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https://www.coursehero.com/file/5906695/Quiz-6/	1,487,816,824,000,000,000	text/html	crawl-data/CC-MAIN-2017-09/segments/1487501171070.80/warc/CC-MAIN-20170219104611-00339-ip-10-171-10-108.ec2.internal.warc.gz	803,460,828	49,214	# Quiz 6 - Version 101 Quiz 6 Laude (53755) This print-out... This preview shows pages 1–2. Sign up to view the full content. Version 101 – Quiz 6 – Laude – (53755) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 5.0 points IF a given system releases 900 J oF heat, what will Δ S oF the surroundings be iF this takes place at 27 C? 1. - 33 . 3 J · K 1 2. 3 J · K 1 correct 3. - 3 J · K 1 4. 33 . 3 J · K 1 Explanation: Heat released by the system will increase the entropy oF the surroundings by an amount equal to q/T , in this case 900 J 300 K = 3 J · K 1 . 002 5.0 points Rank the Following systems in terms oF in- creasing entropy: a) 1 mol oF pure ice b) 1 mol oF water with 1 mol oF salt dissolved in it. c) 1 mol oF pure water 1. b < a < c 2. a < c < b correct 3. a < b < c 4. c < a < b 5. b < c < a 6. c < b < a Explanation: Entropy increases as systems go through endothermic phase transitions and when there is more matter or more dispersed matter present. 003 5.0 points IF you had 1 mole oF H± molecules, each oF which had two possible orientations, what would their total theoretical positional en- This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 06/06/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin. ### Page1 / 3 Quiz 6 - Version 101 Quiz 6 Laude (53755) This print-out... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	519	1,769	{"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}	2.796875	3	CC-MAIN-2017-09	longest	en	0.874282
myqualityday.blogspot.com	1,718,835,724,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00563.warc.gz	371,136,880	24,263	Thursday, December 19, 2013 Puppy Chow Math Test Just a little project this morning to make some fun packages to take to people at work. At some point I realized I was doing several elementary school math story problems to complete the project. (and one a little more complex) I don't understand people who say they never use math. So here's a quiz for you. Answers at end. I decided to make "puppy chow" snack mix for everyone. Q1. How many people actually work in the mail room and on the presses? Q2. Which is the better value? A 3-pack of red curly ribbon for 79¢ or a 6 pack of assorted colors for \$1.35? The spools of ribbon are all the same size. Q3. If the recipe calls for 8 oz each of rice and corn chex, but you want to add Cheerios too, and you are going to double it, and the rice chex is an 18 oz box, and the corn chex is a 14 oz box, and the Cheerios is a 16 oz box, how many boxes should you buy? Q4. If the recipe calls for ½ c. peanut butter, but you are going to double it, and you have an 18 oz jar, how much should you use? (You melt the p.b. with oleo and chocolate chips.) Q5. How big a pan or bowl will you need to mix it all? Q6. The recipe calls for ½ c. powdered sugar, but you are doubling the recipe. How much powdered sugar will you need? Sorry, no picture of the finished product, but here's the bag with sugar in it. You add the chocolate coated cereal and shake it up to give it a sugar coating too. Q7. How many gift bags can you fill with what you just made? Q8. With 6 spools of ribbon, each labeled as 10m long, if you use 2 arm lengths for each bow, will you have enough ribbon? Q9. How many combinations of two colors can you make without repeating any of the combinations? Q10. How much did this gift cost per person? The 3 boxes of cereal cost \$2.99 each, and I had a \$1.00 off coupon (not per box). The 2 bags of chocolate chips were \$1.99 each. The peanut butter was \$1.88 for the 18 oz jar and I had a 75¢ off coupon. The oleo was 88¢ a pound and I used 2 sticks. The sugar was \$1.89 for 2 pounds, and I used about ¼ pound. The plastic bags cost \$2.00 for 50 bags. The paper bags cost about a penny each (had on hand), and the ribbon was \$1.35 for the six spools. ------------------------------------------------------ A1. It would have been intelligent to ask someone, eh? O well, I decided to make a double batch and hope it would be enough. This is called estimating. I think this was an 8th grade lesson. A2. 79/3= 26.3¢ each. 1.35/6= 22.5¢ so the 6 pack is the better value. Quick and dirty method: 792= \$1.58 which is more than \$1.35, so the 6 pack is cheaper. And more colors is nice anyway. A3. One of each, because you are going to just dump it all in and not worry about measuring this stuff. This is not a fussy recipe, and you have a coupon for \$1.00 off if you buy 2 boxes of cereal. A4. You need ½2 = 1 cup peanut butter. There are 8 oz in a cup, and half the jar of p.b. is 9 oz, so just use about half the jar. Please don't mess up a measuring cup to do this. This is not a fussy recipe. A5. This answer requires you to know more about the density and bulk of the product, not the weight. Trial and error works as well as anything. The answer is- the two biggest pans you own, so there's room to mix the chocolate and coat the cereal. A6. You thought this was going to be an easy one, right, as in ½2= 1? Haha. The answer is, at least twice that much to properly coat the cereal. Don't believe everything you read in recipes. A7. Just do it and see. Then please write the answer on the recipe card so you'll know next time you make this. You can make 18 bags with 5 oz of puppy chow in each one (yes I weighed them). That leaves about 3 oz that fell on the floor for the cook to eat (5 second rule). Just in case you are feeling OCD, it's interesting to note that 50 oz cereal + 9 oz peanut butter + 8 oz oleo + 24 oz chocolate chips + 4 oz powdered sugar = 95 oz of ingredients. 95/5= 19, which is almost exactly what I ended up with, allowing for some product loss that stuck to the pan, etc. A8. A meter is about a yard. An arm length is about a yard. You have 6 spools 10 m = 60 meters which equals about 60 yards. You have 18 bags to decorate. 182= 36 yards needed. You have plenty of ribbon. A9. OK, this one isn't a grade school question, it's basic statistics. But the visual answer is in the picture of the bags. The formula is n!/(n-r)!(r)! where n is the number of objects to choose from (6 colors), and r is the number you want to use in each set, which is 2. The ! means multiply the number by every smaller number so n!= 654321. (n-r)! = (6-2)!= 4321, and r!= 21, so you end up with (654321)/(432121). You can cancel everything out except the 65 on top and a 2 on the bottom, so 30/2= 15 different combinations. But the visual answer is seen in the rows. The back row has red combined with every other possible color, so R-L we have red/silver, red/green, red/magenta, red/gold, red/blue. Then the next row forward is blue with every other possible color, so using the same order R-L we have blue/silver, blue/green, blue/magenta, blue/gold. This leaves blue/red, but we've already got that in the back row, so we only have 4 possibilities in this group (It's a whole different math problem if the order matters- in other words, if red/blue is not the same thing as blue/red). The next row forward is gold with every other possible color, so now we have gold/silver, gold/green, and gold/magenta, but that's all we can do that aren't duplicates. Next we do magenta/silver, and magenta/green. Only two new possibilities. Then in the front row is green/silver, and that's all we can do. A10. Are you still here? Who cares? But quizzes have to have 10 questions, right? And I am curious. I snuck one little assumption in the question, and that is that you know there are 4 sticks of oleo in a pound. OK. {[(\$2.993)-1.00] + (21.99) + (1.88-.75)/2 + (.88/2) + [1.89/(2/¼)] + [(2.00/50)18] + (18.01) + [1.35(36/60]}/18 Do each part of the formula {7.97 + 3.98 + .67 + .44 + .24 + .72 + .18 + .81}/18 Continuing 15.01/18 = \$0.83 each. Not bad, not bad... everyone loved it. I got mega brownie points, and there was enough to go around, including for the guys running the presses, who often get overlooked when it comes to treats. I suspect only two of my regular readers actually made it this far. You know who you are! if you like this blog, click the +1 or Loritenor said... Math teachers would have loved given this to their students...oh and "estimating" in math is the big thing now along with compatible numbers...or something like that. There are a zillion new terms for everything. I think I had #3 correct...just dump in the three boxes.... Ann said... My head may be spinning right now but I did make it through the whole post. Now in keeping with the math subject you may need to add 1 to your suspicion if I was not among the two who you thought would make it all the way through....lol Ann said... oh and I forgot to say I like the way you make this, that's the way I do it too The Furry Gnome said... Well I made it to the end too, though I admit to skipping a word or two. I love your answers, but you lost me with Q.9. Meantime, you left out the most important question - how did it taste? Maybe everything can't be reduced to math? vanilla said... My wife always says, "If you can read, you can cook." Apparently math is also required. Oh, well. (I made it to the end. I do like a good math quiz.) Fun. Unknown said... I understand math a lot better now that I have had 52 years of practical experience.	2,113	7,673	{"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-2024-26	latest	en	0.957811
https://east-centricarch.eu/en/to-divide-14-23-divide-by-2-34-erik-multiplied-14-23-times-43-explain-eriks-error.4815573.html	1,669,628,040,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00574.warc.gz	269,875,148	11,220	grapeangela 13 # To divide 14 2/3 divide by 2 3/4, Erik multiplied 14 2/3 times 4/3. Explain Erik's error. mikalaevans9 The error is that he turned 2 3/4 to 4/3 instead of 11/4.	76	180	{"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-2022-49	latest	en	0.759639
https://math.stackexchange.com/questions/2796561/how-do-i-know-when-to-use-u-substitution-when-it-isnt-obvious	1,563,348,284,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00009.warc.gz	473,283,583	36,135	# How do I know when to use u-substitution when it isn't obvious? By "obvious" I mean something like: $$\int {x^2\over x^3+2}dx$$ I know I can use u-sub with $u=x^3+2$ and $du=3x^2dx$ to get: $${\ln(x^3+2)\over3}$$ But what about a problem like: $$\int 35\sqrt x e^\sqrt x dx$$ I first tried integrating by parts because I thought of it as functions $35\sqrt x$ and $e^\sqrt x$ multiplied together. I didn't know to set $u=\sqrt x$ until I got ${70\over 3}x^{3\over2}-35\int xe^\sqrt x$ through integration by parts (and then looked it up on an online calculator), and I'm assuming that would go into another loop of integration by parts because of the way derivatives with $e$ works. How do I know whether to try a u-sub or integration by parts first when there isn't an apparent function I can differentiate and easily replace with in the integral? This includes problems with and without $e$. • On general principles of "substitute so that common terms are unified", I'd have tried the $u = \sqrt{x}$ substitution on that integral. – Patrick Stevens May 26 '18 at 7:47 • Thanks for the reply. Can you clarify which common terms you are talking about? – VolTorian May 27 '18 at 6:07 • In this instance, $x$ appears only in the form $\sqrt{x}$, so I'd try substituting it away. – Patrick Stevens May 27 '18 at 8:27 In $\int \frac{x^2}{x^3+2}$ you need to take the $u$ substitution of a term that cancels the numerator. Here if we take $u=x^3+2$ then $dx=\frac{du}{3x^2}$. So the $x^2$ term in the numerator gets canceled. In the cases like $\int(6x^2)(2x^3+5)^6dx$ it is always advisable to take the number with the highest power. In this case $u=2x^3+5$ In the cases like $\int cos(5x-7)dx$ it is always advisable to take the number in the brackets. In this case $u=5x-7$ In the cases like $\int 35\sqrt{x}e^{\sqrt{x}}$ it is always advisable to take the exponent of $e$ as $u$. In this case $u=\sqrt{x}$ Another example of this is $\int \frac{e^x}{e^x}dx=\int1+e^{-x}dx$, even in this case we take the exponent of e as $u$. $u=-x$ • So in generally unless it's like my mentioned obvious case, I'd want to set the exponent of $e$ as u? Can you think of other integrals that wouldn't work this way? – VolTorian May 29 '18 at 3:06 • Example-$\frac{e^x}{1-e^x}dx$ here we take $u=1-e^x$ – tien lee May 29 '18 at 3:11 • Well I would say that's an obvious one with a clear u and du but thanks for the reply anyway – VolTorian May 29 '18 at 17:36	770	2,452	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2019-30	latest	en	0.904791
https://laurenceanywaysthemovie.com/how-many-beats-is-a-48-time-signature/	1,670,075,020,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00582.warc.gz	383,757,113	12,952	# How many beats is a 4/8 time signature? ## How many beats is a 4/8 time signature? 4/8 means you have four beats counted as eighth notes, each foot tap is an eighth note (a quarter note now takes 2 beats). ## What kind of time signature is 4 8? 4/2 and 4/8 are also simple quadruple. Notice that a time signature in simple meter will always have a 2, 3, or 4 for the top number. While beats in simple meter are divided into two notes, beats in compound meter are divided into three. What is time signature example? There are various types of time signatures, including: simple (such as 3/4 or 4/4), compound (e.g., 9/8 or 12/8), complex (e.g., 5/4 or 7/8), mixed (e.g., 5/8 & 3/8 or 6/8 & 3/4), additive (e.g., 3+2+3/8), fractional (e.g., 2½/4), and irrational meters (e.g., 3/10 or 5/24). What is a 4/4 time signature? 4/4 time signature The most common is 4/4: In a 4/4 time signature, there are four beats per measure and the quarter note receives one beat. A whole note takes up one entire measure in 4/4 time. There are 8, eighth notes in 4/4 time. ### How many beats is a quarter note in 4 8 time? And if the bottom number is an 8, it means the beats are 8th notes. 4/4 means there are 4 beats in each measure and a quarter note receives one count. 2/4 means there are 2 beats in each measure and a quarter note receives one count. ### How many quarter notes are there in a measure of a 4/4 time signature? four quarter note In 4/4, the stacked numbers tell you that each measure contains four quarter note beats. So, to count 4/4 meter, each time you tap the beat, you’re tapping the equivalent of one quarter note. What song has a 6 8 time signature? My favorite 6/8 folk songs are: Farmer In The Dell. Johnny Works With One Hammer. Hickory Dickory Dock. How do you count music in 4 4 Time? In 4/4, the stacked numbers tell you that each measure contains four quarter note beats. So, to count 4/4 meter, each time you tap the beat, you’re tapping the equivalent of one quarter note. ## How many beats are in 6/8 time? In 6/8 time, there are 2 beats per measure and the dotted quarter note (which = 3 eighth notes) has one beat. In 9/8 time, there are 3 beats per measure and the dotted quarter note (which = 3 eighth ## What does 6 8 time signature mean? In the 6/8 time signature in each measure you are going to have 6 eighth notes. In this same time signature you could have any combinations of notes that equals 6 eighth notes. 3 quarter notes, 12 sixteenth notes or any other combination that gets the right amount of time. What is 6 8 time signature? There is one commonly-used time signature that is just wrong. 6/8 literally means that there are six beats in the measure, and an eighth note gets one beat. Almost always there are two beats in the measure and a dotted quarter note gets one beat. What are some common time signatures? The single most common time signature for the last 100 years has been 4/4 time, so called “common time”. 90% or more of “popular” music in the last century has been in this time signature, as well as the majority of orchestral, choral and military band music.	850	3,135	{"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}	3.65625	4	CC-MAIN-2022-49	latest	en	0.931794
http://worldebookfair.org/Results.aspx?PageIndex=1&SearchSubject=Probability+and+combinatorics&SearchCollection=Dependent+events+series&SearchAcademicCollection=Trigonometry&EverythingType=0&TitleType=0&AuthorType=0&SubjectType=1&PublisherType=0&DisplayMode=List	1,582,711,706,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875146341.16/warc/CC-MAIN-20200226084902-20200226114902-00029.warc.gz	151,295,478	30,861	#jsDisabledContent { display:none; } My Account \| Register \| Help Search Results (5 titles) Searched over 21.6 Million titles in 0.42 seconds Federated Media Type Probability and combinatorics (X) Trigonometry (X) Dependent events series (X) 1 Records: 1 - 5 of 5 - Pages: Book Id: WPLBN0002823279 ► Abstract Description DetailsThinking about how the probability of an event can be dependent on another event occuring. Excerpt Details...Basics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in... Table of Contents Detailsprobability, coins, unfair Book Id: WPLBN0002823282 ► Abstract Description DetailsPresentation and analysis of the famous Monty Hall Problem Excerpt Details...Basics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in... Table of Contents Detailsprobability, monty, hall, lets, make, deal Book Id: WPLBN0002823278 ► Abstract Description DetailsExample where the probability of an outcome is dependent on which coin you happen to pick Excerpt Details...Basics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in... Book Id: WPLBN0002823281 ► Abstract Excerpt Details...Basics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in... Table of Contents Detailsdependent, replacement, probability, independent, events Book Id: WPLBN0002823280 ► Abstract Excerpt Details...Basics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in... 1 Records: 1 - 5 of 5 - Pages:	537	2,280	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2020-10	latest	en	0.835252
