reflects the characteristics of the tails of a distribution.
For this purpose, we will use the XLSTAT Descriptive Statistics tools. We will compute and interpret the skewness and the kurtosis on time data for each of the three schools.
## Setting up the dialog box for computing skewness and kurtosis
1. Once XLSTAT is open, select the XLSTAT / Describing data / Descriptive statistics command as shown below.
2. The Descriptive Statistics dialog box appears.
3. In the General tab, select the columns corresponding to the time spent on each assessment in the Quantitative data field.
Then select the column corresponding to the school name in the Subsamples field.
We also want to display Variable-Category labels in the output. These include the variable name as a prefix and the category name as a suffix.
Finally, select the Sheet option in order to display the results on a new sheet and the Sample labels to consider the first row of the data table as labels.
4. In the Options tab, activate the following options.
5. In the Outputs tab, click on the All button to select all the statistics for quantitative data. You can also select the statistics you’re interested in one by one.
## How to interpret Skewness and Kurtosis
The results are displayed on a new sheet named Desc. A full set of descriptive statistics is displayed per school (columns C-H). XLSTAT proposes several coefficients of skewness and kurtosis. In this example, we will be referring to the Fisher coefficients which are not biased on the assumption that the data is normally distributed.
Formulas can be found in the XLSTAT Help menu (click on the Help button in the dialog box).
### a. Interpreting the skewness
Skewness measures the asymmetry of a distribution. A distribution is called asymmetric when one tail is longer than the other. If the skewness is positive, then the distribution is skewed to the right while a negative skewness implies a distribution skewed to the left. A zero skewness suggests a perfectly symmetric distribution.
In this part, we will interpret results related to the maths assessment (see below).
The three samples seem to have contrasted skewness coefficients:
• Sample A has a strong positive skewness (1.42). This reflects a long distribution tail on the right.
• Sample B has a strong negative skewness (-1.63). This reflects a long distribution tail on the left
• A zero skewness is estimated for sample C. In fact, the median of sample C (49.8) is almost identical to the mean value (49.6).
Histograms allow us to confirm the above observations. The top histogram (sample A) shows a distribution skewed to the right, the second one (sample B) is a distribution skewed to the left while the third one (sample C) a symmetric one.
### b. Interpreting the kurtosis
Kurtosis provides information on the tails (the extremes, or outliers) of a distribution. When interpreting kurtosis, the normal distribution is used a reference. A positive kurtosis implies a distribution with more extreme possible data values (outliers) than a normal distribution thus fatter tails (Leptokurtic distributions). A negative kurtosis implies a distribution with less extreme possible data values than a normal distribution thus thinner tails (Platykurtic distributions). Finally, distributions with zero kurtosis have roughly the same outlier character as a normal distribution (Mesokurtic distributions).
In this section, we will interpret the results related to the second assessment (see below). Based on the coefficients above, the shape of the three distributions differ in terms of kurtosis:
• A positive kurtosis is estimated for School A (5.40)
• A negative one is estimated for School B (-1.32)
• A zero kurtosis was detected for School C.
Histograms confirm these observations. The top histogram (sample A) shows a leptokurtic distribution, while the second one (sample B) shows a platykurtic distribution. The third one (sample C) displays a distribution with a shape similar to the shape of a normal distribution.
Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4321753/
## What’s next: How to visualize skewness and kurtosis
The Histogram is commonly used to examine the distribution of numerical data. It allows to observe the tail and the peak of a frequency distribution in a single chart. On the top, it provides information on the central tendency, data dispersion as well as the presence of outliers. Here’s how to generate histograms in XLSTAT
### Contacto
#### Contactar con nuestro soporte técnico : support@xlstat.com
https://cdn.desk.com/
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desk
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hace #{num} segundos
hace un minuto
hace #{num} minutos
hace una hora
hace #{num} horas
hace un día
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/customer/portal/articles/autocomplete
9283 | 1,285 | 5,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-35 | longest | en | 0.809841 |
https://calculatorshub.net/lifestyle-calculators/parking-lot-salt-calculator/ | 1,713,266,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00005.warc.gz | 132,566,271 | 29,971 | Home » Simplify your calculations with ease. » Lifestyle Calculators » Parking Lot Salt Calculator | Maximizing Winter Safety
# Parking Lot Salt Calculator | Maximizing Winter Safety
Parking lot salt is a critical component of winter maintenance. When applied correctly, it prevents the buildup of ice and enhances the safety of the environment. The application of this salt plays a crucial role in commercial, residential, and public parking lots, especially in colder climates.
Now, before we proceed, what is the Parking Lot Salt Calculator? It’s a specialized tool designed to accurately calculate the exact amount of salt needed for a parking lot based on its area. The tool uses a specific formula to estimate the quantity of salt in pounds (lbs) required per square foot (ft²) of parking area. Let’s dive deeper into how this calculator works.
## Working of the Parking Lot Salt Calculator
The Parking Lot Salt Calculator operates on a simple mathematical formula. By inputting the total area of the parking lot into the calculator, it computes the appropriate amount of salt needed for effective ice prevention. This automated tool takes the guesswork out of the process, ensuring accuracy and efficiency in calculating salt requirements.
## Explanation of the Parking Lot Salt Formula and Its Variables
The Parking Lot Salt Calculator uses the formula: PLS = A / 1000 * 2.3. In this formula:
• PLS stands for Parking Lot Salt, represented in pounds (lbs)
• A represents the total parking lot area in square feet (ft²)
To calculate the required Parking Lot Salt, you divide the total parking lot area (A) by 1000 and then multiply the result by 2.3.
## Detailed Example: Calculating Parking Lot Salt
Let’s demonstrate the use of this formula with a real-world example. Suppose you have a parking lot with a total area of 545 ft².
1. Divide the total area by 1000: 545 / 1000 = 0.545
2. Then multiply the result by 2.3: 0.545 * 2.3 = 1.2535 lbs
Therefore, according to the Parking Lot Salt Calculator, you would need approximately 1.25 pounds of salt for a 545-square-foot parking lot.
## Applications of the Parking Lot Salt Calculator
The Parking Lot Salt Calculator has a wide array of applications. It’s primarily used by maintenance personnel responsible for ensuring safe parking conditions during winter. These individuals can range from property managers of commercial and residential buildings to municipal workers in charge of public parking lots.
It’s also a handy tool for companies providing winter maintenance services. By using this calculator, they can provide more accurate estimates to their clients, ensuring proper ice prevention without wastage of resources.
Moreover, it can assist in planning and budgeting for winter maintenance supplies. By knowing how much salt is required, businesses and municipalities can ensure they are sufficiently stocked for the winter season.
Can the Parking Lot Salt Calculator be used for any sized parking lot?
Yes, the Parking Lot Salt Calculator can accommodate any parking lot size, making it a versatile tool for different scenarios.
Is the formula used by the Parking Lot Salt Calculator applicable for extremely cold conditions?
The formula provides a general estimate. For extremely cold conditions where ice is more prevalent, it might be beneficial to apply additional salt.
## Conclusion
Understanding the importance of parking lot salt for winter safety and how to accurately calculate its requirement using the Parking Lot Salt Calculator is essential for efficient winter maintenance. This calculator, with its simple yet effective formula, ensures that you can maintain safe, ice-free parking lots while managing resources effectively. No matter the size or location of your parking lot, this tool is an invaluable asset in your winter maintenance toolkit. | 754 | 3,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-18 | longest | en | 0.88169 |
https://math.stackexchange.com/questions/3742045/how-can-i-find-int-0-infty-ln-n-leftx-right-e-axb-dx | 1,628,051,077,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00662.warc.gz | 371,980,347 | 36,859 | # How can i find $\int _0^{\infty }\ln ^n\left(x\right)\:e^{-ax^b}\:dx$
I tried using certain substitutions like $$u=ax^b$$ but that lead to $$\displaystyle\frac{1}{a^{\frac{1}{b}}b^n}\int _0^{\infty }e^{-u}\:\ln ^n\left(\frac{u}{a}\right)u^{\frac{1}{b}-1}du\:$$ i tried to use special functions to evaluate this but that $$\ln ^n\left(\frac{u}{a}\right)$$ is very annoying, i'd appreciate any help.
• What about $u=\ln(x)$? – EDX Jul 2 '20 at 3:37
You can start using the following identity, $$\int _0^{\infty }x^m\:e^{-ax^b}\:dx=\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}$$ You can now differentiate both sides $$n$$ times with respect to m and then set it to $$0$$, $$\int _0^{\infty }x^m\:\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}$$ $$\boxed{\int _0^{\infty }\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\lim _{m\to 0}\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}}$$ | 418 | 1,022 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-31 | latest | en | 0.546762 |
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# GMAT 2nd time - 690
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Manager
Joined: 05 May 2003
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GMAT 2nd time - 690 [#permalink]
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20 Jul 2003, 11:24
1
KUDOS
690 - 49(Q), 35(V) - 92percentile
Fist time 650 - 44(Q), 35(V) - 85percentile
Took the test today and am very satisfied with my performance. Though mid way through the test I doubted if I could improve on my score from my first take. Quant was pretty easy but I did get bogged down with some calculations... a few were pretty messy. I remember guessing 2 questions which I was having trouble with. Just managed to completed the section before the time expired.
SC was pretty straight forward. I recieved 4 RCs in all and found them to be easy. The first RC came immediately after the 1st question.
CR was alright too. Had one bold face CR question. it was the 40th question.
Comparing the test with my first attempt, I feel I did a better in math. Managed time in the Verbal section better than the first time( I had to guess an entire RC as I ran out of time) , but ended up getting the same score 35 as in the first attement.
If you have any question, I would be happy to answer them.
Special Thanks to Solyar for the exercises on Probability\Perm& Combination, I had no problem in answering the probability question I faced.
Last edited by KSS on 22 Jul 2003, 19:28, edited 1 time in total.
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20 Jul 2003, 11:40
Hi KSS,
Good job on the score improvement!
A couple of questions for you :
1. Do you think familiarity with the test, because of a previous attempt, helped you score better this time in any way?
2. Was the Permutation/Combination questions the hardest type you faced on the Quant section? What about inequalities (I tend to struggle a bit with these )
Thanks!
_________________
Get with the program!
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20 Jul 2003, 13:49
Nice work man.
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20 Jul 2003, 19:11
1
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Good boy!!!
What was your plan, when you were preparing for your second attempt?
how did you estimate your progress before exam?
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### Show Tags
20 Jul 2003, 23:10
1
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What kind of probability/combination questions did you get?
What are the RC topics: Sicene/Micro-biology, social science, business ...
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Re: GMAT 2nd time - 690 [#permalink]
### Show Tags
21 Jul 2003, 00:41
1
KUDOS
Expert's post
KSS wrote:
690 - 49(Q), 35(V) - 92percentile
Fist time 650 - 44(Q), 35(V) - 85percentile
Took the test today and am very satisfied with my performance. Though mid way through the test I doubted if I could improve on my score from my first take. Quant was pretty easy but I did get bogged down with some calculations... a few were pretty messy. I remember guessing 2 questions which I having trouble with. Just managed to completed the section before the time expired.
SC was pretty straight forward. I recieved 4 RCs in all and found them to be easy. The first RC came immediately after the 1st question.
CR was alright too. Had one bold face CR question. it was the 40th question.
Comparing the test with my first attempt, I feel I did a better in math. Managed time in the Verbal section better than the first time( I had to guess an entire RC as I ran out of time) , but ended up getting the same score 35 as in the first attement.
If you have any question, I would be happy to answer them.
Special Thanks to Solyar for the exercises on Probability\Perm& Combination, I had no problem in answering the probability question I faced.
Cheers to the Man!
That's great you are satisfied - there is nothing more pleasing than seeing return on the investment and being pleased with the outcome.
Well, now you are a true certified expert of the GMAT
Your score should be good currency anywhere...
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21 Jul 2003, 03:36
RK73 wrote:
Good boy!!!
What was your plan, when you were preparing for your second attempt?
how did you estimate your progress before exam?
My first attempt was on March 10, I have a post for that too, you could look that up.
I started preparing for the second attempt 3-4 weeks before the test date.
I basically worked on all the OG problems which I had got wrong earlier. I always mark something when i have problems with it. So I went back to the OG and did all those questions again. Took a kaplan and 2 Princeton review tests, and 2 PP tests. I worked on all the Probability questions on this site and that helped me alot. I even remember the question from the actual exam but I would not be wright for me to give it away. Trust me, if you go through the ones posted by Stolyar they should be good enough practice.
Good luck
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### Show Tags
21 Jul 2003, 03:38
What kind of probability/combination questions did you get?
What are the RC topics: Sicene/Micro-biology, social science, business ...
The probability and combination( actually it was permutation ) questions were very simple. Understand the concepts from Stolyars examples and you should be good to go.
RC surprisingly were very interesting and readable from me this time.
One was on science, one on business, one on history, and i dont remember the other.
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21 Jul 2003, 17:03
Good shot Mr. Tendulkar (Sometimeback, you had a avatar of tendulkar on this forumn).
Hard luck you Tendulkar missed his 50... but by a point. But i would say great show since it had the best of shots rom the book.
Well, not dealing too much into cricket. Congrauts KSS.
I should say you have done great job. Hard work pays.
All the best for your apps.
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21 Jul 2003, 20:05
evensflow wrote:
Good shot Mr. Tendulkar (Sometimeback, you had a avatar of tendulkar on this forumn).
Hard luck you Tendulkar missed his 50... but by a point. But i would say great show since it had the best of shots rom the book.
Well, not dealing too much into cricket. Congrauts KSS.
I should say you have done great job. Hard work pays.
All the best for your apps.
Ah. You remember me!!!
I am happy to have performed to my own expectations.
Yep Hard work pays and you need a bit of luck too....
So did you take the test yet?
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21 Jul 2003, 20:05
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# GMAT 2nd time - 690
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## matlab求两直线的夹角
(2011-05-31 20:56:38)
lyqmath
# 1 根据节点计算直线方程并求交点
## 代码
% By lyqmath
% DLUT School of Mathematical Sciences
% BLOG：http://blog.csdn.net/lyqmath
clc; clear all; close all;
%% 符号变量
syms x1 y1 x2 y2 x3 y3 x4 y4
syms C
syms y
%% 计算A1，B1，A2，B2
eq1 = A*x1 + B*y1 + C;
eq2 = A*x2 + B*y2 + C;
sov1 = solve(eq1, eq2, A, B);
eq3 = A*x3 + B*y3 + C;
eq4 = A*x4 + B*y4 + C;
sov2 = solve(eq3, eq4, A, B);
A1 = simplify(sov1.A / C);
B1 = simplify(sov1.B / C);
A2 = simplify(sov2.A / C);
B2 = simplify(sov2.B / C);
% A1 = (y1 - y2)/(x1*y2 - x2*y1);
% B1 = (-x1 + x2)/(x1*y2 - x2*y1);
% A2 = (y3 - y4)/(x3*y4 - x4*y3);
% B2 = (-x3 + x4)/(x3*y4 - x4*y3);
%% 求交点
eq1 = A1*x + B1*y + 1;
eq2 = A2*x + B2*y + 1;
sov = solve(eq1, eq2, x, y);
pretty(eq1)
pretty(eq2)
pretty(sov.x)
pretty(sov.y)
## 结果
% By lyqmath
% DLUT School of Mathematical Sciences
% BLOG：http://blog.csdn.net/lyqmath
function [px, py] = comput_intersec(line1, line2)
p1 = line1.point1; p2 = line1.point2; 第1条边的两点
p3 = line2.point1; p4 = line2.point2; 第2条边的两点
x1 = p1(1); y1 = p1(2);
x2 = p2(1); y2 = p2(2);
x3 = p3(1); y3 = p3(2);
x4 = p4(1); y4 = p4(2);
px = (x1*x3*y2 - x2*x3*y1 - x1*x4*y2 + x2*x4*y1 - x1*x3*y4 + x1*x4*y3 + x2*x3*y4 - x2*x4*y3 )/...
(x1*y3 - x3*y1 - x1*y4 - x2*y3 + x3*y2 + x4*y1 + x2*y4 - x4*y2);
py = (x1*y2*y3 - x2*y1*y3 - x1*y2*y4 + x2*y1*y4 - x3*y1*y4 + x4*y1*y3 + x3*y2*y4 - x4*y2*y3 )/...
(x1*y3 - x3*y1 - x1*y4 - x2*y3 + x3*y2 + x4*y1 + x2*y4 - x4*y2);
# 2 计算直线夹角
theta1 = acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])));
# 4 结论
0
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-5-review-page-331/125 | 1,548,234,339,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584328678.85/warc/CC-MAIN-20190123085337-20190123111337-00237.warc.gz | 807,066,765 | 13,152 | ## Intermediate Algebra (6th Edition)
$x=\frac{1}{3}$ and $x=-7$
$(3x-1)(x+7)=0$ Using the Zero Product Property $3x-1=0$ and $x+7=0$ Solving for x: $3x-1=0$ add $1$ to both sides $3x=1$ Divide by $3$ $x=\frac{1}{3}$ Solving for x: $x+7=0$ subtract $7$ from both sides $x=-7$ | 120 | 276 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-04 | latest | en | 0.615997 |
http://www.absoluteastronomy.com/topics/Chart | 1,550,693,074,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496080.57/warc/CC-MAIN-20190220190527-20190220212527-00322.warc.gz | 311,911,667 | 11,428 | Chart
# Chart
Discussion
Encyclopedia
A chart is a graphical representation of data
Data
The term data refers to qualitative or quantitative attributes of a variable or set of variables. Data are typically the results of measurements and can be the basis of graphs, images, or observations of a set of variables. Data are often viewed as the lowest level of abstraction from which...
, in which "the data is represented by symbol
Symbol
A symbol is something which represents an idea, a physical entity or a process but is distinct from it. The purpose of a symbol is to communicate meaning. For example, a red octagon may be a symbol for "STOP". On a map, a picture of a tent might represent a campsite. Numerals are symbols for...
s, such as bars in a bar chart
Bar chart
A bar chart or bar graph is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally....
, lines in a line chart
Line chart
A line chart or line graph is a type of graph, which displays information as a series of data points connected by straight line segments. It is a basic type of chart common in many fields. It is an extension of a scatter graph, and is created by connecting a series of points that represent...
, or slices in a pie chart
Pie chart
A pie chart is a circular chart divided into sectors, illustrating proportion. In a pie chart, the arc length of each sector , is proportional to the quantity it represents. When angles are measured with 1 turn as unit then a number of percent is identified with the same number of centiturns...
". A chart can represent tabular
Table (information)
A table is a means of arranging data in rows and columns.Production % of goalNorth 4087102%South 4093110% The use of tables is pervasive throughout all communication, research and data analysis. Tables appear in print media, handwritten notes, computer software, architectural...
numeric
Number
A number is a mathematical object used to count and measure. In mathematics, the definition of number has been extended over the years to include such numbers as zero, negative numbers, rational numbers, irrational numbers, and complex numbers....
data, functions
Graph of a function
In mathematics, the graph of a function f is the collection of all ordered pairs . In particular, if x is a real number, graph means the graphical representation of this collection, in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. Graphing on a Cartesian plane is...
or some kinds of qualitative structures.
The term "chart" as a graphical representation of data
Data
The term data refers to qualitative or quantitative attributes of a variable or set of variables. Data are typically the results of measurements and can be the basis of graphs, images, or observations of a set of variables. Data are often viewed as the lowest level of abstraction from which...
has multiple meanings:
• A data chart is a type of diagram
Diagram
A diagram is a two-dimensional geometric symbolic representation of information according to some visualization technique. Sometimes, the technique uses a three-dimensional visualization which is then projected onto the two-dimensional surface...
or graph
Graphics
Graphics are visual presentations on some surface, such as a wall, canvas, computer screen, paper, or stone to brand, inform, illustrate, or entertain. Examples are photographs, drawings, Line Art, graphs, diagrams, typography, numbers, symbols, geometric designs, maps, engineering drawings,or...
, that organizes and represents a set of numerical or qualitative data.
• Map
Map
A map is a visual representation of an area—a symbolic depiction highlighting relationships between elements of that space such as objects, regions, and themes....
s that are adorned with extra information for some specific purpose are often known as charts, such as a nautical chart
Nautical chart
A nautical chart is a graphic representation of a maritime area and adjacent coastal regions. Depending on the scale of the chart, it may show depths of water and heights of land , natural features of the seabed, details of the coastline, navigational hazards, locations of natural and man-made aids...
or aeronautical chart
Aeronautical chart
An aeronautical chart is a map designed to assist in navigation of aircraft, much as nautical charts do for watercraft, or a roadmap for drivers...
.
• Other domain specific constructs are sometimes called charts, such as the chord chart
Chord chart
A chord chart is a form of musical notation that in addition to writing out non-embellished melody, describes harmonic and rhythmic information. It is the most common form of notation used by professional session musicians playing jazz or popular music. It is intended primarily for a rhythm section...
in music notation or a record chart
Record chart
A record chart is a ranking of recorded music according to popularity during a given period of time. Examples of music charts are the Hit parade, Hot 100 or Top 40....
for album popularity.
Charts are often used to ease understanding of large quantities of data and the relationships between parts of the data. Charts can usually be read more quickly than the raw data that they are produced from. They are used in a wide variety of fields, and can be created by hand (often on graph paper
Graph paper
Graph paper, graphing paper, grid paper or millimeter paper is writing paper that is printed with fine lines making up a regular grid. The lines are often used as guides for plotting mathematical functions or experimental data and drawing diagrams. It is commonly found in mathematics and...
) or by computer using a charting application
Charting application
A charting application is a computer program that is used to graphically create a graphical representation based on some non-graphical data that is entered by a user, most often through a spreadsheet application, but also through a dedicated specific scientific application , or...
. Certain types of charts are more useful for presenting a given data set than others. For example, data that presents percentage
Percentage
In mathematics, a percentage is a way of expressing a number as a fraction of 100 . It is often denoted using the percent sign, “%”, or the abbreviation “pct”. For example, 45% is equal to 45/100, or 0.45.Percentages are used to express how large/small one quantity is, relative to another quantity...
s in different groups (such as "satisfied, not satisfied, unsure") are often displayed in a pie chart
Pie chart
A pie chart is a circular chart divided into sectors, illustrating proportion. In a pie chart, the arc length of each sector , is proportional to the quantity it represents. When angles are measured with 1 turn as unit then a number of percent is identified with the same number of centiturns...
, but may be more easily understood when presented in a horizontal bar chart
Bar chart
A bar chart or bar graph is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally....
. On the other hand, data that represents numbers that change over a period of time (such as "annual revenue from 1990 to 2000") might be best shown as a line chart
Line chart
A line chart or line graph is a type of graph, which displays information as a series of data points connected by straight line segments. It is a basic type of chart common in many fields. It is an extension of a scatter graph, and is created by connecting a series of points that represent...
.
## Features of a chart
A chart can take a large variety ; however, there are common features that provide the chart with its ability to extract meaning from data.
Typically a chart is graphical, containing very little text, since humans are generally able to infer meaning from pictures quicker than from text. One of the more important uses of text in a graph is in the title. A graph's title usually appears above the main graphic and provides a succinct description of what the data in the graph refers to.
Dimensions in the data are often displayed on axes. If a horizontal and a vertical axis are used, they are usually referred to as the x-axis and y-axis respectively. Each axis will have a scale, denoted by periodic graduations and usually accompanied by numerical or categorical indications. Each axis will typically also have a label displayed outside or beside it, briefly describing the dimension represented. If the scale is numerical, the label will often be suffixed with the unit of that scale in parentheses. For example, "Distance traveled (m)" is a typical x-axis label and would mean that the distance travelled in metres is related to the horizontal position of the data.
Within the graph a grid of lines may appear to aid in the visual alignment of data. The grid can be enhanced by visually emphasizing the lines at regular or significant graduations. The emphasized lines are then called major grid lines and the rest of the grid lines are minor grid lines.
The data of a chart can appear in all manner of formats, with or without individual labels. It may appear as dots or shapes, connected or unconnected, and in any combination of colors and patterns. Inferences or points of interest can be overlayed directly on the graph to further aid information extraction.
When the data appearing in a chart contains multiple variables, the chart may include a legend. A legend contains a list of the variables appearing in the chart and an example of their appearance. This information allows the data from each variable to be identified in the chart. by: Jemimah
### Common charts
Four of the most common charts are:
This gallery shows:
• A histogram
Histogram
In statistics, a histogram is a graphical representation showing a visual impression of the distribution of data. It is an estimate of the probability distribution of a continuous variable and was first introduced by Karl Pearson...
consists of tabular frequencies, shown as adjacent rectangles, erected over discrete intervals (bins), with an area equal to the frequency of the observations in the interval.
• A bar chart
Bar chart
A bar chart or bar graph is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally....
is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally.
• A pie chart
Pie chart
A pie chart is a circular chart divided into sectors, illustrating proportion. In a pie chart, the arc length of each sector , is proportional to the quantity it represents. When angles are measured with 1 turn as unit then a number of percent is identified with the same number of centiturns...
shows percentage values as a slice of a pie.
• A line chart
Line chart
A line chart or line graph is a type of graph, which displays information as a series of data points connected by straight line segments. It is a basic type of chart common in many fields. It is an extension of a scatter graph, and is created by connecting a series of points that represent...
is a two-dimensional scatterplot of ordered observations where the observations are connected following their order.
Other common charts are:
### Less-common charts
Examples of less common charts are:
This gallery shows:
• A bubble chart
Bubble chart
A bubble chart is a type of chart where each plotted entity is defined in terms of three distinct numeric parameters. Bubble charts can facilitate the understanding of the social, economical, medical, and other scientific relationships.- Overview :...
is a two-dimensional scatterplot where a third variable is represented by the size of the points.
• A Polar area diagram, sometimes called a Coxcomb chart developed by Florence Nightingale
Florence Nightingale
Florence Nightingale OM, RRC was a celebrated English nurse, writer and statistician. She came to prominence for her pioneering work in nursing during the Crimean War, where she tended to wounded soldiers. She was dubbed "The Lady with the Lamp" after her habit of making rounds at night...
is an enhanced form of pie chart.
A radar chart is a graphical method of displaying multivariate data in the form of a two-dimensional chart of three or more quantitative variables represented on axes starting from the same point...
or "spider chart" is a two-dimensional chart of three or more quantitative variables represented on axes starting from the same point.
• A waterfall chart
Waterfall chart
Waterfall Chart is a form of data visualization which helps in determining the cumulative effect of sequentially introduced positive or negative values. The waterfall chart is also known as a Flying Bricks Chart or Mario Chart due to the apparent suspension of columns in mid-air...
also known as a "Walk" chart, is a special type of floating-column chart.
• A Tree Map
Treemapping
In information visualization and computing, treemapping is a method for displaying hierarchical data by using nested rectangles.- Main idea :...
where the areas of the rectangles correspond to values. Other dimensions can be represented with colour or hue.
### Field-specific charts
Some types of charts have specific uses in a certain field
This gallery shows:
• Stock market prices are often depicted with a open-high-low-close chart
Open-high-low-close chart
An open-high-low-close chart is a type of chart typically used to illustrate movements in the price of a financial instrument over time. Each vertical line on the chart shows the price range over one unit of time, e.g. one day or one hour...
with a traditional bar chart of volume at the bottom.
• Candlestick chart
Candlestick chart
A candlestick chart is a style of bar-chart used primarily to describe price movements of a security, derivative, or currency over time.It is a combination of a line-chart and a bar-chart, in that each bar represents the range of price movement over a given time interval. It is most often used in...
s are another type of bar chart used to describe price movements of an equity over time.
• A Kagi chart
Kagi chart
The Kagi chart is a chart used for tracking price movements and to make decisions on purchasing stock. It differs from traditional stock charts, such as the Candlestick chart by being mostly independent of time...
is a time-independent stock tracking chart that attempts to minimise noise.
• Alternatively, where less detail is required, and chart size is paramount, a Sparkline
Sparkline
A sparkline is a type of information graphic characterized by its small size and high data density. Sparklines present trends and variations associated with some measurement, such as average temperature or stock market activity, in a simple and condensed way...
may be used.
Other examples:
• Interest rates, temperatures, etc., at the close of the period are plotted with a line chart.
• Project planners use a Gantt chart
Gantt chart
A Gantt chart is a type of bar chart that illustrates a project schedule. Gantt charts illustrate the start and finish dates of the terminal elements and summary elements of a project. Terminal elements and summary elements comprise the work breakdown structure of the project. Some Gantt charts...
to show the timing of tasks as they occur over time.
### Well-known named charts
Some of the better-known named charts are:
Some specific charts have become well known by effectively explaining a phenomenon or idea.
• An Allele chart
Allele chart
An Allele chart is a type of chart that shows the interaction of two data points at the grid intersection of their respective axes. The most common example of such a chart is from the multiplication table chart taught in many United States elementary schools, though the actual origins of this type...
is a chart originating from the study of genetics
Genetics
Genetics , a discipline of biology, is the science of genes, heredity, and variation in living organisms....
to show the interaction of two data points in a grid.
• A Gantt chart
Gantt chart
A Gantt chart is a type of bar chart that illustrates a project schedule. Gantt charts illustrate the start and finish dates of the terminal elements and summary elements of a project. Terminal elements and summary elements comprise the work breakdown structure of the project. Some Gantt charts...
helps in scheduling complex projects.
• The Nolan chart
Nolan Chart
The Nolan Chart is a political diagram popularized by the American libertarian David Nolan. He reasoned that virtually all human political action can be divided into two general categories: economic and personal...
is a libertarian political chart.
• A PERT chart
Program Evaluation and Review Technique
The Program ' Evaluation and Review Technique, commonly abbreviated PERT, is a statistical tool, used in project management, that is designed to analyze and represent the tasks involved in completing a given project...
is often used in project management
Project management
Project management is the discipline of planning, organizing, securing, and managing resources to achieve specific goals. A project is a temporary endeavor with a defined beginning and end , undertaken to meet unique goals and objectives, typically to bring about beneficial change or added value...
.
• The Pournelle chart
Pournelle chart
The Pournelle chart, developed by Jerry Pournelle , is a 2-dimensional coordinate system which can be used to distinguish political ideologies...
is a political chart to categorize state and rational ideologies.
• The Smith chart
Smith chart
The Smith chart, invented by Phillip H. Smith , is a graphical aid or nomogram designed for electrical and electronics engineers specializing in radio frequency engineering to assist in solving problems with transmission lines and matching circuits...
*For the magazine, see Radio-ElectronicsRadio electronics is the sub-field of electrical engineering concerning itself with the class of electronic circuits which receive or transmit radio signals....
.
• Diagram
Diagram
A diagram is a two-dimensional geometric symbolic representation of information according to some visualization technique. Sometimes, the technique uses a three-dimensional visualization which is then projected onto the two-dimensional surface...
• Edward Tufte
Edward Tufte
Edward Rolf Tufte is an American statistician and professor emeritus of political science, statistics, and computer science at Yale University. He is noted for his writings on information design and as a pioneer in the field of data visualization....
• Exploratory data analysis
Exploratory data analysis
In statistics, exploratory data analysis is an approach to analysing data sets to summarize their main characteristics in easy-to-understand form, often with visual graphs, without using a statistical model or having formulated a hypothesis...
• Information graphics
Information graphics
Information graphics or infographics are graphic visual representations of information, data or knowledge. These graphics present complex information quickly and clearly, such as in signs, maps, journalism, technical writing, and education...
• Graphic organizer
• Mathematical diagram
Mathematical diagram
Mathematical diagrams are diagrams in the field of mathematics, and diagrams using mathematics such as charts and graphs, that are mainly designed to convey mathematical relationships, for example, comparisons over time.- Argand diagram :...
• Official statistics
Official statistics
Official statistics are statistics published by government agencies or other public bodies such as international organizations. They provide quantitative or qualitative information on all major areas of citizens' lives, such as economic and social development, living conditions, health, education,...
• Plot (graphics)
Plot (graphics)
A plot is a graphical technique for representing a data set, usually as a graph showing the relationship between two or more variables. The plot can be drawn by hand or by a mechanical or electronic plotter. Graphs are a visual representation of the relationship between variables, very useful for... | 4,001 | 20,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-09 | latest | en | 0.922269 |
http://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.471207.html | 1,369,305,593,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703227943/warc/CC-MAIN-20130516112027-00054-ip-10-60-113-184.ec2.internal.warc.gz | 325,196,203 | 4,771 | # SOLUTION: Allie has three fewer than twice the number of coins that Jonathan has. If Jonathan gave 2 coins to Allie, she would have three time as many coins as he would. How many coins does
Algebra -> Algebra -> Customizable Word Problem Solvers -> Misc -> SOLUTION: Allie has three fewer than twice the number of coins that Jonathan has. If Jonathan gave 2 coins to Allie, she would have three time as many coins as he would. How many coins does Log On
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Word Problems: Miscellaneous Word Problems Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Miscellaneous Word Problems Question 471207: Allie has three fewer than twice the number of coins that Jonathan has. If Jonathan gave 2 coins to Allie, she would have three time as many coins as he would. How many coins does Allie have? a. 2 b. 3 c. 5 d. 7 e. 9Answer by ankor@dixie-net.com(15652) (Show Source): You can put this solution on YOUR website!Allie has three fewer than twice the number of coins that Jonathan has. If Jonathan gave 2 coins to Allie, she would have three time as many coins as he would. How many coins does Allie have? : Let a = no. of coins that A has Let j = no. of coins that J has : Write an equation for each statement: : "Allie has three fewer than twice the number of coins that Jonathan has." a = 2j - 3 : "If Jonathan gave 2 coins to Allie, she would have three time as many coins as he would." 3(j-2) = a + 2 3j - 6 = a + 2 3j = a + 2 + 6 3j = a + 8 replace a with (2j-3) 3j = (2j-3) + 8 3j - 2j = -3 + 8 j = 5 coins has J originally then a = 2(5) - 3 a = 7 coins has Allie | 514 | 1,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2013-20 | latest | en | 0.968498 |
https://fiixsoftware.com/maintenance-metrics/what-is-asset-turnover/ | 1,713,405,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00562.warc.gz | 235,758,955 | 20,549 | What is asset turnover?
Asset turnover is a measure of the use of assets in a business. It's used to determine how efficiently a company uses its assets, and to measure the financial performance of a business.
How is asset turnover calculated?
Asset turnover is calculated by dividing the net sales by the average total assets. The formula is:
Asset Turnover
=
Net Sales ÷
Average Total Assets
Example of asset turnover
Let's review an example of asset turnover in use. Suppose company A has \$10 million in annual sales and \$10 million worth of assets; its asset turnover is 1.0 (\$10M/\$10M). On the other hand, company B has \$100 million in annual sales and \$20 million worth of assets, so its asset turnover is 5 (\$100M/\$20M). These asset turnover numbers are sometimes referred to as asset turnover ratios or asset turns.
Is having a low or high asset turnover better?
The higher the asset turnover, the better. This means you are getting more from each dollar invested in your business.
For example, a high asset turnover tells investors (and your finance team) you're using your money well. If you have an asset turnover ratio of 2x, it means that for every \$1 spent on assets (like machinery and equipment), they got \$2 worth of sales revenue back in return. That's great news for investors because it means they're getting twice as much back from when they invested in the company than if they'd put their money into another company with a lower ATR (asset turnover ratio).
What are the benefits of tracking asset turnover?
If you run a manufacturing facility, for example, having a low asset turnover will give you a good indication that your assets (like equipment) and liabilities (like loans) are not where they need to be to break even (or profit) in terms of sales. This information can help the team make key changes to get back on track, like increasing production volume to meet financial goals.
Some tips to improve asset turnover
Here are a few tips to follow to improve your overall asset turnover:
• Track the number of assets you have. This will help you determine how many assets are being used and how many are in maintenance, repair, or disposal.
• Keep track of your asset turnover ratio so that if it's low, you can work on improving it by selling off unused equipment and increasing the amount of money spent on maintaining what remains. | 503 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | longest | en | 0.953226 |
https://askdev.io/questions/995511/does-the-closure-of-the-interior-equal-the-interior-of-the | 1,660,535,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00348.warc.gz | 125,694,466 | 7,777 | # Does the 'closure of the interior' equal the 'interior of the closure'?
My solution is no because, $\mathbb{Q}^o = \emptyset$ therefore $\overline{(\mathbb{Q}^o)} = \emptyset$ yet $\overline{\mathbb{Q}} = \mathbb{R}$ therefore $\big(\overline{\mathbb{Q}}\,\big)^o = \mathbb{R}$.
Is my instance deal with?
7
2022-07-25 20:46:51
Source Share | 123 | 344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.540185 |
http://slideplayer.com/slide/2816535/ | 1,571,230,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00268.warc.gz | 171,722,785 | 28,912 | # Chapter 16 Inferential Statistics
## Presentation on theme: "Chapter 16 Inferential Statistics"— Presentation transcript:
Chapter 16 Inferential Statistics
Inferential statistics is defined as the branch of statistics that is used to make inferences about the characteristics of a population based on sample data. • The goal is to go beyond the data at hand and make inferences about population parameters. • In order to use inferential statistics, it is assumed that either random selection or random assignment was carried out (i.e., some form of randomization must is assumed).
Is there a relationship? Answered by Null Hypothesis Significance Tests (NHST) t tests, F tests, χ2, p-values, for example What kind of relationship? (linear, curvilinear?) How strong is the relationship? Answered by effect size measures, not NHST’s
The Logic of Inferential Statistics
Population: the entire universe of individuals we are interested in studying Sample: the selected subgroup that is actually observed and measured (with sample size N) Sampling Distribution of the Statistic: A theoretical distribution that describes how a statistic behaves across a large number of samples
The Three Distributions Used in Inferential Statistics
I. Population Inference Selection III. Sampling Distribution of the Statistic Evaluation II. Sample
The Logic of NHST’s Samples are drawn from the population of interest
Samples are observed and measured and sample statistics are calculated In order to determine whether sample statistics are consistent with a null hypothesis or not, sample statistics are compared to a model that describes what is expected for sample statistics This model is the sampling distribution of the statistic
Sampling Distributions
Sampling distributions allow us to make "probability" statements in inferential statistics. • A sampling distribution is defined as "The theoretical probability distribution of the values of a statistic that result when all possible random samples of a particular size are drawn from a population." • One specific type of sampling distribution is called the sampling distribution of the mean. To generate this distribution by hand (not done in practice), you would randomly select a sample, calculate the mean, randomly select another sample, calculate the mean, and continue this process until you have calculated the means for a very large number of samples. You could then construct a histogram to show the sampling distribution of the mean. This distribution would tell you how likely or probable it is to get any particular value of the mean.
