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https://forums.adobe.com/thread/1870607 | 1,519,439,698,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00145.warc.gz | 630,401,052 | 28,524 | 5 Replies Latest reply on Jun 16, 2015 5:50 AM by rob day
# How do I know the actual size of type in a "web" intent document?
I'm designing in a document set for iPhone 6plus pixel resolution (1334x750) and need to know how the type size translates. For example, when I designate some text to be 30 pt., in the real world when viewed on the iPhone it's something much smaller than that... perhaps 12 pt. This has something to with the number number of pixels in canvas... but how do I convert?
• ###### 1. Re: How do I know the actual size of type in a "web" intent document?
Web intent docs are set to export at 72 ppi which means 1 pt = 1 pixel.
• ###### 2. Re: How do I know the actual size of type in a "web" intent document?
So how do I calculate the real point size of type? I'm not good at math, I need to be told how to do it.
• ###### 3. Re: How do I know the actual size of type in a "web" intent document?
If you're creating for a device then you test it on the device to see how it looks. If it looks too big, make it smaller...too small, make it larger.
There is no rule for this other than, if it looks good it is good.
• ###### 4. Re: How do I know the actual size of type in a "web" intent document?
You mean the regular iPhone6? The 6plus runs at 1920x1080 at 401ppi and the 6 runs at 1334x750 at 336
I'm designing in a document set for iPhone 6plus pixel resolution (1334x750) and need to know how the type size translates
You could get a visual translation by setting your InDesign Zoom Level so the window's physical diagonal dimension matches the iPhone6's 4.7" diagonal.
The percentage for the iPhone6 1334x750 page would be (72 / 326) * 100, or 22%
For the 6plus it would be (72 / 401) * 100, or 18.0%, but you would have to set up your document to be 1920x1080 pixels
• ###### 5. Re: How do I know the actual size of type in a "web" intent document?
So how do I calculate the real point size of type? I'm not good at math, I need to be told how to do it.
For output and display at 100%, InDesign sets the physical dimension of a pixel as 1/72", so the relationship between type's printed size and the iPhone's displayed size is a simple ratio. For the iPhone6 72/326 gets the reduction in relative size from print to display, and 326/72 would be a multiplier. So to display 12pt type the way it would print, set it at 54.4 pt (12 x 4.53). Or, to display a 1 inch square as 1 inch, scale it by 453%.
For the iPhone6 Plus the multiplier is 401/72 or 5.57
Apple - iPhone 6 - Technical Specifications | 702 | 2,540 | {"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-2018-09 | latest | en | 0.875323 |
https://www.jiskha.com/display.cgi?id=1295231464 | 1,516,596,396,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00214.warc.gz | 938,809,163 | 3,607 | # NWFSC
posted by .
A 277 oscillator has a speed of 122 when its displacement is 2.90 and 47.5 when its displacement is 6.80. What is the oscillator's maximum speed?
## Similar Questions
1. ### physics
An oscillator has an amplitude of 3.2. At this instant displacement of the oscillator is 1.4. What are the two possible phases of the oscillator at this instant?
2. ### physics
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5. ### Physics
A 300 g oscillator has a speed of 95.4 cm/s when its displacement is 3.00 cm and 71.4 cm/s when its displacement is 6.00 cm. What is the oscillator's maximum speed?
6. ### Physics
A 395cm oscillator has a speed of 104.5cm/s when its displacement is 2.80cm and 74.5cm/s when its displacement is 6.60cm. What is the oscillator's maximum speed? | 410 | 1,476 | {"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-2018-05 | latest | en | 0.801869 |
https://www.crazy-numbers.com/en/4900 | 1,721,076,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00774.warc.gz | 637,027,599 | 3,594 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 4900: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 4900: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 4900
Is 4900 a prime number? No
Is 4900 a perfect number? No
Number of divisors 27
List of dividers
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1, 2, 4, 5, 7, 10, 14, 20, 25, 28, 35, 49, 50, 70, 98, 100, 140, 175, 196, 245, 350, 490, 700, 980, 1225, 2450, 4900
Sum of divisors 12369
Prime factorization 22 x 52 x 72
Prime factors
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2, 5, 7
## How to write / spell 4900 in letters?
In letters, the number 4900 is written as: Four thousand nine hundred. And in other languages? how does it spell?
4900 in other languages
Write 4900 in english Four thousand nine hundred
Write 4900 in french Quatre mille neuf cents
Write 4900 in spanish Cuatro mil novecientos
Write 4900 in portuguese Quatro mil novecentos
## Decomposition of the number 4900
The number 4900 is composed of:
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
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2 iterations of the number 0 : ... Find out more about the number 0
Other ways to write 4900
In letter Four thousand nine hundred
In roman numeral
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MMMMCM
In binary 1001100100100
In octal 11444 | 917 | 3,039 | {"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} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.692542 |
https://encyclopediaofmath.org/index.php?title=Riesz_representation_theorem&diff=next&oldid=27453 | 1,653,528,804,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00471.warc.gz | 285,822,953 | 8,190 | # Difference between revisions of "Riesz representation theorem"
2010 Mathematics Subject Classification: Primary: 28A33 [MSN][ZBL]
A central theorem in classical measure theory, sometimes called Riesz-Markov theorem, which states the following. Let $X$ be a compact Hausdorff topological space, $C(X)$ the Banach space of real valued continuous functions on $X$ and $L: C(X)\to \mathbb R$ a continuous linear functional which is nonnegative, i.e. such that $L(f)\geq 0$ whenever $f\geq 0$. Then there is a Radon measure $\mu$ on the $\sigma$-algebra of Borel sets $\mathcal{B} (X)$ such that $L (f) = \int_X f\, d\mu \qquad \forall f\in C (X)\, .$ The theorem is often stated in a more general version for locally compact Hausdorff spaces $X$, of which the statement above is a simple corollary (cp, with Section 2.14 of [Ru])
An analogous statement which is commonly referred to as Riesz representation theorem is that, under the assumptions above, the dual of $C(X)$ is the space $\mathcal{M}^b (X)$ of $\mathbb R$-valued measures with finite total variation (cp. with Signed measure for the relevant definitions). Combined with the Radon-Nikodým theorem, this amounts to the following alternative statement: for any element $L\in (C(X))'$ there are a Radon measure $\mu$ and a Borel function $g$ such that $|g|=1$ $\mu$-a.e. and $L (f) = \int_X fg\, d\mu\qquad \forall f\in C(X)\, .$
The statement can be also generalized to a similar description of the dual of $C(X,B)$ when $B$ is Banach space. For instance, if $B$ is a finite-dimensional space, then for any $L\in C(X,B)'$ there is a Radon measure $\mu$ on $X$ and a Borel measurable map $g: X\to B'$ such that $\|g\|_{B'}=1$ $\mu$-a.e. and $L (f) = \int_X g (f)\, d\mu \qquad \forall f\in C(X, B)\, .$
#### References
[AmFuPa] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Bo] N. Bourbaki, "Elements of mathematics. Integration" , Addison-Wesley (1975) pp. Chapt.6;7;8 (Translated from French) MR0583191 Zbl 1116.28002 Zbl 1106.46005 Zbl 1106.46006 Zbl 1182.28002 Zbl 1182.28001 Zbl 1095.28002 Zbl 1095.28001 Zbl 0156.06001 [DS] N. Dunford, J.T. Schwartz, "Linear operators. General theory" , 1 , Interscience (1958) MR0117523 [Bi] P. Billingsley, "Convergence of probability measures" , Wiley (1968) MR0233396 Zbl 0172.21201 [Ma] P. Mattila, "Geometry of sets and measures in euclidean spaces". Cambridge Studies in Advanced Mathematics, 44. Cambridge University Press, Cambridge, 1995. MR1333890 Zbl 0911.28005 [Ru] W. Rudin, "Real and complex analysis". McGraw-Hill Book Co., New York-Toronto, Ont.-London 1966 MR021052 Zbl 0142.01701
How to Cite This Entry:
Riesz representation theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Riesz_representation_theorem&oldid=27453 | 916 | 2,953 | {"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": 2, "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-2022-21 | latest | en | 0.840865 |
https://www.soccerwidow.com/football-gambling/betting-knowledge/value-betting-academy/odds-calculation-en/cluster-tables-guide-over-under-betting/ | 1,571,005,233,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00313.warc.gz | 1,198,401,113 | 30,790 | # How to Use Soccerwidow’s Over/Under Betting Cluster Tables
Soccerwidow’s Cluster Tables are an essential tool for identifying value bets and creating a profitable portfolio in the Over/Under ‘X’ Goals market.
They rely on dividing historical data (previous five complete seasons) into clusters according to the “HO/AO quotient” to provide a reliable comparison with future matches under analysis.
Betting odds are a mixture of statistical fact and public opinion (people voting with their money) as to what the likely outcome of an event will be.
Introducing the HO/AO quotient allows us to to ‘cluster’ groups of past matches and with that, to quantify the mutual relationship between the number of goals scored in matches and the strength of the teams involved. (The HO/AO quotients are a practical application of corellation).
This allows us to put an upcoming game into perspective.
In other words, we use the group of past matches bearing HO/AO quotients most similar to the match under analysis in order to make more accurate assessments about its likely number of goals.
The number of goals scored↔team strength relationship is a hugely strong correlation known to the bookmakers and used to a greater or lesser degree when setting their opening odds.
However, as public opinion (market pressure) leads to ‘errors’ in market pricing (odds), using the knowledge of the correlation allows us to spot ‘value’.
Following on from our Betting with Cluster Tables introductory article, here are the four simple steps needed to calculate pinpoint zero odds for intelligent value betting decisions in the Over/Under ‘X’ Goals market:
1. Find the Home and Away Odds
2. Calculate the HO/AO Quotient
3. Record the Cluster Table Results
4. Perform the Final Calculations
We shall look at each of these steps using the English Premier League as example.
Let’s look in fine detail at the EPL match: West Ham vs. Southampton from 31st March, 2018.
## Step 1 – Find the Home and Away Odds
One of the most important components of the Cluster Tables is the HO/AO quotient (home odds divided by away odds), hence the need for both odds before referring to the tables.
To find the latest, up-to-date odds for any fixture you can employ bookmakers or betting exchanges of your choice, or make use of an odds comparison site. For the sake of our example, we are using OddsPortal.com as they are the only site showing time-stamped odds to support our illustrations.
Oddsportal Ante Post Odds Composite Screenshot – West Ham vs. Southampton 31/03/2018
The screenshot on the left is a composite image showing both the home and away odds just before this game started. Click on the image to enlarge it in a new tab.
Betsafe offered a price of 2.90 on West Ham six minutes before kick-off, whilst 5Dimes gave best price of 2.73 on Southampton seconds before the start.
Despite the multitude of odds movements throughout the entire ante post market, you will find in the vast majority of cases that the relationship between the home and away odds will stay pretty much the same throughout the ante post market.
Usually, the HO/AO quotient locates the match firmly between the two ends of a cluster, and the quotient tends to remain in that same cluster group no matter how the odds move during the lead up to kick-off.
This means that the timing of the analysis is not critical; you can perform it at any period during the ante post market before the match kicks-off. And of course, bet placement timing then also becomes just a matter of finding market odds containing value.
Timing only becomes an issue in the very rare event that the HO/AO quotient places the match very close to one of the ends of the cluster range for either team. It is then always wise to check odds close to kick-off to ensure that you have the match in the right HO/AO clusters for both teams.
Okay, we have our home and away odds – onto the next step…
## Step 2 – Calculate the HO/AO Quotient
Easy! Take a calculator or enter the figures into a spreadsheet and just divide the home odds by the away odds to provide a quotient.
In this case, the quotient is: 2.90 divided by 2.73 = 1.0623 (rounded-up)
## Step 3 – Identify the Relevant Cluster and Percentage Result
Cross-checking any team’s HO/AO quotient against their statistical percentages for any of the over/under 0.5 to 6.5 options in any match under analysis is extremely easy.
Within the Cluster Table for the appropriate league, click on the Betting Tables tab. This reveals a one-touch spreadsheet for obtaining both team’s results.
Here is the table of figures for West Ham (click on the image below to enlarge it in a new tab – and then use the magnifier to enlarge again if necessary):
Over/Under Cluster Table – Betting Table Screenshot
To change the team, simply click on the orange team name in the top left-hand corner to access the drop-down menu of all teams with five-season data sets.
By clicking on the team you are looking for, the figures in the table will automatically revert to those of that team.
The first half of the sheet contains the home figures: Summary at the top, Over ‘X’ Goals, and then Under ‘X’ Goals. The bottom three panels are the away results.
For this example, let’s decide to go for the most popular ‘Over 2.5 Goals’ bet.
For West Ham’s home figures, using the second panel from the top, you can see on the left-hand side in dark blue, their dedicated HO/AO clusters.
The HO/AO quotient we have calculated is 1.0623 and this fits neatly into the third cluster down. Looking under ‘Running Total Probability’, we simply record the percentage figure, in this case, 73.9%.
Here are West Ham’s top two panels with the relevant cluster row and percentage result for Over 2.5 Goals highlighted:
Over/Under Cluster Table – Betting Tables Tab – West Ham’s Cluster Row Highlighted
And after changing the team name, here are Southampton’s away figures in their fourth and fifth panels:
Over/Under Cluster Table – Betting Tables Tab – Southampton’s Cluster Row Highlighted
As you can see, Southampton’s Over 2.5 Goals percentage for an HO/AO quotient of 1.0623 in their away games is shown as 33.3% in the second cluster down.
You will also note that the HO/AO quotient fitted very firmly inside the relevant cluster group of both teams, and not too close to its edges (West Ham’s cluster group was 0.7301-1.9345, whilst Southampton’s was 0.8211-1.3880).
Again, you will rarely encounter situations that will need monitoring – most games will see the same cluster groups used despite the odds movements throughout the ante post market. This means that neither analysing nor placing the bets is time-sensitive, and both exercises need not be performed at the same time either.
Next Page: Step 4 – Do the Maths!
#### learn to think like a bookmaker!### Do You Know our Bestsellers? They Are Helping Thousands Betters World Wide! True Odds & Value Detector: League Games with H2H History £29.90 Fundamentals of Sports Betting £59.00 Over Under Cluster Tables from £7.50
Last Update: 15 September 2018
### 31 Responses to “How to Use Soccerwidow’s Over/Under Betting Cluster Tables”
1. Andy
23 February 2019 at 6:48 pm #
Today I backed o2.5 on st johnstone v aberdeen @ 2.4 I did calculate the zero odds to be 1.71 so it seemed like an extremely good value bet.
I calculated ho/ao as: 3.95/2.18 = 1.8119
However do you think I made some kind of error, I think I may have done seeing as there was such a large discrepancy in available odds and zero odds?
• Soccerwidow
24 February 2019 at 8:20 am #
Hi Andy,
you probably didn’t make a mistake as there are occasional discrepancies like that in the market. However, finding a value bet, of course, doesn’t mean that each bet wins.
2. Andrew Shipley
5 February 2019 at 4:23 pm #
Approximately how many bets would you need to do before you would expect to see a positive ROI?
I have done 20 bets and I am – 3 points, I haven’t become emotional about it yet because I am using very low stakes on purpose.
I am only backing or laying o2.5 at odds about 1.5 to 2.5
I just want to know whats a reasonable expectation?
30 50 75 100 bets?
• Soccerwidow
9 February 2019 at 7:14 am #
Hi Andrew,
you should normally see a positive ROI pretty straight away; of course, only if you have chosen bets with value and you don’t start with a losing streak straight away.
Have a look at my reports I did in real time in 2012 when I was writing match previews for Betfair Germany: Soccerwidow’s Value Betting Results: 274 Bets, from 07/12/2011 to 30/06/2012
Of course, nobody can predict with certainty how bumpy the ride is going to be and it depends a lot of the probabilities you are playing.
The rule of thumb is: The lower the probabilities the bumpier your ride, the more difficult on your nerves.
3. Andy
9 December 2018 at 11:31 am #
Yes I have worked through it all, the only issue is that what it seems to be based on (odds inflation) appears to be contradictory to what I have been used to using in my other betting.
I was always told that “odds inflation” does not exist by the kick off time on an exchange.
The price it kicks off at was always viewed as infallible on the exchange because the free battle between backers and layers always settles it fairly.
It seems to be saying that thousands of people cant get it right between them.
• Soccerwidow
9 December 2018 at 12:06 pm #
Just to name a few factors that influence the prices (odds)…
When odds are initially set, they are based on statistics plus the expected market demand as judged by the bookmakers.
When the exchanges open the markets, traders come in (including bots).
One hour before kick-off, the public emotions become very strong when the team news is being published.
• Andy
9 December 2018 at 2:53 pm #
So you think its best to wait until team news is published before we enter the bets?
• Soccerwidow
10 December 2018 at 6:27 am #
Hi Andy,
Your question was about ‘odds inflation’ and I answered with a few examples that affect odds (= prices of bets) (i.e. rising or falling). The fluctuations go in both directions simultaneously. When one side goes up, the other goes down. Further detail on this topic is contained within the pages of the Odds Calculation coursebook.
4. Andy
9 December 2018 at 1:08 am #
I am still unsure how I can make money long term from this unless I beat the closing line of the exchange or a sharp bookmaker like Pinnacle. I have always been told that beating the closing line of an exchange is the only way to take value since the market knows everything its possible to know about an event and thus it is reflected in the price.
• Soccerwidow
9 December 2018 at 6:48 am #
Hi Andy, how to make profits when using the Cluster Tables is described in minute detail in the Over/Under course book: https://www.soccerwidow.com/products/over-under-goals-fundamentals-sports-betting/
The market ‘knows’ everything possible but the different markets are connected and especially public expectations put a lot of price pressure on the odds. So, for example, if Bayern Munich plays away from home a weaker team at their home grounds, everybody is expecting to see lots of goals in that match; however, analysing the upcoming match statistically, the weaker team may be actually pretty strong in defending their own goal, even against teams like Bayern. In this example, the Over odds would be all inflated and you can recognise that using the Cluster Tables.
Have you worked through your course book from the beginning to the end, page by page?
5. Andy
5 December 2018 at 9:10 pm #
Hi,
I have noticed that sometimes the ho/ao result will be in a certain cluster – however when I look at the odds they will be in another cluster, the “average” odds section I am talking about.
So what should I do when they do not both align?
• Soccerwidow
8 December 2018 at 9:50 am #
Hi Andy, I assume that you’re referring to the ‘home odds (average)’ display on the ‘Betting Tables’ tab?
They are for information purposes only, to help people to get a feeling what the HO/AO clusters mean, especially when looking through upcoming matches, in order to identify quickly matches that may be in the correct cluster. Nothing else.
No, they do not have to align. It’s only the HO/AO quotient that matters.
6. Andy
25 November 2018 at 7:40 pm #
“However, if you bet on portfolios then it is sometimes a good idea to include all bets above ‘zero’ odds, even if they are ‘only’ within the ‘fair’ odds range to spread the risk as much as possible.”
Is what you are saying here one approach is just to take all the bets that are above zero odds and disregard standard deviation in order to smooth out the variance?
• Soccerwidow
26 November 2018 at 12:58 pm #
Hi Andy, you’ve just bought the course book a few days ago. Please be patient and first work through the whole course book; page by page; chapter by chapter. It is a lot to comprehend but once you succeed to understand you will be able to answer your question yourself. To truly master betting a huge learning curve is required. It takes time and a lot of hard work. You are just at the beginning.
Good luck!
7. Andy
25 November 2018 at 2:34 pm #
So on my first attempt to calculate anything, I found that the upcoming 1/12 match Werder vs Bayern the fair odds of o2.5 is exactly 2.0, and the betfair odds is 1.48 which is way out in the wrong way for a value bet.
Anyone can comment if I got it right? TY
• Soccerwidow
26 November 2018 at 1:06 pm #
Andy, please work through the course book! You will find the answer to that there.
It’s an away game for Bayern and they are far less strong when playing away than when playing on home grounds. Therefore, the odds you calculated for O2.5 are probably right. Bayern Munich away games are always excellent examples where the bookmakers take huge advantage of the lack of knowledge to calculate probabilities by the majority of punters.
8. Andrew
25 November 2018 at 12:19 pm #
Anyone been able to get these tables to work on a mac?
I tried numbers, google sheets but they both don’t work properly.
Only other thing is if I get excel for mac.
• Soccerwidow
26 November 2018 at 1:03 pm #
Hi Andrew, I’m sorry if you don’t get the tables to work on mac. Do you have anywhere where you have access to a PC?
Alternatively, just print out the course book. The layout allows for double-sided printing and binding. To work through the course book you don’t necessarily require the Excel tables.
9. AM0751
3 May 2018 at 9:42 am #
Oh, another question popped up, this one seems more legit…
I have noticed that the %CV values for over and under x goals are the same in Japan, Ireland, and Iceland for example.
Yet in South-Korea and Sweden for example, they are different.
am having trouble explaining that to myself. Can you shed some light on this?
;
Thank you 🙂
• Soccerwidow
3 May 2018 at 9:53 am #
Hi AM0751,
the CV% should normally be the same for Over and Under as you see it in Japan, Ireland and Iceland for example.
South Korea changed the format in 2014: from former 266 matches per annum it reduced to 228. Same applies to Sweden; it’s not as massive as in South Korea but there were two void matches in the five years of data.
Therefore the uncertainty (deviation) in South Korea is much bigger than in Sweden; and these two have a higher uncertainty that the ‘regular’ leagues like Japan, Ireland and Iceland.
I hope that makes sense.
• AM0751
3 May 2018 at 7:43 pm #
Thank you very much for your answer. I have one more question.
On another page, right winger says that generally speaking, odds that have smaller yield (Value 2) than %CV values are candidates for lay bets.
Can you maybe expand on that a bit more? Let me give you an example.
For arguments sake, say we have a 75% chance on either over or under (zero odds).
Makes 1.33
The %CV is 2%
So fair odds are 1.3 to 1.37
Do the lay candidates need to be below 1.3 to qualify?
Or is a Value2 smaller than %CV already enough to have it qualify as a lay bet?
• Soccerwidow
4 May 2018 at 9:21 am #
It really all depends on the strategy you’re following and it’s all about risk diversification. The problem with ‘zero odds’ is that it is very unlikely that the observed results of the forthcoming season will be exactly matching the ‘zero’ probability (the mean probability of the last 5 seasons). Even with using the standard deviation to calculate the expected (= most probable) range there is only a 95% confidence level that the results observed in reality will be within the expected range.
Therefore, if ‘Lay’ or ‘Back’ within the range really depends on the strategy you’re following. It’s like throwing a dice… you’ve got there a 50/50 chance to have head or tails. That the distribution will be even 50/50 you will only observe, say, after you’ve thrown the dice ten thousands times. But if you only throw them 100 times (like you do in betting) then the observed distribution is likely to be uneven. It can be something like 55/45 or 45/55. Therefore, it really depends on your strategy if you decide to ‘lay’ heads or if you ‘back’ them during the 100 throws.
Generally speaking, everything below the ‘zero’ odds is likely to be a lay candidate, everything above a back candidate. But I say ‘likely’ not ‘is’.
10. AM0751
2 May 2018 at 9:30 am #
Please disregard my last comment. I have been looking for an answer all over, and at the moment I decide to ask you about it, I see the comment below me addressed the exact same issue. Sorry about that, and have a nice day.
• Soccerwidow
2 May 2018 at 9:36 am #
You too, have a nice day! I still replied to your comment as it was straight to the point and it will hopefully save other people’s time to search for an answer.
11. AM0751
2 May 2018 at 9:25 am #
Hello soccerwidow,
A quick question.
You mention zero odds here, but there is no mention here of deviation at all, while it is prominently mentioned in the course. Should that not be the “extra” final step?
• Soccerwidow
2 May 2018 at 9:35 am #
Hi AM0751,
this article is an abridged version (or call it, a subsequent example) from the course book. Problem is that the deviation part in the course book contains numerous pages; at least 16, to explain the whole concept in minute detail… No chance to be able to summarize that in a short article.
We will certainly at some stage publish mor articles on that subject but good things take their time. 😉
12. Simon
5 April 2018 at 11:49 pm #
Hi RW,
I read and studied the betting over/under course book back in 2016/17 and for a while regularly used the cluster tables for finding over/under bets before focusing on the HDAFU tables.
In the course book I recall that standard deviation was a consideration and that when using the tables to find the percentage/odds for say over 2.5 goals, you would obtain the number for the home and away side, add them together, divide by 2, then apply the relevant standard deviation to that number. I recall the explanation in the course being that having markets odds over and above the zero odds is not in and of itself a marker of value; instead the market odds should be above the zero odds after applying the correct standard deviation.
I couldn’t see any mention of this in the article, so was wondering is this step no longer necessary in the method to find betting opportunities? I remember you guys saying the course was being updated, so am not sure if the method has been revised. It’s been a while since I used these tables and haven’t purchased an updated course, so I may be out of the loop a little!
Thanks.
Simon.
• Soccerwidow
6 April 2018 at 5:17 am #
Hi Simon,
there is still the distinction between ‘zero’ odds and ‘fair’ odds ranges necessary, as explained in the course. And it still applies that odds within the ‘fair’ odds range (‘zero’ probability ± deviation) don’t really contain long-term value.
However, if you bet on portfolios then it is sometimes a good idea to include all bets above ‘zero’ odds, even if they are ‘only’ within the ‘fair’ odds range to spread the risk as much as possible.
The problem with writing articles for the website is that they have to be kept short & sweet as well easily understandable. The course explains value betting in much greater length and covers loads of different topics necessarily to understand odds calculation in order to find an edge in the bookmaker market. That is unfortunately not possible when writing a 1,000 words article for Soccerwidow.
However, what we should probably do is to add a box at the end of these articles with a disclaimer that the information provided in the article is an abridged version and/or a subsequent example from the course book. Thanks for pointing that out! We will keep that in mind. 🙂
By the way, how did you get on with the cluster tables before switching to the HDA tables?
• Simon
22 April 2018 at 6:18 pm #
Hi Soccerwidow,
Well, what I did was to backtest 2 seasons worth of results for 6 top European Leagues:
I decided to just focus on over/under 2.5 goals, no matter what the probability of that outcome would be. So I have an overall set of results for the seasons I tested, and also results when just focusing on bets with probability of success of more than 50%.
2015/16:
Overall results:
349 bets: 170 wins, 179 losses, 48.71% hit rate, £1100 profit
Over 50% chance:
272 bets: 148 wins, 124 losses, 54.41% hit rate , £2526 profit
2016/17:
Overall results:
359 bets: 182 wins, 177 losses, 50.70% hit rate, £4700 profit
Over 50% chance:
291 bets: 149 wins, 142 losses, 51.20% hit rate, £1513
It was strange to see those results because in the 15/16 season the bets with over 50% chance did better than betting on everything. The reverse was true when looking at 16/17 season.
I delved into individual game date to see why that may have been the case, and in the 16/17 season, there were many games that hit with big odds, especially for the under 2.5. There was one game at Barcelona which they won 1-0 and the odds for under 2.5 are around 7!
So it all really depends on what bets you want to take into your portfolio which will shape your final results. I can imagine it would be nerve wracking to watch Barcelona at home playing a minnow and praying for under 2.5 goals!
• Soccerwidow
2 May 2018 at 6:49 am #
Hi Simon, it’s nice to hear that your back testing was so successful. 🙂
Can you perhaps try to avoid watching games that you have placed bets on (e.g. your example with Barcelona)?
I personally find it much more rewarding to carry out the calculations by Friday evening, place the bets, have a nice weekend off, and then check the results on Sunday evening.
This is the main reason why, over the whole website, I specialise in Ante-Post betting and do not indulge trading. This is to stop (or at least, control) these nerve wracking experiences.
But then again I don’t remember the last time I watched a match. 😉
• Right Winger
2 May 2018 at 9:25 am #
Simon, I can confirm that I have never known Soccerwidow to sit through an entire match in all the time I have known her.
She truly is interested solely in the mathematics of predicting outcomes (and the bragging rights when she gets it right!). | 5,453 | 23,314 | {"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} | 2.8125 | 3 | CC-MAIN-2019-43 | longest | en | 0.914964 |
https://www.jiskha.com/display.cgi?id=1298204190 | 1,501,044,380,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425766.58/warc/CC-MAIN-20170726042247-20170726062247-00345.warc.gz | 811,155,333 | 3,850 | # fizik
posted by .
1. A sports car travelling east at 48.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 22 m/s in 4.10 s
a) Find the magnitude and direction of the car’s acceleration as it slows down
b) Calculate the distance traveled during the 4.10 s time interval
• fizik -
(a) a = [(22 - 48)m/s]/4.1s
= -6.3 m/s^2
(b) X = Vaverage*(4.1s) = 35 m/s*4.1s
= 143.5 m
• fizik -
can i know how to get Vaverage?
• fizik -
Vav is the average of the initial and final velocities during deceleration.
It equals (Vinitial + Vfinal)/2 | 184 | 581 | {"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-2017-30 | longest | en | 0.888131 |
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April 1, 2014, 14:38 problem with real number in udf #1 New Member amr mobarez Join Date: Jan 2013 Posts: 9 Rep Power: 4 hi guys i'm making udf for unsteady inlet velocity ( step train ) with certain value with respect to time but i having problem with direction and magnitude as some of the flow goes into the reverse direction here is the code i wrote i hope someone can tell me what is wrong with it #include "udf.h" DEFINE_PROFILE(unsteady_x_velocity, thread, position) { face_t f; real t = CURRENT_TIME; if(t < 0.1) { F_PROFILE (f, thread, position)= 8.479; } if(0.1 < t > 0.2) { F_PROFILE (f,thread,position)= 1.357; } if(0.2 < t <0.3) { F_PROFILE (f,thread,position)= 8.479; } if(0.3 < t > 0.4) { F_PROFILE (f,thread,position)= 1.357; } if(0.4 < t < 0.5) { F_PROFILE (f,thread,position)= 8.479; } if(0.5 < t < 0.6) { F_PROFILE (f,thread,position)= 1.357; } if(t > 0.6) { F_PROFILE (f, thread, position)= 0; } }
April 2, 2014, 16:57 #2 Super Moderator Ghazlani M. Ali Join Date: May 2011 Location: Canada Posts: 1,291 Blog Entries: 23 Rep Power: 20 I think you should have a loop !! __________________ Regards, New to ICEM CFD, try this document --> http://goo.gl/G2gkE Ali
April 3, 2014, 18:14 #3 New Member amr mobarez Join Date: Jan 2013 Posts: 9 Rep Power: 4 you are right diamond x i tried the loop and it corrected the direction , still have problems with values but thanks for helping
April 4, 2014, 06:03 #4 Member Join Date: Jul 2013 Posts: 70 Rep Power: 4 You have ambiguous instructions, for example t>0.4 is defined twice: in (0.3 < t >0.4) and (0.4 < t < 0.5). And t=0.1, t=0.2... is not defined. Check that, is an error you made several times. User diamondx is right, you need a loop. Cheers
April 4, 2014, 17:01 #5 New Member amr mobarez Join Date: Jan 2013 Posts: 9 Rep Power: 4 i fixed it and it worked and here is the right code in case someone needed it #include "udf.h" DEFINE_PROFILE(unsteady_x_velocity12, thread, position) { face_t f; real tm = CURRENT_TIME; begin_f_loop (f,thread) { if(tm < 0.1) { F_PROFILE (f,thread,position)= 8.479; } if(tm > 0.1) { F_PROFILE (f,thread,position)= 1.36; } if(tm > 0.2) { F_PROFILE (f,thread,position)= 8.479; } if(tm > 0.3) { F_PROFILE (f,thread,position)= 1.36; } if(tm > 0.4) { F_PROFILE (f,thread,position)= 8.479; } if(tm > 0.5) { F_PROFILE (f,thread,position)= 1.36; } } end_f_loop (f,thread) }
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# - PowerPoint PPT Presentation
FOCS – Human Computer Interaction. Unit 2 Day 10. Journal Entry: Unit #2 Entry #9. Convert text to binary Convert the message ‘Poly pride’ into binary. Explain the process. Tower Building Activity.
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### FOCS – Human Computer Interaction
Unit 2 Day 10
Journal Entry: Unit #2 Entry #9
Convert text to binary
Convert the message ‘Poly pride’ into binary. Explain the process.
• Donald trump wants to build a 100 meter tall tower as quickly as possible. He has unlimited resources, and an unlimited budget and is willing to spend any amount of money to get the job done.
Constraints:
• He has blocks that are 100 meters wide, and 100 meters long, but only 1 meter high.
• Using special lifters putting one block on top of another block takes one week
I n your groups answer the following question: (solutions due at the end of class). One group will present their solution.
• What is the shortest amount of time he could build a 10 meter tower?
• What is the shortest amount of time he could build a 100 meter tower? | 392 | 1,648 | {"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-2017-51 | latest | en | 0.905378 |
https://mail.python.org/pipermail/python-list/2010-March/570724.html | 1,529,690,271,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864740.48/warc/CC-MAIN-20180622162604-20180622182604-00220.warc.gz | 656,419,851 | 2,687 | Calculating very large exponents in python
casevh casevh at gmail.com
Tue Mar 9 07:39:30 CET 2010
```[also replying to Geremy since the OP's message doesn't appear...]
On Mar 8, 11:05 am, geremy condra <debat... at gmail.com> wrote:
> On Mon, Mar 8, 2010 at 2:15 AM, Fahad Ahmad <miracles... at hotmail.com> wrote:
> > Thanks Geremy,
>
> > That has been an absolute bump........... GOD i cant sit on my chair, it has
> > worked even on 512 bit number and with no time..........
> > superb i would say.
>
> > lastly, i am using the code below to calculate Largest Prime factor of a
> > number:
>
> > print
> > ('''==============================================================================='''
> > ''' CALCULATE HIGHEST PRIME
> > FACTOR '''
>
> > '''===============================================================================''')
>
> > #!/usr/bin/env python
> > def highest_prime_factor(n):
> > if isprime(n):
> > return n
> > for x in xrange(2,n ** 0.5 + 1):
> > if not n % x:
> > return highest_prime_factor(n/x)
> > def isprime(n):
> > for x in xrange(2,n ** 0.5 + 1):
> > if not n % x:
> > return False
> > return True
> > if __name__ == "__main__":
> > import time
> > start = time.time()
> > print highest_prime_factor(1238162376372637826)
> > print time.time() - start
>
> > the code works with a bit of delay on the number : "1238162376372637826" but
> > extending it to
> > (109026109913291424366305511581086089650628117463925776754560048454991130443047109026109913291424366305511581086089650628117463925776754560048454991130443047)
> > makes python go crazy. Is there any way just like above, i can have it
> > calculated it in no time.
>
> > thanks for the support.
>
> If you're just looking for the largest prime factor I would suggest using
> a fermat factorization attack. In the example you gave, it returns
> nearly immediately.
>
> Geremy Condra- Hide quoted text -
>
> - Show quoted text -
For a Python-based solution, you might want to look at pyecm (http://
sourceforge.net/projects/pyecm/)
On a system with gmpy installed also, pyecm found the following
factors:
101, 521, 3121, 9901, 36479, 300623, 53397071018461,
1900381976777332243781
There still is a 98 digit unfactored composite:
60252507174568243758911151187828438446814447653986842279796823262165159406500174226172705680274911
Factoring this remaining composite using ECM may not be practical.
casevh
``` | 734 | 2,522 | {"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-2018-26 | latest | en | 0.592271 |
https://www.meritnation.com/ask-answer/question/numbers-from-1-to-20-are-written-on-20-separate-slips-the-sl/data-handling/16538795 | 1,638,743,170,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00609.warc.gz | 988,547,103 | 8,194 | # Numbers from 1 to 20 are written on 20 separate slips. The slips are put into a box and mixed thoroughly. One slip is drawn from the box randomly. What is the probability of drawing a prime number?
Dear Student,
Total possible outcomes = 20.
Favourable outcomes = Prime numbers between 1 to 20.
Favourable Outcomes = 2,3,5,7,11,13,17,19
Favourable Number of Outcomes = 8
Probability = Favourable Outcomes/Total Outcones = 8/20 =2/5
Regards!
• 2 | 132 | 448 | {"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} | 3.90625 | 4 | CC-MAIN-2021-49 | latest | en | 0.932218 |
https://www.fmaths.com/tips/quick-answer-what-is-venn-diagram-in-discrete-mathematics.html | 1,624,558,460,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00165.warc.gz | 695,211,498 | 11,705 | # Quick Answer: What Is Venn Diagram In Discrete Mathematics?
## What is a Venn diagram in mathematics?
A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while circles that do not overlap do not share those traits. Venn diagrams help to visually represent the similarities and differences between two concepts.
## How do you describe a Venn diagram?
A Venn diagram uses overlapping circles or other shapes to illustrate the logical relationships between two or more sets of items. Often, they serve to graphically organize things, highlighting how the items are similar and different. Venn diagrams show relationships even if a set is empty.
## What are the different types of Venn diagrams?
Types of Venn Diagrams
• Two-Set Diagrams. This type of a Venn diagram uses two circles or ovals to show overlapping properties.
• Three-Set Diagrams. You can always call these three circle diagrams as well.
• Four-Set Diagrams. A four-set Venn diagram is one that’s packed with four, overlapping sets.
• Five-Set Diagram.
You might be interested: What Is The Benefits Of Mathematics?
## What does a ∩ B mean?
In mathematics, the intersection of two sets A and B, denoted by A ∩ B, is the set containing all elements of A that also belong to B (or equivalently, all elements of B that also belong to A).
## What is the middle of a Venn diagram called?
A schematic diagram used in logic theory to depict collections of sets and represent their relationships. in the order three Venn diagram in the special case of the center of each being located at the intersection of the other two is a geometric shape known as a Reuleaux triangle.
## How do you explain Venn diagram to students?
Venn diagrams enable students to organise information visually so they are able to see the relationships between two or three sets of items. They can then identify similarities and differences. A Venn diagram consists of overlapping circles. Each circle contains all the elements of a set.
## How do you show a Venn diagram?
How to Make a Venn Diagram
1. The first step to creating a Venn diagram is deciding what to compare. Place a descriptive title at the top of the page.
2. Create the diagram. Make a circle for each of the subjects.
3. Label each circle.
4. Enter the differences.
5. Enter the similarities.
## How do you introduce a Venn diagram?
Introduce some blue objects and ask where those items could go on the Venn diagram. Encourage the students to think aloud as they work to solve the problem of what to do with the new items and help them to see that they do not fit in the Venn diagram and need to be placed outside of the circles.
You might be interested: Quick Answer: Why Is Mathematics Is Important To Learn?
## How many circles can a Venn diagram have?
They are often confused with Euler diagrams. While both have circles, Venn diagrams show the whole of a set while Euler diagrams can show parts of a set. Venn diagrams can have unlimited circles, but more than three becomes extremely complicated so you’ll usually see just two or three circles in a Venn diagram drawing.
## What is B in Venn diagram?
We use to denote the universal set, which is all of the items which can appear in any set. This is usually represented by the outside rectangle on the venn diagram. A B represents the intersection of sets A and B. This is all the items which appear in set A and in set B.
## How do you show subsets in a Venn diagram?
Venn Diagrams
1. If a set A is a subset of set B, then the circle representing set A is drawn inside the circle representing set B.
2. If set A and set B have some elements in common, then to represent them, we draw two circles which are overlapping.
## What is Venn diagram in logic?
Venn diagrams consist of two or three intersecting circles, each representing a class and each labeled with an uppercase letter. Two-circle Venn diagrams are used to represent categorical propositions, whose logical relations were first studied systematically by Aristotle.
## What does AUB )’ mean?
Definition 1. The union of the sets A and B, denoted by A U B, is the set that contains those elements that are either in A or in B, or in both.
## What is ∈ called?
The relation “is an element of “, also called set membership, is denoted by the symbol ” ∈ “. Writing. means that “x is an element of A”.
You might be interested: Question: Where Is Mathematics?
## What is the U and upside down U in math?
It means the Intersection of a set. For example, IF you have a set of even numbers and a set of odd numbers, the Union ‘ U ‘ of these two sets would be ALL numbers. But, the Intersection ( upside down U ) would mean that NONE of the numbers in Evens are in common with any of the Odds in the second set. | 1,066 | 4,860 | {"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} | 4.53125 | 5 | CC-MAIN-2021-25 | longest | en | 0.949503 |
http://safetoolboxes.com/howto_buildforecastmodel.html | 1,542,188,775,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741764.35/warc/CC-MAIN-20181114082713-20181114104713-00524.warc.gz | 257,235,642 | 7,431 | # Learning Center
SAFE TOOLBOXES® comes with the following forecast models:
# Forecast Model
1
ARIMA (Autoregressive integrated moving average model)
2
AR (Autoregressive model)
3
MA (Moving average model)
4
GARCH (Generalized Autoregressive Conditional Heteroskedasticity model)
5
6
Holt-Winters Multiplicative
7
Holt-Winter Exponential Smoothing
8
Last value
9
Historical mean
10
Moving average
11
Linear time tendency
12
13
Cubic time tendency
Regression models can also be used for forecasting purposes. This can be done separating one part of your data to represent the “past” and another part of your data to represent the “future”. This method may be useful if the explanatory variables are known in advance or if they are easier to predicted than the dependent variable itself. In this approach only the “past” portion of your data is used to fit the parameters of the regression model.
### A short example
Consider the following time series:
A B C 1 Sample database 2 3 Time Value 4 1 103.2 5 2 106.6 6 3 106.6 7 4 106.6 8 5 106.6 9 6 106.6 10 7 108.5 11 8 111.7 12 9 116.2 13 10 117.4 14 11 119.2 15 12 121.0 16 13 122.5 17 14 124.2 18 15 119.1 19 16 117.0 20 17 114.0 21 18 109.0 22 19 107.0 23 20 104.0 24 21 98.2 25 22 93.5 26 23 88.7 27 24 83.9 28 25 80.7 29 26 78.5 30 27 73.9 31 28 72.3 32 29 71.4 33 30 71.6 34 31 71.1 35 32 71.2 36 33 71.5 37 34 72.6 38 35 74.2 39 36 79.0 40 37 82.3 41 38 84.5 42 39 89.0 43 40 92.5 44 41 94.5 45 42 99.0 46 43 101.7 47 44 104.3 48 45 106.5 49 46 108.7 50 47 109.0 51 48 109.0 52 49 109.0 53 50 109.0 54
To forecast a value for dates 51 to 55 of this series using an AR(1) model just follow the steps below:
1. Select the Econometrics Toolbox tab
2. Select any cell containing the series (let’s say, cell B5)
3. Adjust the start sample to 0% and the final sample to 100% (to use all 50 sample points) and the forecast length to 110% (to predict inside the sample and five extras points)
4. Set option Category = “Forecast models” and item = “AR” under the "Analysis" group and click on the button to confirm your choice
5. Specify the order of the AR model changing the text “AR(p)” to “AR(1)” in the command window textbox and click on the button
Now, the results will appear at the bottom of the Econometrics Toolbox tab.
A first check of the goodness of fit of our model can be done by inspecting the line plot chart. A good fit will put the red line (predicted values) near to the blue line (actual values). The dotted line represents the out of sample prediction of our model.
The equation of our AR(1) model is placed in the tab. This table shows the fitted parameters and useful statistics of our model. Not all forecasting models have a representation in the equation tab. This is the case, for instance, for the Holt-Winters models or the moving average model.
You can run different specification for your ARIMA model or use other forecast methodologies. The “Historical results” tab will keep the main statistics of your models. Particularly, in the case of forecasting models, is a good practice to reserve some portion of your data to run out of sample tests. By doing this, the forecast statistics: MAD (mean absolute deviation), MAPE (mean absolute percent error) and RMSE (root of mean square error) available in this tab can help you to rank your models. | 1,036 | 3,359 | {"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} | 3.4375 | 3 | CC-MAIN-2018-47 | longest | en | 0.697607 |
https://crazyproject.wordpress.com/2011/08/07/a-fact-about-the-alt-map-on-a-tensor-power/ | 1,490,681,582,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189680.78/warc/CC-MAIN-20170322212949-00268-ip-10-233-31-227.ec2.internal.warc.gz | 778,638,742 | 18,443 | ## A fact about the Alt map on a tensor power
Let $F$ be a field in which $k!$ is a unit and let $V$ be an $F$-vector space. Recall that $S_k$ acts on the tensor power $\mathcal{T}^k(V)$ by permuting the components, and that $\mathsf{Alt}_k$ is defined on $\mathcal{T}^k(V)$ by $\mathsf{Alt}_k(z) = \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$. Prove that $\mathsf{im}\ \mathsf{Alt}_k$ is the unique largest subspace of $\mathcal{T}^k(V)$ on which each permutation $\sigma \in S_k$ acts as multiplication by $\epsilon(\sigma)$.