https://everything.explained.today/Cryptography/	1,653,090,060,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00274.warc.gz	297,917,668	43,993	# Cryptography Explained Cryptography, or cryptology (from Greek, Ancient (to 1453);: {{wikt-lang\|grc\|κρυπτός "hidden, secret"; and graphein, "to write", or -logia, "study", respectively[1]), is the practice and study of techniques for secure communication in the presence of adversarial behavior.[2] More generally, cryptography is about constructing and analyzing protocols that prevent third parties or the public from reading private messages;[3] various aspects in information security such as data confidentiality, data integrity, authentication, and non-repudiation[4] are central to modern cryptography. Modern cryptography exists at the intersection of the disciplines of mathematics, computer science, electrical engineering, communication science, and physics. Applications of cryptography include electronic commerce, chip-based payment cards, digital currencies, computer passwords, and military communications. Cryptography prior to the modern age was effectively synonymous with encryption, converting information from a readable state to unintelligible nonsense. The sender of an encrypted message shares the decoding technique only with intended recipients to preclude access from adversaries. The cryptography literature often uses the names Alice ("A") for the sender, Bob ("B") for the intended recipient, and Eve ("eavesdropper") for the adversary.[5] Since the development of rotor cipher machines in World War I and the advent of computers in World War II, cryptography methods have become increasingly complex and their applications more varied. Modern cryptography is heavily based on mathematical theory and computer science practice; cryptographic algorithms are designed around computational hardness assumptions, making such algorithms hard to break in actual practice by any adversary. While it is theoretically possible to break into a well-designed system, it is infeasible in actual practice to do so. Such schemes, if well designed, are therefore termed "computationally secure"; theoretical advances (e.g., improvements in integer factorization algorithms) and faster computing technology require these designs to be continually reevaluated, and if necessary, adapted. Information-theoretically secure schemes that cannot be broken even with unlimited computing power, such as the one-time pad, are much more difficult to use in practice than the best theoretically breakable, but computationally secure, schemes. The growth of cryptographic technology has raised a number of legal issues in the Information Age. Cryptography's potential for use as a tool for espionage and sedition has led many governments to classify it as a weapon and to limit or even prohibit its use and export. In some jurisdictions where the use of cryptography is legal, laws permit investigators to compel the disclosure of encryption keys for documents relevant to an investigation.[6] Cryptography also plays a major role in digital rights management and copyright infringement disputes in regard to digital media. ## Terminology The first use of the term "cryptograph" (as opposed to "cryptogram") dates back to the 19th century—originating from "The Gold-Bug," a story by Edgar Allan Poe.[7] Until modern times, cryptography referred almost exclusively to "encryption", which is the process of converting ordinary information (called plaintext) into an unintelligible form (called ciphertext). Decryption is the reverse, in other words, moving from the unintelligible ciphertext back to plaintext. A cipher (or cypher) is a pair of algorithms that carry out the encryption and the reversing decryption. The detailed operation of a cipher is controlled both by the algorithm and, in each instance, by a "key". The key is a secret (ideally known only to the communicants), usually a string of characters (ideally short so it can be remembered by the user), which is needed to decrypt the ciphertext. In formal mathematical terms, a "cryptosystem" is the ordered list of elements of finite possible plaintexts, finite possible cyphertexts, finite possible keys, and the encryption and decryption algorithms that correspond to each key. Keys are important both formally and in actual practice, as ciphers without variable keys can be trivially broken with only the knowledge of the cipher used and are therefore useless (or even counter-productive) for most purposes. Historically, ciphers were often used directly for encryption or decryption without additional procedures such as authentication or integrity checks. There are two main types of cryptosystems: symmetric and asymmetric. In symmetric systems, the only ones known until the 1970s, the same secret key encrypts and decrypts a message. Data manipulation in symmetric systems is significantly faster than in asymmetric systems. Asymmetric systems use a "public key" to encrypt a message and a related "private key" to decrypt it. The advantage of asymmetric systems is that the public key can be freely published, allowing parties to establish secure communication without having a shared secret key. In practice, asymmetric systems are used to first exchange a secret key, and then secure communication proceeds via a more efficient symmetric system using that key.[8] Examples of asymmetric systems include Diffie–Hellman key exchange, RSA (Rivest–Shamir–Adleman), ECC (Elliptic Curve Cryptography), and Post-quantum cryptography. Secure symmetric algorithms include the commonly used AES (Advanced Encryption Standard) which replaced the older DES (Data Encryption Standard).[9] Insecure symmetric algorithms include children's language tangling schemes such as Pig Latin or other cant, and all historical cryptographic schemes, however seriously intended, prior to the invention of the one-time pad early in the 20th century. In colloquial use, the term "code" is often used to mean any method of encryption or concealment of meaning. However, in cryptography, code has a more specific meaning: the replacement of a unit of plaintext (i.e., a meaningful word or phrase) with a code word (for example, "wallaby" replaces "attack at dawn"). A cypher, in contrast, is a scheme for changing or substituting an element below such a level (a letter, a syllable, or a pair of letters, etc.) in order to produce a cyphertext. Cryptanalysis is the term used for the study of methods for obtaining the meaning of encrypted information without access to the key normally required to do so; i.e., it is the study of how to "crack" encryption algorithms or their implementations. Some use the terms "cryptography" and "cryptology" interchangeably in English, while others (including US military practice generally) use "cryptography" to refer specifically to the use and practice of cryptographic techniques and "cryptology" to refer to the combined study of cryptography and cryptanalysis.[10] [11] English is more flexible than several other languages in which "cryptology" (done by cryptologists) is always used in the second sense above. advises that steganography is sometimes included in cryptology.[12] The study of characteristics of languages that have some application in cryptography or cryptology (e.g. frequency data, letter combinations, universal patterns, etc.) is called cryptolinguistics. ## History of cryptography and cryptanalysis See main article: History of cryptography. Before the modern era, cryptography focused on message confidentiality (i.e., encryption)—conversion of messages from a comprehensible form into an incomprehensible one and back again at the other end, rendering it unreadable by interceptors or eavesdroppers without secret knowledge (namely the key needed for decryption of that message). Encryption attempted to ensure secrecy in communications, such as those of spies, military leaders, and diplomats. In recent decades, the field has expanded beyond confidentiality concerns to include techniques for message integrity checking, sender/receiver identity authentication, digital signatures, interactive proofs and secure computation, among others. ### Classic cryptography The main classical cipher types are transposition ciphers, which rearrange the order of letters in a message (e.g., 'hello world' becomes 'ehlol owrdl' in a trivially simple rearrangement scheme), and substitution ciphers, which systematically replace letters or groups of letters with other letters or groups of letters (e.g., 'fly at once' becomes 'gmz bu podf' by replacing each letter with the one following it in the Latin alphabet).[13] Simple versions of either have never offered much confidentiality from enterprising opponents. An early substitution cipher was the Caesar cipher, in which each letter in the plaintext was replaced by a letter some fixed number of positions further down the alphabet. Suetonius reports that Julius Caesar used it with a shift of three to communicate with his generals. Atbash is an example of an early Hebrew cipher. The earliest known use of cryptography is some carved ciphertext on stone in Egypt (ca 1900 BCE), but this may have been done for the amusement of literate observers rather than as a way of concealing information. The Greeks of Classical times are said to have known of ciphers (e.g., the scytale transposition cipher claimed to have been used by the Spartan military).[14] Steganography (i.e., hiding even the existence of a message so as to keep it confidential) was also first developed in ancient times. An early example, from Herodotus, was a message tattooed on a slave's shaved head and concealed under the regrown hair.[15] More modern examples of steganography include the use of invisible ink, microdots, and digital watermarks to conceal information. In India, the 2000-year-old Kamasutra of Vātsyāyana speaks of two different kinds of ciphers called Kautiliyam and Mulavediya. In the Kautiliyam, the cipher letter substitutions are based on phonetic relations, such as vowels becoming consonants. In the Mulavediya, the cipher alphabet consists of pairing letters and using the reciprocal ones. In Sassanid Persia, there were two secret scripts, according to the Muslim author Ibn al-Nadim: the šāh-dabīrīya (literally "King's script") which was used for official correspondence, and the rāz-saharīya which was used to communicate secret messages with other countries.[16] David Kahn notes in The Codebreakers that modern cryptology originated among the Arabs, the first people to systematically document cryptanalytic methods.[17] Al-Khalil (717–786) wrote the Book of Cryptographic Messages, which contains the first use of permutations and combinations to list all possible Arabic words with and without vowels.[18] Ciphertexts produced by a classical cipher (and some modern ciphers) will reveal statistical information about the plaintext, and that information can often be used to break the cipher. After the discovery of frequency analysis, perhaps by the Arab mathematician and polymath Al-Kindi (also known as Alkindus) in the 9th century,[19] nearly all such ciphers could be broken by an informed attacker. Such classical ciphers still enjoy popularity today, though mostly as puzzles (see cryptogram). Al-Kindi wrote a book on cryptography entitled Risalah fi Istikhraj al-Mu'amma (Manuscript for the Deciphering Cryptographic Messages), which described the first known use of frequency analysis cryptanalysis techniques.[19] [20] Language letter frequencies may offer little help for some extended historical encryption techniques such as homophonic cipher that tend to flatten the frequency distribution. For those ciphers, language letter group (or n-gram) frequencies may provide an attack. Essentially all ciphers remained vulnerable to cryptanalysis using the frequency analysis technique until the development of the polyalphabetic cipher, most clearly by Leon Battista Alberti around the year 1467, though there is some indication that it was already known to Al-Kindi.[20] Alberti's innovation was to use different ciphers (i.e., substitution alphabets) for various parts of a message (perhaps for each successive plaintext letter at the limit). He also invented what was probably the first automatic cipher device, a wheel which implemented a partial realization of his invention. In the Vigenère cipher, a polyalphabetic cipher, encryption uses a key word, which controls letter substitution depending on which letter of the key word is used. In the mid-19th century Charles Babbage showed that the Vigenère cipher was vulnerable to Kasiski examination, but this was first published about ten years later by Friedrich Kasiski.[21] Although frequency analysis can be a powerful and general technique against many ciphers, encryption has still often been effective in practice, as many a would-be cryptanalyst was unaware of the technique. Breaking a message without using frequency analysis essentially required knowledge of the cipher used and perhaps of the key involved, thus making espionage, bribery, burglary, defection, etc., more attractive approaches to the cryptanalytically uninformed. It was finally explicitly recognized in the 19th century that secrecy of a cipher's algorithm is not a sensible nor practical safeguard of message security; in fact, it was further realized that any adequate cryptographic scheme (including ciphers) should remain secure even if the adversary fully understands the cipher algorithm itself. Security of the key used should alone be sufficient for a good cipher to maintain confidentiality under an attack. This fundamental principle was first explicitly stated in 1883 by Auguste Kerckhoffs and is generally called Kerckhoffs's Principle; alternatively and more bluntly, it was restated by Claude Shannon, the inventor of information theory and the fundamentals of theoretical cryptography, as Shannon's Maxim—'the enemy knows the system'. Different physical devices and aids have been used to assist with ciphers. One of the earliest may have been the scytale of ancient Greece, a rod supposedly used by the Spartans as an aid for a transposition cipher. In medieval times, other aids were invented such as the cipher grille, which was also used for a kind of steganography. With the invention of polyalphabetic ciphers came more sophisticated aids such as Alberti's own cipher disk, Johannes Trithemius' tabula recta scheme, and Thomas Jefferson's wheel cypher (not publicly known, and reinvented independently by Bazeries around 1900). Many mechanical encryption/decryption devices were invented early in the 20th century, and several patented, among them rotor machines—famously including the Enigma machine used by the German government and military from the late 1920s and during World War II.[22] The ciphers implemented by better quality examples of these machine designs brought about a substantial increase in cryptanalytic difficulty after WWI.[23] ### Computer era Prior to the early 20th century, cryptography was mainly concerned with linguistic and lexicographic patterns. Since then the emphasis has shifted, and cryptography now makes extensive use of mathematics, including aspects of information theory, computational complexity, statistics, combinatorics, abstract algebra, number theory, and finite mathematics generally beginning with the seminal paper, New direction in cryptography.[24] Cryptography is also a branch of engineering, but an unusual one since it deals with active, intelligent, and malevolent opposition; other kinds of engineering (e.g., civil or chemical engineering) need deal only with neutral natural forces. There is also active research examining the relationship between cryptographic problems and quantum physics. Just as the development of digital computers and electronics helped in cryptanalysis, it made possible much more complex ciphers. Furthermore, computers allowed for the encryption of any kind of data representable in any binary format, unlike classical ciphers which only encrypted written language texts; this was new and significant. Computer use has thus supplanted linguistic cryptography, both for cipher design and cryptanalysis. Many computer ciphers can be characterized by their operation on binary bit sequences (sometimes in groups or blocks), unlike classical and mechanical schemes, which generally manipulate traditional characters (i.e., letters and digits) directly. However, computers have also assisted cryptanalysis, which has compensated to some extent for increased cipher complexity. Nonetheless, good modern ciphers have stayed ahead of cryptanalysis; it is typically the case that use of a quality cipher is very efficient (i.e., fast and requiring few resources, such as memory or CPU capability), while breaking it requires an effort many orders of magnitude larger, and vastly larger than that required for any classical cipher, making cryptanalysis so inefficient and impractical as to be effectively impossible. Cryptanalysis of the new mechanical devices proved to be both difficult and laborious. In the United Kingdom, cryptanalytic efforts at Bletchley Park during WWII spurred the development of more efficient means for carrying out repetitious tasks. This culminated in the development of the Colossus, the world's first fully electronic, digital, programmable computer, which assisted in the decryption of ciphers generated by the German Army's Lorenz SZ40/42 machine. Extensive open academic research into cryptography is relatively recent, beginning in the mid-1970s. In the early 1970s IBM personnel designed the Data Encryption Standard (DES) algorithm that became the first federal government cryptography standard in the United States.[25] In 1976 Whitfield Diffie and Martin Hellman published the Diffie–Hellman key exchange algorithm. In 1977 the RSA algorithm was published in Martin Gardner's Scientific American column.[26] Since then, cryptography has become a widely used tool in communications, computer networks, and computer security generally. Some modern cryptographic techniques can only keep their keys secret if certain mathematical problems are intractable, such as the integer factorization or the discrete logarithm problems, so there are deep connections with abstract mathematics. There are very few cryptosystems that are proven to be unconditionally secure. The one-time pad is one, and was proven to be so by Claude Shannon. There are a few important algorithms that have been proven secure under certain assumptions. For example, the infeasibility of factoring extremely large integers is the basis for believing that RSA is secure, and some other systems, but even so, proof of unbreakability is unavailable since the underlying mathematical problem remains open. In practice, these are widely used, and are believed unbreakable in practice by most competent observers. There are systems similar to RSA, such as one by Michael O. Rabin that are provably secure provided factoring n = pq is impossible; it is quite unusable in practice. The discrete logarithm problem is the basis for believing some other cryptosystems are secure, and again, there are related, less practical systems that are provably secure relative to the solvability or insolvability discrete log problem.[27] As well as being aware of cryptographic history, cryptographic algorithm and system designers must also sensibly consider probable future developments while working on their designs. For instance, continuous improvements in computer processing power have increased the scope of brute-force attacks, so when specifying key lengths, the required key lengths are similarly advancing.[28] The potential impact of quantum computing are already being considered by some cryptographic system designers developing post-quantum cryptography. The announced imminence of small implementations of these machines may be making the need for preemptive caution rather more than merely speculative. ## Modern cryptography ### Symmetric-key cryptography See main article: Symmetric-key algorithm. Symmetric-key cryptography refers to encryption methods in which both the sender and receiver share the same key (or, less commonly, in which their keys are different, but related in an easily computable way). This was the only kind of encryption publicly known until June 1976.[29] Symmetric key ciphers are implemented as either block ciphers or stream ciphers. A block cipher enciphers input in blocks of plaintext as opposed to individual characters, the input form used by a stream cipher. The Data Encryption Standard (DES) and the Advanced Encryption Standard (AES) are block cipher designs that have been designated cryptography standards by the US government (though DES's designation was finally withdrawn after the AES was adopted).[30] Despite its deprecation as an official standard, DES (especially its still-approved and much more secure triple-DES variant) remains quite popular; it is used across a wide range of applications, from ATM encryption[31] to e-mail privacy[32] and secure remote access.[33] Many other block ciphers have been designed and released, with considerable variation in quality. Many, even some designed by capable practitioners, have been thoroughly broken, such as FEAL.[34] Stream ciphers, in contrast to the 'block' type, create an arbitrarily long stream of key material, which is combined with the plaintext bit-by-bit or character-by-character, somewhat like the one-time pad. In a stream cipher, the output stream is created based on a hidden internal state that changes as the cipher operates. That internal state is initially set up using the secret key material. RC4 is a widely used stream cipher. Block ciphers can be used as stream ciphers by generating blocks of a keystream (in place of a Pseudorandom number generator) and applying an XOR operation to each bit of the plaintext with each bit of the keystream.