• No matter what the shape of the population distribution or the sample, the sampling distribution of the mean is normally distributed (as long as your sample size is about 30 or more for your sampling). • Note that the mean of the sampling distribution of the mean is equal to the population mean! That tells you that repeated sampling will, over the long run, estimate the correct mean. The variance shows you that sample means will tend to be somewhat different from the true population mean in an individual sample.
Although I just described the sampling distribution of the mean, it is important to remember that a sampling distribution can be obtained for any statistic. For example, you could also obtain the following sampling distributions: • Sampling distribution of the percentage (or proportion). • Sampling distribution of the variance. • Sampling distribution of the correlation. • Sampling distribution of the regression coefficient. • Sampling distribution of the difference between two means.
The standard deviation of a sampling distribution is called the standard error (SE). In other words, the standard error is just a special kind of standard deviation and you learned what a standard deviation was in the last chapter. The smaller the standard error, the more precisely the statistic has been estimated. Sample size has a powerful effect on precision of estimation:
Estimation The key estimation question is "Based on my random sample, what is my estimate of the population parameter?” • The basic idea is that you are going to use your sample data to provide information about the population. There are actually two types of estimation. • They can be first understood through the following analogy: Let's say that you take your car to your local car dealer's service department and you ask the service manager how much it will cost to repair your car. If the manager says it will cost you \$500 then she is providing a point estimate. If the manager says it will cost somewhere between \$400 and \$600 then she is providing an interval estimate.
In other words, a point estimate is a single number, and an interval estimate is a range of numbers.
• A point estimate is the value of your sample statistic (e.g., your sample mean or sample correlation), and it is used to estimate the population parameter (e.g., the population mean or the population correlation) • For example, if you take a random sample from adults living an the United States and you find that the average income is \$45,000, then your best guess or your point estimate for the population of adults in the U.S. will be \$45,000. • Again, whenever you engage in point estimation, all you need to do is to use the value of your sample statistic as your "best guess" (i.e., as your estimate) of the (unknown) population parameter.
Oftentimes, we like to put an interval around our point estimates so that we realize that the actual population value is somewhat different from our point estimate because sampling error is always present in sampling. • An interval estimate (also called a confidence interval) is a range of numbers inferred from the sample that has a known probability of capturing the population parameter over the long run (i.e., over repeated sampling). The probability relates to the area in a normal curve
99.7% 95% 68% μ – 3σ μ – 2σ μ – σ μ μ + σ μ + 2σ μ + 3σ
• The "beauty" of confidence intervals is that we know the probability of the true population parameter. • Specifically, with a 95 percent confidence interval, you are able to be "95% confident" that it will include the population parameter. • For example, you might take the point estimate of annual income of U.S. adults of \$45,000 and surround it by a 95% confidence interval and find the interval is \$43,000 to \$47,000. You can be "95% confident" that the average income is somewhere between \$43,000 and \$47,000.
Hypothesis Testing (NHST) is the branch of inferential statistics concerned with how well the sample data support a null hypothesis. • First note that the null hypothesis is usually the prediction that there is no relationship in the population or no difference between groups • The alternative hypothesis is the logical opposite of the null hypothesis and says there is a relationship in the population or the groups are different • The researcher’s interest is in the alternative hypothesis • The question answered in hypothesis testing: "Is the value of my sample statistic unlikely enough (assuming that the null hypothesis is true) for me to reject the null hypothesis and tentatively accept the alternative hypothesis?“ • Note that it is the null hypothesis that is directly tested in hypothesis testing (not the alternative hypothesis).
You may be wondering, when do you actually reject the null hypothesis and make the decision to tentatively accept the alternative hypothesis? • You reject the null hypothesis when the probability of your result assuming a true null is very small. That is, you reject the null when your sample result would be very unlikely if the null was true. • In particular, you set a significance level (also called the alpha level) to use in your research study to decide when to reject the null as a plausible description of the sample.
By convention, the significance level (alpha) is set at p =
By convention, the significance level (alpha) is set at p = .05 or smaller When your the sample statistic is compared to the sampling distribution, a probability value (p-value) is obtained for your sample statistic. If that p-value is less than the decision rule (.05), the null hypothesis is rejected. If your p-value is larger (more probable) than .05, no conclusion is reached.
• The probability value (p-value) is a number that tells you the probability of your result or a more extreme result when it is assumed that there is no relationship in the population (i.e., when you are assuming that the null hypothesis is true which is what we do in hypothesis testing and in jurisprudence). • The significance level (alpha) is the decision rule, chosen in advance (a priori), that you will use to conclude that the null should be rejected. By convention in education and the social sciences, alpha is usually chosen as .05 or a smaller value. • The logic of the NHST is the following: If my sample result is so rare (p < .05) that is very unlikely to have occurred when the null is true, then the null must be false and something else must be occurring (i.e., there really is a relationship or a difference between groups)
• Statistical significance does not tell you whether you have practical significance.
• If a finding is statistically significant then you can claim that the evidence suggests that the observed result (e.g., your observed correlation or your observed difference between two means) was probably not just due to chance. • An effect size measure can provide additional important information to help interpret the strength of a statistically significant relationship. An effect size indicator is defined as a measure of the strength of a relationship. • A finding is practically significant when the difference between the means or the size of the correlation is big enough, in your opinion, to be of practical use. ((For example, a correlation of .15 would probably not be practically significant, even if it was statistically significant. On the other hand, a correlation of .85 would probably be practically significant. )) • Practical significance requires you to make a non-quantitative decision and to think about what the impact or utility of a result would be in application | 2,015 | 10,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-43 | latest | en | 0.912335 |
https://www.difference.wiki/ancova-vs-anova/ | 1,701,589,587,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00462.warc.gz | 833,448,038 | 24,365 | # ANCOVA vs. ANOVA: What's the Difference?
Edited by Sawaira Riaz || By Sumera Saeed || Updated on October 17, 2023
ANCOVA (Analysis of Covariance) adjusts for covariates, while ANOVA (Analysis of Variance) tests differences among group means. Both are statistical methods.
## Key Differences
ANCOVA and ANOVA are both advanced statistical techniques used to analyze differences among group means. While they share some similarities, they serve distinct purposes in the field of research. ANCOVA, or Analysis of Covariance, integrates ANOVA and regression. ANOVA, or Analysis of Variance, solely assesses the differences among group means.
Sumera Saeed
Oct 17, 2023
The primary distinction between ANCOVA and ANOVA lies in the presence of a covariate. ANCOVA adjusts for one or more covariates that may be influencing the dependent variable, aiming to remove the effect of these covariates to see the effect of the independent variable more clearly. In contrast, ANOVA does not account for these covariates.
Sumera Saeed
Oct 17, 2023
Another essential difference between ANCOVA and ANOVA is in their applications. Researchers use ANCOVA when they suspect an external variable could distort the results, providing a clearer picture of the variables' relationships. On the other hand, ANOVA is employed when there's no need to adjust for other external variables, focusing solely on the differences in means.
Sumera Saeed
Oct 17, 2023
In terms of output, both ANCOVA and ANOVA provide the researcher with F-values, p-values, and other related statistics. However, ANCOVA offers adjusted group means after considering the covariate's influence, whereas ANOVA provides straightforward group means without any adjustment.
Sumera Saeed
Oct 17, 2023
In summary, while both ANCOVA and ANOVA are powerful tools for examining the differences among groups, the choice between them depends on the research question and the presence of potential confounding variables. ANCOVA adjusts for these variables, whereas ANOVA does not.
Sumera Saeed
Oct 17, 2023
## Comparison Chart
### Definition
Analysis of Covariance
Analysis of Variance
Sumera Saeed
Oct 17, 2023
Yes
No
Sumera Saeed
Oct 17, 2023
### Main Purpose
Examine group differences while adjusting for covariates
Examine group differences
Sumera Saeed
Oct 17, 2023
### Output
Provides direct group means
Sara Rehman
Oct 17, 2023
### Application
Used when an external variable might influence the result
Used when group mean differences are the sole focus
Sara Rehman
Oct 17, 2023
## ANCOVA and ANOVA Definitions
#### ANCOVA
With ANCOVA, we obtained means that accounted for income differences.
Sumera Saeed
Oct 17, 2023
#### Anova
Produces F-values and p-values for group comparisons.
The ANOVA results showed significant differences between the three conditions.
Sumera Saeed
Oct 17, 2023
#### ANCOVA
Merges characteristics of ANOVA and regression.
For a comprehensive analysis, he chose ANCOVA over simple ANOVA.
Janet White
Oct 17, 2023
#### Anova
A statistical method for testing group mean differences.
He applied ANOVA to see if different diets led to distinct weight loss results.
Sawaira Riaz
Oct 17, 2023
#### ANCOVA
Assesses the effect of categorical variables with adjustments.
Using ANCOVA, they isolated the treatment effect from the age factor.
Harlon Moss
Oct 17, 2023
#### Anova
Examines if group averages significantly differ.
Through ANOVA, it was clear that the interventions had different outcomes.
Sumera Saeed
Oct 17, 2023
#### ANCOVA
Adjusts for covariates when comparing group means.
ANCOVA helped eliminate the influence of prior experience in the analysis.
Sumera Saeed
Oct 17, 2023
#### Anova
Assesses the impact of one or more factors on a dependent variable.
ANOVA was perfect for studying the influence of various teaching methods on scores.
Sumera Saeed
Oct 17, 2023
#### ANCOVA
A statistical method combining regression and ANOVA.
She used ANCOVA to adjust for participants' age in her study.
Sawaira Riaz
Oct 17, 2023
#### Anova
Commonly used in experimental research.
Her experimental design called for the application of ANOVA.
Aimie Carlson
Oct 17, 2023
#### Anova
A statistical method for making simultaneous comparisons between two or more means; a statistical method that yields values that can be tested to determine whether a significant relation exists between variables
Sumera Saeed
Apr 13, 2016
## FAQs
#### What does ANCOVA stand for?
ANCOVA stands for Analysis of Covariance.
Sumera Saeed
Oct 17, 2023
#### Is ANCOVA a combination of other statistical methods?
Yes, ANCOVA merges features of ANOVA and regression.
Sumera Saeed
Oct 17, 2023
#### When should I use ANOVA?
Use ANOVA when comparing group means without adjusting for covariates.
Aimie Carlson
Oct 17, 2023
#### Which is more complex: ANCOVA or ANOVA?
ANCOVA is generally more complex due to the inclusion of covariates.
Sumera Saeed
Oct 17, 2023
#### What is the main function of ANOVA?
ANOVA tests for differences among group means.
Sawaira Riaz
Oct 17, 2023
#### How does ANCOVA differ from ANOVA?
ANCOVA adjusts for covariates while ANOVA does not.
Sumera Saeed
Oct 17, 2023
#### Is the output of ANOVA and ANCOVA similar?
Both provide F-values and p-values, but ANCOVA offers adjusted means.
Janet White
Oct 17, 2023
#### Can ANOVA handle non-continuous variables?
ANOVA deals with continuous dependent variables but categorical independent ones.
Sara Rehman
Oct 17, 2023
#### How many groups can ANOVA compare?
ANOVA can compare two or more groups.
Sara Rehman
Oct 17, 2023
#### Can ANCOVA adjust for multiple covariates?
Yes, ANCOVA can account for more than one covariate.
Sumera Saeed
Oct 17, 2023
#### Can I use ANCOVA for categorical covariates?
Typically, ANCOVA uses continuous covariates, but dummy coding allows categorical ones.
Sumera Saeed
Oct 17, 2023
#### Why adjust for covariates using ANCOVA?
Adjusting with ANCOVA provides a clearer understanding of variable relationships.
Janet White
Oct 17, 2023
#### Is sample size crucial for ANCOVA and ANOVA?
Yes, adequate sample size is vital for power and reliable results in both tests.
Sumera Saeed
Oct 17, 2023
#### Are there different types of ANOVA?
Yes, there's One-way ANOVA, Two-way ANOVA, etc., based on the number of factors.
Sumera Saeed
Oct 17, 2023
#### Is ANCOVA used in experimental research?
Yes, especially when controlling for confounding variables.
Harlon Moss
Oct 17, 2023
#### How are ANCOVA and ANOVA related to regression?
ANCOVA merges features of regression and ANOVA; ANOVA is related through linear models.
Sumera Saeed
Oct 17, 2023
#### Can I use ANCOVA for non-linear relationships?
ANCOVA assumes a linear relationship between the covariate and dependent variable.
Harlon Moss
Oct 17, 2023
#### What software can conduct ANOVA and ANCOVA?
Both can be conducted using software like SPSS, R, and SAS.
Sara Rehman
Oct 17, 2023
#### Do ANOVA and ANCOVA require normality assumptions?
Yes, both assume that the residuals are normally distributed.
Sumera Saeed
Oct 17, 2023
#### What's the key output metric in ANOVA?
The F-value is the primary output in ANOVA to test group differences.
Harlon Moss
Oct 17, 2023 | 1,901 | 7,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | longest | en | 0.908384 |
https://tex.stackexchange.com/questions/95419/multiple-split-equations | 1,719,176,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00840.warc.gz | 501,682,064 | 38,842 | # Multiple split equations
I find myself again and again using a combination of the gather and aligned environments. So it would be nice to not have to write so much. In particular, it is often only clear after writing the formula, whether it needs splitting. So the new environment should act like gather, only that when lines are separated by a special symbol/macro (\n for example) they should be put together into an aligned-environment. I tried the following
\documentclass{article}
\usepackage[fleqn]{amsmath}
\newenvironment{multeq}{%
\let\n{\\}
\let\\{\end{aligned}\\\begin{aligned}}
\begin{aligned}
}{%
\end{aligned}
\let\\{\n}
}
\begin{document}
\begin{gather}
\begin{aligned}
S(A) &= (I_{n-k, k}AI_{n-k,k})_{i,j} = (I_{n-k,k})_{i,i}A_{i,j}(I_{n-k,k})_{j,j}\\
&= \begin{cases}A_{i,j}&\begin{aligned}&(i\leq n-k\wedge j\leq n-k)\\\vee&(i>n-k\wedge j>n-k)\end{aligned}\\-A_{i,j}&\text{otherwise}\end{cases}
\end{aligned}\\
\Rightarrow H = S(U(n-k)\times U(k))
\end{gather}
\begin{multeq}
S(A) &= (I_{n-k, k}AI_{n-k,k})_{i,j} = (I_{n-k,k})_{i,i}A_{i,j}(I_{n-k,k})_{j,j}\n
&= \begin{cases}A_{i,j}&\begin{aligned}&(i\leq n-k\wedge j\leq n-k)\n\vee&(i>n-k\wedge j>n-k)\end{aligned}\n-A_{i,j}&\text{otherwise}\end{cases}\\
\Rightarrow H = S(U(n-k)\times U(k))
\end{multeq}
\end{document}
where the output should be 2×
but it gives the error:
! LaTeX Error: There's no line here to end.
l.24 \begin{multeq}
How does this error come about and what is a good way to create such an environment?
So using David’s corrections, and adding gather:
\newenvironment{multeq}{\begin{gather}%
\def\n{\\}%
\def\\{\end{aligned}\\\begin{aligned}}%
\begin{aligned}%
}{%
\end{aligned}%
\end{gather}}
I got
! Paragraph ended before \gather was complete.
So I tried using \Collect@Body from environ:
\def\savebody#1{\def\BODY{#1}}
\usepackage{environ}
\makeatletter
\newenvironment{multeq}{
\def\n{\\}%
\def\\{\end{aligned}\\\begin{aligned}}%
\Collect@Body\savebody
}{\begin{gather}%
\begin{aligned}%
\BODY%
\end{aligned}%
\end{gather}}
\makeatother
But now the redefinition of \\ is ignored. What else could I do?
• \let\n{\\} defines \n to be { like \bgroup. You mean \def there. Commented Jan 26, 2013 at 11:04
• @DavidCarlisle: Is \let\n\\ ok? (and why isn't this markedup as code?)
– bodo
Commented Jan 26, 2013 at 11:05
• that defines \n to have the current meaning of \\ (the markup around backticks with backslashes and spaces in comments is a bit odd, I had to edit my previous comment a couple of times to get it to display I know what you mean though:-) Commented Jan 26, 2013 at 11:12
• so \let\n\ is a legal assignment but would define \n to the wrong thing. Commented Jan 26, 2013 at 11:13
• @DavidCarlisle (Not related to the problem) I thought I was the only one having trouble with escaping backslashes and backticks in comment markdowns. Here is an so meta post for backslashes: Backslash escaped with backticks rendered incorrectly in comments as a well related post for backticks :-) Commented Jan 27, 2013 at 4:33
You need \def not \let \nd you need to get into math mode:
\documentclass{article}
\usepackage[fleqn]{amsmath}
\newenvironment{multeq}{% \def\n{\\}% \def\\{\end{aligned}\\\begin{aligned}}% \begin{aligned}% }{% \end{aligned}%}
\begin{document}
\begin{gather}
\begin{aligned}
S(A) &= (I_{n-k, k}AI_{n-k,k})_{i,j} = (I_{n-k,k})_{i,i}A_{i,j}(I_{n-k,k})_{j,j}\\
&= \begin{cases}A_{i,j}&\begin{aligned}&(i\leq n-k\wedge j\leq n-k)\\\vee&(i>n-k\wedge j>n-k)\end{aligned}\\-A_{i,j}&\text{otherwise}\end{cases}
\end{aligned}\\
\Rightarrow H = S(U(n-k)\times U(k))
\end{gather}
\begin{multeq}
S(A) &= (I_{n-k, k}AI_{n-k,k})_{i,j} = (I_{n-k,k})_{i,i}A_{i,j}(I_{n-k,k})_{j,j}\n
&= \begin{cases}A_{i,j}&\begin{aligned}&(i\leq n-k\wedge j\leq n-k)\n\vee&(i>n-k\wedge j>n-k)\end{aligned}\n-A_{i,j}&\text{otherwise}\end{cases}\\
\Rightarrow H = S(U(n-k)\times U(k))
\end{multeq}
\end{document}
You may need some other changes, depending quite wh\t input markup you want, the above produces
• Actually I wanted the new environment to be a short cut for the construction with gather. So to enter math mode I'd rather use gather. (I indeed forgot to put that into my code). Is that possible?
– bodo
Commented Jan 26, 2013 at 11:12
• When I try it with gather, I get ! Paragraph ended before \gather was complete.
– bodo
Commented Jan 26, 2013 at 11:15
• yes just get rid of the I added and replac should work by gather Commented Jan 26, 2013 at 11:15
• Maybe I need to use \Collect@Body from environ?
– bodo
Commented Jan 26, 2013 at 11:33
After looking at some macro expansions, I found the actual problem. amsmath defines \let\\\math@cr when it expands gather. So whatever I do to \\ beforehand, will not matter. Thus I changed \math@cr et voilá, it works. At least if I collect the body with NewEnviron.
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{environ}
\makeatletter
\NewEnviron{multeq}{%
\begingroup
\let\n\math@cr
\begin{gather}
\def\math@cr{\end{aligned}\n\begin{aligned}}
\begin{aligned}
\BODY
\end{aligned}
\end{gather}
\endgroup
}
\makeatother
is the needed definition. It also works with align instead of gather or split instead of aligned. And for a simpler usage example:
\begin{document}
\begin{multeq}
0 + 0 = 0\n
0 * 0 = 0\\
1 + 1 = 2\n
1 * 1 = 1
\end{multeq}
\end{document}
will be printed as | 1,802 | 5,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.665199 |
https://www.physicsforums.com/threads/kinetic-energy.69982/ | 1,545,128,776,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00449.warc.gz | 969,270,074 | 14,250 | # Homework Help: Kinetic energy
1. Apr 4, 2005
### tony873004
A system consists of 10 0.50-kg balls, each moving radially outward from a common center in a symmetrical pattern, all at a speed of 12 m/s. What is the kinetic energy of the system?
My instinct is to say 0, since it it symmetrical and every ball moving one direction will be cancelled by a ball of the same mass and speed moving the opposite direction.
But energy is a scalar, right? And scalars only have magnitude, not direction, right?
$$e_k=10*\frac{1}{2}mv^2$$
$$e_k=10*\frac{1}{2}0.50kg *(12m/s)^2$$
$$e_k=360 J$$
2. Apr 4, 2005
### whozum
My guess is 0, but I'm not sure why.
I think if they all are radiating outward from a common center, then they must have been motionless at some point during which the potential enegry and kinetic energy is zero.
Energy is a scalar though and they don't cancel each other out, I'd still like to know why its not 0 in a definitive fashion though.
3. Apr 4, 2005
### HallsofIvy
No, your "second thought" was correct. E= (1/2)mv2 where you can think of v2 as meaning either the dot product or the square of the magnitude of the v vector.
In any case, kinetic energy, unlike momentum, IS a scalar, not a vector, and is always positive. The kinetic energys of the two objects do not cancel, they add.
No, if they are radiating outward, they are moving. If they were "motionless at some point" there must have been some force acting to cause them to move. That force does work and contributes energy.
4. Apr 4, 2005
### tony873004
I'm guessing that the velocities cancel. Since they're vectors and not scalars, I can cancel them before plugging them into the K equation. Then one of my factors is 0 which makes the whole answer 0.
I think.
5. Apr 4, 2005
### whozum
Well take the balls to be physical objects. If they are all moving outwards at t=t1, then isnt it feasible to say at some t<t1 they were at the same point? If so, motionless?
Its kind of like the big bang.
6. Apr 4, 2005
### tony873004
I posted before I saw your post HallsofIvy. Can't I cancel the v's first?
7. Apr 4, 2005
### tony873004
Thinking about it, I guess it couldn't be 0. I could always spin down this system with a generator and light a light bulb for a moment.
8. Apr 4, 2005
### asrodan
Cancelling out the velocities would impose a direction requirement on the kinetic energy, which we already know it doesn't have. | 675 | 2,433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-51 | latest | en | 0.956679 |
https://www.lexic.us/definition-of/quantics | 1,723,352,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00649.warc.gz | 677,407,108 | 7,267 | ### Definition of Quantics
1. Noun. (plural of quantic) ¹
¹ Source: wiktionary.com
### Lexicographical Neighbors of Quantics
quannetquannetsquantquant fundquant fundsquant polequant polesquantaquantalquantal mitosis quantalityquantallyquantasomequantedquanticquantics (current term)quantifiabilitiesquantifiabilityquantifiablequantifiableness quantifiablesquantifiablyquantificationquantificationalquantificationsquantifiedquantifierquantifiersquantifiesquantify
### Literary usage of Quantics
Below you will find example usage of this term as found in modern and/or classical literature:
1. Lessons Introductory to the Modern Higher Algebra by George Salmon (1885)
"SYSTEMS OF quantics. 176. If still remains to explain a few properties of systems of ... An invariant of a system of quantics of the same degree is called a ..."
2. Lessons Introductory to the Modern Higher Algebra by George Salmon (1885)
"SYSTEMS OF quantics. 176. IT still remains to explain a few properties of systems of ... An invariant of a system of quantics of the same degree is called a ..."
3. Lessons Introductory to the Modern Higher Algebra by George Salmon (1876)
"SYSTEMS OF quantics. 176. IT still remains to explain a few properties of systems of ... An invariant of a system of quantics of the same degree is called a ..."
4. Lessons Introductory to the Modern Higher Algebra by George Salmon (1866)
"SYSTEMS OF quantics. 173. IT still remains to explain a few properties of systems of ... An invariant of a system of quantics is called a combinant if it is ..."
5. The Collected Mathematical Papers of Arthur Cayley by Arthur Cayley (1889)
"68 to 71, although introduced in reference to binary quantics, relate or may be considered as relating to quantics of the like general form. ..."
6. The Encyclopaedia Britannica: A Dictionary of Arts, Sciences, Literature and (1910)
"Moreover, instead of having one pair of varia we ma have several airs \, j-, z¡ •;;... in addition and multipartite. f severa quantics Symbolic Form. ..."
7. Mathematical Papers by William Kingdon Clifford (1882)
"... RELATIONS CONNECTING THE POWERS OF LINEAR quantics. I THINK the first treatment of this subject is to be found in some very interesting articles of M. ..." | 524 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.857437 |
https://www.jiskha.com/display.cgi?id=1510230114 | 1,531,853,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00319.warc.gz | 885,632,643 | 4,126 | # As physics
posted by cylinder
The drag force acting on a car travelling at a speed v is given by the equation F=kAv^2
Where A is the area of the front of the car.
Show that a suitable unit for the quantity k is kgm-3.
F= kgms-2
or
A=m2
V=ms-1
Which is K=kgm-3
I am confused could someone show the working out clearly involving all steps and equations used.
1. Damon
F = mass * acceleration
so
kg m/s^2 or Newtons = Force units
kg m/s^2 = kunits * m^2 *m^2/s^2
or
k units = kg /m^3
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Topic: extract the rows except the specific index
Replies: 3 Last Post: Dec 28, 2012 4:19 PM
Messages: [ Previous | Next ]
uny gg Posts: 45 Registered: 8/4/09
extract the rows except the specific index
Posted: Dec 28, 2012 12:56 PM
Hello, I am newbie.. for matlab..
I would like to extract the rows except the specific index..
For example,
A = (1,2,3;4,5,6;7,8,9;10,11,12;13,14,15;16,17,18) % 6 by 3 matrix.
What I want to do is..
divide this matrix, 3 by 3 and 3 by 3 matrix randomly ...
Here is what I try.
bottom =1;
top =6;
ran_idx = bottom + (top - bottom)*rand([1,3]);
ran_idx = ceil(ran_idx);
train = A(ran_idx,:);
test = A(~ran_idx.:); % this one is not working,, return. empty matrix..
I can extract the train matrix.. but I could not make test..
Date Subject Author
12/28/12 uny gg
12/28/12 dpb
12/28/12 dpb
12/28/12 james bejon | 316 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-30 | latest | en | 0.814338 |
https://quizizz.com/admin/quiz/5552655105e4a1f4062316a4 | 1,603,510,620,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00139.warc.gz | 483,301,025 | 8,417 | Liczby całkowite
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Q. - 8 - 9 =
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17
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-17
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Q. Liczba przeciwna do 5 to | 355 | 814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.447258 |
https://codereview.stackexchange.com/questions/235724/almostincreasingsequence | 1,719,127,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00656.warc.gz | 151,508,513 | 38,040 | # almostIncreasingSequence
This is assignment from codesignal:
Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.
fn almostIncreasingSequence(sequence: Vec<i32>) -> bool {
if sequence.len() == 2
|| sequence.len() == 3 && (sequence[1] > sequence[0] || sequence[2] > sequence[1])
{
true
} else {
let mut no_increase = sequence[1] <= sequence[0];
for i in 2..(sequence.len() - 1) {
//println!("sequence[{}] {}", i, sequence[i]);
//println!("no_increase {:?}", no_increase);
if sequence[i] <= sequence[i - 1] {
if no_increase
|| sequence[i] <= sequence[i - 2] && sequence[i + 1] <= sequence[i - 1]
|| i == (sequence.len() - 2) && sequence[i + 1] <= sequence[i]
{
return false;
}
no_increase = true;
} else if i == (sequence.len() - 2) && no_increase && sequence[i + 1] <= sequence[i] {
return false;
}
}
true
}
}
#[test]
fn test_almostIncreasingSequence() {
let samples: Vec<(Vec<i32>, bool)> = vec![
(vec![3, 6, -2, -5, 7, 3], false),
(vec![1, 3, 2, 1], false),
(vec![1, 2, 1, 2], false),
(vec![0, -2, 5, 6], true),
(vec![10, 1, 2, 3, 4, 5], true),
(vec![40, 50, 60, 10, 20, 30], false),
(vec![3, 6, 5, 8, 10, 20, 15], false),
(vec![1, 1, 2, 3, 4, 4], false),
];
for (inputArray, expected) in samples {
println!("\n{:?}", inputArray);
assert_eq!(expected, almostIncreasingSequence(inputArray));
}
}
Run with cargo test test_almostIncreasingSequence
Edit: Added check for i == (sequence.len() - 2) element
Edit: Added more tests as suggested which showed bug for sequence.len() == 3. Now fixed:
fn almostIncreasingSequence(sequence: Vec<i32>) -> bool {
if sequence.len() < 3 {
true
} else if sequence.len() == 3 {
sequence[1] > sequence[0] || sequence[2] > sequence[1] || sequence[2] > sequence[0]
} else {
let mut no_increase = sequence[1] <= sequence[0];
for i in 2..(sequence.len() - 1) {
//println!("sequence[{}] {}", i, sequence[i]);
//println!("no_increase {:?}", no_increase);
if sequence[i] <= sequence[i - 1] {
if no_increase
|| sequence[i] <= sequence[i - 2] && sequence[i + 1] <= sequence[i - 1]
|| i == (sequence.len() - 2) && sequence[i + 1] <= sequence[i]
{
return false;
}
no_increase = true;
} else if i == (sequence.len() - 2) && no_increase && sequence[i + 1] <= sequence[i] {
return false;
}
}
true
}
}
#[test]
fn test_almostIncreasingSequence() {
let samples: Vec<(Vec<i32>, bool)> = vec![
(vec![], true),
(vec![1], true),
(vec![1, 1], true),
(vec![1, 1, 1], false),
(vec![3, 6, -2, -5, 7, 3], false),
(vec![1, 3, 2, 1], false),
(vec![1, 2, 1, 2], false),
(vec![0, -2, 5, 6], true),
(vec![10, 1, 2, 3, 4, 5], true),
(vec![40, 50, 60, 10, 20, 30], false),
(vec![3, 6, 5, 8, 10, 20, 15], false),
(vec![1, 1, 2, 3, 4, 4], false),
];
for (inputArray, expected) in samples {
println!("\n{:?}", inputArray);
assert_eq!(expected, almostIncreasingSequence(inputArray));
}
}
Edit: Checking for edge case moved out of loop.
fn almostIncreasingSequence(sequence: Vec<i32>) -> bool {
if sequence.len() < 3 {
true
} else if sequence.len() == 3 {
sequence[1] > sequence[0] || sequence[2] > sequence[1] || sequence[2] > sequence[0]
} else {
let mut no_increase = sequence[1] <= sequence[0];
for i in 2..(sequence.len() - 1) {
//println!("sequence[{}] {}", i, sequence[i]);
//println!("no_increase {:?}", no_increase);
if sequence[i] <= sequence[i - 1] {
if no_increase
|| sequence[i] <= sequence[i - 2] && sequence[i + 1] <= sequence[i - 1]
{
return false;
}
no_increase = true;
}
}
if no_increase && sequence[sequence.len() - 1] <= sequence[sequence.len() - 2] {
return false;
}
true
}
}
• Almost increasing? I'd say it's either increasing or it isn't, but I suspect the actual problem assignment would clarify this. Can you post more about what the code is supposed to do? The line "We can remove only one element in the process." isn't clear either.
– Mast
Commented Jan 16, 2020 at 14:17
• I have added original description Commented Jan 16, 2020 at 14:31
• I recommend adding test cases for sequences of length 0, 1, 2, and 3. The first two are common special cases for problems like these, and the latter two are special cased in your implementation. Commented Feb 9, 2020 at 19:43
• @cbojar Thanks. With that I found a bug. Commented Feb 10, 2020 at 13:17 | 1,382 | 4,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.548908 |
https://puzzling.stackexchange.com/questions/tagged/arithmetic?tab=Votes | 1,571,868,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00153.warc.gz | 658,082,064 | 39,234 | Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
# Questions tagged [arithmetic]
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https://www.mathportal.org/calculators/polynomials-solvers/polynomials-operations-calculator.php | 1,722,693,673,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00746.warc.gz | 705,922,803 | 20,071 | Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
# Calculator for polynomial operations
This solver performs arithmetic operations on polynomial, including addition, subtraction, multiplication and division. For multiplication, it uses both the GRID and FOIL methods. The calculator displays complete work as well as a detailed step-by-step explanation for each operation.
Operations on polynomials calculator
add, subtract, multiply and divide polynomials
help ↓↓ examples ↓↓ tutorial ↓↓
(3x^2+4x-3)×(8x+5)
(x^2+2x-4)÷(3x-5)
Add polynomials Subtract polynomials Multiply polynomials (default) Divide polynomials
working...
EXAMPLES
example 1:ex 1:
$(3x-4)+(5+3x-4x^2)$
example 2:ex 2:
$(9x+5)-(3x-3)$
example 3:ex 3:
$(-2x^6 + x^5 - 3x^2 - 4x + 7) - (x^5 + 2x^2 - 4x + 4)$
example 4:ex 4:
$(2x+3)\cdot(5x-3)$
example 5:ex 5:
$(x^2 - x + 3)\cdot(2x^2 + 4x - 3)$
example 6:ex 6:
$\dfrac{x^3 + x^2 + 4}{x + 2}$
Find more worked-out examples in our database of solved problems..
Search our database with more than 300 calculators
TUTORIAL
## Operations on polynomials
In this short tutorial, you will learn how to perform basic operations on polynomials.
The basic operations are 1. addition 2. subtraction 3. FOIL method for binomial multiplication 4. standard multiplication 5. division by monomial and 6. long division. Note that this calculator displays a step-by-step explanation for each of these operations. Nevertheless, let's start with addition.
### 1A: Polynomial addition - horizontal
Example 01: Add $(2a+5) + (4a-3)$
First we will remove the parenthesis because there are no minus sign in front of the brackets:
$$(2a+5) + (4a-3) = 2a + 5 + 4a - 3$$
We'll now group the like terms:
$$2a + 5 + 4a - 3 = 2a + 4a + 5 - 3$$
Finally, we combine like terms:
$$2a + 4a + 5 - 3 = 6a + 2$$
Putting all together we have
\begin{aligned} (2a+5) + (4a-3) \overbrace{=}^{\text{remove par.}}& \color{blue}{2a} + 5 + \color{blue}{4a} - 3 = \\ \overbrace{=}^{\text{group like terms}}& \color{blue}{2a + 4a} + 5 - 3 = \\ \overbrace{=}^{\text{combine like terms}}& \color{blue}{6a} + 2 \end{aligned}
### 1B: Polynomial addition - vertical
Example 02: Add $(5x^3 - 3x^2 - 2x + 5) + (-x^3 + 2x^2 - 7)$
To perform vertical addition, we must arrange like terms one above the other.
$$\begin{array}{rrr} \color{blue}{5x^3} & \color{orangered}{-3x^2} & -2x & \color{purple}{5} \\ \color{blue}{-x^3} & \color{orangered}{2x^2} & & \color{purple}{-7} \\ \hline \end{array}$$
It is now quite simple to combine like terms
$$\begin{array}{rrr} \color{blue}{5x^3} & \color{orangered}{-3x^2} & -2x & \color{purple}{5} \\ \color{blue}{-x^3} & \color{orangered}{2x^2} & & \color{purple}{-7} \\ \hline \color{blue}{4x^3} & \color{orangered}{-x^2} & -2x & \color{purple}{-2} \end{array}$$
$$4x^2 -x^2-2x-2$$
### 2B: Polynomial subtraction – vertical
Example 04: Subtract $(2x^2 + x - 3) - (x^2 - 3x + 5)$
Place like terms one above the other, but in the second polynomial, we must now alter all of the signs.
$$\begin{array}{rrrr} 2x^2 & x & -2 \\ \color{red}{\bf{-}}x^2 & \color{red}{\bf {+}}3x & \color{red}{\bf{-}}5 \\ \hline \end{array}$$
Now we can combine like terms
$$\begin{array}{rrrr} 2x^2 & x & -2 \\ -x^2 & 3x & -5 \\ \hline x^2 & 4x & -7 \\ \end{array}$$
So the answer is $x^2 + 4x - 7$
### 2A: Polynomial subtraction – horizontal
Example 03: Subtract $(5x - 7) - (3x - 3)$
Here we remove parenthesis by changing the sign of every term in the second bracket.
$$(5x - 7) \color{blue}{- (3x - 3)} = 5x - 7 \color{blue}{- 3x + 3}$$
Now, as in previous example, group the like terms ...
$$5x - 7 -3x + 3 = 5x - 3x -7 + 3$$
...and combine them:
$$5x - 3x - 7 + 3 = 2x - 4$$
Putting all together we have
\begin{aligned} (5x-7) - (3x-3) \overbrace{=}^{\text{remove par.}}& 5x - 7 - 3x + 3 = \\ \overbrace{=}^{\text{group like terms}}& 5x - 3x - 7 + 3 = \\ \overbrace{=}^{\text{combine like terms}}& 2x - 4 \end{aligned}
### 3: Polynomial multiplication - FOIL
This method is used to multiply two binomials. The best way to explain the FOIL method is to use an example:
Example 05: Use FOIL method to multiply $(5a + 2) \cdot(2a - 3)$
$$\begin{array}{lcccc} \text{First} & : & 5a & \cdot & 2a & = & 10a^2 \\ \text{Outer} & : & 5a & \cdot & -3 & = & -15a \\ \text{Inner} & : & 2 & \cdot & 2a & = & 4a \\ \text{Last} & : & 2 & \cdot & -3 & = & -6 \\ \hline \end{array}$$
\begin{aligned} (5a + 2) \cdot(2a - 3) &= \underbrace{10a^2}_{F} - \underbrace{15a}_{O} + \underbrace{4a}_{I} - \underbrace{6}_{L} = \\[1em] &= 10a^2 - 11a - 6 \end{aligned}
RESOURCES
440 702 265 solved problems | 1,769 | 4,648 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.698625 |
http://chadorzel.steelypips.org/principles/2012/03/23/scientific-commuting-some-answ/ | 1,513,020,393,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513866.9/warc/CC-MAIN-20171211183649-20171211203649-00159.warc.gz | 62,615,002 | 10,967 | # Scientific Commuting: Some Answers to “How Much Faster?”
So, the previous post poses a physics question based on some previous fooling around with modeling my commute:
A car starts from rest at the beginning of a straight 1km course, accelerates up to some speed, cruises at constant speed for a while, then decelerates to a stop at the end of the course. A second, identical car does the same course, but decelerates to a stop at the halfway point. It then immediately accelerates back to its cruising speed, and then decelerates to a stop at the end of the course.
How much faster does the second car have to go in order to complete the course in the same time as the first car?
This is deliberately somewhat vague, because I want to leave room for some discussion of what are reasonable values for cruising speed and acceleration, and so on. For the purposes of attempting to answer this, I used values that were moderately realistic and also mathematically convenient: A cruising speed for the first car of 16 m/s, and an acceleration (both speeding up and slowing down) of 4 m/s/s.
So, how do you work this out mathematically? Well, if you look back at the first post I did modeling this, you find that the time required to make the trip in N stages is:
Here you see the reason why I didn’t end up doing anything with this last fall: this is an equation that’s a little tricky for intro physics students, whose level of math preparation isn’t always what we would like it to be. To get an algebraic solution, you would need to take this equation with N=1 and set it equal to the equation for N=2 but a different final speed, and solve for that final speed. That’s more math than I really want to wrestle with, and I guarantee it would be a disaster for most of our intro students.