Suppose $z \in \mathcal{T}^k(V)$ such that $\sigma z = \epsilon(\sigma) z$ for all $\sigma \in S_k$. Then $\mathsf{Alt}_k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \epsilon(\sigma) z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} z$ $= z$, so that $z \in \mathsf{im}\ \mathsf{Alt}_k$.
In particular, any subspace of $\mathcal{T}^k(V)$ upon which every permutation $\sigma \in S_k$ acts as scalar multiplication by $\epsilon(\sigma)$ is contained in $\mathsf{im}\ \mathsf{Alt}_k$.
Note also that $\sigma \in S_k$ acts on $\mathsf{im}\ \mathsf{Alt}_k$ as multiplication by $\epsilon(\sigma)$ since $\mathsf{im}\ \mathsf{Alt}_k \cong_F \bigwedge^k(V)$. | 447 | 1,257 | {"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": 29, "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} | 3.21875 | 3 | CC-MAIN-2017-13 | latest | en | 0.60996 |
https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/06%3A_Problems_in_Higher_Dimensions | 1,723,548,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00765.warc.gz | 295,898,901 | 32,710 | # 6: Problems in Higher Dimensions
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"Equations of such complexity as are the equations of the gravitational field can be found only through the discovery of a logically simple mathematical condition that determines the equations completely or at least almost completely."
~ Albert Einstein (1879-1955)
In this chapter we will explore several examples of the solution of initial-boundary value problems involving higher spatial dimensions. These are described by higher dimensional partial differential equations, such as the ones presented in Table 2.1.1 in Chapter 2. The spatial domains of the problems span many different geometries, which will necessitate the use of rectangular, polar, cylindrical, or spherical coordinates.
We will solve many of these problems using the method of separation of variables, which we first saw in Chapter 2. Using separation of variables will result in a system of ordinary differential equations for each problem. Adding the boundary conditions, we will need to solve a variety of eigenvalue problems. The product solutions that result will involve trigonometric or some of the special functions that we had encountered in Chapter 5. These methods are used in solving the hydrogen atom and other problems in quantum mechanics and in electrostatic problems in electrodynamics. We will bring to this discussion many of the tools from earlier in this book showing how much of what we have seen can be used to solve some generic partial differential equations which describe oscillation and diffusion type problems.
As we proceed through the examples in this chapter, we will see some common features. For example, the two key equations that we have studied are the heat equation and the wave equation. For higher dimensional problems these take the form \begin{align} u_{t} &=k \nabla^{2} u,\label{eq:1} \\ u_{t t} &=c^{2} \nabla^{2} u .\label{eq:2} \end{align}
We can separate out the time dependence in each equation. Inserting a guess of $$u(\mathbf{r}, t)=\phi(\mathbf{r}) T(t)$$ into the heat and wave equations, we obtain $T^{\prime} \phi=k T \nabla^{2} \phi \text {, }\label{eq:3}$ $T^{\prime \prime} \phi=c^{2} T \nabla^{2} \phi .\label{eq:4}$
## Note
The Helmholtz equation is named after Hermann Ludwig Ferdinand von Helmholtz (1821-1894). He was both a physician and a physicist and made significant contributions in physiology, optics, acoustics, and electromagnetism.
Dividing each equation by $$\phi(\mathbf{r}) T(t)$$, we can separate the time and space dependence just as we had in Chapter ??. In each case we find that a function of time equals a function of the spatial variables. Thus, these functions must be constant functions. We set these equal to the constant $$-\lambda$$ and find the respective equations \begin{align} \frac{1}{k} \frac{T^{\prime}}{T} &=\frac{\nabla^{2} \phi}{\phi}=-\lambda\label{eq:5} \\ \frac{1}{c^{2}} \frac{T^{\prime \prime}}{T} &=\frac{\nabla^{2} \phi}{\phi}=-\lambda\label{eq:6} \end{align} The sign of $$\lambda$$ is chosen because we expect decaying solutions in time for the heat equation and oscillations in time for the wave equation and will pick $$\lambda>0$$.
The respective equations for the temporal functions $$T(t)$$ are given by \begin{align} T^{\prime} &=-\lambda k T,\label{eq:7} \\ T^{\prime \prime}+c^{2} \lambda T &=0 .\label{eq:8} \end{align} These are easily solved as we had seen in Chapter ??. We have \begin{align} &T(t)=T(0) e^{-\lambda k t},\label{eq:9} \\ &T(t)=a \cos \omega t+b \sin \omega t, \quad \omega=c \sqrt{\lambda},\label{eq:10} \end{align} where $$T(0), a$$, and $$b$$ are integration constants and $$\omega$$ is the angular frequency of vibration.
In both cases the spatial equation is of the same form, $\nabla^{2} \phi+\lambda \phi=0 \text {. }\label{eq:11}$ This equation is called the Helmholtz equation. For one dimensional problems, which we have already solved, the Helmholtz equation takes the form $$\phi^{\prime \prime}+\lambda \phi=0$$. We had to impose the boundary conditions and found that there were a discrete set of eigenvalues, $$\lambda_{n}$$, and associated eigenfunctions, $$\phi_{n}$$.
In higher dimensional problems we need to further separate out the spatial dependence. We will again use the boundary conditions to find the eigenvalues, $$\lambda$$, and eigenfunctions, $$\phi(\mathbf{r})$$, for the Helmholtz equation, though the eigenfunctions will be labeled with more than one index. The resulting boundary value problems are often second order ordinary differential equations, which can be set up as Sturm-Liouville problems. We know from Chapter 5 that such problems possess an orthogonal set of eigenfunctions. These can then be used to construct a general solution from the product solutions which may involve elementary, or special, functions, such as Legendre polynomials and Bessel functions.
We will begin our study of higher dimensional problems by considering the vibrations of two dimensional membranes. First we will solve the problem of a vibrating rectangular membrane and then we will turn our attention to a vibrating circular membrane. The rest of the chapter will be devoted to the study of other two and three dimensional problems possessing cylindrical or spherical symmetry.
Thumbnail: A three dimensional view of the vibrating annular membrane. (CC BY-NC-SA 3.0 Unported; Russell Herman)
This page titled 6: Problems in Higher Dimensions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform. | 3,127 | 9,970 | {"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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.199538 |
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# 20-1
CHAPTER 20
MAGNETIC PROPERTIES
PROBLEM SOLUTIONS
Basic Concepts
20.1 (a) We may calculate the magnetic field strength generated by this coil using Equation 20.1 as
NI
H =
l
(400 turns)(15 A)
= = 24, 000 A - turns/m
0.25 m
(b) In a vacuum, the flux density is determined from Equation 20.3. Thus,
B0 = µ 0 H
## = (1.257 x 10-6 H/m)(24, 000 A - turns/m) = 3.0168 x 10-2 tesla
(c) When a bar of chromium is positioned within the coil, we must use an expression that is a combination
of Equations 20.5 and 20.6 in order to compute the flux density given the magnetic susceptibility. Inasmuch as χm
## = 3.13 x 10-4 (Table 20.2), then
B = µ 0 H + µ 0 M = µ 0 H + µ 0 χ mH = µ 0 H (1 + χ m)
## = 3.0177 x 10-2 tesla
which is essentially the same result as part (b). This is to say that the influence of the chromium bar within the coil
makes an imperceptible difference in the magnitude of the B field.
(d) The magnetization is computed from Equation 20.6:
## M = χ mH = (3.13 x 10-4 )(24, 000 A - turns/m) = 7.51 A/m
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. | 445 | 1,472 | {"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} | 3.09375 | 3 | CC-MAIN-2020-16 | latest | en | 0.897066 |
https://slideplayer.com/slide/691729/ | 1,531,886,546,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590046.11/warc/CC-MAIN-20180718021906-20180718041906-00295.warc.gz | 752,104,482 | 21,536 | # Whole Class Review Activity Directions: Give each student an index card with a number from 1-8. Students are instructed to write a review question, math.
## Presentation on theme: "Whole Class Review Activity Directions: Give each student an index card with a number from 1-8. Students are instructed to write a review question, math."— Presentation transcript:
Whole Class Review Activity
Directions: Give each student an index card with a number from 1-8. Students are instructed to write a review question, math problem, or spelling word on their card. Click the Spin Button. When the wheel hits the number on the index card, the student chooses who will receive the question on their card.
12 3 4 56 7 8 SPIN
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https://www.ozlotteries.com/oz-lotto/systems?utm_source=blog&utm_medium=article&utm_content=25-million-oz-lotto&utm_campaign=blog-jackpot-qa | 1,660,259,388,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571536.89/warc/CC-MAIN-20220811224716-20220812014716-00087.warc.gz | 815,367,042 | 191,224 | • Play Oz Lotto
• Results
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# Oz Lotto Systems
### This Tuesday\$5,000,000
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Select game:
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## Oz Lotto Systems by Oz Lotteries
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A System ticket allows you to play more numbers than the 7 numbers in a standard Oz Lotto game. The system covers all possible 7 number combinations of your 8 or more numbers.
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#### Standard Entry
A Standard entry would be your 7 chosen numbers plus the Powerball, like the example below.
GameMain Numbers
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A System ticket gives you more chances to win. This type of entry allows players to increase their chances of winning a prize by playing a wider range of number combinations.
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Sunday: Closed | 684 | 2,239 | {"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} | 4.0625 | 4 | CC-MAIN-2022-33 | longest | en | 0.857629 |
https://japhr.blogspot.com/2012/01/dart-isolates-for-speed.html | 1,474,848,778,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660467.49/warc/CC-MAIN-20160924173740-00094-ip-10-143-35-109.ec2.internal.warc.gz | 840,366,321 | 26,283 | Monday, January 9, 2012
Dart Isolates for Speed
‹prev | My Chain | next›
Try as I might, I cannot avoid the dreaded fibonacci sequence example when working with Dart Isolates. I need a function that is going to take a little while so that I can compare the timing of a long running series of calculations with and without isolates.
So fibonacci, it is. In Dart, this looks like:
```fib(i) {
if (i < 2) return i;
return fib(i-2) + fib(i-1);
}```
That quite similar to just about every other fibonacci implemented in any other language. Now to make use of it. I take a series of six numbers, calculate the fibonnaci for each and print out the timing results:
```#import('pretty_stopwatch.dart');
main() {
var list = [5, 40, 39, 32, 6, 41];
var timer = new PrettyStopwatch.start();
var answers = {};
list.forEach((i) {
print("fib(" + i + ") = " + answers[i]);
});
timer.stop();
}
fib(i) {
if (i < 2) return i;
return fib(i-2) + fib(i-1);
}```
(try.dartlang.org) The result is:
```➜ command_line git:(master) ✗ dart timer06.dart
fib(5) = 5
fib(40) = 102334155
fib(39) = 63245986
fib(32) = 2178309
fib(6) = 8
fib(41) = 165580141
Elapsed time: 11337ms```
11.3 seconds. Pity my poor MacBook Air for not being faster. But even my early model Air has two cores, which means that, if I split the work across two isolates, I ought to halve the time it takes to calculate all six. I am using some non-DRY code, but it should serve its purpose to generate some numbers:
```#import('pretty_stopwatch.dart');
main() {
final fib_solver = new FibSolver();
var list = [5, 40, 39, 32, 6, 41];
// Spawn isolate #1 for the first 3 elements
var timer1 = new PrettyStopwatch.start();
fib_solver.spawn().then((port) {
print("fib(" + i + ") = " + answer);
});
timer1.stop();
});
});
// Spawn isolate #2 for the last 3 elements
var timer2 = new PrettyStopwatch.start();
fib_solver.spawn().then((port) {
print("fib(" + i + ") = " + answer);
});
timer2.stop();
});
});
}
// Isolate class for solving lists of fibonacci numbers
class FibSolver extends Isolate {
main() {
var answers = {};
message.forEach((i) {
// print("fib(" + i + ") = " + answers[i]);
});
});
}
}
fib(i) {
if (i < 2) return i;
return fib(i-2) + fib(i-1);
}```
This results in:
```➜ command_line git:(master) ✗ dart timer07.dart
fib(32) = 2178309
fib(6) = 8
fib(41) = 165580141
Elapsed time: 6554ms
fib(5) = 5
fib(40) = 102334155
fib(39) = 63245986
Elapsed time: 6856ms```
It looks like both complete in around 6.8 seconds. Rather than run the time for each isolate individually, I would like to know when both have completed. Since I am not DRYing my code well here, I need two booleans—one to track each of the two isolates. When both have completed, I mark a timer Completer as complete, triggering the timer to stop and print out the results:
```main() {
final fib_solver = new FibSolver();
var list = [5, 40, 39, 32, 6, 41];
var timer = new PrettyStopwatch.start()
, completer = new Completer();
completer.future.then((args) {
timer.stop();
});
var done1 = false
, done2 = false;
fib_solver.spawn().then((port) {
print("fib(" + i + ") = " + answer);
});
done1 = true;
if (done2) completer.complete(true);
});
});
fib_solver.spawn().then((port) {
print("fib(" + i + ") = " + answer);
});
done2 = true;
if (done1) completer.complete(true);
});
});
}```
(try.dartlang.org) I won't win any style points for that, but it does the trick, I now have accurate timing information for splitting the sequence of calculations across two isolates (each of which gets a core on my Air). The results:
```➜ command_line git:(master) ✗ dart timer08.dart
fib(32) = 2178309
fib(6) = 8
fib(41) = 165580141
fib(5) = 5
fib(40) = 102334155
fib(39) = 63245986
Elapsed time: 6375ms```
6.4 seconds. Not quite half, but that is to be expected since it was not an exact divide of calculations and there is some overhead involved in building isolates. If nothing else, I finally have a decent measure of how useful Isolates can be. Day #260
1. This comment has been removed by the author.
2. is this working in browsers?
3. @Peter: The pure Dart version won't work unless you download and compile "Dartium": http://japhr.blogspot.com/2012/01/dartium.html
But, just like with coffeescript, you can compile this into Javascript, which will run in any browser. The resultant Javascript is still on the large side, but that'll change at the tool evolves.
Here's an example of converting Dart to JS: http://japhr.blogspot.com/2012/01/converting-complex-dart-apps-to.html
1. This comment has been removed by the author.
4. i tried to get this working via a normal application project with the dart editor (i.e. html file, 2 dart files) which is converted to js anyway (at least the dart files)...doesn't work right now and my main problem is: how can the compiler convert multi-threading to js because js is always single-threaded?
1. Actually, this should be single threaded as well. Dart Isolates come in two varieties: regular and heavy. The default, which I am using here, are single threaded just like JS. Under the covers they rely on evented callbacks to swap execution.
The heavy version of Isolates start in different threads. When compiled into Javascript, these would then have to fall back to the regular, single-threaded version.
I'll investigate this a bit more in a follow-up post. I am curious how the heavy Isolates affect the timing.
2. OK. I looked into it and was surprised at the results. Heavy vs. light Isolates made no difference in Dart. But it *did* make a difference when compiled to Javascript.
The gory details: http://japhr.blogspot.com/2012/01/heavy-vs-light-dart-isolates.html
(includes links to working dart and JS pages)
5. Hi Chris,
I'm making a thesis on Dart and the actor model for my graduation in computer science engineering. I've tried your Fibonacci code with light and heavy isolate on a DartBoard and it gave me different results: heavy about 8sec; light about 16sec. Have you already seen this results?
Thanks for your work, it helped my a lot.
Ciao from Italy
1. Yup, I found similar results the following night when I compiled this into Javascript: http://japhr.blogspot.com/2012/01/heavy-vs-light-dart-isolates.html (Dartboard is also compiled to Javascript).
Glad to hear my stuff has been of use -- good luck with the thesis! | 1,739 | 6,326 | {"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-2016-40 | longest | en | 0.734063 |
https://www.univerkov.com/body-weight-4-kg-determine-the-pressure-on-the-floor-if-the-support-area-is-150-dm2/ | 1,701,590,400,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00013.warc.gz | 1,151,384,426 | 6,225 | # Body weight 4 kg. Determine the pressure on the floor if the support area is 150 dm2.
As we reliably know, in order to calculate the value of pressure, it is possible to use the formula known from the school curriculum:
p = F: S.
Let us determine what value will be equal to the force with which this body presses on the floor, if from the condition of our problem we know for sure that its mass is 4 kilograms:
F = 4 * 10 = 40.
Now we can calculate what the pressure that the specified body exerts on the floor will equal:
150 dm2 is the same as 1.5 m2;
40: 1.5 = 26, (6).
Answer: 26, (6) Pa.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 218 | 914 | {"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} | 3.78125 | 4 | CC-MAIN-2023-50 | latest | en | 0.934541 |
http://www.jiskha.com/display.cgi?id=1370540263 | 1,495,921,684,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609061.61/warc/CC-MAIN-20170527210417-20170527230417-00465.warc.gz | 691,933,316 | 3,829 | # math
posted by on .
15 tv were shipped 4 were defective if 2 tv's were random selected compute the possibility of both tv's working and the probability that at least one of the 2 tv does not work
• math - ,
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
11/15 * 11/15 = ?
That means one or both do not work.
One not work = 4/15 * 11/15 = ?
Both not work = 4/15 * 11/15 = ?
Either-or probabilities are found by adding the individual probabilities. | 136 | 559 | {"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-2017-22 | latest | en | 0.978362 |
http://educators.brainpop.com/bp-jr-topic/basic-subtraction/ | 1,444,726,372,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738004493.88/warc/CC-MAIN-20151001222004-00132-ip-10-137-6-227.ec2.internal.warc.gz | 102,786,895 | 13,191 | # Basic Subtraction
An operation is a way to solve a math problem, such as addition and subtraction. In this movie, you’ll learn how to subtract, or take away, using different strategies. You can count back to find the difference or use fact families and related facts to subtract. You can also use a fact triangle, a number line, or draw pictures to help you subtract. You'll watch as Annie writes and solves subtraction sentences using numbers, the minus sign, and the equal sign. What is 10 – 2?
## Math Facts and Number Sense Lesson Plan: The Jelly Bean Game
In this lesson plan which is adaptable for grades K-4, students will use jelly beans in an online game and real jellybeans as math manipulatives to practice number sense concepts, such as counting, more and less, estimation, algebraic thinking and missing addend equations, and addition and subtraction math facts. This lesson plan is aligned to Common Core State Standards. See more »
## Number Jumble Math Game Lesson Plan: Multiples
In this lesson plan which is adaptable for grades 2-5, students will use BrainPOP Jr. and BrainPOP resources (including an online math game) to practice multiplying whole numbers and/or decimals. Students will identify patterns within a multiplication table and create their own multiplication tables with unique patterns. This lesson plan is aligned to Common Core State Standards. See more »
## Basic Subtraction Lesson Plan: Fun With Fact Families
In this lesson plan, which is adaptable for grades K through 3, students use BrainPOP Jr. resources to learn how to use a variety of strategies to subtract small numbers. They will also learn about fact families while working collaboratively in independent groups. This lesson plan is aligned to Common Core State Standards. See more »
## Basic Subtraction Background Information for Teachers and Parents
This page contains information to support educators and families in teaching K-3 students about basic subtraction. The information is designed to complement the BrainPOP Jr. movie Basic Subtraction. It explains the type of content covered in the movie, provides ideas for how teachers and parents can develop related understandings, and suggests how other BrainPOP Jr. resources can be used to scaffold and extend student learning. See more »
## Basic Subtraction Activities for Kids
In this set of activities adaptable for grades K-3, parents and educators will find ideas for teaching about basic subtraction. These activities are designed to complement the BrainPOP Jr. Basic Subtraction topic page, which includes a movie, quizzes, online games, printable activities, and more. See more » | 531 | 2,661 | {"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} | 3.71875 | 4 | CC-MAIN-2015-40 | latest | en | 0.901024 |
https://byjus.com/question-answer/a-father-is-three-times-as-old-as-his-son-in-12-years-time-he-will-be-twice-as-old-as-his-son-find-the-present-ages-of-father-and-the-son/ | 1,675,209,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00345.warc.gz | 159,297,178 | 17,700 | Question
# A father is three times as old as his son. In $12$ years time, he will be twice as old as his son. Find the present ages of father and the son.
Open in App
Solution
## Let the present age of son be $x$.And the present age of father be $3x$.As given in the question,The age of son after $12$ years = $x+12$And the age of the father after $12$ years = $2\left(x+12\right)$Therefore,$\begin{array}{rcl}3x+12& =& 2\left(x+12\right)\\ & =& 3x+12=2x+24\\ & =& x=12\end{array}$Therefore,The present age of the son is $12$ years.The present age of the father is $12×3=36$ years.
Suggest Corrections
0 | 197 | 607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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} | 4.34375 | 4 | CC-MAIN-2023-06 | latest | en | 0.949839 |
https://www.thehelper.net/threads/need-confirmation-on-damage-calculation.158494/ | 1,702,155,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00572.warc.gz | 1,138,854,854 | 17,880 | # Need confirmation on damage calculation
#### Solu9
##### You can change this now in User CP.
I have an ability which has this damage calculation:
"Unit - Cause (Triggering unit) to damage (Picked unit), dealing ((((Real((Strength of (Triggering unit) (Include bonuses)))) x (1.00 + ((Real((Level of Power Strike for (Triggering unit)))) x 0.10))) x EnhancedDamage[(Player number of (Triggering player))]) x 5.00) damage of attack type Normal and damage type Normal"
If we insert the following values:
Strength = 50
Power Strike level = 5
EnhancedDamage = 1.70
So:
((((50) x (1 +(5 x 0.1))) x 1.70) x 5) =
(((50 x (1.5)) x 1.70) x 5) =
((75 x 1.70) x 5) =
(127.5 x 5) =
637.5
Is this calculated correctly?
#### Maelbog
##### Member
Yes
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Nicola De Noni on 16 May 2022
Commented: Nicola De Noni on 16 May 2022
Hi everyone, I’m trying to solve this differential equation (dT/dt):
dTdt = ((U*A)/(M_mix*cp_mix))*(T_in-T_out)/(log(T_in-T)/(T_out-T));
Where I want to see the temperature profile T as the weather changes. Unfortunately, T_in and T_out are two temperature vectors whose values vary over time. How can I solve this differential equation? This is my code:
function f = CalcoloSperimentale(t,T)
%% Parameters
n = 10; % length of time vector
M_mix = linspace(11379,11379,n); % Weigth [Kg]
cp_mix = linspace(2000,2000,n); % Cp
U = linspace(53,53,n); % Thermal coefficient [W/m2K]
A = linspace(25.2,25.2,n); % Area
xmin_in = 15; % Minimum inlet temperature
xmax_in = 16; % Maximum inlet temperature
T_in = xmin_in+rand(1,n)*(xmax_in-xmin_in);
xmin_out = 20; % Minimum outlet temperature
xmax_out = 40; % Maximum outlet temperature
T_out = linspace(xmax_out,xmin_out,n);
% Differential equation
dTdt(i) = ((U(i)*A(i))/(M_mix(i)*cp_mix(i)))*(T_in(i)-T_out(i))/(log(T_in(i)-T(i))/(T_out(i)-T(i)));
f = dTdt;
end
And the second script is:
n = 10;
timeStart = 0;
timeEnd = 5280;
timespan = linspace(timeStart,timeEnd,n);
% Initial condition
initT = 98.5+273.15;
[timeOut, T] = ode45(@CalcoloSperimentale, timespan, initT);
% Plotting data
plot(timeOut, T)
In order to solve this differential equation i should have to consider also the final condition of temperature, but i don't know hot to implement it.
##### 2 CommentsShowHide 1 older comment
Nicola De Noni on 16 May 2022
Absolutely! T_in is the inlet temperature of a chemical reactor jacket and T_out is the outlet temperature of the reactor jacket. I’m studying the heat transfer in a water-cooled tank. T is the temperature inside the vessel.
I would like to plot the temperature inside the reactor vs time (by knowing the inlet and outlet temperature of the cooling fluid) in order to understand the heat transfer.
Torsten on 16 May 2022
n=10;
xmin_in = 15; % Minimum inlet temperature
xmax_in = 16; % Maximum inlet temperature
T_in = xmin_in+rand(1,n)*(xmax_in-xmin_in);
xmin_out = 20; % Minimum outlet temperature
xmax_out = 40; % Maximum outlet temperature
T_out = linspace(xmax_out,xmin_out,n);
M_mix = 11379; % Weigth [Kg]
cp_mix = 2000; % Cp
U = 53; % Thermal coefficient [W/m2K]
A = 25.2; % Area
timeStart = 0;
timeEnd = 5280;
timespan = linspace(timeStart,timeEnd,n);
T_in = @(t)interp1(timespan,T_in,t);
T_out = @(t)interp1(timespan,T_out,t);
f = @(t,T) U*A/(M_mix*cp_mix)*(T_in(t)-T_out(t))/log((T_in(t)-T)/(T_out(t)-T));
% Initial condition
initT = 98.5+273.15;
[timeOut, T] = ode45(f, timespan, initT);
% Plotting data
plot(timeOut, T)
Nicola De Noni on 16 May 2022
You're right, i'm sorry! Anyway, thanks a lot! | 869 | 2,828 | {"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} | 3.40625 | 3 | CC-MAIN-2022-27 | latest | en | 0.826124 |
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### Lifetime on the Main Sequenc
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https://gis.stackexchange.com/questions/172333/interpreting-gwr-result-issue | 1,718,752,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00532.warc.gz | 237,452,862 | 40,846 | Interpreting GWR Result - Issue
I used the Geographically Weighted Regression (GWR) to find out which independent variables lead to crime and how they vary over the space.
After a literature review, I ran a global Ordinary Least Squares (OLS) model to find the most suitable set of independent variables, then used the local GWR method. For both the OLS and GWR model I then conducted the Moran's I for the standardized residuals, to be sure that they are not spatially autocorrelated. I used ArcGIS for the analysis.
When I investigated the coefficient values within the different regions in the study area, I came to an issue that is unclear for me: For example I have included the risk factor bus stop, given as density per square kilometers, to explain burglaries (given as crime rate per 100 000 inhabitants).
The values for C1_BusDen (Coefficient 1 for bus stop density) ranges from 21 to 39. One region has a very low value with a crime rate of 1100 and bus stop density of 4.5. Another region has a very high value with a crime rate of 1250 and bus stop density of 5.07.
If the density of bus stops is high in both regions and there is also a similar crime rate, shouldn't be the relationship (value) also the same - or at least similar, but not in the lowest and highest class? Also, the region with the highest density of bus stops has at the same time the highest crime rate - the GWR coefficient value is only 26 - shouldn't it be the highest?
Do I interpret the result in the correct way, because it would make sense that a higher value in the independent variable and a higher value in the dependent varialbe lead to a stronger relationship.
• Please expand abbreviations like GWR and OLS the first time you use them in your questions so that they can help serve future readers as learning materials.
– PolyGeo
Commented Dec 2, 2015 at 17:37
• What is your citation in the lit review that indicates that model selection should be based on an OLS? Commonly GWR model selection is performed via AICc and I have not seen support for using a 1st order OLS regression for selecting parameters in a local regression. Mathematically this does not track and is a made up ESRI thing. Commented Dec 2, 2015 at 21:18
• In literature I looked for the theoretical background of GWR - using a global model first was indeed provided by Esri. Commented Dec 3, 2015 at 7:54
Well, it is a locally weighted regression and does not account for first-order effects very well so, this is not really surprising and inherently one of the limitations of the method. Perhaps if you changed the size of the objective kernel function (bandwidth) this would be mitigated to some degree but, I would not hold my breath. I would point out that the specified distribution and bandwidth can have notable effects on the model. If you are using the ESRI implementation, don't. There is no flexibility in specifying the distributional assumption (eg., Gaussian, Poisson, binomial, multinomial) and you are stuck with the canned regression diagnostics with no access to the objects comprising the model.
Unless you have measurable nonstationarity, GWR is a dubious method and you would be better suited by just using an OLS or GLM. Please take some time to review some of the literature relating to GWR. The Wheeler and Tiefelsdorf (2005) paper demonstrates serious bias in the coefficients due to nonsystematic localized collinearity. Páez et al., (2011) showed, through simulations, that GWR did not reliably discriminate spatially varying process and sometimes exhibited spurious correlations in the local fits. I am not saying don't use GWR but, outside of exploratory analysis, it should be used with great caution and extensively tested for validity and bias.
If 1st order spatial autocorrelation is in fact, an issue affecting residuals and iid, a spatial autoregressive or mixed effects model would be in order. You can use the Lagrange diagnostics (Anselin et al., 1996) to test for spatial dependence in linear models. This will indicate if autocorrelation is influencing the residual error. If you would like to formalize a spatial model, where nonstationarity (2nd order autocorrelation) may be an issue, two methods I would recommend investigating would be principal coordinate analysis of neighbor matrices (Dray et al., 2006) or Eigenvector-based spatial filtering (Griffith 2000).
For inferential models, I particularity like spatial filtering approaches. De Jong et al., (1984) showed an empirical relationship between Eigenvectors and Moran's-I. Building on this concept one can use Eigenvector(s), in a semi-parametric approach, to partial out the effects of spatial process on regression estimates. The concept of PCNM's is to quantify multiscale spatial process, thus directly representing the autocorrelation structure, using scaled-Eigenvector matrices. The reason I cite Dray et al., (2006), for PCNM, is that his extension allows for negative autocorrelation structures. To implement a spatial filtering approach for regression models the function SpatialFiltering in the R library spdep formalizes a brute force method that make model specification quite easy. To formally explore the spatial structure of the data, the R library pcnm provides methods for specifying and visualizing principal coordinate neighbor matrices.
References
Anselin, L., A.K. Bera, R. Florax, & M.J. Yoon (1996) Simple diagnostic tests for spatial dependence. Regional Science and Urban Economics (26)77–104.
De Jong, P., C. Sprenger, F. van Veen, (1984) On extreme values of Moran’s I and Geary’s c. Geographic Analysis. 16:17-24.
Dray, S., P. Legendre, & P.R. Peres-Neto, (2006) Spatial modeling: a comprehensive framework for principal coordinate analysis of neighbor matrices (PCNM). Ecological Modeling, 196:483-493.
Griffith, D. A., (2000) A linear regression solution to the spatial autocorrelation problem. Journal of Geographical Systems, 2:141-156.
Páez,A., S. Farber, & D. Wheeler (2011). A simulation based study of geographically weighted regression as a method for investigating spatially varying relationships. Environment and Planning 43(12):2992–3010.
Wheeler, D. & M. Tiefelsdorf, (2005). Multicollinearity and correlation among local regression coefficients in geographically weighted regression. Journal of Geographical Systems, 7:161–187.
• Thank you for your detailed answer! I use ArcGIS for the analysis, and for both OLS and GWR models I tested spatial autocorrelation, which indicates that the residuals are randomly distributed. I do not have time to get again further into detail with GWR, I am just wondering if I can use my model and if the calculation of the coefficient values for the regions, as I indicated above, are correct? Commented Dec 3, 2015 at 7:58
• If you do not have any structure in your residuals, due to autocorrelation, then there is absolutely no support for running a GWR model in the first place and you should just go with the OLS. Commented Dec 3, 2015 at 15:37 | 1,603 | 7,013 | {"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} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.94517 |
https://cs.nyu.edu/pipermail/fom/2002-April/005451.html | 1,708,982,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00306.warc.gz | 185,200,904 | 2,142 | # FOM: A problem in the foundations of statistics
JoeShipman@aol.com JoeShipman at aol.com
Wed Apr 24 16:48:58 EDT 2002
```If, in m=a+b trials, a successes and b failures have been observed, what is the chance that the next n=c+d trials will contain c successes and d failures?
The following assumption is reasonable:
There is an underlying binomial process with parameter p and the trials are independent instances of that process.
The simplest procedure is to assume that p=a/m and calculate the quantity
(n choose c)(p^c)((1-p)^d) =
n!(a^c)(b^d)
------------
c!d!(m^n)
but this assumes too much. In particular, if the first m trials were all successes then this method expresses *certainty* that p=1 and that the next m trials will also be successes. Professional statisticians prefer to be more conservative in their predictions.
This is truly a foundational problem. There is no obvious "right answer". One common solution is to assume a "prior" uniform distribution for p -- that is, p is equally likely to be anywhere between 0 and 1. Then the observation of a successes and b failures leads to a "posterior" distribution for p according to Bayes's theorem, from which may be obtained a probability to be associated with particular values of c and d.
But is this a reasonable way to do it?
Other solution have also been proposed. Even the case a=m, b=0, c=n=1, d=0 is interesting. Having seen m successes in a row, what odds would you be willing to give that the next trial will be a success, assuming no cheating?
The Bayesian calculation gives an interesting answer, but it's not the only reasonable one.
-- Joe Shipman
``` | 393 | 1,652 | {"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-2024-10 | latest | en | 0.942825 |
http://www.03964.com/read/3cb169a37dc7bf5f0af38262.html | 1,532,194,649,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00146.warc.gz | 392,647,490 | 12,255 | # Physics Paper 3 HL May 2000
INTERNATIONAL BACCALAUREATE BACCALAUR?AT INTERNATIONAL BACHILLERATO INTERNACIONAL
M00/430/H(3)
PHYSICS HIGHER LEVEL PAPER 3 Tuesday 9 May 2000 (morning) 1 hour 15 minutes
Name
Number
INSTRUCTIONS TO CANDIDATES ! ! ! ! Write your candidate name and number in the boxes above. Do not open this examination paper until instructed to do so. Answer all of the questions from two of the Options in the spaces provided. At the end of the examination, indicate the letters of the Options answered in the boxes below.
EXAMINER /30 /30 TOTAL /60
TEAM LEADER /30 /30 TOTAL /60 TOTAL
IBCA /30 /30 /60
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27 pages
–2– OPTION D – BIOMEDICAL PHYSICS D1. Forces in the human arm The arm pulls against the strap as shown so that the scale reads a force of 100 N.
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F = 100 N Strap
28 cm 32 cm
biceps
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humerus (a)
elbow joint
State in words the two conditions for a rigid body to be in equilibrium under the action of a number of forces. ......................................................................... ......................................................................... .........................................................................
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–3– (Question D1 continued) (b)
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The forearm is redrawn below. On the diagram, draw in force vectors to represent all the forces acting on the forearm. State what object exerts each force.
[4]
(c)
Calculate the torque about the elbow joint produced on the arm by the strap. Is this torque clockwise or counter clockwise in the diagram? ......................................................................... .........................................................................
[2]
(d)
Will the force exerted by the biceps muscle be greater than, less than or equal to the force exerted by the strap? Explain your reasoning, without calculations. ......................................................................... .........................................................................
[1]
(e)
Calculate the force exerted by the biceps muscle. ......................................................................... .........................................................................
[2]
(f)
Explain why, in this particular situation, the weight of the forearm does not play a role in determining the force exerted by the biceps muscle. ......................................................................... .........................................................................
[2]
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–4– D2. Walking barefoot on gravel
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An adult and a child are walking barefoot over rough gravel. The adult is twice as tall as the child, and they are of similar shape.
For the adult as compared to the child, determine the following: (a) The ratio of their masses. ......................................................................... (b) The ratio of the forces on the soles of their feet. ......................................................................... (c) The ratio of the areas of their feet in contact with the ground. ......................................................................... (d) The ratio of the pressures (force per unit area) on the soles of their feet. ......................................................................... ......................................................................... (e) Which one is likely to find it more painful walking on the rough gravel, or will it be the same for both? Explain briefly. ......................................................................... ......................................................................... [1] [2] [1] [1] [1]
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A chemical compound containing a radioactive isotope is introduced into a patient as a ‘tracer’ to study a physiological process. The radioactive half-life of the isotope is 3 days, and the biological half-life of the chemical compound in the body is 2 days. (a) Explain the terms: (i) Radioactive half-life ..................................................................... ..................................................................... ..................................................................... (ii) Biological half-life ..................................................................... ..................................................................... ..................................................................... (b) If the activity of the tracer sample introduced into the body was A, determine the activity of that portion of tracer remaining in the body 6 days later. ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... (c) For physiological studies, explain why it is desirable to use a tracer substance which has a greater radioactive half-life than biological half-life. ......................................................................... ......................................................................... ......................................................................... [2] [4] [2] [2]
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–6– OPTION E – HISTORICAL PHYSICS E1. Models of the universe
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The Ptolemaic model of the universe was geocentric while the Copernican model was heliocentric. (a) Ptolemaic model How did the Ptolemaic model account for the following observations? (i) The stars move in the sky during the course of the night, while the pattern of stars nevertheless remains unchanged. ..................................................................... ..................................................................... ..................................................................... (ii) The moon also moves across the sky but at a slightly slower rate than the stars, so that its position relative to the stars changes continually. ..................................................................... ..................................................................... ..................................................................... (iii) The ‘wandering stars’ (planets) move gradually with respect to the stars, and periodically reverse their motions before continuing again. Explain this by sketching suitable planetary cycles and epicycles on the diagram below, and tracing out the resulting motion of a planet. [2] [2]
[3]
Earth
. ... .... ... ..... .. . .. . ... . . . ...
Fixed stars
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–7– (Question E1 continued) (b) Copernican model
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On the Copernican (heliocentric) model the planets orbit the sun. The diagram below shows the orbits of the earth and Jupiter around the sun, with the ‘fixed’ stars in the far distance (not to scale).
Sun Orbit of Earth Orbit of Jupiter
. .. .. .. .. . . . .. . . . . . . .. . .. .. . Distant stars . . ..
The positions of earth and Jupiter at one particular time are as shown. The speed of Jupiter in its orbit is less than half that of the earth. (i) (ii) Draw in the approximate position of the earth and Jupiter roughly three months later. With the aid of constructions on the diagram, explain why Jupiter will exhibit ‘retrograde’ motion against the background of the stars. ..................................................................... ..................................................................... ..................................................................... ..................................................................... [1] [2]
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–8– (Question E1 continued) (c)
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Kepler found that the Copernican model could account for the observed motion of Jupiter and most of the other planets. However it could not quite match the motion of Mars, to the precision of Tycho Brahe’s observations, even using subsidiary cycles. What important break with all previous models did Kepler finally make to account for the motion of Mars? ......................................................................... .........................................................................
[2]
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–9– E2. Cannon boring and caloric
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Count Rumford observed the barrels of iron cannons being bored out by cutting tools, producing metal chips. He reported in 1798 that this mechanical process seemed to provide an ‘inexhaustible supply of heat’. (a) How did the ‘caloric’ theory existing at that time account for the production of heat in the cutting process? ......................................................................... ......................................................................... ......................................................................... (b) If the boring tool became blunt so that it did not cut as well, what did the caloric theory predict would happen to the rate of heat production, and why? Was this observed? ......................................................................... ......................................................................... ......................................................................... ......................................................................... (c) Why was the observation that the supply of heat seemed ‘inexhaustible’ a problem for the caloric theory? ......................................................................... ......................................................................... ......................................................................... (d) What new idea did Rumford propose to account for the production of heat in this process? ......................................................................... ......................................................................... ......................................................................... [2] [2] [2] [2]
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– 10 – E3. Virtual particles and forces (a) Explain what is meant by a virtual particle.
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[2]
......................................................................... ......................................................................... ......................................................................... (b) Describe how the electrostatic repulsion between two like charges (Coulomb force) can be explained in terms of the exchange of virtual photons. ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... (c) Complete the following table, listing the four fundamental forces, giving the exchange particles and characterising the ranges. Force 1. 2. 3. 4. ............. Electromagnetic ............. ............. Exchange particle graviton photon W ± , Z0 ............. . Range of force (infinite or short range) ............. ............. ............. ............. [5] [3]
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– 11 – OPTION F – ASTROPHYSICS F1. Stellar distances This question is about determining the distance to a nearby star.
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Two photographs of an area of the night sky are taken through a telescope from earth, one six months after the other. Comparing the photographs, one star seems to have shifted slightly relative to the other stars, as shown in the figure. (The figure is made up of the two photographs overlapping.)
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[1]
(a)
What can we deduce from the fact that one star appears displaced against the others? .........................................................................
(b)
If the observed angular displacement of a star is θ and the diameter of the earth’s orbit is d, show with the aid of a diagram that the distance D to the star is given approximately by the formula D ≈ d / θ , if θ is small and measured in radians.
[4]
......................................................................... ......................................................................... (c) Why can Hubble’s law not be used to determine the distance to the star? For what objects in the universe can Hubble’s law be used to determine distance? ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... [3]
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– 12 – F2. Hertzsprung-Russell diagram A Hertzsprung-Russell (H-R) diagram is shown in the figure below.
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Star magnitude
? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?? ?? ? ? ? ? ?? ? ? * ? ? ? ? ?? ? ? ? ?? ? ? ?? ?? ? ? ?? ? ? ? ? ?? ? our sun ? ? ?? ? ?? ?? ?? ? ? ?? ? ? ??
10 000 7000 5000 Temperature / K 3500
? ? ??
? ? ? ??
?
(a)
The vertical axis gives ‘star magnitude’. Must this be the apparent (observed) magnitude or absolute magnitude? Explain why. ......................................................................... .........................................................................
[2]
(b)
On the H-R diagram, why do we choose to use a logarithmic scale for star magnitudes rather than a linear scale? ......................................................................... ......................................................................... .........................................................................
[2]
(c)
The horizontal axis of the H-R diagram gives star temperature. temperature or the surface temperature of the star? Explain why.
Is this the interior [2]
......................................................................... .........................................................................
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– 13 – (Question F2 continued) (d) How are the values for star temperatures obtained from the earth? procedure.
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Outline a possible [2]
......................................................................... ......................................................................... ......................................................................... (e) Consider a star of similar age to our own sun but of greater mass. (i) How would the luminosity and the colour of this more massive star compare with those of our sun? Explain your reasoning briefly. ..................................................................... ..................................................................... ..................................................................... ..................................................................... (ii) The position of our own sun on the H-R diagram is marked. Where on the diagram, relative to the sun, would the more massive star be located? Mark and label the region where it might occur. [3]
[1]
Star magnitude
? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?? ?? ? ? ? ? ?? ? ? * ? ? ? ? ?? ? ? ? ?? ? ? ?? ?? ? ? ?? ? ? ? ? ?? ? our sun ? ? ?? ? ?? ?? ?? ? ? ?? ? ? ??
10 000 7000 5000 Temperature / K 3500
? ? ??
? ? ? ??
?
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– 14 – (Question F2 continued) (f)
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Our sun will remain on the main sequence of the H-R diagram for about another 5 billion years, and will then become a red giant, following the evolutionary path shown on the previous page. (i) Explain why the sun will leave the main sequence, and describe the processes that occur as it becomes a red giant. ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... (ii) When the sun is in the ‘red giant’ stage, why will it be redder and more luminous than it is now? ..................................................................... ..................................................................... ..................................................................... ..................................................................... (iii) After the red giant stage the sun will evolve further. On the H-R diagram on the previous page trace the evolutionary path it will follow. (iv) What will be the eventual fate of our sun? Describe what it will be like. ..................................................................... ..................................................................... ..................................................................... ..................................................................... [1] [3] [2] [4]
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– 15 – OPTION G – SPECIAL AND GENERAL RELATIVITY G1. Relativity and simultaneity (a) State the two postulates of the special theory of relativity.
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[2]
......................................................................... ......................................................................... ......................................................................... ......................................................................... Einstein proposed a ‘thought experiment’ along the following lines. Imagine a train of proper length 100 m passing through a station at half the speed of light. There are two lightning strikes, one at the front and one at the rear of the train, leaving scorch marks on both the train and the station platform. Observer S is standing on the station platform midway between the two strikes, while observer T is sitting in the middle of the train. Light from each strike travels to both observers.
0.5 c
(b)
If observer S on the station concludes from his observations that the two lightning strikes occurred simultaneously, explain why observer T on the train will conclude that they did not occur simultaneously.
[4]
......................................................................... ......................................................................... ......................................................................... ......................................................................... (This question continues on the following page)
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– 16 – (Question G1 continued) (c) Which strike will T conclude occurred first?