[35] Message authentication codes (MACs) are much like cryptographic hash functions, except that a secret key can be used to authenticate the hash value upon receipt; this additional complication blocks an attack scheme against bare digest algorithms, and so has been thought worth the effort. Cryptographic hash functions are a third type of cryptographic algorithm. They take a message of any length as input, and output a short, fixed-length hash, which can be used in (for example) a digital signature. For good hash functions, an attacker cannot find two messages that produce the same hash. MD4 is a long-used hash function that is now broken; MD5, a strengthened variant of MD4, is also widely used but broken in practice. The US National Security Agency developed the Secure Hash Algorithm series of MD5-like hash functions: SHA-0 was a flawed algorithm that the agency withdrew; SHA-1 is widely deployed and more secure than MD5, but cryptanalysts have identified attacks against it; the SHA-2 family improves on SHA-1, but is vulnerable to clashes as of 2011; and the US standards authority thought it "prudent" from a security perspective to develop a new standard to "significantly improve the robustness of NIST's overall hash algorithm toolkit."[36] Thus, a hash function design competition was meant to select a new U.S. national standard, to be called SHA-3, by 2012. The competition ended on October 2, 2012, when the NIST announced that Keccak would be the new SHA-3 hash algorithm.[37] Unlike block and stream ciphers that are invertible, cryptographic hash functions produce a hashed output that cannot be used to retrieve the original input data. Cryptographic hash functions are used to verify the authenticity of data retrieved from an untrusted source or to add a layer of security. ### Public-key cryptography See main article: Public-key cryptography. Symmetric-key cryptosystems use the same key for encryption and decryption of a message, although a message or group of messages can have a different key than others. A significant disadvantage of symmetric ciphers is the key management necessary to use them securely. Each distinct pair of communicating parties must, ideally, share a different key, and perhaps for each ciphertext exchanged as well. The number of keys required increases as the square of the number of network members, which very quickly requires complex key management schemes to keep them all consistent and secret. In a groundbreaking 1976 paper, Whitfield Diffie and Martin Hellman proposed the notion of public-key (also, more generally, called asymmetric key) cryptography in which two different but mathematically related keys are used—a public key and a private key.[38] A public key system is so constructed that calculation of one key (the 'private key') is computationally infeasible from the other (the 'public key'), even though they are necessarily related. Instead, both keys are generated secretly, as an interrelated pair.[39] The historian David Kahn described public-key cryptography as "the most revolutionary new concept in the field since polyalphabetic substitution emerged in the Renaissance".[40] In public-key cryptosystems, the public key may be freely distributed, while its paired private key must remain secret. In a public-key encryption system, the public key is used for encryption, while the private or secret key is used for decryption. While Diffie and Hellman could not find such a system, they showed that public-key cryptography was indeed possible by presenting the Diffie–Hellman key exchange protocol, a solution that is now widely used in secure communications to allow two parties to secretly agree on a shared encryption key.The X.509 standard defines the most commonly used format for public key certificates.[41] Diffie and Hellman's publication sparked widespread academic efforts in finding a practical public-key encryption system. This race was finally won in 1978 by Ronald Rivest, Adi Shamir, and Len Adleman, whose solution has since become known as the RSA algorithm.[42] The Diffie–Hellman and RSA algorithms, in addition to being the first publicly known examples of high-quality public-key algorithms, have been among the most widely used. Other asymmetric-key algorithms include the Cramer–Shoup cryptosystem, ElGamal encryption, and various elliptic curve techniques. A document published in 1997 by the Government Communications Headquarters (GCHQ), a British intelligence organization, revealed that cryptographers at GCHQ had anticipated several academic developments.[43] Reportedly, around 1970, James H. Ellis had conceived the principles of asymmetric key cryptography. In 1973, Clifford Cocks invented a solution that was very similar in design rationale to RSA.[44] In 1974, Malcolm J. Williamson is claimed to have developed the Diffie–Hellman key exchange.[45] Public-key cryptography is also used for implementing digital signature schemes. A digital signature is reminiscent of an ordinary signature; they both have the characteristic of being easy for a user to produce, but difficult for anyone else to forge. Digital signatures can also be permanently tied to the content of the message being signed; they cannot then be 'moved' from one document to another, for any attempt will be detectable. In digital signature schemes, there are two algorithms: one for signing, in which a secret key is used to process the message (or a hash of the message, or both), and one for verification, in which the matching public key is used with the message to check the validity of the signature. RSA and DSA are two of the most popular digital signature schemes. Digital signatures are central to the operation of public key infrastructures and many network security schemes (e.g., SSL/TLS, many VPNs, etc.). Public-key algorithms are most often based on the computational complexity of "hard" problems, often from number theory. For example, the hardness of RSA is related to the integer factorization problem, while Diffie–Hellman and DSA are related to the discrete logarithm problem. The security of elliptic curve cryptography is based on number theoretic problems involving elliptic curves. Because of the difficulty of the underlying problems, most public-key algorithms involve operations such as modular multiplication and exponentiation, which are much more computationally expensive than the techniques used in most block ciphers, especially with typical key sizes. As a result, public-key cryptosystems are commonly hybrid cryptosystems, in which a fast high-quality symmetric-key encryption algorithm is used for the message itself, while the relevant symmetric key is sent with the message, but encrypted using a public-key algorithm. Similarly, hybrid signature schemes are often used, in which a cryptographic hash function is computed, and only the resulting hash is digitally signed. ### Cryptographic Hash Functions Cryptographic Hash Functions are cryptographic algorithms that are ways to generate and utilize specific keys to encrypt data for either symmetric or asymmetric encryption, and such functions may be viewed as keys themselves. They take a message of any length as input, and output a short, fixed-length hash, which can be used in (for example) a digital signature. For good hash functions, an attacker cannot find two messages that produce the same hash. MD4 is a long-used hash function that is now broken; MD5, a strengthened variant of MD4, is also widely used but broken in practice. The US National Security Agency developed the Secure Hash Algorithm series of MD5-like hash functions: SHA-0 was a flawed algorithm that the agency withdrew; SHA-1 is widely deployed and more secure than MD5, but cryptanalysts have identified attacks against it; the SHA-2 family improves on SHA-1, but is vulnerable to clashes as of 2011; and the US standards authority thought it "prudent" from a security perspective to develop a new standard to "significantly improve the robustness of NIST's overall hash algorithm toolkit."[46] Thus, a hash function design competition was meant to select a new U.S. national standard, to be called SHA-3, by 2012. The competition ended on October 2, 2012, when the NIST announced that Keccak would be the new SHA-3 hash algorithm.[47] Unlike block and stream ciphers that are invertible, cryptographic hash functions produce a hashed output that cannot be used to retrieve the original input data. Cryptographic hash functions are used to verify the authenticity of data retrieved from an untrusted source or to add a layer of security. ### Cryptanalysis See main article: Cryptanalysis. The goal of cryptanalysis is to find some weakness or insecurity in a cryptographic scheme, thus permitting its subversion or evasion. It is a common misconception that every encryption method can be broken. In connection with his WWII work at Bell Labs, Claude Shannon proved that the one-time pad cipher is unbreakable, provided the key material is truly random, never reused, kept secret from all possible attackers, and of equal or greater length than the message.[48] Most ciphers, apart from the one-time pad, can be broken with enough computational effort by brute force attack, but the amount of effort needed may be exponentially dependent on the key size, as compared to the effort needed to make use of the cipher. In such cases, effective security could be achieved if it is proven that the effort required (i.e., "work factor", in Shannon's terms) is beyond the ability of any adversary. This means it must be shown that no efficient method (as opposed to the time-consuming brute force method) can be found to break the cipher. Since no such proof has been found to date, the one-time-pad remains the only theoretically unbreakable cipher. Although well-implemented one-time-pad encryption cannot be broken, traffic analysis is still possible. There are a wide variety of cryptanalytic attacks, and they can be classified in any of several ways. A common distinction turns on what Eve (an attacker) knows and what capabilities are available. In a ciphertext-only attack, Eve has access only to the ciphertext (good modern cryptosystems are usually effectively immune to ciphertext-only attacks). In a known-plaintext attack, Eve has access to a ciphertext and its corresponding plaintext (or to many such pairs). In a chosen-plaintext attack, Eve may choose a plaintext and learn its corresponding ciphertext (perhaps many times); an example is gardening, used by the British during WWII. In a chosen-ciphertext attack, Eve may be able to choose ciphertexts and learn their corresponding plaintexts. Finally in a man-in-the-middle attack Eve gets in between Alice (the sender) and Bob (the recipient), accesses and modifies the traffic and then forwards it to the recipient.[49] Also important, often overwhelmingly so, are mistakes (generally in the design or use of one of the protocols involved).Cryptanalysis of symmetric-key ciphers typically involves looking for attacks against the block ciphers or stream ciphers that are more efficient than any attack that could be against a perfect cipher. For example, a simple brute force attack against DES requires one known plaintext and 255 decryptions, trying approximately half of the possible keys, to reach a point at which chances are better than even that the key sought will have been found. But this may not be enough assurance; a linear cryptanalysis attack against DES requires 243 known plaintexts (with their corresponding ciphertexts) and approximately 243 DES operations.[50] This is a considerable improvement over brute force attacks. Public-key algorithms are based on the computational difficulty of various problems. The most famous of these are the difficulty of integer factorization of semiprimes and the difficulty of calculating discrete logarithms, both of which are not yet proven to be solvable in polynomial time (P) using only a classical Turing-complete computer. Much public-key cryptanalysis concerns designing algorithms in P that can solve these problems, or using other technologies, such as quantum computers. For instance, the best-known algorithms for solving the elliptic curve-based version of discrete logarithm are much more time-consuming than the best-known algorithms for factoring, at least for problems of more or less equivalent size. Thus, to achieve an equivalent strength of encryption, techniques that depend upon the difficulty of factoring large composite numbers, such as the RSA cryptosystem, require larger keys than elliptic curve techniques. For this reason, public-key cryptosystems based on elliptic curves have become popular since their invention in the mid-1990s. While pure cryptanalysis uses weaknesses in the algorithms themselves, other attacks on cryptosystems are based on actual use of the algorithms in real devices, and are called side-channel attacks. If a cryptanalyst has access to, for example, the amount of time the device took to encrypt a number of plaintexts or report an error in a password or PIN character, he may be able to use a timing attack to break a cipher that is otherwise resistant to analysis. An attacker might also study the pattern and length of messages to derive valuable information; this is known as traffic analysis[51] and can be quite useful to an alert adversary. Poor administration of a cryptosystem, such as permitting too short keys, will make any system vulnerable, regardless of other virtues. Social engineering and other attacks against humans (e.g., bribery, extortion, blackmail, espionage, torture, ...) are usually employed due to being more cost-effective and feasible to perform in a reasonable amount of time compared to pure cryptanalysis by a high margin. ### Cryptographic primitives Much of the theoretical work in cryptography concerns cryptographic primitives—algorithms with basic cryptographic properties—and their relationship to other cryptographic problems. More complicated cryptographic tools are then built from these basic primitives. These primitives provide fundamental properties, which are used to develop more complex tools called cryptosystems or cryptographic protocols, which guarantee one or more high-level security properties. Note, however, that the distinction between cryptographic primitives and cryptosystems, is quite arbitrary; for example, the RSA algorithm is sometimes considered a cryptosystem, and sometimes a primitive. Typical examples of cryptographic primitives include pseudorandom functions, one-way functions, etc. ### Cryptosystems See main article: List of cryptosystems. One or more cryptographic primitives are often used to develop a more complex algorithm, called a cryptographic system, or cryptosystem. Cryptosystems (e.g., El-Gamal encryption) are designed to provide particular functionality (e.g., public key encryption) while guaranteeing certain security properties (e.g., chosen-plaintext attack (CPA) security in the random oracle model). Cryptosystems use the properties of the underlying cryptographic primitives to support the system's security properties. As the distinction between primitives and cryptosystems is somewhat arbitrary, a sophisticated cryptosystem can be derived from a combination of several more primitive cryptosystems. In many cases, the cryptosystem's structure involves back and forth communication among two or more parties in space (e.g., between the sender of a secure message and its receiver) or across time (e.g., cryptographically protected backup data). Such cryptosystems are sometimes called cryptographic protocols. Some widely known cryptosystems include RSA, Schnorr signature, ElGamal encryption, and Pretty Good Privacy (PGP). More complex cryptosystems include electronic cash[52] systems, signcryption systems, etc. Some more 'theoretical' cryptosystems include interactive proof systems,[53] (like zero-knowledge proofs),[54] systems for secret sharing,[55] [56] etc. ### Lightweight cryptography Lightweight cryptography (LWC) concerns cryptographic algorithms developed for a strictly constrained environment. The growth of Internet of Things (IoT) has spiked research into the development of lightweight algorithms that are better suited for the environment. An IoT environment requires strict constraints on power consumption, processing power, and security.[57] Algorithms such as PRESENT, AES, and SPECK are examples of the many LWC algorithms that have been developed to achieve the standard set by the National Institute of Standards and Technology.[58] ## Applications ### In general To ensure secrecy during transmission, many systems use private key cryptography to protect transmitted information. With public-key systems, one can maintain secrecy without a master key or a large number of keys.[59] ### In cybersecurity Cryptography can be used to secure communications by encrypting them. Websites use encryption via HTTPS.[60] "End-to-end" encryption, where only sender and receiver can read messages, is implemented for email in Pretty Good Privacy and for secure messaging in general in Signal and WhatsApp.[61] Operating systems use encryption to keep passwords secret, conceal parts of the system, and ensure that software updates are truly from the system maker.[62] Instead of storing plaintext passwords, computer systems store hashes thereof; then, when a user logs in, the system passes the given password through a cryptographic hash function and compares it to the hashed value on file. In this manner, neither the system nor an attacker has at any point access to the password in plaintext.[63] Encryption is sometimes used to encrypt one's entire drive. For example, University College London has implemented BitLocker (a program by Microsoft) to render drive data opaque without users logging in.[64] ## Legal issues ### Prohibitions Cryptography has long been of interest to intelligence gathering and law enforcement agencies. Secret communications may be criminal or even treasonous . Because of its facilitation of privacy, and the diminution of privacy attendant on its prohibition, cryptography is also of considerable interest to civil rights supporters. Accordingly, there has been a history of controversial legal issues surrounding cryptography, especially since the advent of inexpensive computers has made widespread access to high-quality cryptography possible. In some countries, even the domestic use of cryptography is, or has been, restricted. Until 1999, France significantly restricted the use of cryptography domestically, though it has since relaxed many of these rules. In China and Iran, a license is still required to use cryptography.[65] Many countries have tight restrictions on the use of cryptography. Among the more restrictive are laws in Belarus, Kazakhstan, Mongolia, Pakistan, Singapore, Tunisia, and Vietnam.[66] In the United States, cryptography is legal for domestic use, but there has been much conflict over legal issues related to cryptography. One particularly important issue has been the export of cryptography and cryptographic software and hardware. Probably because of the importance of cryptanalysis in World War II and an expectation that cryptography would continue to be important for national security, many Western governments have, at some point, strictly regulated export of cryptography. After World War II, it was illegal in the US to sell or distribute encryption technology overseas; in fact, encryption was designated as auxiliary military equipment and put on the United States Munitions List.[67] Until the development of the personal computer, asymmetric key algorithms (i.e., public key techniques), and the Internet, this was not especially problematic. However, as the Internet grew and computers became more widely available, high-quality encryption techniques became well known around the globe. ### Export controls See main article: Export of cryptography. In the 1990s, there were several challenges to US export regulation of cryptography. After the source code for Philip Zimmermann's Pretty Good Privacy (PGP) encryption program found its way onto the Internet in June 1991, a complaint by RSA Security (then called RSA Data Security, Inc.) resulted in a lengthy criminal investigation of Zimmermann by the US Customs Service and the FBI, though no charges were ever filed.[68] [69] Daniel J. Bernstein, then a graduate student at UC Berkeley, brought a lawsuit against the US government challenging some aspects of the restrictions based on free speech grounds. The 1995 case Bernstein v. United States ultimately resulted in a 1999 decision that printed source code for cryptographic algorithms and systems was protected as free speech by the United States Constitution.