However, this is an easy enough equation to plug into a computer and generate solutions, which you can plot as nice graphs. So here are a few graphs giving the answer for the situations I looked at:
The horizontal line here indicates the time to cover 1km for a car accelerating at 4m/s to a speed of 16 m/s. The points represent the time to complete the trip (vertical axis) for a car with a given cruising speed (horizontal axis). When the two cross each other, that’s the speed at which the travel times are equal, which turns out to be between 17 and 18 m/s.
That was a little surprising to me, as it’s only about a 10% increase in the speed. The difference gets bigger if you increase the cruising speed of the first car to 25 m/s:
In this case, the speed for the second car is about 35 m/s, or 40% faster. You can also see from this that there’s a minimum in the time as a function of speed (which you can find with calculus, if you care), and if you increase the initial speed enough, that minimum will move above the time for the first car, at which point there’s no speed you can accelerate to that will let you catch the first car.
So, the correct answer to the poll depends on what your values for speed and acceleration are. I punched a bunch of different numbers in, though, and didn’t seem to find any where the difference in speeds was greater than about 40% (though, admittedly, I didn’t spend that much time on this….).
The question doesn’t ask this, but you can play with other factors as well. If you increase the number of stops, you increase the required speed for the second car:
This graph is for doing the trip in four segments, that is, with three stops along the way. Again, there’s a minimum time for a particular speed, and this minimum value moves up as the number of stops increases. For five or more segments (four stops), there’s no speed at which the second car catches the first.
You can also vary the length of the course:
This is for a 500m course rather than 1km. This also increases the difference in speeds– the second car needs to be going at 20-21 m/s to match the first car at 16 m/s. (This makes sense– you’ve got less room in which to make up the ground you lose while stopping…)
I suspect this still probably requires more mathematical sophistication than we can rely on in our intro courses, but it’s kind of fun to play around with. And it was surprising to me to see how small the difference was for reasonable values, so whether it ends up being useful in class or not, I learned something. And got two blog posts out of the deal, so, hey, big win for me…
## 3 thoughts on “Scientific Commuting: Some Answers to “How Much Faster?””
1. My intuition was right, for the speeds I had in mind. I don’t think you’d find stop signs on roads where the speed limit is much above 40 mph, in which case the presence of a stop sign doesn’t actually slow you down all that much.
Of course, I tend to accelerate more gradually when the speed limit is lower, so that would affect the result in practice.
Also, you have a unit error below the first graph, where you give acceleration in m/s.
2. No need for any fancy math, just consider the extremes:
1. Acceleration/deceleration is infinitely fast. Ratio of speeds = 1.
2. Acceleration/deceleration is very slow, just fast enough that car 2 reaches cruising speed, then immediately begins decelerating. Car 1’s average speed is 1.5 times that of car 2.
3. joemac53 says:
We would do similar problems with a velocity/time graph, using the area under the curve as the distance traveled. Since distances are the same, areas are equal. Once we made some assumptions about acceleration we could do most straightforward problems. One assumption that we always argued about was how much friction you could get between tires and road. We finally decided that 1 g was the upper limit, although I am still not certain that is the case. | 1,289 | 5,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-51 | latest | en | 0.962804 |
https://community.filemaker.com/thread/75990 | 1,503,195,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105961.34/warc/CC-MAIN-20170820015021-20170820035021-00198.warc.gz | 750,735,612 | 25,008 | 6 Replies Latest reply on May 2, 2014 8:39 PM by user19752
# Can anyone tell me how I can format a number to read as follows:
I have tried everything without success and been unable to find it in the help section of FileMaker.
US Stationing; always has a + sign 2 numbers to the left.
45789 = 457+89
6890445 = 68904+45
Metric Stationing; always has a + sign 3 numbers to the left.
45789 = 45+789
6890445 = 6890+445
I would like to do this in Custom Functions
• ###### 1. Re: Can anyone tell me how I can format a number to read as follows:
First you need to format your field as text and not a number. a plus sign is not numeric.
The custom function would be:
Let([
l = Length(number);
lstr = Case (format = "US" ; left (number ; l - 2 ) ; format = "Metric"; left (number ; l - 3 ) ; "" ) ;
rstr = Case (format = "US" ; right (number ; 2 ) ; format = "Metric"; right (number ; 3 ) ; "" )
];
lstr & "+" & rstr
)
You can then call it via AddPlus( "45789" ; "US") = 457+89
etc..
• ###### 2. Re: Can anyone tell me how I can format a number to read as follows:
I did it slightly different.
/*
Station( StationNum ; SignPosition )
StationNum is your number that has no + sign.
SignPosition is the number of charactors from the right to put the + sign.
*/
Let(
[
LeftNum = Left ( StationNum ; Length ( StationNum )-SignPosition );
RightNum = Right(StationNum;SignPosition)
];
LeftNum & "+" & RightNum
)
• ###### 3. Re: Can anyone tell me how I can format a number to read as follows:
six to one, half dozen to the other
• ###### 4. Re: Can anyone tell me how I can format a number to read as follows:
To combine the best features of Mike's and Kyle's suggestions …
Stationing ( Number ; Format ) =
Let (
pos = Case ( format = "US"; 2 ; format = "Metric" ; 3 ; "Wrong format argument" ) ;
Case (
pos = 2 or pos = 3 ;
Let ( [
n = Trim ( Number ) ;
lstr = Left ( n ; Length ( n ) - pos ) ;
rstr = Right ( n ; pos )
] ;
lstr & "+" & rstr
) ;
pos
)
)
• ###### 5. Re: Can anyone tell me how I can format a number to read as follows:
Thanks!!
• ###### 6. Re: Can anyone tell me how I can format a number to read as follows:
Replace ( str ; pos ; 0 ; str2 ) insert str2 to str at pos. | 649 | 2,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-34 | longest | en | 0.861397 |
https://rip94550.wordpress.com/tag/topology/ | 1,498,745,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329344.98/warc/CC-MAIN-20170629135715-20170629155715-00166.warc.gz | 838,585,718 | 40,330 | ## Introduction and text
The last section of Bloch’s chapter 3 (simplicial surfaces) is a long (and to my mind at this time, uninteresting) proof of the 2D Brouwer fixed point theorem: any continuous map from the disk to the disk has a fixed point. Bloch also proves a corollary, the no-retraction theorem, that there is no continuous map r from the disk to the circle such that r(x) = x for all x on the circle.
That one sounds interesting. We’ve seen in before, with the commentary that you can’t map the surface of a drum onto its rim without tearing it. I still don’t see it that way. But it is rather shocking that the map r cannot preserve all the points on the rim.
Anyway, we’re not going to fight with those. For me, the climax of chapter 3 is the simplicial Gauss-Bonnet theorem. It shows that there is a definition of curvature for simplicial surfaces (in fact, for polyhedra in general) such that the total curvature of a surface is equal to $2\ \pi$ times its Euler characteristic $\chi\$.
(A simplicial surface is a polyhedron all of whose faces are triangles. I expect we’ll see this again in another post.)
That the total Gaussian curvature of a surface is equal to $2\ \pi\ \chi$ is called the Gauss-Bonnet theorem. It is a reasonable culmination of a first course in differential geometry. The simplicial version means that we have a definition of curvature for simplicial surfaces and polyhedra which gives us a form of the Gauss-Bonnet theorem. That says it’s a reasonable definition of curvature.
So what is this marvelous definition of simplicial curvature? It’s also called the angle defect, and goes back to Descartes. Read the rest of this entry »
## Introduction
Putting out the following few books has been far harder than I expected, and has taken a lot more time. There are 6 of them: 3 texts, 1 reference, and 2 small sets of notes.
The fundamental problem is that I haven’t worked thru these books yet. Simply put, I’m effectively a grad student trying to figure out which books to read in order to introduce myself to a new field.
To put it more fancifully, I feel a bit like a wide-eyed urchin looking in a bakery window, trying to figure out what the different pastries will taste like, and I’ve picked out a few of them to try.
That simile fails, of course, because I’m not just looking at the pastries; I’ve held them in my hands and looked inside. I own these books, I’ve read each preface and table-of-contents, and I’ve read further into them. I’ve seen every one of them in other bibliographies; I’ve just read some of the reviews on Amazon….
The problem is, I haven’t gone into these books and come out the other side.
Read the rest of this entry »
## Introduction and K4, the complete graph on 4 vertices
I want to show you something clever, but I’m going to omit the details of how we justify part of it. And I’m going to raise a question about another part of it. But I think this application of the Euler characteristic is interesting, even if I won’t or can’t cross all the t’s.
We can define the Euler characteristic of a graph as $\chi = v - e\$. We can show, for example, that every tree (a graph with no closed paths) has $\chi = 1\$. If a graph is not a tree, then the closed path might create a face, but we don’t count the faces.
One question that arises when we have a graph is: is the graph planar? That is, can it be drawn in the plane so that edges do not have extraneous intersections?
Better to show you. Draw a square (or rectangle), and draw the two diagonals. This is called K4, the complete graph on 4 vertices, because every vertex is connected to every other vertex.
But the diagonals cross each other.
Read the rest of this entry »
## The Euler Characteristic: Teasers
These are things i came across when I first started looking at the Euler characteristic, in fact, when I was looking at triangulations in particular.
## n-manifolds
The Euler characteristic $\chi$ generalizes to dimensions other than 2, and there are at least three noteworthy theorems involving the Euler characteristic. I’m not going to say much about them, because they, like so much else, are still outside my comfort zone. I’ll just barely tell you what they are, and leave you to chase them down if they interest you.
As we’ve seen, the Euler characteristic of a polyhedron is given by
$\chi = v - e + f\$,
where v, e, f are the numbers of vertices, edges, and faces. Homeomorphic polyhedra have the same Euler characteristic, and that means we can define the Euler characteristic of a topological surface as the Euler characteristic of any polyhedron which is homeomorphic to it.
This alternating-sign sum of the numbers of 0-, 1-, and 2- simplices generalizes in the obvious way: for an n-simplex, we take the sum, with alternating signs, of the numbers of k-simplices, for k <= n. As for surfaces, so for n-manifolds: this is a topological invariant, and we want to define the Euler characteristic of an n-manifold as the Euler characteristic of any k-simplex homeomorphic to it.
Read the rest of this entry »
## Introduction
I first came across the separation axioms in a functional analysis text (Bachman & Narici, “Functional Analysis”; Dover 1998, orig. 1966). I really like classification theorems, and these seemed really cool. As I said in the second post about general topology books, there is still not general agreement on the terminology. The mathematics is unambiguous, but there are two sets of intertwined terminology.
For example, the terms T4 and normal (to follow) are combined with the term T1 in either of two ways. T1 is unambiguous, but we either say that a topological space is
normal iff it is T1 and T4
or
T4 iff it is T1 and normal.
That is, there is a property called either T4 or normal. While we can study spaces which have that property alone, it is usually more interesting to study spaces which have that property and the T1 property. Such spaces are called normal or T4, respectively, depending on what name we assigned to the property. That’s the rub: is the property itself called T4 or normal? Then the other term is used for the combination with T1.
I choose to use the terminology typified by
Read the rest of this entry »
## Books Added: general topology 2
OK, I said I wouldn’t go buy more books because mine were old. I didn’t. I bought two more old books. I was looking on the internet for more about the “separation axioms” and came across these two. One was a familiar title that I probably should have gone looking for (“Counterexamples”), but I didn’t know of the other.
(Any discussion of the separation axioms must cope with the fact that there are two distinct sets of terminology. These books were cited as the epitomes of the two terminologies.)
They’re both very well reviewed and, it seems to me, excellent. Quite apart from that, they are also Dover paperbacks, which means they are quite affordable.
Willard is in the same class as Dugundji and Kelley: a textbook which is exhaustive enough to serve as a reference. Like Kelley, it has lots of problems, and many of them investigate auxiliary material. Oh, unlike the other two, Willard has a few pictures.
It is also fun to read. No, he’s not trying to be a stand-up comic, but every once in a while he phrases something nicely. “In the next (and obvious) step to normal spaces, we find ourselves confronted with the real bad boy among the separation axioms.”
Steen and Seebach is a compact presentation of topology (40 pages), beautifully organized counterexamples (120 pages), a summary of metrization theory (24 pages), and a collection of charts and tables for finding a desired example (20 pages). I would think, speaking as an onlooker, that this is an indispensable reference if you do much topology.
Need a reference text? Unless you need something specific from Dugundji or Kelley, I suggest you get Willard.
Doing topology beyond your first course? Get Steen & Seebach on general principles.
Steen, Lynn Arthur and Seebach, J. Arthur Jr., Counterexamples in Topology, Dover, 1995 (orig. 1978),
ISBN 0 486 68735 X
[general topology; 17 Nov 2008]
Reference. Very well organized, with many charts of relationships.
Willard, Stephen. General Topology, Dover, 2004 (orig. 1970).
ISBN 0 486 43479 6.
[general topology; 17 Nov 2008]
Textbook and reference. Well-written. Copious historical references and notes.
## Introduction
Let me discuss my favorite general topology, i.e. “point set topology”, books. I have already discussed “algebraic topology” here.
Like so much other pure mathematics that I do not use professionally (for modeling power plants), topology is not on the tip of my tongue. But it’s fun, so I do it once in a while. And it’s fundamental, so I often have to go back to it when I’m playing with other mathematics.
This is a discouraging review in one respect: 4 of these 10 books are out of print: Kasriel, Dugundji, Sieradski, and Seifert & Threlfall. Heck, if you want Seifert & Threlfall, you should buy it in German! And for two of the books (Naber, Chinn & Steenrod) that Amazon claims to have in stock, there are multiple listings, many of which say the books are not available.
But I’m not going to go buy more books just because the ones I have are out of print. This is what I like, of what I have. | 2,203 | 9,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-26 | longest | en | 0.936354 |
https://www.tentotwelvemath.com/fmp-10/system-of-two-equations-two-unknowns/introduction-to-systems-of-equations/ | 1,716,207,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00746.warc.gz | 900,267,490 | 23,673 | # Introduction to Systems of Equations
Khan Academy: Introduction to Systems of Equations
Stations: Unit 7 Stations
# One unknown, one equation
Suppose a child asks a grown up their age. The grown up replies “I’m not telling you, but if I live for another 72 years then I’ll be 100”.
The age of the grown up is still ‘unknown’ but there is enough information to figure it out. Because we know that we can figure out that the grown up is 28 years old.
Or, using algebra:
# Two unknowns, two equations
Now suppose you ask my childrens’ ages. I have two children. I reply “I’m not telling you but if you add their ages together you get 10, and they are 4 years apart”.
The ages are still ‘unknown’ but because there are two different facts about the two ages, it is possible to figure them out.
To solve, we need to find two integers that add together to give 10, then find the one pair where the difference between them is four. Here are all the options for adding to 10, written :
The answer is as and .
To use algebra, we might say youngest child has age and oldest has age . We have the following information:
Their ages add up to 10. That leads us to the equation , (or ).
They are 4 years apart. That leads us to the equation , (or ).
This unit is about how to find the values of two unknowns when we have two equations about them.
# Unique Solution
In the case about the two children, there was only one value of and that simultaneously solved the first and the second equation. This is called the unique solution. If we graph the two equations between and we have:
Notice that because the point lies on both lines, it is the point of intersection of the two lines. Finding the solution to two equations in two unknowns boils down to finding the point where the two graphs intersect. If graphing is not efficient, an algebraic approach can be used.
# Identifying the unique Solution
Which ordered pair satisfies both equations? Use mental arithmetic to determine which one of the three points lies on both lines.
# Infinitely Many Solutions
When two expressions are equivalent, all ordered pairs will satisfy both equations.
Consider
and
The expression on the right is just a rearrangment of the expression on the left.
Using the relation , we can generate the following ordered pairs:
…etc
Notice that all of those ordered pairs also satisfy .
As there are infinitely many points on any line, there are infinitely many ordered pairs that satisfy the system
If one equation can be rearranged to the other, the equations are equivalent. All ordered pairs satisfying one equation will also satisfy the other.
# No Solutions
If our set of equations leads to two lines that are parallel but not coincident, then there is no ordered pair that will satisfy both relations.
On a graph:
Desmos Activities: Go to student.desmos.com, ask for the class code.
• Polygraph (requires pairs of students)
• Systems of Two Linear Equations (requires at least 4 students at a time)
Maths is Fun Page: | 663 | 3,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-22 | latest | en | 0.951529 |
https://www.hackmath.net/en/example/1525 | 1,561,437,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999787.0/warc/CC-MAIN-20190625031825-20190625053825-00258.warc.gz | 763,999,014 | 6,517 | # Square and circle
Into square is inscribed circle with diameter 10 cm.What is difference between circumference square and circle?
Result
x = 8.584 cm
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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## Next similar examples:
1. Star
Calculate the content of the shaded part of square with side 28 cm.
2. Square
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4. Flowerbed
In the park there is a large circular flowerbed with a diameter of 12 m. Jakub circulated him ten times and the smaller Vojtoseven times. How many meters each went by and how many meters did Jakub run more than Vojta?
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6. Two circles
Two circles with a radius 4 cm and 3 cm have a center distance 0.5cm. How many common points have these circles?
7. Mine
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8. Circle r,D
Calculate the diameter and radius of the circle if it has length 52.45 cm.
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11. Bicycle wheel
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Calculate the area of two circles annulus k1 (S, 3 cm) and k2 (S, 5 cm).
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Perimeter of regular hexagon is 113. Calculate its circumradius (radius of circumscribed circle).
16. Athlete
How long length run athlete when the track is circular shape of radius 120 meters and an athlete runs five times in the circuit?
17. Circle - easy 2
The circle has a radius 6 cm. Calculate: | 594 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-26 | latest | en | 0.916212 |
https://www.enotes.com/homework-help/f-x-e-4x-n-4-find-nth-maclaurin-polynomial-810505 | 1,670,135,393,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00329.warc.gz | 786,508,753 | 18,983 | `f(x)=e^(4x) , n=4` Find the n'th Maclaurin polynomial for the function.
Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:
`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`
or
`f(x)= f(0)+(f'(0)x)/(1!)+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`
To determine the `4th` Maclaurin polynomial from the given function `f(x)=e^(4x)` ,
we may apply derivative formula for...
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Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:
`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`
or
`f(x)= f(0)+(f'(0)x)/(1!)+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`
To determine the `4th` Maclaurin polynomial from the given function `f(x)=e^(4x)` ,
we may apply derivative formula for exponential function:` d/(dx) e^u = e^u * (du)/(dx)`
Let `u =4x` then `(du)/(dx)= 4 `
Applying the values on the derivative formula for exponential function, we get:
`d/(dx) e^(4x) = e^(4x) *4`
Applying `d/(dx) e^(4x)= 4e^(4x)` for each `f^n(x)` , we get:
`f'(x) = d/(dx) e^(4x)`
`=e^(4x) * 4`
`= 4e^(4x)`
`f^2(x) = 4 *d/(dx) e^(4x)`
`= 4*4e^(4x)`
`=16e^(4x)`
`f^3(x) = 16*d/(dx) e^(4x)`
`= 16*4e^(4x)`
`=64e^(4x)`
`f^4(x) = 64*d/(dx) e^(4x)`
`= 64*4e^(4x)`
`=256e^(4x)`
Plug-in `x=0` , we get:
`f(0) =e^(4*0) =1`
`f'(0) =4e^(4*0)=4`
`f^2(0) =16e^(4*0)=16`
`f^3(0) =64e^(4*0)=64`
`f^4(0) =2564e^(4*0)=256`
Note: `e^(4*0)=e^0 =1` .
Plug-in the values on the formula for Maclaurin series.
`f(x)=sum_(n=0)^4 (f^n(0))/(n!) x^n`
`= 1+4/(1!)x+16x^2+64x^3+256/(4!)x^4`
`=1+ 4/1x +16/(1*2)x^2 + 64/(1*2*3)x^3 +256/(1*2*3*4)x^4`
`=1+ 4/1x +16/2x^2 + 64/6x^3 +256/24x^4`
`= 1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`
The 4th Maclaurin polynomial for the given function `f(x)= e^(4x)` will be:
`e^(4x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`
or `P_4(x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`
Approved by eNotes Editorial Team | 971 | 2,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.570665 |
http://tedmcginley.com/lib/functional-programming-international-computer-science-series | 1,508,502,513,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00847.warc.gz | 343,373,967 | 12,779 | Format: Hardcover
Language: English
Format: PDF / Kindle / ePub
Size: 9.21 MB
Note that a value of type M a is just a promise to produce a value of type a, and no effect is yet performed. You want to set up as much computation up front and let the system schedule the most advantageous execution. This practical guide demonstrates the Python implementation of a number of functional programming techniques and design patterns. What happens if you use <- instead of <<-? You are able to create data streams of anything, not just from click and hover events. It is the job of whatever calls the paint method to determine whether and when the Component needs to be painted.
Pages: 616
Publisher: Addison-Wesley (July 1988)
ISBN: 0201192497
Functional Programming, Glasgow 1992: Proceedings of the 1992 Glasgow Workshop on Functional Programming, Ayr, Scotland, 6-8 July 1992 (Workshops in Computing)
A Practical Course in Functional Programming Using ML
HTML / XHTML
Computation and Deduction
But I haven’t found this to be at all distracting so far , cited: Introduction to Functional read pdf read pdf. The advantages of FP are significant: Verification: There is a close correspondence between the mathematical reasoning that justifies the correctness of a program and the program itself. Principles of proof by mathematical induction go hand-in-hand with the programming technique of recursion pdf. MkStack a (Stack a) push :: a -> Stack a -> Stack a push x s = MkStack x s size :: Stack a -> Integer size s = length (stkToLst s) where stkToLst Empty = [] stktoLst (MkStack x s) = x:xs where xs = stkToLst s pop :: Stack a -> (a, Stack a) pop (MkStack x s) = (x, case s of r -> i r where i x = x) top :: Stack a -> a top (MkStack x s) = x module Qs where qs :: [Int] -> [Int] qs [] = [] qs (a:as) = qs [x The implementation of that method could however change from blocking to non-blocking and the code would not need to change. It is not yet feature complete with the. Net version of Rx, but what is implemented has been in use for the past year in production within the Netflix API , e.g. Functional Programming download pdf http://tedmcginley.com/lib/functional-programming-international-computer-science-series. Particular topics of interest include: Language Design: type systems; concurrency and distribution; modules; components and composition; metaprogramming; relations to object-oriented or logic programming; interoperability The conference also solicits Experience Reports, which are short papers that provide evidence that functional programming really works or describe obstacles that have kept it from working in a particular application Functional Programming, read online primaryspeakers.com. As soon as encapsulation is really enforced the object can do what it needs to deal with the design parameters (including keeping histories, etc.). What are your current thoughts on objects and simulation in distributed environments? You have historically worked or advised on projects using approaches that never reached the mainstream epub.
The Functional Approach to Programming
On Meaningful Scientific Laws
Implementation and Application of Functional Languages: 19th International Workshop, IFL 2007, Freiburg, Germany, September 27-29, 2007 Revised Selected Papers (Lecture Notes in Computer Science)
In the previous example is shown how you can transform a function by reducing the number of arguments by one. If you want to break it further you will need to create another higher order function for partitioning , e.g. Clojure Applied: From Practice to Practitioner http://primaryfineart.com/books/clojure-applied-from-practice-to-practitioner. It's a mindset-altering philosophy and/or worldview, not a completely novel theory of computation. It's also the myth that pisses me off the most, because the correct statement looks very similar and says something incredibly important about functional programming. Referential transparency is a subtle concept, which is founded on two other concepts: substitutability and frames of reference , cited: The IT Measurement Compendium: Estimating and Benchmarking Success with Functional Size Measurement http://tedmcginley.com/lib/the-it-measurement-compendium-estimating-and-benchmarking-success-with-functional-size-measurement. It also contains a variety of utilities that will make things actually appear within the Component. Just as you don't have to determine when or whether paint should be invoked, you don't need to provide the Graphics object. Like magic, when paint is invoked, the Graphics object will be there Introduction to Functional Programming using Haskell (2nd Edition) read online. As you can see, it is possible to implement recursion even with the anonymous functions/lambda expressions. Partial functions are functions that reduce the number of function arguments by using default values. If you have a function with N parameters, you can create a wrapper function with N-1 parameters that calls the original function with a fixed (default) argument , cited: Introduction to Functional Programming using Haskell (2nd Edition) http://tedmcginley.com/lib/introduction-to-functional-programming-using-haskell-2-nd-edition. To demonstrate this in terms of formal logic, let me define the following symbols: p1 ::= 'the morning star' equals 'the planet venus' p2 ::= 'the evening star' equals 'the planet venus' p3 ::= 'the morning star' equals 'the evening star' (+) ::= an operator that means 'two things which equal the same thing, equal each other' => ::= an operator which means 'this rule produces this result' With those, we can generate the following statements: which is all well and good , source: Functional Programming: Questions and Answers http://jasperarmstrong.com/?lib/functional-programming-questions-and-answers.
Advanced Functional Programming: First International Spring School on Advanced Functional Programming Techniques, Bastad, Sweden, May 24-30, 1995 (Lecture Notes in Computer Science)
Functional Reactive Programming
Advanced Functional Programming: Third International School, AFP'98, Braga, Portugal, September 12-19, 1998, Revised Lectures (Lecture Notes in Computer Science)
Functional and Object Oriented Analysis and Design: An Integrated Methodology
The Theory of Best Approximation and Functional Analysis (Regional Conference Series in Applied Mathematics - Vol 13)
Approaches and Applications of Inductive Programming: Third International Workshop, AAIP 2009, Edinburgh, UK, September 4, 2009, Revised Papers (Lecture Notes in Computer Science)
Functional and Logic Programming: 8th International Symposium, FLOPS 2006, Fuji-Susono, Japan, April 24-26, 2006, Proceedings (Lecture Notes in Computer Science)
Trends in Functional Programming 10
Foundations of Logic and Functional Programming: Workshop, Trento, Italy, December 15-19, 1986. Proceedings (Lecture Notes in Computer Science)
Programming Scala: Scalability = Functional Programming + Objects
Functional and Constraint Logic Programming: 20th International Workshop, WFLP 2011, Odense, Denmark, July 19, 2011, Proceedings (Lecture Notes in Computer Science)
Lyapunov-Schmidt Methods in Nonlinear Analysis and Applications (Mathematics and Its Applications) (Volume 550)
Introduction to Functional Programming Systems Using Haskell (Cambridge Computer Science Texts)
Advanced Functional Programming: First International Spring School on Advanced Functional Programming Techniques, Bastad, Sweden, May 24 - 30, 1995. Tutorial Text (Lecture Notes in Computer Science)
Programming Clojure
Certified Programming with Dependent Types: A Pragmatic Introduction to the Coq Proof Assistant (MIT Press)
Functional Programming Using F#
Wavelets in Soft Computing (World Scientific Series in Robotics and Intelligent Systems, 25) (v. 25)
Mastering Clojure
Wavelets in Soft Computing (World Scientific Series in Robotics and Intelligent Systems, 25) (v. 25) | 1,690 | 7,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-43 | latest | en | 0.874893 |
https://blog.arkieva.com/safety-stock-process/ | 1,726,866,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00375.warc.gz | 122,121,677 | 26,105 | Do others challenge your safety stock selections? Are you in need of a safety stock process or policy? Want to take the drama out of your safety stock process?
## Use this collaborative six-step process to demystify your safety stock process and create buy-in.
1. Offer multiple calculations of safety stock
2. Add descriptive statistics for the data element of inventory: Item-Location
3. Review with Planners and Sales for realism
4. Negotiate
5. Use Excel Pivot Tables to check levels
6. Rinse and repeat every 90 days
Note: In these examples, I’ll show how these calculations can be achieved using Excel. For larger datasets, using a planning software solution might make the calculation and analysis process less cumbersome.
1) Offer multiple calculations
There are several methods for calculating safety stocks, such as a period of supply and statistical safety stock. These are all sound methods to arrive at a safety stock value, however, think of others that are available.
• Download shipment history to Excel and determine the 12-month MAX. Safety Stock is then equal to MAX 12 – Forecast. When this turns negative with strong sales projected, you turn to the methods above. However, with stable products, this easily understood method gives a reasonable buffer to protect against the highest demand seen in a 12-month period.
• Use the LARGE function in Excel to factor the expression (MAX 12 – Forecast). For 80% of the MAX 12 = 0.8*(LARGE Array, 1) where 1 is the largest value in the array. You can also choose the second largest value and so on as well as change the factor, 0.8 to other numbers.
• MAXddlt : Calculate the maximum demand in a lead time period every day or every week using the sum function in Excel. Continue using MAXddlt – Forecast as above. Note: When the maximum demand is exceeded in a lead time period, you find yourself at the lead time fence for which you can make a change with no lead time compression.
• Choose two or three of these methods that best fits your business and consider taking the minimum calculation to start the conversation as you present to Planners and Sales.
To give meaning to the safety stock presented for review, apply relevant descriptive statistics alongside the safety stock calculation. This provides grounding for the reviewers.
• Show product families, warehouse, who buys, and basic product descriptions. In addition, display these descriptive statistics mentioned in previous blogs, at the data element item-location:
• ABC code from Product/Customer Matrix
• Coefficient of Variation >>COV<<(Std Dev/Mean): from the variability of the demand stream
• Average Demand Interval >>ADI<<(12/No. of months sold in a 12-month period)
• Volume/Variability Matrix: Negotiate a ‘hurdle rate’, or threshold for monthly sales by product family and a COV ‘hurdle rate’ (typically 0.3 or 0.5). See Fig. 1 for an example.
Fig. 1 Volume Variability Matrix to label safety stock quadrants
If you consider the above minimum essential, then customize with the following, at the SKU level:
• Show last three months of shipments alongside next three months of forecast
• List the number of ship to customers
• Show stock out history (keep a record of weekly stock outs and calculate frequency: No. of stock outs/No. of weeks)
• Months of inventory from weekly reporting (decide on what level is ‘too much’, ‘too little’ or ‘just right’.
3) Review with Planner and Sales for realism
Compile the above data in an Excel spreadsheet, create a Sales/Planner team for review, and fend off all challenges with descriptive statistics. Here’s some you might face:
• If the SKU has an ADI of 6, in Quad III, and they want more safety stock, you might want to ask why invest in a product that sells every 6 months with high variability and low volume.
• If the product shows no shipments in the past 12 months, why add safety stock if we hold inventory of the phase-out product?
• For SKUs in Quad I with low variability and steady production, do we really need lots of inventory?
4) Negotiate
No matter how sophisticated your analytics, NEVER ignore Market Intelligence. When Sales wants more safety stock of that SKU with an ADI of 6, it could be because they landed a new contract. Likewise, for an SKU with no sales. And when a Quad I appear steady; the Planner might know the strategic customer will soon be leaving.
5) Use Excel pivot tables to check levels
Become a ‘Pivoteer’ and show safety stocks in volume and dollars (Cost of Goods Sold) by groups of:
• Product family
• Strategic Customers
• High Volume (Quads I & II) – Low Volume (Quads III & IV) – High Variability (Quads II & III) and Low Variability (Quads I & IV)
• Frequency of Stock Outs
• Warehouse
• SKU and Ship to Customer combination
• Your bespoke set of descriptive statistics
6) Rinse and Repeat every 90 days
Things happen as you’ll soon find out from Market Intelligence (its not an oxymoron). You need to keep up with the changes in the market place and create realistic safety stock your team can agree upon. 90 days may not fit your business, so consider every lead time period at least.
Which of the steps is most difficult?
Step 3 by far. You need to create buy-in for this important process. Safety Stocks support sales, so you need their recognition of responsibility. Emphasize the importance by using the language of business (dollars) and create buy-in. Monitor results, stay committed, stay disciplined and others will follow.
Note: Apply these descriptive statistics for your Product Manager to prune the portfolio. Can you quantify what makes a “good SKU” from a “bad SKU”? Stay tuned… | 1,232 | 5,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.911953 |
http://en.m.wikibooks.org/wiki/Electrodynamics/Electric_Charge | 1,432,982,425,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930995.36/warc/CC-MAIN-20150521113210-00275-ip-10-180-206-219.ec2.internal.warc.gz | 82,133,259 | 6,362 | # Electrodynamics/Electric Charge
## Electric ForceEdit
Subatomic particles, such as the electron and the proton, exist that display an ability to affect each other without physical interaction. As an example of the type of interaction we are talking here, consider the pull of gravity between the sun and the earth: the sun and the earth are not touching each other, but they still have a relationship because of the force of gravity. Similar to gravity, the electric force is able to affect objects at a distance.
Some objects are affected by an electric force, and some objects are not. Inside the atom, to continue our example, there are three types of particles: electrons, protons, and neutrons. Of these three, the neutron is not charged and has no electric force. Protons and electrons affect each other in a complicated way, as demonstrated by this table:
Proton Electron
Proton Push Pull
Electron Pull Push
Two charges will exert an electric force on each other. This force can either repel (in the case of like-signs) or can attract (in the case of opposite signs). The strength of the force depends on how far away the two charges are: if they are close the force will be strong, if they are far, the force will be weak. We will see the exact relationship that governs this force in the next chapter on Coulomb's Law.
## Charge ConventionsEdit
As a matter of mathematical interest, it helps to quantify the different charges in some way. We could, in a naive way, say that an electron has a charge value of 1, and a proton has a charge value of 2. However this system would be of very limited usefulness, and it would actually take more work for us to write equations that used this system. Modern science has settled on the convention of assigning a +1 charge to a proton, and a −1 charge to an electron. The system could easily have been reversed and many of our equations would still hold true (or would need a simple sign change). However, as a matter of convention we say that electrons are negatively charged, and protons are positively charged, and that neutrons are neutrally charged (with a charge of 0).
## Point ChargesEdit
When dealing with small electric charges mathematically, it is the most easy for us to consider a point charge. Point charges have mass and charge but no volume and are infinitely small. Of the fundamental particles the electron is perhaps the closest approximation to a point charge as all experimental evidence is consistent with it being a point charge. Protons etc. are believed to be composed of quarks and thus have a finite volume but in most cases can be considered as point charges.
## Charge DensityEdit
We can define a function ρ(x, y, z) to be the charge density. The charge density is a measurement of the average amount of charge per unit volume. If the charge is distributed over a fine layer on a surface, it is also useful to define a superficial or surface charge density σ(x, y, z). The surface charge density measures the amount of charge per unit area.
## UnitsEdit
The amount of charge transferred between two points at 1 ampere over 1 second is the amount of electric charge and its derived SI unit is called a coulomb. It's base SI units are ampere-seconds. | 690 | 3,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2015-22 | latest | en | 0.951076 |
https://www.techprofree.com/learn-essential-math-for-data-science-step-by-step/ | 1,726,370,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00567.warc.gz | 974,155,635 | 19,339 | # Learn Essential Math For Data Science Step by Step
Download Essential Math For Data Science Step by Step PDF Notes totally free. We no longer only apply desktop or laptop computers to access the internet. We now carry it around with ourselves in our cars, smart phones, and home appliances. Over the span of two decades, this has gradually encouraged a transformation. Data currently changed from being a tool for operations to being gathered and analyzed for less clear goals. A smart watch continuously records information about our respiration, strolling distance, and other metrics. The data is then uploaded to a cloud where it can be analyzed by several people. Computerized automobiles are recording our driving behaviors, and manufacturers are using this information to build self-driving cars. Drugstores are already stocking “smart toothbrushes,” which track brushing patterns and store that information in the cloud.
The purpose of these notes is to introduce readers to various mathematical, statistical, & machine learning fields that can be used to solve practical issues. Foundational math ideas are covered in the first four chapters covering statistics, linear algebra, probability, and practical calculus. Machine learning will be introduced in the final three chapters. The ultimate goal of instructing machine learning is for us to connect all we have learned and show how to use machine learning and analytical libraries in ways that go beyond a simple black box comprehension.
Essential Mathematics for Data Science
Mathematics structures the fundamental backbone of contemporary scientific disciplines; data science is no exception. Most contemporary data science methods have strong foundations in mathematics.
Mathematics serves as a vital toolbox. It provides the framework for identifying patterns in data and creating clever algorithms. Our main objective in creating this tutorial is to save you time while you go through the fundamental mathematical topics. Imagine being able to avoid spending months studying machine learning-related math texts cover to cover. Here, we focus exclusively on the essential elements of mathematics for data science, saving you time and letting you focus on the important stuff.
The Importance of Mathematics in Data Science
While proficiency in programming, business insight and analytical thinking are essential for excelling as a data scientist, having a thorough understanding of the underlying mathematics is invaluable. Instead of merely operating as the person behind the wheel, knowing the intricate mechanics behind algorithms provides a significant advantage in the field.
This knowledge becomes especially vital for professionals transitioning to data science from diverse backgrounds such as hardware engineering, retail, chemical processes, medicine, and business management. Although these fields may involve basic numerical calculations, the specific math skills demanded in data science can differ significantly.
Covering topics from algebra to calculus, this guide emphasizes the comprehension of data patterns, the creation of predictive formulas and the evaluation of model accuracy. Whether you’re a beginner or an aspiring expert, this guide empowers you to converse with data, making it an indispensable resource for anyone delving into.
Essential Math for Data Science
Before diving into data science, a solid understanding of fundamental mathematics is essential. A strong grasp of number theory, primes, factors and proficiency in manipulating fractions, decimals and percentages are essential foundations for understanding mathematical concepts in data science.
1. Concepts of Linear Algebra
The basic ideas of linear algebra are presented in this part, beginning with vectors and associated operations as mathematical representations of data and transformations. After that, matrix algebra is covered, including multiplication and inversion, which are essential for handling transformations and data. The investigation includes eigenvalues and eigenvectors, which are essential for reducing the dimensionality of data analysis and a range of data science and machine learning applications. These basic subjects lay the mathematical groundwork required to comprehend intricate data structures.
1. Essential Statistical Concepts
The main statistical ideas that are crucial to data science are covered in this section. It explains measures of central tendency such as mean, median and mode; shedding light on typical values seen in datasets. It also looks at dispersion metrics like variance and standard deviation, which shed light on the variability and spread of the data. The fundamentals of probability, including significant concepts and probability distributions, are explained and serve as the foundation for a number of data science approaches. Furthermore, key descriptive statistics methods are presented, which facilitate effective summarization and the extraction of meaningful information from complex datasets.
1. Fundamental Calculus Principles
Calculus fundamentals that are essential for data science applications are covered in this chapter. The study of differential calculus centers on derivatives and their applications in everyday life. Concepts from integral calculus are discussed, highlighting the importance of integrals in data analysis, especially for topics such as cumulative distribution functions. The subject covers boundaries and continuity, which are essential concepts for comprehending mathematical behavior in modeling and data analysis at crucial junctures.