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[1]
......................................................................... (d) What will be the distance between the scorch marks on the train, according to T and according to S? ......................................................................... ......................................................................... ......................................................................... (e) What will be the distance between the scorch marks on the platform, according to T and according to S? ......................................................................... ......................................................................... ......................................................................... ......................................................................... [2] [3]
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– 17 – G2. Space capsule
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Two space travellers Lee and Anna are put into a state of hibernation in a ventilated capsule in a spaceship, for a long trip to find another habitable planet. They eventually awake, but do not know whether the ship is still travelling or whether they have landed. They feel attracted toward the floor of the capsule, an experience rather like weak gravity. Lee says the spaceship must have landed on a planet and they are experiencing its gravitational attraction. Anna says the spaceship must be accelerating and the capsule floor is pushing on them. (a) Hoping to decide which of them is right, they try an experiment. They release a hammer in mid air, and it accelerates straight to the floor. Does this observation help them decide? How would each of them explain the motion of the hammer? ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... (b) They propose another experiment, namely to shine monochromatic light from the floor to the ceiling of the capsule and use sensitive apparatus to detect any change in frequency. (i) How would Lee explain how a redshift arises, viewing the radiation as photons moving in a gravitational field? ..................................................................... ..................................................................... ..................................................................... (ii) How would Anna explain how a redshift arises, viewing the radiation in terms of wavefronts arriving at a detector whose speed is increasing? ..................................................................... ..................................................................... ..................................................................... ..................................................................... (c) Can Lee and Anna perform any experiment in the capsule which could distinguish whether they are on the surface of a planet or accelerating in space? . . . . . . . . . . . . . . . . . . . . . . . . . . State why: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [1] [2] [2]
[4]
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– 18 – (Question G2 continued) (d)
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Later they notice that the gravitational-like sensation starts diminishing gradually, until they eventually ‘float weightless’ in the capsule. Lee suggests that they must have taken off from the planet, and as they got further away its gravitational attraction diminished until it was negligible. Anna suggests that the spaceship must have gradually reduced its thrust and acceleration to zero. Which explanation is feasible, or is there no way to tell who is right? Explain. ......................................................................... ......................................................................... ......................................................................... ......................................................................... .........................................................................
[3]
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– 19 – G3. Decay in flight
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An unstable nucleus is moving with velocity 0.5 c relative to the laboratory. While moving it decays by emitting an electron in the same direction of travel as the nucleus, at a speed of 0.7 c relative to the nucleus. (a) If one attempted to use Galilean kinematics, what speed would be predicted for the electron relative to the laboratory? ......................................................................... (b) Using relativistic kinematics, calculate the speed of the electron as measured in the laboratory frame. ......................................................................... ......................................................................... ......................................................................... ......................................................................... ......................................................................... (c) Viewed from the laboratory frame, how much faster is the electron travelling than the nucleus? ......................................................................... ......................................................................... [2] [3] [1]
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– 20 – OPTION H – OPTICS H1. Images in a convex lens
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An elderly lady buys a ‘magnifying glass’ to read small print in the telephone directory. To her surprise she finds that if she holds the convex lens fairly close to the page she gets one kind of image, while if she holds it fairly far from the page she gets quite another kind of image. Having studied physics long ago, she wishes to understand this using ray diagrams. (a) Lens close to the page For the lens quite close to the page she draws the ray diagram below. printed page Lens Letter “i” on page P
?
P′
(i)
Where should her eye be located in order to see the image of the letter “i”? Tick the correct answer below. To the left of the lens Anywhere to the right of the lens To the right of the focal point P′
[1]
(ii)
If she looks at letters on a page in this way, how will they appear to her? Right way up or upside down? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Enlarged or diminished? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Behind the lens or in front of the lens? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Should she be able to read the telephone directory using the lens this way? . . . . . . .
[2]
(iii) Instead of looking through the lens to see the image, could she ‘capture’ it by placing a screen or film at the image location? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [1]
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– 21 – (Question H1 continued) (b) Lens further from the page
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The lady now moves the page further from the lens. The diagram below represents the situation where the page is more than twice the focal distance from the lens.
printed page Lens
?
Letter “i” on page
P
P′
?
(i) (ii)
Locate the image of the “i” by tracing suitable rays on the diagram. Where should the lady’s eye be located in order to see the image? Tick the correct answer below. To the left of the lens Anywhere to the right of the lens Between the lens and P′ To the right of the image
[4] [1]
(iii) If she looks at letters on a page in this way, how will they appear to her? Right way up or upside down? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Enlarged or diminished? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Behind the lens or in front of the lens? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nearer or further away than the page? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Will she be able to read the telephone directory using the lens this way? . . . . . . . . . . . (iv) Instead of looking through the lens to see the image, could she ‘capture’ it placing a screen or film at the image location? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
[2]
[1]
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– 22 – H2. Double-slit interference
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In a classroom demonstration, laser light is shone onto two narrow slits S1 and S2 in a dark room, and an interference pattern of bright and dark lines (fringes) appears on a screen, as shown in the figure below. (The fringe spacing is exaggerated for clarity.) Bright Dark Bright P Dark S1 S2 C Bright Dark Bright Dark Bright (a) A fellow student asks you: “How can it be dark at point P? After all, light must be arriving there from both slits”. Is the student correct that light is arriving at point P from both slits? . . . . . . . . . . . . . . . . . . . How would you explain to the student why it is dark at point P? ......................................................................... ......................................................................... ......................................................................... ......................................................................... [2]
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– 23 – (Question H2 continued) (b)
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The situation is shown again in Figure 1. (a) below, and a ‘magnified’ view of the slit region is shown in the circle in Figure 1. (b).
Figure 1. (a)
Bright P Dark
S1 S2
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Dark Bright
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Figure 1. (b) (magnified view)
Show that the angle θ at which the first dark fringe P occurs is given by the expression sin θ = λ / 2d , where λ is the wavelength of the light. Assume the screen is far away. Draw and label any construction needed on Figures 1. (a) and 1. (b), and explain the steps in your derivation. ............................................................................. ............................................................................. ............................................................................. .............................................................................
[4]
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– 24 – (Question H2 continued) (c) Suppose a third slit were opened an equal distance below S2 , as shown.
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S1 S2
Third slit Would point P on the screen remain dark, or not? Explain. ......................................................................... ......................................................................... ......................................................................... ......................................................................... [2]
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– 25 – H3. Optical dipstick This question is about total internal reflection of light and a practical application. (a)
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The diagram below shows light incident on the interface between two transparent media of refractive indices n1 and n 2 . In general the light will be partly reflected and partly refracted. Sketch in the reflected and refracted rays, for the case n 2 < n1 .
[2]
n2
θ
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(b)
Show that the critical angle of incidence θc for total internal reflection of the light is given by n the expression sin θ c = 2 . n1 ......................................................................... .........................................................................
[2]
(This question continues on the following page) Turn over
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– 26 – (Question H3 continued) (c)
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The figures below show an ‘optical dipstick’ used to warn if the level of oil in a storage tank drops too low. A glass rod is bent into a curved U-section as shown. Light enters at one end of the rod and there is a light detector at the other end. Normally the U-section is immersed in oil but if the level drops too low the section is exposed to air. The refractive indices of glass, oil and air are 1.5, 1.3 and 1.0 respectively. source detector source detector
oil
(i)
By completing the diagrams above, illustrate how light is totally internally reflected around the bend in one case and not in the other. Sketch the paths of the light rays in each case.
[2]
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M00/430/H(3)
The figure below shows a ray entering a glass rod of radius r and radius of curvature R. The refractive index of the glass is n1 and that of the surrounding medium is n 2 . The central ray strikes the side of the rod at an angle α as shown. central ray glass rod r r
R α n2 surrounding medium n1 R average radius of curvature
glass rod
The radius of curvature R of the bend cannot be less than a certain minimum value or total internal reflection will not occur. Show that the minimum bend radius R is related to the rod radius r and the refractive indices n1 and n 2 by the equation r . R= n1 ?1 n2 ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... ..................................................................... .....................................................................
[4]
220-228 | 5,920 | 33,411 | {"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} | 3.328125 | 3 | CC-MAIN-2018-30 | latest | en | 0.674798 |
https://www.rocketryforum.com/threads/minuteman-3-scale-scratch-build.9773/ | 1,701,546,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00019.warc.gz | 1,083,646,580 | 29,820 | # Minuteman 3 scale scratch build
### Help Support The Rocketry Forum:
#### Fred22
##### Well-Known Member
Well guys Im going to make a start on this thread with some some general ideas and chime in any time with advice or suggestions Im trying to use model rockets by Timothy Van Milligan. Bought it last year so its time to get some use out of it. Its a great book so dont judge it by what i do for goodness sake Following the suggestted steps for a scale build I selected the Minuteman 3.
I gathered some scale data here
https://www.captainswoop.com/icbm/msls.html
Im going to use a quad 18mm power plant to simulate the configuration of the real one in a BT80 body tube. I figure the scale to be 1/25 by using 2.6 inches for the BT80 body tube and 66 inches or 5.5 feet for the base diameter of the real rocket. Math is not my long suit but here goes. Using the scale of 1/25 then the overall length should be 28.75 inches long. I seem to be missing from my drawing the diameter of the upperstage so I will try to figure that out tommorrow
Cheers
fred
#### Fred22
##### Well-Known Member
Thanks for the info I found the data on another part of the good captains websiteDid a little more math based on scale data. The diameter is expressed as 4.33 feet so my second stage should be about two inches in diameter. That means a BT70 upperstage. Not perfect but I have a pile of stock parts and without a micrometer whos going to notice Next I will try to figure out transitions and a nose cone. Im tempted by plastic as its so easy to add weight to
Cheers
fred
#### SpaceAXEplorer
##### Well-Known Member
Thanks for the link(s) I've been saving up data on the Minutemen to do a build of my own (non-rocketry). Every little bit helps.
Eric
#### Fred22
##### Well-Known Member
Thanks for the link(s) I've been saving up data on the Minutemen to do a build of my own (non-rocketry). Every little bit helps.
Eric
No problem Eric I am learning as I go. I usually just kitbash so this is new ground to me
Cheers
Fred
#### kjohnson
##### mox nix
One of the coolest things in Peter Alway's Art of Scale Modeling book is the tube ratio chart. If you can figure out the ratio between your two tube diameters, this chart will show you all the commercial tubes that fall around that ratio so you can chose a scale based on that.
I don't know if anyone else has put something like that together somewhere else, and AoSM is long out of print and hard to find, but it's really great tool for when you start doing scratch builds and don't want to make your own tubes.
kj
Thanks for the info I found the data on another part of the good captains websiteDid a little more math based on scale data. The diameter is expressed as 4.33 feet so my second stage should be about two inches in diameter. That means a BT70 upperstage. Not perfect but I have a pile of stock parts and without a micrometer whos going to notice Next I will try to figure out transitions and a nose cone. Im tempted by plastic as its so easy to add weight to
Cheers
fred
#### Fred22
##### Well-Known Member
One of the coolest things in Peter Alway's Art of Scale Modeling book is the tube ratio chart. If you can figure out the ratio between your two tube diameters, this chart will show you all the commercial tubes that fall around that ratio so you can chose a scale based on that.
I don't know if anyone else has put something like that together somewhere else, and AoSM is long out of print and hard to find, but it's really great tool for when you start doing scratch builds and don't want to make your own tubes.
kj
Im pretty comfortale with the tube selection and power plant. The size and shape of the fins will be the next puzzler and how mauch weight to put in the nose if any to balance it. I also need to figure out how to attach the BT70 to the bt80? I will use a paper shroud for the midsection as EMRR has a good calc for that and the chute size. I apreciate your interest in my project and i hope you continue to follow it and offer your sdvice. I apreciate it
Cheers
fred
#### dedleytedley
##### Well-Known Member
Cool project Fred! You'll want to use two BT-70-BT-80 rings to place the 70 within 80. The smaller tube should be about three inches into the larger tube. Four 18mm should give a fast exciting boost. Sunward sells a kit the Eruption that has four Canted 18mm in a BT-70. Using canted motors will increase the stability of the rocket reducing the fin area required. Ted
#### Fred22
##### Well-Known Member
Cool project Fred! You'll want to use two BT-70-BT-80 rings to place the 70 within 80. The smaller tube should be about three inches into the larger tube. Four 18mm should give a fast exciting boost. Sunward sells a kit the Eruption that has four Canted 18mm in a BT-70. Using canted motors will increase the stability of the rocket reducing the fin area required. Ted
Good advice Im going with straight motors though as thats the mount I have. Means I still have to figure the fins though
The cone I want to use is an about an inch to long maybe for scale but that depends on how much goes into the tube.
https://www.semroc.com/Store/scripts/prodView.asp?idproduct=2146
Found those rings sweet
https://www.semroc.com/Store/scripts/prodView.asp?idproduct=1902
Im think A transition may save a bit of work with paper as well and i could use it for the recovery and attach the nose
https://www.semroc.com/Store/Scripts/prodView.asp?idproduct=915
This is getting very interesting. Semroc sure has a great selection of stuff
Cheers
fred
#### luke strawwalker
##### Well-Known Member
Good advice Im going with straight motors though as thats the mount I have. Means I still have to figure the fins though
The cone I want to use is an about an inch to long maybe for scale but that depends on how much goes into the tube.
https://www.semroc.com/Store/scripts/prodView.asp?idproduct=2146
Found those rings sweet
https://www.semroc.com/Store/scripts/prodView.asp?idproduct=1902
Im think A transition may save a bit of work with paper as well and i could use it for the recovery and attach the nose
https://www.semroc.com/Store/Scripts/prodView.asp?idproduct=915
This is getting very interesting. Semroc sure has a great selection of stuff
Cheers
fred
Course you could always turn the cone down an inch by chucking it in a drill and putting some sandpaper to it, like I did the transition for my 1/100 Ares I scratchbuild over in the archives... It's not the best solution, but it works, and it's easier than making a cone from scratch from a block of balsa...
Later and good luck! OL JR
#### Marlin523
##### Well-Known Member
Interesting project. If you go with BT-70 and BT-80, what lengths will they be?
#### Fred22
##### Well-Known Member
Course you could always turn the cone down an inch by chucking it in a drill and putting some sandpaper to it, like I did the transition for my 1/100 Ares I scratchbuild over in the archives... It's not the best solution, but it works, and it's easier than making a cone from scratch from a block of balsa...
Later and good luck! OL JR
I apreciate the suggestion but the extra inch vs what I would do to the cone is not worthwhile Oh well sport scale here we come
Cheers
fred
#### Fred22
##### Well-Known Member
Interesting project. If you go with BT-70 and BT-80, what lengths will they be?
Given the extra big nosecone 7.5 inches for BT70 10.37 for BT80. The is will allow a 2 inch transition and a seven inch nose cone aproximately.
Cheers
fred
fins?
#### Fred22
##### Well-Known Member
Gotta have them to make it stable as I seem to have left my manual for gymballing LPR engines somewhere lol. They will be clear plastic to minimise the effect on the appearance
Cheers
fred
#### Fred22
##### Well-Known Member
Well I orderred the bits and bobs from Semroc for my minuteman today
Here's the list
BNC-70AJ Balsa Nose Cone BT-70 4.25" Ogive
2 BT-70 Body Tube BT-70 18.0" Long
Centering Ring BT-70 to BT-80 (Pkg of 6)
1 TA-7080 Tube Adapter BT-70 to BT-80
Im really excited about orderring this stuff as Ive heard SEMROC service is the best and maybe my little bit of coin will help cheer Carl. I figure the next thing will be a detailled scale drawing. next its fin city and how much do I need. Im thinking I will build them right into the rocket possibly attaching them right to a stuffer tube assembly buts all the skull sweat i have availible for now
Cheers
fred
#### Fred22
##### Well-Known Member
Okay I got my bits and bobs from SEMROC. They got my stuff up here is a very reasonable amount of time for a great price including price My nosecone is not to scale but is going to be 10.8 cm long. Should be more then that but thats as close as i could get in my budget. Upper stage will be 36.16 cm long to compensate for short nose and transition. The transition will be 5.5 cm long and the first stage will 36.72 cm long. The raceway on the outside will be 62.48 cm long. I will attch the chute to the transition. Next I need to figure the fins again I also need to figure out how big a chute
Cheers
fred
#### Fred22
##### Well-Known Member
Well I've still gotta build a tunel and fins but it's shaping up My hobby area is kindly described as somewhat chaotic.
Cheers
Fred
#### mperdue
##### Well-Known Member
Ive seen a lot of bone white minuteman 3's in photos as gate guards or displays. Does this photo show what one looked like when they had it in a silo?
https://www.nasm.si.edu/collections/artifact.cfm?id=A19761115000
Cheers
fred
Great build. The white missiles were for display. The one in the silo look like this https://www.fas.org/nuke/guide/usa/icbm/us_nuke_minuteman3_01.jpg.
There were a normally differences in the paint scheme from ones missile to another but tha photo is close to the "standard" look.
FWIW, I worked on Minuteman III when I was in the USAF.
Mario
#### Fred22
##### Well-Known Member
Hey Mario thanks for the picture What colour from the model master lineup of spray cans do you think I should use? Working on the real missile must have been interesting work. I toured a LF in North Dakota and a silo trainer at ellesworth which is as close as I got. Kind of means I should pay attention to your suggestions lol I apreciate your help
Cheers
Fred
#### mperdue
##### Well-Known Member
Hey Mario thanks for the picture What colour from the model master lineup of spray cans do you think I should use? Working on the real missile must have been interesting work. I toured a LF in North Dakota and a silo trainer at ellesworth which is as close as I got. Kind of means I should pay attention to your suggestions lol I apreciate your help
Cheers
Fred
Colors varied with the individual missile. The USAF only paints for protection, not for cosmetics. The white areas were flat white. The dark areas were either flat black or flat dark olive drab. The light area is a difficult color to match. It's a flat yellow zinc chromate primer with a bit of a green tint. I've never seen an exact match for it anywhere but it could be mixed pretty easily.
FWIW, I was stationed at Grand Forks AFB in North Dakota.
##### Well-Known Member
TRF Supporter
Do you have clear plastic fins for this model?
#### Fred22
##### Well-Known Member
Colors varied with the individual missile. The USAF only paints for protection, not for cosmetics. The white areas were flat white. The dark areas were either flat black or flat dark olive drab. The light area is a difficult color to match. It's a flat yellow zinc chromate primer with a bit of a green tint. I've never seen an exact match for it anywhere but it could be mixed pretty easily.
FWIW, I was stationed at Grand Forks AFB in North Dakota.
It never ceases to amaze me how much firepower is buried in the ground so close to the border. Saw lots of active silos on my last vacation as well. I've got an airbrush coming so a mixing we will go It maymake you laugh but Im going to swing test it I'm a bit of a dinosaur that way.
Cheers
fred
#### Fred22
##### Well-Known Member
Do you have clear plastic fins for this model?
Not yet but Im going to make them and attch them after painting
Cheers
fred
#### dpan
##### New Member
Colors varied with the individual missile. The USAF only paints for protection, not for cosmetics. The white areas were flat white. The dark areas were either flat black or flat dark olive drab. The light area is a difficult color to match. It's a flat yellow zinc chromate primer with a bit of a green tint. I've never seen an exact match for it anywhere but it could be mixed pretty easily.
FWIW, I was stationed at Grand Forks AFB in North Dakota.
I just found this thread. Great topic.
I was the officer in charge of the team training branch at the 91 SMW, Minot AFB, ND about 25 years ago. The missiles we had (MM IIIs) were not the color of that FAS picture. I believe that pic was taken at the USAF Museum at Wright Patterson AFB, OH. All the missiles we had in the silos had the following color scheme - 1st stage - kind of glossy green/lime green with the name "Thiokol" in the company's script logo going down the length of the stage (in red). Second and third stages were a more olive drab type of paint. All the stages had white on brown rectangular stencils with the word "LOADED" on them painted horizontally. The post-boost vehicle and guidance unit (the Post-Boost Control System) were painted white. The Re-entry system (nose cone) shell was bare metal / aluminum. That's about it. I spent enough time inside the LFs (I was an MMT officer) to have had that missile permanently planted in my brain
#### Fred22
##### Well-Known Member
I just found this thread. Great topic.
I was the officer in charge of the team training branch at the 91 SMW, Minot AFB, ND about 25 years ago. The missiles we had (MM IIIs) were not the color of that FAS picture. I believe that pic was taken at the USAF Museum at Wright Patterson AFB, OH. All the missiles we had in the silos had the following color scheme - 1st stage - kind of glossy green/lime green with the name "Thiokol" in the company's script logo going down the length of the stage (in red). Second and third stages were a more olive drab type of paint. All the stages had white on brown rectangular stencils with the word "LOADED" on them painted horizontally. The post-boost vehicle and guidance unit (the Post-Boost Control System) were painted white. The Re-entry system (nose cone) shell was bare metal / aluminum. That's about it. I spent enough time inside the LFs (I was an MMT officer) to have had that missile permanently planted in my brain
Thanks Its very cool talking to folks who were there as it were . Have you got a picture? Nothing that would require you to be imprisoned anyway
Cheers
fred
#### dpan
##### New Member
Thanks Its very cool talking to folks who were there as it were . Have you got a picture? Nothing that would require you to be imprisoned anyway
Cheers
fred
I just did a quick image search on Google. Here are a few interesting pics:
https://i63.photobucket.com/albums/h146/boeing722/MMII.jpg
That's of a MM II, but the paint scheme is pretty accurate. Though I'm more used to a lighter shade of green on the first stage with the Thiokol logo (https://www.ask.com/wiki/Thiokol).
This shot of a MMIII inside the LF has a good look at the PBCS and the RS shell. (https://www.corbisimages.com/Enlargement/UG004029.html)
This one of the MMIII in the Smithsonian Air and Space Museum is pretty accurate for the first stage color scheme. (https://www.flickr.com/photos/nostri-imago/3095747545/) Though I've never seen an operation missile with that Re-entry System paint scheme. It's always the bare metal one seen previously.
Hope that helps.
#### mperdue
##### Well-Known Member
It's interesting to see how the paint schemes changed over the years. The scheme I referenced was accurate for the missiles I worked on in the mid 1970s.
#### Fred22
##### Well-Known Member
It's interesting to see how the paint schemes changed over the years. The scheme I referenced was accurate for the missiles I worked on in the mid 1970s.
With the military it's usually a different shade of green
Cheers
fred
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663 | 3,961 | 16,252 | {"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-2023-50 | latest | en | 0.903424 |
http://mathhelpforum.com/calculus/171373-show-there-no-tangents-curve-y-x-2-3x-4-positive-slope.html | 1,529,901,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00470.warc.gz | 195,210,447 | 9,921 | # Thread: Show that there are no tangents to the curve y = (x+2)/(3x+4) with a positive slope.
1. ## Show that there are no tangents to the curve y = (x+2)/(3x+4) with a positive slope.
Show that there are no tangents to the curve $\displaystyle y=\frac{x+2}{3x+4}$ with a positive slope.
How would I solve this? All help appreciated!
2. Originally Posted by youngb11
Show that there are no tangents to the curve $\displaystyle y=\frac{x+2}{3x+4}$ with a positive slope.
How would I solve this? All help appreciated!
Calculate the numerator of the derivative with the Quotient Rule
and show that it cannot be >0
(the denominator of the derivative is a square)
$\displaystyle v\frac{du}{dx}-u\frac{dv}{dx}=(3x+4)-3(x+2)$
3. Originally Posted by Archie Meade
Calculate the numerator of the derivative with the Quotient Rule
and show that it cannot be >0
(the denominator of the derivative is a square)
$\displaystyle v\frac{du}{dx}-u\frac{dv}{dx}=(3x+4)-3(x+2)$
Numerator worked out to be $\displaystyle -2$. Any tips on how you would show it (format) if you were doing it? Thanks again!
4. The derivative gives you the slope of the tangent, in terms of x.
Since the denominator is a square, it's positive.
Then, since the numerator is free of x and negative, the derivative is always negative,
which means that all tangents have negative slopes
(no tangent has a positive slope). | 378 | 1,389 | {"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} | 4.03125 | 4 | CC-MAIN-2018-26 | latest | en | 0.935805 |
https://projectiot123.com/2019/03/06/laplace-transform/ | 1,632,004,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00517.warc.gz | 531,424,513 | 54,968 | Select Page
[otw_is sidebar=otw-sidebar-1]
In this post I will discuss about (introduction to Laplace Transform)the Laplace Transform Definition and Inverse Laplace Transform. In the last post I have discussed about the transfer function and bode plot in matlab and there I have mentioned that the Transfer Function in MATLAB and bode plot in matlab
After reading this post you will learn about the laplace transformation,and different examples of laplace transformation. So sit back, keep reading and enjoy learning.
[otw_is sidebar=otw-sidebar-2]
#### Laplace Transform:
A signal or a system can be represented in either the time domain or frequency domain. The representation of the signal or the system in time domain is a function of time and the representation of the signal or system in frequency domain is a function of frequency. The representation of the system in terms of the frequency variable is commonly called the transfer function. The transfer function of the system compliances the knowledge about the behavior of the system in the frequency domain. Thus the transfer function gives the knowledge about the magnitude and phase relationship between the input and output of the system for a particular range of frequency thus the transfer function of the system can be thought of as the frequency response of the system.
The concept of transform is very important in Mathematics with the help of which one can alter the representation of data without affecting the specific nature of the data. In engineering circuit analysis and design there are various transform tools available some important and relevant ones are mentioned below:
• Laplace Transform.
• Fourier Transform.
• Z-Transform.
[otw_is sidebar=otw-sidebar-3]
##### introduction to Laplace Transform:
The discussion here is oriented on Laplace Transform and Fourier transform and Z- transform are out of the scope of this discussion.
With the help of this transform technique one can alter the representation of data for example the concept of Laplace transform is used to alter the representation of system or signal from time domain to frequency domain. Thus the Laplace transform is a tool for transforming the time domain representation of the system or signal to the frequency domain representation. It must be noted here that the Laplace transform of the system represents the frequency response of the system and in the similar way the Laplace Transform of the signal in time domain represents the spectrum of the signal. The representation of the Laplace transform is as shown in the following figure:
The expression on the right hand side shows the Laplace transform of the time domain representation of the system. The symbol ‘s’ in this representation is used to denote the complex frequency variable.
#### Laplace Transform Definition:
[otw_is sidebar=otw-sidebar-2]
The Laplace transform is defined by the following equation;
The Integral transform shown in the above equation converts the time domain representation of the system into the frequency domain representation of the system.
#### s-Domain Circuit Analysis:
The concept of Laplace transform is also used in the circuit analysis. The Laplace transform of the time domain representation of the system gives the different perspective of the system. Anyhow the use of the concept of Laplace transform of the system is depicted in the following figure:
The circuit analysis can be performed in two ways:
1. Classical Approach for circuit analysis.
2. Laplace method for circuit analysis.
##### introduction to Laplace Transform
In the classical approach for circuit analysis the mathematical operations are performed on the time domain representation of the system. As can be seen in the above image that the input and output relationship of the system can be represented in the time domain using the differential equation. In order to solve these differential equations one has to involve in critical calculations before they get the required information. The Laplace transform method for circuit analysis and design bypasses these hectic calculations. The Laplace transform of the differential equation converts the representation of the system in the frequency domain and also converts the differential equations into the simple algebraic equations with variable ‘s’ (complex frequency variable) which can be solved using relatively simple algebraic manipulations. Once the algebraic manipulations have been carried out on the Laplace Transform of the system and have got the final response of the system in s domain the inverse Laplace transform is carried out to obtain the final response of the system in the time domain. Thus in this way Laplace transform bypasses the hectic calculations involved in the time domain calculations.
#### Inverse Laplace Transform:
The Laplace transform of the system or signal can be converted back to the time domain representation of the system with the help of the Inverse Laplace Transform. The Inverse Laplace transform is represented as follows:
The definition of the Inverse Laplace transform is shown in the following figure:
#### Laplace Transform Pairs:
The Laplace transform of the commonly known functions are listed in the following table:
#### Laplace Transform Properties:
[otw_is sidebar=otw-sidebar-3]
The Laplace transform follow commonly known properties. The following list shows some of the properties that the Laplace Transform posses:
That is all for now I hope this article (introduction to Laplace Transform)would be helpful for you. In the next post I will come up with more interesting educational topics. Till then stay connected, keep reading and enjoy learning. | 1,067 | 5,706 | {"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} | 3.71875 | 4 | CC-MAIN-2021-39 | longest | en | 0.88195 |
https://study.com/academy/topic/decimals-for-elementary-school.html | 1,563,339,883,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525046.5/warc/CC-MAIN-20190717041500-20190717063500-00189.warc.gz | 560,720,678 | 24,233 | # Ch 21: Decimals for Elementary School
These online text lessons are a great resource for helping elementary school students learn about decimals. Students can use them to improve their skills in working with decimals in mathematics, help them with their homework or to help them prepare for a test.
## Decimals for Elementary School - Chapter Summary
If you are the parent or teacher of an elementary school child who is learning about decimals, use these online lessons to help them with their studies or math homework. Each lesson provides an age-appropriate overview of different decimal concepts that they will cover in their math class. The chapter also includes practice quizzes students can use to ensure they have understood the main points from the lessons.
## Who It's For
Kids who need extra help with decimals will benefit from the lessons in this chapter. There's no faster or easier way to learn about decimals.
• Students who have fallen behind in understanding math problems involving decimals
• Students who have learning disabilities or need a little extra help learning about decimals
• Students who prefer multiple ways of learning math (visual or auditory)
• Students who have missed class and need to catch up
• Students who need an efficient and convenient way to learn about decimals
• Students who struggle to understand their math teachers
• Students who attend schools without extra math resources
## How It Works
• Find lessons in the chapter that cover decimal topics that your student needs to learn or review.
• Press play to watch the video lesson with your student or read through the text lesson.
• Review the lesson or video transcripts, emphasizing the highlighted vocab words to reinforce learning about decimals.
• Test your student's understanding of each lesson with short quizzes.
• Verify your student understands decimals by completing the Decimals for Elementary School chapter exam.
4 Lessons in Chapter 21: Decimals for Elementary School
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 497 | 2,606 | {"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-2019-30 | longest | en | 0.921676 |
http://www.transum.org/software/sw/starter_of_the_day/Similar.asp?ID_Topic=15 | 1,519,127,664,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00381.warc.gz | 566,127,670 | 12,963 | There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Factors. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Factors Starters:
Abundant Buses: A game based around the concept of abundant numbers.
Factuples: Spot the factors and the multiples amongst the numbers in the grid.
Flabbergasted: If each number in a sequence must be a factor or multiple of the previous number what is the longest sequence that can be made from the given numbers?
Four Factors: Find four single digit numbers that multiply together to give 120. How many different ways are there of answering this question?
Hotel Digital: A puzzle about the lifts in a hotel which serve floors based on the day of the week.
Verruca Value: The Verruca Value of a word is the number of vowels multiplied by the number of consonants. How many words can you find with Verruca Value of 24?
#### Four Colour Theorem
This mathematical activity involves colouring in! That makes a change. Using the paint bucket tool can you flood fill the regions so that no two adjacent regions are the same colour?
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Could we have some on angles too please?"
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### Notes:
A factor is a whole number that divides exactly into another whole number. We say the first number is a factor of the second number. Prime numbers only have two factors, one and themselves.
After becoming familiar with times tables pupils then practice using this knowledge by recognising factors of numbers. There are well known and some less well known divisibility tests that are of some use in solving more complex number problems.
Pupils need to know how to find the highest common factor (HCF) of two or more numbers either mentally or using a pen and paper strategy so that they can correctly manipulate fractions and algebraic expressions.
### Factors Teacher Resources:
Number Grids: Investigate the properties of number with these interactive number grids.
Prison Cell Problem: A number patterns investigation involving prisoners and prison guards.
Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers.
### Factors Activities:
Connect 4 Factors: This a game for one or two players. The winner is the first to line up four numbers with a common factor.
Delightfully Divisible: Arrange the digits one to nine to make a number which is divisible in the way described.
Divisibility Test: Practise using the quick ways to spot whether a number is divisible by the digits 2 to 9.
Factor Trees: Create factor trees to find the prime factors of the given numbers.
Factorising: Practise the skills of algebraic factorisation in this structured online self marking exercise.
Fizz Buzzer: The digital version of the popular fizz buzz game. Press the buzzers if they are factors of the counter.
HCF and LCM: Practise finding the highest common factor (H.C.F), sometimes called the greatest common divisor, and the lowest common multiple (L.C.M) of two numbers.
Number Grids: Investigate the properties of number with these interactive number grids.
Prime Labyrinth: Find the path to the centre of the labyrinth by moving along the prime numbers.
Prison Cell Problem: A number patterns investigation involving prisoners and prison guards.
Satisfy: Place the nine numbers in the table so they obey the row and column headings.
Scallywags and Scoundrels: Arrange the scallywags and scoundrels on the chairs so that the numbers of any two sitting next to each other add up to a prime number.
Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers.
Stamp Sticking: Drag stamps onto the envelopes to make the exact postage as shown at the top left of each envelope.
Three Ways: Find three different ways of multiplying four different digits together to get the given target number. There are nine levels for this online challenge.
Times Square: Practise your times tables with this self-checking multiplication grid
### Factors Investigations:
Aunt Sophie's Post Office: Investigate the ways of making up various postage amounts using 3p and 8p stamps. An online stamp calculator is provided for you to check your working.
### Factors Videos:
Finding Prime Factors: A straight forward explanation from SLEP
HCF and LCM explained: This video from Mathsmaster.org shows very clearly the step by step method of finding the LCM and HCF of two numbers.
HCF and LCM explained part 2: This video from Mathsmaster.org shows very clearly the step by step method of finding the LCM and HCF of two numbers using Prime Factorisation.
Links to other websites containing resources for Factors are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
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#### Measuring Angles
Measure the size of the given angles to within two degrees of their actual value. So far this activity has been accessed 5205 times and 1178 people have earned a Transum Trophy for completing it.
### Homepage
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Tuesday, August 29, 2017
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http://www.ibpsguide.com/2016/09/ibps-clerk-2016-practice-reasoning-questions-set-3_7.html | 1,498,350,043,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320368.57/warc/CC-MAIN-20170624235551-20170625015551-00683.warc.gz | 544,647,497 | 84,346 | ## 8 Sep 2016
### IBPS Clerk 2016- Practice Reasoning Questions [Answers Updated]
IBPS Clerk 2016- Practice Reasoning Questions Set-3:
Dear Readers, Important Practice Reasoning Questions with explanation for Upcoming IBPS Clerk Exams was given here with explanation, candidates those who are preparing for those exams can use this practice questions.
Directions (Q. 1-5): Study the following information to answer the given questions.
Twelve people are sitting in two parallel rows containing six people each such that they are equidistant from each other.
In row 1, Abhinav, Abhiraam, Abhiraj, Abhirath, Abhirup and Abhishek are sitting facing South. In row 2, Abhati, Abhidha, Abhidhya, Abhijna, Abhilasa and Abhinithi are sitting facing North.
Therefore, in the given seating arrangement, each member sitting in a row faces another member of the other row.
Three persons sit between Abhiraj and Abhirath. Either Abhiraj or Abhirath sits at an extreme end of the line. The one who faces Abhirath sits third to the left of Abhidhya. Abhijna faces the one who sits third to the left of Abhinav and he cannot sit adjacent to Abhidhya. The immediate neighbour of Abhidha faces the immediate neighbour of Abhinav. Only one person sits between Abhati and Abhilasa, who is facing the one sitting on the immediate right of Abhirup. Neither Abhirup nor Abhishek faces Abhidhya. Abhinithi and Abhidha cannot sit adjacent to each other.
1). Who among the following faces Abhiraam?
a) Abhati
b) Abhidha
c) Abhilasa
d) Abhidhya
e) Abhijna
2). Who among the following sit at the extreme ends of the rows?
a) Abhirath, Abhidha
b) Abhilasa, Abhishek
c) Abhirup, Abhijna
d) Abhiraam, Abhilasa
e) Abhishek, Abhidha
3). If Abhirup is related to Abhati in the same way as Abhiraj is related to Abhijna, which of the following is Abhinav related to, following the same pattern?
a) Abhinithi
b) Abhidhya
c) Abhidha
d) Abhilasa
e) Cannot be determined
4). How many persons are sitting between Abhirup and Abhishek?
a) None
b) One
c) Two
d) Three
e) Four
5). Which of the following is true regarding Abhati?
a) Abhidha is an immediate neighbour of Abhati.
b) Abhati faces the immediate neighbour of Abhiraam.
c) Abhati sits second from the left end of the row.
d) Abhati sits at one of the extreme ends.
e) Abhati faces the one who is second from the right end of the row.
Directions (Q. 6-10): Below is given a passage followed by several possible inferences which can be drawn from the facts stated in the passage. You have to examine each inference separately in the context of the passage and decide upon its degree of truth or falsity. Mark answer
a) if the inference is “definitely true”, ie it properly follows from the statement of facts given.
b) if the inference is “probably true” though not “definitely true” in the light of the facts given.
c) if the “data are inadequate”, ie from the facts given you cannot say whether the inference is likely to be true or false.
d) if the inference is “probably false” though not “definitely false” in the light of the facts given.
e) if the inference is “definitely false”, ie it cannot possibly be drawn from the facts given or it contradicts the given facts.
Pharmaceutical science involves the synthesis and mass manufacture of drugs from natural and synthetic sources for the treatment and prevention of diseases. Doctors rely on pharmaceutical companies for the development of new and more effective drugs to combat diseases. Today, pharmaceutical sector is a highly competitive sector with lots of companies competing for supremacy and market control. Pharmacists are concerned with production, productive processes, quality control and testing of drugs. According to Indian government statistics, today the market comprises over 25000 manufacturing units; one-fifth of them involved in making bulk drugs. The fact that the industry has been showing nearly 10 per cent growth, indicates the potential it has. With liberalisation in practice, there are many job opportunities opening up, especially in multinational pharmaceutical companies.
6). Below 5000 manufacture units are involved in making bulk drugs.
7). An interest in life science is an essential qualification to specialise in the field of pharmaceuticals.
8). There is a good range of employment opportunities for specialisers.
9). Indigenous companies are accumulating huge profits as compared to multinational companies.
10). All those who want to become a doctor but fail, end up in pharmacy.
1)d 2)b 3)b 4)d 5)b 6)e 7)b 8)a 9)d 10)c
Explanation:
Directions (Q.1-5):
6). According to the passage, more than 5000 manufacturing units are involved in making bulk drugs.
7). Service provided through pharmaceuticals and its nature of job make it probably true.
8). Passage says that pharmacists are concerned with different types of work such as production, productive processes, quality control and testing of drugs etc. And the last sentence of the passage says that there are many job opportunities opening up.
9). Job prospects are better for an individual in MNCs. It is likely MNCs are in a stronger position than indigenous companies. Hence, the inference drawn seems false. | 1,283 | 5,265 | {"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} | 3.25 | 3 | CC-MAIN-2017-26 | latest | en | 0.906101 |
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LinuxQuestions.org A formula for 'applicable figure' in IRS form 8962
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12-18-2018, 11:17 AM #1 RandomTroll Senior Member Registered: Mar 2010 Distribution: Slackware Posts: 1,090 Rep: A formula for 'applicable figure' in IRS form 8962 I do my taxes with a spreadsheet. I want a formula to calculate 'applicable figure' (stupid name!) in IRS form 8962. Has anyone done this? I found a page in Google Groups that isn't good enough. I decided it was hopeless, created a lookup table. I made a CSV version, good for 2018's taxes, as of 2018 December 18. You can find it at https://pastebin.com/4UZN0tJg . I use sc; if anyone wants the sc version, ask. Last edited by RandomTroll; 12-18-2018 at 01:35 PM. Reason: Answered my own question.
12-18-2018, 01:38 PM #2 michaelk Moderator Registered: Aug 2002 Posts: 17,934 Rep: Been a few years since I have had to do logical formulas. Untested but here is an Excel example. Code: `=IF(L5>=300,0.0956,IF(L5>132,L5*0.0006-0.0496,0.0201))` Similar for libreoffice except it uses ; instead of , I think. Code: `=IF(L5>=300;0.0956,IF(L5>132;L5*0.0006-0.0496;0.0201))` L5 is your 8962 line five cell value. Crap... After I paid more attention to the table I noticed that it isn't a simple first order equation... You might be able to graph the values and see what a higher order equation looks like and replace it with mine. Last edited by michaelk; 12-18-2018 at 01:48 PM. Reason: oops...
12-18-2018, 02:00 PM #3 MensaWater LQ Guru Registered: May 2005 Location: Atlanta Georgia USA Distribution: Redhat (RHEL), CentOS, Fedora, CoreOS, Debian, FreeBSD, HP-UX, Solaris, SCO Posts: 7,561 Blog Entries: 15 Rep: So I went off and decided there is no obvious math for the values in the table at and IRS 8962 Instructions. Accordingly I created the list of values with vim and sort: (Start with 0 since any number below 133 = 0.0201, end at 300 since any number at or above [to 400] = 0.0956. [The PDF doesn't say whether it is possible to go above 400 and if so what to do.]) 0 0.0201 133 0.0302 134 0.0308 135 0.0314 136 0.0320 137 0.0326 138 0.0332 139 0.0338 140 0.0344 141 0.0350 142 0.0355 143 0.0361 144 0.0367 145 0.0373 146 0.0379 147 0.0385 148 0.0391 149 0.0397 150 0.0403 151 0.0408 152 0.0412 153 0.0417 154 0.0421 155 0.0426 156 0.0431 157 0.0435 158 0.0440 159 0.0445 160 0.0449 161 0.0454 162 0.0458 163 0.0463 164 0.0468 165 0.0472 166 0.0477 167 0.0482 168 0.0486 169 0.0491 170 0.0495 171 0.0500 172 0.0505 173 0.0509 174 0.0514 175 0.0519 176 0.0523 177 0.0528 178 0.0532 179 0.0537 180 0.0542 181 0.0546 182 0.0551 183 0.0555 184 0.0560 185 0.0565 186 0.0569 187 0.0574 188 0.0579 189 0.0583 190 0.0588 191 0.0592 192 0.0597 193 0.0602 194 0.0606 195 0.0611 196 0.0616 197 0.0620 198 0.0625 199 0.0629 200 0.0634 201 0.0638 202 0.0641 203 0.0645 204 0.0648 205 0.0652 206 0.0655 207 0.0659 208 0.0662 209 0.0666 210 0.0669 211 0.0673 212 0.0676 213 0.0680 214 0.0683 215 0.0687 216 0.0690 217 0.0694 218 0.0697 219 0.0701 220 0.0704 221 0.0708 222 0.0711 223 0.0715 224 0.0718 225 0.0722 226 0.0726 227 0.0729 228 0.0733 229 0.0736 230 0.0740 231 0.0743 232 0.0747 233 0.0750 234 0.0754 235 0.0757 236 0.0761 237 0.0764 238 0.0768 239 0.0771 240 0.0775 241 0.0778 242 0.0782 243 0.0785 244 0.0789 245 0.0792 246 0.0796 247 0.0799 248 0.0803 249 0.0806 250 0.0810 251 0.0813 252 0.0816 253 0.0819 254 0.0822 255 0.0825 256 0.0828 257 0.0830 258 0.0833 259 0.0836 260 0.0839 261 0.0842 262 0.0845 263 0.0848 264 0.0851 265 0.0854 266 0.0857 267 0.0860 268 0.0863 269 0.0865 270 0.0868 271 0.0871 272 0.0874 273 0.0877 274 0.0880 275 0.0883 276 0.0886 277 0.0889 278 0.0892 279 0.0895 280 0.0898 281 0.0901 282 0.0903 283 0.0906 284 0.0909 285 0.0912 286 0.0915 287 0.0918 288 0.0921 289 0.0924 290 0.0927 291 0.0930 292 0.0933 293 0.0936 294 0.0938 295 0.0941 296 0.0944 297 0.0947 298 0.0950 299 0.0953 300 0.0956 I imported that into an Excel workbook splitting it into 2 columns (using text import with space as delimiter) in one sheet I called "Lookup". In that the actual values above are in columns B and C and rows 5 through 173. In another sheet in same work book I input a field for line 5 of 8962 (in cell C5 of that other sheet) then in the column next to it placed the following formula: =VLOOKUP(C5,Lookup!\$B\$5:Lookup!\$C\$173,2,TRUE) For any value below 133 it will give the Applicable Figure for "0" as it is 0 to 132. For any value above 300 it will give the Applicable Figure for 300. One could modify the formula to give an error if the value is above 400 given that the instruction doesn't define that - I didn't. This site has a tutorial for VLOOKUP (with a slight error - it shows the lookup range as one column but that gives a #REF if you do that - you have to tell it both columns in your lookup table). By the time I got back to post this I see OP already figured out to do a lookup table and the first responder figured out it wasn't a simple formula. I figured I'd share just as another example. Last edited by MensaWater; 12-18-2018 at 02:05 PM.
12-18-2018, 04:04 PM #4
RandomTroll
Senior Member
Registered: Mar 2010
Distribution: Slackware
Posts: 1,090
Original Poster
Rep:
Quote:
Originally Posted by michaelk Crap... After I paid more attention to the table I noticed that it isn't a simple first order equation... You might be able to graph the values and see what a higher order equation looks like and replace it with mine.
It's worse than that: the increment isn't monotonic: from 160 to 161 it goes up by .0005, from 161 to 162 it goes up by .0004, from 162 to 163 it goes up by .0005. It's hopeless.
Quote:
Originally Posted by MensaWater [The PDF doesn't say whether it is possible to go above 400 and if so what to do.])
There is no applicable figure above 400 because there's no PTC if you make that much.
The table you have is what I put in my pastebin 'cause I didn't want to clutter up LQ.