[70] In 1996, thirty-nine countries signed the Wassenaar Arrangement, an arms control treaty that deals with the export of arms and "dual-use" technologies such as cryptography. The treaty stipulated that the use of cryptography with short key-lengths (56-bit for symmetric encryption, 512-bit for RSA) would no longer be export-controlled.[71] Cryptography exports from the US became less strictly regulated as a consequence of a major relaxation in 2000;[72] there are no longer very many restrictions on key sizes in US-exported mass-market software. Since this relaxation in US export restrictions, and because most personal computers connected to the Internet include US-sourced web browsers such as Firefox or Internet Explorer, almost every Internet user worldwide has potential access to quality cryptography via their browsers (e.g., via Transport Layer Security). The Mozilla Thunderbird and Microsoft Outlook E-mail client programs similarly can transmit and receive emails via TLS, and can send and receive email encrypted with S/MIME. Many Internet users don't realize that their basic application software contains such extensive cryptosystems. These browsers and email programs are so ubiquitous that even governments whose intent is to regulate civilian use of cryptography generally don't find it practical to do much to control distribution or use of cryptography of this quality, so even when such laws are in force, actual enforcement is often effectively impossible. ### NSA involvement See also: Clipper chip. Another contentious issue connected to cryptography in the United States is the influence of the National Security Agency on cipher development and policy. The NSA was involved with the design of DES during its development at IBM and its consideration by the National Bureau of Standards as a possible Federal Standard for cryptography.[73] DES was designed to be resistant to differential cryptanalysis,[74] a powerful and general cryptanalytic technique known to the NSA and IBM, that became publicly known only when it was rediscovered in the late 1980s.[75] According to Steven Levy, IBM discovered differential cryptanalysis, but kept the technique secret at the NSA's request. The technique became publicly known only when Biham and Shamir re-discovered and announced it some years later. The entire affair illustrates the difficulty of determining what resources and knowledge an attacker might actually have. Another instance of the NSA's involvement was the 1993 Clipper chip affair, an encryption microchip intended to be part of the Capstone cryptography-control initiative. Clipper was widely criticized by cryptographers for two reasons. The cipher algorithm (called Skipjack) was then classified (declassified in 1998, long after the Clipper initiative lapsed). The classified cipher caused concerns that the NSA had deliberately made the cipher weak in order to assist its intelligence efforts. The whole initiative was also criticized based on its violation of Kerckhoffs's Principle, as the scheme included a special escrow key held by the government for use by law enforcement (i.e. wiretapping). ### Digital rights management See main article: Digital rights management. Cryptography is central to digital rights management (DRM), a group of techniques for technologically controlling use of copyrighted material, being widely implemented and deployed at the behest of some copyright holders. In 1998, U.S. President Bill Clinton signed the Digital Millennium Copyright Act (DMCA), which criminalized all production, dissemination, and use of certain cryptanalytic techniques and technology (now known or later discovered); specifically, those that could be used to circumvent DRM technological schemes.[76] This had a noticeable impact on the cryptography research community since an argument can be made that any cryptanalytic research violated the DMCA. Similar statutes have since been enacted in several countries and regions, including the implementation in the EU Copyright Directive. Similar restrictions are called for by treaties signed by World Intellectual Property Organization member-states. The United States Department of Justice and FBI have not enforced the DMCA as rigorously as had been feared by some, but the law, nonetheless, remains a controversial one. Niels Ferguson, a well-respected cryptography researcher, has publicly stated that he will not release some of his research into an Intel security design for fear of prosecution under the DMCA.[77] Cryptologist Bruce Schneier has argued that the DMCA encourages vendor lock-in, while inhibiting actual measures toward cyber-security.[78] Both Alan Cox (longtime Linux kernel developer) and Edward Felten (and some of his students at Princeton) have encountered problems related to the Act. Dmitry Sklyarov was arrested during a visit to the US from Russia, and jailed for five months pending trial for alleged violations of the DMCA arising from work he had done in Russia, where the work was legal. In 2007, the cryptographic keys responsible for Blu-ray and HD DVD content scrambling were discovered and released onto the Internet. In both cases, the Motion Picture Association of America sent out numerous DMCA takedown notices, and there was a massive Internet backlash[79] triggered by the perceived impact of such notices on fair use and free speech. ### Forced disclosure of encryption keys See main article: Key disclosure law. In the United Kingdom, the Regulation of Investigatory Powers Act gives UK police the powers to force suspects to decrypt files or hand over passwords that protect encryption keys. Failure to comply is an offense in its own right, punishable on conviction by a two-year jail sentence or up to five years in cases involving national security.[80] Successful prosecutions have occurred under the Act; the first, in 2009,[81] resulted in a term of 13 months' imprisonment.[82] Similar forced disclosure laws in Australia, Finland, France, and India compel individual suspects under investigation to hand over encryption keys or passwords during a criminal investigation. In the United States, the federal criminal case of United States v. Fricosu addressed whether a search warrant can compel a person to reveal an encryption passphrase or password.[83] The Electronic Frontier Foundation (EFF) argued that this is a violation of the protection from self-incrimination given by the Fifth Amendment.[84] In 2012, the court ruled that under the All Writs Act, the defendant was required to produce an unencrypted hard drive for the court.[85] In many jurisdictions, the legal status of forced disclosure remains unclear. The 2016 FBI–Apple encryption dispute concerns the ability of courts in the United States to compel manufacturers' assistance in unlocking cell phones whose contents are cryptographically protected. As a potential counter-measure to forced disclosure some cryptographic software supports plausible deniability, where the encrypted data is indistinguishable from unused random data (for example such as that of a drive which has been securely wiped). • Book: Arbib . Jonathan . Dwyer . John . Discrete Mathematics for Cryptography . Algana Publishing . 1 . 31 January 2011 . 978-1-907934-01-8 . • Book: Becket, B . Introduction to Cryptology . Blackwell Scientific Publications . 1988 . 978-0-632-01836-9 . 16832704. Excellent coverage of many classical ciphers and cryptography concepts and of the "modern" DES and RSA systems. • Cryptography and Mathematics by Bernhard Esslinger, 200 pages, part of the free open-source package CrypTool, Web site: PDF download . 23 December 2013 . 22 July 2011 . https://web.archive.org/web/20110722183013/http://www.cryptool.org/download/CrypToolScript-en.pdf . bot: unknown . . CrypTool is the most widespread e-learning program about cryptography and cryptanalysis, open source. • In Code: A Mathematical Journey by Sarah Flannery (with David Flannery). Popular account of Sarah's award-winning project on public-key cryptography, co-written with her father. • James Gannon, Stealing Secrets, Telling Lies: How Spies and Codebreakers Helped Shape the Twentieth Century, Washington, D.C., Brassey's, 2001, . • Oded Goldreich, Foundations of Cryptography, in two volumes, Cambridge University Press, 2001 and 2004. • Alvin's Secret Code by Clifford B. Hicks (children's novel that introduces some basic cryptography and cryptanalysis). • Introduction to Modern Cryptography by Jonathan Katz and Yehuda Lindell. • Ibrahim A. Al-Kadi, "The Origins of Cryptology: the Arab Contributions," Cryptologia, vol. 16, no. 2 (April 1992), pp. 97–126. • Christof Paar, Jan Pelzl, Understanding Cryptography, A Textbook for Students and Practitioners. Springer, 2009. (Slides, online cryptography lectures and other information are available on the companion web site.) Very accessible introduction to practical cryptography for non-mathematicians. • Web site: Max Planck Encyclopedia of Public International Law., giving an overview of international law issues regarding cryptography. • Introduction to Modern Cryptography by Phillip Rogaway and Mihir Bellare, a mathematical introduction to theoretical cryptography including reduction-based security proofs. PDF download. • Book: Stallings, William . William Stallings . Cryptography and Network Security: Principles and Practice . Prentice Hall . March 2013 . 6th . 978-0-13-335469-0. • Tenzer, Theo (2021): SUPER SECRETO – The Third Epoch of Cryptography: Multiple, exponential, quantum-secure and above all, simple and practical Encryption for Everyone, Norderstedt, . • Johann-Christoph Woltag, 'Coded Communications (Encryption)' in Rüdiger Wolfrum (ed) Max Planck Encyclopedia of Public International Law (Oxford University Press 2009). ## Notes and References 1. Book: Liddell. Henry George. Henry George Liddell. Scott. Robert. Jones. Henry Stuart. Henry Stuart Jones. McKenzie. Roderick. A Greek-English Lexicon. Oxford University Press. 1984. A Greek-English Lexicon. 2. Book: Rivest, Ronald L.. Ron Rivest. J. Van Leeuwen. Handbook of Theoretical Computer Science. Cryptography. 1. Elsevier. 1990. 3. Book: Mihir. Bellare. Phillip. Rogaway. Introduction to Modern Cryptography. Introduction. 10. 21 September 2005. 4. Book: Handbook of Applied Cryptography. Menezes. A.J.. van Oorschot. P.C.. Vanstone. S.A.. 1997. 978-0-8493-8523-0. 5. 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Berlin. 567365751. 36. Notices. Federal Register. 72. 212. 2 November 2007. Web site: Archived copy . 27 January 2009 . https://web.archive.org/web/20080228075550/http://csrc.nist.gov/groups/ST/hash/documents/FR_Notice_Nov07.pdf . 28 February 2008 . bot: unknown. 37. Web site: NIST Selects Winner of Secure Hash Algorithm (SHA-3) Competition. Tech Beat. National Institute of Standards and Technology. October 2, 2012. 26 March 2015. 38. Whitfield. Diffie. Whitfield Diffie. Martin. Hellman. Martin Hellman. Multi-user cryptographic techniques. AFIPS Proceedings. 45. 109–112. 8 June 1976. 10.1145/1499799.1499815. 13210741. 39. [Ralph Merkle] 40. David. Kahn. Cryptology Goes Public. Foreign Affairs. 58. 141–159. 1. Fall 1979. 10.2307/20040343. 20040343. 41. Web site: Using Client-Certificate based authentication with NGINX on Ubuntu - SSLTrust. SSLTrust. 13 June 2019. 42. Ronald L.. Rivest. Ronald L. Rivest. A.. Shamir. L.. Adleman. A Method for Obtaining Digital Signatures and Public-Key Cryptosystems. Communications of the ACM. 21. 2. 120–126. 1978. 10.1145/359340.359342. 10.1.1.607.2677. 2873616. Web site: Archived copy . 20 April 2006 . https://web.archive.org/web/20011116122233/http://theory.lcs.mit.edu/~rivest/rsapaper.pdf . 16 November 2001 . dead. Previously released as an MIT "Technical Memo" in April 1977, and published in Martin Gardner's Scientific American Mathematical recreations column 43. News: Wayner. Peter. British Document Outlines Early Encryption Discovery. 24 December 1997. 2015-03-26. The New York Times. 44. A Note on 'Non-Secret Encryption'. Clifford. Cocks. CESG Research Report. 20 November 1973. 45. Book: Singh, Simon . The Code Book . registration . . 1999 . 279–292. 9780385495318 . 46. Notices. Federal Register. 72. 212. 2 November 2007. 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United States District Court for the District of Colorado. 26 March 2015.	14,976	67,386	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2022-21	longest	en	0.909674
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/SK90/4.48.trs.Thm12:POLO_7172_DP:NO.html.lzma	1,718,636,738,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00331.warc.gz	86,948,935	1,729	Term Rewriting System R: [x, y, z, u, v] f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v)) f(x, y, y) -> y f(x, y, g(y)) -> x f(x, x, y) -> x f(g(x), x, y) -> y Termination of R to be shown. ` R` ` ↳Dependency Pair Analysis` R contains the following Dependency Pairs: F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v)) F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v) Furthermore, R contains one SCC. ` R` ` ↳DPs` ` →DP Problem 1` ` ↳Polynomial Ordering` Dependency Pairs: F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v) F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v)) Rules: f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v)) f(x, y, y) -> y f(x, y, g(y)) -> x f(x, x, y) -> x f(g(x), x, y) -> y The following dependency pairs can be strictly oriented: F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v) F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v)) Additionally, the following rules can be oriented: f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v)) f(x, y, y) -> y f(x, y, g(y)) -> x f(x, x, y) -> x f(g(x), x, y) -> y Used ordering: Polynomial ordering with Polynomial interpretation: POL(g(x1)) = 0 POL(f(x1, x2, x3)) = 1 + x1 + x3 POL(F(x1, x2, x3)) = x1 resulting in one new DP problem. ` R` ` ↳DPs` ` →DP Problem 1` ` ↳Polo` ` →DP Problem 2` ` ↳Dependency Graph` Dependency Pair: Rules: f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v)) f(x, y, y) -> y f(x, y, g(y)) -> x f(x, x, y) -> x f(g(x), x, y) -> y Using the Dependency Graph resulted in no new DP problems. Termination of R successfully shown. Duration: 0:00 minutes	690	1,626	{"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-2024-26	latest	en	0.579283
https://gmatclub.com/forum/if-a-then-b-if-b-then-c-if-c-then-d-if-all-of-the-42529.html?fl=similar	1,508,234,280,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00506.warc.gz	694,665,346	44,734	It is currently 17 Oct 2017, 02:58 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If A, then B. If B, then C. If C, then D. If all of the Author Message Director Joined: 09 Aug 2006 Posts: 521 Kudos [?]: 110 [0], given: 0 If A, then B. If B, then C. If C, then D. If all of the [#permalink] ### Show Tags 24 Feb 2007, 04:32 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 1 sessions ### HideShow timer Statistics 7. If A, then B. If B, then C. If C, then D. If all of the statements above are true, which of the following must also be true? (A) If D, then A. (B) If not B, then not C. (C) If not D, then not A. (D) If D, then E. (E) If not A, then not D. Kudos [?]: 110 [0], given: 0 Director Joined: 28 Dec 2005 Posts: 917 Kudos [?]: 57 [0], given: 0 ### Show Tags 24 Feb 2007, 05:04 I will go with C. If not D, then not A. If D has not occured, we can trace it all the way back to A not occuring. Kudos [?]: 57 [0], given: 0 Director Affiliations: FRM Charter holder Joined: 02 Dec 2006 Posts: 726 Kudos [?]: 96 [0], given: 4 Schools: Stanford, Chicago Booth, Babson College ### Show Tags 24 Feb 2007, 07:31 hsampath wrote: I will go with C. If not D, then not A. If D has not occured, we can trace it all the way back to A not occuring. One more C for the same reason. Kudos [?]: 96 [0], given: 4 Director Joined: 24 Aug 2006 Posts: 743 Kudos [?]: 202 [0], given: 0 Location: Dallas, Texas ### Show Tags 24 Feb 2007, 11:50 elementary watson (C) _________________ "Education is what remains when one has forgotten everything he learned in school." Kudos [?]: 202 [0], given: 0 Senior Manager Joined: 19 Jul 2006 Posts: 358 Kudos [?]: 8 [0], given: 0 ### Show Tags 24 Feb 2007, 11:59 C Kudos [?]: 8 [0], given: 0 Manager Joined: 12 Feb 2007 Posts: 167 Kudos [?]: 1 [0], given: 0 ### Show Tags 24 Feb 2007, 12:43 Yes, C Kudos [?]: 1 [0], given: 0 Director Joined: 09 Aug 2006 Posts: 521 Kudos [?]: 110 [0], given: 0 ### Show Tags 25 Feb 2007, 07:15 Hi Guys.. Thanks for u r reponses. The OA is indeed C But, I argue why not B ? It is clearly said in the question statement tht "if B then C" that is C is there coz of B hence if there is no B then no C. can any one please enlighten ? Thanks Kudos [?]: 110 [0], given: 0 Director Joined: 24 Aug 2006 Posts: 743 Kudos [?]: 202 [0], given: 0 Location: Dallas, Texas ### Show Tags 25 Feb 2007, 14:00 If A, then B. If B, then C. If C, then D. Then if A then D Contrapositive of this is also true: if Not D then Not A Deductive logic is fun and much more interesting than inductive logic. But most GMAT prep books do not address deductive logic . Some LSAT books do. Alternatively, You can borrow the Copi-Cohen's book from local library and spend a weekend over it. I'm sure you will significantly improve your reasoning ability. _________________ "Education is what remains when one has forgotten everything he learned in school." Kudos [?]: 202 [0], given: 0 Manager Joined: 09 Jan 2007 Posts: 240 Kudos [?]: 39 [0], given: 0 ### Show Tags 25 Feb 2007, 19:07 If not A then not B If not B then not C If not C then not D. Therefore, if not A then not D is true. Kudos [?]: 39 [0], given: 0 Director Joined: 24 Aug 2006 Posts: 743 Kudos [?]: 202 [0], given: 0 Location: Dallas, Texas ### Show Tags 25 Feb 2007, 20:32 rdg wrote: If not A then not B If not B then not C If not C then not D. Therefore, if not A then not D is true. If A then B is given , you CAN NOT conclude If Not A then not B. I'll recommend Copi-Cohen's book to you also. _________________ "Education is what remains when one has forgotten everything he learned in school." Kudos [?]: 202 [0], given: 0 Manager Joined: 09 Jan 2007 Posts: 240 Kudos [?]: 39 [0], given: 0 ### Show Tags 25 Feb 2007, 21:28 Swagatalakshmi wrote: rdg wrote: If not A then not B If not B then not C If not C then not D. Therefore, if not A then not D is true. If A then B is given , you CAN NOT conclude If Not A then not B. I'll recommend Copi-Cohen's book to you also. Thanks Swagatalakshmi. Kudos [?]: 39 [0], given: 0 Senior Manager Joined: 19 Feb 2007 Posts: 325 Kudos [?]: 66 [0], given: 0 ### Show Tags 28 Feb 2007, 04:34 Amit05 wrote: Hi Guys.. Thanks for u r reponses. The OA is indeed C But, I argue why not B ? It is clearly said in the question statement tht "if B then C" that is C is there coz of B hence if there is no B then no C. can any one please enlighten ? Thanks It's not B because it says if B then C and not if B then only C. C may be because of lets say X. If not C then not B would have been correct. Kudos [?]: 66 [0], given: 0 28 Feb 2007, 04:34 Display posts from previous: Sort by	1,711	5,256	{"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-2017-43	latest	en	0.798776