Expert Tip
It is not necessary to be an expert in complex mathematics to start working with data science. Prioritize the aforementioned fundamentals before delving into machine learning. As your machine learning skills develop, familiarize yourself with new mathematical ideas as they arise. Consider it as an organic way to improve your arithmetic skills while focusing on what’s required for the current activity. Beginning with the fundamentals and filling in the blanks when you come across them in practical applications can help you quickly understand and efficiently apply mathematical concepts.
Summary
Understanding essential mathematics is the cornerstone of proficiency in data science. It provides the fundamental framework for uncovering patterns in data, constructing intelligent algorithms and making informed decisions. This guide has been meticulously crafted to streamline your learning process, allowing you to concentrate on the critical mathematical aspects necessary for data science without getting lost in extensive textbooks.
In a nutshell, mastering fundamental mathematics is pivotal to excelling in data science. While programming and analytical skills are crucial, a deep understanding of mathematics provides a competitive edge. This article aims at key areas like linear algebra, statistics and calculus, laying the foundation for interpreting complex data patterns and creating effective models. For those transitioning from diverse fields, these mathematical concepts bridge the gap, enabling a smooth entry into data science. Expertise in intricate mathematics isn’t mandatory initially; start with basics and progress organically, aligning learning with practical applications. This approach allows efficient understanding and application of mathematical principles. Whether you’re a novice or aspiring expert, this guide empowers you to navigate the world of data science mathematics, offering a pathway to unlock the field’s vast potential.
Following topics are covered in these notes
• Basic Math and Calculus Review
• Probability
• Descriptive and Inferential Statistics
• Linear Algebra
• Linear Regression
• Logistic Regression and Classification
• Neural Networks
• Career Advice and the Path Forward | 1,304 | 7,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.919443 |
https://mathoverflow.net/questions/327672/number-of-edge-invariant-walks-in-complete-graph | 1,679,615,162,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00642.warc.gz | 439,847,331 | 27,745 | # Number of edge-invariant walks in complete graph
Given a complete graph $$(V,E)$$ with $$n$$ vertices $$V$$ and walks $$p \in V^{l+1}$$ of length $$l$$. We say the edges of walks $$p$$ are the multiset
$$e_p = \{ (p_i,p_{i+1}) \mid 1 \leq i \leq l \}.$$ Also the frequency of vertex $$v \in V$$ in walk $$p$$ is $$p[v] = |\{ i \mid p_i = v \}|$$. The frequency of all vertices $$f_p \in \mathbb{N}^n$$ in walk $$p$$ is then given by $$f_p = [ p[v_1], p[v_2], \dots, p[v_n] ].$$
Now, two walks $$p$$ and $$q$$ are called edge-invariant if $$f_p = f_q$$ and $$e_p = e_q$$. How many different walks of length $$l$$, which are not edge-invariant, exist? More precisely, i.e., $$max_{P \subset V^{l+1}} | \{ P \mid \forall p, q \in P: e_p \neq e_q \textrm{ or } f_p \neq f_q \textrm{ for } p \neq q \}|.$$
For example, $$n = 2$$, it's $$4, 8, 14, 22, 32, 44, 58, 74, 92, 112, 134, 158$$ for $$l = 1,\dots,12$$. Specifically for $$l = 3$$, out of the 16 possible walks, two are edge-invariant: $$e_{[1,1,2,1]} = e_{[1,2,1,1]} \textrm{ and } e_{[2,2,1,2]} = e_{[2,1,2,2]}.$$
For $$n = 3$$, it's given here as $$9, 27, 75, 186, 414, 840, 1578, 2784, 4662, 7476, 11556$$ for $$l = 1,\dots,11$$.
It seems to be related to k-abelian equivalence classes, where the general problem seems to be rather difficult, but maybe the 2-abelian cases is simpler?
• What do you mean by complete directed graph? Tournament? Next, what is "edge invariant" - a binary relation on pairs? Then what does "how many edge-invariant paths" mean? Apr 10, 2019 at 14:30
• @FedorPetrov Sorry, I meant simply a complete graph (with edges from each vertices to any other) and not a tournament. I also clarified the edge-invariant property.
– Jiro
Apr 10, 2019 at 14:55
• @MaxAlekseyev oh yes, I'm sorry, it should be trails not paths. I will correct this
– Jiro
Apr 10, 2019 at 15:20
• @MaxAlekseyev In fact, it's even walks
– Jiro
Apr 10, 2019 at 15:27
• @MaxAlekseyev Thanks for the clarification. $e_p$ are multisets.
– Jiro
Apr 10, 2019 at 15:31
I'm not sure if the formula below is useful, but at least it can be used to numerically count edge-invariant walks classes for small $$n,l$$.
Let's assign a unique variable to each edge and each vertex of the complete graph, say, $$x_{ij}$$ to an edge $$(i,j)$$ and $$y_i$$ to a vertex $$i$$. Then, we assign to each edge $$(i,j)$$ the weight $$x_{i,j}y_j$$ and consider the graph adjacency matrix $$A$$. For example, for $$n=4$$ the matrix is $$A = \begin{bmatrix} x_{11}y_1 & x_{12}y_2 & x_{13}y_3 & x_{14}y_4\\ x_{21}y_1 & x_{22}y_2 & x_{23}y_3 & x_{24}y_4\\ x_{31}y_1 & x_{32}y_2 & x_{33}y_3 & x_{34}y_4\\ x_{41}y_1 & x_{42}y_2 & x_{43}y_3 & x_{44}y_4 \end{bmatrix}.$$
Then the number of classes of edge-invariant of length $$l$$ is given by the number of distinct monomial terms in $$[y_1,y_2,\dots,y_n]\cdot A^l\cdot [1,1,\dots,1]^T.$$
For example, for $$n=2$$ and $$l=3$$ we get the polynomial: $$x_{11}^3y_1^4 + x_{11}^2x_{12}y_1^3y_2 + x_{11}^2x_{21}y_1^3y_2 + 2x_{11}x_{12}x_{21}y_1^3y_2 + x_{11}x_{12}x_{21}y_1^2y_2^2 + x_{12}^2x_{21}y_1^2y_2^2 + x_{12}x_{21}^2y_1^2y_2^2 + x_{11}x_{12}x_{22}y_1^2y_2^2 + x_{11}x_{21}x_{22}y_1^2y_2^2 + x_{12}x_{21}x_{22}y_1^2y_2^2 + 2x_{12}x_{21}x_{22}y_1y_2^3 + x_{12}x_{22}^2y_1y_2^3 + x_{21}x_{22}^2y_1y_2^3 + x_{22}^3y_2^4,$$ where the monomials describe equivalence classes of walks (more specifically, $$x$$'s describe $$e_p$$ and $$y$$'s describe $$f_p$$) and the coefficients give the size of each class. There are $$14$$ distinct monomials here and only two of them have coefficient $$2$$, as expected.
Here is my SAGE implementation of this formula:
# generator for variables
class VariableGenerator(object):
def __init__(self, prefix):
self.__prefix = prefix
@cached_method
def __getitem__(self, key):
return SR.var("%s%s"%(self.__prefix,key))
def NumEIClasses2(n,l):
x = VariableGenerator('x')
y = VariableGenerator('y')
R = PolynomialRing(QQ,[x[i] for i in range(n*n)] + [y[i] for i in range(n)])
A = matrix([[x[n*i + j]*y[j] for j in range(n)] for i in range(n)])
u = vector([1 for i in range(n)])
Y = vector([y[i] for i in range(n)])
P = R( (Y.row() * A^l * u.column())[0,0] )
#print P # print the resulting polynomial
return P.hamming_weight()
For example, for $$n=4$$ it gives counts 16, 64, 244, 856, 2728, 7892, 20876, 51020, 116408, ...
• that's certainly very helpful! Also I can confirm the solutions to $n=4$.
– Jiro
Apr 10, 2019 at 22:16
• Although, ideally I would prefer a bit more closed Form solution. I had the impression that the generating functions for this problem are rather compact, although I don't know exactly how to get there.
– Jiro
Apr 10, 2019 at 22:24
• @SebastianSchlecht: I doubt there is a simple general formula. But for a fixed $n$, it seems to be a polynomial in $l$. For $n=2$ and $n=3$, we have second- and sixth- degree polynomials. Apr 10, 2019 at 22:31
I believe, I have found an asymptotic answer here.
Define our problem in terms of k-abelian equivalence classes. Two words $$u$$ and $$v$$ are said to be $$k$$-abelian equivalent if, for each word $$x$$ of length at most $$k$$, the number of occurrences of $$x$$ as a factor of $$u$$ is the same as for $$v$$. It is easy to see the correspondence between words and walks (used in the question), and the subwords of length 1 ($$f_p$$) and length ($$e_p$$). Thus, our problem is to find the number of 2-abelian equivalence classes with length $$l$$ and alphabet size $$n$$.
In the mentioned paper (Theorem 5.1), it is shown that the number of $$k$$-abelian equivalence classes is asymtotically equal to $$C l^{n^{k}(n-1)}$$, where $$C$$ is a rational number dependent on $$k$$ and $$n$$. | 2,086 | 5,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 66, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-14 | latest | en | 0.864958 |
https://philoid.com/question/17647-the-radius-of-the-internal-and-external-surfaces-of-a-hollow-spherical-shell-are-3-cm-and-5-cm-respectively-if-it-is-melted-and- | 1,725,926,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00169.warc.gz | 427,623,958 | 8,098 | ##### The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height cm. Find the diameter of the cylinder.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of height cm.
Volume of material = Volume of the solid cylinder
r2 = 49
r = 7 cm
Diameter = 14 cm
20 | 139 | 537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.878371 |
mrswilliamsmath.wordpress.com | 1,498,670,228,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323721.80/warc/CC-MAIN-20170628171342-20170628191342-00564.warc.gz | 796,948,436 | 37,194 | ## Factoring Foldable
My 8th graders really got factoring this year! Because of all of the time lost due to snow days, I was attempting to compact the curriculum, and decided to teach factoring and solving quadratics at the same time (instead of factoring first, and then solving afterwards). We also used Desmos to graph the equations as we went. As a result, students appreciated the point of factoring as a tool to find solutions and graph quadratics.
If you’ve read any of my previous blog posts, you will know I am a huge fan of foldables. I really believe these teach students how to organize and summarize notes, and most importantly, then refer BACK to their notes as a resource. So many 8th graders will dutifully take notes and then never look at them again, but I find they will refer to their foldables because they are organized and contain the most important topics.
For factoring we created a “waterfall” foldable using three sheets of paper. Stack the three sheets of paper and stagger them so there is about a half-inch band of color before the next paper starts. Fold the top three sheets down and crease (the middle section will have the same color twice). Staple across the top (make sure you get all three pieces of paper in the stapling). We wrote Factoring on the front, then labeled the tabs at the bottom, “GCF, 2 terms, 3 terms x^2+bx+c, 3 terms ax^2+bx+c and 4 terms.
We reviewed GCF’s last week and put these notes in the foldable on the GCF tab. I chose to skip over the 2 terms tab at this point and came back to that after factoring trinomials. We did finish the foldable (pictures at the bottom), but what made this lesson so effective this year was integrating the solving and graphing as I taught factoring.
We started with the x-puzzles that @jreulbach talks about in this post. Students love these – it usually doesn’t take long for them to get the hang of them, and soon someone says, “well these are fun, but what does it have to do with Algebra?” Yaaas.
I told students we were going to use their awesome puzzle solving skills to learn how to “undo” multiplying binomials. They multiplied (x+3)(x+4), and then we talked about how we could possibly go backwards from x^2 +7x + 12 to find (x+3)(x+4). They brainstormed a bit and then I drew the x for an x-factor and put the 7 and 12 into it. Immediately there were shouts of “positive 3 and positive 4!!” We did a few more examples, working some with negative numbers. Then we talked about what y=(x+3)(x+4) and y=x^2+7x+12 would look like if graphed. The majority of students believed they would be the same, so we graphed them in Desmos and talked about how graphing this way is a good way to check their factoring.
Next, I projected two graphs and we did a little, “what do you notice, what do you wonder?” This was followed by a great discussion about factors versus solutions, and the relationship between the two. From there we talked about the zero product property, and in one block period students had factored and solved a quadratic, and seen the relationship between factors and intercepts.
For homework they practiced their factoring skills, and the next day we expanded factoring to include quadratics with a leading coefficient greater than 1. Anyone who has taught Algebra knows this is fairly challenging for some students. Our district strongly encourages factoring by grouping as the primary method, so students took notes on how to split the middle term, and then factor out the GCF of both groups. Again, we graphed the expanded quadratic and the factored quadratic in Desmos to ensure we had factored correctly. Some students in each class noticed that the intercepts were not necessarily integers, which was a lovely tie-in to taking each of the factors, setting them equal to zero and solving for x. Here are pics of the two pages on trinomials we added to the foldable:
By the next day, I had a few students coming in just grinning that their older siblings had shown them “a much easier” way to factor. This is always so cute – I just love that A) there is a discussion about MATH going on at home (win! win!), and B) that students are excited to think they’ve outsmarted me (again, win! win!). At this point we talk about the Divide and Slide method. I personally do not care which method they use, but am just happy they have found a method they can rally behind and not get to high school pretending they’ve never even heard of factoring.
Here are the final two pages for the foldable. We went back and revisited factoring 4 terms (they already knew how to do this because of factoring by grouping), but we also talked about how a cubic function can have 3 intercepts, and how they need to look for an x^2 term to potentially factor as a difference of two perfect squares (I include an example of this on the top of that tab). Finally, we finished with the 2 term tab:
The proof will come when we come back from Spring Break on Monday when I see if they remember how to factor after a week of not doing ANY math! | 1,221 | 5,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-26 | longest | en | 0.963926 |
https://stats.stackexchange.com/questions/504049/multiple-testing-for-bayesian-revenue-models | 1,721,276,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00533.warc.gz | 474,786,097 | 40,371 | # Multiple testing for Bayesian revenue models
Following this paper from VWO I have implemented the following model for revenue in a subscription business:
The revenue generated by user $$i$$ is given by:
$$\alpha_i \leftarrow Bernoulli\left(\lambda\right)$$ $$r_i \leftarrow Expon\left(\theta\right) %$$ $$v_i \leftarrow \alpha_i r_i$$
where $$\alpha_i \in \{0, 1\}$$ is an r.v. representing whether user $$i$$ bought anythign and $$r_i \in \mathrm{R}^+$$ is the size of the purchase. Finally $$v_i$$ is the size of the purchase.
In a given A/B test there $$n_A$$ users in variation A, and $$c_A$$ sales.
$$s_A = \frac{1}{c_a}\sum_{k=1}^{c_A} s_A^K$$
Is the revenue per sale in variation A
We assume a $$\mathrm{Beta}\left(a, b\right)$$ prior on $$\lambda$$ and a $$\mathrm{Gamma}\left(k, \Theta\right)$$ on $$\theta$$. Then the total posterior on $$\left(\lambda, \theta\right)$$ is
$$P\left(\lambda, \theta \mid n, c, s\right) = \mathrm{Beta}\left(a+c, b+n-c\right) \mathrm{Gamma}\left(k+c, \frac{\Theta}{1+\Theta c s}\right)$$
From this posterior you can calculate things like:
$$P\left( \frac{\lambda_A}{\theta_A} > \frac{\lambda_B}{\theta_B}\right)$$
by direct simulation.
However, I am interested in comparing multiple hypotheses. In this case, whether
$$P\left( \lambda_A > \lambda_B \right)$$
and
$$P\left( \theta_A > \theta_B\right)$$
and
$$P\left( \frac{\lambda_A}{\theta_A} > \frac{\lambda_B}{\theta_B}\right)$$
Simultaneously.
Andrew Gelman in this paper suggest using a heirarchical regression model to soove the multiple comparisons problem. So would a model where I put a hyperprior over $$(\lambda_A, \theta_A)$$ and $$(\lambda_B, \theta_B)$$ be sufficient?
Something like:
The revenue generated by user $$i$$ in group $$j$$ is given by:
$$\alpha_{ij} \leftarrow Bernoulli\left(\lambda_j\right)$$ $$r_{ij} \leftarrow Expon\left(\theta_j\right)$$ $$v_{ij} \leftarrow \alpha_{ij} r_{ij}$$ $$\lambda_j \sim \mathcal{D}\left(\lambda, \sigma_{\lambda}\right)$$ $$\theta_j \sim \mathcal{D}\left(\theta, \sigma_{\theta}\right)$$
where $$\mathcal{D}$$ is some suitable distribution I haven't figured out yet.
Do you really mean an A/B test (so A and B are the only variations)? I'll assume you do, and answer accordingly.
The hierarchical regression modelling approach advocated by Gelman is intended for situations where you have a large number of variations, $$A, B, ..., Z$$, and wish to compare them pairwise: $$P(\lambda_A > \lambda_B), P(\lambda_A > \lambda_C), ..., P(\lambda_Y > \lambda_Z)$$. It works by modelling all of the $$\lambda$$ parameters as samples from a population, and shrinking noisy, too-high or too-low estimates towards the population mean.
It doesn't really make sense here, where you have only two variations, and three tests. I don't think it's sensible to assume that estimate of $$\lambda_A - \lambda_B$$ has any bearing on the estimate of $$\theta_A - \theta_B$$, and even if it did, with only three tests you don't have enough information to start estimating the distribution of the parameters.
I think there are three more sensible things you can do here.
## Stop worrying
You're running three tests, on different parameters, so multiple comparisons aren't such a big concern here (unlike the situation described above, with a large number of variations).
If you have good reason to want to be conservative in your decisions, you can always estimate, e.g., $$P(\lambda_A > \lambda_B)$$ normally, and then adjust your estimate to be more conservative. A natural way to do this is by multiplying the odds: $$\frac{P(\lambda_A > \lambda_B)}{P(\lambda_A < \lambda_B)}_{\text{Adjusted}} = \frac{P(\lambda_A > \lambda_B)}{P(\lambda_A < \lambda_B)} \times \frac{1}{3}$$ if $$P(\lambda_A > \lambda_B)$$, and $$\times \frac{3}{1}$$ otherwise.
(Please check I have the signs right here!)
## Use conservative priors
Uninformative, high-variance priors indicate that $$A$$ and $$B$$ can vary widely, so are likely to be dissimilar, increasing the chance of finding a large posterior difference. Informative, low-variance priors strongly constrain the values of $$\lambda$$, reducing the difference between them. The variance of the Beta prior depends on $$a+b$$. I can't remember offhand what dictates the variance of the Gamma.
• Thanks! I concetrated on the simple 2 variation case here, but in reality there will 3+ variations, and we would like to run comparisons of conversion rate and mean spend against for all variations. Commented Jan 11, 2021 at 15:39
• Ah. In that case, yes, it does make sense to model $\lambda$ and $\theta$ as samples from a population (and it sound like you know how). Modelling $[log(\lambda), log(\theta)]$ as samples from a bivariate Normal distribution would probably be a good place to start.
– Eoin
Commented Jan 11, 2021 at 18:09
• Out of interest, is it still called "A/B testing" when you have more than two conditions?
– Eoin
Commented Jan 11, 2021 at 18:09
• Yes - in the industry an A/B test can have more than 2 variations. Commented Mar 10, 2021 at 17:19 | 1,406 | 5,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.644883 |
https://www.instructables.com/id/Make-a-Whistle-With-Tinkercad/ | 1,571,538,339,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00121.warc.gz | 945,282,997 | 21,592 | # Make a Whistle With Tinkercad
2,370
4
3
Create your own whistle with this tutorial in less than 1 hour (of which approximately 20 minutes is in 3D printing) and for less than 1\$.
This whistle can become a perfect and inexpensive toy for children, who can play with them imagining their favorite professions, referee a football match in schoolyard ... But it can also be useful to both adults and children hiking in the mountains, in the case where we get lost.
At the end of these 32 simple steps and adding a small chickpea (or any object similar to a ball) at the time of 3D printing, we will have our custom whistle :).
You can view and take the original design from Neotekna Tinkercad Account.
You will need:
- A computer and internet connection.
- A 3D printer.
- A small ball.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Create a Cylinder
Create a cylinder of 20 mm diameter x 7.5 mm height.
## Step 2: Create a Hollow Cylinder
Create a hollow cylinder of 17 mm diameter x 8.5 mm height. Height does not matter, but the diameter will give you a 3 mm wall. It is important to lift 1.5 mm from the workplane.
## Step 3: Align and Group It
Align both cylinders on the X axis and on the Y axis. Not on the Z axis. Group it.
## Step 4: Create a Cube
Make a "cube" (actually it is a prism) of 24 x 7 x 7.5 mm.
## Step 5: Align Cube and Cylinder
Align the cube with the cylinder, on the Y axis.
## Step 6: Put the Prism Tangent to the Cylinder
Move the prism with the arrows on the keyboard and put tangent to the cylinder.
## Step 8: Group All
Group cylinder, hollow cylinder and prism.
## Step 9: Create a Hollow "cube"
Create a hollow cube of dimensions 24 x 7.5 x 2 mm. Raise it 3.75mm from the workplane.
## Step 10: Place the Hollow Prism
Align hollow prism tangent with whistle. It may be necessary to adjust grid to 0.1mm.
## Step 11: Duplicate Hollow Prism
Duplicate hollow prism and reserve it for next steps.
## Step 12: Group Whistle Body
Group whistle to make the inner hole.
## Step 13: Rotate Hollow Prism
Rotate -22.5º the hollow prism you have doubled before.
## Step 14: Place the Rotated Hollow Prism
Put that hollow prism tangent to the hollow interior of the whistle. It may be necessary to adjust the grid to 0.1mm.
## Step 15: Create Anothes Hollow Prism
Create a hollow prism of 24 x 3 x 7.5 mm, raised 3.75mm from the workplane.
## Step 16: Place the New Prism
Match this thicker prism in the following way, in that blue point.
## Step 17: Group All
Group all and you have one half of the whistle.
## Step 18: Duplicate It
Duplicate it to get the other half of the whistle.
## Step 22: Raise the Upper Half
Mount one half on the other by raising it 7.5 mm.
## Step 24: Create a Toroid
Create a 12 x 12 x 3 mm toroid raised 1.5 mm from the workplane.
## Step 25: Align Toroid and Whistle
Align the toroid with the whistle on the Y axis and on the Z axis. Not on the X axis.
## Step 26: Duplicate the Whistle
Duplicate the whistle and make it hollow.
## Step 27: Cut the Toroid
Put approximately half of the toroid in the hollow whistle and group it.
## Step 28: Remove One Half of the Toroid
Remove half of the toroid you do not want with a hollow cube.
## Step 29: Place Toroid in Whistle
Place that half of the toroid to the body of the whistle. Check that nothing protrudes from the toroid inside.
## Step 31: Customize It
To finish you can put a name or whatever you want to the whistle.
## Step 32: Add a Small Ball Inside
To facilitate the printing of the whistle it is better to add a small ball inside when it is being printed instead of printing it (it would be quite complicated). If you do not have a small ball or you can not think you can get in, before buying anything try a small chickpea, it works well.
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• ### Mold Making & Casting Class
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## 3 Discussions
Math is not correct will give a 1.5 mm wall which is hard to center
Hi Audrey! Thanks!
We created this design last month in our academy together with the children and we have printed it several times for them. Kids customized their own whistles and love them. It is a good project to do in class: it is simple, fast and needs little material.
Go ahead and do it! If you follow the tutorial I guarantee that your whistle will sound too (and very loud! :)) | 1,151 | 4,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.886251 |
https://www.tutorvista.com/content/math/exponents-with-variables/ | 1,569,215,458,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576047.85/warc/CC-MAIN-20190923043830-20190923065830-00435.warc.gz | 1,064,056,007 | 9,563 | Top
# Exponents with Variables
Exponent of a variable is the power raised to it. Exponent implies the repeated multiplication to itself. The number of times a variable is multiplied, is raised as its exponent or power. We can define it by the following relation:
Where, x is a variable and n is its exponent.
$x^{n}$ is read as "x raised to the exponent of n", "x raised to the power of n" or more casually "x to the power n". Exponent of 2 and 3 over a variable are pronounced differently, such as $x^{2}$ and $x^{3}$ are read as "x squared" and "x cubed" respectively.
There are few important laws that are to be followed while dealing with exponents. These laws are as follows (Let us consider two variables x and y and exponents m and n):
1. $x^{0}=1$
2. $x^{-m}$=$\frac{1}{x^{m}}$
3. $x^{m}.x^{n}=x^{m+n}$
4. $\frac{x^{m}}{x^{n}}$=$x^{m-n}$
5. $(x^{m})^{n}=x^{mn}$
6. $x^{m}y^{m}=(xy)^{m}$
7. $\frac{x^{m}}{y^{m}}$=($\frac{x}{y}$)$^{m}$
8. $x^{\frac{m}{n}}$=$\sqrt[n]{x^{m}}$
An example based on exponents of variables and their laws is given below:
Example: Solve $(-2x^{-1}y^{3})^{2}.(-5x^{2}y^{-2}z)$.
Solution: $(-2x^{-1}y^{3})^{2}.(-5x^{2}y^{-2}z)$
= $(-2)^{2}.(x^{-1})^{2}.(y^{3})^{2}.(-5).x^{2}.y^{-2}.z$
= $4.x^{-2}.y^{6}.(-5).x^{2}.y^{-2}.z$
= $-20.\frac{1}{x^{2}}.y^{6}.x^{2}.\frac{1}{y^{2}}.z$
= $-20.x^{2-2}.y^{6-2}.z$
= $-20x^{0}y^{4}z$
= $-20y^{4}z$
Related Calculators Simplifying Fractions with Variables and Exponents Calculator Calculating Exponents Calculator Variable Fractional Exponents Calculator
*AP and SAT are registered trademarks of the College Board. | 579 | 1,608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-39 | longest | en | 0.839405 |
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Questions on the evaluation of derivatives or problems involving derivatives (for example, use of the mean value theorem).
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### How to derive Marshallian demand function, indirect utility function and the expenditure function for cobb douglas utility function
I am having a problem in deriving marshallian demand function, indirect utility function and expenditure function from following cobb-douglas utility function, U(X,Y)=A.X^alpha.Y^beta
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### Interesting Integral including $\ln x$
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Let $f:[0,1]\to \mathbb{R}$ be a continuous function. If $\int_{0}^{1}f(x)g(x)=0$ for all continuous functions $g(x)$, then $f(x)=0 \ \forall x \in [0,1]$. I would like to know if my proof holds, ...
### Find right and left sided derivative of $|2^x - 2|$ at $x = 1$
Find right and left sided derivative of $|2^x - 2|$ at $x = 1.$ Right sided: $\lim_{x\to1^{+}}\frac{2^x-2-0}{x-1}$ Left sided: $\lim_{x\to1^{-}}\frac{2-2^x-0}{x-1}$ I don't know how to continue to ... | 823 | 2,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-17 | longest | en | 0.91093 |
http://slideplayer.com/slide/3525942/ | 1,524,667,785,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947822.98/warc/CC-MAIN-20180425135246-20180425155246-00630.warc.gz | 283,531,087 | 23,163 | Longitudinal and Multilevel Methods for Models with Discrete Outcomes with Parametric and Non-Parametric Corrections for Unobserved Heterogeneity David.
Presentation on theme: "Longitudinal and Multilevel Methods for Models with Discrete Outcomes with Parametric and Non-Parametric Corrections for Unobserved Heterogeneity David."— Presentation transcript:
Longitudinal and Multilevel Methods for Models with Discrete Outcomes with Parametric and Non-Parametric Corrections for Unobserved Heterogeneity David K. Guilkey
Focus of this talk: Binary dependent variables Unordered categorical dependent variables Models will be logit based – will not discuss probit, poisson or negative binomial models although STATA has methods for these estimators as well Empirical example uses data from the Indonesian Family Life Survey: Two outcomes: Binary indicator for whether the respondent uses contraception Unordered categorical variable for method choice
Data Set Overview Four waves of data: 1993, 1997, 2000, and 2007 Individual level information on fertility, education, migration Community and facility level data on health and family planning providers Data from 321 enumeration areas – we will consider these communities IFLS Longitudinal Sample Size Initial Participation Cohort Survey Year 1993199720002007total Wave 1 Cohort352028732684149810575 Wave 2 Cohort2207174211525101 Wave 3 Cohort14669332399 Wave 4 Cohort 2287 total observations 20362
Basic Model for Longitudinal Logit: Where: Y ti : observed binary variable (respondent i from time period t) X ti : time varying explanatory variables (age and education level) P ti : time varying program variable (posyandus) Z i : time invariant regressors (Muslim) i=1,2,…N (individuals) t=1,2,…T i (observations per individual -- unbalanced panel)
Assumptions: for the parametric logit in STATA (xtlogit, melogit, and one variant of GLLAMM) and: Note that observations for the same individual will be correlated because of the time invariant error – sometimes referred to as unobserved heterogeneity Given the assumptions, estimation options are: 1. Simple logit yields consistent point estimates but incorrect SE’s 2. Simple logit with cluster option corrects SE’s 3. Parametric or semi-parametric maximum likelihood
The likelihood function for this model is derived as follows: This is the probability that individual i at time t is using contraception conditional on time invariant heterogeneity. For individual i, we observe T i binary responses that we can write as: Y i = (1,0,0,1) for a woman that is observed for 4 time periods and used contraception at times 1 and 4.
Let Y i be the set of observed outcomes for individual i, then: Joint probability must be approximated -- approximating the area under a curve. With the assumption of normality the approximation method is Gaussian Quadrature or Hermite integration Points: 1. More accurate with more Hermite points – but execution time is longer. 2. You need more points as T i gets larger.
Hermite integration replaces the integral with a sum: where the weights (w m ’s) and the masspoints (μ m ’s) are known because of the assumption of normality Alternative: The discrete factor approximation searches over weights and mass points along with the other parameters of the model. Must impose a normalization; 1. Weights sum to one 2. Either set one mass point to zero (fortran program) or set mean of distribution to zero (GLLAMM)
Simple Logit Simple Logit with Corrected Se’s
Parametric Maximum Likelihood
Semi-Parametric Maximum Likelihood
Multilevel Panel Models Basic Form of the model: where j=1,2,…,J (communities) i=1,2,…,N j (individuals from community j) t=1,2,…,T ij (observations for person i for community j)
X tij : individual level variables (some could be fixed through time) P tij : time varying program variable Z j : time invariant community level variables μ ij : time invariant individual level unobserved heterogeneity λ j : time invariant community level unobserved heterogeneity This model allows observations on the same individual to be correlated and observations from the same community to be correlated.
Assumptions: 1. Simple logit yields consistent point estimates but incorrect SE’s 2. Simple logit with cluster option corrects SE’s (at community level) 3. Parametric or semi-parametric maximum likelihood Maximum likelihood estimator is a straight forward extension of the longitudinal data model:
You need the unconditional joint probability of the observed set of outcomes for the set of individuals in each community: Conditional on the unobservables at the community level, the probability of the set of observed outcomes for person i from community j are: The unconditional joint probability of the set of observed outcomes for all individuals in community j is then: We then either use Hermite integration or the discrete factor method to approximate the integral.
Simple logit Simple Logit with Corrected SE’s
Parametric Maximum Likelihood
Non-parametric Maximum Likelihood
Testing for Program Targeting Programs may target high need areas or areas where they feel residents would be receptive to family planning For example: family planning programs may concentrate on high fertility areas Result is that simple methods may understate or overstate program impact Statistical Implication of program targeting:
Solutions: Explicitly model program placement and estimate placement simultaneously with program impact equations (Angeles, Guilkey, and Mroz, 1998) Treat as fixed effects and include dummies for communities or some other fixed effects method (Gertler and Molyneau, 1994) Angeles, Guilkey, and Mroz show that the joint modeling approach yields smaller standard errors in Tanzania but the two methods gave similar results
Example (fixed effects) plus Hausman Test for endogenous placement: Efficient estimator under the null of no endogeneity (random effects):
Consistent estimator under the alternate (fixed effects):
Hausman test results:
State Dependence and Unobserved Heterogeneity Consider the simple model: Note: Implies: Unless (no time invariant unobserved heterogeneity) Now consider: Now: Very difficult to distinguish between the two models
Same problem would exist if the unobserved heterogeneity were at the community level Solution is to estimate a comprehensive model: Initial conditions problem: Must either be able to set or jointly estimate the equation of interest with an equation of the form:
Often it is reasonable to set the initial value: Observations start at the beginning of the woman’s child bearing years In this example, it is not since women enter the year one data set at different ages Joint estimation is basically a simultaneous equations problem subject to standard identification issues. However, time varying exogenous variables provide identification (age and education in this case) Example follows:
Estimation with no controls for unobserved heterogeneity and initial conditions:
Estimation with Controls:
Estimation with Controls (continued)
Basic Model Longitudinal Multinomial Logit with 3 Choices: Individual i at time t time makes choice 3 (for example) if : If we assume that the ε’s follow independent extreme value distributions and impose the restriction that:
So that the probabilities sum to one then: for k=2,3. The discrete factor model allows a more general pattern of correlation: for m=1,2…,M and a common set of weights: allows for correlation in the μ’s
Unfortunately, GLLAMM estimates a needlessly restrictive version of the model: Parametric: If there are more than 3 choices, all ρ’s are restricted Non-parametric: for all m.
Extension to Multilevel Panel Model: Parametric: Semi-parametric:
The empirical example estimates a model with four choices: 1= Non use 2=Temporary Methods (pill, condom, injection) 3=Long Lasting Methods (IUD, sterilization) 4=Traditional Methods We show the complete results for the most general model and then report partial results for other models:
Comparison of Posyandu effects across estimation methods:
Download ppt "Longitudinal and Multilevel Methods for Models with Discrete Outcomes with Parametric and Non-Parametric Corrections for Unobserved Heterogeneity David."
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## How do you get a vector or function to be recalculated?
I have a function that produce a series of Pi using Monte Carlo.
Is there a way I can use it in a loop so that it recalculates all its values of Pi each iteration?
Thanks
Tags (2)
1 ACCEPTED SOLUTION
Accepted Solutions
## Re: How do you get a vector or function to be recalculated?
The function argument must not becessarily be a dummy argument. It may determine how many iterations the Montecarlo simulation should go through.
maybe something like the following.
5 REPLIES 5
## Re: How do you get a vector or function to be recalculated?
If its really a function (of a dummy argument) and you call it within a loop with, lets say, the loop counter as argument, the function will create new random number every time and give you a different result.
For more help you'll have to show what you've done so far by attaching your worksheet.
## Re: How do you get a vector or function to be recalculated?
The function argument must not becessarily be a dummy argument. It may determine how many iterations the Montecarlo simulation should go through.
maybe something like the following.
## Re: How do you get a vector or function to be recalculated?
I needed to generate three data sets of 100 determinations of Pi but using increasing number of darts thrown each time (I used 100, 1000, 10000). The problem I was having is that the CalcPi(N) function I used wouldn't update its values when I tried to use it in my loop within my function PiDS(N,D) (Data Sets of Pi where N is number of darts thrown and D is number of determinations of Pi to make and parse into array). The way I fixed it is by using a temp array and set it to CalcPi(1) before setting it to CalcPi(N) each iteration, that way it made the function generate a new array each time. Not sure if there is a proper way of doing it. Hope I made sense...
Thanks for the help.
## Re: How do you get a vector or function to be recalculated?
Not sure if there is a proper way of doing it.
Hard to say without seeing what exactly you have done.
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https://math.stackexchange.com/questions/2240338/if-the-magnitude-of-the-resultant-of-two-equal-vectors-is-equal-to-that-of-eithe | 1,701,806,323,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00078.warc.gz | 452,985,916 | 37,053 | # If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.
If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.
My Attempt: Let $\vec {a}$ and $\vec {b}$ be two vectors such that $\|\vec {a}\|=\|\vec {b}\|$
Magnitude of Resultant: $$=\sqrt {a^2+b^2+2ab\cos \theta}$$ $$=\sqrt {2a^2+2ab\cos \theta}$$
How do I proceed further?
• Hint: which rhombus has a diagonal equal to the sides?
– dxiv
Apr 18, 2017 at 16:57
• It is better to say the original two vectors $a,b$ have equal magnitudes than to say these vectors are equal (which we reserve for $a=b$). Apr 18, 2017 at 17:23
From $\|a+b\|^2=\|a\|^2$ and $\|a\|=\|b\|$ we have $$\|a\|^2=\|a+b\|^2=\|a\|^2+\|b\|^2+2\langle a,b\rangle=2\|a\|^2+2\langle a,b\rangle,$$ so $\langle a,b\rangle=-\|a\|^2/2$. Hence the cosine of the angle between $a$ and $b$ using $\|a\|=\|b\|$ again is $$\frac{\langle a,b\rangle}{\|a\|\|b\|}=\frac{-1}{2}.$$
• How did you get $(a,b)=-|b|^2/2$?
– pi-π
Apr 19, 2017 at 1:31
• See my edit, please. Apr 19, 2017 at 6:54
The resultant vector $\mathbf{OC}$ is the diagonal of the rhombus whose adjacent sides are $\mathbf{OA}, \mathbf{OB}$. Since its length is equal to either, we have an equilateral triangle $\mathbf{OBC}$. Thus the angle between the vectors is $120^\circ$
Let,$R$ be the resultant of the sum of the two vectors $\vec A$ and $\vec B$ such that $|\vec A|=|\vec B|$.So,$$R=\sqrt{|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta}$$
where $\theta$ is the angle between $\vec A$ and $\vec B$.So,
$$R=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$.
Now magnitude of resultant $R$ is equal to either of the vectors $\vec A$ and $\vec B$.So,
$$|A|=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$ $$\implies|\vec A|^2=2|\vec A|^2+2|\vec A|^2\cos\theta$$ $$\implies\frac{-|\vec A|^2}{2|\vec A|^2}=\cos \theta$$ $$\implies\theta=\cos^{-1}-\frac{1}{2}$$.
Hope this helps!! | 733 | 1,984 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-50 | latest | en | 0.722003 |
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### Making Maths: Rolypoly
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Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour.
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Here are some ideas to try in the classroom for using counters to investigate number patterns. | 2,051 | 8,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-16 | longest | en | 0.871513 |
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Classical Light Waves
Consider a classical, monochromatic, linearly polarized, plane light wave, propagating through a vacuum in the -direction. It is convenient to characterize a light wave (which is, of course, a type of electromagnetic wave) by specifying its associated electric field. Suppose that the wave is polarized such that this electric field oscillates in the -direction. (According to standard electromagnetic theory, the magnetic field oscillates in the -direction, in phase with the electric field, with an amplitude which is that of the electric field divided by the velocity of light in vacuum.) Now, the electric field can be conveniently represented in terms of a complex wavefunction:
(46)
Here, , and are real parameters, and is a complex wave amplitude. By convention, the physical electric field is the real part of the above expression. Suppose that
(47)
where is real. It follows that the physical electric field takes the form
(48)
where is the amplitude of the electric oscillation, the wavenumber, the angular frequency, and the phase angle. In addition, is the wavelength, and the frequency (in hertz).