Last edited by RandomTroll; 12-18-2018 at 09:23 PM. Reason: Fixed mistake
12-18-2018, 04:43 PM #5
MensaWater
LQ Guru
Registered: May 2005
Location: Atlanta Georgia USA
Distribution: Redhat (RHEL), CentOS, Fedora, CoreOS, Debian, FreeBSD, HP-UX, Solaris, SCO
Posts: 7,561
Blog Entries: 15
Rep:
Quote:
Originally Posted by RandomTroll The table you have is what I put in my pastebin 'cause I didn't want to clutter up LQ.
I hate relying on external links for key data because the external link locations may have changed or gone defunct by the time someone sees the post. There is many a historical post I've gone to with bad links that made it useless because the important details weren't in it. Worse yet other places screen scrape to make their own posts and often don't even have the links when you find them.
One of my blog posts relates to a post I found about an error after time change and it was maddening because I appeared to find multiple hits related to the error in a web search but every hit link I opened turned out just to be a copy of the original post which didn't answer the question.
I wasn't trying to steal your thunder. I noted in my post that you'd resolved it and said what you did before I posted - I just felt like, having gone through the trouble, I should share the details including the formula and some discussion.
12-19-2018, 01:31 AM #6
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Originally Posted by MensaWater I hate relying on external links for key data...
Yeah, those pastebin links have a limited life. Back when I had my own website I would have posted it there and alerted the search engines. I always thought I was the only person writing his/her own spreadsheet to do taxes (+my friend Rick), but if there are others, it's useful to have it in a permanent place.
To document what I did:
1) pdftexted i8962.pdf
2) edited out everything but the table's entries
3) grepped on 0.0 for applicable figures
4) grep -ved on 0.0 for income ratios
5) imported those into those into a spreadsheet; because that's sc for me, a short script that incremented the row and set that cell's value to each entry in the file, and allowed me to combine the 2 and have sc put them parallel.
6) then I merged it into Taxes2018.sc, used an @index to point to it.
Quote:
Originally Posted by MensaWater I wasn't trying to steal your thunder. I noted in my post that you'd resolved it and said what you did before I posted - I just felt like, having gone through the trouble, I should share the details including the formula and some discussion.
I wasn't slanging you; it's the right thing to do.
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Contact Us - Advertising Info - Rules - LQ Merchandise - Donations - Contributing Member - LQ Sitemap - | 3,253 | 9,553 | {"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} | 3.109375 | 3 | CC-MAIN-2019-04 | latest | en | 0.899149 |
http://www.cs.unibo.it/~sangio/Book_Bis_Coind.html | 1,695,817,366,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00138.warc.gz | 59,099,090 | 2,449 | Bisimulation and Coinduction
Introduction to Bisimulation and Coinduction Advanced Topics in Bisimulation and Coinduction Davide Sangiorgi Davide Sangiorgi and Jan Rutten (eds)
The two books treat bisimulation and coinduction. They books analyse the most fundamental aspects of bisimulation and coinduction, exploring concepts and techniques that can be transported to many areas. Bisimulation is a special case of coinduction, by far the most studied coinductive concept. Bisimulation has been discovered in Concurrency Theory and processes remain the main application area. This explains the special emphasis on bisimulation and processes that one finds throughout the two volumes. The introduction volume treats basic topics. It explains coinduction, and its duality with induction, from various angles, starting from some simple results of fixed-point theory. It then goes on to bisimulation, as a tool for defining behavioural equality among processes (bisimilarity), and for proving such equalities. It compares bisimulation to other notions of behavioural equivalence. It also presents a simple process calculus, both to show algebraic techniques for bisimulation and to illustrate the combination of inductive and coinductive reasoning. The advanced volume deals with more specialised topics. A chapter recalls the history of the discovery of bisimulation and coinduction. Another chapter unravels the duality between induction and coinduction, both as defining principles and as proof techniques, in terms of the duality between the mathematical notions of algebra and coalgebra and properties such as initiality and finality. A third chapter analyses the profound implications of the concept of bisimulation in modal logics, with some beautiful results on the expressiveness of the logics. Two further chapters are devoted to the bisimulation proof method, a major ingredient for success of bisimulation: the algorithmic content of the method, showing striking separation results between bisimilarity and other behavioural equivalences; and enhancements of the bisimulation proof method, whose goal is to further facilitate the proof of bisimilarity results. Finally, separate chapters discuss two important refinements of bisimulation, which have to do with probabilities and higher-order linguistic constructs. Bisimulation and coinduction offer us powerful tools for defining, understanding, and reasoning about objects and structures that are common in Computer Science. Today, bisimulation and coinduction are also used in other fields, e.g., Artificial Intelligence, Cognitive Science, Mathematics, Modal Logics, Philosophy, Physics. Follow the links above for details on the single books, including order information. | 517 | 2,738 | {"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} | 3.109375 | 3 | CC-MAIN-2023-40 | latest | en | 0.926069 |
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# Hypothesis testing and confidence interval problem
A. *HT* A test was conducted to compare the wearing quality of the tires produced by two tire companies. A random sample of 16 cars is equipped with one tire of Brand X and one tire of Brand Y (the other two tires on each car are not part of the test), and driven for 30 days. The following table gives the amount of wear in thousandths of an inch that resulted (the smaller the number the better the wear quality of the tire). Assume tire wear is normally distributed. At the 1% level of significance, do Brand X tires have better wear quality than Brand Y tires?
B. *CI* Calculate the 98% confidence interval estimate for the mean difference. Does this interval give EXACTLY the same information as the hypothesis test in Part A? EXPLAIN why - be very specific!
Car X Y
1 92 95
2 86 86
3 85 84
4 76 80
5 97 99
6 89 92
7 88 91
8 86 89
9 82 83
10 81 79
11 79 81
12 83 84
13 94 92
14 93 97
15 83 84
16 84 82
#### Solution Summary
Step by step method for testing the hypothesis under 5 step approach and construction of confidence interval is discussed here. Excel template for each problem is also included. This template can be used to obtain the answers of similar problems.
\$2.19 | 338 | 1,264 | {"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} | 3.25 | 3 | CC-MAIN-2016-44 | latest | en | 0.927294 |
https://gohighbrow.com/the-golden-ratio/ | 1,695,744,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00454.warc.gz | 317,338,331 | 16,484 | # The Golden Ratio
21.06.2015
1.6180339887…
Episode #3 of the course “Most important numbers in the world”
In ancient mathematics, especially geometry, mathematicians ran across an important relationship between line segments and two-dimensional straight-lined shapes. In configuring quadrilaterals, pentagons, pentagrams, and other shapes, the proportion between certain lengths of the lines was not only most pleasing to the eye, but most structurally and mathematically sound. First described by Euclid in ancient Greece, the ratio now known as the “golden mean” is the sectioning of a line into an “extreme” mean of a specific proportion. Since the first descriptions and definitions of this predictable, repeatable relationship, mathematicians have been fascinated by its simplicity and its permeance through nature.
Also known as the “divine ratio” or the “golden number,” the golden ratio is often symbolized by the Greek letter “phi” ($\varphi$ or $\phi$). It describes a relationship between a line and two of its segments, if the larger of the two segments is 1.6180 times the length of the smaller. When extrapolated into two-dimensional and three-dimensional shapes, the “golden ratio” becomes highly important for creating aesthetically and mathematically pleasing rectangles, columns, cylinders, and numerous polyhedra.
Artists and architects throughout history have been fascinated with the golden ratio, utilizing its proportions to create balance and perspective, movement, and enhanced geometrical emphasis in numerous paintings, sculptures, and buildings around the world. From classical Greek works of art and iconic architectural style came a number of revivals throughout history, each of which reappropriated the golden mean with renewed vigor. As mathematics grows to encompass more complex theories founded in the golden ratio, contemporary artists will continue to utilize this ancient constant relationship is new ways.
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### Help in understanding table.copyrange
Posted: Thu Jan 18, 2018 7:08 am
Hi,
I would like to copy the range of one table to another using a loop.
tableT_{%a}_{%x}_{%j}.copyrange A2 N23 tableALL A2 ;
the first table will get copied into tableALL at A2 , the second will be at A2+ 22 ( number of row inserted = 22)
Can I used the following formula in my loop.
number of columns in tableT = c = 14
number of rows in tableT = r = 22
tableT_{%a}_{%x}_{%j}.copyrange 1 1 c r tableALL (1+ i)
Any help will be greatly appreciated.
Thanks
### Re: Help in understanding table.copyrange
Posted: Thu Jan 18, 2018 9:46 am
Copyrange unfortunately does not take expressions. So you will have to precalculate the source row and column as well as the destination row.
Here is an example:
Code: Select all
!c=24
!r=22
' copy 4 tables to tableAll
for !i=0 to 3
!dest_r = (!i*!r) + 2
tableT_{%a}_{%x}_{%j}.copyrange 1 1 !r !c tableAll !dest_r 1
next
### Re: Help in understanding table.copyrange
Posted: Thu Jan 18, 2018 10:45 am
Thanks alots, it's working. | 346 | 1,069 | {"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-2019-13 | latest | en | 0.810065 |
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# PS-Power Prep
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### Show Tags
21 Feb 2006, 18:22
E.
Range of S will always be either equal or greater than T.
Take example
S = {1,2,3,4,5,6,7,8,9} and T = {1,2,3,4,6,7,8,9}
Then mean, median and range of S and T are equal. So A,B and C are out.
Mean of S can be greater than that of T for this set
S = {1,2,3,4,5,6,7,8,9} and T = {1,2,3,4,5,6,7,8}
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21 Feb 2006, 18:22
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# PS-Power Prep
Powered by phpBB © phpBB Group and phpBB SEO 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®. | 606 | 2,043 | {"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-2017-04 | latest | en | 0.900432 |
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# Adjacent Angles -Definition, Properties, Examples
Adjacent Angles are one of the types of angles in Geometry. Adjacent Angles are two angles that share a common side and a common vertex (Corner point) but do not overlap. When adjacent angles have a common vertex and side, they can be complementary angles or supplementary angles. Let us learn about the Adjacent angle’s definition, properties, Examples, and many more.
As the name “Adjacent Angles” Suggests, Angles are always situated next to one another. When two angles have a common vertex and side, they are referred to as adjacent angles. The vertex of an angle is the point at which the rays that make up its sides come to an end. When two straight lines cross, four angles are formed, with the intersection point serving as the vertex of each. The term “adjacent angles” refers to two angles that have a common vertex and side.
Check: What is Scalene Angles?
When two angles share a common vertex and a common side, those angles are known as Adjacent Angles. When two rays intersect at a shared endpoint, Adjacent Angles are formed that are always positioned next to one another positioned but they don’t cross over each other. Observe the below figure of adjacent angles for better understanding. In the below figure, ∠a and ∠b are adjacent angles that share a common vertex and a common side.
## Adjacent Angles Examples in Real Life
There are numerous examples of adjacent angles in everyday life.
1. Most of us loved to eat pizza. Adjacent angles are formed when two pizza slices are placed next to one another. This is the most typical real-world example of adjacent angles.
2. When all three are separated from one another in a wall clock. The minute hand and second hand of the clock make one angle, and the hour hand and second hand make another angle. These two simultaneous angles are referred to as adjacent angles because they are close to one another.
The properties of adjacent angles are listed below to make it simple to recognize them. The properties of adjacent angles are as follows
1. A common arm is always shared by adjacent angles.
2. They are associated with a common vertex.
3. They don’t cross over each other or overlap.
4. On either side of the common arm, they have a non-common arm.
5. They are not connected by an interior point.
6. Depending on the total of the individual angles’ measures, two adjacent angles can be supplementary or complementary.
## How to Identify Adjacent Angles?
Earlier we know from the properties of adjacent Angles, that adjacent angles are always shared with a common vertex and a common side. we can easily identify adjacent angles using these properties. Any two angles are not regarded as adjacent angles if they only meet one of these criteria. Both of these properties must be satisfied by the angles.
If the sum of two adjacent angles is 180 degrees, then those angles are shown to be supplementary angles. Two supplementary angles are considered to be linear pairs if they are next to one another. The non-common arms form a line if the sum of two adjacent angles is 180°.
Sum of two adjacent supplementary angles = 180º
If the sum of the Adjacent angles that share a common vertex and a common Side is 90º, then the adjacent angles are called Complementary Adjacent Angles.
In the below the figure, two angles that measure 60º and 30º respectively id known as Complementary Adjacent Angles.
## Adjacent angles of a parallelogram are
A quadrilateral with two sets of parallel sides is referred to as a parallelogram. In a parallelogram, the opposing sides are of equal length, and the opposing angles are of equal size.is the sum of all interior angles.is 360 degrees. Additionally, the interior angles that are supplementary to the transversal on the same side.
In the below parallelogram named ABCD. Which adjacent angles are ∠A and ∠B, ∠B and ∠C,∠C and ∠D.and ∠A and ∠D
As the properties of a parallelogram, the interior adjacent angles are supplementary, meaning the sum of two adjacent angles is 180 degrees. Consequently, a parallelogram’s adjacent angles are supplementary.
Q.What do you mean by Adjacent Angles?
When two angles share a common vertex and a common side, those angles are known as Adjacent Angles. When two rays intersect at a shared endpoint, Adjacent Angles are formed that are always positioned next to one another positioned but they don’t cross over each other.
Q.What is required for an adjacent angle?
Adjacent Angles are two angles that share a common side and a common vertex (Corner point) but do not overlap.
Answer: From the properties of adjacent Angles, those adjacent angles are always shared with a common vertex and a common side. we can easily identify adjacent angles using these properties. Any two angles are not regarded as adjacent angles if they only meet one of these criteria. Both of these properties must be satisfied by the angles.
Ans: When two straight lines cross, four angles are formed, with the intersection point serving as the vertex of each. The term “adjacent angles” refers to two angles that have a common vertex and side.
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## FAQs
### What do you mean by Adjacent Angles?
When two angles share a common vertex and a common side, those angles are known as Adjacent Angles. When two rays intersect at a shared endpoint, Adjacent Angles are formed that are always positioned next to one another positioned but they don't cross over each other.
### What is required for an adjacent angle?
Adjacent Angles are two angles that share a common side and a common vertex (Corner point) but do not overlap.
### How do you identify adjacent?
Answer: From the properties of adjacent Angles, those adjacent angles are always shared with a common vertex and a common side. we can easily identify adjacent angles using these properties. Any two angles are not regarded as adjacent angles if they only meet one of these criteria. Both of these properties must be satisfied by the angles.
### How do adjacent angles formed?
Ans: When two straight lines cross, four angles are formed, with the intersection point serving as the vertex of each. The term "adjacent angles" refers to two angles that have a common vertex and side. | 1,350 | 6,338 | {"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} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.940242 |
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## Minkowski’s criterion
Here is a related post that I find interesting.
In linear algebra, Minkowski‘s criterion states the following.
Theorem (Minkowski’s Criterion). Let $A$ be an $n\times n$ matrix with real entries such that the diagonal entries are all positive, off diagonal entries are all negative, and the row sums are all positive. Then $\det(A)\neq 0$.
This is a nice criterion and is not very difficult to prove, but for a random matrix it is asking too much. To decide whether a matrix is singular one usually looks for a row/column consisting of zeros or adding up to zero. The following result gives sufficient conditions for this to work. Unfortunately, it does not generalise Minkowski’s result.
Theorem. Let $A$ be a $n\times n$ matrix with real entries such that its row sums are all $\ge 0$, its lower diagonal entries are $\ge 0$ and its upper diagonal entries are $\le 0$. Then $\det(A)=0$ if and only if $A$ has either a row consisting entirely of zeros or all the row sums equal to zero.
Proof. Suppose that $Ab=0$, where $b=(b_1,\dots,b_n)^T\neq 0$. Assume that $b_1\ge\cdots\ge b_n$. Then there exists $1\le m such that $a_{1,1}',\dots,a_{1,m}'\ge 0$ and $a_{1,m+1}',\dots,a_{1,n}'\le 0$. Hence
\begin{aligned}0=\sum_{j=1}^na_{1,j}'b_j&\ge b_m\sum_{j=1}^ma_{1,j}'+b_{m+1}\sum_{j=m+1}^na_{1,j}\\&\ge (b_m-b_{m+1})\sum_{j=1}^ma_{1,j}'\ge 0.\end{aligned}
So we must have (i) $b_1=\cdots=b_m$, (ii) $b_{m+1}=\cdots=b_n$, (iii) $a_{1,1}'+\cdots+a_{1,n}'=0$ and (iv) either $b_m=b_{m+1}$ or $a_{1,1}'+\cdots+a_{1,m}'=0$. These boil down to having either $b_1=\cdots=b_n$ or $a_{1,1}'=\cdots=a_{1,n}'=0$. Apply this argument to each row of $A$ to obtain the desired conclusion. $\square$ | 610 | 1,750 | {"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": 26, "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} | 4.09375 | 4 | CC-MAIN-2019-30 | latest | en | 0.667472 |
http://mathhelpforum.com/advanced-statistics/167114-liminf-product.html | 1,516,146,544,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00433.warc.gz | 207,080,676 | 11,459 | # Thread: liminf of a product
1. ## liminf of a product
I am trying to prove the following:
$\underbrace{(\lim \inf f_n)}_f\underbrace{(\lim\inf g_n)}_g\leq \underbrace{\lim\inf (f_ng_n)}_k$.
Here's what I have so far:
FTSOC assume $fg>k$. We know that $f_n and $g_n for only finitely many $n$. Then $f_ng_n for only finitely many $n$. This implies $f_ng_n\geq fg-\epsilon(f+g-\epsilon)$ almost always. So, $f_ng_n\geq k-\epsilon(f+g-\epsilon)$ almost always.
But this doesn't get me anywhere. Where am I going wrong? If I let $\epsilon=\min\{f,g\}$, I just get $f_ng_n\geq 0$. So, this tells me nothing.
Any help would be appreciated. (Even though it doesn't state it, I am assuming that all $f_n, g_n$ are nonnegative.)
2. I don't think that is true.
Consider $f_n$ the sequence $-1,0,-1,0,...$
and $g_n$ the sequence $2,0,2,0,...$
so $f_ng_n$ is $-2, 0,-2, 0,...$
Now $(\lim \inf f_n)(\lim\inf g_n) = -1 \times 0 = 0$
$\lim\inf (f_ng_n) = -2$
3. Originally Posted by amheissan
...I am assuming that all $f_n, g_n$ are nonnegative.
You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.
Any more thoughts on proving this property? Thanks!
4. Originally Posted by amheissan
You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.
Any more thoughts on proving this property? Thanks!
Note that by definition $\displaystyle \liminf_{n\to\infty}f_n(x)=\inf \left\{\lim_{n\to\infty}f_{n_k}(x):f_{n_k}(x)\text { is a subsequence of }f_n(x)\right\}$ and similarly for $\displaystyle \liminf_{n\to\infty}g_n(x)$. Use the fact then that $\inf\left(A\right)\inf\left(B\right)\leqslant \inf\left(AB\right)$.
5. Okay, but how would you prove $\inf(A)\inf(B)\leq \inf (AB)$? I'm really just not seeing this property.
6. For only nonnegative sequences, Drexel28 is giving you the hint to only look at the inf of the set of values at any subsequence (ignore the order of the sequence).
Let $AB = \{ab | a\in A, b\in B\}$.
Consider any element $a\in A$ and $b\in B$, $ab \geq \inf(A)\inf(B)$ (by definition since everything is nonnegative).
Also, remember that the inf of a subset of AB must be larger than the inf of AB
### liminf product
Click on a term to search for related topics. | 746 | 2,353 | {"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": 29, "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} | 3.5 | 4 | CC-MAIN-2018-05 | longest | en | 0.902174 |
https://mathoverflow.net/questions/tagged/differential-equations?sort=unanswered&pageSize=50 | 1,547,829,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00520.warc.gz | 578,093,133 | 32,918 | # Questions tagged [differential-equations]
Ordinary or partial differential equations. Delay differential equations, neutral equations, integro-differential equations. Well-posedness, asymptotic behavior, and related questions.
315 questions
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I feel like the above must be true but embarrassingly cannot seem to prove it. Take linearly independent, commuting vector fields $X$ and $Y$ on a manifold and corresponding flows $\Phi^t_X$, \$\Phi^... | 2,590 | 9,650 | {"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} | 2.65625 | 3 | CC-MAIN-2019-04 | latest | en | 0.874353 |
https://docs.sympy.org/1.1/modules/physics/units/quantities.html | 1,550,867,254,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247526282.78/warc/CC-MAIN-20190222200334-20190222222334-00089.warc.gz | 530,220,423 | 3,905 | # Physical quantities¶
Physical quantities.
class sympy.physics.units.quantities.Quantity[source]
Physical quantity.
abbrev
Symbol representing the unit name.
Prepend the abbreviation with the prefix symbol if it is defines.
convert_to(other)[source]
Convert the quantity to another quantity of same dimensions.
Examples
>>> from sympy.physics.units import speed_of_light, meter, second
>>> speed_of_light
speed_of_light
>>> speed_of_light.convert_to(meter/second)
299792458*meter/second
>>> from sympy.physics.units import liter
>>> liter.convert_to(meter**3)
meter**3/1000
scale_factor
Overall magnitude of the quantity as compared to the canonical units.
## Conversion between quantities¶
Several methods to simplify expressions involving unit objects.
sympy.physics.units.util.convert_to(expr, target_units)[source]
Convert expr to the same expression with all of its units and quantities represented as factors of target_units, whenever the dimension is compatible.
target_units may be a single unit/quantity, or a collection of units/quantities.
Examples
>>> from sympy.physics.units import speed_of_light, meter, gram, second, day
>>> from sympy.physics.units import mile, newton, kilogram, atomic_mass_constant
>>> from sympy.physics.units import kilometer, centimeter
>>> from sympy.physics.units import convert_to
>>> convert_to(mile, kilometer)
25146*kilometer/15625
>>> convert_to(mile, kilometer).n()
1.609344*kilometer
>>> convert_to(speed_of_light, meter/second)
299792458*meter/second
>>> convert_to(day, second)
86400*second
>>> 3*newton
3*newton
>>> convert_to(3*newton, kilogram*meter/second**2)
3*kilogram*meter/second**2
>>> convert_to(atomic_mass_constant, gram)
1.66053904e-24*gram
Conversion to multiple units:
>>> convert_to(speed_of_light, [meter, second])
299792458*meter/second
>>> convert_to(3*newton, [centimeter, gram, second])
300000*centimeter*gram/second**2
Conversion to Planck units:
>>> from sympy.physics.units import gravitational_constant, hbar
>>> convert_to(atomic_mass_constant, [gravitational_constant, speed_of_light, hbar]).n()
7.62950196312651e-20*gravitational_constant**(-0.5)*hbar**0.5*speed_of_light**0.5 | 541 | 2,181 | {"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} | 2.59375 | 3 | CC-MAIN-2019-09 | latest | en | 0.696698 |
https://www.mathworks.com/matlabcentral/cody/problems/11-back-and-forth-rows/solutions/144128 | 1,508,340,827,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822992.27/warc/CC-MAIN-20171018142658-20171018162658-00677.warc.gz | 971,249,140 | 11,474 | Cody
# Problem 11. Back and Forth Rows
Solution 144128
Submitted on 3 Oct 2012 by Akiva Gordon
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
n = 4; a = [ 1 2 3 4; 8 7 6 5; 9 10 11 12; 16 15 14 13]; assert(isequal(a,back_and_forth(n)));
2 Pass
%% n = 5; a = [ 1 2 3 4 5; 10 9 8 7 6; 11 12 13 14 15; 20 19 18 17 16; 21 22 23 24 25]; assert(isequal(a,back_and_forth(n))); | 207 | 496 | {"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 | 0.660416 |
https://socratic.org/statistics/linear-regression-and-correlation/least-squares-regression-line-lsrl | 1,726,494,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00719.warc.gz | 468,181,545 | 11,119 | # Least Squares Regression Line (LSRL)
## Key Questions
• Equation for least-squares linear regression:
$y = m x + b$
where
$m = \frac{\sum \left({x}_{i} {y}_{i}\right) - \frac{\sum {x}_{i} \sum {y}_{i}}{n}}{\sum {x}_{i}^{2} - \frac{{\left(\sum {x}_{i}\right)}^{2}}{n}}$
and
$b = \frac{\sum {y}_{i} - m \sum {x}_{i}}{n}$
for a collection of $n$ pairs $\left({x}_{i} , {y}_{i}\right)$
This looks horrible to evaluate (and it is, if you are doing it by hand); but using a computer (with, for example, a spreadsheet with columns :$y , x , x y , \mathmr{and} {x}^{2}$) it isn't too bad.
• The primary use of linear regression is to fit a line to 2 sets of data and determine how much they are related.
Examples are:
2 sets of stock prices
rainfall and crop output
With respect to correlation, the general consensus is:
Correlation values of 0.8 or higher denote a strong correlation
Correlation values of 0.5 or higher up to 0.8 denote a weak correlation
Correlation values less than 0.5 denote a very weak correlation\f
All this means is the minimum between the sum of the difference between the actual y value and the predicted y value.
$\min {\sum}_{i = 1}^{n} {\left({y}_{i} - \hat{y}\right)}^{2}$
#### Explanation:
Just means the minimum between the sum of all the resuidals
$\min {\sum}_{i = 1}^{n} {\hat{u}}_{i}^{2}$
all this means is the minimum between the sum of the difference between the actual y value and the predicted y value.
$\min {\sum}_{i = 1}^{n} {\left({y}_{i} - \hat{y}\right)}^{2}$
This way by minimizing the error between the predicted and error you get the best fit for the regression line. | 492 | 1,632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.804366 |
https://epsg.io/9647-method | 1,653,520,297,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00772.warc.gz | 287,489,039 | 7,163 | # EPSG:9647
## General polynomial of degree 4
### Attributes
Data source: OGP
Information source: EPSG guidance note #7-2, http://www.epsg.org
Revision date: 2002-12-21
#### Export
Definition: OGP XML
```<?xml version="1.0" encoding="UTF-8"?>
<epsg:informationSource>EPSG guidance note #7-2, http://www.epsg.org</epsg:informationSource>
<epsg:revisionDate>2002-12-21</epsg:revisionDate>
<epsg:isDeprecated>false</epsg:isDeprecated>
<epsg:isOperationReversible>false</epsg:isOperationReversible>
<epsg:example>(none)</epsg:example>
<gml:identifier codeSpace="OGP">urn:ogc:def:method:EPSG::9647</gml:identifier>
<gml:name>General polynomial of degree 4</gml:name>
<gml:formula>Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided.
The simplest of all polynomials is the general polynomial function. In order to avoid problems of numerical instability this type of polynomial should be used after reducing the coordinate values in both the source and the target coordinate reference system to ‘manageable’ numbers, between –10 and +10 at most. This is achieved by working with offsets relative to a central evaluation point, scaled to the desired number range by applying a scaling factor to the coordinate offsets.
Hence an evaluation point is chosen in the source coordinate reference system (XS0, YS0) and in the target coordinate reference system (XT0, YT0). Often these two sets of coordinates do not refer to the same physical point but two points are chosen that have the same coordinate values in both the source and the target coordinate reference system. (When the two points have identical coordinates, these parameters are conveniently eliminated from the formulas, but the general case where the coordinates differ is given here).
The selection of an evaluation point in each of the two coordinate reference systems allows the point coordinates in both to be reduced as follows:
XS - XS0
YS - YS0
and
XT – XT0
YT – YT0
These coordinate differences are expressed in their own unit of measure, which may not be the same as that of the corresponding coordinate reference system. )
A further reduction step is usually necessary to bring these coordinate differences into the desired numerical range by applying a scaling factor to the coordinate differences in order to reduce them to a value range that may be applied to the polynomial formulae below without introducing numerical precision errors:
U = mS.(XS - XS0)
V = mS.(YS - YS0)
where
XS , YS are coordinates in the source coordinate reference system,
XS0 , YS0 are coordinates of the evaluation point in the source coordinate reference system,
mS is the scaling factor applied the coordinate differences in the source coordinate reference system.
The normalised coordinates U and V of the point whose coordinates are to be transformed are used as input to the polynomial transformation formula. In order to control the numerical range of the polynomial coefficients An and Bn the output coordinate differences dX and dY are multiplied by a scaling factor, mT.
mT.dX = A0 + A1.U + A2.V + A3.U^2 + A4.U.V + A5.V^2 + A6.U^3 + A7.U^2.V + A8.U.V^2 + A9.V^3
+ A10.U^4 + A11.U^3.V + A12.U^2.V^2 + A13.U.V^3 + A14.V^4
mT.dY = B0 + B1.U + B2.V + B3.U^2 + B4.U.V + B5.V^2 + B6.U^3 + B7.U^2.V + B8.U.V^2 + B9.V^3
+ B10.U^4 + B11.U^3.V + B12.U^2.V^2 + B13.U.V^3 + B14.V^4
from which dX and dY are evaluated. These will be in the units of the target coordinate reference system.
The polynomial coefficients are given as parameters of the form Aumvn and Bumvn, where m is the power to which U is raised and n is the power to which V is raised. For example, A13 is represented as coordinate operation parameter Au1v3.
The relationship between the two coordinate reference systems can now be written as follows:
(XT - XT0) = (XS – XS0) + dX
(YT - YT0) = (YS – YS0) + dY
or
XT = XS – XS0 + XT0 + dX
YT = YS – YS0 + YT0 + dY
where:
XT , YT are coordinates in the target coordinate reference system,
XS , YS are coordinates in the source coordinate reference system,
XS0 , YS0 are coordinates of the evaluation point in the source coordinate reference system,
XT0 , YT0 are coordinates of the evaluation point in the target coordinate reference system,
dX, dY are derived through the scaled polynomial formulas.</gml:formula> | 1,151 | 4,469 | {"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-2022-21 | latest | en | 0.728441 |
https://ericlippert.com/2020/05/28/life-part-13/?replytocom=113524 | 1,632,301,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00337.warc.gz | 289,332,287 | 24,057 | # Life, part 13
Source code for this episode is here.
Just as a reminder: I am developing my C# version of Stafford’s “QLIFE” algorithm by taking Abrash’s “keep the neighbour counts around” algorithm and gradually adding optimizations. That’s much easier to understand than trying to explain the whole complicated thing at once.
So far, we’ve got the following algorithm:
• Keep up-to-date neighbour counts of all cells in a byte array
• Maintain a list of cells which changed on the previous tick
• On the current tick, examine only the cells which changed on the previous tick and their neighbours. Update the ones that changed, and create a deduplicated change list for the next tick.
However, we are still using a similar technique as we used in our naïve algorithm to ensure that we never write a cell before we need to read its value: we make a complete copy of all the cells on each tick, read from the copy, and write to the original. Making that copy is fast, but it is O(n) in the total number of cells; it seems plausible that we ought to be able to come up with an algorithm that is O(changed cells).
Here’s what we’re going to do.
Recall that we are spending one byte per cell, but are only using five of the eight bits available in a byte: four for the neighbour count, one for the state. We’re going to use one of the extra bits to store the next state of a cell that is about to change.
Why do we need that? Surely if we already know what cells are about to change, then the next state is just the opposite of the current state, right? But we need it because we are trying to solve two problems at once here: compute the new state, and deduplicate the new change list without using a hash table. Since we are looking at all neighbours of previously-changed cells, we will frequently end up looking at the same cells multiple times, but we do not want the change list to become more and more full of duplicates as time goes on. If the “next state” and the “current state” bits are the same, then we already updated the current state, so we can skip adding it to the new change list, and thereby deduplicate it. (As I noted last time, we’ll get to algorithms that use automatically-deduplicated sets in future episodes.)
As always I want the bit twiddling to be isolated into methods of their own and structs to be immutable. Adding accessors for another bit in my Cell struct is straightforward:
```private const int next = 5;
private const int nextm = 1 << next;
public bool Next => (cell & nextm) != 0;
public Cell NextAlive() => new Cell((byte)(cell | nextm));
public Cell NextDead() => new Cell((byte)(cell & ~nextm));```
Easy peasy. I’m going to change the Step algorithm but everything else — BecomeAlive and BecomeDead in particular — stays exactly the same. As does most of Step:
```public void Step()
{
```
We’re going to create a new, temporary list of the cells that are about to change on this tick. This list is allowed to have duplicates.
``` var currentChanges = new List<(int, int)>();
```
There is a line of code conspicuous by its absence here. We are not cloning the array!
Once again we loop over the cells which changed last time and their neighbours; this is all the same:
``` foreach ((int cx, int cy) in changes)
{
int minx = Max(cx - 1, 1);
int maxx = Min(cx + 2, width - 1);
int miny = Max(cy - 1, 1);
int maxy = Min(cy + 2, height - 1);
for (int y = miny; y < maxy; y += 1)
{
for (int x = minx; x < maxx; x += 1)
{
```
Once again we have the Life condition that you are now familiar with:
``` Cell cell = cells[x, y];
int count = cell.Count;
bool state = cell.State;
bool newState = count == 3 | count == 2 & state;
```
Is this cell going to change? If so, record the new state in bit 5 and add it to the current changes list:
``` if (state & !newState)
{
}
else if (!state & newState)
{
cells[x, y] = cell.NextAlive();
}
}
}
}```
Again, yes, I could do the mutation of the bit in-places since we have a collection of variables, but I can’t bring myself to mutate a value type.
We’re done our first pass; the list of previous changes is now useless to us:
``` changes.Clear();
```
The second pass is much simpler than the first. For all the cells that need changing, use idempotent helper functions from last time to record the new neighbour counts and update the change list for the next tick:
``` foreach ((int x, int y) in currentChanges)
{
if (cells[x, y].Next)
BecomeAlive(x, y);
else
}
}
```
And we’re done.
If you look at the asymptotic performance, plainly it is O(changed cells), and not O(total number of cells), which is great! This means that we could have extremely large boards with lots of empty space or still lifes, and we only pay a per-tick time price for the evolving or oscillating cells that change.
Our “static” memory consumption, for the array, is still O(total cells) of course, but our dynamic burden on the garbage collector is also O(changed cells) per tick, which seems like it ought to be a win.
What about our actual performance of computing acorn for 5000 cycles on an 8-quad?
```Algorithm time(ms) ticks size(quad) megacells/s
Naïve (Optimized): 4000 5K 8 82
Scholes 3250 5K 8 101
Frijters SIMD 1000 5M 4 1200
Abrash 550 5K 8 596
Abrash w/ changes 190 5K 8 1725
Abrash, O(c) 240 5K 8 1365
```
I was slightly surprised to discover that it is around 20% slower! As I have pointed out a number of times, copying a 64K array of bytes is astonishingly fast. That’s the thing to always remember about asymptotic performance: it is about how the performance cost changes as the problem becomes large. It would be interesting to do some scaling experiments and discover when the cost of copying the array becomes the dominating cost, but I’m not going to; if anyone wants to do the experiment please do and report back.
Update: Reader jaloopa has done the experiment on their machine; when bumping up from an 8-quad to a 10-quad — so, just 16 times bigger — the O(c) algorithm is ten times faster than the O(n) algorithm! So this is actually a big win once the board sizes get significantly larger than 64KB. I was closer to the break-even point than I thought.
Next time on FAIC: I’ve been talking a lot about Life algorithms but very little about Life itself. Let’s digress for an episode or two and explore some interesting basic patterns.
After that: We are now using six bits per cell to store the neighbour count, current state and next state. If we can get that back down to five bits per cell then we can fit three cells into a two-byte short. That’s slightly more memory efficient, but at a cost of greatly increasing the amount of bit twiddling we need to do. This seems like a performance-killing change; will it eventually pay for itself by enabling further optimizations, or is it a big waste of effort? We’ll find out!
## 15 thoughts on “Life, part 13”
• Thanks! I am enjoying it too, and there’s a lot to go still.
1. I also want to leave a note that I love this series (like most other content on this blog). Thank you!
• Thanks, I appreciate it!
2. Obviously, different algorithms perform differently under different conditions. What do you think about the idea of swapping out the algorithms based on heuristics? Say, you find that “Abrash w/ changes” performs better until you start copying an array over 1M in size? Would you then be able to change over to “Abrash, O(c)” at that point?
Taking this further, what if these algorithms that work in O(changed) start to perform worse than O(total) when greater than 50% (or some other point). That’d be something interesting to look at, I think.
I’m trying to come up with some other heuristics that might be interesting to look at.
• Absolutely, there are many real-world systems that use heuristics to make composite algorithms. In a recent post I was discussing string searching algorithms and how the naive string searching algorithm is almost always better performing even though it has a bad asymptotic worst case behaviour. We almost never hit the worst case in real life! But there are some string searching libraries that detect when the naive algorithm is likely to do poorly and switch over to a better performing algorithm based on some heuristic.
We could certainly do the same thing with Life. As we’ll see later on in this series, there are some algorithms that are extremely good at large, long-lived patterns but require a lot of “set up” time, so they perform relatively poorly on small, simple patterns. A heuristic that switched off between implementations depending on how it was used would be pretty smart.
3. >It would be interesting to do some scaling experiments and discover when the cost of copying the array becomes the dominating cost, but I’m not going to; if anyone wants to do the experiment please do and report back.
Turns out it didn’t take long, at least on my hardware
Release build, running directly from the execuatble.
29/05/2020 01:16:33 PM:ConwaysLife.Scholes:3614
29/05/2020 01:16:34 PM:ConwaysLife.Abrash:556
29/05/2020 01:16:34 PM:ConwaysLife.AbrashChangeList:229
29/05/2020 01:16:34 PM:ConwaysLife.AbrashOneArray:260
Changing from sizes of 256 to 512 (514 for the Abrash ones)
29/05/2020 01:15:21 PM:ConwaysLife.Scholes:19552
29/05/2020 01:15:23 PM:ConwaysLife.Abrash:2359
29/05/2020 01:15:23 PM:ConwaysLife.AbrashChangeList:439
29/05/2020 01:15:24 PM:ConwaysLife.AbrashOneArray:289
I ran a few times, and this was the biggest difference between the change list and one array versions. All others were in the region of 100ms difference
• Getting bigger the savings get very obvious very quickly. Excluding Scholes because I didn’t want to be waiting ages, here’s the 1026 results
29/05/2020 01:28:37 PM:ConwaysLife.Abrash:10035
29/05/2020 01:28:40 PM:ConwaysLife.AbrashChangeList:2599
29/05/2020 01:28:40 PM:ConwaysLife.AbrashOneArray:266
• That’s great; thanks for taking the time to take the times. 🙂
4. Probably you can come back to 5 bits per cell when you store the next state in `currentChanges`, right? So that would be `var currentChanges = new List();`
• That’s an excellent guess and an interesting idea, but not actually where we’re going next. It’s a little weirder, where we’re going.
I may explore ideas related to yours in a later episode if I do one on sparse array approaches.
• Oh, I forgot that generic arguments aka. HTML tags are swallowed for comments, but obviously you knew what I mean. I’ll try it again anyways:
var currentChanges = new List();
Or is it
var currentChanges = new List< (int, int, bool) >();
• Thanks for the link! There has been some interesting work done on application of continuous math techniques to discrete problems like this one; I had not seen this particular application before though. | 2,715 | 10,998 | {"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} | 3.4375 | 3 | CC-MAIN-2021-39 | latest | en | 0.951957 |
http://www.velocityreviews.com/forums/t344929-multiple-cmp-s-chained-one-after-another.html | 1,394,367,164,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999677605/warc/CC-MAIN-20140305060757-00014-ip-10-183-142-35.ec2.internal.warc.gz | 601,935,305 | 12,763 | Velocity Reviews > Multiple "cmp"s chained one after another
# Multiple "cmp"s chained one after another
Volker Grabsch
Guest
Posts: n/a
05-14-2005
Hello!
Ich just found a very nice 'pythonic' solution for an often appearing
problem. I didn't find it documented anywhere, so I'm posting it here.
If this isn't new in any way, I'd really like to get to know it.
Example problem:
I have some "datetime" objects and want to sort them, as here:
birthdays = [d1,d2,d3,d4]
birthdays.sort()
However, I don't want to sort them the default way. These are birthdays,
so only the month and day do matter, not the year. E.g.:
2003-01-01 should be smaller than 1984-05-01
So I have to write the comparison on my own, e.g.
def cmp_birthdays(d1,d2):
if d1.month > d2.month: return 1
if d1.month < d2.month: return -1
if d1.day > d2.day: return 1
if d1.day < d2.day: return -1
return 0
...
birthdays.sort(cmp_birthdays)
This implementation of cmp_birthdays is very ugly. Image you want to
chain more than 2 values in that "cmp_birthdays". I also want to use the
builtin "cmp" function, not ">" and "<".
After thinking some minutes about it, I found a very nice solution:
I have some "cmp"s one aftter another. If one if them return 1 oder -1,
it sould be returned. If it returns 0, the next "cmp" is used. In other
words: I have a sequence of numbers, and want to get the first one that
is not 0. (or return 0, if all numbers were 0)
But this is exactly what the "or" operator does, due to short-circuit
evaluation. In this example, that means:
def cmp_bithdays(d1,d2):
return cmp(d1.month,d2.month) or cmp(d1.day,d2.day)
The generic pattern is:
return cmp(...) or cmp (...) or cmp(...) or ...
I'm not sure whether this pattern is already a "common recipe", but
I found it to be a very nice idea.
Any opinions?
Greets,
Volker
--
Volker Grabsch
---<<(())>>---
\frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
[G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
vincent wehren
Guest
Posts: n/a
05-14-2005
"Volker Grabsch" <(E-Mail Removed)> schrieb im
Newsbeitrag news:d64io9$s6o$05$(E-Mail Removed)-online.com... | Hello! | | Ich just found a very nice 'pythonic' solution for an often appearing | problem. I didn't find it documented anywhere, so I'm posting it here. | If this isn't new in any way, I'd really like to get to know it. | | Example problem: | I have some "datetime" objects and want to sort them, as here: | | birthdays = [d1,d2,d3,d4] | birthdays.sort() | | However, I don't want to sort them the default way. These are birthdays, | so only the month and day do matter, not the year. E.g.: | | 2003-01-01 should be smaller than 1984-05-01 | | So I have to write the comparison on my own, e.g. | | def cmp_birthdays(d1,d2): | if d1.month > d2.month: return 1 | if d1.month < d2.month: return -1 | if d1.day > d2.day: return 1 | if d1.day < d2.day: return -1 | return 0 | | ... | birthdays.sort(cmp_birthdays) If you don't care about the year, why not just "normalize" the year to all be the same using the replace method of the date instance? Something like: d1 = datetime.date(2004, 12, 2) d2 = datetime.date(2001, 12, 3) d3 = datetime.date(2002, 12, 6) d4 = datetime.date(1977, 12, 7) dates =[d1,d2,d3,d4] datesNorm = [obj.replace(year=1900) for obj in (dates)] datesNorm.sort() print datesNorm # etcetera HTH, --- Vincent | | This implementation of cmp_birthdays is very ugly. Image you want to | chain more than 2 values in that "cmp_birthdays". I also want to use the | builtin "cmp" function, not ">" and "<". | | After thinking some minutes about it, I found a very nice solution: | I have some "cmp"s one aftter another. If one if them return 1 oder -1, | it sould be returned. If it returns 0, the next "cmp" is used. In other | words: I have a sequence of numbers, and want to get the first one that | is not 0. (or return 0, if all numbers were 0) | | But this is exactly what the "or" operator does, due to short-circuit | evaluation. In this example, that means: | | def cmp_bithdays(d1,d2): | return cmp(d1.month,d2.month) or cmp(d1.day,d2.day) | | The generic pattern is: | | return cmp(...) or cmp (...) or cmp(...) or ... | | I'm not sure whether this pattern is already a "common recipe", but | I found it to be a very nice idea. | | Any opinions? | | | Greets, | | Volker | | -- | Volker Grabsch | ---<<(())>>--- | \frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt | [G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}} Robert Kern Guest Posts: n/a 05-14-2005 Volker Grabsch wrote: > Hello! > > Ich just found a very nice 'pythonic' solution for an often appearing > problem. I didn't find it documented anywhere, so I'm posting it here. > If this isn't new in any way, I'd really like to get to know it. > > Example problem: > I have some "datetime" objects and want to sort them, as here: > > birthdays = [d1,d2,d3,d4] > birthdays.sort() > > However, I don't want to sort them the default way. These are birthdays, > so only the month and day do matter, not the year. E.g.: > > 2003-01-01 should be smaller than 1984-05-01 [snip] > Any opinions? I find that using the "key" argument to sort is much nicer than "cmp" for these tasks. In [5]:L = [datetime.date(2005,5,2), datetime.date(1984,12,15), datetime.date(1954,1,1)] In [7]:L.sort(key=lambda x: (x.month, x.day)) In [8]:L Out[8]: [datetime.date(1954, 1, 1), datetime.date(2005, 5, 2), datetime.date(1984, 12, 15)] -- Robert Kern http://www.velocityreviews.com/forums/(E-Mail Removed) "In the fields of hell where the grass grows high Are the graves of dreams allowed to die." -- Richard Harter Peter Hansen Guest Posts: n/a 05-14-2005 vincent wehren wrote: > "Volker Grabsch" <(E-Mail Removed)> schrieb im > Newsbeitrag news:d64io9$s6o$05$(E-Mail Removed)-online.com...