https://www.justintools.com/unit-conversion/pressure.php?k1=kilopascals&k2=poundal-per-square-meter	1,586,445,601,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00117.warc.gz	940,437,832	28,156	Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :) # PRESSURE Units Conversionkilopascals to poundal-per-square-meter 1 Kilopascals = 7233.0114643232 Poundal Per Square Meter Category: pressure Conversion: Kilopascals to Poundal Per Square Meter The base unit for pressure is pascals (Non-SI Unit) [Kilopascals] symbol/abbrevation: (kPa) [Poundal Per Square Meter] symbol/abbrevation: (pdl/m2) How to convert Kilopascals to Poundal Per Square Meter (kPa to pdl/m2)? 1 kPa = 7233.0114643232 pdl/m2. 1 x 7233.0114643232 pdl/m2 = 7233.0114643232 Poundal Per Square Meter. Always check the results; rounding errors may occur. Definition: In relation to the base unit of [pressure] => (pascals), 1 Kilopascals (kPa) is equal to 1000 pascals, while 1 Poundal Per Square Meter (pdl/m2) = 0.138255 pascals. 1 Kilopascals to common pressure units 1 kPa = 1000 pascals (Pa) 1 kPa = 0.0098692326671601 atmosphere atm standard (atm) 1 kPa = 7.5006375541921 millimeter of mercury (mmHg) 1 kPa = 0.01 bars (bar) 1 kPa = 7.5006168271009 torrs (Torr) 1 kPa = 0.00014503773772954 ksi (ksi) 1 kPa = 0.14503773772954 psi (psi) 1 kPa = 101.97162129779 kilogram force per square meter (kgf/m2) 1 kPa = 0.010197162129779 atmosphere at technical (at) 1 kPa = 1.0000000001E+21 attopascals (aPa) Kilopascalsto Poundal Per Square Meter (table conversion) 1 kPa = 7233.0114643232 pdl/m2 2 kPa = 14466.022928646 pdl/m2 3 kPa = 21699.03439297 pdl/m2 4 kPa = 28932.045857293 pdl/m2 5 kPa = 36165.057321616 pdl/m2 6 kPa = 43398.068785939 pdl/m2 7 kPa = 50631.080250262 pdl/m2 8 kPa = 57864.091714585 pdl/m2 9 kPa = 65097.103178909 pdl/m2 10 kPa = 72330.114643232 pdl/m2 20 kPa = 144660.22928646 pdl/m2 30 kPa = 216990.3439297 pdl/m2 40 kPa = 289320.45857293 pdl/m2 50 kPa = 361650.57321616 pdl/m2 60 kPa = 433980.68785939 pdl/m2 70 kPa = 506310.80250262 pdl/m2 80 kPa = 578640.91714585 pdl/m2 90 kPa = 650971.03178909 pdl/m2 100 kPa = 723301.14643232 pdl/m2 200 kPa = 1446602.2928646 pdl/m2 300 kPa = 2169903.439297 pdl/m2 400 kPa = 2893204.5857293 pdl/m2 500 kPa = 3616505.7321616 pdl/m2 600 kPa = 4339806.8785939 pdl/m2 700 kPa = 5063108.0250262 pdl/m2 800 kPa = 5786409.1714585 pdl/m2 900 kPa = 6509710.3178909 pdl/m2 1000 kPa = 7233011.4643232 pdl/m2 2000 kPa = 14466022.928646 pdl/m2 4000 kPa = 28932045.857293 pdl/m2 5000 kPa = 36165057.321616 pdl/m2 7500 kPa = 54247585.982424 pdl/m2 10000 kPa = 72330114.643232 pdl/m2 25000 kPa = 180825286.60808 pdl/m2 50000 kPa = 361650573.21616 pdl/m2 100000 kPa = 723301146.43232 pdl/m2 1000000 kPa = 7233011464.3232 pdl/m2 1000000000 kPa = 7233011464323.2 pdl/m2 (Kilopascals) to (Poundal Per Square Meter) conversions	1,178	2,941	{"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}	3.296875	3	CC-MAIN-2020-16	latest	en	0.663058
http://wiki.orthogonaldevices.com/index.php?title=ER-301/Sine_Osc&oldid=2245	1,642,866,938,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00711.warc.gz	74,699,310	7,074	# ER-301/Sine Osc ## Applications • Synth voice • FM/PM synthesis • LFO with sync capabilities • Pass through a limiter to create a square wave. • Waveshaper ## Parameters ### V/oct Control Type Has Sub-chain? Sub-chain Parameters Fader Scale Pitch yes transpose (-3600¢ to 3600¢) logarithmic ratio This parameter is used to transpose the base frequency up or down according to an incoming modulation value that is V/oct calibrated (typically one of the inputs from the ABCD matrix). In other words, $\displaystyle \text{f} = \text{f0} * 2 ^{\frac{x}{1200}}$ where $\displaystyle x$ is the value of this parameter (which is in cents). For example, if the base frequency (f0) is equal to 55Hz and this V/oct parameter is set to 2400¢, then the resulting frequency will be 220Hz (i.e. 2400¢ = 2 octaves which is 4x). ### f0 Control Type Has Sub-chain? Sub-chain Parameters Fader Scale Gain/Bias yes gain (-10,000 to 10,000) bias (10 octaves above and below 27.5Hz) frequency in Hertz This parameter sets the fundamental frequency that is subsequently transposed by the V/oct parameter. Modulating this parameter corresponds to (thru-zero) linear FM. ### phase Control Type Has Sub-chain? Sub-chain Parameters Fader Scale Gain/Bias yes gain (-10 to 10) bias (-1 to 1) linear Sets the phase offset of the oscillator. Modulating this parameter is equivalent to PM (phase modulation). You can even set the f0 parameter to zero and put the oscillator completely under manual control via this phase parameter. This last case is also called waveshaping. ### feedback Control Type Has Sub-chain? Sub-chain Parameters Fader Scale Gain/Bias yes gain (-10 to 10) bias (-1 to 1) linear The feedback parameter determines how much of the unit's (pre-level) output should be used to self-modulate the phase offset. Increasing amounts of feedback causes the sine wave to morph into a saw-like or ramp-like waveform. FYI: For low frequencies (<55Hz) and high feedback amounts (>0.95), the oscillator will turn chaotic briefly as it passes through zero. ### sync Control Type Has Sub-chain? Sub-chain Parameters Behavior Threshold yes threshold (-1 to 1) trigger A trigger into this parameter will cause the oscillator to reset its phase accumulator to the value of the phase parameter. This is equivalent to the hard-sync on many oscillators. This sync operation is not anti-aliased. So audio-rate triggering of this parameter will exhibit strong anti-aliasing artifacts which depending on the desired result could be a good thing or a bad thing. ### level Control Type Has Sub-chain? Sub-chain Parameters Fader Scale Gain/Bias yes gain (-10 to 10) bias (-1 to 1) linear gain This parameter controls an internal bipolar linear VCA on the output of the unit. It behaves exactly the same as the level parameter of a Linear Bipolar VCA unit.	675	2,851	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 2}	2.578125	3	CC-MAIN-2022-05	latest	en	0.737499
https://www.physicsforums.com/threads/displacement-average-speed-and-average-velocity.769038/	1,685,571,624,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00069.warc.gz	1,044,969,008	14,904	# Displacement, Average Speed, and Average Velocity • person100 #### person100 Okay so I don't understand how to do the displacement of distance and time on a graph. I also need help with figuring out the average speed and average velocity. The graph deals with a Turtle's journey. On the Y-axis of the graph is distance in meters and the x-axis is time in minutes. The y-axis goes from 1-23 and the x going by threes all the up to 66. So the coordinates to the dots on the graph are the following: (3,2) (6,4) (12,6) (18,8) (24,10) (28,8) (30,7) (35,10) (40,12) (46,14) (51,14) (54,17) (57,20) (65,20) (66,23) *Also the total distance the turtle traveled was 29 m. The following question's are: 1. What was the turtle's displacement? 2. What was the turtle's average speed? 3. What was the turtle's average velocity?	241	821	{"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}	3.53125	4	CC-MAIN-2023-23	latest	en	0.939274
https://practice.geeksforgeeks.org/problems/number-of-compositions-of-a-natural-number/0	1,597,469,225,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439740679.96/warc/CC-MAIN-20200815035250-20200815065250-00378.warc.gz	446,662,820	24,049	Timer is Running Number of compositions of a natural number Submissions: 483 Accuracy: 29.45% Difficulty: Basic Marks: 1 Given a natural number N, find the number of ways in which N can be expressed as a sum of natural numbers when order is taken into consideration. Two sequences that differ in the order of their terms define different compositions of their sum. Input: The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains a number N as input. Output: For each test case, print the total number of ways in new line. Constraints: 1 <= T <= 50 1 <= N <= 64 Example: Input: 2 2 4 Output: 2 8 Explanation: ```Input : 4 Output : 8 All 8 position composition are: 4, 1+3, 3+1, 2+2, 1+1+2, 1+2+1, 2+1+1 and 1+1+1+1``` For More Input/Output Examples Use 'Expected Output' option Contributor: Arun Tyagi Author: arun03 If you have purchased any course from GeeksforGeeks then please ask your doubt on course discussion forum. You will get quick replies from GFG Moderators there.	302	1,076	{"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-2020-34	latest	en	0.837411
https://doubtnut.com/question-answer/there-is-a-slide-in-a-park-one-of-its-side-walls-has-been-painted-in-some-colour-with-a-message-keep-3862	1,553,230,554,000,000,000	text/html	crawl-data/CC-MAIN-2019-13/segments/1552912202628.42/warc/CC-MAIN-20190322034516-20190322060516-00516.warc.gz	465,644,515	18,794	X Doubtnut Math Doubt App Click photo & get instant FREE video solution Install in 1 Sec Now! (9,487+) COURSE EXAM STUDY MATERIALS # There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. Question from Class 9 Chapter HERONS FORMULA NCERT Class 9 AREAS OF PARALLELOGRAMS AND TRIANGLES Maths Solutions NCERT Class 9 CIRCLES Maths Solutions NCERT Class 9 COORDINATE GEOMETRY Maths Solutions NCERT Class 9 HERONS FORMULA Maths Solutions NCERT Class 9 INTRODUCTION TO EUCLIDS GEOMETRY Maths Solutions NCERT Class 9 LINEAR EQUATIONS IN TWO VARIABLES Maths Solutions NCERT Class 9 LINES AND ANGLES Maths Solutions NCERT Class 9 NUMBER SYSTEMS Maths Solutions NCERT Class 9 POLYNOMIALS Maths Solutions NCERT Class 9 PROBABILITY Maths Solutions NCERT Class 9 QUADRILATERALS Maths Solutions NCERT Class 9 STATISTICS Maths Solutions NCERT Class 9 SURFACE AREAS AND VOLUMES Maths Solutions NCERT Class 9 TRIANGLES Maths Solutions Get here the Maths video solutions of all the questions of of NCERT book. We provide you step by step solution for all the questions given in NCERT Maths book as per CBSE guidelines .The topics and subtopics of are as follows - Click on the link to Learn any Topic for Free: INTRODUCTION • Play • Play • Play HERONS FORMULA • Play • Play • Play Herons formula • Play	390	1,465	{"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-2019-13	latest	en	0.76408
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/prerequisites-p-2-exponents-and-radicals-exercises-page-25/44	1,723,239,413,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00720.warc.gz	636,280,503	12,461	Algebra and Trigonometry 10th Edition (a) $\frac{2\sqrt[3] 2}{3}$ (b) $\frac{5\sqrt 3}{2}$ (a) $\sqrt[3] {\frac{16}{27}}=\frac{\sqrt[3] {16}}{\sqrt[3] {27}}=\frac{\sqrt[3] {8\times2}}{\sqrt[3] {3^3}}=\frac{\sqrt[3] {2^3\times2}}{3}=\frac{2\sqrt[3] 2}{3}$ (b) $\sqrt {\frac{75}{4}}=\frac{\sqrt {75}}{\sqrt 4}=\frac{\sqrt {25\times3}}{\sqrt {2^2}}=\frac{\sqrt {5^2\times3}}{2}=\frac{5\sqrt 3}{2}$	197	395	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2024-33	latest	en	0.317152
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.18432.1/holomorphic/4.4.18432.1-7.2-e	1,606,594,130,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141195745.90/warc/CC-MAIN-20201128184858-20201128214858-00325.warc.gz	137,015,054	5,521	# Properties Label 4.4.18432.1-7.2-e Base field 4.4.18432.1 Weight $[2, 2, 2, 2]$ Level norm $7$ Level $[7,7,-\frac{1}{3}w^{2} + w + 1]$ Dimension $8$ CM no Base change no # Related objects • L-function not available ## Base field 4.4.18432.1 Generator $$w$$, with minimal polynomial $$x^{4} - 12x^{2} + 18$$; narrow class number $$2$$ and class number $$1$$. ## Form Weight: $[2, 2, 2, 2]$ Level: $[7,7,-\frac{1}{3}w^{2} + w + 1]$ Dimension: $8$ CM: no Base change: no Newspace dimension: $14$ ## Hecke eigenvalues ($q$-expansion) The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial: $$x^{8} - 13x^{6} + 54x^{4} - 71x^{2} + 1$$ Norm Prime Eigenvalue 2 $[2, 2, -\frac{1}{3}w^{3} - \frac{1}{3}w^{2} + 3w + 4]$ $\phantom{-}e$ 7 $[7, 7, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 3w - 3]$ $-\frac{1}{2}e^{6} + 4e^{4} - 8e^{2} + \frac{9}{2}$ 7 $[7, 7, -\frac{1}{3}w^{2} + w + 1]$ $-1$ 7 $[7, 7, \frac{1}{3}w^{2} + w - 1]$ $-\frac{1}{2}e^{6} + 5e^{4} - 12e^{2} + \frac{5}{2}$ 7 $[7, 7, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 3w + 3]$ $\phantom{-}e^{2} - 3$ 9 $[9, 3, w - 3]$ $-\frac{1}{2}e^{7} + 7e^{5} - 31e^{3} + \frac{87}{2}e$ 41 $[41, 41, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 3w + 1]$ $\phantom{-}\frac{3}{2}e^{7} - 20e^{5} + 84e^{3} - \frac{215}{2}e$ 41 $[41, 41, -\frac{1}{3}w^{2} + w + 3]$ $-\frac{1}{2}e^{7} + 8e^{5} - 38e^{3} + \frac{105}{2}e$ 41 $[41, 41, \frac{1}{3}w^{2} + w - 3]$ $-e^{7} + 12e^{5} - 47e^{3} + 60e$ 41 $[41, 41, -\frac{1}{3}w^{3} - \frac{1}{3}w^{2} + 3w + 1]$ $-\frac{3}{2}e^{7} + 17e^{5} - 60e^{3} + \frac{131}{2}e$ 47 $[47, 47, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 4w - 1]$ $\phantom{-}2e^{5} - 16e^{3} + 26e$ 47 $[47, 47, -\frac{1}{3}w^{3} + \frac{1}{3}w^{2} + 2w - 3]$ $\phantom{-}4e^{5} - 33e^{3} + 63e$ 47 $[47, 47, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 2w - 3]$ $\phantom{-}e^{7} - 13e^{5} + 54e^{3} - 70e$ 47 $[47, 47, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 4w + 1]$ $-\frac{1}{2}e^{7} + 8e^{5} - 42e^{3} + \frac{141}{2}e$ 89 $[89, 89, -\frac{1}{3}w^{3} + \frac{2}{3}w^{2} + 3w - 3]$ $\phantom{-}2e^{7} - 23e^{5} + 83e^{3} - 92e$ 89 $[89, 89, \frac{2}{3}w^{2} + w - 5]$ $-3e^{7} + 34e^{5} - 118e^{3} + 121e$ 89 $[89, 89, \frac{2}{3}w^{2} - w - 5]$ $\phantom{-}2e^{7} - 25e^{5} + 100e^{3} - 129e$ 89 $[89, 89, \frac{1}{3}w^{3} + \frac{2}{3}w^{2} - 3w - 3]$ $\phantom{-}\frac{1}{2}e^{7} - 2e^{5} - 13e^{3} + \frac{97}{2}e$ 97 $[97, 97, \frac{2}{3}w^{3} - \frac{1}{3}w^{2} - 6w + 5]$ $-2e^{6} + 19e^{4} - 45e^{2} + 6$ 97 $[97, 97, -w^{3} - \frac{5}{3}w^{2} + 10w + 15]$ $-e^{4} + 8e^{2} - 11$ Display number of eigenvalues ## Atkin-Lehner eigenvalues Norm Prime Eigenvalue $7$ $[7,7,-\frac{1}{3}w^{2} + w + 1]$ $1$	1,535	2,687	{"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}	2.9375	3	CC-MAIN-2020-50	latest	en	0.177068
https://ask.csdn.net/questions/250118	1,531,777,544,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00334.warc.gz	597,296,559	16,211	lvyouzi 于 2016.04.16 14:47 提问 #include main() { int a,b,c,d,e,s,w,y1,y2,y3,year; printf("请输入任意年、月、日（以空格隔开）\n"); scanf("%d%d%d",&a,&b,&c); if((year%4==0 && year%100==0) \|\| year%400==0) { s=a-1; y1=s/4; y2=s/100; y3=s/400; w=(s-y1+y2-y3)365+(y1-y2+y3)366+1; if(b==1) {c=31; e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==2) {c=28; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==3) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==4) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==5) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==6) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==7) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==8) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==9) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==10) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==11) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==12) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } `````` } else `````` { s=a-1; y1=s/4; y2=s/100; y3=s/400; w=(s-y1+y2-y3)365+(y1-y2+y3)366+1; printf("请输入月份："); scanf("%d",&b); if(b==1) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==2) {c=29; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==3) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==4) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==5) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==6) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==7) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==8) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==9) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==10) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==11) {c=31; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } if(b==12) {c=30; printf("请输入日期："); scanf("%d",&c); e=w+b+c; printf("俩年份之间差%d天",&e); } `````` } `````` } 2个回答 qq423399099 2016.04.16 15:12 if((year%4==0 && year%100==0) \|\| year%400==0) 应该改成if((year%4==0 && year%100!=0) \|\| year%400==0) qq423399099 printf("俩年份之间差%d天",&e);应该是printf("俩年份之间差%d天",e); 2 年多之前回复 u012155923 2016.04.16 16:19 C语言实验：输入任意一个日期的年、月、日的值，求出从公元1年1月1日到这一天总共有多少天，并求出这一天是星期几。 16.根据输入的日期输出星期几（已知公元1年1月1日星期一） #include #include using namespace std; int months[]={31,28,31,30,31,30,31,31,30,31,30,31}; int IsLeap( int year) { if((year%4==0&&year%100!=0)\|\|(year%100==0&&year%400==0)) { return 1;//闰年 C 已知1900年1月1日是星期一，输入某年某月某日，求星期几 #include void main() { int y,m,d,n; int a[13],i,sum1,sum2,sum; scanf("%d%d%d",&y,&m,&d); sum1=0;sum2=d;sum=0; for(i=1900;i { if(i%4==0&&i%100!=0\|\|i%400==0) n=366; else n=365; sum1=su /#include #include using namespace std; bool isLeapYear(int year) //判断是否为闰年 { return (year % 4 == 0 && year % 100 != 0) \|\| (year % 400 == 0); } int main() ural1759_计算从公元元年1月1日到某年某月某日的天数 <br />#include <iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; struct node { int y,m,d; }; struct live { node birth,death; }p[101]; int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; bool leap(int y) / c++实现万年历，从公元1年1月1日开始，这一天是星期一 //从公元元年一月一日是星期一开始计算 //用类实现求万年历的算法 #include #include #include using namespace std; class Calendar { private: int month; int date; string Day[7];//char Day[7][10];//私有成员不能初始化，而且字符数组要想 #include int main() { int year=0,month=0,weekday=0,sum=0; printf("请输入年和月：\n"); scanf("%d %d",&year,&month); for(int y=1900;y if((y%4==0 && y%100!=0)\|\|y%40 2. 计算出1900年1月1日至当前日期之间总共间隔多少天，将其除以7取余数，该结果即为当月第一天是星期几，按日历格式输出 //输入一个日期，2010-10 输出该月的月历： //******************************** // 10月 //******************************** //日一二三四五六 // 1 2 //3 4 5 6 7 8 9 //10 11 12 13 14 15 16 //17 18 19 20 21 22 23 / JAVA代码计算1900年到所求年份共多少天。 import java.util.Scanner;public class test11{public static void main(String[]args){ Scanner in=new Scanner(System.in); System.out.println("请输入所查找的年份："); int year=in.nextInt(); int sum=0; for(int i =1... Java实现1900年1月1日到2016年5月7日一共多少天？ public class Date2 { public static void main(String[] args){ int sum=0; for(int year=1900;year<=2015;year++){ if(year%4==0 && year%100!=0 \|\| year%400==0){	2,196	4,689	{"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-2018-30	longest	en	0.135298
http://www.chegg.com/homework-help/questions-and-answers/suppose-you-put-100-into-a-savings-account-today-the-account-pays-a-nominal-annual-interes-q3472066	1,368,963,837,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368697442043/warc/CC-MAIN-20130516094402-00067-ip-10-60-113-184.ec2.internal.warc.gz	370,537,669	8,044	Suppose you put $100 into a savings account today, the account pays a nominal annual interest rate of 6%, but compounded semiannually, and you withdraw$100 after 6 months. What would your ending balance be 20 years after the initial $100 deposit was made? Answer A)$115.35 B)$62.91 C)$226.20 D)$36.97 E)$9.50 F)$3.00 ## Answers (7) • a • E)$9.50 • E)$9.50 • after 6 months balance =103-100=$3 after 20 yrs 31.03^39 =$9.5 ans:E • E)$9.50 • E)$9.50 • balance after 6 months = 100+0.06100/2 =$103 hence balance remaining after withdrwal of $100 =$3 remaining periods = 20*2-1 = 39 balance after 20 years =FV(0.06/2,39,0,-3) = \$9.50 ---option (E) Get homework help	238	670	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2013-20	latest	en	0.845625
https://bytes.com/topic/c/answers/479039-empty-class	1,560,679,682,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00082.warc.gz	376,242,360	8,107	424,948 Members \| 795 Online Need help? Post your question and get tips & solutions from a community of 424,948 IT Pros & Developers. It's quick & easy. # empty class P: n/a can any budy tell me why the size of the empty class object is 1 thank you kapil kaushik Apr 11 '06 #1 7 Replies P: n/a dost wrote: can any budy tell me why the size of the empty class object is 1 Because in C++, every object has a unique address. Now consider what happens if you make an array of zero-sized objects. Apr 11 '06 #2 P: n/a In article <11********************@g10g2000cwb.googlegroups .com>, "dost" wrote: can any budy tell me why the size of the empty class object is 1 Objects exist, therefore they take up space. sizeof tells you how big they are. If an object were to have sizeof == 0 then it would not exist. -- Magic depends on tradition and belief. It does not welcome observation, nor does it profit by experiment. On the other hand, science is based on experience; it is open to correction by observation and experiment. Apr 11 '06 #3 P: n/a Just imagine how bad a programmers life will be if size of empty class is 0. U created a many objects of an empty classes and all are having same address!!! Link to wht stroustrup says... http://public.research.att.com/~bs/b...l#sizeof-empty ~K2G dost wrote: can any budy tell me why the size of the empty class object is 1 thank you kapil kaushik Apr 11 '06 #4 P: n/a if the size of empty class object is "1"....and if we declare some..variable..in it suppose int then its..size change....only the size of the variable will matter...where the difault size gone example class empty { int a;} its object size will be 2.....where the difault size gone thanks Apr 11 '06 #5 P: n/a On 11 Apr 2006 03:16:37 -0700, "dost" wrote in comp.lang.c++: can any budy tell me why the size of the empty class object is 1 The size of an empty class object is not necessarily 1. It is merely required to be greater than 0. -- Jack Klein Home: http://JK-Technology.Com FAQs for comp.lang.c http://c-faq.com/ comp.lang.c++ http://www.parashift.com/c++-faq-lite/ alt.comp.lang.learn.c-c++ http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html Apr 12 '06 #6 P: n/a In article <11********************@j33g2000cwa.googlegroups .com>, "dost" wrote: if the size of empty class object is "1"....and if we declare some..variable..in it suppose int then its..size change....only the size of the variable will matter...where the difault size gone example class empty { int a;} its object size will be 2.....where the difault size gone thanks Who says there ever was a default size? The size of an object is either 1 or the sum of the sizeof each contained element (possibly plus some padding,) whichever is greater. -- Magic depends on tradition and belief. It does not welcome observation, nor does it profit by experiment. On the other hand, science is based on experience; it is open to correction by observation and experiment. Apr 12 '06 #7 P: n/a Jack Klein wrote: On 11 Apr 2006 03:16:37 -0700, "dost" wrote in comp.lang.c++: can any budy tell me why the size of the empty class object is 1 The size of an empty class object is not necessarily 1. It is merely required to be greater than 0. -- Size of an empty class object is >= 1. However, an empty base-class component of a derived classes can be size 0. Apr 12 '06 #8 ### This discussion thread is closed Replies have been disabled for this discussion.	911	3,445	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2019-26	latest	en	0.844989