According to standard electromagnetic theory, the frequency and wavelength of light waves are related according to the well-known expression
(49)
or, equivalently,
(50)
where . Equations (48) and (50) yield
(51)
Note that depends on and only via the combination . It follows that the wave maxima and minima satisfy
(52)
Thus, the wave maxima and minima propagate in the -direction at the fixed velocity
(53)
An expression, such as (50), which determines the wave angular frequency as a function of the wavenumber, is generally termed a dispersion relation. As we have already seen, and as is apparent from Eq. (51), the maxima and minima of a plane wave propagate at the characteristic velocity
(54)
which is known as the phase velocity. Hence, the dispersion relation (50) is effectively saying that the phase velocity of a plane light wave propagating through a vacuum always takes the fixed value , irrespective of its wavelength or frequency.
Now, from standard electromagnetic theory, the energy density (i.e., the energy per unit volume) of a light wave is
(55)
where is the permittivity of free space. Hence, it follows from Eqs. (46) and (48) that
(56)
Furthermore, a light wave possesses linear momentum, as well as energy. This momentum is directed along the wave's direction of propagation, and is of density
(57)
Next: Photoelectric Effect Up: Wave-Particle Duality Previous: Representation of Waves via
Richard Fitzpatrick 2010-07-20 | 588 | 2,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-09 | latest | en | 0.925272 |
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1. ## Math
I have 13 tens who am I?
2. ## math
mulitply -7X(-62)= ???
3. ## math
how to factor 4x^2+4x-1 and x^2-2x+1=5
4. ## math
2y = 4/3 simplify
5. ## math
1367.00+3.6%=
6. ## math
{-(2m^-2m^o)/(4m2n^-1)}^(-3)
7. ## math
let f(x)=3x-2 and g(x)=x squared+6 use them (f-g)(x)
8. ## Math
5 divided by 1/3
9. ## math
let f(x)=3x-2 and g(x)=x squared+6 use them (f-g)(x)
10. ## Math
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11. ## math
5x+negative13x+6x+4x
12. ## Math
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13. ## math
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14. ## math
evaluate |-8| + |2|
7rs&s
16. ## math
Solve (x - 3)(x – 4)/(x – 5)(x – 6)^2 >or= 0 any help?
17. ## Math
13% of what number is 16.25?
18. ## math
let f(x)=3x-2 and g(x)=x squared+6 use them (f-g)(x)
19. ## math
If x=a^2/3-a^-2/3 then find x^3 3x=?
4xsquared=32
21. ## math
All of the following are equilavent to 3:8 = x:32 except 8:3=32:x 3:8=8:32 x:8=3:32 8:32=3:x
22. ## math
what is the function if x is 3,0,7,5 and the y being 16,10,24,20?
23. ## Math
- 1/4 Is Greater Than -8
factor 6-4t
26. ## Math
for all integers a, b, c, If a/b - 2c and a/2b + 3c then a/b and a/c Thanks for your help.
27. ## math
Antiderivativeofe.5x-0.1x
28. ## Math
15 1/6 - 6 5/12 ok so I know 15 1/6 = 15 2/12 right? but you can't subtract 5 from 2?
29. ## Math
Solve. 10/(3x) + 4/3 = (7 + x)/2x A) x = 1/3 B) x = 17/5x C) x = 1/5 D) x = 1/6 help??
30. ## math
60 is 25% of what number
31. ## math
40% of what number is 82
32. ## Math
Solve for x and y X=1/6y Y=x-5
33. ## math
Which one is associative? 6(3) + 6(7) = 6(10) or 1/2(8*9)=4*9
34. ## Math
-6 ? |-8| what symbol > < =
35. ## Math
1/2 divided by 1/2=1?
36. ## Math
Let f(a,b) = 2a - 3b^2 + 7. If f(b,3) = 90, then what is b? Thank you for helping!
37. ## Math
what is a example of interest?
38. ## math
73 is 20% of what number
39. ## math
Simplify: 2x^3y^2+ 4x^2z / 2x^4
40. ## Math
If f(x)=5x-12, find a value for x so that f^-1(x)=f(x+1). Thank you so much
41. ## Math
-5 1/3 divided by 2/3
42. ## Math
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43. ## Math
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44. ## math
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45. ## Math
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46. ## math
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47. ## math
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48. ## Math
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49. ## math
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50. ## Math
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51. ## Math
Solve. 10/3x + 4/3 = 7 + x/2x A) x = 1/3 B) x = 17/5x C) x = 1/5 D) x = 1/6
52. ## Math
is 2/3 larger than 4/7
53. ## Math
Solve. 10/(3x) + 4/3 = (7 + x)/2x A) x = 1/3 B) x = 17/5x C) x = 1/5 D) x = 1/6 help??
54. ## Math
Help please, one third of negative 18
55. ## math
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56. ## Math
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57. ## math
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58. ## math
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59. ## math
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60. ## math
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62. ## math
solve for x if 1/15(6+x)=1/6(x+3)
63. ## math
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64. ## math
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65. ## Math
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66. ## math
solve x x-4y=-14
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68. ## Math
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69. ## math
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70. ## math
3.5% of d is 0.105
71. ## Math
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72. ## math
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73. ## Math
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74. ## math
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75. ## math
57% of 109 is what #
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78. ## Math
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79. ## math
Solve for x:x + 5 < 1
80. ## math
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81. ## math
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82. ## math
solve. x(1 +2x) = (2x - 1)(x - 2)
85. ## math
Help! Solve: 1. 3x + 15 = 0 2. 6x + 18 =0 thanks!
86. ## math
Add. 2 1/2+ 3 1/4 + 3 5/8
87. ## math
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88. ## Math
What is greater -19 or -17?
89. ## MATH
-1 3/4 = p - 1/2 HOW DO I FIGURE THIS OUT
90. ## MATH
Factor 5A^2 - 5A
91. ## Math
estimate 4 1/9+ 7 9/10+ 2 1/7 + 5 7/8=
92. ## Math
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93. ## math
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94. ## math
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96. ## math
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97. ## math
how would you solve this 31=2*x+c+1 | 2,066 | 4,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-39 | latest | en | 0.59464 |
https://www.physicsforums.com/threads/spin-angular-momentum-bullet-hitting-bottom-of-a-thin-rod.746740/ | 1,508,465,220,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00334.warc.gz | 963,745,784 | 16,070 | # Spin Angular Momentum - Bullet hitting bottom of a thin rod?
1. Apr 3, 2014
### mintsnapple
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Angular momentum is the only parameter conserved. This is because there are no external torques acting on the system. Energy is not conserved because the collision is inelastic. Finally, linear momentum is not conserved because there is an external force acting on the rod's center of mass that prevents the system from moving forward after the collision.
Now, for part d, I am not so sure what to do. Should I equate the angular momentum of the system before the bullet hits with the final angular momentum of the system, solve for angular velocity, and turn that into linear velocity? Though I have no idea how g would come into play..
2. Apr 3, 2014
### BvU
Your plan to get going is good. After the embedding of the bullet, the thing is just a kind of pendulum (a variation on the ballistic pendulum).
If you fill in the relevant equations under 2, you automatically get to see the role of g. | 255 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | longest | en | 0.881587 |
http://www.sciencemadness.org/talk/viewthread.php?action=printable&tid=72593 | 1,656,550,131,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103646990.40/warc/CC-MAIN-20220630001553-20220630031553-00017.warc.gz | 100,036,339 | 11,231 | ## High Voltage Generation
Invictos - 1-3-2017 at 21:15
Good Evening everyone,
I need to generate about 1.5 million volts for an electrostatics experiment (I need positrons, if you're curious...)
I currently have a 200 KeV Van De Graaff generator at my disposal. I have two options, wondering if y'all would have any input:
1.) This one's the simpler of the two: I buy another 7 Van De Graaff generators, and put them in series, or I get bigger spheres and get say 4 extra, or some variant of that.
2.) I use the generator to charge a Leyden Jar. I disconnect the generator once it's fully charged. Then, in order to increase voltage, I increase distance between the plates, lowering capacity, increasing voltage. This will almost certainly be done mechanically, with a lot of rubber insulation. See the attached diagram for this one if it'd help!
Thoughts? I'm leaning option 2 because of how expensive option 1 will be, though 2 has its distinctive sketchiness...
Thanks for your time, other ideas are welcome!
macckone - 1-3-2017 at 23:41
This won't create positrons, but it will give you the voltage
you need.
Option 3:
Charge the leyden jars in parallel, then with great care
not to electrocute yourself change them to a series configuration.
Option 4:
Marx Generator and ditch the Van DeGraaff.
This is what is used for really high voltage experimentation.
The capacitors need to be 10KV variety as do the
resistors. With a 10K DC power supply you will need 150
stages but it is straight forward. A narrow beam UV diode
can be used as a trigger to get instant flow across all
spark gaps.
Sulaiman - 2-3-2017 at 01:39
I expect that you will have serious corona problems,
charge can ionise air once the field gradient is sufficient, for rounded objects use 3 kV per mm of radius.
So 1.5 MV = 1500/3 = 500mm
therefore, to avoid breakout at 1.5 MV, the minimum diameter of spheres or cylinders is 1 metre.
One common 'trick' is to use +0.5.V and -0.5.V = +/- 750 kV in this case.
I remember a tall brick building at Harwell, built to house a tandem VDG to generate 1 - 2 MeV particles by this method, it was over 100' tall.
You would probably find a microwave LINAC or synchrotron more practical.
(I helped to commission a 55 MeV microwave LINAC at Harwell in the '70's)
The classic h.v. generator is https://en.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_gener...
though I suspect that 1.5 MV will require a lot of input power just due to leakage/corona.
As above, a Marx generator for +/- 750 kV may be doable at hobby scale https://en.wikipedia.org/wiki/Marx_generator
P.S. due to the very high dV/dt generated by a Marx generator, EXTREMELY long arcs through the air can be formed
(you thought chemistry was a good mix of fun & danger ?)
IF you do generate 1.5 Mv (or +/- 750 kV) the machine will have to be quite large to withstand such voltages,
so the mean free paths will be measured in metres - quite a hard vacuum
(fingerprints on the inside of the machine can add many hours to pump-down time, that was using roughing-, turbo- and ion-pumps)
[Edited on 2-3-2017 by Sulaiman]
wg48 - 2-3-2017 at 03:25
Invictos: Your option 2 is a form of influence machine. The changing of capacitance can be done with rotating discs. One form of influence machine is a self starting Wimshurst machine that would not require your Van De Graaff generator. The one below has 7 feet discs allegedly the largest in the world. Yours would be huge as it might need 50ft discs unless you put it in a pressure tank of .sulfur hexafluoride but the discs would still have to be say 3ft thick.
Start the crowd funding now and call Guinness LOL
Check out Antonio Carlos M De Queiroz (a mad science type of guy) site http://www.coe.ufrj.br/~acmq/bigwin.html with links to lots of HV machines and info.
If you don't mind the high voltage being alternating you could do it with a big Tesla coil like the one below.
From https://www.quora.com/What-does-a-Tesla-coil-do
[Edited on 2-3-2017 by wg48]
Sulaiman - 2-3-2017 at 04:51
I remember Greg Leyh sharing the development of the Electrum on pupman.com
where there is enough information to reproduce an Electrum,
a fantastic project, not for the faint-hearted.
.........................................................................................
for option 2, the voltage will rise proportional to distance (ideally)
e.g. capacitor plates 5mm apart charged to 15 kV would only need to be increased to 500 mm separation for 1.5 MV
- why have I never tried this !
insulation will be 'interesting'
and 'stray/parasitic' capacitance reduction will be challenging.
[Edited on 2-3-2017 by Sulaiman]
wg48 - 2-3-2017 at 06:18
Quote: Originally posted by Sulaiman I remember Greg Leyh sharing the development of the Electrum on pupman.com where there is enough information to reproduce an Electrum, a fantastic project, not for the faint-hearted. [Edited on 2-3-2017 by Sulaiman]
Yes almost anything to do with 2,000,000V is not for the faint of heart lol. It would sort the real mad scientists from the tinkers lol.
Even for an amateur mad scientist
Morgan - 2-3-2017 at 10:46
I always liked this old gizmo.
"Note the deliberately indirect steam path. Quite what this had to do with charging the steam is at present obscure."
"While the Armstrong boilers were considered to be the most powerful electrostatic machines of their day, there were problems. In operation they inevitably filled the room with steam, producing a humid environment hardly suitable for electrostatic experiments which depend on dryness to maintain insulation. One also wonders how the machine could be fuelled while in operation without giving the stoker a severe shock; possibly the boiler would be grounded while this was done."
http://www.aqpl43.dsl.pipex.com/MUSEUM/POWER/staticboil/stat...
http://physics.kenyon.edu/EarlyApparatus/Static_Electricity/...
[Edited on 2-3-2017 by Morgan]
Invictos - 2-3-2017 at 23:27
You're all amazing, wow is all I can say!
The biggest thing I've pulled from this is that these things don't quite scale
The biggest, most out there end goal is to accelerate electrons in a vacuum tube by using this voltage differential, use the Bremsstrahlung radiation in order to convert the energy in the electrons to photons, and then use the photons in pair production. Pair production for positron/electron pairs occurs at 1.1 MeV, so I was shooting for the 1.5 MeV range to account for the energy lost in Bremsstrahlung. Given the scaling difficulty, I may need to aim closer to 1.1 MeV apparently! If I shoot enough electrons down the pipe, some of them should make it at just the right angle to get that energy, right?
Note, I crunched the numbers at 100 KeV for now, and found the power given off by Bremsstrahlung to be 3.31e-20 Watts, while the energy of the particle itself is 1.59e-14 Joules, so it does seem remotely possible. However, 1.1 MeV is really cutting it!
As for some of these ideas...
Quote: "I've always liked this old gizmo"
I can see why, that's certainly a unique way to go about it!
Quote: Option 3: Charge the leyden jars in parallel, then with great care not to electrocute yourself change them to a series configuration.
This is an idea...I'd bet if I used some industrial level transistors, it could be done, right? That would increase the voltage an insane amount too, because you're switching between inverse addition to regular...I don't even think you could get a transistor that big, I'd probably be easier off hooking up servo motors to metal rods to use as switches!
Quote: Your option 2 is a form of influence machine. The changing of capacitance can be done with rotating discs. One form of influence machine is a self starting Wimshurst machine that would not require your Van De Graaff generator. The one below has 7 feet discs allegedly the largest in the world. Yours would be huge as it might need 50ft discs unless you put it in a pressure tank of .sulfur hexafluoride but the discs would still have to be say 3ft thick.
Hey I did a thing! But wow those are some numbers The disk approach is so majestic I love it. Going off of that, one could, in theory, use some sort of conductive fluid for the dialectric, and drain and fill the fluid in order to change capacitance, right? It'd be clunky, but it's a thought!
Quote: Option 4: Marx Generator and ditch the Van De Graaff.
This has been suggested a couple of times, and I certainly see why! Looking into it, I see a ton of people have gotten quite far with this approach!
Quote: One common 'trick' is to use +0.5.V and -0.5.V = +/- 750 kV in this case.
That's clever...very, very clever... which leads me into my next silly question: Could I use some weird macgyver style hybrid of the above? Could a Marx generator be hooked up in series to the Van De Graaff somehow? If not, could one at least create the two set ups, one for positive and one for negative...?
I have some resources here, just trying to squeak out as much as I can!
Thanks again everyone!
Sulaiman - 3-3-2017 at 01:32
The problem with putting one voltage source on top of another is insulation
... how do you supply power to something that is at e.g. 100 kV ?
(Cockroft-Walton / Villard / Greinacher) voltage multipliers and Marx generators are fairly easy to configure for +ve and/or - ve voltages
A VDG can be configured to produce +ve or -ve voltage if a h.v. supply is used to spray charge onto the belt.
Tesla coils can also be bipolar, but not useful here I think.
P.S. the idea of charging leyden jars (capacitors) in parallel then stacking them up in series
. is incredibly hazardous.
. is a manually switched Mark Generator
Invictos - 4-3-2017 at 15:57
Quote: Originally posted by Sulaiman The problem with putting one voltage source on top of another is insulation ... how do you supply power to something that is at e.g. 100 kV ? (Cockroft-Walton / Villard / Greinacher) voltage multipliers and Marx generators are fairly easy to configure for +ve and/or - ve voltages A VDG can be configured to produce +ve or -ve voltage if a h.v. supply is used to spray charge onto the belt. Tesla coils can also be bipolar, but not useful here I think. P.S. the idea of charging leyden jars (capacitors) in parallel then stacking them up in series . is incredibly hazardous. . is a manually switched Mark Generator
I UNDERSTAND MARX GENERATORS NOW
I was looking over the circuit drawings but it never really clicked on why it worked...I see...
Thanks for your time, I've got plenty to look through and consider at this point!
unionised - 5-3-2017 at 02:17
Quote: Originally posted by Invictos I UNDERSTAND MARX GENERATORS NOW
That's probably going to be the greatest achievement of this thread.
Sulaiman - 5-3-2017 at 02:47
The more that I think of generating 1.5 MV the less practical I consider it;
constant potential difference (d.c.) causes air to ionise at about 3 kV/mm,
1.5 MV is therefore equivalent to a minimum gap in air of 500mm
or a curved conductor of 500 mm min. radius.
Due to inevitable surface contamination, monoatomic layers of water etc.
Electricity will arc along surfaces at much much lower v/m.
with high dV/dt the breakdown of air is much more complicated,
related to streamer growth at the tip of the arc.
e.g. you may have seen Steve Ward's tesla guns https://www.youtube.com/watch?v=R2jlgTA0QtI
< 100 kV in bursts produced those arcs if I remember correctly !
When Marx generators are used, if the risetime is just right ...
http://teslamania.delete.org/frames/longarc.jpg
In short, I doubt that the simple approach of accelerating charged particles in a linear electric field is a viable method of producing 1.5 MeV electrons.
(with the possible exception of bipolar VDG, BIG, EXPENSIVE, but doable)
I think that in an amateur setting, the most likely to succed is a cyclotron https://en.wikipedia.org/wiki/Cyclotron
A LINAC using a microwave oven magnetron may be another viable option ...
http://hpschapters.org/florida/13PPT.pdf
http://users.ox.ac.uk/~atdgroup/referencematerial/The%20novi...
P.S. it seems to me that a 1.5 MeV accelerator would become your only hobby for the forseeable future if you go that way.
Is there a university near you ?
Academics (desperate to publish a paper), and facility operators (I suggest 1 beer per MeV.minute) can be incredibly helpful !
Once an Academic has introduced you to the operator/technician, access is easy (in my case it was for S.E.M. time)
[Edited on 5-3-2017 by Sulaiman]
[Edited on 5-3-2017 by Sulaiman]
wg48 - 5-3-2017 at 06:03
Sulairman is correct the technical difficulties will be a challenge to say the least and the machine will be large so you will need a lot space if its air insulated. I looked for a paper I read about HV construction techniques and generators but failed to find it. Some experimenters at Imperial College London built a million/s volt discharge machine from the info. They used rolls of polyethylene film and aluminium foil to build the caps of a Marx. They used dilute cooper sulphate solution in plastic tubes for resistors. The Marx was about 8 feet high. The Marx output voltage was stored briefly in a very pure water insulated transmission line/s and doubled twice to get to the million/s of volts and increase the discharge speed.
A large Tesla coil can be used I recall an article in Scientific America. It used a Tesla coil to power an evacuated long glass tube to accelerate electrons. A filament at one end of the glass tube rectified the alternating voltage, an x-ray tube. It could sterilize seeds. I don’t recall if it was the x-rays or the electrons that achieved that. You could build a bipolar coil and mount the glass tube between them. The coils will have to be long perhaps 8ft or more.
An other problem from your description of what you’re doing is: Brem photons are broad band. Meaning the electrons kinetic energy is converted in to several photons and only very occasionally in to a single one and then that would probably require very high deceleration acceleration. Probably the level of deceleration/acceleration you get when they hit/interact with a proton/electron or even better the nucleus of a heavy atom.
I assumed you where going to fire the electrons at a suitable target that would generate the positrons. But I don’t think that will work or is very improbable. You would need to hit nucleus that absorbs the negative electron then decays and emits a positive positron. Do any do that? Yes if they are traveling very fast you can make a shower of partcles but thats a lot higher than 1 to 2 MV of acceleration.
[Edited on 5-3-2017 by wg48]
[Edited on 5-3-2017 by wg48]
unionised - 5-3-2017 at 06:13
There are not many projects where a DIY cyclotron looks like the sensible option: this is one of them.
WGTR - 5-3-2017 at 07:21
How much beam current do you need (at 1.5MV)?
There are a number of ways to make low power, fairly compact modules at these high voltages, but the materials are not cheap, and the whole high voltage section would need to be encapsulated in a dielectric.
Invictos - 5-3-2017 at 14:41
Quote: Originally posted by WGTR How much beam current do you need (at 1.5MV)? There are a number of ways to make low power, fairly compact modules at these high voltages, but the materials are not cheap, and the whole high voltage section would need to be encapsulated in a dielectric.
Whatever current we could reasonably get! We're trying to get pair production to occur, and I don't believe current is a factor in that (correct me if I'm wrong)
As for the target, we were planning on using tungsten.
Also, as this thread is taking a bit of a...pessimistic.. turn (darn Sulaiman, you and your knowledge...pft...), I've got a question about plan B that I'm going to put in another thread in a minute here...
Again, y'all are perfect
WGTR - 5-3-2017 at 16:32
Is this a personal curiosity, or a funded project? It's not a trick question; it would help you the most to have a solution tailored to your needs and resources.
Much of my electronics work has been in general power electronics design, up to about 50kV, but others in our facility work specifically with high voltage. The power side sounds doable on a benchtop to me, but without understanding all the details of your application it's impossible to say for sure. It would require some rather cleverly enforced isolation techniques.
The beam current question comes back to me not understanding the application very well. Are we talking about pA, or several amps? Does the current need to be continuous, or pulsed? If pulsed, then with what pulse length? If current requirements are in the pA or nA range, then power supply design becomes much simpler (and safer for you to use).
Dow Corning is my go-to place for dielectric potting gels, but generally their stuff is going to run several \$100 per gallon. If it's what the Dr. orders, though...
Sulaiman - 6-3-2017 at 06:20
even if a potting gell is the ultimate goal,
unless you are confident that no future repairs or modifications may be required
Invictos - 6-3-2017 at 19:09
Quote: Originally posted by WGTR Is this a personal curiosity, or a funded project? It's not a trick question; it would help you the most to have a solution tailored to your needs and resources. Much of my electronics work has been in general power electronics design, up to about 50kV, but others in our facility work specifically with high voltage. The power side sounds doable on a benchtop to me, but without understanding all the details of your application it's impossible to say for sure. It would require some rather cleverly enforced isolation techniques. The beam current question comes back to me not understanding the application very well. Are we talking about pA, or several amps? Does the current need to be continuous, or pulsed? If pulsed, then with what pulse length? If current requirements are in the pA or nA range, then power supply design becomes much simpler (and safer for you to use). Dow Corning is my go-to place for dielectric potting gels, but generally their stuff is going to run several \$100 per gallon. If it's what the Dr. orders, though...
Sure! This is a hobby grade project, so our budget is pretty much duct taped together...Also, our requirement as far as current goes is as low as we can reasonably expect to detect! And Sulaiman, a mineral oil or something is probably a smart idea to try it all!
Invictos - 11-3-2017 at 17:12
Hey everyone, reopening this can of worms...silly question: What would happen if you were to make a massive Marx generator while using Leyden jars as the capacitors?
The dielectric breakdown of something macroscopic like glass or plastic is probably stronger than the breakdown of a store bought capacitor, right? We have absolutely plenty of space, so things like 20cm spark gaps aren't an issue, and assuming we can charge them faster than the corona puts out, wouldn't we be alright?
Isn't that what this is, albeit on a less industrial scale? Would this be able to scale? : https://www.youtube.com/watch?v=5-nmtNg1i10
[Edited on 3-12-2017 by Invictos]
jpsmith123 - 11-3-2017 at 17:53
I'm wondering, are you just looking to make big sparks, or do you need an electron beam?
[Edited on 12-3-2017 by jpsmith123]
Invictos - 12-3-2017 at 05:38
Quote: Originally posted by jpsmith123 I'm wondering, are you just looking to make big sparks, or do you need an electron beam? [Edited on 12-3-2017 by jpsmith123]
We just need a high voltage in order to get a short burst of an electron beam. (I think that is a valid answer to that question )
unionised - 12-3-2017 at 05:45
Quote: Originally posted by Invictos H The dielectric breakdown of something macroscopic like glass or plastic is probably stronger than the breakdown of a store bought capacitor, right? [Edited on 3-12-2017 by Invictos]
If it was, guess what they would make store bought caps from...
Invictos - 12-3-2017 at 06:10
Quote: Originally posted by unionised
Quote: Originally posted by Invictos H The dielectric breakdown of something macroscopic like glass or plastic is probably stronger than the breakdown of a store bought capacitor, right? [Edited on 3-12-2017 by Invictos]
If it was, guess what they would make store bought caps from...
I'm sure they use Kapton or something similar, but I bet half an inch of plastic is stronger than the hair's thickness they try to cram into those caps!
After all, how many people would be buying five gallon buckets for hobby electronics?
Morgan - 12-3-2017 at 09:56
There are these film caps that probably wouldn't have a long life or duty cycle but might work for some things. I bought a dozen just to fiddle with.
"30kV DC 1000pF High Voltage Film Capacitor" for \$1.84 a piece on eBay.
It's interesting how tricky some of the designs are for "film capacitors" on the Wiki page.
jpsmith123 - 12-3-2017 at 13:04
High voltage will make sparks, but that's not the same thing as an electron beam.
You might have to do something like what's described in this paper:
But you'd have to scale it up to get the voltage that you want, which would probably involve replacing the spiral generator with an appropriately designed and driven air core transformer, and increasing the size of the device to prevent HV breakdown across insulator surfaces.
Quote: Originally posted by Invictos
Quote: Originally posted by jpsmith123 I'm wondering, are you just looking to make big sparks, or do you need an electron beam? [Edited on 12-3-2017 by jpsmith123]
We just need a high voltage in order to get a short burst of an electron beam. (I think that is a valid answer to that question )
Invictos - 13-3-2017 at 04:29
Quote: Originally posted by jpsmith123
High voltage will make sparks, but that's not the same thing as an electron beam.
You might have to do something like what's described in this paper:
But you'd have to scale it up to get the voltage that you want, which would probably involve replacing the spiral generator with an appropriately designed and driven air core transformer, and increasing the size of the device to prevent HV breakdown across insulator surfaces.
Quote: Originally posted by Invictos
Quote: Originally posted by jpsmith123 I'm wondering, are you just looking to make big sparks, or do you need an electron beam? [Edited on 12-3-2017 by jpsmith123]
We just need a high voltage in order to get a short burst of an electron beam. (I think that is a valid answer to that question )
I'll look into that! And we have electron emitters and a tube, we're just using the voltage difference to actually accelerate them
macckone - 20-3-2017 at 09:33
https://www.sciencenews.org/article/signature-antimatter-det...
Spark created positrons.
jpsmith123 - 20-3-2017 at 20:12
According to the following linked paper, relativistic electron beams have also been detected during storms; so maybe it's not the "sparks", per se, that are making positrons, but bremsstrahlung from the high energy electrons?
http://www.atmos-chem-phys.net/11/7747/2011/acp-11-7747-2011... | 5,723 | 23,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.912742 |
https://brainsanswers.co.uk/mathematics/at-the-end-of-a-snow-storm-allison-14726783 | 1,632,623,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00587.warc.gz | 190,328,589 | 32,929 | , 11.02.2020 04:29 knela
# At the end of a snow storm, Allison saw there was a lot of snow on her front lawn. The temperature increased and the snow began to melt at a steady rate. After the storm, the snow started to melt at a rate of 0.75 inches per hour and is known that 4 hours after the storm ended, the depth of snow was down to 9inches. Write an equation for the function S(t),representing the depth of snow on Allison lawn, in inches, t hours after the snow stopped falling
### Another question on Mathematics
Mathematics, 04.02.2019 05:43
Terri makes a quilt using three sizes of fabric squares the side lenght of each fabric square is the square root of the area | 172 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | latest | en | 0.937115 |
https://www.physicsforums.com/threads/point-charge-and-charged-sphere.795831/ | 1,508,834,247,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00286.warc.gz | 940,710,017 | 15,886 | # Point Charge and Charged Sphere
1. Feb 3, 2015
### skycap
1. The problem statement, all variables and given/known data
A point charge q1 = -6.1 μC is located at the center of a thick conducting shell of inner radius a = 2.8 cm and outer radius b = 4.8 cm, The conducting shell has a net charge of q2 = 2.6 μC.
2. Relevant equations
E = (kQ)/r2
F = (kq1q2)/r2
3. The attempt at a solution
I honestly do not know how to approach this problem mostly because I've been dealing with point charges. I'm not sure how to find the field of a charged sphere. I'm sure once I get this part down, the rest will follow. Thanks to any and all that can help, I really appreciate it; This is probably the most difficult time I've had with physics (to date).
2. Feb 3, 2015
### Staff: Mentor
What is the question?
Do you know Gauß' law? | 241 | 830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-43 | longest | en | 0.959314 |
http://blog.csdn.net/u011151784/article/details/46834413 | 1,503,151,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105451.99/warc/CC-MAIN-20170819124333-20170819144333-00392.warc.gz | 52,523,633 | 15,985 | # ZOJ 3757 Alice and Bob and Cue Sports
311人阅读 评论(0)
ZOJ Problem Set - 3757
Alice and Bob and Cue Sports
Time Limit: 2 Seconds Memory Limit: 65536 KB
Alice and Bob both love playing games. Now, Alice and Bob are playing a cue sport like Nine-ball Pool. Two players move in turn. A move is that one player use the cue ball to hit the target ball for pocketing the target ball. At the beginning of game, there are one cue ball and n object balls. Each object ball has a number (a positive integer) on it, and no two object balls have the same number. In a move, the target ball is a object ball still on the table with the smallest number. If the player does the following things in a move, a foul is called and a penalty point is added to the opposite's point:
• Cue ball do not hit any object ball. Penalty: the number of target ball.
• Cue ball is not pocketed and hit at least one ball, but do not hit target ball first or hit more than one object ball first at the same time. Penalty: the largest number of ball in the first hit balls.
• Cue ball is pocketed and hit at least one ball. Penalty: the largest number of ball in the first hit balls.
If the player pockets the target ball without a foul, a point will be added to the player's points, which is the sum of numbers of the ball pocketed in this move, and the player can make another move. But, if the player pockets the target ball with a foul or pockets at least one object ball not including the target ball, this point will be added to the opposite's points. If cue ball is pocketed, it will be pull out of the pocket and put on the table again while object balls will not be put on the table again after they are pocketed, even if the player pocketed them with a foul.
Now given n object balls and m moves, Bob wants to write a program to calculate the final points of Alice and him after these m moves. Because of Lady's first Rule, Alice always moves first.
#### Input
There are multiple cases. The first line of each case consists of two integers, n and m (1 ≤ n ≤ 1000, 0 ≤ m ≤ 1000), as described above. The second line consists of n positive integers. Each integer ai means the number of ith object ball (1 ≤ ai ≤ 10000). There are 2m lines followed. The (2i-1)th and (2i)th lines describe the ith move. The (2i-1)th line first includes an integer p, meaning cue ball hits p balls first. There are p integers followed in the same line, describing the hit balls. The (2i)th line first includes an integer q, meaning there are q balls pocketed. There are q integers followed in the same line, describing the pocketed balls (0 means cue ball). It's guaranteed that there is no impossible data.
#### Output
For each case, output one line as the format: "AP : BP", where AP means Alice's points and BP means Bob's points.
#### Sample Input
3 3
2 3 5
1 2
1 2
1 3
1 5
1 3
1 3
#### Sample Output
2 : 8
Author: YU, Xiaoyao
Source: ZOJ Monthly, March 2014
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
using namespace std;
long long num[200000];
long long hit[101000];
long long in[101000];
long long w[101000];//每个球的重量
int main()
{
long long n,m;
while(cin >> n >> m)
{
for(int i=1;i<=n;i++)
cin >> w[i];
long long ans[2]={0};
long long now = 0;
sort(w+1,w+1+n);//排序后找出目标球
long long pot = 1;//刚开始的目标球就是W【POT】
memset(num,0,sizeof(num));
while(m--)
{
long long len_hit,len_in;
cin >> len_hit;
long long sum=0;
bool flag1=0;
bool flag2 = 0;
for(int i=0;i<len_hit;i++)
{
cin >> hit[i];
if(hit[i]==w[pot] && len_hit==1)flag1 = 1;
}
cin >> len_in;
for(int i=0;i<len_in;i++)
{
cin >> in[i];
if(in[i]==0)
flag2 = 1;
num[in[i]]=1;
sum+=in[i];
}
sort(hit,hit+len_hit);
sort(in,in+len_in);
if(len_hit==0) ans[1-now]+=w[pot];
else if(flag2==0&&flag1==0)
{
ans[1-now] +=sum;
ans[1-now]+=hit[len_hit-1];
}
else if(flag2&&len_hit)
{
ans[1-now] +=sum;
ans[1-now]+=hit[len_hit-1];
}
//以上是犯规的处理。
else if(num[w[pot]]==0) ans[1-now]+=sum;
//如果不犯规而且目标球进洞了，如下else
else
{
ans[now]+=sum;
now = 1-now;
}
while(num[w[pot]])pot++;
now = 1-now;
}
cout<<ans[0]<<" : "<<ans[1]<<endl;
}
return 0;
}
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最新评论 | 1,349 | 4,424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-34 | longest | en | 0.927805 |
https://mathematica.stackexchange.com/questions/184709/black-hole-orbit-in-mathematica | 1,721,275,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00243.warc.gz | 345,309,538 | 40,642 | Black hole orbit in Mathematica
I am currently investigating the motion of a particle around a black hole. The Lagrangian for the system is
$$\mathcal{L}=F(r)\dot{t}^2-\frac{\dot{r}^2}{F(r)}-(r\dot{\phi})^2$$
Where $$F(r) = 1 - 1/r$$. Dots represent differentiation wrt a parameter $$s$$. This gives constants of motion:
$$F(r)\dot{t} = E$$ $$r^2\dot{\phi} = L$$
The Lagrangian can be re-written in terms of $$u=1/r$$ and becomes
F[u_] := 1 - u;
L[u_, t_, p_] := F[u]*D[t, s]^2 - D[u, s]^2/(F[u]*u^4) - (D[p, s]/u)^2;
Solving the Euler-Lagrange equation can be done in one line as
Solve[D[D[L[u[s], t[s], p[s]], u'[s]], s] == D[L[u[s], t[s], p[s]], u], u''[s]]
Which gives $$\ddot u = \frac{(5 u-4) \dot u^2}{(u-1)u}$$ Solving numerically in Mathematica is done below (u has been denoted w here)
solu = NDSolve[{w''[s] == ((-4 + 5 w[s])(w'[s])^2)/((-1 + w[s]) w[s]), w'[0] == 0, w[0] == 1/6}, {w, w'}, {s, 0, 10}]
but this gives a constant w (so u) for the whole range. Why is this? Obviously the particle is going to move if we put it around a black hole.
• One possible solution of your ode is w[s]==1/6! Commented Oct 26, 2018 at 18:27
• Can you please tell me where you got this equation from?
– ayr
Commented Jun 12, 2023 at 8:17
If you change the intial condition to w'[0]=.01 you get a variing solution
W = NDSolveValue[{w''[s] == ((-4 + 5 w[s]) w'[s]^2)/((-1 + w[s]) w[s]),
w'[0] == 1/100, (* small initial velocity*)
w[0] == 1 /6}, w , {s, 0, 10}]
Plot[W[s], {s, 0, 10}, PlotRange -> {0, 1.1}, AxesLabel -> {s, w[s]}]
Decreasing w'[0] furthermore moves the significant change of w to greater values of s. The limit case w'[0]->0 moves the step to infinity. That's what NDSolve calculated in your numerical solution w[s]==1/6~constant!
I wonder if the problem lies in the numerical solution of the differential equation. With all due respect, I question the differential equation itself you obtained. Because this might tell a different story:
<< VariationalMethods
Clear[F, ℒ]
F[r_] := 1 - 1/r;
ℒ[r_, t_, ϕ_][s_] := F[r[s]] (t'[s])^2 - (r'[s])^2/F[r[s]] - (r[s] ϕ'[s])^2
EulerEquations[ℒ[r, t, ϕ][s], {r[s], t[s], ϕ[s]}, s] // FullSimplify
`
\begin{align} r \left(\frac{2 \ddot r}{r-1}-2 \dot\phi^2\right) + \frac{\dot t^2}{r^2} &= \frac{\dot r^2}{(r-1)^2}, \\ 0 = \frac{\dot r \dot t}{r} + (r-1) \ddot t &= r\frac{\mathrm d}{\mathrm ds}\left(F(r)\dot t\right), \\ 0 = r \left(2 \dot r \dot\phi + r \ddot\phi\right) &= \frac{\mathrm d}{\mathrm ds}\left(r^2\dot\phi\right). \end{align}
The last two equations establish the constants of motion, indeed, as you said. By using the constants of motion, however, the first equation becomes
$$\ddot r = \frac{2L^2(r-1)^2+r^3(\dot r^2-E^2)}{2r^4(r-1)},$$
or in terms of $$u = 1/r$$
$$\ddot u = \frac{u^5 \left[2 L^2 (u-1)^2 u-E^2\right] + (5u-4) \dot u^2}{2(u-1) u}.$$
It is this equation that is worthy of solving, I suppose. | 1,107 | 2,907 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.797574 |
https://socratic.org/questions/what-is-the-arc-length-of-f-t-t-2-4t-5-1-t-over-t-in-3-4 | 1,670,404,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00193.warc.gz | 547,462,881 | 6,152 | # What is the arc length of f(t)=(t^2-4t,5-1/t) over t in [3,4] ?