> | However, I don't want to sort them the default way. These are birthdays,
> | so only the month and day do matter, not the year. E.g.:
> |
....
> If you don't care about the year, why not just "normalize" the year
> to all be the same using the replace method of the date instance?
> Something like:
>
> d1 = datetime.date(2004, 12, 2)
> d2 = datetime.date(2001, 12, 3)
> d3 = datetime.date(2002, 12, 6)
> d4 = datetime.date(1977, 12, 7)
> dates =[d1,d2,d3,d4]
> datesNorm = [obj.replace(year=1900) for obj in (dates)]
> datesNorm.sort()
> print datesNorm # etcetera
Or just use the .timetuple() method on datetime objects and sort on the
8th element of the 9-element tuple, which is the day-of-the-year.
-Peter
Steven Bethard
Guest
Posts: n/a
05-14-2005
Robert Kern wrote:
> I find that using the "key" argument to sort is much nicer than "cmp"
>
> In [5]:L = [datetime.date(2005,5,2), datetime.date(1984,12,15),
> datetime.date(1954,1,1)]
>
> In [7]:L.sort(key=lambda x: (x.month, x.day))
>
> In [8]:L
> Out[8]:
> [datetime.date(1954, 1, 1),
> datetime.date(2005, 5, 2),
> datetime.date(1984, 12, 15)]
Yes, definitely. Also worth noting in Robert Kern's solution is that
def mycmp(d1, d2):
return cmp(d1.month,d2.month) or cmp(d1.day,d2.day)
you can write:
def mycmp(d1, d2):
return cmp((d1.month, d1.day), (d2.month, d2.day))
or if you're using the key= argument (like you probably should):
def mykey(d):
return (d.month, d.day)
The point here is that rather than chaining cmp() calls with ors, you
should just use a tuple -- the standard comparison order in tuples is
exactly what you're looking for.
STeVe
Volker Grabsch
Guest
Posts: n/a
05-14-2005
Steven Bethard wrote:
> Robert Kern wrote:
> def mykey(d):
> return (d.month, d.day)
>
> The point here is that rather than chaining cmp() calls with ors, you
> should just use a tuple -- the standard comparison order in tuples is
> exactly what you're looking for.
That's an excellent idea! Thanks a lot. I really didn't think of the "key="
argument.
Greets,
--
Volker Grabsch
---<<(())>>---
\frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
[G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
Volker Grabsch
Guest
Posts: n/a
05-14-2005
vincent wehren wrote:
>
> If you don't care about the year, why not just "normalize" the year
> to all be the same using the replace method of the date instance?
That's a very bad idea. In my example, this would work, but in "reality"
I don't sort datetime objects, of course! (Is there any real application
where you want to do that?)
Instead, I'm sorting "Person" objects using a "birthday" attribute.
Since I use these Person objects also in other places, they should never
be modified without just to be sorted. In general, the number side effects
should always be minimized.
> datesNorm = [obj.replace(year=1900) for obj in (dates)]
> datesNorm.sort()
This code would go bad especially in my situation, where my "Person"
objects are SQLObjects, thus the "normalisation" would be written into
the database. Okay, one could use transactions and rollback", but I
think, my point is clear now.
Nevertheless, I think you idea is very interesting. Is there any "real"
application where normalizing just for sorting would be reasonable?
Greets,
--
Volker Grabsch
---<<(())>>---
\frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
[G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
Volker Grabsch
Guest
Posts: n/a
05-14-2005
Peter Hansen wrote:
>
> Or just use the .timetuple() method on datetime objects and sort on the
> 8th element of the 9-element tuple, which is the day-of-the-year.
An interesting idea. But wouldn't sorting by (dd.month,dd.day) be more
effective?
In other words: Does .timetuple() create a tuple, or does it just return
a tuple which is present anyway?
Greets,
--
Volker Grabsch
---<<(())>>---
\frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
[G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
vincent wehren
Guest
Posts: n/a
05-14-2005
"Volker Grabsch" <(E-Mail Removed)> schrieb im
Newsbeitrag news:d65h73$fgr$02\$(E-Mail Removed)-online.com...
| vincent wehren wrote:
| >
| > If you don't care about the year, why not just "normalize" the year
| > to all be the same using the replace method of the date instance?
|
| That's a very bad idea. In my example, this would work, but in "reality"
| I don't sort datetime objects, of course! (Is there any real application
| where you want to do that?)
|
| Instead, I'm sorting "Person" objects using a "birthday" attribute.
| Since I use these Person objects also in other places, they should never
| be modified without just to be sorted. In general, the number side effects
| should always be minimized.
Can you explain where you see a modification to the orginal object
happening?
(or in any of the other solutions proposed for that matter...)
Not here I hope:
|
| > datesNorm = [obj.replace(year=1900) for obj in (dates)]
| > datesNorm.sort()
If you print datesNorm, you'll see:
[datetime.date(1900, 12, 2), datetime.date(1900, 12, 3), datetime.date(1900,
12, 6), datetime.date(1900, 12, 7)]
However, dates is still the same:
[datetime.date(2004, 12, 2), datetime.date(2001, 12, 3), datetime.date(2002,
12, 6), datetime.date(1977, 12, 7)]
|
| This code would go bad especially in my situation, where my "Person"
| objects are SQLObjects, thus the "normalisation" would be written into
| the database. Okay, one could use transactions and rollback", but I
| think, my point is clear now.
Since you wouldn't need to change the attribute object to
perform a sort (on the instances) using it (or portions of it) as key, it's
not.
Or is there something fundamental I am missing about your particular use
case?
| Nevertheless, I think you idea is very interesting. Is there any "real"
| application where normalizing just for sorting would be reasonable?
How about a case-insensitive sort of strings? (uppering being the
normalization step)
Or getting rid of accented / special characters before sorting.
These sound like fairly straight-forward use cases to me
Anyway, I was doing some SQL this morning where getting rid of portions of
'datetime' fields in
a WHERE clause is sometimes just the ticket - hence the connection.
--
Vincent
|
| --
| Volker Grabsch
| ---<<(())>>---
| \frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
| [G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
Volker Grabsch
Guest
Posts: n/a
05-14-2005
vincent wehren wrote:
>
>| > If you don't care about the year, why not just "normalize" the year
>| > to all be the same using the replace method of the date instance?
>|
>| That's a very bad idea. In my example, this would work, but in "reality"
>| I don't sort datetime objects, of course! (Is there any real application
>| where you want to do that?)
>|
>| Instead, I'm sorting "Person" objects using a "birthday" attribute.
>| Since I use these Person objects also in other places, they should never
>| be modified without just to be sorted. In general, the number side effects
>| should always be minimized.
>
> Can you explain where you see a modification to the orginal object
> happening?
> (or in any of the other solutions proposed for that matter...)
Sorry, my fault. I didn't read carefully enough. X-)
> Not here I hope:
>
>| > datesNorm = [obj.replace(year=1900) for obj in (dates)]
>| > datesNorm.sort()
While you don't change the original objects, there's still a problem since
you're sorting the normalized values. However, I want to sort the original
list (i.e. the list of "Person" objects).
But that's not a real problem if one normalizes in a key function:
def key_birthday(d):
return d.replace(year=1900)
...
dates.sort(key=key_birthday)
...as suggested in other followups of my posting.
>| Nevertheless, I think you idea is very interesting. Is there any "real"
>| application where normalizing just for sorting would be reasonable?
>
> How about a case-insensitive sort of strings? (uppering being the
> normalization step)
> Or getting rid of accented / special characters before sorting.
> These sound like fairly straight-forward use cases to me
For your solution these are good examples. But my question was, whether
normalizing first, and just sorting the normalized values (not the original
values) is reasonable.
I.e., when I sort some strings case-insensitive, I don't want my resulting
(sorted) list to contain only lowercase string. But that's what I would
get if I used the algorithm you described above.
Greets,
--
Volker Grabsch
---<<(())>>---
\frac{\left|\vartheta_0\times\{\ell,\kappa\in\Re\} \right|}{\sqrt
[G]{-\Gamma(\alpha)\cdot\mathcal{B}^{\left[\oint\!c_\hbar\right]}}}
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# MTH501 - Linear Algebra GDB NO 2 Solution & Discussion Fall 2015
MTH501 - Linear Algebra GDB NO 2 Solution & Discussion Fall 2015
Solve the following system of linear equations by inversion method:
Type all the steps to reach the solution of the above linear system.
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Muhammad Umar Rashid thanks for sharing
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AoA math type main to matrix main 4 value nhn lehke jat ehay keonek space allow nhn hay ap ne kse kea hay plzz tell me?
bahi jan main nay video upload ke hai ap dakh lay na kah kasy kia hai
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Rashid g! ap ne simple Method kya hy ... pr Inversion Method tu is mn nzr hi ni aa rha hy ..???
Ye math type method hai inversion mathod nahe
Not sure
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5 | 620 | 2,216 | {"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} | 2.859375 | 3 | CC-MAIN-2021-04 | longest | en | 0.765814 |
https://www.saleh-theory.com/article/accurate-calculation-of-the-volume-and-density-of-the-big-bang | 1,713,791,877,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00515.warc.gz | 860,169,383 | 8,551 | # Accurate Calculation of the Volume and Density of the Big Bang
## Accurate Calculation of the Volume and Density of the Big Bang
Considering that the smallest, fastest, and lightest object in the universe is photon and the universe mass is about 1053ππ, if we consider the photon as the basis of the Big Bang, the volume and density equations are as follows:
Where ππ is the total mass of universe, π is the number of photon, ππ, ππ, ππ πππ ππ are the mass, radius, volume and density of the photon and ππ΅π΅, ππ΅π΅ πππ ππ΅π΅ are the radius, volume and density of the Big Bang sphere.
According to the data, it can be said that if the existing sphere at the moment of the Big Bang was made of photons, by such a volume and mass, it is by far different from the information that obtained for the Big Bang before. In other words, the photon is not the desired particle that could have formed the Big Bang sphere.
Therefore, we define a special particle called βsub-photonβ with dimensions of one billionth of a photon (in terms of radius value).
By using this particle, the calculations are as follows:
Where ππ π πππ ππ π are the radius and volume of the βsub-photonβ.
Due to obtained values, this fundamental particle is capable of defining the Big Bang phenomenon. It can be said that for the Big Bang, a density higher than 1042 ππ/π3 and a volume less than 1010 π3 cannot be imagined.
Therefore, the density of the Big Bang is definitely less than 1042 ππ/π3 and its volume is undoubtedly more than 1010 π3.
Photon Sub-photon big bang phenomena density of the big bang Volume of the big bang | 532 | 1,741 | {"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} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.896477 |
https://www.physicsforums.com/threads/mathematical-definition-of-energy.591622/ | 1,519,532,249,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816094.78/warc/CC-MAIN-20180225031153-20180225051153-00177.warc.gz | 888,716,768 | 15,683 | # Mathematical Definition of Energy
1. Mar 29, 2012
### azabak
What is the general definition of energy? I already know that it means ability to perform work and that Work = ∫Force d(displacement) = Δ Kinetic Energy = -Δ Potential Energy ( in a conservative field "a closed path integral of the force = 0"), Σ Kinetic-Potential = constant, ∫Kinetic-Potential d(time) = minimum action... so just cut to the chase. None of those concepts define energy in a general mathematical sense. More precisely I'm asking IF there's any definition or not.
Last edited: Mar 29, 2012
2. Mar 29, 2012
### Khashishi
Energy is sort of a bunch of different abstract concepts that all happen to work out to be the same thing. But none of these concepts can really be thought of as the most fundamental or definitive definition of energy. Energy is something we use in physical models--it's not something we measure directly--so any definition is going to seem like a mathematical contrivance or bookkeeping mechanism. The most concrete definition is probably found in relativity. Energy is mass times a unit conversion factor. Energy also can be interpreted as a generator of infinitesimal time displacements.
3. Mar 30, 2012
### K^2
Well, what you stated is basically the definition. It's ability to do work. What probably seems confusing about it is that it's hard to define the point where the body can no longer do work. For example, we define gravitational potential energy to be zero at ground level. But what if I dig a hole and lower the object further. It has negative potential energy now. What's that all about?
And the answer to this is that as far as mechanics goes, there is no absolute zero for energy. You can define whatever state of the system you like as zero energy, and then see how much work needs to be done on the system to get it to other states. You call that work the energy of that new state. This is sufficient because in mechanics, only relative energy is important. And so defining energy as ability to work is entirely sufficient as far as all the math goes.
This also holds true for Quantum Mechanics and Thermodynamics. Though, for later, in a really roundabout way, quite often. Where it all starts getting a little crazy is General Relativity. But I wouldn't worry about any of it for now.
4. Mar 30, 2012
### azabak
Although I'm asking this I know one definition of energy. If you define work as W=∫F dr, energy will arise "naturally" as the factor that changes in relation of the speed (kinetic) or the position (potential). Defining both kinetic and potential energy from work leads to its conservation, so that: W = ΔK = -ΔU therefore ΔK+ΔU = 0, Ki-Kf+Ui-Uf = 0, Ki+Ui = Kf+Uf = constant. And since none of those factors are time dependent the constant neither increases nor decreases as time passes, showing an "intuitive" proof the the Noether's Theorem that conservation of energy imply symmetry in time translation. The constant is the total energy, the general definition, but since potential energy is by convention negative the constant is K-U, or the Lagrangian. The closest I can get to define this constant is as the time derivative of the action. The problem is that unless I define action from other quantity (as momentum or angular momentum) this definition will be circular. | 757 | 3,322 | {"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} | 3.75 | 4 | CC-MAIN-2018-09 | longest | en | 0.942246 |
https://customwritingspapers.com/elementary-calculus/ | 1,600,509,253,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00353.warc.gz | 357,293,281 | 7,433 | # Elementary Calculus
## Question 1
Write an equation to embody the administration from the forthcoming table of prizes:
A.y = -2x B.y = 2x C.y = x + 1 D.y = x + 2
## Question 2
Write an equation to embody the administration from the forthcoming table of prizes:
A.y = x + 3 B.y = 3x C.y = 4x D.y = x – 3
## Question 3
Which one of the forthcoming kinsfolk is not a administration?
A.A B.B C.C D.D
## Question 4
A = {-3, -2, -1, 0, 1, 2, 3} f is a administration from A to the set of complete gum as defined in the forthcoming table: What is the lordship of f?
A. The set of Integers B. The set of Complete gum
C. {-3, -2, -1, 0, 1, 2, 3} D. {0, 1, 4, 9}
## Question 5
Which one of these graphs does not make-clear a administration?
A. A B. B C. C D. D
## Question 6
Which one of the forthcoming is not a administration?
A B
C D
## Question 7
Which relative is not a administration?
A. F(x) = √x B. F(x) = -√x C. F(x) = ±√x D. F(x) = √x – 1
## Question 8
The administration f is defined on the developed gum by f(x) = 2 + x - x2 What is the prize of f(-3)?
A. -10 B. -4 C. 8 D. 14
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http://lispforum.com/viewtopic.php?f=33&t=4611&view=print | 1,563,451,747,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00418.warc.gz | 99,439,428 | 2,542 | Page 1 of 1
### Help for a tree manipulations.
Posted: Mon Nov 28, 2016 9:48 am
Hello, first sorry about my English, I am French.
I am in computer science and I have a duty in Clisp.
This is an alphabetic sort, I block on a function.
The "ajoute" function, which adds a word to a tree, presents three cases:
The tree is empty: we create a tree with just the word as root: ( "word");
The word is greater than the root: we recurse with the right branch as root;
We recurse with the left branch as root.
Here are the correct tests :
(setq arbre (ajoute "k" nil)) => ("k")
(setq arbre (ajoute "a" arbre)) => ("k" ("a"))
(setq arbre (ajoute "m" arbre)) => ("k" ("a") ("m"))
(setq arbre (ajoute "n" arbre)) => ("k" ("a") ("m" nil ("n")))
(setq arbre (ajoute "b" arbre)) => ("k" ("a" nil ("b")) ("m" nil ("n")))
And here are my tests
(setq arbre (ajoute "k" nil)) => ("k")
(setq arbre (ajoute "a" arbre)) => ("k" ("a"))
(setq arbre (ajoute "m" arbre)) => ("k" ("a") ("m"))
(setq arbre (ajoute "n" arbre)) => ("k" ("a") ("m") ("m" ("n")))
Here is my function :
Code: Select all
`(defun ajoute (mot arbre) (print arbre) (cond ((atom arbre) (list mot . nil)) ; NOT LIST => ADDS AT TREE ((listp (car arbre)) (ajoute mot (car arbre))) ((equal (french-string mot (car arbre)) t) (append arbre (list (ajoute mot (cdr (cdr arbre)))))) ; RIGT TREE ((append arbre (list (ajoute mot (car (cdr arbre)))))) ) ) ; LEFT TREE`
french-string is a function that returns true if argument 1 is a word after argument 2 in alphabetical order.
exemple :
[182]> (french-string "a" "b")
nil
[183]> (french-string "b" "a")
t
I think the problem with my function "adds" comes from my use of list and append, but I can not solve the problem. Do you have any leads to help me?
`(list (car arbre) (cadr arbre) (ajoute mot (caddr arbre)))` | 598 | 1,822 | {"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-2019-30 | latest | en | 0.624306 |
https://digitalcommons.northgeorgia.edu/ungauthors/2021/bibliography/44/ | 1,618,545,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00287.warc.gz | 244,518,386 | 8,261 | ## 2020-2021
Gainesville
7-9-2020
#### Publisher
the Teaching Statistics: An International Journal for Teachers
#### Book or Journal Information
the Teaching Statistics: An International Journal for Teachers, https://onlinelibrary.wiley.com/doi/abs/10.1111/test.12233
#### Abstract
College‐level statistics courses emphasize the use of the coefficient of determination, R‐squared, in evaluating a linear regression model: higher R‐squared is better. This often gives students an impression that higher R‐squared implies better predictability since textbooks tend to use sample data to support the theory and students rarely have an opportunity to work on real data. In this paper, health care stocks are used as predictors and the result demonstrates that high R‐squared does not necessarily mean high predictability and that multiple linear regression can be used in the study of data behavior. In particular, by learning the pattern of the near and far out‐of‐sample‐prediction errors for different time periods throughout a dataset, the near out‐of‐sample prediction errors can be used to control the prediction errors and identify a subset of predictors that can well reflect the trend of S&P 500.
#### Share
COinS
Multiple Linear Regression: Identify Potential Health Care Stocks for Investments Using Out-of-Sample Predictions
College‐level statistics courses emphasize the use of the coefficient of determination, R‐squared, in evaluating a linear regression model: higher R‐squared is better. This often gives students an impression that higher R‐squared implies better predictability since textbooks tend to use sample data to support the theory and students rarely have an opportunity to work on real data. In this paper, health care stocks are used as predictors and the result demonstrates that high R‐squared does not necessarily mean high predictability and that multiple linear regression can be used in the study of data behavior. In particular, by learning the pattern of the near and far out‐of‐sample‐prediction errors for different time periods throughout a dataset, the near out‐of‐sample prediction errors can be used to control the prediction errors and identify a subset of predictors that can well reflect the trend of S&P 500. | 459 | 2,264 | {"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-2021-17 | latest | en | 0.863744 |
https://www.deskera.com/blog/solvency-ratio-vs-liquidity-ratios/ | 1,656,432,948,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103556871.29/warc/CC-MAIN-20220628142305-20220628172305-00545.warc.gz | 773,502,056 | 15,827 | # Solvency Ratio vs. Liquidity Ratios: What's the Difference?
As an investor, you would certainly want to assess a firm’s liquidity ratios as well as its solvency ratios which are important parameters to give a go-ahead for an investment decision.
It is essential for the investors to know whether or not the investment will be beneficial. For a clear understanding, the article will elaborate upon the following points:
• Definition of Liquidity
• Definition of Solvency
• What is Solvency Risk?
• Comparison between Liquidity and Solvency
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Investing decisions are based on liquidity ratios and solvency ratios. An organization's liquidity ratio measures its ability to translate its assets into cash. In contrast, the solvency ratio determines whether a company is capable of meeting its financial obligations.
## Definition of Liquidity
Liquidity in accounting refers to a company's ability to pay its liabilities as due, in a timely manner. An abundance of current assets and cash indicates high liquidity.
The ease with which an asset can be converted into cash quickly and at a minimal discount is also considered while estimating liquidity. A liquid asset is one that has an active market with many buyers and sellers. It is still possible for companies that lack the liquidity to go bankrupt despite being solvent.
## Definition of Solvency
The business' long-term financial stability is called solvency. Solvency refers to the total assets being greater than the total liabilities of a company. An assessment of solvency is based on solvency ratios. By measuring these ratios, we can determine if the business can repay its long-term debts and interest.
Solvency Ratio Formula
The formula for the Solvency Ratio is given by:
(Net after-tax income + Non-cash expenses) / (Short-term liabilities + Long-term liabilities)
The steps to calculate the solvency ratio are as follows:
• Adding all non-cash expenses to the after-tax business income will provide the approximate cash flow generated by the business
• Then, liabilities for both short- and long-term are added together
• Total liabilities divided by adjusted net income
## What is Solvency Risk?
Despite disposing of its assets, an organization faces the risk of not being able to meet its financial obligations at full value. Insolvent businesses are unable to pay their debts and will be forced to file for bankruptcy. A company's financial statements should be examined thoroughly to ensure the business is solvent and is in a profitable condition.
## Comparison between Liquidity and Solvency
The following table gives away the differences between liquidity and solvency:
Factors for Comparison Liquidity Solvency Definition Liquidity in accounting refers to a company's ability to pay its liabilities as due, in a timely manner. The business' long-term financial stability is called solvency. Solvency refers to the total assets being greater than the total liabilities of a company. Purpose Provides insight into whether the firm’s current assets can be turned into cash. Lack of solvency could be detrimental to a firm’s routine operations. Obligations Addresses short-term obligations Addresses long-term obligations Risk Although the risk is low, it can potentially impact the company’s reputation Risk is high as insolvency could result in the company filing for banruptcy Functional Challenges When a firm is unable to convert its assets into cash quickly When a firm lacks enough number of assets for the repayment of their due bills Ratios Current ratio, cash ratio, quick ratio/acid test ratio Debt-to-equity ratio, interest coverage ratio, debt-to-asset ratio Basic functionality Using all the current assets to settle all outstanding debts To determine a company's ability to meet its debt obligations on time by measuring its financial health Ratio Analysis Higher ratios result in better outcomes for the company High ratio of interest coverage and low ratio of debt-to-equity is indicative of the firm not defaulting on its long term obligations Impact on Each Other High solvency can help in achieving liquidity in a short period of time High liquidity may or may not help in achieving solvency in a short span of time
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## Key Takeaways
It is important to grasp the fundamentals of solvency and liquidity and misinformation about either of them could have serious repercussions on business operations.
• Despite measuring the ability of a company to pay its obligations, they are not interchangeable since both attribute to different ranges and objectives
• Liquidity in accounting refers to a company's ability to pay its liabilities as due, in a timely manner
• The business' long-term financial stability is called solvency. Solvency refers to the total assets being greater than the total liabilities of a company
• Financially troubled companies may have difficulty attracting customers and vendors. Unexpected bankruptcy can happen to a business in certain extreme cases
Hey! Try Deskera Now! | 1,215 | 6,287 | {"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-2022-27 | longest | en | 0.951917 |
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String art projects for geometry use a technique known as curve stitching, which creates circles and curves from straight lines. According to the Agnes Scott College website, Mary Everest Boole, a British educator and author, invented this technique to enable children to comprehend the mathematics of angles and spaces. Using string, thread or floss, children can begin stitching angles and then graduate to more complex circles and triangles.
## Basic Technique to Create Angles
Gather a ruler, needle, scissors, push pins, small carpet squares, thread and a 6” square poster board. Draw an angle, such as right, acute or obtuse, on the poster board. Make sure the arms are of equal length. Mark each arm in 1/2-inch increments. Slide a carpet square under the board so you can prick holes into the board with the pins. Number the marks, starting with #1 on one arm located at the angle’s vertex. Reverse the numbering on the other arm so that you end with the highest number on the vertex. Thread the needle and knot the end. Wind the string around #1 on the bottom arm, coming up to #1 on the other arm, then coming down to loop around #2 on the bottom arm and so forth. This basic advancing weave will result in a smooth Belzier curve.
## Next Stage: A Circle
Use a compass to draw a circle on a 6-inch-square poster board. Use a protractor to mark the circle’s circumference into equal parts of either 5 or 10 degrees. Slide a carpet square under the board. Use a pin to make a hole at each mark on the circumference. Number the holes, starting with #1 at the topmost hole. Thread a needle. Pull the thread through hole #1 and stitch across the circle to any other hole, such as hole #6. Move to the next hole, or hole #7. Stitch back to one hole from where you started, or hole #2. Move over one hole to hole #3, and then stitch over to hole #7. Repeat this procedure until you finish the circle. The succession of equal chords will forge a concentric circle within the drawn circle.
## Variation on a Circle
Create a circle burst as you grow more familiar with stitching techniques. Use the same setup as you would for a circle, except run a thread from a pin placed in the center of a circle to every pin along the circle’s perimeter. Knot the thread around the center pin to start. Loop the thread around the topmost perimeter pin and then wrap it around the center pin. Wrap the thread clockwise around the second pin and return the stitch to the center pin. Continue until you have wrapped all of the pins along the perimeter. Push the string around the center nail down so that each new loop falls on top of the last.
## Complex Shape: Icosigenagon
Gather push pins, foam board and spool of thread. Use a compass, ruler and pencil to draw 21 equidistant radial lines that form a ring on tracing paper. Place the design on the board. Mark and pin 21 points at the end of the lines, then pull the paper off. Knot the end of the threat around the top pin, which is zero on the count. Move clockwise to the next pin, or the first of ten steps in each round, and wrap the thread around the pin. Each time you wrap a pin, you have a new starting point. Count for 2 pins, and loop the thread around the end pin. Count for 3 pins and wrap the end pin. Repeat this procedure for the counts of 4 to 10. Start the next 10-step round from the pin you landed on in step 10 of the first round. Repeat all ten steps for each pin on the perimeter of the circle, or 21 times. Knot the thread on the final pin to complete a complex 21-sided polygon, or the icosihenagon. | 836 | 3,681 | {"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} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.890755 |
http://en.wikipedia.org/wiki/Proleptic_Gregorian_calendar | 1,432,384,737,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927592.52/warc/CC-MAIN-20150521113207-00160-ip-10-180-206-219.ec2.internal.warc.gz | 78,887,942 | 14,391 | Proleptic Gregorian calendar
The proleptic Gregorian calendar is produced by extending the Gregorian calendar backward to dates preceding its official introduction in 1582.
Usage
The proleptic Gregorian calendar is explicitly required for all dates before 1582 by ISO 8601:2004 (clause 4.3.2.1 The Gregorian calendar) if the partners to information exchange agree. It is also used by most Maya scholars,[1] especially when converting Long Count dates (1st century BC – 10th century). However, neither astronomers nor non-Maya historians generally use it.
For these calendars we can distinguish two systems of numbering years BC. Bede and later historians did not use the Latin zero, nulla, as a year (see 0 (year)), so the year preceding AD 1 is 1 BC. In this system the year 1 BC is a leap year (likewise in the proleptic Julian calendar). Mathematically, it is more convenient to include a year 0 and represent earlier years as negative, for the specific purpose of facilitating the calculation of the number of years between a negative (BC) year and a positive (AD) year. This is the convention used in astronomical year numbering and in the international standard date system, ISO 8601. In these systems, the year 0 is a leap year.[2]
Although the nominal Julian calendar began in 45 BC, leap years between 45 BC and 1 BC were irregular (see Leap year error). Thus the Julian calendar with quadrennial leap years was only used from the end of AD 4 until 1582 or later, so historians and astronomers prefer to use the actual Julian calendar during that period (see From Julian to Gregorian). But when seasonal dates are important, the proleptic Gregorian calendar is sometimes used, especially when discussing cultures that did not use the Julian calendar.
The proleptic Gregorian calendar is sometimes used in computer software to simplify the handling of older dates. For example, it is the calendar used by MySQL,[3] SQLite,[4] PHP, CIM, Delphi, Python[5] and COBOL.
Difference between Julian and proleptic Gregorian calendar dates
Before the introduction of the Gregorian calendar, the differences between Julian and proleptic Gregorian calendar dates were as follows:
The table below assumes a Julian leap day of 29 February, but the Julian leap day (the bissextile day) was ante diem bis sextum Kalendas Martias in Latin or 24 February (see Julian reform), so dates between 24 and 29 February in all leap years were irregular.
When converting a date in a year which is leap in one calendar but not the other include 29 February in the calculation when the conversion crosses the border between February and March.
Julian range Proleptic Gregorian range Gregorian ahead by:
to 1 March 100
to 28 February 100
−2 days
From 2 March 100
to 29 February 200
From 1 March 100
to 28 February 200
−1 days
From 1 March 200
to 28 February 300
From 1 March 200
to 28 February 300
0 days
From 29 February 300
to 27 February 500
From 1 March 300
to 28 February 500
1 day
From 28 February 500
to 26 February 600
From 1 March 500
to 28 February 600
2 days
From 27 February 600
to 25 February 700
From 1 March 600
to 28 February 700
3 days
From 26 February 700
to 24 February 900
From 1 March 700
to 28 February 900
4 days
From 25 February 900
to 23 February 1000
From 1 March 900
to 28 February 1000
5 days
From 24 February 1000
to 22 February 1100
From 1 March 1000
to 28 February 1100
6 days
From 23 February 1100
to 21 February 1300
From 1 March 1100
to 28 February 1300
7 days
From 22 February 1300
to 20 February 1400
From 1 March 1300
to 28 February 1400
8 days
From 21 February 1400
to 19 February 1500
From 1 March 1400
to 28 February 1500
9 days
From 20 February 1500
to 4 October 1582
From 1 March 1500
to 14 October 1582
10 days | 1,009 | 3,742 | {"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-2015-22 | latest | en | 0.899652 |
https://forums.eveonline.com/t/launcher-cycle-time-missile-distance-and-ticks/349317 | 1,653,562,746,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00249.warc.gz | 340,364,255 | 5,993 | # Launcher cycle time, missile distance and ticks
Let’s say I’m trying to not waste missiles so for distant targets I’m manually stopping the launchers before the target goes boom. However I’m way too lazy to do this all the time, so I’m calculating the distance a missile can travel before the launcher cycles. No need to manually stop if the target goes boom before the cycle ends.
Current numbers: 14603 m/s missile speed, 4.05 s launcher cycle time. 4*14.6 is 58.4, so I don’t need to stop the launcher manually if the target is 58 km away or closer. This is visible in practice.
Now as far as I know, everything in Eve happens on 1 second ticks. My question is then: what happens if i bring the cycle time under 4 seconds (can still upgrade an implant)? Do I need to multiply the missile speed by just 3 seconds then?
Anything from 3.01 to 4.00 seconds is still 4 seconds as far as the server is concerned. Only 2.99 seconds and bellow count as 3 seconds. The difference between 3.01 and 4 seconds is only the relative frequency between 3 second and 4 second cycles because some cycles times are rounded down while others are rounded up if the cycle times are not full seconds.
Oh… so you’re saying my 4.05 current seconds count as 5? I.e. about 73 km travel distance?
In some cases it does, in many cases it doesn’t. The time after the full second determines which percentage of cycles gets rounded up or down. That’s at least how it works with ship hulls.
That’s wrong.
A tick is an update of the server sent to the clients.
many actions are done in the order they are sent to the server. However a few things only happen on ticks :
• movement change (going in warp, direction, speed, MJD, leaving grid, …)
• grid entity changes (creation/destruction/stances)
So your launcher DOES cycle with 4.05 s. The initial launch is created the tick after the server received the initial activation(on your position with maximum speed towards the target) , and the next cycles will happen 4.05s after this, and create a new launch on the tick after that .
After a while, because your cycle is not 4.0, your cycle will restart just after a tick, and so the delay difference between launch will be not 4 but 5 ticks. That will happen statistically one in 20 launches.
so in order to be sure your missiles hit before the next launch, your target must be closer than 4.0×14603 = 58 km. One on 20 will also hit before the next out of average 20 launches.
If you go to 3.75, then 0.75 = 3/4 launches will be 3.0 tick delay, and 0.25 = 1/4 launches will be 4.0 delay.
Thanks everyone. I upgraded the implant and my cycle time went to 3.97 seconds and indeed the only thing that happened is that my “fire and forget” distance went from ~58 to ~56 km.
This topic was automatically closed 90 days after the last reply. New replies are no longer allowed. | 695 | 2,856 | {"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} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.931476 |
https://tradeciety.com/the-truth-about-the-stop-loss-why-only-losing-traders-dont-use-stops | 1,725,945,281,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00283.warc.gz | 534,849,907 | 17,370 | I understand that this is a controversial topic but I fully stand to my word and after working with hundreds and thousands of traders, the losing traders have one thing in common: they don’t use stops. Of course, there is often more that separates the profitable from the unprofitable traders, but stops are a big thing.
Here are the main reasons why trading profitably without a stop loss is impossible:
## 1. No money management
If a trader doesn’t use a stop loss, then he has no money management in his trading. The stop loss distance is used to calculate how many contracts you have to buy or sell in order to achieve a certain position size.
For example, a 10 pip stop loss requires a very different size than a 30 pip stop loss. People always say that “you should risk 1% of your capital per trade” but without having a fixed stop loss, you cannot determine how much this 1% is going to be for your trades and your numbers will be all over the place.
And no, a mental stop doesn’t cut it. Just executing a mental stop a few pips later will completely throw off your money management.
## 2. No risk management
…this brings us to the next point. The Reward:Risk Ratio and the R-Multiple are arguably the most important metrics for any trader and only a trader who can compare those two metrics to his winrate, can estimate if he can be profitable.
For example, if you know that your trades usually roughly have a 1.5:1 Reward:Risk ratio, it means that your winrate only has to be 40% to at least break even. But if your winrate or Reward:Risk ratio don’t add up, you won’t be able to make any money – ever!
Your historical winrate Minimum reward:risk ratio 25% 3 : 1 33% 2 : 1 40% 1.5 : 1 50% 1 : 1 60% 0.7 : 1 75% 0.3 : 1
However, you can only determine the Reward:Risk ratio and the R-Multiple if you have a stop loss. The stop loss is a mandatory number in the Reward:Risk ratio and without it, it’s simply impossible to calculate it.
And then, you are just stumbling around, taking random trades, not knowing if the winners are large enough, if you can keep the losses small and profitable trading becomes an impossible task.
## 3. You are the victim of your emotions
The two common arguments against the first two points mostly are:
3.1 My broker hunts my stops: dead wrong! Your broker really doesn’t care about your o.1 lot trade and the \$10 he can make by looking at your stop. What happens is that you place your stop like everyone else and then you just get taken out.
3.2 I use mental stops: as I said above, that doesn’t quite cut it. Even executing the mental stop a few pips later will completely throw off your position size metrics and your Reward:Risk ratio and you are back to complete amateur trading.
A stop loss is your insurance in the market. A stop loss doesn’t think and it doesn’t care about your PnL, unlike you. Traders often try to rationalize their emotionally driven (dumb) impulses with excuses. A fixed stop loss, if you don’t tinker with it, makes sure that you don’t fall prey to your emotions.
Thus, it’s so important to set a FIXED stop loss upon entering a trade and then stand by it. If price reaches the stop loss, you’re out. You don’t widen the stop, you don’t add to the loss and you don’t revenge trade. Your stop is your red flag and it signals that your trade idea didn’t work out.
I know that many people don’t agree with me, but this is the truth and looking at the arguments for position size and risk management should make it obvious that a stop loss is so much more than you think.
#### How To Trade The Engulfing Candlestick Pattern
When trading single candlestick patterns, no pattern is more powerful than the engulfing candlestick pattern. You can create strategies with only... | 878 | 3,761 | {"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-2024-38 | latest | en | 0.929959 |
https://brainly.ph/question/15008 | 1,485,080,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281421.33/warc/CC-MAIN-20170116095121-00461-ip-10-171-10-70.ec2.internal.warc.gz | 810,124,757 | 9,743 | # What is the solution of this equation? 3x + 7 = 1
1
by MarkdelaCruz
3x = 1 - 7
3x = -6
x = 2
3x/3 = -6/3 = x = -2
x = -2 final ans | 70 | 133 | {"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} | 3.8125 | 4 | CC-MAIN-2017-04 | latest | en | 0.822907 |
https://crystalholidaystravel.com/1761ab/878868-how-much-is-4-grams-of-ginger-in-teaspoons | 1,618,761,942,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00611.warc.gz | 302,137,421 | 12,218 | Dessertspoon How much is 4 grams of ginger powder in terms of teaspoon? I use fresh ginger and don't have access to standardized extracts or anything like that. Use this page to learn how to convert between teaspoons and grams. of ground ginger are in a teaspoon. of ginger. Inputs? ›› Quick conversion chart of grams to teaspoons. 2 / 3 teaspoon of ginger = 1.17 grams of ginger. 0.68 grams of ground sage fit into one teaspoon. of ginger. I have no idea how much that is. During Pregnancy. of ginger. - 6 grams is equal to 1.20 teaspoons. How many US teaspoons of baking powder in 4 grams? 20 grams to teaspoons = 4.69421 teaspoons 5 grams to teaspoons = 1.17355 teaspoons. How much does does a 100 dollar roblox gift card get you in robhx? 2 / 3 teaspoon of cinnamon = 1.76 grams of cinnamon. 2 teaspoons of ginger = 3.52 grams of ginger. How many tsp is a "slice" of ginger? The conversion factors are approximate once it is intended for recipes measurements. ½ teaspoon of paprika = 1.15 grams of paprika. 2 teaspoons of paprika = 4.60 grams of paprika.
of ginger. I have a stir-fry recipe here calling for 5 slices of ginger. 1 cubic meter is equal to 200000 teaspoons, or 1000000 grams. 3 Metric cups of ginger = 9.30 oz. ½ teaspoon of ginger = 0.88 grams of ginger. 50g (1 oz/28.349523125 g) = 1.763698097479021 ounces exactly Convert 4 Grams to Teaspoons. Converting 30 grams to teaspoons is not as straightforward as you might think. 3 teaspoons of ginger = 0.18 oz. ¼ teaspoon of ginger = 0.44 grams of ginger. how many teaspoons is 2 inches of ginger, And be careful processing it! Grams are a mass unit while teaspoons are a volume unit. ¾ teaspoon of ginger = 1.32 grams of ginger. How many teaspoons is 6 grams? Notes: the results in this calculator are rounded (by default) to 3 significant figures. 4 grams of oil equals 0.86 ( ~ 3 / 4) US teaspoon (*) 'Weight' to Volume Converter. How many grams of ground ginger are in 1 teaspoons? Notes: the results in this calculator are rounded (by default) to 2 significant figures. How many teaspoons are 14 grams? 4 grams = 0.843 teaspoons (a little less than a teaspoon) A good rule of thumb is that there are 5 grams per teaspoon of most cooking ingredients. 50 grams is the metric measure of mass (like weight in US). Maxfield said, “On our first batch of Pedal Hard Ginger Beer, we juiced over 900 lbs. Use this page to learn how to convert between grams and teaspoons. Further, teaspoon is a considered to be a measurement of volume, while a gram is considered to be a measurement of weight. 1 oz sliced Ginger = 25g = 1/4 cup. How to Convert Grams to Teaspoons. 3 teaspoons of sage = 2.04 grams of sage. I've been using ginger to control morning sickness with great results, but I've read that I shouldn't take more than 1 gram per day. But by using 50 grams of sugar instead of 12 teaspoons, you can't go wrong. How many grams of ground ginger are in one Metric cup? Note that rounding errors may occur, so always check the results. The answer to this question totally depends upon the substance that is to be measured. Grams Teaspoon. of ginger. Volume ⇀ Weight Weight ⇀ Volume. 3 teaspoons of ginger = 5.28 grams of ginger. Please note that all conversion figures given are for level teaspoons and not heaped teaspoons. As an example, a teaspoon of sugar weighs about 4.2 grams, while a teaspoon of salt weighs about 6 grams.
Dessertspoon How much is 4 grams of ginger powder in terms of teaspoon? I use fresh ginger and don't have access to standardized extracts or anything like that. Use this page to learn how to convert between teaspoons and grams. of ground ginger are in a teaspoon. of ginger. Inputs? ›› Quick conversion chart of grams to teaspoons. 2 / 3 teaspoon of ginger = 1.17 grams of ginger. 0.68 grams of ground sage fit into one teaspoon. of ginger. I have no idea how much that is. During Pregnancy. of ginger. - 6 grams is equal to 1.20 teaspoons. How many US teaspoons of baking powder in 4 grams? 20 grams to teaspoons = 4.69421 teaspoons 5 grams to teaspoons = 1.17355 teaspoons. How much does does a 100 dollar roblox gift card get you in robhx? 2 / 3 teaspoon of cinnamon = 1.76 grams of cinnamon. 2 teaspoons of ginger = 3.52 grams of ginger. How many tsp is a "slice" of ginger? The conversion factors are approximate once it is intended for recipes measurements. ½ teaspoon of paprika = 1.15 grams of paprika. 2 teaspoons of paprika = 4.60 grams of paprika.
of ginger. I have a stir-fry recipe here calling for 5 slices of ginger. 1 cubic meter is equal to 200000 teaspoons, or 1000000 grams. 3 Metric cups of ginger = 9.30 oz. ½ teaspoon of ginger = 0.88 grams of ginger. 50g (1 oz/28.349523125 g) = 1.763698097479021 ounces exactly Convert 4 Grams to Teaspoons. Converting 30 grams to teaspoons is not as straightforward as you might think. 3 teaspoons of ginger = 0.18 oz. ¼ teaspoon of ginger = 0.44 grams of ginger. how many teaspoons is 2 inches of ginger, And be careful processing it! Grams are a mass unit while teaspoons are a volume unit. ¾ teaspoon of ginger = 1.32 grams of ginger. How many teaspoons is 6 grams? Notes: the results in this calculator are rounded (by default) to 3 significant figures. 4 grams of oil equals 0.86 ( ~ 3 / 4) US teaspoon (*) 'Weight' to Volume Converter. How many grams of ground ginger are in 1 teaspoons? Notes: the results in this calculator are rounded (by default) to 2 significant figures. How many teaspoons are 14 grams? 4 grams = 0.843 teaspoons (a little less than a teaspoon) A good rule of thumb is that there are 5 grams per teaspoon of most cooking ingredients. 50 grams is the metric measure of mass (like weight in US). Maxfield said, “On our first batch of Pedal Hard Ginger Beer, we juiced over 900 lbs. Use this page to learn how to convert between grams and teaspoons. Further, teaspoon is a considered to be a measurement of volume, while a gram is considered to be a measurement of weight. 1 oz sliced Ginger = 25g = 1/4 cup. How to Convert Grams to Teaspoons. 3 teaspoons of sage = 2.04 grams of sage. I've been using ginger to control morning sickness with great results, but I've read that I shouldn't take more than 1 gram per day. But by using 50 grams of sugar instead of 12 teaspoons, you can't go wrong. How many grams of ground ginger are in one Metric cup? Note that rounding errors may occur, so always check the results. The answer to this question totally depends upon the substance that is to be measured. Grams Teaspoon. of ginger. Volume ⇀ Weight Weight ⇀ Volume. 3 teaspoons of ginger = 5.28 grams of ginger. Please note that all conversion figures given are for level teaspoons and not heaped teaspoons. As an example, a teaspoon of sugar weighs about 4.2 grams, while a teaspoon of salt weighs about 6 grams. | 1,749 | 6,792 | {"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-2021-17 | latest | en | 0.904633 |
https://www.physicsforums.com/threads/clock-of-observers.725995/ | 1,527,109,120,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00156.warc.gz | 839,453,675 | 15,483 | Clock of observers
1. Dec 1, 2013
Hepic
We have three observers. The A is in earth,B is into a car that runs with big speed,C is out of galaxy.
All observesr see that their clock runs normal. A will see that clock of B run slower,and B will see the same for A.
1)
Logically, C will see A's and B's clock run slower from his clock,but which clock will run slower? A's or B's (I think A's,because has more speed,I am right?)
2)
And from the opposite,who from both(A,B) will see C's clock run slower?(I think again A,due of speed).
I am correct to 1) and 2)
Thanks!!!