https://aptitude.gateoverflow.in/3855/cat1996-183?show=3903	1,508,325,575,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00419.warc.gz	652,133,166	16,282	13 views How old is Sachin in 1997? 1. Sachin is 11 years younger than Anil whose age will be prime number in $1998$. 2. Anil’s age was a prime number in $1996$. 1. if both the statements together are insufficient to answer the question. 2. if any one of the two statements is sufficient to answer the question. 3. if each statement alone is sufficient to answer the question. 4. if both the statements together are sufficient to answer the question, but neither statement alone is sufficient. recategorized \| 13 views 1) Using Statement (1), We cannot find the age of Sachin in 1997.NOT SUFFICIENT 2) Using Statement (2), We cannot find the age of Sachin in 1997 ;NOT SUFFICIENT taking 1 and 2 together Since range of prime numbers is not given, we cannot calculate the age of Sachin even using both the statement together ;NOT SUFFICIENT Statement (1) and Statement(2) together is not sufficient. Hence option A is correct. answered by (2.5k points) 1 7 28 +1 vote	254	978	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.359375	3	CC-MAIN-2017-43	latest	en	0.905997
https://transum.org/Maths/Exam/Question.asp?Q=689	1,721,864,045,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00349.warc.gz	494,424,788	7,378	# Exam-Style Question on Correlation ## A mathematics exam-style question with a worked solution that can be revealed gradually ##### List Of QuestionsExam-Style QuestionMore Correlation QuestionsMore on this Topic Question id: 689. This question is similar to one that appeared on a GCSE Higher paper in 2022. The use of a calculator is allowed. In the Bubbleless Bandits free-diving club, members practice breath-holding techniques as part of their routine. In this club, the older members have been practicing these techniques for many years, leading to increased breath-holding capacity through disciplined training and improved technique. The older members tend to be the more experienced free-divers. They have mastered relaxation and breath control, contributing to their ability to hold their breath for longer periods. The times, $$t$$ seconds, that club members could hold their breath while diving and their ages, $$x$$ years, are recorded. The results are shown in the table. Age, $$x$$ years Time, $$t$$ seconds 19 37 31 44 55 101 33 63 29 55 61 110 40 78 35 71 (a) Draw a scatter diagram to represent this data. (b) What type of correlation is shown on the scatter diagram? (c) Find the equation of the regression line giving your answer in the form $$t = ax + b$$. (d) Use your regression equation to estimate the time a person aged 45 could hold their breath. (e) Give a reason why you should not use the regression equation to estimate the time it would take a person aged 14 to do the task. The worked solutions to these exam-style questions are only available to those who have a Transum Subscription. Subscribers can drag down the panel to reveal the solution line by line. This is a very helpful strategy for the student who does not know how to do the question but given a clue, a peep at the beginnings of a method, they may be able to make progress themselves. This could be a great resource for a teacher using a projector or for a parent helping their child work through the solution to this question. The worked solutions also contain screen shots (where needed) of the step by step calculator procedures. A subscription also opens up the answers to all of the other online exercises, puzzles and lesson starters on Transum Mathematics and provides an ad-free browsing experience. Drag this panel down to reveal the solution If you need more practice try the self-checking interactive exercises called Estimating Correlation. The exam-style questions appearing on this site are based on those set in previous examinations (or sample assessment papers for future examinations) by the major examination boards. The wording, diagrams and figures used in these questions have been changed from the originals so that students can have fresh, relevant problem solving practice even if they have previously worked through the related exam paper. The solutions to the questions on this website are only available to those who have a Transum Subscription. Exam-Style Questions Main Page Search for exam-style questions containing a particular word or phrase: To search the entire Transum website use the search box in the grey area below. ## Comments: Do you have any comments about these exam-style questions? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments. For Students: For All: ©1997-2024 WWW.TRANSUM.ORG ©1997 - 2024 Transum Mathematics :: For more exam-style questions and worked solutions go to Transum.org/Maths/Exam/	742	3,602	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.390625	3	CC-MAIN-2024-30	latest	en	0.947298
https://www.usefullinks.org/cat/Discrete_groups.html	1,723,159,221,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00409.warc.gz	819,612,403	4,873	# Category: Discrete groups Lattice (group) In geometry and group theory, a lattice in the real coordinate space is an infinite set of points in this space with the properties that coordinate wise addition or subtraction of two points in the la Schottky group In mathematics, a Schottky group is a special sort of Kleinian group, first studied by Friedrich Schottky. Crystallographic point group In crystallography, a crystallographic point group is a set of symmetry operations, corresponding to one of the point groups in three dimensions, such that each operation (perhaps followed by a transl Layer group In mathematics, a layer group is a three-dimensional extension of a wallpaper group, with reflections in the third dimension. It is a space group with a two-dimensional lattice, meaning that it is sym Automorphic function In mathematics, an automorphic function is a function on a space that is invariant under the action of some group, in other words a function on the quotient space. Often the space is a complex manifol Discrete group In mathematics, a topological group like G is called a discrete group if there is no limit point in it (i.e., for each element in G, there is a neighborhood which only contains that element). Equivale Space group In mathematics, physics and chemistry, a space group is the symmetry group of an object in space, usually in three dimensions. The elements of a space group (its symmetry operations) are the rigid tra Arithmetic Fuchsian group Arithmetic Fuchsian groups are a special class of Fuchsian groups constructed using orders in quaternion algebras. They are particular instances of arithmetic groups. The prototypical example of an ar Frieze group In mathematics, a frieze or frieze pattern is a two-dimensional design that repeats in one direction. Such patterns occur frequently in architecture and decorative art. Frieze patterns can be classifi Fibrifold In mathematics, a fibrifold is (roughly) a fiber space whose fibers and base spaces are orbifolds. They were introduced by John Horton Conway, Olaf Delgado Friedrichs, and Daniel H. Huson et al., who Fuchsian group In mathematics, a Fuchsian group is a discrete subgroup of PSL(2,R). The group PSL(2,R) can be regarded equivalently as a group of isometries of the hyperbolic plane, or conformal transformations of t Non-Euclidean crystallographic group In mathematics, a non-Euclidean crystallographic group, NEC group or N.E.C. group is a discrete group of isometries of the hyperbolic plane. These symmetry groups correspond to the wallpaper groups in Kleinian group In mathematics, a Kleinian group is a discrete subgroup of the group of orientation-preserving isometries of hyperbolic 3-space H3. The latter, identifiable with PSL(2, C), is the quotient group of th Congruence subgroup In mathematics, a congruence subgroup of a matrix group with integer entries is a subgroup defined by congruence conditions on the entries. A very simple example would be invertible 2 × 2 integer matr Paramodular group In mathematics, a paramodular group is a special sort of arithmetic subgroup of the symplectic group. It is a generalization of the Siegel modular group, and has the same relation to polarized abelian Wallpaper group A wallpaper is a mathematical object covering a whole Euclidean plane by repeating a motif indefinitely, in manner that certain isometries keep the drawing unchanged. To a given wallpaper there corres Rod group In mathematics, a rod group is a three-dimensional line group whose point group is one of the axial crystallographic point groups. This constraint means that the point group must be the symmetry of so Index group In operator theory, a branch of mathematics, every Banach algebra can be associated with a group called its abstract index group. Line group A line group is a mathematical way of describing symmetries associated with moving along a line. These symmetries include repeating along that line, making that line a one-dimensional lattice. However Ahlfors finiteness theorem In the mathematical theory of Kleinian groups, the Ahlfors finiteness theorem describes the quotient of the domain of discontinuity by a finitely generated Kleinian group. The theorem was proved by La Mahler's compactness theorem In mathematics, Mahler's compactness theorem, proved by Kurt Mahler, is a foundational result on lattices in Euclidean space, characterising sets of lattices that are 'bounded' in a certain definite s Superrigidity In mathematics, in the theory of discrete groups, superrigidity is a concept designed to show how a linear representation ρ of a discrete group Γ inside an algebraic group G can, under some circumstan Local rigidity Local rigidity theorems in the theory of discrete subgroups of Lie groups are results which show that small deformations of certain such subgroups are always trivial. It is different from Mostow rigid Affine root system In mathematics, an affine root system is a root system of affine-linear functions on a Euclidean space. They are used in the classification of affine Lie algebras and superalgebras, and semisimple p-a Geometric group action In mathematics, specifically geometric group theory, a geometric group action is a certain type of action of a discrete group on a metric space. Ping-pong lemma In mathematics, the ping-pong lemma, or table-tennis lemma, is any of several mathematical statements that ensure that several elements in a group acting on a set freely generates a free subgroup of t	1,211	5,498	{"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}	3.265625	3	CC-MAIN-2024-33	latest	en	0.91574
https://www.jiskha.com/questions/27566/can-some-one-please-help-with-this-question-find-the-probability-that-in-four-tosses-of-a	1,642,436,345,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00697.warc.gz	887,009,945	4,903	# statistics Can some one please help with this question? Find the probability that in FOUR tosses of a fair die a 3 appears a)exactly no (zero)time b)exactly three times 1. 👍 2. 👎 3. 👁 1. a) (5/6)^4 = 625/1296 = 0.48225 b) (1/6)^3 (5/6)4 = 20/1296 = 0.01543 (The 4 comes from the fact that there are 4 possibilities for which toss is not a 3) 1. 👍 2. 👎 ## Similar Questions 1. ### Math 10. A coin is loaded so that the probability of heads is 0.55 and the probability of tails is 0.45, Suppose the coin is tossed twice and the results of tosses are independent. a.What is the probability of obtaining exactly two 2. ### Math The table shows the results of rolling a number cube labeled one through six 50 times. Number Rolled Frequency 1 7 2 9 3 11 4 6 5 9 6 8 1. What is the experimental probability of rolling a 3? 0.11 0.22 0.30 ** 0.27 2. Based on the 3. ### Math A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads? b) use your answer in a) to find 4. ### Math "The probability of getting heads on a biased coin is 1/3. Sammy tosses the coin 3 times. Find the probability of getting two heads and one tail". I thought that all you have to do is: (1/3)(1/3)(2/3) It makes sense to me, but 1. ### Math A number cube is rolled 100 times. The results are shown in the table below. Outcome: 1, 2, 3, 4, 5, 6. Number of times rolled: 22, 18, 9, 11, 19, 21. Find the experimental probability and express it as a percent. P(even) = ? A. 2. ### Math The table below shows the results of flipping two coins. how does the experimental probability of getting at least one tails compare to the theoretical probability of getting at least one tails Outcome: HH HT TH TT # of times 3. ### Math Determine the probability of 3 sixes in 5 tosses on a fair die. 4. ### Math There is a 100 question multiple choice test each with 5 choices. You randomly guess each question. Set up the binomial probability formula to find the probability of getting exactly 80 correct out of 100. Check to see if you can 1. ### statistics an experiment consists of 4 tosses of a coin.denoting the outcomes HHTH,THTT,..and assuming that all 16 outcomes are equally likely, find the probability distribution for the total number of heads. 2. ### probability As in the previous exercise, let È be the bias of a coin, i.e., the probability of Heads at each toss. We assume that È is uniformly distributed on [0,1]. Let K be the number of Heads in 9 independent tosses. We have seen that 3. ### math I do not know how to answer this question. Would someone please help me with each of the four question below? Thank you. Question 1. Write the sample space for the outcomes of tossing three coins using H for heads and T for tails. 4. ### Math A number cube is rolled 100 times. The results are shown in the table below. Outcome: 1, 2, 3, 4, 5, 6. Number of times rolled: 22, 18, 9, 11, 19, 21. Find the experimental probability and express it as a percent. P(even) = ? A.	897	3,111	{"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-2022-05	latest	en	0.91616
https://www.tothesquareinch.com/products/7th-grade-math-worksheets	1,725,998,911,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00309.warc.gz	978,572,812	51,902	All To The Square Inch Resources can be purchased on TPT or on this site. # 7th Grade Math Worksheets ## 7th Grade Math Worksheets Regular price \$20.00 Regular price Sale price \$20.00 Sale Sold out Shipping calculated at checkout. Couldn't load pickup availability 7th Grade Math Worksheets by Topic This set of 60 printable worksheets is a perfect way to give your students additional practice on all 7th grade math topics. Each page has 10 practice questions. This aligns with the To the Square Inch 7th Grade Math Curriculum and follows the Common Core Standards. It is easy to mix and match sets to meet your individual standards. Topics covered in this product: 1. Adding Integers with the Same Sign* 2. Adding Integers with Different Signs* 3. Subtracting Integers* 4. Multiplying and Dividing Integers* 5. Terminating and Repeating Decimals* 6. Adding Rational Numbers* 7. Subtracting Rational Numbers* 8. Multiplying and Dividing Rational Numbers* 9. Ratios & Rates* 10. Proportions* 11. Graphing Proportional Relationships* 12. Writing and Solving Proportions * 13. Slope* 14. Direct Variation* 15. Percents, Decimals, and Fractions* 16. Percent Proportion* 17. Percent of Increase and Decrease* 18. Discounts and Mark-Ups* 19. Simple Interest* 20. Tax, Tip and Commissions* 21. The Percent Equation* 22. Adjacent and Vertical Angles* 23. Complementary and Supplementary Angles* 24. Types of Triangles* 25. Constructing Triangles 26. Angle Measures of Triangles* 27. Types of Quadrilaterals* 28. Angle Measures of Quadrilaterals * 29. Similar Shapes* 30. Scale Drawings* 31. Circle Vocabulary* 32. Circumference* 33. Area of Circles* 34. Area of Composite Figures Involving Circles* 35. Surface Area of Prisms* 36. Surface Area of Pyramids* 37. Surface Area of Cylinders* 38. Volume of Prisms* 39. Volume of Pyramids* 40. Volume of Cylinders* 41. Algebraic Expressions* 42. Adding and Subtracting Linear Expressions* 43. Solving Equations Using Addition and Subtraction* 44. Solving Equations Using Multiplication and Division* 45. Solving Multi-Step Equations* 46. Writing and Graphing Inequalities 47. Solving Inequalities Using Addition and Subtraction* 48. Solving Inequalities Using Multiplication and Division* 49. Writing and Solving 2-Step Inequalities * 50. Populations and Samples* 51. Estimating Populations* 52. Measures of Central Tendency* 53. Measures of Variation * 54. Outcomes and Events* 55. Intro to Probability* 56. Experimental Probability* 57. Theoretical Probability* 58. Fundamental Counting Principle * 59. Compound Events* 60. Independent and Dependent Events* Answer keys are included for each worksheet. Copyright © To The Square Inch LLC All rights reserved by the author. Permission to copy for single classroom use only. ## Shipping All products are delivered digitally via email after checkout. View full details	681	2,868	{"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-2024-38	latest	en	0.737627