Nov 16, 2016
Define $x \left(t\right) , y \left(t\right)$ as follows
$\left\{\begin{matrix}x \left(t\right) = {t}^{2} - 4 t \\ y \left(t\right) = 5 - \frac{1}{t}\end{matrix}\right. \implies f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$
Differentiating $x \left(t\right) , y \left(t\right)$ wrt $t$ we get:
$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - 4$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{t} ^ 2$
Then the arc length of $f \left(t\right)$ over $t \in \left[\alpha , \beta\right]$ is given by;
$L = {\int}_{\alpha}^{\beta} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}}$
So over $\left[3 , 4\right]$ we have
$L = {\int}_{3}^{4} \sqrt{{\left(2 t - 4\right)}^{2} + {\left(\frac{1}{t} ^ 2\right)}^{2}}$
$\therefore L = {\int}_{3}^{4} \sqrt{4 {t}^{2} - 16 t + 16 + \frac{1}{t} ^ 4}$
This definite integral does not have an intrinsic solution and would need to be solved numerically, using either a computer or estimated using the Trapezium Rule or Simpson's Rule | 441 | 1,126 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.474065 |
http://mathoverflow.net/questions/33977/poincare-disk-model-is-this-locus-a-known-curve/86601 | 1,397,718,882,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00439-ip-10-147-4-33.ec2.internal.warc.gz | 159,291,757 | 14,547 | # Poincaré disk model: is this locus a known curve?
Please, consider a line segment $AB$ in the Poincaré disk model. Now, consider the set $S$ of all point $P$ in the disk such that the angle $\angle APB$ is constant.
Question: is $S$ a known curve?
Thanks!
-
Found something. See this related question, perhaps strongly related. mathoverflow.net/questions/24307/… Note that, in the upper half plane model, if you take your points $A,B$ as being $0, \infty$ along the imaginary axis, you and they get the same answer: the locus is a slanted ray beginning at $0.$ – Will Jagy Jul 31 '10 at 18:27
Anyway, here is a link to the first page of the article that answers the other question (constant area instead of constant opposite angle) springerlink.com/content/h031vf8c7ydfaxrf – Will Jagy Jul 31 '10 at 21:56
The next easiest, fixing $A = i$ and $B = \infty$ along the imaginary axis, forcing a right angle at $P = x + i y$ with $x \geq 0$ gives the hyperbola $y = \sqrt{1 + x^2}.$ So the answer to your question is at least slightly different from that of question 24307. If the answer with $A = i$ and $B = \lambda i$ with finite real $\lambda > 0$ is a conic section other than a circular arc, well, I do not know the meaning of those in the upper half plane model, but I bet there are books that say. – Will Jagy Aug 1 '10 at 1:31
Google books pulled up the following page from Richter-Gebert's recent book "Perspectives on Projective Geometry", though I can't access most of the discussion: books.google.com/… – j.c. Jan 25 '12 at 10:33
In the Klein model, one may see that this is also a circle. Consider a line segment with one point on the center of the disk. One side of the triangle goes through the center. Then orthogonal lines to a line through the center are also orthogonal in the hyperbolic metric, e.g. since they are preserved by reflection. So one sees that a circle is traced out which goes through the origin. If you'd rather center the curve at the origin, then it will be an ellipse, since hyperbolic isometries of the Klein model are projective transformations.
To convert to the Poincare model, take a hemisphere sitting over the disk, and project vertically. The projection of the circle is given by the intersection of a cylinder over the circle with the upper hemisphere. This upper hemisphere is conformally equivalent to the Poincare model, e.g. by inversion through a sphere centered at the south pole of the lower hemisphere. I haven't computed the curve this traces out though.
-
This appears to be Theorem 26.1 in the book by Richter-Gebert (the reference pointed out by jc). For angles other than 90 degrees, he points out that the curve is quartic. dl.dropbox.com/u/8592391/Ch26RichterGebert.pdf – Ian Agol Jan 25 '12 at 17:41
Following on from Will Jagy's comment, for a fixed angle $\theta$ at $P = x + iy$ and with the other two vertices in the upper half-plane placed at $i$ and $\infty$, the locus $S$ for the vertex $P$ is $$y^2 = 1 + x^2 - 2xy\cot{\theta}.$$ We can use this to extend to the case where the two fixed vertices are at $i$ and $e^h i$, where $h$ is the distance between the two points $A$ and $B$. Suppose $P$ makes the fixed angle $\theta$ with these two vertices, and denote by $\psi$ the angle at $P$ of the ideal triangle with vertices at $e^h i$ and $\infty$. We then have the two equations $$y^2 = e^{2h} + x^2 - 2xy\cot{\psi}$$ $$y^2 = 1 + x^2 - 2xy\cot{(\psi + \theta)}.$$ Using trig identities to remove $\psi$, from this we get $$y^4 - (e^{2h} + 1)y^2 + x^4 + (e^{2h} + 1)x^2 + 2xy(xy + \cot{\theta} - e^{2h}\cot{\theta}) = e^{2h}.$$ I am not personally aware of any particular significance of this locus, though I'd be interested to hear if there is one. | 1,045 | 3,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2014-15 | latest | en | 0.92034 |
https://www.physicsforums.com/threads/solve-ac-voltage-form-rlc-series-circuit-power.434019/ | 1,726,657,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00370.warc.gz | 863,450,594 | 20,010 | # Solve AC Voltage Form: RLC Series Circuit Power
• Acuben
In summary, an AC voltage of the form V= (100 v) sin (1 000t) is applied to a series RLC circuit, where V is the peak voltage and ω is the angular frequency. The voltage at any instant T is V sin (ωT). The voltage can range from plus 100 volts to minus 100 volts, including zero. To find ω, compare the terms in the equation V=V0sin(ωt). The angular velocity is also related to frequency, where ω = 2πf.
Acuben
## Homework Statement
You don't even have to read the whole thing... just the red part is sufficient.
I saw this on the homework: V= (100 v) sin (1 000t)
What does this mean?
-----
Can I treat it as having 100 real magnitude and 0 imaginary magnitude (and therefore 0 phase angle)? or does sin(1 000t) tells something about the phase angle?
or does sin (1 000t) tells something about angular velocity or frequency?
I can't solve this problem without knowing frequency or angular velocity
just for side note: let...
V= delta voltage
v= unit volts
H= Henries
RLC= i believe it stands for Resistance, Inductor, Capacitor
An AC voltage of the form V= (100 v) sin (1 000t) is applied to a series RLC circuit. Assume the resistance is 400 Ohms, the capacitance is 5.00 uF, and the inductance is 0.500 H. Find the average power delivered to the circuit
## Homework Equations
w= angular velocity (supposed to be omega)
L= inductor
C= capacitor
j= complex coefficent = sqrt(-1)
Zl= resistance of inductor
Zc= resistance of capacitor
Ztot= total resistance
R= resistance of resistor
P= Power
Xl=wL
Xc=1/wc
Zl=j*Xl
Zc=-j*Xc
Ztot= R + Zl + Zc = R + j(Xl-Xc)
P=I*V*(P.F)
P.F= Power Factor. Is is cos $$\varphi$$
the angle between voltage and current.
by default everything is in RMS (rootmeansquare), but it shouldn't matter in calculation)
## The Attempt at a Solution
well it's easy except I don't know frequency nor angular velocity.
Otherwise finding Ztot will be easy and finding power is easy as well
Last edited:
Your voltage is V= (100 v) sin (1 000t), which is in the form V=V0sin(ωt).
hence the amplitude is V0 and the angular frequency is ω. Compare the terms and get ω.
The voltage at any instant T is V sin (wT) where w is (should be) omega, the angular velocity.
So, V is the peak voltage.
Omega = 2 * PI * F
In this case, omega = 1000 = 2 * PI * F ... so F = 159.154 Hz. ( ie 1000 / (2 * pi) )
V is the maximum voltage, but the actual voltage depends on the sine function, so the actual voltage can be anywhere between plus 100 volts and minus 100 volts, including zero.
For example what would the voltage be after 0.2 seconds?
V = 100 * sin (2 * pi * 159.154 * 0.2 ) or -34.2 volts
rock.freak667 said:
...
which is in the form V=V0sin(ωt).
...
hence the amplitude is V0 and the angular frequency is ω. Compare the terms and get ω.
vk6kro said:
The voltage at any instant T is V sin (wT) where w is (should be) omega, the angular velocity.
So, V is the peak voltage.
V is the maximum voltage,
Thank you very much. I understood it now =D
.
To solve this problem, we first need to find the impedance (Ztot) of the series RLC circuit. This can be done using the formula Ztot = R + j(Xl-Xc), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.
To find Xl and Xc, we use the formulas Xl = wL and Xc = 1/wC, where w is the angular velocity and L and C are the inductance and capacitance of the circuit, respectively.
However, since the frequency (f) is not given, we cannot directly calculate the angular velocity using the formula w = 2πf. Instead, we can use the given AC voltage form V = (100 v) sin (1 000t) to find the frequency.
In this equation, 100 v is the maximum voltage (amplitude), and 1 000t represents the angular velocity (2πf) multiplied by time. Therefore, we can rearrange the equation to find the frequency: f = 1 000t / 2π.
Substituting this value of frequency into the formula for angular velocity, we can now calculate the impedance Ztot and then find the average power delivered to the circuit using the formula P = I*V*(P.F).
In conclusion, the given AC voltage form provides information about the amplitude, frequency, and angular velocity of the circuit, which are all necessary to solve the problem and find the average power delivered to the circuit.
## 1. What is an RLC series circuit?
An RLC series circuit is a type of electrical circuit that consists of a resistor (R), inductor (L), and capacitor (C) connected in a series configuration. This means that the components are connected end-to-end, so that the same current flows through each component.
## 2. How do you solve for AC voltage in an RLC series circuit?
To solve for AC voltage in an RLC series circuit, you can use the formula V = IZ, where V is the voltage, I is the current, and Z is the impedance. The impedance can be calculated using the formula Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.
## 3. What is the power in an RLC series circuit?
The power in an RLC series circuit is the rate at which energy is transferred or used. It can be calculated using the formula P = VI, where P is the power, V is the voltage, and I is the current. In an AC circuit, the power is given by the formula P = VrmsIrmscosφ, where Vrms and Irms are the root mean square values of the voltage and current, and φ is the phase angle between them.
## 4. How does changing the value of the components affect the AC voltage in an RLC series circuit?
Changing the value of a component in an RLC series circuit can affect the AC voltage in different ways. For example, increasing the resistance will decrease the voltage, while increasing the inductance or capacitance will increase the voltage. The exact effect will depend on the specific values of the components and the frequency of the AC source.
## 5. What are some real-world applications of RLC series circuits?
RLC series circuits have many real-world applications, including in electrical filters, resonance circuits, and power supplies. They are also commonly used in electronic devices such as radios, televisions, and computers. In power systems, RLC series circuits are used for power factor correction and voltage regulation.
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2K | 1,726 | 6,487 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.888965 |
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04-21-13, 09:31 AM #1 skrzacik0 Newbie Thread Starter Join Date: Apr 2013 Bikes: Posts: 2 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Type of tyre Hi people, I am new here and I am a newbie with bikes etc. I would like to know what size (diameter) or type I have in my bike. It's Trek 3700 and the name of type is: Bontrager LT3 26x2.0". I would like to know it, as I want to set up my new bike computer. I can choose or 26" (650A) or ATB26x2 (650B) or I can calculate it (tyre circumference) by multiplying my diameter by 3.1416, but I wasn't sure if my tyre is 26" or ATB26" (I'm not sure what ATB means, exactly). Please help me, thanks!
04-21-13, 10:01 AM #2 Retro Grouch Senior Member Join Date: Feb 2004 Location: St Peters, Missouri Bikes: Catrike 559 I own some others but they don't get ridden very much. Posts: 27,764 Mentioned: 4 Post(s) Tagged: 0 Thread(s) Quoted: 564 Post(s) That's a more difficult question than you probably think. The most precise and most reliable way to set your bike computer is to draw a pencil line on the floor of your garage or driveway. Set your tire's valve stem on that line and roll it forward 1 revolution. Draw another line there. Measure the distance between the 2 lines in millimeters. Program that number into your computer. Works for every size tire.
04-21-13, 10:47 AM #3
skrzacik0
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Originally Posted by Retro Grouch That's a more difficult question than you probably think. The most precise and most reliable way to set your bike computer is to draw a pencil line on the floor of your garage or driveway. Set your tire's valve stem on that line and roll it forward 1 revolution. Draw another line there. Measure the distance between the 2 lines in millimeters. Program that number into your computer. Works for every size tire.
Thanks! It was really cleaver
04-21-13, 12:12 PM #4 jeffpoulin Senior Member Join Date: Jul 2008 Bikes: Posts: 2,226 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 26 Post(s) Start with 210. That's what Schwalbe lists the circumference of their 26x2.00 tires (in centimeters, which my bike computers use. Multiply by 10 if your computer uses millimeters.). However, when I measured mine over 3 complete revolutions, I ended up with 205.6cm per revolution, so you can't always trust printed specs. Doing a road test over a known distance is a good way to check how accurate it is.
04-21-13, 01:22 PM #5 HvPnyrs Senior Member Join Date: Jan 2011 Location: In The Middle Of "Out There" / Downtown "Lost Angels" Bikes: 2001 Trek 520 - Hvy Hauler, Epic Adventure Bike / 2011 Fuji Newest 1.0 - Sporty Quick Bike Posts: 257 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 15 Post(s) Ideally one would do the rolling test While Sitting on the Saddle and with any normally carried load on racks (back and/or front) handlebar or seatpacks, rack trunks, frame bags and etc. Weight on the bike compresses the tire, effectively reducing the size of the wheel, wheel covers less distance per rotation. Example (using nominally identical rims, tires, and pressures): 130 lb. Racer Boi, calibrating computer on 17 lb. Road bike carrying small seat wedge and 2 water bottles would have longest measured distance per rotation. 250+lb. Clydesdale, calibrating computer on 27 lb. Road commuter bike carrying a 17 inch lap top, Hard cover engineering school books, lunch, thermos of hot beverage, and etc. totaling an additional 23 lb. would get notably less measured distance per rotation.
04-21-13, 02:34 PM #6 fietsbob coprolite Join Date: Jun 2010 Location: NW,Oregon Coast Bikes: 8 Posts: 27,264 Mentioned: 49 Post(s) Tagged: 0 Thread(s) Quoted: 2625 Post(s) The ETRO (metric) number is the disambiguation 559-50 is Mountain Bike 26"x2".. other numbers is Not. drop some white paint on the street, ride through it, (dog poo, may do ,and already be there.) then measure the distance between the marks, on the street or sidewalk , those being the actual circumference of the rolling wheel with you sitting on it.
04-21-13, 05:57 PM #7 furballi Senior Member Join Date: Mar 2010 Bikes: Posts: 919 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) The most accurate method is to pump up the tires to the correct pressure. Put a piece of tape on the front tire's tread area. Sit on the bike and line up the tape with the ground. This is point A. While sitting on the bike, slowly rotate the front tire exactly one revolution so that the tape is again lined up with the ground. This is point B. Measure the distance from point A to point B and enter this value (in mm) to the bike's computer. Try not to wiggle the front wheel when moving from point A to point B. A 700c tire should measure around 2100 mm. Repeat 3x to obtain an average reading.
04-21-13, 06:22 PM #8
JanMM
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Quote:
Originally Posted by skrzacik0 Hi people, I am new here and I am a newbie with bikes etc. I would like to know what size (diameter) or type I have in my bike. It's Trek 3700 and the name of type is: Bontrager LT3 26x2.0". I would like to know it, as I want to set up my new bike computer. I can choose or 26" (650A) or ATB26x2 (650B) or I can calculate it (tyre circumference) by multiplying my diameter by 3.1416, but I wasn't sure if my tyre is 26" or ATB26" (I'm not sure what ATB means, exactly). Please help me, thanks!
That's pretty confusing. Your bike does not (should not) have either 650A 650B or any variation of 650 tires. http://sheldonbrown.com/650b.html
In the early days of mountainbikes, they were also called ATB's (all terrain bikes). That lost out to MTB in popular usage.
When in doubt, can't beat measuring tire roll-out, as previously suggested.
__________________
RANS V3 Ti, RANS V3 Cromo, RANS Screamer
04-21-13, 06:35 PM #9 MichaelW Senior Member Join Date: Feb 2001 Location: England Bikes: Posts: 12,951 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 13 Post(s) is it a 650B? FYI, this is a very niche size, originally a French utility wheel, occasionally used for smaller custom touring bikes in 1980s (now superseded by MTB 26") but making a small comeback through niche frame companies. It is a bit larger than the std MTB rim but still classed as one of the 26" sizes.
04-21-13, 07:17 PM #10 JanMM rebmeM roineS Join Date: Jan 2006 Location: Metro Indy, IN Bikes: RANS V3 ti, RANS V3 cromo, RANS Screamer Posts: 14,280 Mentioned: 8 Post(s) Tagged: 0 Thread(s) Quoted: 235 Post(s) Would an entry-level Trek bike (3700) have 650B wheels? __________________ RANS V3 Ti, RANS V3 Cromo, RANS Screamer
04-21-13, 09:31 PM #11 Jim Kukula Senior Member Join Date: Oct 2010 Location: Utah Bikes: Thorn Nomad Mk2, 1996 Trek 520, Workcycles Transport, Brompton Posts: 589 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Those choices ("26" (650A) or ATB26x2 (650B)) must come out of the manual for the bike computer. Who knows what the person writing the manual was thinking! Here is the Bontrager tire http://bontrager.com/model/08058 but it doesn't give such precise information. Probably it is a 559-50 tire. Here are the circumferences that Schwalbe gives: http://www.schwalbetires.com/tech_in...#circumference but those are just approximate. I ride on Schwalbe Marathon Supreme 26x2 = 559-50 tires and I measured more like 2010 mm circumference. It does depend on pressure, load, etc.
04-24-13, 07:08 AM #12
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Originally Posted by Jim Kukula Those choices ("26" (650A) or ATB26x2 (650B)) must come out of the manual for the bike computer. Who knows what the person writing the manual was thinking!
Hi,
They are probably presets for a given tyre circumference,
and will equate to a specific sized tyre on standard rims.
rgds, sreten.
04-24-13, 08:57 AM #13
erig007
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Originally Posted by HvPnyrs Ideally one would do the rolling test While Sitting on the Saddle and with any normally carried load on racks (back and/or front) handlebar or seatpacks, rack trunks, frame bags and etc. Weight on the bike compresses the tire, effectively reducing the size of the wheel, wheel covers less distance per rotation. Example (using nominally identical rims, tires, and pressures): 130 lb. Racer Boi, calibrating computer on 17 lb. Road bike carrying small seat wedge and 2 water bottles would have longest measured distance per rotation. 250+lb. Clydesdale, calibrating computer on 27 lb. Road commuter bike carrying a 17 inch lap top, Hard cover engineering school books, lunch, thermos of hot beverage, and etc. totaling an additional 23 lb. would get notably less measured distance per rotation.
To add to that weight fluctuations occurs dynamically when riding depending on wind, inclination of the road, sitting position etc... so that more weight is on the rear tyre or on the front tyre, tyre pressure change each day due to leaks, weight distribution change as the items carried in bags change etc | 2,720 | 9,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | latest | en | 0.933817 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Time_Dependent_Quantum_Mechanics_and_Spectroscopy_(Tokmakoff)/03%3A__Time-Evolution_Operator/3.05%3A_Schrodinger_and_Heisenberg_Representations | 1,685,827,564,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00173.warc.gz | 191,245,733 | 30,840 | # 3.5: Schrödinger and Heisenberg Representations
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
The mathematical formulation of quantum dynamics that has been presented is not unique. So far, we have described the dynamics by propagating the wavefunction, which encodes probability densities. Ultimately, since we cannot measure a wavefunction, we are interested in observables, which are probability amplitudes associated with Hermitian operators, with time dependence that can be interpreted differently. Consider the expectation value:
\begin{align} \langle \hat {A} (t) \rangle & = \langle \psi (t) | \hat {A} | \psi (t) \rangle = \left\langle \psi ( 0 ) \left| U^{\dagger} \hat {A} U \right| \psi ( 0 ) \right\rangle \\[4pt] & = ( \langle \psi ( 0 ) | U^{\dagger} ) \hat {A} ( U | \psi ( 0 ) \rangle ) \label{S rep} \\[4pt] & = \left\langle \psi ( 0 ) \left| \left( U^{\dagger} \hat {A} U \right) \right| \psi ( 0 ) \right\rangle \label{2.76} \end{align}
The last two expressions are written to emphasize alternate “pictures” of the dynamics. Equation \ref{S rep} is known as the Schrödinger picture, refers to everything we have done so far. Here we propagate the wavefunction or eigenvectors in time as $$U | \psi \rangle$$. Operators are unchanged because they carry no time-dependence. Alternatively, we can work in the Heisenberg picture (Equation \ref{2.76}) that uses the unitary property of $$U$$ to time-propagate the operators as $$\hat {A} (t) = U^{\dagger} \hat {A} U,$$ but the wavefunction is now stationary. The Heisenberg picture has an appealing physical picture behind it, because particles move. That is, there is a time-dependence to position and momentum.
## Schrödinger Picture
In the Schrödinger picture, the time-development of $$| \psi \rangle$$ is governed by the TDSE
$i \hbar \frac {\partial} {\partial t} | \psi \rangle = H | \psi \rangle \label{2.77A}$
or equivalently, the time propagator:
$| \psi (t) \rangle = U \left( t , t _ {0} \right) | \psi \left( t _ {0} \right) \rangle \label{2.77B}$
In the Schrödinger picture, operators are typically independent of time, $$\partial A / \partial t = 0$$. What about observables? For expectation values of operators
$\langle A (t) \rangle = \langle \psi | A | \psi \rangle$
\begin{align} i \hbar \frac {\partial} {\partial t} \langle \hat {A} (t) \rangle & = i \hbar \left[ \left\langle \psi | \hat {A} | \frac {\partial \psi} {\partial t} \right\rangle + \left\langle \frac {\partial \psi} {\partial t} | \hat {A} | \psi \right\rangle + \cancel{\left\langle \psi \left| \frac {\partial \hat {A}} {\partial t} \right| \psi \right\rangle} \right] \\[4pt] & = \langle \psi | \hat {A} H | \psi \rangle - \langle \psi | H \hat {A} | \psi \rangle \\[4pt] & = \langle [ \hat {A} , H ] \rangle \label{2.78} \end{align}
If$$\hat{A}$$ is independent of time (as we expect in the Schrödinger picture), and if it commutes with $$\hat{H}$$, it is referred to as a constant of motion.
## Heisenberg Picture
From Equation \ref{2.76}, we can distinguish the Schrödinger picture from Heisenberg operators:
$\hat {A} (t) = \langle \psi (t) | \hat {A} | \psi (t) \rangle _ {S} = \left\langle \psi \left( t _ {0} \right) \left| U^{\dagger} \hat {A} U \right| \psi \left( t _ {0} \right) \right\rangle _ {S} = \langle \psi | \hat {A} (t) | \psi \rangle _ {H} \label{2.79}$
where the operator is defined as
\left.\begin{aligned} \hat {A} _ {H} (t) & = U^{\dagger} \left( t , t _ {0} \right) \hat {A} _ {S} U \left( t , t _ {0} \right) \\[4pt] \hat {A} _ {H} \left( t _ {0} \right) & = \hat {A} _ {S} \end{aligned} \right. \label{2.80}
Note, the pictures have the same wavefunction at the reference point $$t_0$$. Since the wavefunction should be time-independent, $$\partial | \psi _ {H} \rangle / \partial t = 0$$, we can relate the Schrödinger and Heisenberg wavefunctions as
$| \psi _ {S} (t) \rangle = U \left( t , t _ {0} \right) | \psi _ {H} \rangle \label{2.81}$
So,
$| \psi _ {H} \rangle = U^{\dagger} \left( t , t _ {0} \right) | \psi _ {S} (t) \rangle = | \psi _ {S} \left( t _ {0} \right) \rangle \label{2.82}$
As expected for a unitary transformation, in either picture the eigenvalues are preserved:
\begin{align} \hat {A} | \varphi _ {i} \rangle _ {S} & = a _ {i} | \varphi _ {i} \rangle _ {S} \\[4pt] U^{\dagger} \hat {A} U U^{\dagger} | \varphi _ {i} \rangle _ {S} & = a _ {i} U^{\dagger} | \varphi _ {i} \rangle _ {S} \\[4pt] \hat {A} _ {H} | \varphi _ {i} \rangle _ {H} & = a _ {i} | \varphi _ {i} \rangle _ {H} \end{align} \label{2.83}
The time evolution of the operators in the Heisenberg picture is:
\begin{aligned} \frac {\partial \hat {A} _ {H}} {\partial t} & = \frac {\partial} {\partial t} \left( U^{\dagger} \hat {A} _ {s} U \right) = \frac {\partial U^{\dagger}} {\partial t} \hat {A} _ {s} U + U^{\dagger} \hat {A} _ {s} \frac {\partial U} {\partial t} + U^{\dagger} \cancel{\frac {\partial \hat {A}} {\partial t}} U \\[4pt] &= \frac {i} {\hbar} U^{\dagger} H \hat {A} _ {S} U - \frac {i} {\hbar} U^{\dagger} \hat {A} _ {S} H U + \left( \cancel{\frac {\partial \hat {A}} {\partial t}} \right) _ {H} \\[4pt] &= \frac {i} {\hbar} H _ {H} \hat {A} _ {H} - \frac {i} {\hbar} \hat {A} _ {H} H _ {H} \\[4pt] &= - \frac {i} {\hbar} [ \hat {A} , H ] _ {H} \end{aligned} \label{2.84}
The result
$i \hbar \frac {\partial} {\partial t} \hat {A} _ {H} = [ \hat {A} , H ] _ {H} \label{2.85}$
is known as the Heisenberg equation of motion. Here I have written the odd looking $$H _ {H} = U^{\dagger} H U$$. This is mainly to remind one about the time-dependence of $$\hat{H}$$. Generally speaking, for a time-independent Hamiltonian $$U = e^{- i H t / h}$$, $$U$$ and $$H$$ commute, and $$H_H =H$$. For a time-dependent Hamiltonian, $$U$$ and $$H$$ need not commute.
## Classical equivalence for particle in a potential
The Heisenberg equation is commonly applied to a particle in an arbitrary potential. Consider a particle with an arbitrary one-dimensional potential
$H = \frac {p^{2}} {2 m} + V (x) \label{2.86}$
For this Hamiltonian, the Heisenberg equation gives the time-dependence of the momentum and position as
$\dot {p} = - \frac {\partial V} {\partial x} \label{2.87}$
$\dot {x} = \frac {p} {m} \label{2.88}$
Here, I have made use of
$\left[ \hat {x}^{n} , \hat {p} \right] = i \hbar n \hat {x}^{n - 1} \label{2.89}$
$\left[ \hat {x} , \hat {p}^{n} \right] = i \hbar n \hat {p}^{n - 1} \label{2.90}$
Curiously, the factors of $$\hbar$$ have vanished in Equations \ref{2.87} and \ref{2.88}, and quantum mechanics does not seem to be present. Instead, these equations indicate that the position and momentum operators follow the same equations of motion as Hamilton’s equations for the classical variables. If we integrate Equation \ref{2.88} over a time period $$t$$ we find that the expectation value for the position of the particle follows the classical motion.
$\langle x (t) \rangle = \frac {\langle p \rangle t} {m} + \langle x ( 0 ) \rangle \label{2.91}$
We can also use the time derivative of Equation \ref{2.88} to obtain an equation that mirrors Newton’s second law of motion, $$F=ma$$:
$m \frac {\partial^{2} \langle x \rangle} {\partial t^{2}} = - \langle \nabla V \rangle \label{2.92}$
These observations underlie Ehrenfest’s Theorem, a statement of the classical correspondence of quantum mechanics, which states that the expectation values for the position and momentum operators will follow the classical equations of motion. | 2,900 | 8,445 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.520962 |
https://www.dieselgeneratortech.com/generator-sets/what-is-synchronous-generator-oscillation.html | 1,721,781,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00638.warc.gz | 647,676,642 | 11,347 | # What is Synchronous Generator Oscillation?
Dec. 30, 2020
When generator working normally, the connection between the stator pole and the rotor pole can be regarded as an elastic magnetic line of force. When increasing the load of generator, power angle will increase, which is equivalent to lengthening the magnetic line of force.
When reducing the load of generator, power angle will reduce, which is equivalent to shortening the magnetic line of force. When the load of generator changing suddenly, the rotor power angle can not be stabilized at a new value immediately due to the inertia of the rotor. However, it goes through several oscillations around the new stable value, which is called the oscillation of synchronous generator.
Oscillation is divided into two types:
1.The amplitude of oscillation becomes smaller and smaller, the swing of power angle gradually attenuates, and finally stabilizes at a new power angle, and still operates stably at synchronous speed, which is called synchronous oscillation.
2.The amplitude of oscillation becomes larger and larger, and the power angle increases continuously until it is out of the stable range, which makes the generator out of step and the generator goes into asynchronous operation, which is called asynchronous oscillation.
Why synchronous generator occurs oscillation?
1.Stator ammeter indication exceeds the normal value and moves violently.
Because the angle between the parallel electromotive forces changes, resulting in electromotive force difference, which makes the generators flow through the circulating current. Due to the swing of the rotor speed, the angle between the electromotive forces is larger and smaller at sometimes, and the torque and power are also larger and smaller at sometimes, so the circulating current is larger and smaller at sometimes, so the pointer of the stator current swings back and forth. This circulating current, together with the original load current, may exceed the normal value.
2.The pointer indication of stator voltmeter and other bus voltmeters is lower than the normal value and oscillates back and forth. This is because the angle between out of step generator and other generators is changing, which causes voltage swing. Because the current is larger than normal, the voltage drop is also large, resulting in low voltage.
3.The active load and reactive load swing sharply. The reason is that the output power of the generator in the oscillation process without out of step is large or small, and sometimes the active power is sent out or absorbed when out of step.
4.The pointer of rotor voltage and current meter swings around the normal value. When the generator is oscillating or out of step, the alternating current will be induced in the rotor winding and fluctuate with the fluctuation of the stator current. The current is superimposed on the original excitation current, which makes the pointer of the rotor ammeter swing near the normal value.
5.The generator gives out a rhythmic sound, which is in step with the swing rhythm of the meter pointer.
6.The current and power indication of unit and line change periodically, but the fluctuation is small, and the active power output of generator is no more than zero.
7.The frequency of the system and generator does not change much, the frequency of the whole system does not increase in one part and decrease in the other part; the noise of the generator is small, and the opening of the guide vane has no obvious change.
How to deal with oscillation of synchronous generator?
1.If it is not caused by the loss of excitation of generator, the excitation current of the generator should be increased immediately to increase the electromotive force of the generator, increase the power limit and improve the stability of the generator. This is due to the increase of excitation current, which increases the tension between the stator and rotor poles, weakens the inertia of the rotor, and pulls the generator into synchronization when it reaches the equilibrium point. At this time, if the generator excitation system is in the forced excitation state, it should not intervene within 1 minute.
2.If it is caused by high power factor of single machine, the active power should be reduced and the excitation current should be increased. This can not only reduce the rotor inertia, but also increase the stable operation capacity of the unit due to the increase of the power limit.
3.After judging that the system is synchronous oscillation, exit the AGC device of the generator set and increase the reactive power output of the generator appropriately.
4.If the oscillation is caused by the fault of the speed regulation system or excitation system of the unit in the plant, the active power output of the unit in fault shall be reduced immediately and the equipment fault shall be eliminated. If the fault cannot be eliminated in a short time, the unit shall be disconnected.
5.If the oscillation is caused by the accident trip of the output line in the plant, the active power output of the unit shall be reduced immediately until the oscillation is eliminated.
6.If the synchronous oscillation is changed into asynchronous oscillation, it will be treated as asynchronous oscillation.
Hope the article is helpful to you when you meet oscillation of synchronous generator problems. We are also a manufacturer of diesel generator set in Jiangsu China, if you are also interested, please contact us by email sales@dieselgeneratortech.com or call us by phone number +8613481024441. | 1,083 | 5,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.937336 |
http://www.onlinemathlearning.com/subtract-using-ten.html | 1,493,173,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121121.5/warc/CC-MAIN-20170423031201-00208-ip-10-145-167-34.ec2.internal.warc.gz | 632,060,721 | 9,132 | # Subtraction Problems using Ten as a Unit
Videos to help Grade 1 students learn how to solve subtraction problems using ten as a unit, and write two-step solutions.
Common Core Standards: 1.OA.1, 1.NBT.2a, 1.NBT.2b, 1.NBT.5
Related Topics:
Lesson Plans and Worksheets for Grade 1
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 1
Common Core For Grade 1
New York State Common Core Math Module 2, Grade 1, Lesson 29
Lesson 29 Problem Set
Solve the problems. Write your answers to show how many tens and ones.
Show your solution in two steps:
Step 1: Write one number sentence to subtract from ten.
Step 2: Write one number sentence to add the remaining parts
1. 14 - 5 =
Lesson 29 Problem Set
4. This week, Maria ate 5 yellow plums and some red plums. If she ate 11 plums in all,
how many red plums did Maria eat?
Lesson 29 Homework
Solve the problems. Write your answers to show how many tens and ones
Solve. Write the two number sentences for each step to show how you take from ten. Remember to put a box around your solution and write a statement.
4. Eli read some science magazines. Then he read 9 sports magazines. If he read 18 magazines altogether, how many science magazines did Eli read?
6. Some children are at the park playing soccer. Seven are wearing white shirts. If there are 14 children playing soccer in all, how many children are wearing shirts that are another color?
Lesson 29 Word Problems
Isabelle had some presents left to open. If she received 13 presents and had already opened 8. How many presents does she still need to open? | 391 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-17 | longest | en | 0.937735 |
https://thetextchemistry.org/qa/question-what-is-the-difference-between-mg-and-mcg.html | 1,611,477,909,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00189.warc.gz | 584,173,916 | 6,882 | # Question: What Is The Difference Between MG And MCG?
## Which is stronger mg or mcg?
1 milligram (mg) is equal to 1,000 micrograms (mcg).
In layman’s terms, a milligram is 1,000 times more powerful..
## Is 1mg the same as 1000 mcg?
1mg = 1000mcg. As there are 1,000 micrograms (mcg) in 1 milligram (mg), you can work out your answer by multiplying your mg figure by 1000.
## How many teaspoons is 3 milligrams?
mg to tsp conversion table:100 mg = 0.02 tsp2100 mg = 0.42 tsp7000 mg = 1.4 tsp900 mg = 0.18 tsp2900 mg = 0.58 tsp15000 mg = 3 tsp1000 mg = 0.2 tsp3000 mg = 0.6 tsp16000 mg = 3.2 tsp1100 mg = 0.22 tsp3100 mg = 0.62 tsp17000 mg = 3.4 tsp1200 mg = 0.24 tsp3200 mg = 0.64 tsp18000 mg = 3.6 tsp13 more rows
## How many milligrams are in a tablespoon?
a density conversion tablegram per US tablespoonmilli- gram per US tablespoon1100022000330004400016 more rows
## How many milligrams is a teaspoon?
The unit is abbreviated as tsp. Convert Milligrams (mg) to Teaspoons (tsp): 1 mg is approximately equal to 0.0002 tsps. One milligram is a relatively small quantity of table salt. A mere one-teaspoon of granulated salt contains some 2325 milligrams.
## Is it OK to take 1000 mcg of b12 a day?
When taken at appropriate doses, vitamin B-12 supplements are generally considered safe. While the recommended daily amount of vitamin B-12 for adults is 2.4 micrograms, you can safely take higher doses. Your body absorbs only as much as it needs, and any excess passes through your urine.
## What is the difference between MG and MCG in vitamins?
In the metric system, 1000 milligrams (mg) is a unit of mass equal to 1 gram and 1000 micrograms (mcg) is equal to 1 milligram (mg).
## Is 400 mcg the same as 1 mg?
A woman of childbearing age needs 400 micrograms of folic acid every day. But she should not take more than 1 milligram (1000 mcg = 1 mg.) daily unless your doctor or health care provider tells you so.
## Is 5mg equal to 400 mcg?
Make sure you are clear about what dose to take – the recommended dose is 400 micrograms once daily. The 5 mg tablet has more than ten times as much folic acid as this.
## How many milligrams are in a teaspoon of salt?
Consider that a single teaspoon of table salt, which is a combination of sodium and chloride, has 2,325 milligrams (mg) of sodium — more than the daily amount recommended in the Dietary Approaches to Stop Hypertension (DASH) diet.
## Is 25 mcg the same as 1000 mg?
How many mg is 25 mcg? – 25 micrograms is equal to 0.03 micrograms. … To convert 25 micrograms to milligrams, divide by 1,000.