2. Dec 1, 2013
Mentz114
What do you mean by 'see' in this
If you mean 'by visual contact', then clocks moving away appear slower and approaching clocks appear faster.
3. Dec 1, 2013
Hepic
That for observer A,clock of B run slower,and for B,clock of A run slower from his clock.
4. Dec 1, 2013
Mentz114
Sure, but this is not a directly observable physical effect. It appears when either time is converted from its rest frame coordinates to the moving frame coordinates.
5. Dec 1, 2013
Staff: Mentor
You still have to specify C's velocity relative to the other two before we can answer the question. Just because he's a looooong ways away doesn't mean he cannot be at rest relative to one, the other, or neither.
(In special relativity, that is. In the general relativity case, we have to be a lot more careful about specifying C's state of motion).
6. Dec 1, 2013
Hepic
C is standing about A and B.
7. Dec 1, 2013
Staff: Mentor
What does that mean?
8. Dec 1, 2013
Staff: Mentor
C cannot be at rest relative to both A and B. They're moving relative to one another, so anything at rest relative to one is moving relative to the other. | 471 | 1,726 | {"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} | 3.1875 | 3 | CC-MAIN-2018-22 | latest | en | 0.942614 |
http://48.klmty.live/coordinate-plane-drawings-worksheets/ | 1,603,490,782,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00427.warc.gz | 1,405,925 | 31,023 | # Coordinate Plane Drawings Worksheets
In Free Printable Worksheets255 views
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Coordinate Plane Drawings Worksheets
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# It is an oversimplified view of cattle raising to say that
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It is an oversimplified view of cattle raising to say that [#permalink]
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29 Mar 2009, 17:04
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11. It is an oversimplified view of cattle raising to say that all one has to do with cattle is leave them alone while they feed themselves, corral them and to drive them to market when the time is ripe.
(A) all one has to do with cattle is leave them alone while they feed themselves, corral them, and to
(B) all one has to do with cattle is to leave them alone to feed themselves, to corral them, and
(C) all one has to do with cattle is leave them alone while they feed themselves and then corral them and
(D) the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them, and
(E) the only thing that has to be done with cattle is to leave them alone while they feed themselves, to corral them, and
If you have any questions
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30 Mar 2009, 12:36
Underline is better than Bold
11. It is an oversimplified view of cattle raising to say that all one has to do with cattle is leave them alone while they feed themselves, corral them and to drive them to market when the time is ripe.
(A) all one has to do with cattle is leave them alone while they feed themselves, corral them, and to
(B) all one has to do with cattle is to leave them alone to feed themselves, to corral them, and
(C) all one has to do with cattle is leave them alone while they feed themselves and then corral them and
(D) the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them, and
(E) the only thing that has to be done with cattle is to leave them alone while they feed themselves, to corral them, and
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30 Mar 2009, 12:46
If there is no typo
then D
issue is parallelism(A B E OUT)
also X,Y and Z type of format(C OUT)
it cant be X and Y and Z
right side of underlined portion is
"drive them to market when the time is ripe."
(A) all one has to do with cattle is leave them alone while they feed themselves, corral them, and to
(B) all one has to do with cattle is to leave them alone to feed themselves, to corral them, and
(C) all one has to do with cattle is leave them alone while they feed themselves and then corral them and
(D) the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them, and
(E) the only thing that has to be done with cattle is to leave them alone while they feed themselves, to corral them, and
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30 Mar 2009, 12:49
This is an awesome Question
sc-oversimplified-view-70342.html
if I can quote SC Guru Bob here
"D has a couple of problems. It says "the only thing that has to be done..." Then it lists three verbs. "Only thing" implies one, not three. And verbs are not things. You can use an infinitive or a gerund as a noun, but not the base form of the verb."
My Question "Is C perfect with that extra AND" or "better among the choices"?
Kudo function is not working(as bb has confirmed)
otherwise +1 for you
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30 Mar 2009, 16:56
what about if choice D were..
D) all one has to do with cattle is leave them alone while they feed themselves, corral them, and drive them to market when the time is ripe.
would it still be correct ( and hence will be the best choice)
Just want to find out whether this is the only problem with D or there are other problems too...
"D has a couple of problems. It says "the only thing that has to be done..." Then it lists three verbs. "Only thing" implies one, not three. And verbs are not things. You can use an infinitive or a gerund as a noun, but not the base form of the verb."
Re: Sentence Correction .....2 [#permalink] 30 Mar 2009, 16:56
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# Challenge description
We've had a few challenges involving the Look-and-say sequence. Quick reminder:
• The sequence starts with 1,
• Subsequent terms of this sequence are generated by enumerating each group of repeating digits in the previous term,
So the first few terms are:
1 "one"
11 "one one" (we look at the previous term)
21 "two ones"
1211 "one two, one one"
111221 "one one, one two, two ones"
312211 "three ones, two twos, one one"
Now let's do the same thing, but use Roman Numerals instead. We start with I and follow the same rules (we apply the digit-counting rule to characters instead, so we read IVX as one one, one five, one ten instead of one four, one ten or some other way):
I "one"
II "one one"
III "two ones" = "II" + "I"
IIII "three ones" = "III" + "I"
IVI "four ones" = "IV" + "I"
IIIVII "one one, one five, one one"
IIIIIVIII "three ones, one five, two ones" = ("III" + "I") + ("I" + "V") + ("II" + "I")
Given a positive integer N, either:
• Output first N numerals of this sequence (any reasonable separator is fine, as well as ["I", "II", "III", ...]
• Output Nth term of this sequence (it may be 0-indexed).
Remember to make your code as short as possible, since this is a challenge!
EDIT: I believe that there is always one standard/preferred way of expressing integers as roman numerals, (like 95 -> XCV instead of VC). Couple of Roman numeral converters I found online corroborate my opinion. If in doubt, use an online converter, as listing all the possible edge-cases and specific rules of writing Roman numerals is not the point of this challenge.
EDIT2: @PeterTaylor and @GregMartin pointed out that only numbers less or equal to 5 appear in the sequence, so you don't have to worry about the ambiguity of Roman numerals (numbers 1 - 8 are I, II, III, IV, V, VI, VII, and VIII)
There isn't a unique Roman numeral expression for each integer. Which numbers might it be necessary to express, and which expressions of those numbers are valid? – Peter Taylor – 2016-08-30T15:13:47.410
What do you mean by "there isn't a unique Roman numeral expression for each integer"? Like 4 / IV / IIII ? Or 95 / XCV / VC ? There might not always be a unique way to express an integer, but I'm pretty sure there's always a preferred (standard) one - correct me if I'm wrong. – shooqie – 2016-08-30T15:17:52.813
1how far do we have to go with our roman numberals? – Maltysen – 2016-08-30T15:19:30.157
Yes, both of those cases. In the second case, I think it's very much a matter of opinion which is preferable. – Peter Taylor – 2016-08-30T15:21:59.777
@Maltysen I think the number of equal digits that can appear in succession is limited to 7. – Martin Ender – 2016-08-30T15:22:21.457
@shooqie I agree with you on this one, but you might just want to solidify the rules for roman numerals in OP. – Maltysen – 2016-08-30T15:26:17.573
@Maltysen: Like, this is something that bothers me the most about this site. Make a challenge a tiny bit more complicated and you get questions like what counts as X? or what about the edge-case #98752314? – shooqie – 2016-08-30T15:35:35.460
9@shooqie if these details weren't clarified, how would you compare answers? If there are certain edge cases left up to interpretation the actual scores become meaningless because they might make a bigger difference than any golfing tricks you could come up with. – Martin Ender – 2016-08-30T16:41:14.420
@MartinEnder: Well, I did my best at clearing up the confusion about the Roman numerals, but I feel like I shouldn't have to say how to write them - we've had challenges with Roman numerals before and there didn't seem to be any confusion whatsoever. – shooqie – 2016-08-30T17:00:28.490
I don't understand the output format - are we converting to roman numerals, grouping by character type, counting the length of each group and turning the whole thing into roman numerals again? – Blue – 2016-08-30T17:30:15.797
@muddyfish: Yeah, pretty much. – shooqie – 2016-08-30T17:32:30.953
I think one can mathematically prove that numbers greater than 8 never appear in the sequence starting at I. If so, then programs need only consider Is and Vs. – Greg Martin – 2016-08-30T18:46:46.700
@GregMartin, it's possible to prove the stronger statement that no run-length greater than 5 appears in the sequence whose first generation is I. – Peter Taylor – 2016-08-30T21:56:00.033
1k rep! Congrats :) – ThreeFx – 2016-08-30T23:51:18.757
@GregMartin: Well, that certainly makes things easier. – shooqie – 2016-08-31T07:57:21.007
Surprisingly roman numerals are in fact a decimal notation where a number is split into powers of 10. So 999 = 900+90+9 = CM + XC + IX = CMXCIX and not IM (This is the modern standard form. The romans themselves were not completely consistent). But as noted this is not needed here. – Ton Hospel – 2016-09-01T06:02:06.530
17
# Perl, 49 bytes
Includes +1 for -p
Run with the 0-based index on STDIN, e.g.
ecce.pl <<< 14
ecce.pl:
#!/usr/bin/perl -p
s,(.)\1*,$&/$1%182 .$1,eg for($_=/$/)x$;y;19;IV
Magic formulas are so magic.
Normally I would use ($_=//)x$' to make the loop control one byte shorter, but scoring on this site gives that a handicap of 2 so it ends up 1 byte longer. On older perls you can drop the space before for. Some versions of perl force you to add a final ; to close the transliteration. But what is given above is the code that works on my system.
## Explanation
Working backwards from solution to code:
The string transformations we need:
I -> II
II -> III
III -> IIII
IIII -> IVI
IIIII -> VI
V -> IV
VV -> IIV
Each replacement ends with the repeated character. I will get a sequence of the same characters using regex /(.)\1*/, so this can be done by appending $1. The part before the -> is in $&. With that I still need:
I -> I
II -> II
III -> III
IIII -> IV
IIIII -> V
V -> I
VV -> II
Write I as 1 and V as 9:
1 -> 1
11 -> 11
111 -> 111
1111 -> 19
11111 -> 9
9 -> 1
99 -> 11
By dividing the part before -> by the repeated digit this becomes:
1 -> 1
11 -> 11
111 -> 111
1111 -> 19
11111 -> 9
1 -> 1
11 -> 11
So now the original repeated V is not an exception anymore. So I want an expression that makes this happen:
1 -> 1
11 -> 11
111 -> 111
1111 -> 19
11111 -> 9
And this can be done by a simple modulo 182:
1 % 182 = 1
11 % 182 = 11
111 % 182 = 111
1111 % 182 = 19
11111 % 182 = 9
(this even gets IIIIII to VI right though it isn't needed here)
All that is left is initializing the working variable to 1 for index 0, repeat this transformation in a loop and at the end replace 1 by I and 9 by V
1, 9 and 182 is the only parameter combination for which this simple formula works.
2This is genius! :) – Lynn – 2016-09-02T11:15:59.837
10
# Mathematica, 11390 83 bytes
Thanks to Martin Ender for suggestions that reduced the length by over 25%!
Showing off the high-level commands in Mathematica.
Nest[Flatten[Characters@{RomanNumeral@#,#2}&@@@Reverse@@@Tally/@Split@#]&,{"I"},#]&
A pure function, taking an argument N and outputting the Nth element of this (0-indexed) sequence, as a list of characters. Spread out a bit:
Nest[
Flatten[
Characters @ {RomanNumeral@#,#2}& @@@
Reverse @@@ Tally /@ Split@ #
]& ,
{"I"}, #]&
The outer Nest iterates the middle four-line function, starting on {"I"}, N times. Line 4 splits the character list of the input Roman numeral into runs of like characters, counts each run with Tally, and puts the counts before the characters they're counting. Line 3 renders the counts as Roman numerals, then splits those Roman numerals up into lists of characters. The Flatten command reduces the whole list-of-lists to a one-dimensional list.
Here's the initial version:
Nest[
"" <> Flatten[{RomanNumeral@#[[1]], #[[2]]} & /@
(Reverse@#[[1]] & /@
Tally /@
Split@Characters@#)] &,
"I", #] &
3Grrr Mathematica ;) – Beta Decay – 2016-08-30T18:50:27.283
If you use @@@ instead of /@ you can use # and #2 instead of #[[1]] and #[[2]]. Also, lists of characters are acceptable string types, so you can work with those and avoid using Characters@. – Martin Ender – 2016-08-30T21:11:27.567
@MartinEnder Aha, I knew there must have been a @@@-like shortcut! As for lists of characters being acceptable string types (which I agree would shorten the code): is there a post on this site you can point me to that describes the community standard(s)? – Greg Martin – 2016-08-30T22:22:54.370
2http://meta.codegolf.stackexchange.com/a/2216/8478 – Martin Ender – 2016-08-30T22:24:09.907
1A few more savings: Characters threads automatically so you can use @, Reverse@#& is of course the same as plain Reverse, in which case you also don't need those parentheses. And prefix notation (in the case of Flatten) doesn't save anything if you need to add parentheses to make it work. Combining all of those: Nest[Flatten[Characters@{RomanNumeral@#,#2}&@@@Reverse@@@Tally/@Split@#]&,{"I"},#]& – Martin Ender – 2016-08-31T07:58:49.063
8
## CJam (33 30 bytes)
"I"{e{(4md1$^'I*\'V*@}%e_}ri* Online demo Key to the correctness of the implementation is the following theorem: ### If the first generation is I, no run length is ever greater than five Lemma: if the first generation is I, no string ever contains VVV. Proof is by contradiction. Suppose that there is a first index n for which the nth generation contains VVV. If that VVV breaks down as (a)V VV then the conversion from the previous generation is bad: it should have been (a+5)V. So it must be VV V(d), and the previous generation contained VVVVV, contradicting the choice of n. Now, suppose there is a first index m for which the mth generation contains ...IIIIII.... Note that there can be no digits other than I and V in the string, because no previous generation has had a run of nine Is or nine Vs. At most four of the Is come from a run of Is in the previous string, so the corresponding section of the previous string must be ...IIIVV... giving ... IIII IIV .... Since the VV in generation m-1 doesn't come from VVVVV (see lemma), the second V must be a run-length of digit I, so in generation m-1 we have ...IIIVVI.... And since we want the initial Is to give IIII and not IVI or VI, it is preceded either by the start of the string or by a V. If we have (...V)?IIIVVI... in generation m-1, what do we have in generation m-2? We've already observed that the VV of gen. m-1 must be parsed as (a)V V(I). Suppose we take a=2: (...V)?I IIV VI... Actually it must be ...VI IIV VI..., although that leading V might be part of IV; so in the previous generation we have either (...V)? IIII VV IIIII... or (...V)? IIIII VV IIIII. Either way we run into trouble with VVIIIII: the second V must be a run-length, but then ...VI IIII... requires a following (run-length, digit) pair with the same digit. So it must be a=1: (...V)?II IV VI.... Since generation m is the first with a run of six Is, that must be (...V)? II IV VI..., so that generation m-2 is (...V)? I V IIIII.... ...VIVIIIII... is impossible: however we choose to interpret the second V we end up with two consecutive (run-length, digit) pairs with the same digit. Therefore generation m-2 must be ^IVIIIII..., parsed as ^IV IIII I(V)... or ^IV III II(V).... These give respectively generation m-3 as ^V III V ... or ^V II VV.... But if we look at the start of the strings beginning with the first one that starts with V, we get a cycle: VI IV I... IV III IV ... II IV IVI ... IIII IV II IV ... and so no generation ever starts with either VIIIV or VIIVV. We must conclude that there is no such m. ### Dissection "I" e# Initial generation { e# Loop... e e# Run-length encode { e# Foreach [run-length char] pair... ( e# Extract the run-length r 4md1$^ e# Get the number of Vs and the number of Is
e# The number of Vs is r/4 ; the number of Is is (r%4)^(r/4)
'I*\'V*@ e# Repeat strings the appropriate number of times and reorder
}%
e_ e# Flatten to a simple string
}ri* e# ... n times, where n is taken from stdin
6
# Python 3, 195 bytes
There's a lot of bytes wasted on the roman numerals, so there's likely some golfing to be done there.
Thanks to @El'endiaStarman, @Sherlock9 and @Shooqie
import re
def f(x,r=""):
for v,i in(5,"V"),(4,"IV"),(1,"I"):a,x=divmod(x,v);r+=i*a
return r
s="I"
for i in[0]*int(input()):print(s);s=re.sub(r'(.)\1*',lambda m:f(len(m.group()))+m.group()[0],s)
Ideone it!
You can omit square brackets: for v,i in(5,"V"),(4,"IV"),(1,"I") – shooqie – 2016-08-30T17:40:34.977
@shooqie I had no idea that you could do that :D – Beta Decay – 2016-08-30T17:53:00.883
for v,i in(5,"V"),(4,"IV"),(1,"I"):a,x=divmod(x,v);r+=i*a saves a byte. – Sherlock9 – 2016-08-30T18:12:15.947
@βετѧΛєҫαγ: Also, you're not seem to be using i (as in for i in range(...)). I tried dabbling with exec but this escaped 1 in the 'sub' method seems to be messing up the code, I haven't been able to find a workaround. – shooqie – 2016-08-30T18:22:01.213
@shooqie I shortened it a bit by getting rid of range – Beta Decay – 2016-08-30T18:26:31.453
Hi, next time you might want to cross-out the previous byte-counts when improvements are made. I.e. Python 3, <s>314</s> <s>210</s> <s>202</s> <s>200</s> <s>199</s> 196 bytes. A.f.a.i.k. it's kinda standard to do so here on PPCG. – Kevin Cruijssen – 2016-09-01T09:23:41.783
@KevinCruijssen Nah, there isn't a standard on PPCG, and I prefer to keep it neater – Beta Decay – 2016-09-01T09:33:06.877
4
## R, 110 107 Bytes
as.roman combined with rle makes this easy. Scoping abuse and built in cat behavior of <<- saves a few bytes.
x="I"
replicate(scan(),{r=rle(strsplit(x,"")[[1]])
x<<-paste(rbind(paste(as.roman(r$l)),r$v),collapse="")})
Takes in N from console. Outputs first 2 to N terms of sequence (which I believe is within spec...)
[1] "II"
[2] "III"
[3] "IIII"
[4] "IVI"
[5] "IIIVII"
[6] "IIIIIVIII"
[7] "VIIVIIII"
[8] "IVIIIIVIVI"
[9] "IIIVIVIIVIIIVII"
[10] "IIIIIVIIIVIIIIVIIIIIVIII"
[11] "VIIVIIIIIVIVIIVVIIVIIII"
[12] "IVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[13] "IIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[14] "IIIIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[15] "VIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[16] "IVIIIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[17] "IIIVIVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[18] "IIIIIVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[19] "VIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[20] "IVIIIIVVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[21] "IIIVIVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[22] "IIIIIVIIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIIVIVIIVIIIIIVVIIVVIIVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[23] "VIIVIIIIIVVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVVIIVIIIVIIIIVVIIIVIIIIIVIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[24] "IVIIIIVVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIIVIIIIVIIIIIVIVIIIVIIIIIVVIIVIVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[25] "IIIVIVIIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIIVIVIIVVIIVIIIVIIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
1
# Perl 6, 62 bytes
{("I",{S:g/(.)$0*/{<I II III IV V>[$/.chars-1]~$0}/}...*)[$_]}
Anonymous function that accepts a zero-based index.
Makes use of the fact that roman numbers higher than 5 aren't needed, because the only groups of repeating digits that can occur, are:
I -> II
II -> III
III -> IIII
IIII -> IVI
IIIII -> VI
V -> IV
VV -> IIV
1
# JavaScript (ES6), 107
Recursive function returning the Nth term 0 based
f=(n,r='I')=>n?f(n-1,r.match(/I+|V+/g).map(x=>((n=x.length)-4?'VIII'.slice(n<5,1+n%5):'IV')+x[0]).join):r
Test
f=(n,r='I')=>n?f(n-1,r.match(/I+|V+/g).map(x=>((n=x.length)-4?'VIII'.slice(n<5,1+n%5):'IV')+x[0]).join):r
function update() {
O.textContent=f(I.value)
}
update()
<input id=I value=25 type=number oninput='update()'><pre id=O></pre>` | 5,399 | 16,577 | {"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} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.909952 |
https://personalstatementhelp.online/creative-writing/problem-solving-algorithm-pdf | 1,680,001,435,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00114.warc.gz | 521,994,644 | 7,843 | ## Genius Foods: Become Smarter, Happier, and More Productive While Protecting Your Brain for Life
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#### IMAGES
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3. 😂 Algorithmic problem solving. Algorithmic Problem Solving Free Pdf Download. 2019-03-06
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5. (PDF) Fundamentals of Algorithm
6. How To Problem-Solve With An Algorithm: Psychology And This Approach
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6. uHunt Introduction + Solving a problem in UVa [বাংলা]
1. Principles of Algorithmic Problem Solving
basic algorithm design, and some standard algorithms and data structures. They seldom include as much problem solving as this book does.
2. Problem Solving with Algorithms and Data Structures
To review the ideas of computer science, programming, and problem-solving. • To understand abstraction and the role it plays in the problem-
3. Chapter 7 Ch.7 Problem Solving and Algorithms
Example: Algorithm to add two arrays of integers. Both arrays have N integers. Take the basic operation to be the addition of two integers. Cost = N.
4. kecs104.pdf
Steps for Problem. Solving. » Algorithm. » Representation of. Algorithms. » Flow of Control. » Verifying Algorithms. » Comparison of. Algorithm. » Coding.
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up on your discrete mathematics, data structures, and problem solving skills. Baase [7] A good algorithms text at the upper-division undergraduate level.
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An algorithm must be general. This means that it must solve every instance of the problem. For example, a program that computes the area of a rectangle should
7. Algorithmic Problem Solving by Roland C. Backhouse
Algorithmic Problem Solving ; Calculus Practice Problems For Dummies. 626 Pages · 2014 ; 3,000 Solved Problems in Physics (Schaum's Solved Problems) (Schaum's
8. Algorithmic Problem Solving with Python
For example, the algorithm for calculating the greatest common ... learning and problem solving, but much of what we will do with Python
9. Algorithms analysis Algorithm • problem solving method suitable for
Pseudo-code is a structured description of an algorithm: not as formal as a programming language. Example: find the maximum element of an array. Algorithm Max(A
10. Problem Solving & Algorithm Design
Problem Solving. &. Algorithm Design. Page 2. 01-2. Problem solving ... An Example Algorithm. How to prepare Hollandaise sauce. Page 26. 01-26. | 648 | 2,771 | {"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} | 3.40625 | 3 | CC-MAIN-2023-14 | latest | en | 0.813535 |
teerstoday.in | 1,702,227,587,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00841.warc.gz | 624,895,008 | 16,096 | # How Do Teer Lottery Winners Get Paid?
With our article on how lottery winners get paid, we hope you can learn more about the lottery industry and its prize. In the lottery game called the teer, the winning prize is determined by the number of arrows drawn. Then, the winning ticket is released, and the prize amount is cash. The promotional value of the teer is derived from the fact that you can win big at the teer. But, first, we will look at how winners are paid in the teer. It is said that the game was first invented in the year 1770, while it was still under British rule; the game then became popular in the 19th century when it became a part of the British curriculum.
The game is also called Suksil Charn. The name “teer” comes from the word “teer,” which means “arrow.” All the game players use long arrows to send a target towards the center of the board. It is said that the game was first invented in the year 1770, while it was still under British rule; the game then became popular in the 19th century when it became a part of the British curriculum. It is also called Suksil Charn.
### How Does The Teer Lottery Work?
The teer lottery works like this. There are two arrows in a teer lottery. Win arrows indicate a win, while loss arrows indicate a loss. In the teer lottery, the winner has the most wins. The number of arrows released determines the winner of the teer lottery. Participants use arrows and bows to play.
Counting the number of arrows determines the winner after the arrows are released simultaneously. Whoever fires their arrows farthest away from the target is the winner. The first to shoot the target wins the match. Slingshots are used for playing the game. The person who is farthest away from the target wins. The player who hits the target the fastest wins.
### What Are The Rules Of The Shillong Teer Lottery?
The Shillong Teer Lottery has simple rules: the winner is determined by the number of arrows. How does a winner get paid? The rules of the Shillong Teer Lottery state that the winner gets INR. 7,000,000 cash and other accessories.
## How Are The Prizes Decided?
When you play a lottery game, you’re not only hoping to win the prize money – you’re also hoping to win the grand prize. You must shoot arrows at the bullseye to win the prize, but what if you miss it? What happens if you can’t shoot any of the arrows?
If you’re lucky enough to hit the bull’s eye, you win all the prizes, which are decided based on the number of arrows. In this case, if you score two arrows, you win the first prize and the second prize. If you score three arrows, you win the second and third prizes, and if you score four arrows, you win the first, second, and third prizes.
## How Are Winners Decided?
If you think about how winners are chosen for the lottery, remember that many factors go into the decision. When deciding the lottery winners, the lottery system will consider population size, income, and age. If the population is large and the majority is struggling financially, the lottery will choose lower on the socioeconomic scale. It is the same if most people are young.
### Conclusion
With our article on how lottery winners get paid, we hope you can learn more about the lottery industry and its prizes. There are many different types of prizes given out to winners, but the two main types are cash and prizes. If you have any further questions, please don’t hesitate to contact us. | 769 | 3,447 | {"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-2023-50 | latest | en | 0.981046 |
https://physics.stackexchange.com/questions/594626/heat-radiation-by-a-body-in-a-hotter-surrounding | 1,709,292,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00572.warc.gz | 442,569,054 | 38,464 | # Heat radiation by a body in a hotter surrounding
Lets say if I have a body at temperature T and the surroundings temperature is t (t>T). Does the body radiate heat to the surroundings or it just absorbs heat?
It does both. It radiates heat to the surroundings and the surroundings radiate to it. Each of these effects will be proportional to the temperature to the fourth power: $$P_{emitted}\propto T_{object}^4$$ and $$P_{received}\propto T_{environment}$$. As such, the net radiative losses or gains are proportional to $$T_{object}^4 - T_{environment}^4$$ | 137 | 563 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.896621 |
https://www.wyzant.com/resources/answers/3831/help_me_solve_8_divided_2_2_3_to_the_2nd_power | 1,506,432,552,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695726.80/warc/CC-MAIN-20170926122822-20170926142822-00536.warc.gz | 895,139,808 | 13,203 | 0 0
# help me solve 8 divided 2.2 - 3 to the 2nd power
Today i went to class and was given another sheet with 33 more problems on them, I recieved a 85% on my homework....these last seven questions has me stuck again can anyone out there help me by not just given me the answers but helping me explain the problem of how I got the answers
thank you
Janice
Problem 1. Simplify 8 divided by 2 times 2 minus 3 to the 2nd power
Problem 2. simplify -4 + 2 3rd power / 4 - 3. 2
Problem 3. simplify 5 - 3 . 4 to the 2nd power + 3(-4 - (-6)
Problem 4. if X = -3 then 4X to the second power - 3x - 10 =
Problem 5. A 8 foot board is cut into two pieces. One piece is 2 feet longer than the other. How long is the smaller piece?
Problem 6. Simplify (3x to the 2nd power y)( 2x to the 3rd power y )
Problem 7. simplify the expression 5a to the 3rd power - 2a to the 2nd power -( 4a to the 2nd power - 3 ) I came up with 5a to the 3rd power - 6a to the 2nd power - 3 for my answer am i right?
Problems 1, 2, and 3: I'm not sure I can rewrite these correctly, so I'll answer generally. It looks like each of these involve the order of operations, which is:
Parentheses, Exponents, Multiplication/Division (from left to right), Addition/Subtraction (from left to right).
So if problem 1 is 8 divided by 2 times 2 minus 3 to the 2nd power
this would look like 8 ÷ 2 * 2 - 32
By order of operations, we do the exponents (32) first, and we get
8 ÷ 2 * 2 - 9
Now we do multiplication and division as we encounter them from left to right. So first we tackle 8 ÷ 2 and rewrite our expression with the result
4 * 2 - 9
Now we have 4 * 2 (which equals 8), and we rewrite our expression again:
8 - 9
Now, there is nothing left but subtraction, and 8-9 = -1.
Given that explanation, see if you can apply this process to problems 2 and 3
You can remember the order of operations with the saying "Please Excuse My Dear Aunt Sally":
Excuse = Exponents
My = Multiplication
Dear = Division
Sally = Subtraction
Problem 4: If x = -3, then 4x2 - 3x - 10 = ?
Replace x in the expression with (-3), to get
4(-3)2 - 3(-3) - 10 = 4(9) + 9 - 10 = (I'll leave the final answer for you)
Problem 5: Since the board is cut into two pieces, let's let x = length of first piece and y = length of second piece. We then know that x + y = 8 feet
Since one piece is 2 feet longer than the other, let's assume that the first piece is longer than the second, so we now have a second equation, x = y + 2
Replace x in the first equation with y+2 (from the second equation), and we now have
(y+2) + y = 8
2y + 2 = 8
2y = 6
y = 3 . Since we assumed that y was the shorter board, this is our answer.
Problem 6: I'm reading this as (3x2y)(2x3y).
If so, we multiply common factors together:
(3*2)(x2x3)(yy) = 6 x2+3y1+1 = 6x5y2
Problem 7: If looks like the problem is 5a3 - 2a2 - (4a2 - 3). If so, then your answer is close, with one change:
The negative in front of (4a2 - 3) gets distributed across both terms within the parentheses. So if we cleared the parens, our expression would look like
5a3 - 2a2 + (-1)( 4a2) + (-1)(- 3) = 5a3 - 6a2 + 3
It would be easier if you wrote like this : (8/2)(2-(3^2))? | 1,046 | 3,197 | {"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} | 4.5625 | 5 | CC-MAIN-2017-39 | latest | en | 0.92727 |
https://www.physicsforums.com/threads/thermodynamics-question-calculating-change-in-temperature.795855/ | 1,695,383,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00598.warc.gz | 1,067,405,274 | 15,930 | # Thermodynamics Question Calculating change in temperature
• Gaith
## Homework Statement
[/B]
50 moles of an ideal gas for which Cp= 5/2R and Cv=3/2R initially has a pressure of 1x10^5 Pa, and a volume of 1.0m^3. It undergoes a process where the pressure and volume both double while they stay proporitional to each other: P=constV. What is the change in temperature for this process?
ΔU= Q-W
PV=nRΔT
ΔU= nCv(ΔT) [/B]
## The Attempt at a Solution
So I started by trying to use PV=nRΔT and calculating T, however I wasnt successful. The answer in my book is in Kelvins, so I tried converting the temperature that I calculated to Kelvins, however it was still wrong.
I'm thinking that I might need to take the integral: ∫ PVdv but I am not sure if that's correct either.
Can somone please help me out, It would be greatly appreciated.
The correct answer is: 720K
Last edited:
Hello.
PV=NRΔT
The above equation is not correct. The ideal gas law is PV = nRT, not PV = nRΔT.
From PV = nRT can you calculate the initial temperature? How about the final temperature?
From the ideal gas law, what is the initial temperature? From the ideal gas law, what is the final temperature?
Chet | 321 | 1,191 | {"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-2023-40 | latest | en | 0.945685 |
https://vbaconsultants.com/tag/conditional-formatting/ | 1,675,625,667,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00741.warc.gz | 629,448,601 | 14,004 | ## Automatically Shade Excel Rows
Before laser printers became common place (yes there was life before laser printers), people working with computer generated reports were used to green-bar paper. While the paper comes in various sizes, the most common is 14 7/8 x 11. The paper is in a carton of five thousand continuous sheets. The left and right sides of the paper have sprocket holes that align with the sprockets on an impact printer. The sprockets actually pull the paper past the print head as opposed to a laser jet’s method of feeding the paper in to the print mechanism. If you pay a little extra you can get the paper with perforations by the sprocket holes so that they can be cleanly removed from the report.
It was called green-bar paper because it had alternating sections of light green coloring imprinted on it. Depending on the point size of the report, each section might contain from three to five rows of print. These alternating colored sections allow the reader of the report to ‘line things up’ as the reader’s eyes travel across the report.
The rise of the laser jet heralded in the demise of the impact printers and with it green-bar paper. However, people still need to read landscape reports with compressed print that might squeeze the same amount of information onto a smaller page. If that is your situation, this post will allow you to mimic green-bar paper using two Excel functions along with the Conditional Formatting command.
ROW Function
The ROW function returns the row number of the current cursor location. As seen in the following screen shot, entering =ROW() in cell E2 returns 2, the row of the cursor when the formula was entered. This function by itself is not such a big deal. You could tell that the formula is in row 2 just by moving your eyes a few inches to the left. This function is the most useful when nested inside another function, or in some VBA code.
MOD Function
The MOD function returns the modulus of one number being divided by another – in plain English, the Remainder. If you enter the values and MOD function as shown in the next screen shot, the MOD function returns the value 1. Six goes into twenty-five four times with one left over.
A formula such as =MOD(ROW(),2) will return a zero whenever the row number is evenly divisible by two, i.e. an even number. Knowing this, you can shade every other row of a selected area using the Conditional Formatting command.
Select a section of the worksheet, or the entire worksheet, and then make the following choices from the Home tab of the Excel ribbon. You enter the formula in the New Format Rule dialog box and then click on the Format button to pick the fill color (i.e. shading). If you are printing on a black and white laser, very light gray will probably work best. If you are printing on a color laser, you can let your imagination run wild.
After clicking the OK button a couple of times, you should see a result similar to the following.
What if you want to change the shading to every third row, or every fifth row? You have to edit the rule for the selected cells using the following options from the same Conditional Formatting icon on the ribbon. The key is that you want to edit the rule, not delete the old rule and create a new rule. You also want to remember to change the formula to reflect how often you want the shading applied.
After clicking the OK, Apply, OK buttons, you should see a result similar to the following.
Although the screen shots in this post are from Excel 2010, the procedure will be similar in Excel 2003, 2007, and 2013.
Copyright 2014 – VBA Consultants Ltd
## Excel’s Conditional Formatting
Microsoft Excel’s Conditional Formatting allows you to create rules whereby the format of one or more cells changes based on the data in your worksheet. You would want to do this so that you can focus on the items that need review or correction. Examples could be employees with excess overtime, departments with high absenteeism or injuries, products with a high number of returns, or salesmen failing to make sales goals. Starting with the 2007 version Excel has not only made it easier to apply conditional formatting, but it also has made more formatting options available. | 886 | 4,240 | {"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-2023-06 | latest | en | 0.941327 |
https://codecov.io/gh/ewenharrison/finalfit/src/bbc50b1a43bc719dfab9a585c532f1c1ba13eea5/R/boot_predict.R | 1,628,218,980,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00638.warc.gz | 189,514,046 | 19,257 | ewenharrison / finalfit
1 ```#' Bootstrap simulation for model prediction ``` 2 ```#' ``` 3 ```#' Generate model predictions against a specified set of explanatory levels with ``` 4 ```#' bootstrapped confidence intervals. Add a comparison by difference or ratio of ``` 5 ```#' the first row of \code{newdata} with all subsequent rows. ``` 6 ```#' ``` 7 ```#' To use this, first generate \code{newdata} for specified levels of ``` 8 ```#' explanatory variables using \code{\link{finalfit_newdata}}. Pass model ``` 9 ```#' objects from \code{lm}, \code{glm}, \code{\link{lmmulti}}, and ``` 10 ```#' \code{\link{glmmulti}}. The comparison metrics are made on individual ``` 11 ```#' bootstrap samples distribution returned as a mean with confidence intervals. ``` 12 ```#' A p-value is generated on the proportion of values on the other side of the ``` 13 ```#' null from the mean, e.g. for a ratio greater than 1.0, p is the number of ``` 14 ```#' bootstrapped predictions under 1.0, multiplied by two so is two-sided. ``` 15 ```#' ``` 16 ```#' @param fit A model generated using \code{lm}, \code{glm}, ``` 17 ```#' \code{\link{lmmulti}}, and \code{\link{glmmulti}}. ``` 18 ```#' @param newdata Dataframe usually generated with ``` 19 ```#' \code{\link{finalfit_newdata}}. ``` 20 ```#' @param type the type of prediction required, see ``` 21 ```#' \code{\link[stats]{predict.glm}}. The default for glm models is on the ``` 22 ```#' scale of the response variable. Thus for a binomial model the default ``` 23 ```#' predictions are predicted probabilities. ``` 24 ```#' @param R Number of simulations. Note default R=100 is very low. ``` 25 ```#' @param estimate_name Name to be given to prediction variable y-hat. ``` 26 ```#' @param confint_sep String separating lower and upper confidence interval ``` 27 ```#' @param condense Logical. FALSE gives numeric values, usually for plotting. ``` 28 ```#' TRUE gives table for final output. ``` 29 ```#' @param boot_compare Include a comparison with the first row of \code{newdata} ``` 30 ```#' with all subsequent rows. See \code{\link{boot_compare}}. ``` 31 ```#' @param compare_name Name to be given to comparison metric. ``` 32 ```#' @param comparison Either "difference" or "ratio". ``` 33 ```#' @param ref_symbol Reference level symbol ``` 34 ```#' @param digits Rounding for estimate values and p-values, default c(2,3). ``` 35 ```#' @return A dataframe of predicted values and confidence intervals, with the ``` 36 ```#' option of including a comparison of difference between first row and all ``` 37 ```#' subsequent rows of \code{newdata}. ``` 38 ```#' ``` 39 ```#' @seealso \link{finalfit_newdata} ``` 40 ```#' ``` 41 ```#' /code{finalfit} predict functions ``` 42 ```#' @export ``` 43 ```#' @importFrom broom tidy ``` 44 ```#' @importFrom boot boot ``` 45 ```#' @importFrom stats predict ``` 46 ```#' ``` 47 ```#' @examples ``` 48 ```#' library(finalfit) ``` 49 ```#' library(dplyr) ``` 50 ```#' ``` 51 ```#' # Predict probability of death across combinations of factor levels ``` 52 ```#' explanatory = c("age.factor", "extent.factor", "perfor.factor") ``` 53 ```#' dependent = 'mort_5yr' ``` 54 ```#' ``` 55 ```#' # Generate combination of factor levels ``` 56 ```#' colon_s %>% ``` 57 ```#' finalfit_newdata(explanatory = explanatory, newdata = list( ``` 58 ```#' c("<40 years", "Submucosa", "No"), ``` 59 ```#' c("<40 years", "Submucosa", "Yes"), ``` 60 ```#' c("<40 years", "Adjacent structures", "No"), ``` 61 ```#' c("<40 years", "Adjacent structures", "Yes") ``` 62 ```#' )) -> newdata ``` 63 ```#' ``` 64 ```#' # Run simulation ``` 65 ```#' colon_s %>% ``` 66 ```#' glmmulti(dependent, explanatory) %>% ``` 67 ```#' boot_predict(newdata, estimate_name = "Predicted probability of death", ``` 68 ```#' compare_name = "Absolute risk difference", R=100, digits = c(2,3)) ``` 69 ```#' ``` 70 ```#' # Plotting ``` 71 ```#' explanatory = c("nodes", "extent.factor", "perfor.factor") ``` 72 ```#' colon_s %>% ``` 73 ```#' finalfit_newdata(explanatory = explanatory, rowwise = FALSE, newdata = list( ``` 74 ```#' rep(seq(0, 30), 4), ``` 75 ```#' c(rep("Muscle", 62), rep("Adjacent structures", 62)), ``` 76 ```#' c(rep("No", 31), rep("Yes", 31), rep("No", 31), rep("Yes", 31)) ``` 77 ```#' )) -> newdata ``` 78 ```#' ``` 79 ```#' colon_s %>% ``` 80 ```#' glmmulti(dependent, explanatory) %>% ``` 81 ```#' boot_predict(newdata, boot_compare = FALSE, R=100, condense=FALSE) -> plot ``` 82 ```#' ``` 83 ```#' library(ggplot2) ``` 84 ```#' theme_set(theme_bw()) ``` 85 ```#' plot %>% ``` 86 ```#' ggplot(aes(x = nodes, y = estimate, ymin = estimate_conf.low, ``` 87 ```#' ymax = estimate_conf.high, fill=extent.factor))+ ``` 88 ```#' geom_line(aes(colour = extent.factor))+ ``` 89 ```#' geom_ribbon(alpha=0.1)+ ``` 90 ```#' facet_grid(.~perfor.factor)+ ``` 91 ```#' xlab("Number of postive lymph nodes")+ ``` 92 ```#' ylab("Probability of death")+ ``` 93 ```#' labs(fill = "Extent of tumour", colour = "Extent of tumour")+ ``` 94 ```#' ggtitle("Probability of death by lymph node count") ``` 95 96 ```boot_predict = function (fit, newdata, type = "response", R = 100, ``` 97 ``` estimate_name = NULL, ``` 98 ``` confint_sep = " to ", condense=TRUE, boot_compare = TRUE, ``` 99 ``` compare_name = NULL, comparison = "difference", ref_symbol = "-", ``` 100 ``` digits = c(2, 3)){ ``` 101 1 ``` fit_class = attr(fit, "class")[1] ``` 102 103 ``` # Ensure lmlist | glmlist objects are length == 1 ``` 104 1 ``` if(fit_class %in% c("lmlist", "glmlist") & length(fit) > 1){ ``` 105 0 ``` stop("Multiple models in fit, must be single model")} ``` 106 107 ``` # Unlist lmlist & glmlist objects ``` 108 0 ``` if(fit_class %in% c("lmlist", "glmlist")) fit = fit[[1]] ``` 109 1 ``` fit_class = attr(fit, "class")[1] ``` 110 111 ``` # Stop if not lm | glm ``` 112 0 ``` if(!any(fit_class %in% c("lm", "glm"))) stop("fit must contain an lm or glm model") ``` 113 114 ``` # Stop if newdata not dataframe ``` 115 0 ``` if(!is.data.frame(newdata)) stop("Must provide dataframe with new data, see examples") ``` 116 117 1 ``` if(is.null(estimate_name)) estimate_name = "estimate" ``` 118 119 1 ``` formula = fit\$terms ``` 120 1 ``` family = fit\$family\$family ``` 121 1 ``` link = fit\$family\$link ``` 122 1 ``` if(is.null(family)) { ``` 123 0 ``` family = substitute(gaussian) ``` 124 ``` }else{ ``` 125 1 ``` family = match.fun(family) ``` 126 1 ``` family = substitute(family(link=link)) ``` 127 ``` } ``` 128 1 ``` .data = fit\$model ``` 129 130 ``` # Statistic function to bootstrap ``` 131 1 ``` statistic = function(formula, family, .data, indices) { ``` 132 1 ``` d = .data[indices, ] ``` 133 1 ``` fit_boot = glm(formula=formula, family = eval(family), data = d) ``` 134 1 ``` out = predict(fit_boot, newdata=newdata, type=type) ``` 135 1 ``` return(out) ``` 136 ``` } ``` 137 138 ``` # Run bootstrap ``` 139 1 ``` bs.out = boot::boot(data = .data, statistic = statistic, R = R, ``` 140 1 ``` formula = formula, family = family) ``` 141 142 1 ``` bs.tidy = broom::tidy(bs.out, conf.int = TRUE, conf.level = 0.95, conf.method = "perc") #Future options ``` 143 1 ``` bs.tidy = data.frame(bs.tidy) ``` 144 145 146 1 ``` if(!condense){ ``` 147 1 ``` df.out = bs.tidy[, c(1, 4,5)] ``` 148 1 ``` colnames(df.out) = c(estimate_name, paste0(estimate_name, "_conf.low"), paste0(estimate_name, "_conf.high")) ``` 149 ``` } else { ``` 150 1 ``` bs.tidy %>% ``` 151 1 ``` dplyr::mutate_all(round_tidy, digits = digits[1]) -> df.out ``` 152 1 ``` df.out = data.frame( ``` 153 1 ``` paste0(df.out\$statistic, " (", df.out\$conf.low, confint_sep, df.out\$conf.high, ")")) ``` 154 1 ``` colnames(df.out) = estimate_name ``` 155 ``` } ``` 156 157 158 1 ``` if(boot_compare){ ``` 159 1 ``` bc.out = boot_compare(bs.out, confint_sep = confint_sep, comparison = comparison, ``` 160 1 ``` condense = condense, ``` 161 1 ``` compare_name = compare_name, digits=digits, ``` 162 1 ``` ref_symbol = ref_symbol) ``` 163 1 ``` df.out = cbind(df.out, bc.out) ``` 164 ``` } ``` 165 166 167 ``` # Final table ``` 168 1 ``` if(condense){ ``` 169 1 ``` labels = extract_variable_label(newdata) ``` 170 1 ``` names(newdata) = labels ``` 171 ``` } ``` 172 1 ``` df.out = cbind(newdata, df.out) ``` 173 1 ``` return(df.out) ``` 174 ```} ```
Read our documentation on viewing source code . | 2,471 | 8,302 | {"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-2021-31 | latest | en | 0.691071 |
https://justaaa.com/statistics-and-probability/128721-situation-hypothesis-testing-tool-i-work-at-a | 1,709,302,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00770.warc.gz | 326,916,281 | 11,416 | Question
# SITUATION HYPOTHESIS TESTING TOOL: I work at a radio station that also has an app, where...
SITUATION HYPOTHESIS TESTING TOOL:
I work at a radio station that also has an app, where you can upgrade to a premium product. The premium app makes it possible to swipe away more songs you don’t like. The data will be relating to the age of the costumers buying premium packages. (The age is very important for our radio station, as we are targeting a very young audience).