http://www.mathisfunforum.com/viewtopic.php?id=17315&p=1	1,495,783,586,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00120.warc.gz	728,174,254	16,674	Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° You are not logged in. ## #1 2012-01-20 09:53:18 wintersolstice Real Member Registered: 2009-06-06 Posts: 125 ### Super Happy Numbers I've been studying these numbers for some time (though not that much) I am atm only looking by exhaustion techniques aided only by spreadsheet formulae apart from the trivial number trap of I only know of two cycles and have so far only tested up to 12! the two cycles are 5365 7034 6056 6736 5785 10474 5493 11565 4451 4537 3394 9925 10426 693 8685 14621 2558 3989 9442 10600 37 1369 4930 3301 1090 8200 6724 5065 6725 5114 2797 10138 1446 2312 673 and 7450 7976 12017 690 8136 7857 9333 9738 10853 2874 6260 7444 7412 5620 3536 2521 1066 4456 5072 7684 12832 1809 405 41 1681 6817 4913 2570 5525 3650 3796 10585 7251 7785 13154 3878 7528 6409 4177 7610 5876 9140 9881 16165 7947 8450 9556 12161 4163 5650 5636 4432 2960 4441 3617 1585 there could be others. Definition: Step 1. You take a number and check the parity (state of being odd or even) of its digits, Step 2a if it has an even number of digits then leave it alone Step 2b if it has an odd number of digits put a zero at the front (left hand side) btw this won't change the value! Step 3 now that the number has an even number of digits split it up in to sets of two Step 4a if any of the sets has a zero as the left digit remove the zero so it becomes single digit number Step 4b if the left hand digit is non-zero leave it as a two digit number Step 5 this list of 1 and 2 digit number are "intervals, square the "intervals" and write them dowm step 6 add the squares together to make a new number Step 7 go back to Step 1 with this new number. If the number iterates to 1 it is "Super Happy" e.g. here are a set of number and there "intervals 123456 - 12,34,56 12345 - 1,23,45 120345 - 12,3,45 123045 - 12,30,45 hopefully that makes sense apart from the trivial cases: 1 and the powers of 10 I haven't found any yet! Why did the chicken cross the Mobius Band? To get to the other ...um...!!! Offline ## #2 2012-01-20 10:30:09 wintersolstice Real Member Registered: 2009-06-06 Posts: 125 ### Re: Super Happy Numbers I've just found 4 Super Happy Numbers 1030 1826 3010 2618 and another cycle! 2228 1268 4768 6833 5713 3418 1480 6596 13441 2838 2228 1268 4768 6833 5713 3418 1480 6596 13441 2838 Edit: 608 and 806 are Super happy! Last edited by wintersolstice (2012-01-20 10:33:45) Why did the chicken cross the Mobius Band? To get to the other ...um...!!! Offline ## #3 2012-01-30 01:25:28 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi wintersolstice, I've just found 4 Super Happy Numbers 1030 1826 3010 2618 If I understand the problem correctly, numbers that have an even number of digits (Step 2a) are invalid...which would count those four numbers out. And if that rule also applies to cycles, that would invalidate the cycles you found (I think). I found 635 Super Happy Numbers below 10,000,000. are the first 100 of them. I haven't looked for any cycles. Last edited by phrontister (2012-01-31 00:54:20) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #4 2012-02-01 00:25:59 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi wintersolstice, I couldn't find any cycles if step 2a applies. I forget what number I tested up to, but I think it was 100,000 or so. However, I found 125 cycles (including your three...2228, 5365 & 7450) for numbers under 10000 if step 2a doesn't apply. they are. I used a search depth of 100, but the deepest was 56 for #41. 1233 & 8833 only needed a depth of 1. I also tried increasing the search depth to 1000, but that didn't find any more cycles. Last edited by phrontister (2012-02-01 00:52:23) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #5 2012-02-01 08:38:43 wintersolstice Real Member Registered: 2009-06-06 Posts: 125 ### Re: Super Happy Numbers You've slightly misunderstood step 2a it doesn't mean "even digits are invalid" I'll reword it Step 2 if the number has an even number of digits go straight to step 3, otherwise add a "0" at the beginning (so there are an even number of digits) then go to step 3 does that make sense?:D btw what method are you using how did you work them out so fast? and you've made a massive mistake with cycles! Shuffling them about like this 1,2,3,4,5 and 2,3,4,5,1 and 3,4,5,1,2 etc are not not classed as new cycles:D different order but the same numbers so your 41 cycle is identical to my second cycle have a look it's the same numbers just shuffled about and 37 is part of my first cycle so thats also been counted I think you need to filter out duplicates because the same cycle reordered doesn't count as different btw "super happy numbers" were my own invention:D whether they've been studied I don't know Last edited by wintersolstice (2012-02-01 08:59:06) Why did the chicken cross the Mobius Band? To get to the other ...um...!!! Offline ## #6 2012-02-01 12:47:09 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi wintersolstice, Looks like I've misunderstood a couple of things...I'll check them out and get back to you. I've found 12 different cycles, with each cycle starting with its lowest number: Have to go...back later. Last edited by phrontister (2012-02-01 12:50:51) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #7 2012-02-01 21:54:08 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi wintersolstice Ok...I think I've got the idea now re the steps. I've trimmed them down a bit to make it easier for me to read (I hope I haven't changed any rules by doing that!).... Step 1. If n has an odd number of digits, add leading zero to n. Step 2. Split n into sets ("intervals") of 2-digit numbers. Step 3. Remove any leading zeros from the intervals. Step 4. Square the intervals. Step 5. Sum the squares. Step 6. Repeat from Step 1, using this sum. If the number iterates to 1 it is "Super Happy" Using an iteration limit of 10, I found 244 Super Happy Numbers below 100,000:- Increasing the number of iterations to 50 produced no more answers. btw what method are you using how did you work them out so fast? I wrote a program in BASIC using the LibertyBASIC software program, testing all numbers below 100,000 against the rules. Running time to find those 244 numbers was about 90 seconds, which would be pretty slow compared to other languages and programs like Mathematica - none of which I know. I'm sure a good BASIC programmer would do better (I just potter around), and running it through a compiler would improve the speed too. Last edited by phrontister (2012-02-01 21:59:43) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #8 2012-02-02 01:38:42 wintersolstice Real Member Registered: 2009-06-06 Posts: 125 ### Re: Super Happy Numbers phrontister wrote: Hi wintersolstice btw what method are you using how did you work them out so fast? I wrote a program in BASIC using the LibertyBASIC software program, testing all numbers below 100,000 against the rules. Running time to find those 244 numbers was about 90 seconds, which would be pretty slow compared to other languages and programs like Mathematica - none of which I know. I'm sure a good BASIC programmer would do better (I just potter around), and running it through a compiler would improve the speed too. well I know absolutly nothing about programming whatsoever:(:( but have longed to learn how to do it. this is just one of hundreds of things I can't do because I don't even know the basics of pogramming:( can you offer any help? (anything would be appreciated:D) I had the idea of extending to "n-Hyper Happy numbers" aswell Why did the chicken cross the Mobius Band? To get to the other ...um...!!! Offline ## #9 2012-02-02 04:59:00 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers I bought a Sharp PC1500A hand-held computer years ago just as a hobby thing. Nothing fancy (no games or apps, except very basic ones)...mainly number manipulation to help me solve tricky (for me) maths puzzles (but not too advanced, because I flunked high school). Its language is BASIC, which I learnt from the user manual and by trial and error...but only ever just enough to get by. I sometimes still use it, but for my PC I use LibertyBASIC. There is a scaled-down freeware version of it on the net called JustBASIC (google it), which you could download and install if you want. If you do, here's my program for the "Super Happy Numbers" part of your puzzle: Just highlight my program, copy it and paste it into JustBASIC. Press the blue 'Run' arrow to start, and in the popup window enter the largest number you want to test and the number of iterations (ie, search depth) you want it to perform. If you want to understand my program you'll have to do some learning. JustBASIC has a tutorial and quite good help files that should get you through...but it will take a while to get there. For the cycles part of your puzzle I run a separate BASIC program that has just a few small changes from the SHNumbers program, but I want to refine it a bit before posting it. Also, I had to copy/paste the results into Excel to sort it and assess the outcome because I was running short of time, and it was quicker for me to use Excel for that bit than to work out what code to write. BASIC may or may not be the way to go for you...depending on what it is you want to do. All programming languages have their good and bad points and application strengths and weaknesses (etc, etc), and I can't advise you on that score because I just don't know enough about it. I don't know if there's any info here on MIF that could help you, but if not, there's scads of it on the web. Btw, what are "n-Hyper Happy numbers"? Last edited by phrontister (2012-02-03 10:55:42) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #10 2012-02-02 08:41:58 wintersolstice Real Member Registered: 2009-06-06 Posts: 125 ### Re: Super Happy Numbers well I'll try and find some info on "JustBasic" (never heard of it!LOL) it doesn't matter that you can't help very much:D btw a "n-Hyper Happy number" is a generalisation I invented just after the "Super Happy numbers" basically you break the number down into intervals of "n" (normal happy numbers n=1, super happy numbers n=2) you would need to add zeros to the beginning of a number (if necessary) to make sure it has a number of digits which is a multiple of "n" the same rules apply:D Why did the chicken cross the Mobius Band? To get to the other ...um...!!! Offline ## #11 2012-02-03 00:35:53 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers I've made a couple of changes to the program in my last post, and I've updated that post. 1. I added an option to enter an iteration number of your choice. 2. After the program has run you're now given an option to run further tests, which saves having to run a new program each time. Also, I've finished refining the cycles program, and here it is... It doesn't filter out duplicates, so that would have to be done by hand (eg, in Excel, which is what I used). I still haven't figured out an automated way for my program to do that, but my initial thought is that it would be very tricky and time-consuming to work out and so I might let that go. What I did in Excel: - Copied the LibertyBASIC output into Excel in separate columns (had to copy/paste cycles individually). - Sorted each one from low to high with 'MultiSort' (a macro I downloaded from here that sorts multiple columns individually). - Sorted the columns from low to high with Excel's Sort function and manually deleted all duplicates. That then left me the results shown in post #6. Last edited by phrontister (2012-02-03 12:26:28) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #12 2012-02-06 04:24:57 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi wintersolstice, After I'd said that running the program through a compiler would improve the speed I thought I might as well try it. So I ran it in QB64 (a freeware BASIC compiler), which blitzed through it in less than an 11th of the original time! That 90 seconds I quoted in post #7 shrank to just 8 seconds. I'll only be using QB64 for programs where I want to shorten the running time, though, because it doesn't have debugging capabilities...which would make it difficult to iron out programming errors, particularly in complex programs. A couple of other downsides to QB64 are: - it only prints the result as an image (like some other compilers do) - which makes using the output rather difficult - but I fixed that by printing the result to a text file; and - it is compatible with QBASIC but not LiberyBASIC, which means I had to change some code...but not much. Last edited by phrontister (2012-02-06 05:13:08) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #13 2012-02-06 17:00:19 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Super Happy Numbers Hi, I found 24718 such numbers below 10 000 000 Here are the first 10 and last 10: 10 100 103 301 367 608 806 1000 1030 1826 9996183 9996217 9997037 9997637 9998002 9998316 9998361 9998790 9999087 9999737 I used C and gcc. "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #14 2012-02-06 23:13:13 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi gAr, I also found 24718, without the 'trivial' 1...which I think is a SHN because it satisfies all the rules. My program (LibertyBASIC + QB64) took 299 seconds to find them. How does your C + gcc speed compare? I have no idea about other languages than BASIC regarding their efficiency, capabilities and speed etc. Before I found QB64 I installed GLBCC, which is a gcc specifically for LibertyBASIC. However, although it seemed to install ok (I followed the installation instructions to the letter) I can't get it to open any of my .bas programs, no matter what I try. Repeating the installation didn't work either...so I gave up on it. Does your gcc print results as a copyable text file or as an image? Last edited by phrontister (2012-02-06 23:14:32) "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #15 2012-02-06 23:25:46 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Super Happy Numbers Hi phrontister, Okay, I haven't tried BASIC. The C program took 42.817 seconds to get the output, but it can be made faster with optimization. I printed to the terminal and saved it to a file using "redirection" "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #16 2012-02-06 23:43:29 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Hi gAr, As each solution was found I printed it directly to a file on my desktop, but at the same time I also got it to display on QB64's output screen so that I could check progress. Would you mind posting your code? I'd be interested to see if your speed gain is due to your better code and/or C vs BASIC. Probably both! I can't promise to make sense of your C code, but I am hopeful (maybe too optimistically) because someone once gave me some C code for a program I'd written in BASIC, and I was able to adapt and improve my program from that. My BASIC code is in post #9. "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #17 2012-02-06 23:57:52 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Super Happy Numbers Hi phrontister, Yeah, sure, here's the code. ``````#include <stdio.h> int countDigits(long n) { int cnt = 0; while(n) { n /= 10; cnt++; } return cnt; } int shn(long n) { long orig = n, cycle = 50, temp = 0; int dgtcnt; while(cycle) { dgtcnt = countDigits(n); while(!(dgtcnt == 0 \|\| dgtcnt == -1)) { temp += (n%100)*(n%100); n /= 100; dgtcnt -= 2; } n = temp; temp = 0; if (n == orig) { return 0; } if (n == 1) { printf("%ld\t\t%ld\n", orig, 50-cycle); return 1; } cycle--; } return 0; } int main() { long i; for(i=2;i<10000000;i++) { shn(i); } return 0; }`````` If you know one procedural language, you can easily learn another without much effort! "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #18 2012-02-07 00:29:57 phrontister Real Member From: The Land of Tomorrow Registered: 2009-07-12 Posts: 4,568 ### Re: Super Happy Numbers Thanks for that, gAr. I'll look at that more closely when I get some more time...getting ready for bed soon. I've always been disappointed with the lack of digit and string manipulation and checking functions in BASIC, and as my main use for writing code is to solve puzzles I've been wondering if I should try learning another language that would suit that purpose better. Do you think that C would be a good choice for puzzle solving? Or some other language maybe? I don't want to depart too much from what I already know and to have to spend a lot of time in learning something that's very different from BASIC...if I can get away with it! "The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson Offline ## #19 2012-02-07 01:20:41 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Super Happy Numbers Hi phrontister, You're welcome. Have a good night. If you like syntax of BASIC, you may also like python. Start with v2.7 Also, it's a lot easier than C. You need not worry about data types, and has a lot of useful packages. I use C only when there's a need for speed, when it contains lot of function calls and loops, since python's not a compiled language. "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #20 2012-02-08 10:18:58 GrimReaper Banned Registered: 2012-02-08 Posts: 10 ### Re: Super Happy Numbers gAr BASIC is Much better than python python is Kinda Hard to Learn. Offline ## #21 2012-02-08 10:20:05 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Super Happy Numbers Hi; Not really, python is easy too. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #22 2012-02-08 10:21:14 GrimReaper Banned Registered: 2012-02-08 Posts: 10 Offline ## #23 2012-02-08 10:22:03 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Super Happy Numbers No, BASIC is pretty much dead. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #24 2012-02-08 10:23:28 GrimReaper Banned Registered: 2012-02-08 Posts: 10 ### Re: Super Happy Numbers Ha Ha Ha Loads of Normal Peopl Use it...Do you av a Problem with that ? Offline ## #25 2012-02-08 10:24:47 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Super Happy Numbers Hi GiB; Of course not. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline	6,108	21,441	{"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-2017-22	latest	en	0.750213
http://forums.eviews.com/viewtopic.php?f=4&t=1862&start=15&view=print	1,601,016,263,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400222515.48/warc/CC-MAIN-20200925053037-20200925083037-00094.warc.gz	46,888,494	3,501	Page 2 of 2 ### Re: Seemingly Unrelated Regressions and robust covariance matrix Posted: Fri Aug 27, 2010 9:56 am One possibility is that you are using automatic bandwidth selection. The equation by equation automatic selection will probably give you different lags than doing the lag selection on the system of equations. ### Re: Seemingly Unrelated Regressions and robust covariance ma Posted: Wed Sep 26, 2012 3:47 pm EViews Glenn wrote:I guess I need to be a bit clearer on what you want (and what I mean ). The seemingly unrelated refers to the fact that you have a set of equations with no apparent cross-equation restrictions, but with non-zero off-diagonals. For the purposes of this discussion there are couple of ways to proceed: . You can estimate the specification using a GLS approach which corrects for cross-sectional heteroskedasticity and contemporaneous correlation (but not for general heteroskedasticity and serial correlation). In principle, you could follow this with a robust standard estimator. . Alternately, you can estimate using system least squares without correlation correction and then compute with a robust standard estimator, for example a system HAC estimator. EViews does not allow you to take the former approach, but does allow you to do the latter using the GMM tools. Note that the equivalence results from treating all of the explanatory variables in your specification as exogenous. Just as TSLS using the original regressors as instruments yields the least squares estimator, so too does GMM with the appropriate orthogonality conditions and weighting matrix (what is termed in the dialog TSLS weighting), yield the system least squares estimator. Then all you have to do is to select the appropriate robust covariance option. Glenn, thanks for your suggestion but I have encountered to another problem when I tried to do system GMM. In short, suppose I have a system of two equations (SUR setup) and when I do system GMM (with 2SLS & GMM Robust SE), the p-value of every estimated coefficient exploded to 0.9+. Which is odd to me because when I do either single equation OLS with HAC or single equation GMM with HAC individually (as expected, they matched almost exactly), many coefficients are significant with practically 0 p-value. Does it imply something about the cross-sectional correlation between residuals? Any suggestions of how to proceed? ### Re: Seemingly Unrelated Regressions and robust covariance ma Posted: Mon Oct 22, 2012 6:11 pm john_stranger wrote:Glenn, thanks for your suggestion but I have encountered to another problem when I tried to do system GMM. In short, suppose I have a system of two equations (SUR setup) and when I do system GMM (with 2SLS & GMM Robust SE), the p-value of every estimated coefficient exploded to 0.9+. Which is odd to me because when I do either single equation OLS with HAC or single equation GMM with HAC individually (as expected, they matched almost exactly), many coefficients are significant with practically 0 p-value. Does it imply something about the cross-sectional correlation between residuals? Any suggestions of how to proceed? I'm encountering exactly the same issue. As soon as I switch to a GMM system, the p-values explode into insignificance. Whereas every other estimation technique delivers estimates where the coefficients and p-values aren't dissimilar. Very frustrating. ### Re: Seemingly Unrelated Regressions and robust covariance ma Posted: Tue Oct 23, 2012 1:30 pm HAC for the system is different than HAC equation by equation. As these posts have all noted, there are now cross-equation covariances to worry about. How many observations do you have? ### Re: Seemingly Unrelated Regressions and robust covariance ma Posted: Tue Oct 23, 2012 4:47 pm EViews Glenn wrote:HAC for the system is different than HAC equation by equation. As these posts have all noted, there are now cross-equation covariances to worry about. How many observations do you have? Aah, probably not enough. 30 for 1 equation, 32 for the other. ### Re: Seemingly Unrelated Regressions and robust covariance matrix Posted: Mon Apr 17, 2017 4:22 pm Hi, I am trying to estimate a two equation SUR system sys1 as follows y c1 x1 z c2 x2 how do I specify the system and estimate sys1.sur? I can't find a sample program in the manual to run sur. thank you so much, Howard ### Re: Seemingly Unrelated Regressions and robust covariance matrix Posted: Mon Apr 17, 2017 4:51 pm Did you type "sur" into the help system? ### Re: Seemingly Unrelated Regressions and robust covariance matrix Posted: Tue Apr 18, 2017 10:56 am yes. all I see in the help manual are running sur interactively. I have many equations and must write a program. I am looking for a sample program in the manual illustrating the codes for specifying a simple two equation system of sur. thank you very much. ### Re: Seemingly Unrelated Regressions and robust covariance matrix Posted: Tue Apr 18, 2017 11:09 am Typing sure into the index in the help system gives you the command to estimate a system by sur. The help under system gives the various commands needed to set up the system. Look at Declare and Append specification line.	1,198	5,218	{"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-2020-40	latest	en	0.920705