## What is 2000 IU in MCG?
2000 IU (50 mcg) per day increases vitamin D blood levels 20 ng/ml (50 nmol/L). | 789 | 2,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-04 | latest | en | 0.864352 |
https://www.mapleprimes.com/users/Hamid/questions | 1,718,329,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00688.warc.gz | 831,734,296 | 17,621 | ## 28 Reputation
17 years, 61 days
## How to solve a boundary value problem!...
Maple
I need to solve following ODE: ODE:= y(f)*(diff(y(f), f, f))+50.0*f -50.0*f^2= 0 Subject to boundary conditions: BC:= (D(y))(.5) = 0, y(0) = 0 I used BVP method : dsol2 := dsolve({ODE, BC}, numeric, method = bvp[midrich], y(f), range = 0 .. .5) But I got " Error, (in dsolve/numeric/bvp) division by zero" Can someone please tell me what I can do. Thanks, http://maplenet.maplesoft.com/maplenet/worksheet/mapleprimes/4384_BVP.mw
## Hoe to solve a boundary value problem...
Maple
I need to solve following ODE: ODE:= y(f)*(diff(y(f), f, f))+50.0*f -50.0*f^2= 0 Subject to boundary conditions: BC:= (D(y))(.5) = 0, y(0) = 0 I used BVP method : dsol2 := dsolve({ODE, BC}, numeric, method = bvp[midrich], y(f), range = 0 .. .5) But I got " Error, (in dsolve/numeric/bvp) division by zero" Can someone please tell me what I can do. Thanks, http://maplenet.maplesoft.com/maplenet/worksheet/mapleprimes/4384_BVP.mw
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| 356 | 1,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.812865 |
https://questioncove.com/updates/516b7315e4b063fcbbbc1a56 | 1,503,351,621,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00017.warc.gz | 825,587,939 | 2,413 | OpenStudy (anonymous):
What is the slope of the line through the points (−6,−4) and (−3,−6)
OpenStudy (anonymous):
$m=\frac{ y_2-y_1 }{ x_2-x_1 }$
OpenStudy (anonymous):
y2=-6 y1=-4 x2=-3 x1=-6 you can go from there
OpenStudy (anonymous):
$m=\frac{ -6+4 }{ -3+6 }$
OpenStudy (anonymous):
do the math | 112 | 308 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-34 | latest | en | 0.847385 |
http://sparse.tamu.edu/Pajek?filterrific%5Bsorted_by%5D=name_asc | 1,713,983,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00385.warc.gz | 26,468,477 | 32,971 | ## Group Pajek
Group Description ```Pajek data sets, V. Batagelj and A. Mrvar. Pajek data sets, from http://vlado.fmf.uni-lj.si/pub/networks/data/, Vladimir Batagelj and Andrej Mrvar (2006): Pajek datasets. If the source of the data set is not specified otherwise, these data sets are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 2.5 License. Converted to sparse adjacency matrix format by Tim Davis, October 2006. A(i,j) is the edge from node i to node j in the graph. If a graph is not listed as weighted, and yet has non-binary entries, then the entry a(i,j) reflects the number of edges (i,j) in the original data (the graph is a multigraph). In some cases, it was clear that the extra entries were duplicates, and the graph should not be a multigraph (problems GD97, EAT_RS, and patents). All data was converted without loss of information, except where intentional. Details are given below. The largest graph is "patents", with over 3.7 million nodes and nearly 15 million entries. "Pajek" is Slovenian for "spider" and is roughly pronounced "Pie yeck". The Pajek networks are all 1-based (nodes are numbered starting at node 1). Some networks when included in the Pajek Dataset were converted from 0-based to 1-based. This was done by renaming node 0 as a new node n, and leaving all other nodes unchanged. This differs from the MATLAB convention, which is to add 1 to all row/column indices. This conversion was left as-is (networks: USpowerGrid, EPA, and Kleinberg) when included in the UF Sparse Matrix Collection. Remember that in MATLAB, A(i,:) is slow to compute; A(:,i) is much faster. If you want row i of a sparse matrix, access the ith column of the transpose instead. ================================================================================ Summary: ================================================================================ CSphd: PhD's in computer science 1882-by-1882 with 1740 nonzeros kind: directed graph with auxiliary node data California: Kleinberg's web search of "California" 9664-by-9664 with 16150 nonzeros kind: directed graph Cities: www.lboro.ac.uk/gawc, data set 6 55-by-46 with 1342 nonzeros kind: weighted bipartite graph with auxiliary node data EAT_RS: Edinburgh Associative Thesaurus (response-stimulus) 23219-by-23219 with 325592 nonzeros kind: directed weighted graph EAT_SR: Edinburgh Associative Thesaurus (stimulus-response) 23219-by-23219 with 325589 nonzeros kind: directed weighted graph EPA: Kleinberg, pages linking to www.epa.gov 4772-by-4772 with 8965 nonzeros kind: directed graph EVA: EVA, corporate inter-relationships 8497-by-8497 with 6726 nonzeros kind: directed graph Erdos02: Erdos collaboration network 6927-by-6927 with 16944 nonzeros kind: undirected graph Erdos971: Erdos collaboration network 472-by-472 with 2628 nonzeros kind: undirected graph Erdos972: Erdos collaboration network 5488-by-5488 with 14170 nonzeros kind: undirected graph with auxiliary node data Erdos981: Erdos collaboration network 485-by-485 with 2762 nonzeros kind: undirected graph Erdos982: Erdos collaboration network 5822-by-5822 with 14750 nonzeros kind: undirected graph with auxiliary node data Erdos991: Erdos collaboration network 492-by-492 with 2834 nonzeros kind: undirected graph Erdos992: Erdos collaboration network 6100-by-6100 with 15030 nonzeros kind: undirected graph with auxiliary node data FA: USF Free (word) Association Norms 10617-by-10617 with 72176 nonzeros kind: directed weighted graph with auxiliary node data GD00_a: Graph Drawing contest 2000 352-by-352 with 458 nonzeros kind: directed graph GD00_c: Graph Drawing contest 2000 638-by-638 with 1041 nonzeros kind: directed multigraph GD01_A: Graph Drawing contest 2001 953-by-953 with 645 nonzeros kind: directed multigraph GD01_a: Graph Drawing contest 2000 311-by-311 with 645 nonzeros kind: directed weighted graph GD01_b: Graph Drawing contest 2001 18-by-18 with 37 nonzeros kind: directed graph GD01_c: Graph Drawing contest 2001 33-by-33 with 135 nonzeros kind: directed multigraph with auxiliary node data GD02_a: Graph Drawing contest 2002 23-by-23 with 87 nonzeros kind: directed graph GD02_b: Graph Drawing contest 2002 80-by-80 with 232 nonzeros kind: directed graph GD06_Java: Graph Drawing contest 2006 1538-by-1538 with 8032 nonzeros kind: directed graph GD06_theory: Graph Drawing contest 2006 101-by-101 with 380 nonzeros kind: undirected graph GD95_a: Graph Drawing contest 1995 36-by-36 with 57 nonzeros kind: directed graph GD95_b: Graph Drawing contest 1995 73-by-73 with 96 nonzeros kind: directed graph GD95_c: Graph Drawing contest 1995 62-by-62 with 287 nonzeros kind: directed graph GD96_a: Graph Drawing contest 1996 1096-by-1096 with 1677 nonzeros kind: directed multigraph GD96_b: Graph Drawing contest 1996 111-by-111 with 193 nonzeros kind: directed graph GD96_c: Graph Drawing contest 1996 65-by-65 with 250 nonzeros kind: undirected graph GD96_d: Graph Drawing contest 1996 180-by-180 with 229 nonzeros kind: directed graph GD97_a: Graph Drawing contest 1997 84-by-84 with 332 nonzeros kind: directed graph GD97_b: Graph Drawing contest 1997 47-by-47 with 264 nonzeros kind: undirected weighted graph GD97_c: Graph Drawing contest 1997 452-by-452 with 460 nonzeros kind: directed multigraph GD98_a: Graph Drawing contest 1998 38-by-38 with 50 nonzeros kind: directed graph GD98_b: Graph Drawing contest 1998 121-by-121 with 207 nonzeros kind: directed graph GD98_c: Graph Drawing contest 1998 112-by-112 with 336 nonzeros kind: undirected graph GD99_b: Graph Drawing contest 1999 64-by-64 with 252 nonzeros kind: undirected multigraph GD99_c: Graph Drawing contest 1999 105-by-105 with 149 nonzeros kind: directed graph with auxiliary node data GlossGT: graph and digraph glossary 72-by-72 with 122 nonzeros kind: directed graph HEP-th-new: High Energy Physics literature 27770-by-27770 with 352807 nonzeros kind: directed graph with auxiliary node data HEP-th: High Energy Physics literature 27240-by-27240 with 342437 nonzeros kind: directed graph IMDB: IMDB movie/actor network, www.imdb.com 428440-by-896308 with 3782463 nonzeros kind: bipartite graph with auxiliary node data Journals: Slovenian journals 1999-2000 124-by-124 with 12068 nonzeros kind: undirected weighted graph with auxiliary node data Kohonen: Kohonen citation network 4470-by-4470 with 12731 nonzeros kind: directed graph with auxiliary node data Lederberg: Lederberg citation network 8843-by-8843 with 41601 nonzeros kind: directed multigraph with auxiliary node data NotreDame_actors: Barabasi's actor network (of www.imdb.com) 392400-by-127823 with 1470404 nonzeros kind: bipartite multigraph NotreDame_www: Barabasi's web page network of nd.edu 325729-by-325729 with 929849 nonzeros kind: directed graph NotreDame_yeast: Barabasi's yeast protein interaction 2114-by-2114 with 4480 nonzeros kind: undirected graph ODLIS: online dictionary of library & inf. sci 2909-by-2909 with 18246 nonzeros kind: directed multigraph Ragusa16: Ragusa set 24-by-24 with 81 nonzeros kind: directed weighted graph Ragusa18: Ragusa set 23-by-23 with 64 nonzeros kind: directed weighted graph Reuters911: Reuters news, Sept 11 to Nov 15, 2001 13332-by-13332 with 296076 nonzeros kind: temporal undirected weighted graph Roget: Roget's Thesaurus, 1879 1022-by-1022 with 5075 nonzeros kind: directed graph Sandi_authors: Klavzar bibliography 86-by-86 with 248 nonzeros kind: undirected weighted graph Sandi_sandi: Klavzar bibliography 314-by-360 with 613 nonzeros kind: bipartite graph SciMet: SciMet citation network 3084-by-3084 with 10413 nonzeros kind: directed multigraph with auxiliary node data SmaGri: SmaGri citation network 1059-by-1059 with 4919 nonzeros kind: directed multigraph with auxiliary node data SmallW: SmallW citation network 396-by-396 with 994 nonzeros kind: directed multigraph with auxiliary node data Stranke94: Slovene Parliamentary Parties 1994 10-by-10 with 90 nonzeros kind: undirected weighted graph Tina_AskCal: student govt, Univ. Ljubljana, 1992 (ask opin., recall) 11-by-11 with 29 nonzeros kind: directed graph Tina_AskCog: student govt, Univ. Ljubljana, 1992 (ask, recognized) 11-by-11 with 36 nonzeros kind: directed graph Tina_DisCal: student govt, Univ. Ljubljana, 1992 (discuss, recall) 11-by-11 with 41 nonzeros kind: directed graph Tina_DisCog: student govt, Univ. Ljubljana, 1992 (discuss, recog.) 11-by-11 with 48 nonzeros kind: directed graph USAir97: US Air flights, 1997 332-by-332 with 4252 nonzeros kind: undirected weighted graph USpowerGrid: US power grid 4941-by-4941 with 13188 nonzeros kind: undirected multigraph Wordnet3: Wordnet3 dictionary network 82670-by-82670 with 132964 nonzeros kind: directed weighted graph with auxiliary node data WorldCities: world city network 315-by-100 with 7518 nonzeros kind: weighted bipartite graph with auxiliary node data Zewail: Zewail citation network 6752-by-6752 with 54233 nonzeros kind: directed multigraph with auxiliary node data dictionary28: dictionary 52652-by-52652 with 178076 nonzeros kind: undirected graph divorce: divorce laws in the 50 US states 50-by-9 with 225 nonzeros kind: bipartite graph foldoc: free on-line dictionary of computing 13356-by-13356 with 120238 nonzeros kind: directed weighted graph football: World Soccer, Paris 1998 35-by-35 with 118 nonzeros kind: directed weighted graph geom: collaboration in computational geometry 7343-by-7343 with 23796 nonzeros kind: undirected weighted graph internet: connectivity of internet routers 124651-by-124651 with 207214 nonzeros kind: directed weighted graph patents: NBER US Patent Citations, 1963-1999, cites 1975-1999 3774768-by-3774768 with 14970767 nonzeros kind: directed graph with auxiliary node data patents_main: main NBER US Patent Citations, 1963-1999, cites 1975-1999 240547-by-240547 with 560943 nonzeros kind: directed weighted graph with auxiliary node data yeast: yeast protein interaction network 2361-by-2361 with 13828 nonzeros kind: undirected graph with auxiliary node data ================================================================================ ==> Citations/Cite2001.txt <== ================================================================================ Citation networks Graph Drawing 2001 Contest - Graph A Graph Drawing 2001 Contest task description for Graph A. Graph A in Pajek's format. Selected citation networks from Garfield's collection Citation networks in Pajek's format obtained from the Garfield's collection of citation network datasets produced using HistCite software. All of these networks are the result of searches in the WebofScience and are used with the permission of ISI of Philadelphia. Please acknowledge this when publishing results based on these data. SmallW: Papers that cite S Milgram's 1967 Psychology Today paper or use Small World in title, Tue Jul 23 13:35:11 2002 SmaGri: Citations to Small & Griffith and Descendants, Thu Nov 8 10:40:55 2001 SciMet: Articles from or citing Scientometrics, 1978-2000, Wed Jun 12 16:39:51 2002 Kohonen: Articles with topic "self-organizing maps" or references to "Kohonen T", Tue Jun 18 10:39:51 2002 Zewail: Articles citing and by AH Zewail, 1970-2002, Wed Jul 31 15:46:38 2002 Lederberg: Articles by and citing J Lederberg, 1945-2002, Wed Jul 31 13:40:22 2002 Some references * Batagelj V.: Some Mathematics of Network Analysis. Network Seminar, Department of Sociology, University of Pittsburgh, January 21, 1991. * Batagelj V., Mrvar A.: Graph Drawing Contest 2001 Layouts. * Garfield E, Sher IH, and Torpie RJ.: The Use of Citation Data in Writing the History of Science. Philadelphia: The Institute for Scientific Information, December 1964. * Garfield E.: From Computational Linguistics to Algorithmic Historiography, paper presented at the Symposium in Honor of Casimir Borkowski at the University of Pittsburgh School of Information Sciences, September 19, 2001. * Hummon N.P., Doreian P.: Connectivity in a Citation Network: The Development of DNA Theory. Social Networks, 11(1989) 39-63. Network Data, Pajek, Vlado ================================================================================ ==> Cities/Url.txt <== ================================================================================ http://www.lboro.ac.uk/gawc/data.html World City Relation Data This is the raison d'être of GaWC. For an introduction, see A Brief Guide to Quantitative Data Collection at GaWC, 1997-2001. INVITATION World city researchers with relational data on world cities are encouraged to post it here. Depositors may apply their own protocol for use by others. (Contact p.j.taylor@lboro.ac.uk) PROTOCOL for using GaWC data As part of our mission to promote the study of non-state relations we wish to encourage people to use the data posted here in their research and teaching. However, we do ask that, as a matter of courtesy bordering on ethics, the research project and the information gatherers are properly acknowledged along with GaWC itself. Each data set will have a statement as to whom to acknowledge and we ask that this be reproduced in every public use of the data. Data Set 1: US Cities: Surrogate Measures of Relations, 1990 (T.R. Longcore, C. McWilliams and P.J. Taylor) Data Set 2: London and New York: Surrogate Measures of Relations, 1997 Data Set 3: Randstad Cities: Surrogate Measures of Relations, 1970-95 (A.M. Beerda) Data Set 4: London's Relations with other Cities Using Producer Service Office Geographies (J.V. Beaverstock, R.G. Smith and P.J. Taylor) Data Set 5: US Cities: Law Firm Office Geographies, 1998 (J.V. Beaverstock, R.G. Smith and P.J. Taylor) Data Set 6: World Cities and Global Firms (P.J. Taylor and D.R.F. Walker) Data Set 7: Inter-City Matrices (P.J. Taylor and D.R.F. Walker) Data Set 8: World Cities: Regional Dimensions (P.J. Taylor, D.R.F. Walker and M. Hoyler) Data Set 9: Cities Mentioned in Advertisements in The Economist (May 2000 - January 2001) (P.J. Taylor) Data Set 10: The Relative Centrality of Cities Based upon Air Passenger Travel, 1977-1997 (M. Timberlake and D.A. Smith with K.-H. Shin) Data Set 11: World City Network: The Basic Data (P.J. Taylor and G. Catalano) Data Set 12: Global Network Service Connectivities for 315 Cities in 2000 (P.J. Taylor) DATA TOOLS Data Tool 1: Macro for Calculating Connectivities (max 254 cities x 255 firms) (E.C. Rossi and C.C.C. Rossi) Data Tool 2: Macro for Calculating Connectivities (max 1100 cities x 255 firms) (R. Aranya) World Cities and Global Firms P.J. Taylor and D.R.F. Walker These data consist of the distribution of offices for 46 'global' advanced producer service firms over 55 world cities. Global firms are defined by having offices in at least 15 different cities. World cities are from the GaWC inventory of world cities (see GaWC Research Bulletin 6). Service values for a firm in a city are given as 3, 2, 1 or 0 as defined in Data Set 4. These data are an experimental set of data derived from Data Set 4 (43 of the firms qualify as global) but with three additional law firms added which do not have London offices. For publications that make use of these data, see GaWC Research Bulletin 13 and GaWC Research Bulletin 17. Key to Data Set 6: ADVANCED PRODUCER SERVICE FIRMS (SECTOR, TABLE CODE AND FIRM) Accountancy KP KPMG CL Coopers & Lybrand EY Ernst & Young International AA Arthur Andersen PW Price Waterhouse Advertising GR Grey Worldwide DM DMB&B (MacManus Group) LH Lowe Howard -Spink SS Saatchi and Saatchi TH JWT (Thompson) OM Ogilvy & Mather Direct Worldwide DE Dentsu YR Young & Rubicam TM TMP PU Publicis AM Abbott Mead Vickers (BBDO) Banking and Finance HS HSBC BA Barclays NW NatWest Group SC Standard Chartered Group CR Creditanstalt-Bankverein DR Dresdner Bank Group MO J P Morgan PA Compagnie Financière de Paribas SA CS Credit Suisse BB BBV Group BT Banker's Trust UB UBS AB ABN-AMRO CB Citibank Law BM Baker & McKenzie WC White & Case LL Leboeuf, Lamb, Greene & MacRae CO Coudert Brothers SK Skadden Arps BC Bryan Cave DW Dorsey & Whitney GJ Graham & James HH Hogan & Hartson JD Jones, Day, Reavis & Pogue MC Miller, Canfield, Paddock & Stone SQ Squire, Sanders & Dempsey WE Wilson, Elser, Moskowitz, Edelman & Dicker AO Allen & Overy CC Clifford Chance FF Freshfields As per our data protocol, the following acknowledgement should accompany any public use of the data: ACKNOWLEDGEMENT: The data used is from Data Set 6 from the GaWC Study Group and Network (http://www.lboro.ac.uk/gawc/). It was created by P.J. Taylor and D.R.F. Walker as part of their project "World City Network: Data Matrix Construction and Analysis" and is based on primary data collected by J.V. Beaverstock, R.G. Smith and P.J. Taylor (ESRC project "The Geographical Scope of London as a World City" (R000222050)). ================================================================================ ==> Days/Days.txt <== ================================================================================ Pajek datasets Reuters terror news network (NOTE: renamed Reuters911 for UF sparse collection) Dataset days Description days.net undirected temporal network with 13332 vertices and 243447 edges. Days, DaysAll. Background Reuters terror news network Days.net in Pajek's format obtained from the CRA networks produced by Steve Corman and Kevin Dooley at Arizona State University. Please acknowledge this when publishing results based on these data. The Reuters terror news network is based on all stories released during 66 consecutive days by the news agency Reuters concerning the September 11 attack on the U.S., beginning at 9:00 AM EST 9/11/01. The vertices of a network are words (terms); there is an edge between two words iff they appear in the same text unit (sentence). The weight of an edge is its frequency. The network has n = 13332 vertices (different words in the news) and m = 243447 edges, 50859 with value larger than 1. There are no loops in the network. The network DaysAll.net contains the main connected component of the network obtained by transforming the Reuters terror news network into a combined network for all 66 days (union of all time points). It has 13308 vertices. The Reuters terror news network was used as a case network for the Viszards visualization session on the Sunbelt XXII International Sunbelt Social Network Conference, New Orleans, USA, 13-17. February 2002. History 1. 5-12. Dec 2001, networks constructed by S. Corman and his group 2. December 2001: CRA data transformed in Pajek format by V. Batagelj. References 1. Steven R. Corman, Timothy Kuhn, Robert D. Mcphee and Kevin J. Dooley (2002): Studying Complex Discursive Systems: Centering Resonance Analysis of Communication. (PDF) 2. Crawdad Technologies 3. Batagelj, V., & Mrvar, A. (2003): A density based approaches to network analysis: Analysis of Reuters terror news network, Ninth Annual ACM SIGKDD, Washington, D.C. (PDF); ( SVG) 4. Jeffrey C. Johnson and Lothar Krempel (2004): Network Visualization: The "Bush Team" in Reuters News Ticker 9/11-11/15/01. The Journal of Social Structure's, Vol. 5, No. 1. ( HTML) Pajek Data; 20. April 2006 / 27. January 2004 ------------------------------------------------------------------------------- When converted to a sparse adjacency matrix for the UF Sparse Matrix Collection, Day{i} is the graph of the ith day. The diagonal entry Day{i}(k,k) is 1 if word k appears in any news on the ith day. Note that it may not appear in conjunction with other words in the same sentence on that day. The sum of nnz(tril(Day{i})) for i=1:66 is 243,447. The overall matrix A is the sum of the Day{i} matrices. A(i,j) is the number of times words i and j appear in same sentence (for i not equal to j). A(k,k) is the number of days the word k appears in any news report. Note that this network has been renamed to Reuters911 here. ================================================================================ ==> EAT/EAT.txt <== ================================================================================ Pajek datasets EAT The Edinburgh Associative Thesaurus Dataset eat Description eatRS.net directed network with 23219 vertices and 325624 arcs (564 loops); stimulus X is associated with response Y N times. eatSR.net directed network with 23219 vertices and 325589 arcs (564 loops); response X is associated with stimulus Y N times. It seems that the SR network is incomplete and that it should be the inverse of RS network. EAT response-stimulus (ZIP, 1321K) EAT stimulus-response (ZIP, 1306K) Background The Edinburgh Associative Thesaurus (EAT) is a set of word association norms showing the counts of word association as collected from subjects. This is not a developed semantic network such as WordNet (3), but empirical association data. The traditional way to collect word association norms is to show or say a word to several people and ask them to say the word which first comes to their minds upon receiving the stimulus. The link established between the stimulus and the response is not semantically labelled (e.g. as synonym, antonym or by a case relation) and can only be regarded as an association. The Edinburgh association norms were collected by growing the network from a nucleus set of words. Responses were collected to words in this nucleus set, then these responses were used to obtain further responses, and so on. In fact the cycle was repeated about three times since by then the number of different responses was so large that they could not be re-used as stimuli. Data collection stopped when 8400 stimulus words had been used. Each stimulus word was presented to 100 different subjects, each of whom received 100 words. This gave rise to a total of 55732 nodes in the Thesaurus network. The subjects were mostly undergraduates from a wide variety of British universities. The age range of the subjects was from 17 to 22 with a mode of 19. The sex distribution was 64 per cent male and 36 per cent female. The data was collected between June 1968 and May 1971. The database consists of two files. The SR (stimulus-response) file, and the RS (response-stimulus) file. Where words have been truncated to 19 characters to save space the per cent character (%) has been placed as the 20th. The EAT here is that included in the MRC Psycholinguistic Database (4), for use with the other measures available there. EAT Data Collection Procedure (1) Stimulus words Since the objective was to obtain a reasonably large complete mapping of the associative network for a large set of words, a systematic procedure of 'growing' the network from a small nucleus was followed. At first responses were obtained from this nucleus set, then these responses were used as stimuli to obtain further responses, and so on. In fact, this cycle was repeated about three times, since by then the number of different responses was so large that they could not all be re-used as stimuli. The nucleus set was derived from (a) the 200 stimuli used in the Palermo and Jenkins (1964) normq (b) the 1,000 most frequent words of the Thorndike and Lorge (1944) word frequency count and (c) the basic English vocabulary of Ogden (1954). Data collection was stopped when 8,400 stimulus words had been used. Only a minimal amount of selection of stimuli was applied in each cycle of the data collection. Effectively all responses which were English words or meaningful verbal units were included, including some phrasal forms and numerals. The data cover a wide range of grammatical form classes and inflexional forms. Procedure Each stimulus word was presented to 100 different subjects. Each subject recieved a computer-printed sheet with 100 stimuli in randomised arrangement (to minimize priming effects). The total contribution of each subject was thus 100 responses. The verbal environment of each word for each subject was different. The instructions asked the subject to write down against each stimulus the first word it made him think of, working as quickly as possible. the total time spent on this task was measured, and most subjects completed the sheet in five to ten minutes. Most of the data was collected in a classroom setting under supervision. Sheets which had more than 25 percent blank responses were rejected and fresh data was collected. History 1. Original EAT: George Kiss, Christine Armstrong, Robert Milroy and J.R.I. Piper (collected between June 1968 and May 1971). 2. MRC Psycholinguistic Database Version modified by: Max Coltheart, S. James, J. Ramshaw, B.M. Philip, B. Reid, J. Benyon-Tinker and E. Doctor; made available by: Philip Quinlan. 3. The present version was re-structured and documented by Michael Wilson at the Rutherford Appleton Laboratory in 1988 (2). 4. transformed in Pajek format: V. Batagelj, 31. July 2003. References 1. Kiss, G.R., Armstrong, C., Milroy, R., and Piper, J. (1973) An associative thesaurus of English and its computer analysis. In Aitken, A.J., Bailey, R.W. and Hamilton-Smith, N. (Eds.), The Computer and Literary Studies. Edinburgh: University Press. 2. The present version of The Edinburgh Associative Thesaurus (ZIP, 2.7M) 3. WordNet 4. MRC Psycholinguistic Database 5. Coltheart, M. (1981) MRC Psycholinguistic Database. Quarterly Journal of Experimental Psychology, 3A, 497-505. 6. MRC Psycholinguistic Database 2 Pajek Data; 31. July 2003 ------------------------------------------------------------------------------- NOTE regarding conversion for UF sparse matrix collection: in the original data there are 325,624 weighted edges. Of those only 32 edges are duplicates, and all of them have identical edge weights as the edges they are duplicates of These extraneous edges have been removed, since this this appears to be a graph, not a multigraph. ================================================================================ ==> EVA/EVA.txt <== ================================================================================ Pajek datasets EVA Extraction, Visualization & Analysis of corporate inter-relationships Dataset EVA Description EVA.net directed network with 8343 vertices and 6726 arcs. EVA.net (ZIP, 204K); included also original files names.txt and ownership.txt. Background EVA / Denali is a multidisciplinary research project combining information extraction, information visualization, and social network analysis techniques to bring greater transparency to the public disclosure of inter-relationships between corporations. The project is described in the paper [1]. Abstract: We present EVA, a prototype system for extracting, visualizing, and analyzing corporate ownership information as a social network. Using probabilistic information retrieval and extraction techniques, we automatically extract ownership relationships from heterogeneous sources of online text, including corporate annual reports (10-Ks) filed with the U.S. Securities and Exchange Commission (SEC). A browser-based visualization interface allows users to query the relationship database and explore large networks of companies. Applying the system and methodology to the telecommunications and media industries, we construct an ownership network with 6,726 relationships among 8,343 companies. Analysis reveals a highly clustered network, with over 50% of all companies connected to one another in a single component. Furthermore, ownership activity is highly skewed: 90% of companies are involved in no more than one relationship, but the top ten companies are parents for over 24% of all relationships. We are also able to identify the most influential companies in the network using social network analysis metrics such as degree, betweenness, cutpoints, and cliques. We believe this methodology and tool can aid government regulators, policy researchers, and the general public to interpret complex corporate ownership structures, thereby bringing greater transparency to the public disclosure of corporate inter-relationships. Note that we do not have ownership relationships for all companies, so there will be companies without links. An arc (X,Y) from company X to company Y exists in the network if in the company X is an owner of company Y. Copyright 2002 by Denali Project. If you use this dataset in your research, please use the citation to paper [1] as the source of the data. "Denali" is the Native American name for the tallest peak in North America. It means "the Great One." If you have any questions, please contact: John Chuang, Mike Gebbie, Gabe Lucas, Kim Norlen. History 1. 2002 collection of original data by the EVA group; 2. March 6, 2004: original data transformed into Pajek format EVA.net by V. Batagelj. References 1. Kim Norlen, Gabriel Lucas, Mike Gebbie, and John Chuang. EVA: Extraction, Visualization and Analysis of the Telecommunications and Media Ownership Network. Proceedings of International Telecommunications Society 14th Biennial Conference, Seoul Korea, August 2002. (paper berkeley / local; slides berkeley / local) Pajek Data; 6. March 2004 ================================================================================ ==> Foldoc/Foldoc.txt <== ================================================================================ Pajek datasets FOLDOC Free On-line Dictionary of Computing Dataset Foldoc Description foldoc.net valued directed network with 13380 vertices and 120700 arcs, value is the multiplicity of arc. foldoc.net (ZIP, 517K) Background FOLDOC is a searchable dictionary of acronyms, jargon, programming languages, tools, architecture, operating systems, networking, theory, conventions, standards, mathematics, telecoms, electronics, institutions, companies, projects, products, history, in fact anything to do with computing. The dictionary has been growing since 1985 and now contains over 13000 definitions totalling nearly five megabytes of text. Entries are cross-referenced to each other and to related resources elsewhere on the net. An arc (X,Y) from term X to term Y exists in the network iff in the FOLDOC dictionary the term Y is used to describe the meaning of term X. Copyright 1993 by Denis Howe. Permission is granted to copy, distribute and/or modify the FOLDOC dictionary under the terms of the GNU Free Documentation License, Version 1.1 or any later version published by the Free Software Foundation. Please refer to the dictionary as "The Free On-line Dictionary of Computing, http://www.foldoc.org/, Editor Denis Howe" or similar. History 1. FOLDOC started in 1985 by Denis Howe; 2. in 1990 put on FTP, other sources included; 3. from 1994 available on the WWW, hundreds of contributors; 4. February/June 2002: Foldoc.net transformed in Pajek format and 'cleaned' by V. Batagelj and A. Mrvar. References 1. Denis Howe, Editor: FOLDOC (2002): Free on-line dictionary of computing. 2. V. Batagelj, A. Mrvar, M. Zavešnik: Network analysis of texts. Language Technologies 2002, Ljubljana, 14 - 15th October 2002, p. 143-148. 3. V. Batagelj, A. Mrvar, M. Zavešnik: Network analysis of dictionaries. Language Technologies 2002, Ljubljana, 14 - 15th October 2002, p. 135-142. Pajek Data; 30. January 2004 ================================================================================ ==> Football/Football.txt <== ================================================================================ Pajek datasets World Soccer Data Paris 1998 Dataset Football Description football.net - valued network with 35 vertices complete dataset Background Our network example describes the 22 soccer teams which participated in the World Championship in Paris, 1998. Players of the national team often have contracts in other countries. This constitutes a players market where national teams export players to other countries. Members of the 22 teams had contracts in altogether 35 countries. Counting which team exports how many players to which country can be described with a valued, asymmetric graph. The graph is highly unsymmetric: some countries only export players, some countries are only importers. History 1. Data collected by Lothar Krempel, October 5, 1999 2. Transformed in Pajek format by V. Batagelj, February 9, 2001. References 1. Dagstuhl seminar: Link Analysis and Visualization, Dagstuhl 1-6. July 2001 Pajek Data; 21. September 2005 ================================================================================ ==> Geom/Geom.txt <== ================================================================================ Pajek datasets Geom Collaboration network in computational geometry Dataset Geom Description Geom.net valued undirected network with 7343 vertices and 11898 edges; author X wrote a joint work with author Y; value is the number of joint works. Geom.net (ZIP, 139K) Background The network Geom.net is based on the file geombib.bib that contains Computational Geometry Database, version February 2002. The authors collaboration network in computational geometry was produced from the BibTeX bibliography [Beebe, 2002] obtained from the Computational Geometry Database geombib, version February 2002 [Jones, 2002]. Two authors are linked with an edge, iff they wrote a common work (paper, book, ...). The value of an edge is the number of common works. Using a simple program written in programming language Python, the BibTeX data were transformed into the corresponding network, and output to the file in Pajek format. The obtained network has 9072 vertices (authors) and 22577 edges (common papers or books) / 13567 edges as a simple network - multiple edges between a pair of authors are replaced with a single edge. The problem with the obtained network is that, because of non standardized writing of the author's name, it contains several vertices corresponding to the same author. For example: R.S. Drysdale, Robert L. Drysdale, Robert L. Scot Drysdale, R.L. Drysdale, S. Drysdale, R. Drysdale, and R.L.S. Drysdale; or: Pankaj K. Agarwal, P. Agarwal, Pankaj Agarwal, and P.K. Agarwal that are easy to guess; but an 'insider' information is needed to know that Otfried Schwarzkopf and Otfried Cheong are the same person. Also, no provision is made in the database to discern two persons with the same name. We manually produced the name equivalence partition and then shrank (in Pajek) the network according to it. The reduced simple network contains 7343 vertices and 11898 edges. It is a sparse network - its average degree is 2m/n = 3.24. History 1. Computational Geometry Database started in 1986 by merging two lists of references - one compiled by Edelsbrunner and van Leeuwen and the other by Guibas and Stolfi; 2. Computational Geometry Database, February 2002 Edition; 3. March-April 2002: Geom.bib transformed in Pajek format and 'cleaned' by V. Batagelj and M. Zaveršnik. References 1. Beebe, N.H.F. (2002): Nelson H.F. Beebe's Bibliographies Page. 2. Jones, B., Computational Geometry Database, February 2002; FTP / HTTP Pajek Data; 27. January 2004 ================================================================================ ==> GlossGT/GlossGT.txt <== ================================================================================ Pajek datasets Graph and Digraph Glossary Dataset glossGT Description glossGT.net mixed network with 72 vertices and 114 arcs and 4 edges; word X is related to word Y. glossTG.paj (4K) Background The network GlossGT.net is based on the file glossary.html containing Bill Cherowitzo's Graph and Digraph Glossary. An arc (X,Y) from term X to term Y exists in the network iff in the Graph and Digraph Glossary the term Y is used to describe the meaning of term X. History 1. 1998-2001, Bill Cherowitzo prepared the glossary. 2. Graph and Digraph Glossary transformed in Pajek format: Barbara Zemlji"c, 2. nov 2003. References 1. Bill Cherowitzo: Graph and Digraph Glossary, version 03-Feb-2001; (Course page) Pajek Data; 25. January 2004 ================================================================================ ==> HEP-th/HEP.txt <== ================================================================================ Pajek datasets KDD Cup 2003 High Energy Particle Physics (HEP) literature Dataset hep-th Description hep-th.net directed network with 27240 vertices and 342437 arcs (39 loops). hep-th-new.net directed network with 27770 vertices and 352807 arcs (39 loops). date-new.vec integer vector on 27770 vertices. year-new.vec integer vector on 27770 vertices. complete dataset (ZIP, 2607K) Background Citation data from KDD Cup 2003, a knowledge discovery and data mining competition held in conjunction with the Ninth Annual ACM SIGKDD Conference. The Stanford Linear Accelerator Center SPIRES-HEP database has been comprehensively cataloguing the High Energy Particle Physics (HEP) literature online since 1974, and indexes more than 500,000 high-energy physics related articles including their full citation tree. The network contains a citation graph of the hep-th portion of the arXiv. The units names are the arXiv IDs of papers; the relation is X cites Y . Note that revised papers may have updated citations. As such, citations may refer to future papers, i.e. a paper may cite another paper that was published after the first paper. The SLAC/SPIRES dates for all hep-th papers are given. Some older papers were uploaded years after their intial publication and the arXiv submission date from the abstracts may not correspond to the publication date. An alternative date has been provided from SLAC/SPIRES that may be a better estimate for the initial publication of these old papers. The first version of data was updated on May 12, 2003. hep-th.net X cites Y relation, first version. hep-th-new.net X cites Y relation, updated version. date-new.vec SLAC date of paper was transformed to the number of days since August 1, 1991, updated version. year-new.vec year from the SLAC date of paper, updated version. References 1. KDD Cup 2003 2. arXiv Transformed in Pajek format by V. Batagelj, 26. July 2003 ================================================================================ ==> IMDB/IMDB.txt <== ================================================================================ Pajek data set: IMDB, the Internet Movie Database, http://www.imdb.com. -------------------------------------------------------------------------------' Pajek network converted to sparse adjacency matrix for inclusion in UF sparse matrix collection, Tim Davis. For Pajek datasets, See V. Batagelj & A. Mrvar, http://vlado.fmf.uni-lj.si/pub/networks/data/. ------------------------------------------------------------------------------- A(i,j)=1 if actor j played in movie i. colname(j,:) is the name of the actor. Column j = 362,181 is Kevin Bacon. Year of movie i is year(i). category(i) gives the category of movie i, use code(category(i),:). Note that movie names are not provided. 1: Drama 2: Short 3: Documentary 4: Comedy 5: Western 6: Family 7: Mystery 8: Thriller 9: - 10: Music 11: Crime 12: Sci-Fi 13: Horror 14: War 15: Fantasy 16: Romance 17: Adventure 18: Animation 19: Action 20: Musical 21: Film-Noir 99: Unknown. Remember that in MATLAB, A(i,:) is slow to compute; A(:,i) is faster. If you want row i of a sparse matrix, access the ith column of the transpose instead. ------------------------------------------------------------------------------- A "Bacon number" is a measure of separation in the graph. Kevin Bacon has a Bacon number of zero. Any actor who played in a movie with Kevin Bacon has a Bacon number of 1. In general, an actor has a Bacon number equal to 1 + the minimum Bacon number of any actor he or she has been in a movie with. A similar definition can be extended to movies; a movie in which Kevin Bacon appears has a Bacon number of 0. In general, the Movie-Bacon number is the smallest Bacon number of any actor in that movie. The following code was used to compute the Bacon numbers: Bacon = Problem.aux.KevinBacon ; A = Problem.A ; [m n] = size (A) ; C = [speye(m) A ; A' speye(n)] ; x = zeros (m+n,1) ; B = inf * ones (m+n,1) ; x (m + Bacon) = 1 ; B (m + Bacon) = 0 ; tlen = 1 ; for k = 1:m+n x = x + C*x ; t = find (x) ; if (tlen == length (t)) break end tlen = length (t) ; B (t) = min (B (t), k) ; end MovieBacon = (B (1:m) - 1) / 2 ; ActorBacon = B (m+1:end) / 2 ; Note that the movie names have been intentionally omitted from this version of the data in the UF Sparse Matrix Collection, as has the name of movie code 9. The above MATLAB code is not terribly efficient; it makes k passes over the matrix, each taking O(nnz(A)) time. Fortunately, k is a small constant (8). A proper breadth-first search would take O(nnz(A)), regardless of k. ================================================================================ ==> Journals/Journals.txt <== ================================================================================ Pajek datasets Slovenian magazines and journals 1999 and 2000 Dataset Journals Description Revije.net - valued network with 124 vertices Revije.clu - partition with 124 vertices Revije.paj - Pajek project file with complete dataset. complete dataset (ZIP, 3K) Background Over 100.000 people have been asked which magazines and journals they read (survey conducted in 1999 and 2000, source CATI Center Ljubljana). They listed 124 different magazines and journals. The collected data can be represented as 2-mode network: Delo Dnevnik Sl.novice ... Reader1 X X ... Reader2 X ... Reader3 X ... ............ ..... ....... ......... ... Obtaining 1-mode network From 2-mode network reader/journal we generated ordinary network, where the vertices are journals * undirected edge with value a between journals means the number of readers * of both journals. loop on selected journal means the number of all * readers that read this journal. Obtained matrix (A): Delo Dnevnik Sl.novice ... Delo 20714 3219 4214 ... Dnevnik 15992 3642 ... Sl. novice 31997 ... ........ ...... ..... ..... ... The second ordinary network on readers would be huge (more than 100.000 vertices) containing large cliques (readers of particular journal). History 1. Transformed in journal X journal matrix, 26. December 2000. 2. Transformed in Pajek format by V. Batagelj, 21. December 2003. References 1. Dagstuhl seminar: Link Analysis and Visualization, Dagstuhl 1-6. July 2001 Pajek Data; 21. December 2003 ================================================================================ ==> KEDS/KEDS.txt <== ================================================================================ NOTE: the KEDS data has not been included. Pajek datasets KEDS The Kansas Event Data System Dataset KEDS Description GulfLDays.net directed multirelational temporal network with 174 vertices and 57131 arcs. From 'leads' Gulf event data, granularity is 1 day. GulfLMonths.net directed multirelational temporal network with 174 vertices and 57131 arcs. From 'leads' Gulf event data, granularity is 1 month. GulfLDow.net directed multirelational temporal network with 174 vertices and 57131 arcs. From 'leads' Gulf event data, in day of the week classes. GulfADays.net directed multirelational temporal network with 202 vertices and 304401 arcs. From Gulf event data, granularity is 1 day. GulfAMonths.net directed multirelational temporal network with 202 vertices and 304401 arcs. From Gulf event data, granularity is 1 month. LevantDays.net directed multirelational temporal network with 485 vertices and 196364 arcs. From Levant event data, granularity is 1 day. LevantMonths.net directed multirelational temporal network with 485 vertices and 196364 arcs. From Levant event data, granularity is 1 month. BalkanDays.net directed multirelational temporal network with 325 vertices and 78667 arcs. From Balkan event data, granularity is 1 day. BalkanMonths.net directed multirelational temporal network with 325 vertices and 78667 arcs. From Balkan event data, granularity is 1 month. GulfLDays.net (ZIP, 239K) GulfLMonths.net (ZIP, 197K) GulfLDow.net (ZIP, 213K) GulfADays.net (ZIP, 1078K) GulfAMonths.net (ZIP, 941K) LevantDays.net (ZIP, 855K) LevantMonths.net (ZIP, 735K) BalkanDays.net (ZIP, 335K) BalkanMonths.net (ZIP, 288K) Background KEDS - The Kansas Event Data System uses automated coding of English-language news reports to generate political event data focusing on the Middle East, Balkans, and West Africa. These data are used in statistical early warning models to predict political change. The ten-year project is based in the Department of Political Science at the University of Kansas; it has been funded primarily by the U.S. National Science Foundation. KEDS data sets from KEDS data collection. Gulf data set: This data set covers the states of the Gulf region and the Arabian peninsula for the period 15 April 1979 to 31 March 1999. The source texts prior to 10 June 97 were located using a NEXIS search command specifically designed to return relevant data. There are two versions of the data: a set coded from the lead sentences only (57,000 events), and a set coded from full stories (304,000 events). There are some errors in the GulfAll data. Events 118196 and 118197 have REUT-0 in place of the date; and in event 173526 the first actor is missing. In Gulf99All.dat the wrong dates are replaced with 890319, and the incomplete event is skiped. Levant data set: Folder containing WEIS-coded events (N=196,337) for dyadic interactions within the following set of countries: Egypt, Israel, Jordan, Lebanon, Palestinians, Syria, USA, and USSR/Russia. Coverage is April 1979 to June 2004. TABARI coding dictionaries are also included. There are some errors (333) in data set - relation codes 012], O24, O53, 213] are replaced with 012, 024, 053, 213 in Levant.dat. Some events don't have description codes - they are marked with *** in relation labels in *.net files. Balkans data set, 1989-2003: Folder containing WEIS-coded events (N = 78,667) for the major actors (including ethnic groups) involved in the conflicts in the former Yugoslavia. Coverage is April 1989 through July 2003. TABARI coding dictionaries are included in the folder. There are some errors (197) in data set - relation codes ---], O24, O53 are replaced with 000, 024, 053 in Balkan.dat. Some events don't have description codes - they are marked with *** in relation labels in *.net files. The original data sets are on MAC files. They should be saved as PC files before processing. History 1. Program Recode (in Delphi) by Vladimir Batagelj, Ljubljana, July 25, 2003 2. Program KEDSrec adapted for KEDS from Recode; Gulf (leads) data recoded into Pajek's format, by Vladimir Batagelj, Ljubljana, November 3, 2003 3. Support for multiple relations networks added to program KEDSrec by Vladimir Batagelj, Ljubljana, November 22, 2004 4. KedsR - functionality of KEDSrec implemented in R; Gulf (leads) data recoded into Pajek's format (days, months, day of the week), by Vladimir Batagelj, Ljubljana, November 27, 2004 5. WeisR - commands from KedsR adapted for WEIS format (similar to KEDS but TAB delimited); Balkan and Levant data recoded into Pajek's format (days, months), by Vladimir Batagelj, Ljubljana, November 28, 2004 6. Gulf99All is a large data set - sapply commands in KedsR had to be replaced by while loops; Gulf (all) data recoded into Pajek's format (days, months), by Vladimir Batagelj, Ljubljana, November 29, 2004 References 1. StuffIt - uncompress program for SIT files Pajek Data; 29. November 2004 / 24. November 2004 ================================================================================ ==> Mixed/Mixed.txt <== ================================================================================ Pajek datasets from different sources US power grid Dataset: USpowerGrid US power grid - unweighted network from Panayiotis Tsaparas' page. Adapted for Pajek by V. Batagelj, March 19, 2006 File: USpowerGrid.net - undirected network with 4941 vertices and 6594 edges Kleinberg's web graphs Dataset: California - Pages matching the query "California". This graph was constructed by expanding a 200-page response set to a search engine query 'California', as in the hub/authority algorithm. Obtained from Jon Kleinberg's page. Adapted for Pajek by V. Batagelj, March 19, 2006 File: California.net - directed network with 9664 vertices and 16150 arcs. References: Amy N. Langville and Carl D. Meyer: A Reordering for the PageRank problem (March 2004) PDF J. Kleinberg. Authoritative sources in a hyperlinked environment. Proc. 9th ACM-SIAM Symposium on Discrete Algorithms, 1998. Extended version in Journal of the ACM 46(1999). PDF Dataset: Epa - Pages linking to www.epa.gov. This graph was constructed by expanding a 200-page response set to a search engine query, as in the hub/authority algorithm. Obtained from Jon Kleinberg's page. Adapted for Pajek by V. Batagelj, March 19, 2006 File: Epa.net - directed network with 4772 vertices and 8965 arcs. Stanford web graphs Dataset: StanfordWeb - Stanford Web Matrix. This graph was obtained from a September 2002 crawl (281903 pages, 2382912 links). The matrix rows represent the inlinks of a page, and the columns represent the outlinks. Downloaded from Sepandar D. Kamvar's page. Adapted for Pajek by V. Batagelj, March 19, 2006 File: StanfordWeb.net - directed network with 281903 vertices and 2382912 arcs. References: Sepandar D. Kamvar, Taher H. Haveliwala, Christopher D. Manning, and Gene H. Golub, "Exploiting the Block Structure of the Web for Computing PageRank." Preprint (March, 2003). PDF Dataset: StanfordBerkeleyWeb - Stanford-Berkeley Web Matrix. This graph was obtained from a December 2002 crawl (685230 pages, 8006115 links). The matrix rows represent the inlinks of a page, and the columns represent the outlinks. Downloaded from Sepandar D. Kamvar's page. Adapted for Pajek by V. Batagelj, March 19, 2006 NOTE by Tim Davis: nodes 683,447 to 685,230 in the Stanford Berkeley web data, discussed above, are not part of the true results. Kamvar's MATLAB script for processing the data deletes those nodes. This graph is already in the UF Sparse Matrix Collection, in the correct size. Also note that Kamvar doesn't consider multiple links from page i to j to be significant. Thus, duplicate edges (i,j) in this graph should be ignored. The graph in the UF collection is thus binary. File: StanfordBerkeleyWeb.net - directed network with 685230 vertices and 8006115 arcs. References: Sepandar D. Kamvar, Taher H. Haveliwala, Christopher D. Manning, and Gene H. Golub, "Exploiting the Block Structure of the Web for Computing PageRank." Preprint (March, 2003). PDF World City Network Dataset: WorldCities These data constitute the empirical basis for measuring the world city network as described in P.J. Taylor (2004) World City Network: A Global Urban Analysis (London: Routledge). The data were produced by P.J. Taylor and G. Catalano. These data were collected in 2000. They are the service values (indicating the importance of a city in the office network of a firm) of 100 global service firms distributed across 315 cities worldwide. All firms supply advanced producer services (accountancy, advertising, banking/finance, insurance, law, and management consultancy) through offices in at least 15 cities (including at least one in each of Pacific Asia, western Europe and northern America). The following coding information is given to help understand and evaluate the data. Downloaded from GaWC Data Set 11 page. Transformed in Pajek format by V. Batagelj, March 20, 2006 Files: WorldCities.net - two-mode network with 415=315+100 vertices and 7518 arcs. WorldCities.mat - same network in Pajek matrix format. WorldCities.net - type of service partition: 0 - city; 1 - accountancy; 2 - advertising; 3 - banking/finance; 4 - insurance; 5 - law; 6 - management consultancy. References: Taylor, P.J. (2004): World city network: a global urban analysis. London and New York: Routledge. Pajek Data; 19. March 2006 ================================================================================ ==> ND/ND.txt <== ================================================================================ Pajek datasets Notre Dame Self-Organized Networks Database Datasets NDwww, NDactors, NDyeast Description Notre Dame Self-Organized Networks: 1. NDwww.net directed network with 325729 vertices and 1497135 arcs (27455 loops); page X is linked to page Y. 2. NDactors.net undirected two-mode network with 520223 vertices (392400 players, 127823 movies) and 1470418 edges; player X plays in movie Y. 3. NDyeast.net undirected network with 2114 vertices and 2277 edges (74 loops); protein X interacts with protein Y. NDwww.net (ZIP, 2050K) NDactors.net (ZIP, 4150K) NDyeast.net (ZIP, 7K) Background The networks ND*.net are based on the files from Notre Dame Self-Organized Networks Database. To transform the data into Pajek format: vertex 0 was replaced by the vertex number equal to the number of vertices in a network; Pajek keywords were inserted; and the network was saved in the short (as lists of neighbors) format. 1. World-Wide-Web:: Each number represents webpage within nd.edu domain. Arcs: From page -> To page Réka Albert, Hawoong Jeong and Albert-László Barabási: Diameter of the World Wide Web, Nature 401, 130 (1999) [ PDF ] See also a decompostion of this network in V. Batgelj, A. Mrvar: How to analyze large networks with Pajek? 2. Actor: Actor network data: (based on www.imdb.com) In the original ND network file: each line corresponds to one movie, each number represents actor: number_1 number_2 ... number_k (k actors who play in the same movie). Albert-László Barabási, Réka Albert: Emergence of scaling in random networks, Science 286, 509 (1999) [ PDF ] 3. Protein Interaction Network for Yeast: Each number represents protein in protein interaction network of yeast. Edges: From protein -> To protein. For other datasets used in supplementary material, please refer indicated references. Hawoong Jeong, Sean Mason, Albert-László Barabási and Zoltán N. Oltvai: Centrality and lethality of protein networks, Nature 411, 41 (2001) [ PDF ] See also Yeast data History 1. Notre Dame Networks Database put on WWW by the Notre Dame team, 2001; 2. 23-25. July 2001: ND nets transformed in Pajek format by V. Batagelj. 3. 23. May 2004: ND nets in Pajek format transformed in short (lists of neighbors) Pajek format by V. Batagelj. References 1. Self-Organized Networks Database, University of Notre Dame. Copyright Extract from the Notre Dame Networks Database page: "... Feel free to use these data in your research." Mail to Hawoong Jeong (author of original ND networks). Pajek Data; 23. May 2004 ================================================================================ ==> ODLIS/ODLIS.txt <== ================================================================================ Pajek datasets ODLIS Online Dictionary of Library and Information Science Dataset odlis Description odlis.net directed network with 2909 vertices and 18419 arcs (5 loops). odlis.net (ZIP, 62K) Background The network Odlis.net is based on the ODLIS: Online Dictionary of Library and Information Science. version December 2000. ODLIS is designed to be a hypertext reference resource for library and information science professionals, university students and faculty, and users of all types of libraries. The primary criterion for including a new term is whether a librarian or other information professional might reasonably be expected to encounter it at some point in his (or her) career, or be required to know its meaning in the course of executing his or her responsibilities as a librarian. The vocabulary of publishing, printing, book history, literature, and computer science has been included when, in the author's judgment, a definition might prove helpful, not only to library and information professionals, but also to laypersons. An arc (X,Y) from term X to term Y exists in the network iff in the ODLIS dictionary the term Y is used to describe the meaning of term X. ODLIS is the work of Joan M. Reitz, Assistant Professor/Instruction Librarian at the Ruth A. Haas Library, Western Connecticut State University (WCSU) in Danbury, CT. Ms. Reitz holds an M.L.I.S. degree (1991) from the University of Washington in Seattle and an M.A. degree (1998) in European History from Western Connecticut State University. Her primary research interests are the history of the book and history of political and social revolutions. History 1. ODLIS began at the Haas Library in 1994 as a five-page photocopied handout intended for undergraduates not fluent in English, and students with limited exposure to library terminology. 2. In 1996, it was expanded and converted to HTML format for installation on the WCSU Libraries HomePage under the title Hypertext Library Lingo: A Glossary of Library Terminology. 3. In 1997, many more hypertext links were added and the format improved in response to suggestions from users. 4. During the summer of 1999, several hundred terms and definitions were added, and a generic version created which omitted all references to the specific conditions and practices at the Haas Library. 5. In the fall of 1999, the glossary was expanded to 1,800 terms, renamed to reflect its extended scope, and copyrighted. 6. In February, 2000, ODLIS was indexed in Yahoo! under "Reference - Dictionaries - Subject." It is also indexed in the WorldCat database in OCLC FirstSearch. 7. During the year 2000, the dictionary was expanded to 2,600 terms. On average, it has received over 6,200 visits per month since January 2, 2000. 8. December 2000: ODLIS transformed in Pajek format and 'cleaned' by A. Mrvar and V. Batagelj. References 1. Joan M. Reitz (2002): ODLIS: Online Dictionary of Library and Information Science. Pajek Data; 27. January 2004 ================================================================================ ==> Patents/Patents.txt <== ================================================================================ Pajek datasets Patents The NBER U.S. Patent Citations Data File Dataset patents Description patents.net directed network with 3,774,768 vertices and 16,522,438 arcs (1 loop). NET, NAM, Year, Date, Cat, Class, Country, Subcat, AppYear Patents NET / main subnetwork Background The network Patents is based on the The NBER U.S. Patent Citations Data File, version 2001. These data comprise detail information on almost 3 million U.S. patents granted between January 1963 and December 1999, all citations made to these patents between 1975 and 1999 (over 16 million), and a reasonably broad match of patents to Compustat (the data set of all firms traded in the U.S. stock market). These data are described in detail in Hall, B. H., A. B. Jaffe, and M. Tratjenberg (2001). "The NBER Patent Citation Data File: Lessons, Insights and Methodological Tools." NBER Working Paper 8498. ALL USERS OF THESE DATA SHOULD READ THIS PAPER, AND SHOULD CITE IT AS THE SOURCE OF THE DATA Further documentation on uses of the patent citation data, including the methodology paper and a CD containing the complete dataset itself, is available in the book Patents, Citations and Innovations: A Window on the Knowledge Economy by Adam Jaffe and Manuel Trajtenberg, MIT Press, Cambridge (2002). The book may be ordered from MIT Press. ISBN 0-262-10095-9. History 1. Data produced by United States Patent and Trademark Office 2. Hall, B. H., A. B. Jaffe, and M. Tratjenberg prepared the NBER dataset, 2001 3. July 10, 2003: NBER data transformed in Pajek format by V. Batagelj. References 1. Hall, B. H., A. B. Jaffe, and M. Tratjenberg (2001): The NBER U.S. Patent Citations Data File. 2. United States Patent and Trademark Office, Patent Number Search 3. Batagelj V. (2003): Efficient Algorithms for Citation Network Analysis. Pajek Data; 27. January 2004 ------------------------------------------------------------------------------- NOTE regarding conversion for UF sparse matrix collection: in original the data there are 14,973,817 edges (unweighted). Of this, 3050 are duplicates This graph is binary; the duplicates have been removed. Also, the original data has auxiliary data for all 6,009,554 US Patents in this time period. This patent network has only 3,774,768 patents, and the auxiliary data (appyear, class, etc.) is matched here to the nodes of the graph. ================================================================================ ==> Roget/Roget.txt <== ================================================================================ Pajek datasets Roget Roget's Thesaurus, 1879 Dataset Roget Description roget.net directed network with 1022 vertices and 5075 arcs (1 loop); word X is related to word Y. Roget.net (ZIP, 17K) Background The network Roget.net is based on the file roget.dat from the Stanford GraphBase that contains cross-references in Roget's Thesaurus, 1879. Dr. Peter Mark Roget (1779-1869) philologist, scientist, physician. The name Roget could soon become a virtual synonym for the word "synonym". For those who use Roget's Thesaurus it is one of the three most important books ever printed...along with The Bible and Webster's Dictionary. In order to communicate one's exact intention...or one's precise meaning, the Thesaurus, being a list of synonyms or verbal equivalents, is a necessary tool. The first draft of the Thesaurus was written in 1805, two years before Webster started on his dictionary. However for a period of 47 years Dr. Roget used his manuscript as his personal, secret, treasure trove. Not until he was 73 years old did he decide to reveal and publish this great manuscript. Since 1852, Roget's Thesaurus has never been out of print. In fact, each succeeding edition has increased the popularity of the work. The original 15,000 words included in the 1805 manuscript has increased to over a quarter of a million in the 1992 edition (the tenth printing). With such an increase in size, it is encouraging to notice that the basic content still remains intact..... for example, where the 1805 Thesaurus traces the word: existence: "Ens, entity, being, existence, essence...", the 1992 Thesaurus contains existence: "existence, being, entity, ens,...essence..." Each vertex of the graph corresponds to one of the 1022 categories in the 1879 edition of Peter Mark Roget's Thesaurus of English Words and Phrases, edited by John Lewis Roget. An arc goes from one category to another if Roget gave a reference to the latter among the words and phrases of the former, or if the two categories were directly related to each other by their positions in Roget's book. For example, the vertex for category 312 (`ascent') has arcs to the vertices for categories 224 (`obliquity'), 313 (`descent'), and 316 (`leap'), because Roget gave explicit cross-references from 312 to 224 and 316, and because category 312 was implicitly paired with 313 in his scheme. History 1. Original Roget's Thesaurus was published in 1852. 2. Peter's son John Luis Roget published the second, improved edition in 1879. 3. Project Gutenberg Roget's Thesaurus (1911 edition) put into electronic format in 1991. 4. Graph Roget.dat of cross-references based on the second edition was produced for Stanford GraphBase (SGB) in 1992/3. 5. MICRA (Pat Cassidy) prepared the electronic version of the 1911 Roget's Thesaurus that is widely available on the internet. 6. SGB Roget.dat transformed in Pajek format: A. Mrvar, 5. December 1996. References 1. Peter Mark Roget: Roget's Thesaurus of English Words and Phrases 2. Project Gutenberg: Roget's Thesaurus 3. Donald E. Knuth: The Stanford GraphBase: A Platform for Combinatorial Computing . New York: ACM Press, 1993 4. The Stanford GraphBase: roget.dat, version 15.6.1993 5. Pat Cassidy: MICRA / Factotum 6. CIDE (Collaborative International Dictionary of English), GNU 1996-2002 Pajek Data; 23. January 2004 ================================================================================ ==> Sandi/Sandi.txt <== ================================================================================ REFERENCES DATASET SANDI DESCRIPTION 2-mode 674×314 network. BACKGROUND These data were obtained from the bibliography of the book Imrich W, Klavžar S. (1999) Graph products. The result is a author-by-paper 2-mode network: arc (i,j) - author i is the (co)author of the paper j. DERIVED DATA AUTHORS authors network REFERENCES * Imrich W, Klavžar S. (1999). Graph products. References. ================================================================================ ==> Stranke94/Slovene_Parties.txt <== ================================================================================ Pajek datasets Slovene Parliamentary Parties 1994 Dataset Stranke94 Description Stranke94.net - valued signed network with 10 vertices Stranke94.net Background Relations between Slovene parliamentary political parties: * SKD - Slovene Christian Democrats; * ZLSD - Associated List of Social Democrats; * SDSS - Social Democratic Party of Slovenia; * LDS - Liberal Democratic Party; * ZSESS - first of two Green Parties, separated after 1992 elections; * ZS - second Green Party; * DS - Democratic Party; * SLS - Slovene People's Party; * SNS - Slovene National Party; * SPS SNS - a group of deputies, former members of SNS, separated after * 1992 elections were estimated by the members of the Slovene National Parliament. So the respondents were well informed and competent to give such estimations. In the questionnaire designed by a group of experts on Parliament activities, some questions about the political space and its dimensions were included and the following question about relations between parliamentary political parties: If various criteria (or various dimensions of the political space) are taken into account, some parties are by average closer than others. How would you personally estimate distances between pairs of parties in the political space? Please, estimate the distance between each pair of parties on the scale from -3 to +3, where: -3 means that parties are very dissimilar; -2 means that parties are quite dissimilar; -1 means that parties are dissimilar; 0 means that parties are neither dissimilar nor similar (somewhere in between); +1 means that parties are similar; +2 means that parties are quite similar; +3 means that parties are very similar. To collect estimations each respondent was given a 10-party by 10-party table with empty cells in the upper triangle. The diagonal and the lower triangle were coloured in black. Each respondent had to estimate relations between 45 pairs of parties. The measures of central tendency were computed on the basis of the estimations given by 72 out of 90 members of the Parliament. 17 members of the Parliament were not available at the time of interviewing and one refused respond. 64 respondents out of the 72 estimated all 45 requested party relations. Only two out of 8 respondents with missing values have a large number of missing values (namely 40), the rest of them have 5 to 10 missing values. As far as the parties (variables) are considered, there are from 0 to 8 missing values and a recognisable pattern: relations involving ZS (which had no representatives in the Parliament at the time of interviewing) include from 6 to 8 missing values, others from 0 to 3 missing values. The weights of arcs in the network are averages of values multiplied by 100 and rounded. The missing values were excluded. History 1. Stran format by V. Batagelj, 19. October 1994. 2. Transformed in Pajek format by V. Batagelj, 15. February 2004. References 1. Samo Kropivnik and Andrej Mrvar: An Analysis of the Slovene Parliamentary Parties Network. Developments in Statistics and Methodology. (A. Ferligoj, A. Kramberger, editors) Metodološki zvezki 12, FDV, Ljubljana, 1996, p. 209-216. 2. Patrick Doreian and Andrej Mrvar(1996): A Partitioning Approach to Structural Balance. Social Networks, 18, p. 149-168. Pajek Data; 15. February 2004 ================================================================================ ==> Tina/Tina.txt <== ================================================================================ Pajek datasets Student Government of the University of Ljubljana / 1992 Dataset Tina Description DisCal.net - network with 11 vertices and 41 arcs DisCog.net - network with 11 vertices and 48 arcs AskCal.net - network with 11 vertices and 29 arcs AskCog.net - network with 11 vertices and 36 arcs AnsCalT.net - network with 11 vertices and 41 arcs AnsCogT.net - network with 11 vertices and 42 arcs DisCalSn.net - network with 11 vertices and 41 arcs Tina.paj - Pajek project file with complete dataset. complete dataset (ZIP, 3K) DisCalSn.net Background In the experiment two alternative methods were used for collection of network data: Recall: Members of the group were asked to identify the members of their egocentric networks by memory. The criteria for enumeration was frequency of the recalled communications. Recognition: The list of all members of the group was given to each member. They were first asked to identify who they communicate with and than to select the persons they communicate with most often. The number of listed persons was not limited in any method. The analyzed network consisted of communication interactions among twelve members and advisors of the Student Government at the University in Ljubljana. The results of the measurement are not real interactions among actors but cognition about communication interactions. Data were collected with face to face interviews which lasted from 20 to 30 minutes. Interviews were conducted in May 1992. Communication flow among actors was identified by three questions: Who of the members and advisors of the Student government do you (most often) informally discuss with? Which members and advisors of the Student Government do you (most often) ask for an opinion? Which of the members and advisors of the Student Government (most often) ask you for an opinion? The content of the communication flow was limited to the matters of the Student Government. The time frame was also defined: questions were referred to the six months period (from the formation of the government to the day of the interview). Only one respondent listed all the others under the recognition method for two relations (discussion, asking for an opinion). For that respondent the first group was defined arbitrary. The cut point was determined by the average number of selected persons at the recognition method (3.5 for the first relation and 4.5 for the second). One respondent refused to cooperate in the experiment. As he was not considered in the analysis, the network consists of eleven actors. History 1. Transformed in Stran format by V. Batagelj, 28. July 1993. 2. Transformed in Pajek format by V. Batagelj, 8. August 2003. Files DisCal.net - discussion, recall DisCog.net - discussion, recognition AskCal.net - asking for an opinion, recall AskCog.net - asking for an opinion, recognition AnsCalT.net - being asked for an opinion, recall (transposed) AnsCogT.net - being asked for an opinion, recognition (transposed) DisCalSn.net - discussion, recall / short names Tina.paj - Pajek project file with complete dataset. References 1. Valentina Hlebec: Recall Versus Recognition: Comparison of the Two Alternative Procedures for Collecting Social Network Data. Developments in Statistics and Methodology. (A. Ferligoj, A. Kramberger, editors) Metodološki zvezki 9, FDV, Ljubljana, 1993, p. 121-129. Pajek Data; 8. August 2003 ================================================================================ ==> WassermanFaust/WassermanFaust.txt <== ================================================================================ NOTE: the Wasserman/Faust networks have not been included. Pajek datasets from the book Social Network Analysis: Methods and Applications Wasserman and Faust, 1994 Wasserman and Faust datasets Dataset WaFa There are five network examples: Krackhardt's high-tech managers, Padgett's Florentine families, Freeman's EIES network, Countries trade network, and Galaskiewicz's CEOs and clubs network. Description HighTech.paj - multirelational network with 21 vertices and 190+102+20 arcs Padgett.paj - multirelational network with 16 vertices and 30+40 arcs EIES.paj - multirelational network with 48/32 vertices and 695+830/460 arcs Trade.paj - multirelational network with 24 vertices and 307+307+310+135+369 arcs CEOs.net - two-mode network with 41 vertices and 98 edges. Background Complete descriptions of these data, including references for the original sources of the data, can be found in Chapter 2 (pages 59- 66) and Appendix B (pages 738-755) of Wasserman and Faust. The original data in ASCII, UCINET and KrackPlot formats are available at INSNA. History 1. INSNA page. 2. Transformed in Pajek format by V. Batagelj, 15. January 2005. Files HighTech.paj - Krackhardt's High-tech managers Padgett.paj - Padgett's Florentine Families EIES.paj - Freeman's EIES network Trade.paj - Countries trade data CEOs.net - Galaskiewicz's CEOs and clubs. References 1. Stanley Wasserman, Katherine Faust: Social Network Analysis: Methods and Applications. CUP, 1994. Pajek Data; 15. January 2005 ================================================================================ ==> Wordnet/Wordnet.txt <== ================================================================================ NOTE: this is a binary graph in the Pajek dataset, but where each edge has a label (not a weight) in the range 1 to 9. The following labels are used: 1 hypernym pointer 2 entailment pointer 3 similar pointer 4 member meronym pointer 5 substance meronym pointer 6 part meronym pointer 7 cause pointer 8 grouped pointer 9 attribute pointer This is not a multigraph. There are no edges (i,j) between the same nodes with the same label. Thus, in the sparse matrix, the edge weight A(i,j) represents the label 1 through 9 of edge (i,j). No loss of information occurs in this translation. The above table is in aux.edgecode(1:9,:). Each node is a word in a dictionary. aux.category(i) gives the category of the word: 1: n (noun?) 63099 words 2: v (verb?) 4496 words 3: a (adjective?) 5501 words 4: r (?) 2846 words 5: s (?) 6728 words. ================================================================================ ==> yeast/yeast.txt <== ================================================================================ Pajek datasets Protein-protein interaction network in budding yeast Dataset Yeast Description yeastS.net network with 2361 vertices and 7182 edges (536 loops). yeastL.net network with 2361 vertices and 7182 edges (536 loops). yeast.clu partition of vertices. yeast.paj Pajek project file with complete dataset. complete dataset (ZIP, 134K) Background Interaction detection methods have led to the discovery of thousands of interactions between proteins, and discerning relevance within large-scale data sets is important to present-day biology. The dataset consists of protein-protein interaction network described and analyzed in (1) and available as an example in the software package - PIN (2). PIN class encoding: 1 - T, 2 - M, 3 - U, 4 - C, 5 - F, 6 - P, 7 - G, 8 - D, 9 - O, 10 - E, 11 - R, 12 - B, 13 - A. yeastS.net X interacts with Y relation, short names. yeastL.net X interacts with Y relation, long labels. yeast.clu PIN class partition of vertices, see encoding. yeast.paj Pajek project file with complete dataset. References 1. Shiwei Sun, Lunjiang Ling, Nan Zhang, Guojie Li and Runsheng Chen: Topological structure analysis of the protein-protein interaction network in budding yeast. Nucleic Acids Research, 2003, Vol. 31, No. 9 2443-2450 (PDF). 2. Software package Protein Interaction Network PIN Transformed in Pajek format by V. Batagelj, 25. July 2003 ================================================================================ ==> GD problems <== ================================================================================ These graphs are from the Graph Drawing 1995-2006 Contests. GD97_b: NOTE regarding conversion for UF sparse matrix collection: in original the data every edge appears exactly twice, with the same edge weight. It could be a multigraph, but it looks more like a graph. The duplicate edges are removed in this version. You can always add them back in yourself; just look at 2*A.```
Displaying collection matrices 1 - 20 of 76 in total
Id Name Group Rows Cols Nonzeros
1456 California Pajek 9,664 9,664 16,150
1457 Cities Pajek 55 46 1,342
1458 CSphd Pajek 1,882 1,882 1,740
1459 dictionary28 Pajek 52,652 52,652 178,076
1460 divorce Pajek 50 9 225
1461 EAT_RS Pajek 23,219 23,219 325,592
1462 EAT_SR Pajek 23,219 23,219 325,589
1463 EPA Pajek 4,772 4,772 8,965
1464 Erdos02 Pajek 6,927 6,927 16,944
1465 Erdos971 Pajek 472 472 2,628
1466 Erdos972 Pajek 5,488 5,488 14,170
1467 Erdos981 Pajek 485 485 2,762
1468 Erdos982 Pajek 5,822 5,822 14,750
1469 Erdos991 Pajek 492 492 2,834
1470 Erdos992 Pajek 6,100 6,100 15,030
1471 EVA Pajek 8,497 8,497 6,726
1472 FA Pajek 10,617 10,617 72,176
1473 foldoc Pajek 13,356 13,356 120,238
1474 football Pajek 35 35 118
1475 GD00_a Pajek 352 352 458 | 18,595 | 76,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.900513 |
https://abc7.com/mega-millions-results-megamillions-million/2335791/ | 1,713,929,410,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00273.warc.gz | 61,992,977 | 57,768 | # Mega Millions and Powerball 2019: How do the jackpots grow?
Thursday, March 28, 2019
Each jackpot starts at \$40 million, but how does it grow to hundreds of millions or more?
Four nights a week, lotto players around the country have the opportunity to become instant millionaires through either the Powerball (on Wednesdays and Saturdays) or the Mega Millions (on Tuesday and Fridays).
Between December and March, the Powerball jackpot continued to grow until a lucky single ticket won \$768.4 million on Wednesday.
It's the third largest jackpot in the country's history. The largest, a \$1.586 billion Powerball jackpot that was split three ways, was won in 2016. The second largest, a \$1.537 billion Mega Millions jackpot, was won by a single anonymous winner in South Carolina in 2018.
RELATED: \$1.5B lottery winner vows to donate to South Carolina charities, including Hurricane Florence relief
Here's a look at how each game's jackpot grows to such a large sum.
How to play
In both games, tickets cost \$2 and each ticket has six numbers. Players can either pick out the numbers themselves or have a machine generate random numbers.
To win the Mega Millions jackpot, players must match the white balls (which are numbered 1-70) and the yellow Mega Ball, which can be any number 1-25. Chances of hitting all six numbers are 1 in 302,575,350.
To win the Powerball jackpot, winners must guess the five white ball numbers from a drum of 69 numbers, and one red ball out of a drum of 26 numbers. Chances of hitting all six numbers are 1 in 292,201,338.
Why they both get so big
RELATED: How much can \$1.6 billion buy?
For both games, each jackpot starts out at \$40 million. Each time a drawing is made without a jackpot winner, the pot rolls over to the next drawing. When more tickets are purchased, the jackpot grows.
In 2015, Powerball added more numbers to the drawing, making it easier to win smaller prizes, but also less likely to win the jackpot. The odds of winning the jackpot went from 1 in 175 million to 1 in 292 million.
"More play happens at the higher level so your odds of sharing it are greater at the greater jackpot levels," Powerball chairman Charlie McIntyre told ABC News.
Similarly, in 2017, Mega Millions expanded the possible numbers that could be picked for the jackpot but decreased the possible numbers for the first five. This meant that the odds got better for winning a lower tier prize but worse for winning the jackpot.
For both games, since it's harder to win, the jackpot can grow to large sums as no one wins drawing after drawing. This prompts more people to buy tickets, leading to the ballooning jackpot prizes.
RELATED: What to consider if you win the lottery
RELATED: The largest Mega Millions, Powerball jackpots in history | 631 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.95973 |
http://www.caraudio.com/forums/archive/index.php/t-471478.html | 1,519,235,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813691.14/warc/CC-MAIN-20180221163021-20180221183021-00654.warc.gz | 405,085,284 | 3,196 | PDA
View Full Version : round port to slot port conversion
THATpurpleKUSH
04-14-2010, 12:35 AM
I am trying to design a ported box for my subs but I'm having some trouble. The manufacturer requires a 4" x 14" round port. When converting from round to slot port, do I only need to worry about surface areas of the port? So each 4"w port would have 12.57^3 in of surface area, so I would need 25.14^3 of port area for 2 woofers in 1 chamber? Then the length would determine the tuning freq?
Fuknmovin
04-14-2010, 04:29 PM
The box's net size determines the port, you need a certain port area for the given box dimensions, you figure it like this
Net box x 16 = port area necessary for max output (or close to it). This is the formula Digital Designs uses and it works very well for most applications. I never really had great luck going by the manufacturers recommendations, they tend to leave a little on the table most of the time...
So basically for a 2.0 ^3 chamber after displacements of woofer and port, you need about 32 SQ" of port opening to be in the upper efficiency range of that particular box.
pawn man
04-14-2010, 05:24 PM
THATpurpleKUSH
04-14-2010, 06:22 PM
Could anyone answer the question better? And are there any differences in port/sub placement on the enclosure? (rear subs and port vs. up subs rear port or up port rear subs)
Fuknmovin
04-14-2010, 06:29 PM
The location of the ports is pretty car specific. In a trunk car (which I think yours is) unless you wall off the trunk behind the seat completely usually the best placement is sub forward port back, but sometimes sub forward port forward will work better, you kind of need to play around with your particular car unless someone here has done that specific car....
basebalz13
04-14-2010, 06:48 PM
trunk car, sub back port back. Lol.
Also when you say a "round" port do you mean a Aero port? If so then you dont need as much port area compared to a slot port
THATpurpleKUSH
04-14-2010, 11:04 PM
The owners manual only says 4"x14" but nothing about an aero port. I would like to go with a slot port, but just needed clarification on conversion of the given dimensions for a round port to a slot setup.
SCcobra4me
04-15-2010, 12:28 AM
Not to hijack but I have a very similar question. I'm getting two subs and the factory specs recommend 20-30 inches of port, what does this mean??
Fuknmovin
04-16-2010, 10:42 AM
That's the "area" of the port opening....if you had a port that was 4"x5" then you would have 20" of port area....
91caprice
04-17-2010, 12:26 AM
use the calculator on the 12volt. just remember to use 14 - 16^in of port per cubic foot of air space(so 14 X net cubic feet). multiply height times width of the slot port opening to get square inches. input your width and height and net box volume and desired tuning freq and it will give you the required port length.
Subwoofer Enclosure Calculators, Fraction to Decimal, Parallel, Series, Port Length and Volume Calculators (http://www.the12volt.com/caraudio/boxcalcs.asp) | 812 | 3,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-09 | latest | en | 0.909713 |
https://projecteuler.net/problem=361 | 1,566,682,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00352.warc.gz | 586,372,737 | 3,000 | ## Subsequence of Thue-Morse sequence
Published on Sunday, 4th December 2011, 04:00 am; Solved by 256;
Difficulty rating: 90%
### Problem 361
The Thue-Morse sequence {Tn} is a binary sequence satisfying:
• T0 = 0
• T2n = Tn
• T2n+1 = 1 - Tn
The first several terms of {Tn} are given as follows:
01101001100101101001011001101001....
We define {An} as the sorted sequence of integers such that the binary expression of each element appears as a subsequence in {Tn}.
For example, the decimal number 18 is expressed as 10010 in binary. 10010 appears in {Tn} (T8 to T12), so 18 is an element of {An}.
The decimal number 14 is expressed as 1110 in binary. 1110 never appears in {Tn}, so 14 is not an element of {An}.
The first several terms of An are given as follows:
n 0 1 2 3 4 5 6 7 8 9 10 11 12 … An 0 1 2 3 4 5 6 9 10 11 12 13 18 …
We can also verify that A100 = 3251 and A1000 = 80852364498.
Find the last 9 digits of $\sum \limits_{k = 1}^{18} {A_{10^k}}$. | 341 | 969 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-35 | longest | en | 0.908599 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-10-change-expressed-as-a-percent-practice-and-problem-solving-exercises-page-148/12 | 1,537,476,455,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156613.38/warc/CC-MAIN-20180920195131-20180920215531-00525.warc.gz | 746,497,272 | 14,581 | ## Algebra 1
There was a decrease of about $2\%$.
Change in value: $=\text{new amount} - \text{original amount} \\= 1975-2008 \\= -33$ There was a decrease. Percent Change: $=\dfrac{\text{change in value}}{\text{original amount}} \\=\dfrac{-33}{2008} \\\approx -0.02 \\\approx -2\%$ There was a decrease of around $2\%$. | 110 | 321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-39 | longest | en | 0.762046 |
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