Null Hypothesis & Alternate Hypothesis:
H0 : μ = 20
H1: μ ≠ 20
Type 1 error could be that we conclude that the age is greater than 20, when it is actually not.
Type 2 error could be that we conclude that the age is greater than 20, when it actually is.
Imagine the p-value is 0.01, I would conclude that there is substantial evidence against H0 and that it can be rejected. The result is very efficient and my assumption about the audience is not true.
Imagine the p-value is 0.20, I would conclude that H0 is ok but still not super sure. With the p-value of 0.01, the probability was only 1 % and therefore made it very unlikely but with a p-value of 0.20, there is 20% probability. The H0 is ok to be rejected by mistake with a 20 % probability. The result is not very efficient.
Feedback:
If young audience is important, your hypothesis should be \mu>=20 \mu<20. If the test result is to reject the null, then you know you do have a young audience. Your interpretations of type 1 and II errors are not consistent with your hypotheses. And they are opposite with the above hypotheses - how can I change it?
Questions
Reformulate your hypothesis test from above incorporate a 2-sample hypothesis test. What would be your data? What is your null hypothesis? What is your alternate hypothesis? What would be your Type 1 and Type 2 errors relative to your decision? Suppose you have a p-value of 0.01, what does this mean relative to your problem and decision? Suppose the p-value is 0.20, what does this mean relative to your problem and decision?
If you reformulated your design for 3 or more samples, what would be the implications of interaction? When would you use Tukey-Kramer test?
Solution :
Given that
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# Finance Question and Problem Sets
• From Business, Finance
• Due on 23 Jul, 2017 12:00:00
• Asked On 20 Jul, 2017 02:12:00
• Due date has already passed, but you can still post solutions.
Question posted by
Students should be able to calculate time value of money problems including solving for; present value, future value, rate and payment, determine the value and yield of corporate bonds, and use the dividend discount model to calculate the value and expected return of a common stock.
Assignment Steps NEED TO SHOW ALL WORK DETAIL
Resources: Tutorial help on Excel® and Word functions can be found on the Microsoft® Office website. There are also additional tutorials via the web that offer support for office products.
Complete the following Questions and Problems from each chapter as indicated.
Show all work and analysis.
Prepare in Microsoft® Excel® or Word.
• Ch. 5: Questions 3 & 4 (Question and Problems section): Microsoft® Excel® templates provided for Problems 3 and 4
3. Calculating Present Values [LO2] For each of the following, compute the present value:
SEE ATTACHMENT FOR TABLES
4. Calculating Interest Rates [LO3] Solve for the unknown interest rate in each of the following:
SEE ATTACHMENT FOR TABLES
• Ch. 6: Questions 2 & 20 (Questions and Problems section)
2. Present Value and Multiple Cash Flows [LO1] Investment X offers to pay you \$4,700 per year for eight years, whereas Investment Y offers to pay you \$6,700 per year for five years. Which of these cash flow streams has the higher present value if the discount rate is 5 percent? If the discount rate is 15 percent?
20. Calculating Loan Payments [LO2, 4] You want to buy a new sports coupe for \$79,500, and the finance office at the dealership has quoted you an APR of 5.8 percent for a 60-month loan to buy the car. What will your monthly payments be? What is the effective annual rate on this loan?
• Ch. 7: Questions 3 &11 (Questions and Problems section)
3. Valuing Bonds [LO2] Even though most corporate bonds in the United States make coupon payments semiannually, bonds issued elsewhere often have annual coupon payments. Suppose a German company issues a bond with a par value of €1,000, 23 years to maturity, and a coupon rate of 5.8 percent paid annually. If the yield to maturity is 4.7 percent, what is the current price of the bond?
11. Valuing Bonds [LO2] Union Local School District has a bond outstanding with a coupon rate of 3.7 percent paid semiannually and 16 years to maturity. The yield to maturity on this bond is 3.9 percent, and the bond has a par value of \$5,000. What is the price of the bond?
• Ch. 8: Questions 1 & 6 (Questions and Problems section): Microsoft® Excel® template provided for Problem 6
1. Stock Values [LO1] The Jackson–Timberlake Wardrobe Co. just paid a dividend of \$1.95 per share on its stock. The dividends are expected to grow at a constant rate of 4 percent per year indefinitely. If investors require a return of 10.5 percent on The Jackson–Timberlake Wardrobe Co. stock, what is the current price? What will the price be in three years? In 15 years?
6. Stock Valuation [LO1] Suppose you know that a company’s stock currently sells for \$63 per share and the required return on the stock is 10.5 percent. You also know that the total return on the stock is evenly divided between a capital gains yield and a dividend yield. If it’s the company’s policy to always maintain a constant growth rate in its dividends, what is the current dividend per share?
Format your assignment consistent with APA guidelines if submitting in Microsoft® Word.
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\$ 629.35 | 1,137 | 4,876 | {"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} | 3.703125 | 4 | CC-MAIN-2021-31 | latest | en | 0.894443 |
https://answerofmath.com/solved-generating-random-samples-with-bivariate-t-copula/ | 1,696,175,674,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00521.warc.gz | 102,350,428 | 19,954 | # Solved – Generating random samples with bivariate t-copula
I'm trying to generate a bivariate random sample of the t-copula (using rho = 0.8), without using the "copula" package and its function "rCopula" with method "tCopula". I'm using the following R-code:
``N <- 10000 R <- array(c(1,0.8,0.8,1), dim=c(2,2)) L <- t(chol(R)) Z <- rbind(rnorm(N),rnorm(N)) X <- L%*%Z df <- 2 W <- df/rchisq(N,df) Y <- sqrt(W)*X plot(Y[1,],Y[2,]) U <- pt(Y,df) plot(U[1,],U[2,]) ``
But the plot does not look like random points from a t-copula:
Does anyone know if I'm making a conceptual error or a mistake in the code?
It should look more like this: (generated using the copula package and its inbuilt functions)
Contents
``#a tcopula using rho = 0.8 #done from first principles require(mvtnorm) numObs <- 10000 #NX2 shaped matrix initialObservations <- rmvnorm(numObs,mean=rep(0,2)) #this is a 2X2 symmetric matrix based off of rho=0.8 and is positive definite psdRhoMatrix <- matrix(c(1,0.8,0.8,1),2,2) #the transpose of the cholesky decomposition of the psdRhoMatrix #which gives us a lower triangle matrix for some particular reason lowerTriangleCholesky <- chol(psdRhoMatrix) #this lower triangle matrix (for whatever reason) is able to make #the observations in each column correlated # NX2 = NX2 %*% 2X2 correlatedObservations <- initialObservations %*% lowerTriangleCholesky degreesOfFreedom <- 2 #the meaning of this step eludes me, it's a vector of random chi-square observations. #Maybe something to do with applying the inverse CDF. randomChiSqrStep <- degreesOfFreedom/rchisq(numObs,degreesOfFreedom) #transforming the correlated variables #the random chiSquaredStep is applied to each column of the correlatedObservations #for element wise multiplication to be properly applied the data needs to be #sorted into columns rather than rows. NX2 = NX1 * NX2 penultimateTransformation <- sqrt(randomChiSqrStep) * correlatedObservations plot(penultimateTransformation[,1],penultimateTransformation[,2]) #run the fully transformed and correlated observations through the t-dist PDF and that's it tCopulaOutPut <- pt(penultimateTransformation,degreesOfFreedom) plot(tcopulaOutPut[,1],tcopulaOutPut[,2]) `` | 624 | 2,216 | {"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} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.764666 |
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David1132
Joined: 12 Dec 2005
Posts: 51
Posted: Tue Jul 11, 2006 7:06 pm Post subject: Advice on messy integral?
I'm struggeling with a weird triple integral:
z / (1 + x^2 + y^2) dxdydz
over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2)
(Which is the volume that is inside the unit sphere, and also inside the
upside-down flipped cone with the pointy end at origo and it's base at
z=1/sqrt(2) )
I've tried to approach this in several ways, but every time i end up with
_really_ messy solutions, like integrating complex ln functions to hell when
I try cylindrical coordinates and even worse situations with spherical
coordinates.
How would I do to solve this problem in the best, nicest and simplest way?
I really hope anyone can help... I would be very grateful for advice.
/ David
Lynn Kurtz
science forum Guru
Joined: 02 May 2005
Posts: 603
Posted: Tue Jul 11, 2006 8:36 pm Post subject: Re: Advice on messy integral?
On Tue, 11 Jul 2006 21:06:20 +0200, "David" <dani4965@student.uu.se>
wrote:
Quote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David
The two surfaces intersect where their z values are equal:
2(x^2 + y^2) = 1, or a circle of radius 1/sqrt(2)
Just use cylindrical coordinates:
x^2 + y^2 = r^2, dx dy dz = r dr dtheta dz
where r and theta traverse the circle. You do get a ln in the answer
but there's nothing difficult about it.
--Lynn
Lynn Kurtz
science forum Guru
Joined: 02 May 2005
Posts: 603
Posted: Tue Jul 11, 2006 8:44 pm Post subject: Re: Advice on messy integral?
On Tue, 11 Jul 2006 20:36:32 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:
Quote: The two surfaces intersect where their z values are equal: 2(x^2 + y^2) = 1, or a circle of radius 1/sqrt(2) Just use cylindrical coordinates: x^2 + y^2 = r^2, dx dy dz = r dr dtheta dz where r and theta traverse the circle. You do get a ln in the answer but there's nothing difficult about it.
Woops, that answer may have been a bit hasty and optimistic....
--Lynn
matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
Posted: Tue Jul 11, 2006 8:47 pm Post subject: Re: Advice on messy integral?
David wrote:
Quote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David
As far as I can see, the whole of the cone you describe is inside the
unit sphere, so I don't see why the unit sphere is relevant. There
isn't a typo somewhere is there? Or maybe I'm misunderstanding
something...
David1132
Joined: 12 Dec 2005
Posts: 51
Posted: Tue Jul 11, 2006 9:00 pm Post subject: Re: Advice on messy integral?
Yes, the cone is inside the unit sphere, so the volume can be said to
consist of the cone plus the upper slice of the sphere; e.g. the cone has a
convex base (or is it concave? being from sweden, my english isn't too
good...)
<matt271829-news@yahoo.co.uk> skrev i meddelandet
Quote: David wrote: I'm struggeling with a weird triple integral: z / (1 + x^2 + y^2) dxdydz over the volume defined as x^2 + y^2 + z^2 <= 1 and z >= sqrt(x^2 + y^2) (Which is the volume that is inside the unit sphere, and also inside the upside-down flipped cone with the pointy end at origo and it's base at z=1/sqrt(2) ) I've tried to approach this in several ways, but every time i end up with _really_ messy solutions, like integrating complex ln functions to hell when I try cylindrical coordinates and even worse situations with spherical coordinates. How would I do to solve this problem in the best, nicest and simplest way? I really hope anyone can help... I would be very grateful for advice. / David As far as I can see, the whole of the cone you describe is inside the unit sphere, so I don't see why the unit sphere is relevant. There isn't a typo somewhere is there? Or maybe I'm misunderstanding something...
Lynn Kurtz
science forum Guru
Joined: 02 May 2005
Posts: 603
Posted: Tue Jul 11, 2006 9:08 pm Post subject: Re: Advice on messy integral?
On Tue, 11 Jul 2006 20:44:56 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:
Quote: Woops, that answer may have been a bit hasty and optimistic.... --Lynn
Here's your integral (Maple format) in spherical coordinates, but I'm
guessing you knew that:
2*Pi*int(rho^3*cos(p)*sin(p)/(1 + rho^2*sin(p)^2),[rho,p] =
Region(0..Pi/4,0..1));
Maple gives an explicit answer about 3 lines of output full of ln's
and sin(1) and cos(1) terms. For what it's worth, here it is but I
have no idea whether it will be readable even with monospaced font:
-1/32*Pi*(-Pi^2-64*ln(2)-Pi^2*cos(1)^2*ln(Pi^2-Pi^2*cos(1)^2+16)+Pi^2*
cos(1)^2*ln(1-cos(1))+Pi^2*cos(1)^2*ln(cos(1)+1)+Pi^2*ln(Pi^2-Pi^2*cos
(1)^2+16)-Pi^2*ln(1-cos(1))-Pi^2*ln(cos(1)+1)-4*Pi^2*ln(2)+16*ln(Pi^2-
Pi^2*cos(1)^2+16)+Pi^2*cos(1)^2+2*ln(sin(1))*Pi^2+4*Pi^2*cos(1)^2*ln(2
Not particularly simple, eh?
--Lynn
science forum beginner
Joined: 07 May 2006
Posts: 38
Posted: Tue Jul 11, 2006 9:23 pm Post subject: Re: Advice on messy integral?
On Tue, 11 Jul 2006 21:08:01 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:
Quote: On Tue, 11 Jul 2006 20:44:56 GMT, "[Mr.] Lynn Kurtz" kurtzDELETE-THIS@asu.edu> wrote: Woops, that answer may have been a bit hasty and optimistic.... --Lynn Here's your integral (Maple format) in spherical coordinates, but I'm guessing you knew that: 2*Pi*int(rho^3*cos(p)*sin(p)/(1 + rho^2*sin(p)^2),[rho,p] >Region(0..Pi/4,0..1)); Maple gives an explicit answer about 3 lines of output full of ln's and sin(1) and cos(1) terms. For what it's worth, here it is but I have no idea whether it will be readable even with monospaced font: -1/32*Pi*(-Pi^2-64*ln(2)-Pi^2*cos(1)^2*ln(Pi^2-Pi^2*cos(1)^2+16)+Pi^2* cos(1)^2*ln(1-cos(1))+Pi^2*cos(1)^2*ln(cos(1)+1)+Pi^2*ln(Pi^2-Pi^2*cos (1)^2+16)-Pi^2*ln(1-cos(1))-Pi^2*ln(cos(1)+1)-4*Pi^2*ln(2)+16*ln(Pi^2- Pi^2*cos(1)^2+16)+Pi^2*cos(1)^2+2*ln(sin(1))*Pi^2+4*Pi^2*cos(1)^2*ln(2 Not particularly simple, eh? --Lynn
Appears the OP multiposted this question to sci.math where its been
answered a couple of times now.
matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
Posted: Tue Jul 11, 2006 9:42 pm Post subject: Re: Advice on messy integral?
David wrote:
Quote: Yes, the cone is inside the unit sphere, so the volume can be said to consist of the cone plus the upper slice of the sphere; e.g. the cone has a convex base (or is it concave? being from sweden, my english isn't too good...)
Convex.
Right, I get it now. The confusion stemmed from the wording "the volume
that is inside the unit sphere, and also inside the upside-down flipped
cone", specifically the word "also" which to me implies the
intersection of the two figures. But no matter!
matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
Posted: Tue Jul 11, 2006 11:05 pm Post subject: Re: Advice on messy integral?
Quote: David wrote: Yes, the cone is inside the unit sphere, so the volume can be said to consist of the cone plus the upper slice of the sphere; e.g. the cone has a convex base (or is it concave? being from sweden, my english isn't too good...) Convex. Right, I get it now. The confusion stemmed from the wording "the volume that is inside the unit sphere, and also inside the upside-down flipped cone", specifically the word "also" which to me implies the intersection of the two figures. But no matter!
Well, I see it's been answered at sci.math, but now that I've done it
anyway I may as well tell you that using cylindrical coordinates I get
the same answer of 3/2*pi*Log(3/2) - pi/2.
Interestingly, the value of the integral over the cone (with plane
base) looks to be exactly the same as the value over the "cap" (the
region between the surface of the sphere and the plane base of the
cone), both being 3/4*pi*Log(3/2) - pi/4.
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Topic Author Forum Replies Last Post Similar Topics ? break one integral into parts Cheng Cosine Math 0 Fri Jul 21, 2006 8:54 am Help with weird integral? Sarah B. Math 2 Thu Jul 20, 2006 3:15 pm Curve integral - correct or not? Daniel Nierro Undergraduate 2 Thu Jul 20, 2006 2:47 pm sum((1+r)^(-t))=integral(exp(-r t)) vjp2.at@at.BioStrategist. Math 0 Mon Jul 17, 2006 6:29 am A contour integral - sin(x)/sqrt(x) Juryu Math 3 Sun Jul 16, 2006 9:09 pm | 3,298 | 10,453 | {"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} | 3.4375 | 3 | CC-MAIN-2019-18 | latest | en | 0.900689 |
http://stackoverflow.com/questions/13530115/error-iterating-pairs-in-sequences?answertab=active | 1,462,390,830,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860123845.65/warc/CC-MAIN-20160428161523-00217-ip-10-239-7-51.ec2.internal.warc.gz | 273,611,813 | 18,957 | # Error iterating pairs in sequences
After reading this question, I would expect the following to work:
``````Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose()
``````
but alas:
``````error: not enough arguments for method transpose: (implicit asTraversable:
Seq[Int] => scala.collection.GenTraversableOnce[B])Seq[Seq[B]].
Unspecified value parameter asTraversable.
Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose()
``````
Also, I can't seem to find any reference to transpose on the scala docs, although Seq refers it
Providing the identity, it seems to work somehow:
``````scala> Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose( a => a)
res10: Seq[Seq[Int]] = List(List(1, 4), List(2, 5), List(3, 6))
``````
But still returns List instead of Seq
-
Just use it without parentheses:
``````Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose
//res0: Seq[Seq[Int]] = List(List(1, 4), List(2, 5), List(3, 6))
``````
But still returns List instead of Seq
Well, actually List is inheritor of Seq, so after all you got a Seq (look at the left part of result).
The reason of such behaviour is that transpose actually defined as a function with argument, but since it's argument defined as implicit you have an option to delegate work of substituting argument to scala compiler (it will perform compile-time lookup for you).
If you writing parentheses, either function has to have overloaded form with no arguments, e.g.
``````def transpose() = ...
``````
or you have to write something inside them (it's actually matter of syntax).
-
Oh, I see! I still don't understand exactly how calling without braces works. Is scala providing the default argument in this case? Why doesn't it do so when using braces? – scala_newbie Nov 23 '12 at 13:40 | 480 | 1,718 | {"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-2016-18 | latest | en | 0.826155 |
http://thisthread.blogspot.com/2014/11/codility-stonewall.html | 1,553,608,613,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00256.warc.gz | 179,078,665 | 17,465 | ## Pages
### Codility StoneWall
We are given a sequence of integers representing the height of a wall that we have to build. We should return the minimum number of rectangular blocks needed to build it.
A good hint is that this problem has been included in the Stacks and Queues codility section. If you want more clues, you could have a look at the codility blog, where this problem is discussed and a solution is given.
Actually, I found the discussion in there more confusing than helping. Still I got a couple of elements to think on. I pondered on them, and in the end I came out with a sort of pseudo-code.
I use a stack to keep track of the current status of the wall, each element in it represent a block I am using to reach the current level.
Let's say that I am at the i-th step. The stack reflects the status of the previous step.
Maybe I don't have to do anything. The current level is the same. OK, just move to the next step.
Maybe I have to increase the wall height. That's easy, just put a new block of the right size on top the stack.
Maybe I have to decrease the wall height. That's a bit more complicate. I should remove enough blocks to get the right size, or something less than it. And then add a new block, if required.
And, well, that's it.
Then I wrote a few test cases, C++ for GoogleTest:
```TEST(StoneWall, Given) // 1
{
std::vector<int> input { 8, 8, 5, 7, 9, 8, 7, 4, 8 };
ASSERT_EQ(7, solution(input));
}
TEST(StoneWall, Simple) // 2
{
std::vector<int> input { 1, 2, 1 };
ASSERT_EQ(2, solution(input));
}
{
std::vector<int> input { 0 };
ASSERT_EQ(-1, solution(input));
}
```
1. Provided by codility.
2. A sort of minimal pyramid.
3. As usual, Codility does not care about bad input data. I decided to add a basic check to avoid non-positive input. A more robust solution would have been using assertions (if the data check would be a Somebody Else's Problem) or exceptions. Here I decided to keep it simpler and just return a minus one.
And here is my solution:
```int solution(std::vector<int>& input)
{
if(std::find_if_not(input.begin(), input.end(), [](int cur){ return cur > 0; }) != input.end()) // 1
return -1;
int blocks = 0;
std::stack<int> buffer;
std::for_each(input.begin(), input.end(), [&](int cur) // 2
{
while(!buffer.empty() && cur < buffer.top()) // 3
{
buffer.pop();
}
if(buffer.empty() || cur > buffer.top()) // 4
{
buffer.push(cur);
++blocks;
}
});
return blocks;
}
```
1. The STL find_if_not() is used to ensure all the element in input are positive. Paranoid check not required by codility.
2. Loop on all the elements. Notice that the lambda captures by reference the external variables so that it can modify them in its body.
3. If the previous wall height is taller than the current request, remove block(s) from the stack.
4. Add a new block to the stack to match the current request. Besides, increase the block counter. | 739 | 2,891 | {"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} | 3.1875 | 3 | CC-MAIN-2019-13 | latest | en | 0.900817 |
https://www.studypool.com/discuss/253252/math-is-difficult?free | 1,480,817,103,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541170.58/warc/CC-MAIN-20161202170901-00270-ip-10-31-129-80.ec2.internal.warc.gz | 1,025,308,381 | 14,012 | ##### math is difficult!!!!!!!!
Mathematics Tutor: None Selected Time limit: 1 Day
a building with a height of 14m casts a shadow that is 16m long while a taller building casts a 24m long shadow. What is the height of a taller building
Oct 12th, 2014
Height of a building is propotional to the lenght of a shadow.
That means that if the height of a taller building is x, then :
14 / 16 = x / 24.
Multiply both sides of equation on 24:
14*24/16 = x*24/24 = x
From here we use that 14 = 2*7, 24 = 3*8 and 16 = 2*8 :
x = 2*7*3*8 / 2*8 = 7*3 = 21.
Oct 12th, 2014
...
Oct 12th, 2014
...
Oct 12th, 2014
Dec 4th, 2016
check_circle | 223 | 637 | {"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} | 3.984375 | 4 | CC-MAIN-2016-50 | longest | en | 0.886194 |
https://drorbn.net/index.php?title=06-1350/Class_Notes_for_Tuesday_October_10 | 1,721,252,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00469.warc.gz | 187,161,982 | 7,762 | 06-1350/Class Notes for Tuesday October 10
Some Questions
Plastic trinions
Question 1. Can you embed a trinion (a.k.a. a sphere with three holes, a pair of pants, or a band theta graph) in ${\displaystyle {\mathbb {R} }^{3}}$ so that each boundary component would be unknotted yet each pair of boundary components would be knotted? How about, so that at least one pair of boundary components would be knotted?
Dror's Speculation. Yes and yes.
Question 2. A trinion ${\displaystyle \gamma }$ is embedded in ${\displaystyle {\mathbb {R} }^{3}}$ so that its boundary is the trivial 3-component link. Does it follow that ${\displaystyle \gamma }$ is trivial?
Dror's Speculation. No.
Question 3. Suppose two trinions ${\displaystyle \gamma _{1}}$ and ${\displaystyle \gamma _{2}}$ are knotted so that the pushforwards ${\displaystyle \gamma _{1\star }L}$ and ${\displaystyle \gamma _{2\star }L}$ are equal for any link ${\displaystyle L}$ which is "drawn" on the parameter space ${\displaystyle \Gamma }$ of ${\displaystyle \gamma _{1}}$ and ${\displaystyle \gamma _{2}}$. Does it follow that ${\displaystyle \gamma _{1}}$ and ${\displaystyle \gamma _{2}}$ are equivalent?
Dror's Speculation. I'm clueless.
The standardly embedded strapped trinion
Question 4. A trinion ${\displaystyle \gamma }$ is embedded in ${\displaystyle {\mathbb {R} }^{3}}$ so that its "strapped boundary" is equivalent to the strapped boundary of the trivially embedded trinion. Does it follow that ${\displaystyle \gamma }$ is trivial?
Dror's Speculation. If yes, it will have terrific consequences. If no, it will explain some of the misery we encounter when we deal with "associators". I would really like to understand this one.
Also see Some Questions About Trinions. | 471 | 1,754 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "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} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.892063 |
https://www.answers.com/Q/Do_all_quadralaterals_have_parallel_sides | 1,610,992,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00436.warc.gz | 664,862,425 | 29,812 | Geometry
# Do all quadralaterals have parallel sides?
###### Wiki User
no some shapes like the rhombus have 4 sides making it a quadrilateral but 2 of its sides are not parallel
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## Related Questions
it is a trapezoid. Quadralaterals can be groups according to how many pairs of parallel sides they have. A square and a rectangle have all 4 sides parallel since they are parallelograms and by definition a prallelogram has two sets of parallel sides. A kite has no parallel sides.
quadralaterals have 4 equal sides and angles
Quadrilaterals are shapes or figures that has four sides.
Any shapes that have 4 sides are quadrilaterals
A square's opposite sides are parallel and all the sides are congruent.
A square or a rhombus has two pairs of parallel sides and all sides are equivalent.
There are 2 pairs of parallel sides in a parallelogram. Each pair of opposite sides are parallel. Adjacent sides are not parallel (they intersect).
no, but all parallelograms are quadralaterals.
No, and all parrallelograms are not quadralaterals.
All of them have parallel sides except for the trapezoid.
Yes opposite sides are parallel in a rectangle
All sides of an octagon are parallel to one other side. There are 4 sets of parallel sides.
There are 3 pairs of parallel sides. All 6 sides are parallel to 1 other side.
All 4 sides are congruent and opposite sides are parallel to each other.
All regular polygons with an even number of sides. Irregular polygons with an odd number of sides can have parallel sides. There are also non polygonal shapes that can have parallel sides.
none, the three sides meet so there are no parallel sides but all the sides are equal.
No. A trapezoid only has one pair of parallel sides. A chevron has no parallel sides.
No it doesnt, they are all adjacent but not parallel.
A four-sided polygon with opposite sides parallel and all sides congruent is a square.
All sides of a trapezoid cannot be congruent, all sides of a rhombus are congruent. All opposite sides of a rhombus are parallel, only 1 pair of opposite sides of a trapezoid are parallel, the other pair are cannot be parallel.
Yes it does. All the four sides have an opposite. The opposites are parallel so in a parallelogram there are two pairs of parallel sides.
No. Circles have absolutely no sides at all.
Two parallel sides is part of the definition for a rhombus.
###### HomeschoolingSchool SubjectsMath and ArithmeticGeometryAlgebra
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https://www.beddingandbeyond.club/products/lego-avengers-theme-wallpaper-for-childrens-room-cartoon-mural | 1,563,584,919,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526401.41/warc/CC-MAIN-20190720004131-20190720030131-00294.warc.gz | 617,829,897 | 22,236 | ## Lego Avengers Theme Wallpaper for Children's Room Cartoon Mural
• \$39.95
IMPORTANT: PLEASE READ! \$39.95 is for 1 piece. only. It is very small, about 4'7" x 2'4". It will NOT cover a full wall. It's more like a large poster. To cover a complete wall you will need more pieces. The example pictures show a wall that is fully covered with wallpaper. You need to measure your wall and calculate how many pieces you need. If you need help, please contact us.
Product Details
• Function: Waterproof, Smoke-Proof, Moisture-Proof, Fireproof, Mold Proof, Soundproof, Sound-Absorbing, Heat Insulation, Anti-static
• Materials Available: 3D relief, Smooth non-woven, Straw, Silk
• Pattern: Lego Avengers
How to calculate how many pieces you need:
1). Measure your wall. Convert your measurements to centimeters (you can use any online metric converter). 2). If your wall is 360cm width x 240cm height for example, multiply those numbers: 360 x 240 = 8.6. So you would need to buy 9 pieces.
Or, if you prefer, you can choose one of our fixed sizes:
Quantity 1 : 1 square meter = 140cm(W) x 70cm(H) (4'7" x 2'4") \$39.95
Quantity 2 : 2 square meter = 200cm(W) x 100cm(H) (6'7" x 3'3") \$79.90
Quantity 3 : 3 square meter = 220cm(W) x 140cm(H) (7'3" x 4'7") \$119.85
Quantity 4 : 4 square meter = 250cm(W) x 160cm(H) (8'2" x 5'3") \$159.80
Quantity 5 : 5 square meter = 280cm(W) x 180cm(H) (9'2" x 5'11") \$199.75
Quantity 6 : 6 square meter = 300cm(W) x 200cm(H) (9'10" x 6'7") \$239.70
Quantity 7 : 7 square meter = 330cm(W) x 210cm(H) (10'10" x 6'11") \$279.65
Quantity 8 : 8 square meter = 360cm(W) x 230cm(H) (11'10" x 7'6") \$319.60
Quantity 9 : 9 square meter = 380cm(W) x 240cm(H) (12'5" x 7'10") \$359.55
Quantity 10 : 10 square meter = 400cm(W) x 250cm(H) (13'1" x 8'2") \$399.50
Quantity 11 : 11 square meter = 420cm(W) x 260cm(H) (13'9'' x 8'6'') \$439.45
Quantity 12 : 12 square meter = 440cm(W) x 270cm(H) (14'5" x 8'10") \$479.40
Quantity 13 : 13 square meter = 460cm(W) x 280cm(H) (15'1'' x 9'2'') \$519.35
Quantity 14 : 14 square meter = 480cm(W) x 290cm(H) (15'9'' x 9'6'') \$559.30
Quantity 15 : 15 square meter= 500cm(W) x 300cm(H) (16'5'' x 9'10'') \$599.25
Quantity 16 : 16 square meter=500cm(W) x 320cm(H) (16'5" x 10'6") \$639.20
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https://www.sawaal.com/profit-and-loss-questions-and-answers/if-the-cost-price-is-25-of-selling-price-then-what-is-the-profit-percent_6434 | 1,537,846,988,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00199.warc.gz | 854,570,286 | 16,165 | 39
Q:
# If the cost price is 25% of selling price. Then what is the profit percent.
A) 150% B) 200% C) 300% D) 350%
Explanation:
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = (75/25) * 100 = 300%
Q:
A guy walks into a store and steals Rs. 100 from the store without the owner's knowledge.
He then buys Rs. 60 worth of goods using that Rs. 100 and the owner gies Rs. 40 in change.
How much money did the owner lose?
A) RS. 200 B) Rs. 160 C) Rs. 100 D) Rs. 40
Explanation:
The first lost Rs. 100, but after the thief bought Rs. 60 goods, he get that Rs. 100 back but he lost Rs. 60 value of goods and Rs. 40 in change.
So, a total of 60 + 40 = 100 Rs. the owner lose.
0 34
Q:
A bottle is sold at a loss of 9%. Had it sold for Rs. 15 more, a profit of 25/2% would have gained. What is the cost price of the bottle (approx)?
A) Rs. 32 B) Rs. 35 C) Rs. 36 D) Rs. 33
Explanation:
Let the cost price of the bottle = Rs. 100
Selling price of the bottle = Rs. 100 - Rs. 9 = Rs. 91
Then, from the given data,
=> 112 (1/2) - 91 = 15
=> 21 (1/2) = 15
=> 43 = 15
Then 100 = ?
1500/43 = 34.885 =~ 35
Hence, the cost price of the bottle = Rs. 35
1 86
Q:
Selling price of a cellphone is increased by 20%. For this sale is decreased by 30%. Find the effect of total sales?
A) 16% B) 14.5% C) 12% D) 10.5%
Explanation:
Let cost of each cell phone = Rs.100 & Sale = 100 phones.
Money receipt = Rs.(100 x 100) = Rs.10,000
New cost per cellphone = Rs.120 and New sale = 70 phones
New Money Receipt = Rs.(70 x 120) = Rs.8400
Then the effect of sales = decrease in money receipt =
3 101
Q:
'X' sells fruits at 21% profit. If X bought it for 9% less and sold it for Rs.29 less, he would have gained 25%. The cost price of fruits is?
A) Rs. 400 B) Rs. 420 C) Rs. 460 D) Rs. 480
Explanation:
Let the cost price of the fruits be 'C.P'
From the given data, after analysis it can be solved as
(121% of C.P) - (125% of 91% of C.P) = 29
Hence, the cost price of the fruits = Rs. 400.
7 819
Q:
Anil purchases two books for Rs.100. He sells first at loss of 5% and second al 20% profit. If he gets Overall profit of 5% then what is cost of first book?
A) 58 B) 60 C) 55 D) 45
Explanation:
-5-----------------------5-------------------20
5-(-5) = 10
20-5= 15
Ratio of cost price of book1 and book2 = 3:2
Then cost price of book 1 is given by
(3/5) x 100 = Rs. 60.
14 1163
Q:
A shop keeper sells an article at a loss of 8%, but when he increases the selling price of the article by Rs. 164 he earns a profit of 2.25% on the cost price. If he sells the same article at Rs. 1760, what is his profit percentage?
A) 2.5% B) 5% C) 10% D) 7.5%
Explanation:
According to the given data,
Let Cost price of the article be 'cp'
Then,
102.25 cp - 92 cp = 164 x 100
10.25 cp = 16400
cp = 1600
Now, if he sells at Rs. 1760
Profit = 1760 - 1600 = 160
Profit% = 160/1600 x 100 = 10%.
6 868
Q:
An item is sold at a loss of 10%. Had it sold for Rs. 9 more, a profit of 25/2% would have gained. What is the cost price of the item?
A) Rs. 90 B) Rs. 75 C) Rs. 55 D) Rs. 40
Explanation:
Let the cost price of the item = Rs. 50
Sold at 10% loss => for Rs. 50 loss S.P = Rs. 45
From the given data,
25/2 % gain if it is sold at Rs. 9 more
56.25 - 45 = 9
=> 11.25 = 9
=> 22.5 = 18
=> 45 = 36
=> 50 = ?
=>
Hence, the Cost price of the item = Rs. 40.
8 2518
Q:
The marked price of an article is increased by 25% and the selling price is increased by 16.66%, then the amount of profit doubles. If the original marked price be Rs. 400 which is greater than the corresponding cost price by 33.33%, what is the increased selling price?
A) Rs. 380 B) Rs. 420 C) Rs. 460 D) Rs. 440
Explanation:
As given in the question, Marked price is 25% more than the Cost price.
=> C.P of the article =
Now,
Let the original S.P of the article be Rs. P
Now the new S.P = P +
=> S.P = $7P6$
According to the question,
=> 5P = 1800
=> P = Rs. 360
Hence, the increased S.P = 360 x 7/6 = Rs. 420. | 1,405 | 4,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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} | 4.15625 | 4 | CC-MAIN-2018-39 | longest | en | 0.930229 |
http://gpuzzles.com/mind-teasers/tough-age-riddle/ | 1,516,200,366,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00789.warc.gz | 146,351,235 | 11,097 | • Views : 30k+
• Sol Viewed : 10k+
# Mind Teasers : Tough Age Riddle
Difficulty Popularity
The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member.
Find how much younger is the new member ?
Discussion
Suggestions
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Guard Door Riddle
Difficulty Popularity
You are in a strange place which is guarded by two guards.One of the guard always say truth while other always lies.You don't know the identity of the two.You can ask only one question to go out from there.
What should you ask?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Famous Priest Well Puzzle
Difficulty Popularity
An old priest fell on a bottom of the well of a church.
The well is 20 inches deep.
At day time, the old priest climbs 5-inches but when at dark he slip back 4 inches.
In how many days priest will come out of the well ?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Awesome Clock Puzzle
Difficulty Popularity
Time 12:21 is a palindrome as it reads the same forwards or backwards.
Whats the Whats the shortest interval between two palindromic times ?
example => 11:11 and 12:21 has interval of 1 hr 10 minutes.
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : How Many Seconds In A Year Riddle
Difficulty Popularity
Without the use of calculation, can you find how many seconds are there in a year ?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Tricky Logical Puzzle
Difficulty Popularity
There are two arch enemy Dhuryodhan and Bheem.
One day two are going together and a god appeared in from of them.
The God grant three wishes to Bheem and one to Dhuryodhan.
Dhuryodhan replied smartly 'Give me twice of whatever Bheem ask for'.
For 1st wish Bheem asked the god 'Give me a room full of money'. Soon Dhuryodhan gets two room full of money.
For 2nd wish Bheem asked the god 'give me 100 slaves'. Soon Dhuryodhan gets 200 slaves.
Whats should be Bheem third wish ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Math Picture Brain Teaser
Difficulty Popularity
In the photo attached with this question, you have to place the numbers from 1 to 19 in the circles such that if we add up any side of the triangle, the sum is 17.
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Castle History Puzzle
Difficulty Popularity
A square island comprised of a square castle. A 14m wide trench surrounded the island from everywhere. Roman Empire wanted to invade the castle and gain the loot as well as possession of the island. They brought along wooden planks to cross the trench. However, they realized that the planks were just 13m long.
How did they use those planks to invade as well as capture the island ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : 3 Gallon Brain Teaser
Difficulty Popularity
You have been given three jars of 3 liters, 5 liters and 8 liters capacity out of which the 8 liters jar is filled completely with water. Now you have to use these three jars to divide the water into two parts of 4 liters each.
How can you do it making the least amount of transfers?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Difficult Brain Twister
Difficulty Popularity
A strange tradition is followed in an orthodox and undeveloped village. The chief of the village collects taxes from all the males of the village yearly. But it is the method of taking taxes that is interesting.
The taxes paid in the form of grains and every male should pay equal pounds corresponding to his age. In simpler terms, a man aged 10 years will have to pay 10 pounds of grains and a 20 years old will be paying 20 pounds of grain.
The chief stands on a riser containing 7 different weights next to a large 2 pan scale. As per the interesting custom, the chief can only weigh using three of the seven weights.
In such a scenario, can you calculate what must be the weights of the seven weights each and who is the oldest man the chief can measure using those weights?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : James Bond - Mini Mystery Story Riddle
Difficulty Popularity
James Bond was relaxing in his hotel room in Lyon when he heard a knock at his door. Bond opened the door and see a beautiful woman whom he had never seen in his life. She said that she is so sorry and she thought that this was her room. She was about to leave the room when the Bond takes out his pistol and ask the girl to stop.
What made Bond suspicious of the girl?
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https://nrich.maths.org/public/leg.php?code=-39&cl=1&cldcmpid=9523 | 1,527,225,713,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00099.warc.gz | 615,272,721 | 9,554 | # Search by Topic
Filter by: Content type:
Stage:
Challenge level:
### There are 111 results
Broad Topics > Decision Mathematics and Combinatorics > Combinations
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Triangle Animals
##### Stage: 1 Challenge Level:
How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### Three Ball Line Up
##### Stage: 1 Challenge Level:
Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land.
### The Twelve Pointed Star Game
##### Stage: 2 Challenge Level:
Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win?
### Cuisenaire Counting
##### Stage: 1 Challenge Level:
Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods?
### Finding Fifteen
##### Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Robot Monsters
##### Stage: 1 Challenge Level:
Use these head, body and leg pieces to make Robot Monsters which are different heights.
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### What Shape and Colour?
##### Stage: 1 Challenge Level:
Can you fill in the empty boxes in the grid with the right shape and colour?
##### Stage: 1 Challenge Level:
In Sam and Jill's garden there are two sorts of ladybirds with 7 spots or 4 spots. What numbers of total spots can you make?
### Sealed Solution
##### Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### A City of Towers
##### Stage: 1 Challenge Level:
In this town, houses are built with one room for each person. There are some families of seven people living in the town. In how many different ways can they build their houses?
### Mixed-up Socks
##### Stage: 1 Challenge Level:
Start with three pairs of socks. Now mix them up so that no mismatched pair is the same as another mismatched pair. Is there more than one way to do it?
### Two Dice
##### Stage: 1 Challenge Level:
Find all the numbers that can be made by adding the dots on two dice.
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Noah
##### Stage: 1 Challenge Level:
Noah saw 12 legs walk by into the Ark. How many creatures did he see?
### Let's Investigate Triangles
##### Stage: 1 Challenge Level:
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Plants
##### Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Those Tea Cups
##### Stage: 2 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Button-up
##### Stage: 1 Challenge Level:
My coat has three buttons. How many ways can you find to do up all the buttons?
### More and More Buckets
##### Stage: 2 Challenge Level:
In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
### We'll Bang the Drum
##### Stage: 1 Challenge Level:
How many different rhythms can you make by putting two drums on the wheel?
### Play a Merry Tune
##### Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
### Sounds Great!
##### Stage: 1 Challenge Level:
Investigate the different sounds you can make by putting the owls and donkeys on the wheel.
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### Four-triangle Arrangements
##### Stage: 1 Challenge Level:
How many different shapes can you make by putting four right- angled isosceles triangles together?
### Combining Cuisenaire
##### Stage: 2 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### Two and One
##### Stage: 1 Challenge Level:
Terry and Ali are playing a game with three balls. Is it fair that Terry wins when the middle ball is red?
### Nineteen Hexagons
##### Stage: 1 Challenge Level:
In this maze of hexagons, you start in the centre at 0. The next hexagon must be a multiple of 2 and the next a multiple of 5. What are the possible paths you could take?
### Butterfly Cards
##### Stage: 2 Challenge Level:
Four children were sharing a set of twenty-four butterfly cards. Are there any cards they all want? Are there any that none of them want?
### The Money Maze
##### Stage: 2 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
### Sorting Symmetries
##### Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Chocoholics
##### Stage: 2 Challenge Level:
George and Jim want to buy a chocolate bar. George needs 2p more and Jim need 50p more to buy it. How much is the chocolate bar?
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Here to There 1 2 3
##### Stage: 1 Challenge Level:
Move from the START to the FINISH by moving across or down to the next square. Can you find a route to make these totals?
### Triangle Edges
##### Stage: 1 Challenge Level:
How many triangles can you make using sticks that are 3cm, 4cm and 5cm long?
### Hamilton's Puzzle
##### Stage: 2 Challenge Level:
I start my journey in Rio de Janeiro and visit all the cities as Hamilton described, passing through Canberra before Madrid, and then returning to Rio. What route could I have taken?
### The Secret World of Codes and Code Breaking
##### Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
### Semaphore Signals
##### Stage: 2 Challenge Level:
Semaphore is a way to signal the alphabet using two flags. You might want to send a message that contains more than just letters. How many other symbols could you send using this code?
### Chocs, Mints, Jellies
##### Stage: 2 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it? | 2,165 | 9,381 | {"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} | 3.921875 | 4 | CC-MAIN-2018-22 | latest | en | 0.89932 |
https://codegolf.meta.stackexchange.com/questions/7623/does-randomization-count-as-encryption-or-hashing | 1,719,167,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00301.warc.gz | 157,066,973 | 31,610 | # Does randomization count as encryption or hashing?
Inspired from this question, I decided to submit an interweaved program (now deleted) written in Seed. This program uses a Mersenne twister to generate programs by entering a program that consists of two numbers. Using these two numbers, Seed generates a Befunge program. This is created by the randomization of the seed. After looking up what the definition was for cryptography and hashing I'm not really sure if it counts as hashing or cryptography. Seed only uses the randomization of the seed, but does not change anything else to the data received from the seed.
Although I only used Seed as an example, there certainly are more programming languages which use roughly the same algorithm. Do these programming languages fall under the category encryption or hashing?
Edit, for the ones interested in my post:
• Sometimes, even valid answers are downvoted if people feel that they abuse loopholes. Commented Nov 30, 2015 at 19:46
• @MartinBüttner, so this is considered a loophole? Commented Nov 30, 2015 at 19:49
• I'd consider it a loophole, because the submission didn't take any effort, has a high score and is practically impossible to crack. If any submission ticks all of those checkboxes it likely wasn't in the spirit of the challenge. Commented Nov 30, 2015 at 19:50
• @MartinBüttner, I disagree with the fact that this is practically impossible to crack, since there are maximally 16384 possible combinations. Also, why is having a high score against the spirit of the challenge? Commented Nov 30, 2015 at 20:22
• "If any submission ticks all of those checkboxes..." Also I count a lot more combinations than that. There are at least 163840 ways to distribute the digits before and after the spaces (minus multiplicities due to repeated characters), times the ways to distribute the four numbers in between without causing either of the other two numbers to be empty. Commented Nov 30, 2015 at 20:29
• @MartinBüttner, yeah sorry. I made a miscalculation, but this still gives a maximum of 262144 combinations (2^4 * 2^14), and it is highly likely to be less than this amount. This would still be easily cracked. Commented Nov 30, 2015 at 20:38
• Whether it's actually hard to crack or not, it doesn't make the answer any more interesting. As xnor's answer states, I think it's a cheap solution. In a completely different context Geobits said in chat yesterday "Here I judge [answers] more as useful to the site, as in, "do I want to see more of these here"." (when asked about when downvoting is appropriate). I definitely don't want to see the CnR tag derail into people posting quasi-arbitrary code that didn't take any effort and can only be cracked by brute force. Commented Dec 1, 2015 at 10:36
• And if the algorithm was chosen because of its flaws? (so that seeding can be reversed in small time (some seconds)) Commented Jul 28, 2017 at 20:44
It's more about the spirit of the rules.
Cops and robbers have had a problem of cop submissions that just do convoluted arithmetic. They are not solved by thinking about code, but by thinking numbers. This is boring because the structure of the code and language often matters little, and the robber is just trying to find the pre-image to some arithmetical function. With a convoluted enough operation, there's nothing better to do that brute-force search all possibilities.
This is what the rule is trying to avoid when it bans hashing or cryptography -- built ins that just give you a mangling function. Random number generators are much the same. Even if those were specifically banned though, there are many ways to home-brew something similar. We've seen lots of them, and feel boring and samey.
So, yes, the rules are vague and many people have "won" using the same idea. Even if your submission was legal, it may draw downvotes because it's cheap. It's the question's fault though that such cheap answers are a good path to winning by the rules as written.
Note that many top-voted cops give an interesting challenge that is meant to be cracked in a fun way, rather than trying to be uncrackable. Maybe it's best to think of cops as puzzles posed rather than code made to win an adversarial game.
## Encryption: no, hashing: maybe
Wikipedia on encryption:
In cryptography, encryption is the process of encoding messages or information in such a way that only authorized parties can read it.
Typically how this works is, you encrypt some data using a key, pass the encrypted data to the other party, and they decrypt it with a key (which may or may not be the same key, depending on if the encryption algorithm uses symmetric or asymmetric keys). The defining feature of encryption is, without the decryption key, a third party would not be able to retrieve the original data. Some algorithms are more secure than others, but that's not a relevant discussion.
Wikipedia on hash functions:
A hash function is any function that can be used to map data of arbitrary size to data of fixed size.
The key difference between hash functions and encryption is that encryption is designed to be reversible if you have the proper key. A hash function may or may not be reversible; a certain class of functions called cryptographic hash functions are designed to be irreversible, outside of brute-force attacks.
A Seed program is definitely not an encrypted form of a Befunge-98 program. There is no key that one could use to efficiently compute a Seed program from a Befunge-98 program.
As for hashing, possibly, depending on how you look at it. If you view the Seed program as being the input, and the Befunge-98 program as being the hash, the definition sort of works. Though it makes use of an RNG (specifically the Mersenne Twister algorithm), computing a Befunge-98 program from a Seed program is wholly deterministic, and can be done in an efficient manner. On the flipside, computing a Seed program from a Befunge-98 program (or even just a Seed program from a given expected output) can only be done by brute-forcing the seed space (unless the Mersenne Twister algorithm has some major flaws in its design).
As for the challenge in question, though submitting two interwoven Seed programs is not explicitly against the rules, some might interpret it as being against the spirit of the challenge, because it involves an RNG. That's barely relevant though, because Seed programs are wholly deterministic. Though the crack would likely come about through the use of brute force, I don't see a problem with that, as the brute forcing is not unbounded; there are a finite number of pairs of valid Seed programs that could be formed from your entry. I suspect the downvotes were from ignorance; people saw "RNG" in the language description and hastily downvoted without reading further and seeing that Seed is actually deterministic.
• One of the downvotes was mine and I can assure you I was aware that the answer was deterministic. Commented Nov 30, 2015 at 20:02
• In general, seeded RNGs are deterministic. That doesn't stop them being RNGs, or stop them being unfun in cops-and-robbers. Commented Nov 30, 2015 at 22:52
• Generating a program from a seeded RNG given the seed and length is conceptually no different than transpiling a program.
– user45941
Commented Nov 30, 2015 at 23:17
• @PeterTaylor, so what you're saying is that my submission is "unfun"? Why? Because the only way to crack it is by using a brute force algorithm? Commented Dec 1, 2015 at 0:39
• @Adnan, yes. One of the lessons which came out of the first ever cops-and-robbers question was that future cops-and-robbers questions should prohibit cops which just throw the hidden state through a trapdoor function. Commented Dec 1, 2015 at 10:39
• @PeterTaylor A Seed program is a trapdoor function, but the difficulty is in the Befune98-to-Seed direction. That's not relevant here.
– user45941
Commented Dec 1, 2015 at 11:05
• @Mego, if the cops weren't required to give the output of their programs, you might have a point. But as it is, the output of the two programs is derived from the output of the trapdoor function, so it's highly relevant. Commented Dec 1, 2015 at 11:57
• @PeterTaylor I believe (please correct me if I am mistaken) that you are assuming that the method of solving the cop answer would involve finding two Befunge98 programs that produce the given output and whose Seed representations can be interwoven to form the cop code. That is incorrect; there is no need to reverse the trapdoor function. Simply iterating over permutations of the interwoven code, splitting on the second space, and running each of the outputs until a suitable match is found is sufficient. No brute-forcing a seed from a program.
– user45941
Commented Dec 1, 2015 at 12:06
• I'm not assuming that at all. The fundamental problem is that because a trapdoor function is involved, it is impossible to say by "inspection" whether or not one possible solution is more likely than another: the only way to solve is brute force of the input space. That is against the spirit of cops-and-robbers, and in this case it's against the letter of the question as well. Commented Dec 1, 2015 at 12:58
To answer the question posed in your subject: Randomization is not hashing nor encryption. Hashing and Encrypting are both linear transformations. Randomization is not.
• Neither hashing nor encrypting are linear transformations. Commented Dec 2, 2015 at 22:03 | 2,173 | 9,461 | {"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-2024-26 | latest | en | 0.95285 |
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# A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
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Solution
## Let l, b, and h represent the length, breadth, and height of the tank respectively. Then, we have height (h) = 2 m Volume of the tank = 8m3 Volume of the tank = l × b × h ∴ 8 = l × b × 2 Now, area of the base = lb = 4 Area of the 4 walls (A) = 2h (l + b) However, the length cannot be negative. Therefore, we have l = 4. Thus, by second derivative test, the area is the minimum when l = 2. We have l = b = h = 2. ∴Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280 Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2) = Rs 8 (90) = Rs 720 Required total cost = Rs (280 + 720) = Rs 1000 Hence, the total cost of the tank will be Rs 1000.
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Location: [CPSC 333] [Ongoing Examples] Cartography System
This page was most recently modified on March 7, 1997
## Problem Statement
This system will be used to maintain information about a set of maps of (parts of) the world. Each map has a unique map number. The part of the world that is displayed by a map is given by the easternmost and westernmost longitudes, and the northernmost and southernmost latitudes, that are displayed.
The system also maintains a small amount of information about countries and cities of the world. In particular, the (unique) name, total population, and total area of each country is maintained, as is the name of its capital city. The name, latitude, and longitude of each city are maintained by the system. No two cities in the same country have the same name.
Finally, the system can be used to determine which maps show any given city, and to determine where a city is shown on a given map, if that city is shown on it. One can also determine which maps display (at least a part of) any given country.
## References in Assignments
This system will be considered in several assignments.
## Solutions for Assignment #1
For this example, a few things that wouldn't necessarily be considered as part of a requirements specification will be shown, because they form part of the solutions for Assignment #1.
### Entity-Relationship Diagram
This certainly isn't the only acceptable ERD that could be given as a solution for Assignment #1. Some groups may have noticed (or made the reasonable assumption that) a city's longitude and latitude could form a key for it. If they realized this, and used these as a key, then ``City'' would be shown as an entity instead of a weak entity, ``city name'' would become a ``non-key attribute,'' and it would be necessary to introduce one more relationship, ``is contained in,'' between City and Country (of type 1:Mc - each city would presumably be contained in exactly one country, while each country could have zero or more cities contained in it).
I've chosen not to use the city's latitude and longitude as its key, because it seems unlikely that a user of the system would have this information available, when that user wants to obtain more information about the city. It seems more likely that the user would know the name of the city and the country in which it lies, instead.
If a different (reasonable) ERD was given, then of course a different data dictionary, etc., would be correct. Later parts of the assignment submissions will be checked for consistency with the beginning of the submission (namely, the ERD), rather than agreement with what appears below this.
One solution for the rest of the assignment - one that agrees with the ERD shown above - will be given here.
### Attributes of Entities, etc.
#### Attributes of ``Map''
• map number
• easternmost longitude
• westernmost longitude
• northernmost latitude
• southernmost latitude
• number of vertical subdivisions
• first letter used
• number of horizontal subdivisions
• first number used
The ``map number'' would be the only attribute in the primary key for this entity.
#### Attributes of ``Country''
• country name
• population
• area
The ``country name'' would be the only attribute in the primary key for this entity.
#### Attributes of ``City''
• city name
• citys latitude
• citys longitude
A primary key for this weak entity would be obtained by including the attribute ``city name'' along with the primary key (``country name'') for ``Country.''
### Types of Relationships
The relationship ``shows'' has type Mc:Mc - each map can show zero or more cities on it, and each city can be shown on zero or more maps.
The relationship ``displays'' has type Mc:Mc - each map can display (a part of) zero or more countries, and (part of) each country can be shown on zero or more maps.
The relationship ``is capital of'' has type 1:1c - each city is the capital of at most one country, and each country has exactly one city as its capital.
Furthermore, the capital city of any country is one of the cities that is contained in it. This fact will be used to produce a data dictionary definition for the relationship ``is capital of'' that might be slightly different than expected.
### Data Management Subsystem
A complete (and detailed) list of the operations corresponding to one entity and one relationship were requested, along with a less detailed list of the names associated with the rest. In the solutions given below, the operations associated with the entity ``Map'' and the relationship ``shows'' will be described reasonably completely.
#### Operations Associated with Maps
Operations that might be associated with the entity ``Map'' include the following.
Inputs: All of the attributes of ``Map''
Effects: If the inputs are syntactically correct and the given ``map number'' isn't already in use, then a new map is created and added to the system's data base
Correct Map Information
Inputs: All of the attributes of ``Map''
Effects: If the inputs are syntactically correct and the given ``map number'' is in use, then the corresponding instance of map is modified, by using the rest of the inputs to reset the values for the instance's non-key attributes
This could be done differently: For example, you could have a separate operation used to change the value of each one of the non-key attributes, instead - with names like ``Correct easternmost longitude,'' ``Correct westernmost longitude,'' etc., and each taking two inputs - a map number, and a new value for the non-key attribute mentioned in the operation's name.
Delete Map
Inputs: map number
Effects: If the input is syntactically correct, corresponds to a map that's in the system, and there are no (parts of countries) listed as displayed on it, or cities listed as shown on it, then the corresponding map is deleted.
This could also be done differently: For example, a less cautious approach would be to have the the system go ahead and delete all the corresponding instances of the relationships ``shows'' and ``displays'' when a map is deleted, instead (as above) of refusing to perform the deletion when any of these instances exist. One way or the other, though, you should ensure that the data base isn't ``corrupted'' (by having instances of ``shows'' or ``displays'' that refer to deleted maps, left in the system).
Display Map Information
Inputs: map number
Effects: If the input is syntactically correct, and corresponds to a map that's in the system, then the values of all the non-key attributes of this map are displayed to the user.
#### Operations Associated with Other Entities and Weak Entities
Operations associated with the entity ``Country'' would probably include the following.
• Correct Country Information
• Delete Country
• Display Country Information
Creation of a ``Country'' would be a bit trickier than creation of a ``Map:'' you'd likely need to create at least one ``City'' that's located in the country (namely, its capital city), at the same time as you ``create'' the country (because of the ``type'' of the relationship, ``is capital of,'' that's listed above).
Deletion of a ``Country'' would be a bit tricker than deletion of a map, as well: While you might refuse to delete a country if any cities except for its capital presently exist, you'd likely need to delete the capital city at the same time as the country.
Operations associated with the entity ``City'' would probably include the following.
• Correct City Information
• Delete City
• Display City Information
Again, you'd probably need to do something like refuse to delete a ``capital city'' unless this is done at the same time as its country is deleted, in order to avoid corruption of the data base.
#### Operations Associated with ``shows''
Operations that might be associated with the relationship ``shows'' might include the following.
Enter Display of City
Inputs: map number, country name, city name
Effects: If the inputs are all syntactically correct, correspond to an existing map and city, and if it's possible for the city to be displayed on the map (because the country containing the city is shown on it, and the city's latitude falls between the map's northernmost and southernmost latitude, and the city's longitude falls between the map's easternmost and westernmost longitude), then the fact that the city is shown on the map should be entered into the system's data.
One might reasonably argue that it shouldn't be a responsibility of the ``data management subsystem'' to compare the city's latitude and longitude to the map's, etc.; if you feel that this is the case, then you'd leave part of the ``error checking'' (suggested above) out of the description of the operation.
Delete Display of City
Inputs: map number, country name, city name
Effects: If the inputs are syntactically and correspond to an existing city that's shown on an existing map, then the information that the city appears on the map is deleted from the system.
List Cities Shown
Inputs: map number
Effects: If the input is syntactically correct and corresponds to an existing map, then a list of the keys (country name and city name) for each of the cities shown on that map is produced.
List Maps Showing City
Inputs: country name, city name
Effects: If the inputs are syntactically correct and correspond to an existing city, then a list of the map numbers of maps showing that city is returned.
One more operation might feasibly be included,
Is City Shown on Map?
Inputs: country name, city name, map number
Effects: If the inputs are syntactically correct and correspond to an existing city and map, then a Yes/No answer to the above question is returned.
#### Operations Associated with Other Relationships
Operations corresponding to the relationship ``displays'' might include the following.
• Add Display of Country
• Delete Display of Country
• List Countries Displayed (on Map)
• List Maps Displaying (Given) Country
• Is Country Shown on Map?
Finally, the operations corresponding to the relationship ``is capital of'' might include the following.
• Change Capital City
• Report Capital City
Because of the ``type'' for the relationship that was given above, you can't add and delete capitals as freely as you can add and delete other things; instead, you can really only replace the city listed as capital for a country, by another city in the same country. See the discussion of addition and deletion of countries, given above, if you wish to think about this some more.
### Data Dictionary for Entity-Relationship Diagram
This uses ``auxiliary definitions'' to define some attributes with common data types (which was discussed via email and the newsgroup, during work on the first assignment). It also ``bends the rules,'' by treating latitudes and longitudes as single attributes, even though they neither have ``elementary'' data types or represent ``enumerated sets.''
Name: area
Kind: At
Type: integer
Description: Non-key attribute of Country (Note: It might be appropriate to list a unit of measure, such as ``square miles'' here, but no one asked about this, so one won't be specified.)
Name: City
Kind: WE
Type: @country name + @city name + citys latitude + citys longitude
Description: Represents a city that might be shown on one or more maps
Name: city name
Kind: At
Type: string
Description: Key attribute of City
Name: citys latitude
Kind: At
Type: latitude
Description: Non-key attribute of City
Name: citys longitude
Kind: At
Type: longitude
Description: Non-key attribute of City
Name: Country
Kind: E
Type: @country name + population + area
Description: Represents a country that might be displayed on one or more maps
Name: country name
Kind: At
Type: string
Description: Key attribute of Country
Name: displays
Kind: R - Mc:Mc
Type: @map number + @country name
Description: Relationship between Map and Country; each map can display (part of) zero or more countries, and (part of) each country can be displayed on zero or more maps
Name: easternmost longitude
Kind: At
Type: longitude
Description: Non-key attribute of Map
Name: first letter used
Kind: At
Type: string
Description: Non-key attribute of Map; represents the label of the easternmost vertical subdivision on the map
Name: first number used
Kind: At
Type: integer
Description: Non-key attribute of Map; represents the label of the northernmost horizontal subdivision on the map
Name: is capital of
Kind: R - 1:1c
Type: @country name + city name
Description: Relationship between City and Country; each city can be the capital of at most one country (and, this must be the country that the city is contained in), and each country has exactly one city as its capital
Name: latitude
Kind: H
Type: latitude distance + latitude direction
Direction: Used to define citys latitude, northernmost latitude, and southernmost latitude
Name: latitude direction
Kind: H
Type: [ `N' | `S' ]
Description: Used to define latitude
Name: latitude distance
Kind: H
Type: real
Description: Unit of measure is ``degrees;'' should be between 0 and 90, and should represent numbers to an accuracy of at least one digit after the decimal point. Used to define latitude
Name: longitude
Kind: H
Type: longitude distance + longitude direction
Description: Used to define citys longitude, easternmost longitude, and westernmost longitude
Name: longitude direction
Kind: H
Type: [ `E' | `W' ]
Description: Used to define longitude
Name: longitude distance
Kind: H
Type: real
Description: Unit of measure is ``degrees;'' should be between 0 and 180, and should represent numbers to an accuracy of at least one digit after the decimal point. Used to define longitude
Name: Map
Kind: E
Type: @map number + easternmost longitude + westernmost longitude + northernmost latitude + southernmost latitude + number of vertical subdivisions + first letter used + number of horizontal subdivisions + first number used
Description: Represents a map of part of the world
Name: map number
Kind: At
Type: integer
Description: Key attribute of Map
Name: northernmost latitude
Kind: At
Type: latitude
Description: Non-key attribute of Map
Name: number of horizontal subdivisions
Kind: At
Type: integer
Description: Non-key attribute of Map; should be greater than zero
Name: number of vertical subdivisions
Kind: At
Type: integer
Description: Non-key attribute of Map; should be greater than zero
Name: population
Kind: At
Type: integer
Description: Non-key attribute of Country
Name: shows
Kind: R - Mc:Mc
Type: @map number + @country name + @city name
Description: Relationship between Map and City; each map can show zero or more cities, and each city can be shown on zero or more maps
Name: southernmost latitude
Kind: At
Type: latitude
Description: Non-key attribute of Map
Name:westernmost longitude
Kind: At
Type: longitude
Description: Non-key attribute of Map
## Solutions for Assignment #2, Question #2
### Class Diagram
One possible class diagram for this system is as follows.
Several variations are possible.
In the above diagram, a ``whole-part structure'' including ``Country'' and ``City'' is used to model the requirement that each city be part of exactly one country (and each country can have zero or more cities in it), while an instance connection between the classes is used to model the requirement that each country has exactly one city as its capital, and each city is the capital of either zero or one country. An alternative diagram might use instance connections to model both requirements (replacing the whole-part structure by an instance connection); in this case, you'd clearly need to label both instance connections so that they could be distinguished (and, it's perfectly acceptable to label the one shown in the above diagram, with a name like, ``has capital,'' if you wanted to do that too).
You might choose to replace the instance connections between ``Map'' and ``Country'' and between ``Map'' and ``City'' with whole-part structures as well, but I think that the argument for using these here is less clear than it is for using a structure between ``Country'' and ``City.''
Finally, a class ``Clerk'' is included here because of an assumption made about the system's interface with the user - namely, that it will be useful to have a class responsible for prompting the user for commands, and then sending messages to other classes to have the commands executed. If there's no need for a class with this responsibility, then there's no reason to include a ``Clerk'' class here (since it will have no attributes in the version of the system that's being considered now).
### Attributes for Classes
``Clerk'' has no attributes. One could use the same set of attributes for the classes ``Map,'' ``Country,'' and ``City'' as were used for the corresponding entities.
Since ``normalization rules'' are less important for object-oriented analysis than for information modeling, one might also choose sets of attributes in the above list that are going to be used altogether, and replace each set with a single attribute (having the attributes in the corresponding set as ``components.''). In particular, one might group together the current attributes ``easternmost longitude,'' ``westernmost longitude,'' ``northernmost latitude,'' and ``southernmost latitude'' of ``Map'' and replace them with a single attribute, ``area displayed,'' for the class ``Map.'' Similarly, one might group together the attributes ``number of vertical subdivisions,'' ``first letter used,'' ``number of horizontal subdivisions,'' and ``first number'' used into an attribute, ``grid information,'' that happens to have four components (it seems likely that, for any given service, you'll either need all four of these values, or you won't need any of them at all, so that grouping them together makes sense). Finally, one could pair up the attributes ``city's latitude'' and ``city's longitude'' of ``City,'' to produce a single attribute called ``location.''
It isn't necessary or desirable to store either more, or less, information for this version of the system than it is for the one specified using ``function-oriented'' (that is, ``structured'') techniques, so these are the only changes in attributes that come to mind.
Here is a revised list of attributes that incorporates these changes.
#### Attributes of ``Map''
• map number
• area displayed
• grid information
• country name
• population
• area
• city's name
• location
### Message Threads and Services
Some assumptions must be made before this part of the question can be answered. There will be lots of other assumptions that one might plausibly make, as well - so there will be lots of different ``correct answers'' for this part of the question.
An assumption about the interface has been already been discussed, and was used to justify the inclusion of a class called ``Clerk.'' Due to this assumption, ``Clerk'' will receive the request to ``List Maps for a Given City.''
Another assumption deals with which class - ``Map'' or ``City'' - keeps track of the instance connection between them. This is relevant, since the information that must be returned is essentially a report of a particular subset of the connections that this ``instance connection'' represents. For this answer, I'll assume that ``Map'' keeps track of these. I'll also assume that ``Map'' is the class that will provide the ``service'' that the ``Clerk'' invokes, in order to satisfy the user's request. This makes the message threads for this operation a bit longer than they otherwise might be, but it would simplify the message threads needed for a request to display information about a given map.
Finally, we must make some assumptions about how an existing city will be specified, and what errors (or relevant decisions that the user can make) are possible. For this answer, I'll assume that the ``City'' class provides a service that can be used to ``Get an Existing City.'' The class will display a list of the cities known to the system in some way, so that syntax errors, selections of nonexistent cities, etc., aren't possible.
While no ``errors'' are possible, I'll assume that the user can choose to cancel the operation instead of selecting a city.
Under all these assumptions, only two ``threads of execution'' related to this operation seem possible.
#### First Thread: Successful Operation
1. The ``Clerk'' object is informed of a user's request to ``List all Maps for a Given City.''
2. The ``Clerk'' object sends a message, ``List Maps for City,'' to the ``Map'' class.
3. The ``Map'' class sends a message, ``Identify Existing City,'' to the ``City'' class.
4. After interaction with the user, the ``City'' class returns a city (probably, as a country name and a city name) to the ``Map'' class as its reply to the request it received.
5. After displaying a report to the user (possibly interacting with individual ``Map'' objects in order to obtain information to display), the ``Map'' class sends a ``finished'' reply back to the ``Clerk.''
#### Second Thread: User Cancels Operation
The first three steps here are identical to the first three given above. For the fourth step, a ``cancel'' signal is returned as a reply from ``City'' back to ``Map,'' since the user requested cancellation of the operation instead of selecting a city. As above, the ``Map'' class sends a ``finished'' reply to the ``Clerk'' class in the fifth step.
### Class Diagram with Message Connections
Only two message connections - from ``Clerk'' to ``Map,'' and from ``Map'' to ``City,'' are used above (recall that we don't introduce message connections to show transmissions of replies to requests). A class diagram that shows these is as follows.
Location: [CPSC 333] [Ongoing Examples] [Cartography System]
Department of Computer Science
University of Calgary
Office: (403) 220-5073
Fax: (403) 284-4707
eberly@cpsc.ucalgary.ca | 4,764 | 21,966 | {"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} | 3.140625 | 3 | CC-MAIN-2017-43 | latest | en | 0.950915 |
https://www.studypool.com/discuss/1003629/need-an-answer-for-this-can-t-seem-to-figure-it-out-1?free | 1,481,113,971,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542112.77/warc/CC-MAIN-20161202170902-00365-ip-10-31-129-80.ec2.internal.warc.gz | 1,002,956,620 | 14,239 | ##### Need an answer for this, can't seem to figure it out.
Mathematics Tutor: None Selected Time limit: 1 Day
Jackie wants to make a mixture of nuts to sell in her store consisting of hazelnuts and cashews. Hazelnuts cost \$6.50 per pound and cashews cost \$4.50 per pound. If Jackie wants 60 pounds total of mixture and the cost to be \$5.10 per pound, how many pounds of each will she need?
May 27th, 2015
x=the cashews at 4.5
y=the hazelnuts at 6.5
x+y=60
4.5x+6.5y / 60 = 5.1
y=60-x rearrange
4.5x+6.5(60-x) /60 = 5.1
then solve for x
4.5x+390-6.5x =306
-2x=-84
x=42
then we plug into the simple equation x+y=60 with x=42 and get y=18 then to verify we use our second equation to see if it equals
4.5(42)+6.5(18) /60 =5.1
189+117 / 60 =5.1
306/60=5.1
5.1=5.1
if you need any more help just ask
You need 42 poudns of cashew and 18 of hazelnuts
May 27th, 2015
...
May 27th, 2015
...
May 27th, 2015
Dec 7th, 2016
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Mark as Final Answer
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Unmark as Final Answer
check_circle | 386 | 1,027 | {"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-2016-50 | longest | en | 0.876666 |
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posted by .
what is the molality of a 10.5% by mass glucose solution?. density of the solution is 1.03g/ml
can someone please explain how to do this question. please and thank you
• chemistry -
10.5% by mass means 10.5 g glucose/100 g solution. That is
10.5 g glucose/(10.5 g glucose + 89.5 g H2O).
moles glucose = 10.5/180 = ??
molality = moles glucose/kg solvent =
0.0583/0.0895 = ??
• chemistry -
thank you!
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Q: More stars than sand particles? ( Answered , 11 Comments )
Question
Subject: More stars than sand particles? Category: Science > Astronomy Asked by: dimator-ga List Price: \$3.50 Posted: 02 Jul 2005 02:05 PDT Expires: 01 Aug 2005 02:05 PDT Question ID: 539329
```I remember hearing or reading that there are more stars in the universe than there are particles of sand on Earth. I was at the beach the other day, and there seemed to be a *LOT* of sand particles on the beach, not even considering the rest of the planet. Is such a statement accurate? How can such an estimate be made logically? I want estimates of how many sand particles there are in one cubic inch, foot, mile, ... planet Earth, and also, the latest, most scientifically valid estimates of how many stars are in the universe.```
Subject: Re: More stars than sand particles? Answered By: richard-ga on 03 Jul 2005 21:10 PDT Rated:
```Hello and thank you for your question. "So how many grains of sand are there in the world? You could start off by trying to guess how many grains of sand there are in a spoon of sand. Use a magnifying glass to count how many grains fit in a small section. Then, count how many of those sections fit in your spoon. Multiply the two numbers together to get an estimate. "Using this same principle, plus some additional information, mathematicians at the University of Hawaii tried to guess how many grains of sand are on the world's beaches. They came up with 7,500,000,000,000,000,000, or seven quintillion five quadrillion grains of sand." How many grains of sand are in the world? http://www.miamisci.org/tripod/whysand.html The calculation is detailed here: http://www.hawaii.edu/suremath/jsand.html That number is 7.5 x 10^18 or 7.5 billion billion. How many stars, galaxies, clusters, QSO's etc. in the Universe? http://www.faqs.org/faqs/astronomy/faq/part8/section-3.html "To get the total stellar population in the Milky Way [that is, in our galaxy alone], we must take the number of luminous stars that we can see at large distances and assume that we know how many fainter stars go along with them. Recent numbers give about 400,000,000,000 (400 billion) stars, but a 50% error either way is quite plausible." So in our galaxy alone, there might be between 2 x 10^11 and 6 x 10^11 stars How many galaxies in the Universe? http://www.faqs.org/faqs/astronomy/faq/part8/section-4.html "the Hubble telescope is capable of detecting about 80 billion galaxies (although not all of these within the foreseeable future!). In fact, there must be many more than this, even within the observable Universe, since the most common kind of galaxy in our own neighborhood is the faint dwarfs which are difficult enough to see nearby, much less at large cosmological distances. For example, in our own local group, there are 3 or 4 giant galaxies which would be detectable at a billion light-years or more (Andromeda, the Milky Way, the Pinwheel in Triangulum, and maybe the Large Magellanic Cloud). However, there are at least another 20 faint members, which would be difficult to find at 100 million light-years, much less the billions of light years to which the brightest galaxies can be seen." So the lower end estimate for the number of galaxies is 8 x 10^10 If we accept even the lower end of these Hubble figures, and if our Milky Way has a typical number of stars in it, that puts the number of stars in the universe to be at least (2 x 10^11) x (8 x 10^10) = 16 x 10^ 21 So if we round the number of sand grains to, say, 10^20 and round the number of stars to, say 10^22 then there are at least 100 stars in the universe for every grain of sand on earth. As you say, that's a *LOT* Search terms used beach sand particles cubic "number of stars in the universe Thanks again for your interesting question Richard-ga```
dimator-ga rated this answer: `What a great response, I'm glad I brought my question here!`
Subject: Re: More stars than sand particles? From: iang-ga on 03 Jul 2005 01:02 PDT
```Unfortunately, noone knows how how many stars or sand grains there are. If you search for grains, sand, Earth, stars & universe you'll find a lot of suggestions though. Most of them are good at explaining the assumptions they've made in counting grains of sand, but state the number of stars as a matter of fact. In reality there are a lot of assumptions made in counting stars, too - these links might be useful:- http://ganymede.nmsu.edu/astro/a110labs/labmanual/node14.html http://sciastro.astronomy.net/sci.astro.8.FAQ Ian G.```
Subject: Re: More stars than sand particles? From: sparkmencer-ga on 03 Jul 2005 14:02 PDT
```There are more stars than grains of sand on the earth. With sand, there is a definite number of particles. The cosmos are infinite...the further you look, the more will be found, forever.```
Subject: Re: More stars than sand particles? From: singularity360cubed-ga on 03 Jul 2005 22:39 PDT
```I would refer you to the Deep Field Image, that the Hubble came up with. You will soon find that the count of stars has to be of greater number, given the small patch that the Deep Field Image represents, once it is factored...! Steve```
Subject: Re: More stars than sand particles? From: pinkfreud-ga on 04 Jul 2005 11:43 PDT
```What a great answer, Richard! I'd have given it five stars. Or 500 grains of sand. :-D```
Subject: Re: More stars than sand particles? From: dprk007-ga on 04 Jul 2005 20:22 PDT
```sparkmencer-ga Modern theories of cosmology contend that our Universe is FINITE both in size, when it was born and total number of GALAXIES and STARS. The researcher has done a calculaton to indicate a lower limit on the number of stars. However there would be an actual total number of stars (i.e. which is not infinite) DPRK007```
Subject: Re: More stars than sand particles? From: pugwashjw-ga on 04 Jul 2005 20:47 PDT
```Genesis 1;16-18 " And God proceeded to make the two great luminaries, the greater luminary for dominating the day and the lesser luminary for dominating the night, AND ALSO THE STARS. 17. Thus God put them in the expanse of the heavens to shine upon the earth. 18. And to dominate by day and by night and to make a division between the light and the darkness. Then God saw that it was good". The scripture implies that Earth is a very special place. More important even than the stars in the universe. The words 'and also the stars' seem to relegate them to a lesser importance. Maybe we have to learn to look after this place before we are allowed out THERE.```
Subject: Re: More stars than sand particles? From: dprk007-ga on 07 Jul 2005 19:22 PDT
```pugwashjw-ga As I do not read the bible that much, I am a little bit unfamiliar with the biblical jargon. So perhaps you can confirm what you have said can be interpeted thus in pure layman's terms: 1. God Said Let there be a Planet earth and a planet earth appeared. 2. One or two days later God said let there be the sun and and a G type yellow dwarf appeared on the scene. 3. A couple of days later God said let there be stars and 10^22 stars appear (as per Richard-ga estimation) along with Planets , moons of planets, asteriods, pulsars, quasars, galaxies,brown dwarfs, planetary nebula and many other types of astronomical bodies which we may never have heard of. 4. And then god said let there be man and a man appeared on October 23rd 4004 BC (was a pretty hectic week all round!!) Regards DPRK007```
Subject: Re: More stars than sand particles? From: dprk007-ga on 07 Jul 2005 19:25 PDT
```oops regarding my point 4 I meant to say: And then god said let there be man and a man appeared on October 23rd 4004 BC at nine o'clock !!. DPRK007```
Subject: Re: More stars than sand particles? From: nemtudom-ga on 25 Jul 2005 04:25 PDT
```7,500,000,000,000,000,000 is seven quintillion five *hundred* quadrillion. It's interesting to note that ancient people, without telescopes, somehow had the idea that the number of stars was so great as to be in the same league with the grains of sand, e.g., "I will multiply thy seed as the stars of the heaven, and as the sand which is upon the sea shore" (Genesis 22:17). It's interesting to note that most statements beginning "it's interesting to note" are not interesting to note.```
Subject: Re: More stars than sand particles? From: noseallinit-ga on 10 Jan 2006 21:35 PST
```http://www.newton.dep.anl.gov/askasci/ast99/ast99215.htm Ask A Scientist© Astronomy Archive Stars or Sand Question: Are there more stars or are there more grains of sand? m french Answer: Well, if every star came with a planet like earth that had billions of sand grains, there would have to be more sand grains than stars! The number of stars in our galaxy is around 100 billion or 10^11 in exponential notation. The number of sand grains on earth is probably somewhere between 10^20 and 10^24. Th e number of sand grains on earth is therefore much greater than the number of stars in our galaxy. However, our galaxy is only one of about 100 billion in the visible universe, and so the total number of stars we know about is around 10^22, which is kind of in the same ballpark as the number of grains of sand on earth. Of course, those numbers are much bigger than we can count - I'm not sure whether better estimates of both numbers would give more stars, or more sand grains - there's a lot of both! Asmith```
Subject: Re: More stars than sand particles? From: richard-ga on 07 Jun 2006 07:31 PDT
```According to this recent article, http://www.newscientistspace.com/article/dn9282-andromeda-galaxy-hosts-a-trillion-stars.html the Andromeda galaxy has a trillion stars, that is, 10^12 So if that's the size of a typical galaxy, there are closer to 10^23 stars in the universe, or about 1,000 stars for each grain of sand. By the way, if you recall Avogadro's number from chemistry (6 x 10^23), the number of stars in the universe neatly matches the number of molecules in 22.4 liters (6 gallons) of uncompressed air or other gas at standard temperature (0 celsius or 32 fahrenheit) and pressure (1 atmosphere). -R``` | 2,570 | 10,098 | {"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} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.919023 |
https://www.physicsforums.com/threads/frictionless-pulley-with-two-weights.331893/ | 1,566,262,984,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00045.warc.gz | 934,283,579 | 20,097 | # Frictionless Pulley with two weights
#### Jotun.uu
Firstly, nice forum!
Secondly, I am Swedish but I will try to use the correct scientific terms in explaining the problem and my proposed solution.
1. Homework Statement
There is a string of negligible mass which "hangs" over a pulley attached to a frictionless horizontal axis. On each end of the string there is a attached a weight. One of the weights (i've chosen the one on the right hand side if drawn the "normal way") is twice the mass of the weight of the other.
The system starts out at rest.
[pulley] = Radius R, and Moment of Inertia I = (1/5)*m*R².
[weights] = A = m and B = 2m.
[string tension] = left side (A): T ; right side (B): S.
[gravity] = As always g denoted the pull of the earth (9.82 m/s² here in Sweden).
When the weights are released from rest (and left to fall freely), what will be the tension on the string on the two sides of the pulley?
2. Homework Equations
In my attempted solution below I used:
Newtons II equation where the sum of the forces are equal to the mass times the acceleration (F = m*a).
I also used Eulers II law for momentum (torque) which in this case would be that the momentum is equal to the moment of inertia times the angular acceleration (M = I*alpha)
I also used that alpha = R * a (along the x-axis (which I set to be positive upward))
3. The Attempt at a Solution
So, as listed above I will call the momentum about the z-axis of the pulley M, the angular acceleration alpha. I called the "left" pulley A and the "right" pulley B.
I set the x-axis to the positive in the upward direction. And I assumed that the string was not elastic in any way.
First I used newtons second law to describe the forces acting on pulley A to be
F = T - mg = m * a
and respectively for B:
F = S - 2mg = m * a
Then I used Eulers II with the moments origo in the center of the pulley to be
M = RT - RS = I * alpha (and since alpha = R * a) = T-S = I * a.
I then simply solved the equations using both NII and EII to get my expressions for what the tensions on S and T are.
I got that T =m(a((R²/5)-1)+2g)
and S = m(a(1-(R²/5))+g)
It should be added I am at the moment of typing a little tired. Also I should probably have made a picture, but I hope my explanation of the problem will be enough.
Usually I would be satisfied with my answer but the "hint" to the question was that I could use that acceleration (dv/dt) could be written as v*dv/ds (using the chain rule)! Why and how could that be used? I am now worried that I have made some very basic assumption or something wrong. Anyways help much appreciated!
Related Introductory Physics Homework Help News on Phys.org
#### Doc Al
Mentor
First I used newtons second law to describe the forces acting on pulley A to be
F = T - mg = m * a
OK.
and respectively for B:
F = S - 2mg = m * a
Two errors on the right hand side:
(1) You have m, but should have 2m.
(2) Careful with signs: If the acceleration of A is +a, what must be the acceleration of B?
Then I used Eulers II with the moments origo in the center of the pulley to be
M = RT - RS = I * alpha (and since alpha = R * a) = T-S = I * a.
Two problems:
(1) Careful with signs. S > T, so use RS - RT = I*alpha.
(2) alpha = a/R, not a*R.
Usually I would be satisfied with my answer but the "hint" to the question was that I could use that acceleration (dv/dt) could be written as v*dv/ds (using the chain rule)! Why and how could that be used?
That hint doesn't seem to apply to this problem. No calculus is needed here.
#### Jotun.uu
Doc Al said:
OK.
Two errors on the right hand side:
(1) You have m, but should have 2m.
(2) Careful with signs: If the acceleration of A is +a, what must be the acceleration of B?
then the acceleration of B must of course be negative! thanks!
So the new NII for B is: S - 2mg = -2ma
Doc Al said:
Two problems:
(1) Careful with signs. S > T, so use RS - RT = I*alpha.
(2) alpha = a/R, not a*R.
Gah! stupid mistake! I misinterpreted some old notes of mine. Of course alpha = a/R!
I am heading out to a wedding now (my girlfriends sister is getting married) I will be sure to post the completed solution when I get back tomorrow! Thank you so much for the help!
#### ideasrule
Homework Helper
I am heading out to a wedding now (my girlfriends sister is getting married) I will be sure to post the completed solution when I get back tomorrow! Thank you so much for the help!
Best of luck for your gf's sister and her husband!
#### Jotun.uu
[pulley] = Radius R, and moment of inertia I = (1/5)mR².
[weights] = A = m and B = 2m.
[string tension] = left side (A): T ; right side (B): S.
[gravity] = As always g denotes the pull of the earth (9.82 m/s² here in Sweden).
The (all new!) attempt at a solution
So, as listed above I will call the momentum about the z-axis of the pulley M, the angular acceleration alpha. I called the "left" pulley A and the "right" pulley B.
I set the x-axis to the positive in the upward direction. And I assumed that the string was not elastic in any way. Also I have come to realise that the main reason to use Euler II in this question is to solve so we get an expression that doesn't assume that the acceleration is known, so that the tension of the strings will only depend on the weight (m) (and g) of the pulley and weights on the string.
First I used Newtons second law to describe the forces acting on pulley A to be:
F = T - mg = ma
and respectively for B:
F = S - 2mg = -2ma
Then I used Eulers II law with the moments origo in the center of the pulley (my y-axis is positive going upwards, and the pulley is rotating clockwise!) to be:
M = RT - RS = I * alpha (and since alpha = a/R) = R(T - S) = mR²a/(5R) = T-S=ma/5.
I now solved the NII equations for T and S which made them:
T = ma + mg and S = 2mg - 2ma
In combination with EII:
ma+mg - 2mg+2ma = ma/5
Solving for a:
15a-5g = a -> 14a = 5g -> a = 5g/14
Thus I inserted the acceleration a into my equations for T and S:
T = 5mg/14 + 14mg/14 = 19mg/14
S = 28mg/14 - 10mg/14 = 9mg/7
Could this be the correct answer?
#### Doc Al
Mentor
First I used Newtons second law to describe the forces acting on pulley A to be:
F = T - mg = ma
and respectively for B:
F = S - 2mg = -2ma
OK.
Then I used Eulers II law with the moments origo in the center of the pulley (my y-axis is positive going upwards, and the pulley is rotating clockwise!) to be:
M = RT - RS = I * alpha (and since alpha = a/R) = R(T - S) = mR²a/(5R) = T-S=ma/5.
You still have a sign problem; see my previous post.
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• Solo and co-op problem solving | 1,908 | 6,848 | {"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-2019-35 | longest | en | 0.915603 |
https://inmobiliariaweb.online/73-in-to-feet/ | 1,696,221,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00392.warc.gz | 343,901,729 | 19,871 | # 73 In To Feet
73 In To Feet. To convert from meters to feet and inches, multiply the. So, the calculation leaves us with 5 whole feet plus the extra 11. Web feet to inches how to convert inches to feet. In) is a unit of length in the imperial and us customary systems of measurement. Web convert 73.81 millimeters to feet (73.81 mm to ft) with our length converter. How many feet in 73.81 mm. Web what is 71 inches in feet?
73 in to ft conversion. Web a scaffolding collapse from 70 feet at a construction site killed 3 people in charlotte on monday morning. Web 73 inches is equivalent to 6.08333333333333 feet. To convert 1.73 inches to feet you have to multiply 1.73 by 0.083333333333333, since 1 inch is 0.083333333333333 feet. 1 in ≈ 0.083333333 ft. We decided to round some conversion factors to fit this table. 1.73 m equals 5 feet and 8.1 inches.
## Web convert 73.81 millimeters to feet (73.81 mm to ft) with our length converter.
The foot is a unit of length used in the imperial and u.s. Man, 73, struck in head with screwdriver in lakeview. We can set up a. Web convert inches to feet. 1.73 m equals 5 feet step 1 convert from meters to feet. We decided to round some conversion factors to fit this table. How many feet in 73.81 mm. Web 73 inches equal 6.0833333333 feet (73in = 6.0833333333ft).
### We Decided To Round Some Conversion Factors To Fit This Table.
71 inches equals 5 feet, 11 inches (or 5.9167 feet). Web 9 rows 6.08333 feet (ft) visit 73 feet to inches conversion.
## You Can Also Find The Answer By Dividing 73.5 In By 12 And Therefore, Convert 73.5 Inches.
Web feet to inches how to convert inches to feet. 1″ = 1/12ft = 0.083333ft.
## Conclusion of 73 In To Feet.
Web value in feet = 73 × 0.083333333333333 = 6 1 / 12 feet definition of inch an inch is a unit of length or distance in a number of systems of measurement, including in the us. Here is the complete solution: You can also find the answer by dividing 73.5 in by 12 and therefore, convert 73.5 inches. Web to calculate 73 inches to the corresponding value in feet, multiply the quantity in inches by 0.083333333333333 (conversion factor).. Web 10.73 inches to feet. 1 meter = 3.28 x feet, so, 1.73 x 1 meter = 1.73 x 3.28 feet, or;
Source | 654 | 2,248 | {"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} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.804665 |
https://learn.careers360.com/school/question-need-solution-for-rd-sharma-maths-class-12-chapter-14-mean-value-theoram-exercise-multiple-choice-question-11/?question_number=11.0 | 1,716,515,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00461.warc.gz | 303,998,770 | 36,453 | #### need solution for rd sharma maths class 12 chapter 14 Mean Value Theoram exercise multiple choice question 11
Option (b)
Hint:
You must know about the concept of Rolle’s Theorem.
Given:
$f(x)=e^{x}\sin x,x\in \left [ 0,\pi \right ]$
Solution:
$f(x)=e^{x}\sin x$
$\Rightarrow f^{'}(x)=e^{x}\cos x-\sin x\left ( e^{x} \right )$
$\Rightarrow f^{'}(x)=e^{x}(\cos x -\sin x )$
$\Rightarrow f^{'}(c)=e^{c}(\cos c -\sin c )$
As Rolle’s Theorem,
$\Rightarrow f^{'}(c)=0$
$\Rightarrow e^{c}(\cos c -\sin c )=0$
$\Rightarrow \cos c -\sin c =0$
$\Rightarrow \cos c=\sin c$
$\Rightarrow \sin \left ( \frac{\pi }{2}-c \right )=\sin c$
$\Rightarrow c=\frac{\pi }{2}-c$
$\Rightarrow 2c=\frac{\pi }{2}$
$\Rightarrow c=\frac{\pi }{4}$
Option (b) is correct. | 276 | 765 | {"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": 13, "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-2024-22 | latest | en | 0.380877 |
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