https://m.hanspub.org/journal/paper/27073	1,638,057,527,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00609.warc.gz	458,520,147	17,923	一种短时序Kalman滤波决策优化预测新方法 # 一种短时序Kalman滤波决策优化预测新方法A New Decision Optimization Prediction Method Based on Short-Term Time Series and Kalman Filter Abstract: In order to make scientific and agile decision, a new short-term prediction TS_KF (Time Se-ries-Kalman Filter) method is proposed based on the combination of time series analysis and Kalman Filter. Aiming to improve the prediction precision and decrease the calculation complexity, a prediction model is built using Kalman filter to describe the prediction process, and the auto regression for time series analysis is utilized to renew and optimize the state transfer matrix, which is the key parameter for Kalman Filter. A coal production prediction test is conducted by comparison with some typical time series prediction method, and the results show that the TS_KF prediction method in this paper has significantly enhanced the prediction precision while keeping the same low calculation complexity. The result gives a strong proof for the effectiveness of the new TS_KF prediction method. 1. 引言 2. TS_KF预测方法 2.1. 预测模型框架 TS_KF预测方法以Kalman滤波预测模型为基础，预测模型能保持Kalman滤波本身的迭代收敛特性，在使用过程中通过样本的积累实现逐步收敛。预测模型框架如图1 Figure 1. Predictive model framework Table 1. Comparison of different prediction methods • 不必进行前期大量样本值的训练，而将此过程融入到运行过程中，减少了模型应用过程中的前期准备工作量； • 在使用过程中可以对样本值进行滤波，减小样本本身的误差，这是其它模型没有实现的。 2.2. 预测模型的构建 $\left\{\begin{array}{l}{x}_{{}_{k}}=A{x}_{k-1}+B{u}_{k}+{w}_{k-1}\\ {z}_{k}=H{x}_{k}+{v}_{k}\end{array}$ (1) $\left\{\begin{array}{l}{w}_{k-1}~N\left(0,Q\right)\\ {v}_{k}~N\left(0,R\right)\end{array}$ (2) $\left\{\begin{array}{l}{x}_{k}=a{x}_{k-1}+{w}_{k-1}\\ {z}_{k}={x}_{k}+{v}_{k}\end{array}$ (3) ${\stackrel{˜}{y}}_{t}={\varphi }_{1}{\stackrel{˜}{y}}_{t-1}+{\varphi }_{2}{\stackrel{˜}{y}}_{t-2}+\cdot \cdot \cdot +{\varphi }_{p}{\stackrel{˜}{y}}_{t-p}+\xi$ (4) ${\stackrel{˜}{y}}_{k}={\varphi }_{1}{\stackrel{˜}{y}}_{k-1}+\xi$ (5) $\mu$${y}_{k}$ 的期望水平的情况下，将 ${\stackrel{˜}{y}}_{k}={y}_{k}-\mu$ 代入到(5)，可得AR(1)的以下等效过程： ${y}_{k}={\varphi }_{1}{y}_{k-1}+\tau +\xi$ (6) $\left\{\begin{array}{l}\stackrel{¯}{y}=\frac{1}{S}{{\sum }_{i=0}^{S}y}_{k-i}\\ {\varphi }_{1}=\frac{{{\sum }_{i=0}^{S-1}{y}_{k-i}y}_{k-i-1}-\left(S-1\right){\stackrel{¯}{y}}^{2}}{{\sum }_{i=0}^{S-1}{y}_{k-i-1}^{2}-\left(S-1\right){\stackrel{¯}{y}}^{2}}\\ \tau =\left(1-{\varphi }_{1}\right)\stackrel{¯}{y}\end{array}$ (7) $\left\{\begin{array}{l}{x}_{{}_{k}}=a{x}_{k-1}+\beta +{w}_{k-1}\\ {z}_{k}={x}_{k}+{v}_{k}\end{array}$ (8) $\left\{\begin{array}{l}{\sigma }^{2}={{\sum }_{i=0}^{S}\left[{\stackrel{^}{y}}_{i}-\left({\varphi }_{1}{\stackrel{^}{y}}_{i-1}+\tau \right)\right]}^{2}\\ {R}_{k}=diag\left({\sigma }^{2}\right)={\sigma }^{2}\end{array}$ (9) 2.3. 算法描述 Figure 2. Predictive algorithm flow 3. 实验研究 Table 2. Comparison experimental parameter setting of different prediction methods Table 3. A detailed list of predicted statistical data Table 4. Error statistics of different prediction models Figure 3. Comparison of forecast effect of coal yield using various predictive methods 4. 结论 1) 综述了决策中现有的典型预测方法，阐述了不同预测方法的应用场合和相互关系。统计时间序列方法主要用于短期预测，是决策敏捷性的重要支撑，因此着重对短期时间序列预测进行了研究。 2) 为提升决策的敏捷性和科学性，以短时段数据的时间序列为研究对象，提出了将时间序列模型和Kalman滤波模型相结合的TS_KF预测方法，并从预测准确性和计算复杂度上优化预测方法。TS_KF方法具有模型简洁、有效、计算复杂度低等特点。 3) 以煤产量数据作为时间序列样本，采用与典型的时间序列预测方法(MA和ES)对比的方法，设计了实验验证算例，以验证TS_KF方法的有效性。结果表明，在保持计算复杂度同水平的前提下，TS_KF方法能极大的提高预测的准确性。 NOTES *通讯作者。 [1] 杨庆芳, 赵小辉, 郑黎黎. 基于模型预测控制的环形交叉口信号配时方法[J]. 浙江大学学报(工学版), 2018, 52(1): 117-124. [2] Pinto, R. and Gaiardelli, P. (2013) Setting Forecasting Model Parameters Using Unconstrained Direct Search Methods: An Empirical Evaluation. Expert Systems With Applications, 40, 5331-5340. https://doi.org/10.1016/j.eswa.2013.03.044 [3] 埃文斯. 数据、模型与决策[M]. 北京: 中国人民大学出版社, 2012. [4] Dunn, D.W., Buelow, J.M., Austin, J.K., et al. (2004) Development of Syndrome Severity Scores for Pediatric Epilepsy. Epilepsia, 45, 661-666. https://doi.org/10.1111/j.0013-9580.2004.53903.x [5] Whitehead, D. (2008) An International Delphi Study Examining Health Promotion and Health Education in Nursing Practice, Education and Policy. Journal of Clinical Nursing, 17, 891-900. https://doi.org/10.1111/j.1365-2702.2007.02079.x [6] Weiss, M., Junginger, M., Patel, M.K., et al. (2010) A Review of Expe-rience Curve Analyses for Energy Demand Technologies. Technological Forecasting and Social Change, 77, 411-428. https://doi.org/10.1016/j.techfore.2009.10.009 [7] Hsu, L. and Wang, C. (2007) Forecasting the Output of Integrated Circuit Industry Using a Grey Model Improved by the Bayesian Analysis. Technological Forecasting and Social Change, 74, 843-853. https://doi.org/10.1016/j.techfore.2006.02.005 [8] Yuan, M. and Lin, Y. (2006) Model Selection and Estimation in Regression with Grouped Variables. Journal of The Royal Statistical Society Series B-Statistical Methodology, 68, 49-67. https://doi.org/10.1111/j.1467-9868.2005.00532.x [9] Yu, H.L. and Macgregor, J.F. (2003) Multivariate Image Analysis and Regression for Prediction of Coating Content and Distribution in the Production of Snack Foods. Chemometrics and Intelligent La-boratory Systems, 67, 125-144. https://doi.org/10.1016/S0169-7439(03)00065-0 [10] Atiya, A.F. (2001) Bankruptcy Prediction for Credit Risk Using Neural Networks: A Survey and New Results. IEEE Transactions on Neural Networks, 12, 929-935. https://doi.org/10.1109/72.935101 [11] Choi, T.M., Yu, Y. and Au, K.F. (2011) A Hybrid SARIMA Wavelet Transform Method for Sales Forecasting. Decision Support Systems, 51, 130-140. https://doi.org/10.1016/j.dss.2010.12.002 [12] Min, S.H., Lee, J. and Han, I. (2006) Hybrid Genetic Algorithms and Support Vector Machines for Bankruptcy Prediction. Expert Systems with Ap-plications, 31, 652-660. https://doi.org/10.1016/j.eswa.2005.09.070 [13] Sheikhan, M. and Mohammadi, N. (2012) Neural-Based Electricity Load Forecasting Using Hybrid of GA and ACO for Feature Selection. Neural Computing & Applications, 21, 1961-1970. https://doi.org/10.1007/s00521-011-0599-1 [14] Hadavandi, E., Shavandi, H., Ghanbari, A., et al. (2012) Developing a Hybrid Artificial Intelligence Model for Outpatient Visits Forecasting in Hospitals. Applied Soft Computing, 12, 700-711. https://doi.org/10.1016/j.asoc.2011.09.018 [15] Carpinone, A., Giorgio, M., Langella, R. and Testa, A. (2015) Markov Chain Modeling for Very-Short-Term Wind Power Forecasting. Electric Power Systems Research, 122, 152-158. https://doi.org/10.1016/j.epsr.2014.12.025 [16] Chudik, A., Grossman, V. and Hashem Pesaran, M. (2016) A Multi-Country Approach to Forecasting Output Growth Using PMIs. Journal of Econometrics, 192, 349-365. https://doi.org/10.1016/j.jeconom.2016.02.003 [17] 宗群, 赵占山, 商安娜. 基于季节自回归单整移动平均模型的电梯交通流递归预测方法[J]. 天津大学学报, 2008, 41(8): 653-659. [18] 吴德会. 动态指数平滑预测方法及其应用[J]. 系统管理学报, 2008, 17(2): 151-155. Top	2,654	6,829	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2021-49	latest	en	0.500808
https://cheapcarinsurancekey.com/carinsurance2/car-insurance-most-affordable-usaa-car-insurance-qa.html	1,581,920,166,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00200.warc.gz	321,247,904	3,297	Calculable loss: There are two elements that must be at least estimable, if not formally calculable: the probability of loss, and the attendant cost. Probability of loss is generally an empirical exercise, while cost has more to do with the ability of a reasonable person in possession of a copy of the insurance policy and a proof of loss associated with a claim presented under that policy to make a reasonably definite and objective evaluation of the amount of the loss recoverable as a result of the claim. Health insurance is now available to more Americans than ever before. Subsidized options are easily available to low-income individuals and families. In the past, many people took the risk of not being insured, but with the Affordable Care Act (ACA) you can be fined if you don't have qualified health care insurance. Instead of paying a fine, people who have not been able to afford insurance before are looking for affordable medical insurance options. ###### An insurance underwriter's job is to evaluate a given risk as to the likelihood that a loss will occur. Any factor that causes a greater likelihood of loss should theoretically be charged a higher rate. This basic principle of insurance must be followed if insurance companies are to remain solvent.[citation needed] Thus, "discrimination" against (i.e., negative differential treatment of) potential insureds in the risk evaluation and premium-setting process is a necessary by-product of the fundamentals of insurance underwriting.[citation needed] For instance, insurers charge older people significantly higher premiums than they charge younger people for term life insurance. Older people are thus treated differently from younger people (i.e., a distinction is made, discrimination occurs). The rationale for the differential treatment goes to the heart of the risk a life insurer takes: Old people are likely to die sooner than young people, so the risk of loss (the insured's death) is greater in any given period of time and therefore the risk premium must be higher to cover the greater risk.[citation needed] However, treating insureds differently when there is no actuarially sound reason for doing so is unlawful discrimination. Car insurance helps provide financial protection for you, and possibly others, if you are involved in an accident. Auto insurance for your vehicle is usually required by all states. Review our state car insurance guides to see the different laws and auto insurance minimum required coverages. You can customize your auto policy to fit your needs. There are different coverages and limits you can choose to create your auto insurance policy. Check out our car insurance coverage page to learn more about each type of coverage.	514	2,741	{"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}	2.65625	3	CC-MAIN-2020-10	latest	en	0.958882
https://www.edaboard.com/threads/pulse-motor-self-start-on-a-certain-direction.402422/	1,701,932,623,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00582.warc.gz	819,528,322	18,495	Continue to Site # Pulse motor self start on a certain direction? Status Not open for further replies. #### neazoi Hi I have seen this video on youtube (I do not know how else to attach this here) Which shows a so called "pulse motor" with self-starting capability. The schematic is shown inside the video. I think what the schematic does is to somehow set the current applied to the coil into plus or minus alternatively, until the motor starts. However the motor starts at a random direction based on the rotor stationary position. I wonder if you have any ideas to make the motor start at a certain direction only? Interference is a thing I want to tackle, so this motor seems to produce no serious RF interference? Motors for radio-controlled craft often are called 3-turn, 5-turn, etc. With an odd number (not just 1 or 2) of coils, it ensures proper spin direction. ### neazoi Points: 2 Agreed, just reversing polarity gives equal pull in opposite directions so unless you give momentum it would just sit there or could even flip back and forth. Adding another force 'off axis' give the phase shift needed to provide a rotating rather than alternating magnetic field. Something to try: start a microwave oven a few times and see which direction the food rotates. Brian. ### neazoi Points: 2 How about placing the coil a bit off center? And flip polarity on the begining? The field runs through the coil core so it doesn't matter if you move it off center, by misaligning them you just reduce the pull it gives the rotor. You can do it easily if you add a second coil at some angle to the original and pulse them out of phase and it also gives you greater torque. Brian. ### neazoi Points: 2 You can do it easily if you add a second coil at some angle to the original and pulse them out of phase and it also gives you greater torque. That's very interesting. I wonder though if you can really start always in one direction with this. I mean, doesn't the starting direction depend on the resting point of the magnet in this case as well? I would love to see a simple discrete circuit on this phase pulsating for 2 coils. Maybe a two series transistors amplifiers, one triggering the other but some kind of small capacitor in the base of the second one, to somehow delay the second pulse? Status Not open for further replies.	516	2,347	{"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-2023-50	latest	en	0.933132
https://magoosh.com/hs/sat/sat-video-post/2016/5-sat-math-tricks-every-student-should-know/	1,590,992,906,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00532.warc.gz	437,713,455	28,480	Math can be tricky. Here’s your chance to turn the tables and apply these easy-to-apply tricks that will let you rule the SAT math section as opposed to letting the SAT math section rule you. While you’re at it, you should really take the time to dig through the the Magoosh High School Blog, as it is chock full of refreshers of foundational concepts that you learned in your algebra and geometry classes as well as SAT-specific strategies that you probably didn’t learn in your algebra or geometry classes. Don’t worry if you come across content that was written before the SAT redesign – a lot of it still applies! At the same time, keep checking this blog for updates, as we are working hard to make sure you get the most up-to-date SAT info! 1. Cross-multiply to find the greater fraction If you’re presented with finding the greater of two fractions and you’re not sure which one it is, cross-multiplying can ease that confusion. Draw an X from the numerator of each fraction to the denominator of the other fraction, multiply, and write the answer next to the corresponding numerator; the fraction with the greater value is bigger. In the graphic below, we use this method to determine that 5/7 is greater than 2/3. 2. “Greater than” vs. “Less than” I’m slightly embarrassed to admit this but to this day, I have to use a mnemonic to remember which direction goes the “less than” sign and which direction goes “greater than.” If you are in the same boat (no shame if you are), here are two ways you can remember: • The alligator always eats the bigger number. This makes sense if you turn the inequality sign into an alligator (see below). I learned this in the 5th grade and it has never, ever let me down. • Or you can just rely on the perks of being a millennial and growing up with “less-than-three.” By simply remembering that <3 uses the “less than” sign, you should be good to go! Wow, that 3 is a jerk. 3. Divisibility rules Knowing divisibility rules – that is, patterns that all multiples of a given number share – can save you time, particularly on the no-calculator section, when you need to find factors of larger numbers. Without even realizing it, you probably already knew the divisibility rules for the numbers 2 (its multiples are always even), 5 (its multiples always end in 5 or 0), and 10 (its multiples always end in 0). But what happens if you come across a problem like this on the SAT: Find a, if b is a positive integer greater than 10 and ab = 57 Assuming you don’t immediately recognize potential factors for 57, this is where the 3-9 divisibility test can come in handy. To check if a number is divisible by 3 or 9, all you have to do is mentally add up its digits and see if they add up to a smaller multiple of 3 or 9. If they add up to a multiple of 9, then the number is divisible by both 3 and 9; if they add up to a multiple of 3, then the number is only divisible by 3. For example: 57: 5 + 7 = 12 (multiple of 3, therefore divisible by 3, but not by 9) Then you would divide 57 by 3 to see that it’s 19. Because b is greater than 10, b has to be 19, making a = 3. Or you use logic to figure out a has to be 3 just by doing the divisibility test. Note: if 57 didn’t pass the 3-9 divisibility test, chances are it’s probably a multiple of 7. Although there are enough divisibility rules to make your head spin, the ones we mention in this post will be sufficient for you to make your factoring that much easier on the SAT. A lot of students have a tough time remembering the quadratic formula and unfortunately, it’s not one of the formulas that the SAT Math test has in the beginning of the two math sections. Fortunately, there’s a nifty song to the tune of “Pop Goes The Weasel” that has saved countless students on their math tests (see for yourself in the comments): And in case “Pop Goes The Weasel” is not exactly your jam, there are so many versions on YouTube that are more hip with the times. Just make sure you’re singing in your head on test day. 5. How to remember trig functions SOHCAHTOA. If don’t know what this is, you’re probably wondering why I’m yelling in a foreign language. SOHCAHTOA is actually a mnemonic that can help you remember how to figure out the sine, cosine, and tangent of an angle from the sides of a right triangle – or vice versa. Now that trig concepts are being tested on the SAT, check out how to use SOHCAHTOA, which will really come in handy for questions where you’re expected to use these functions to determine the length of a missing side. Know any other cool math tricks, shortcuts, or mnemonics that help you on SAT Math? Share them in the comments below! Anika is one of Magoosh’s Blog Editors. She makes sure the content across our blogs is error-free, easy to read, pleasing to the eye, and Google-friendly. Anika has ten years of experience in teaching and facilitating. She has taught English to language learners of all ages in places like Ecuador and Malaysia, has tutored high schoolers in SAT prep, and has led several youth empowerment programs. Anika earned her B.A. in Gender, Women's and Sexuality Studies from Grinnell College and her Masters in Public Policy from Harvard University. When she’s not scouring the web for the perfect gif for the blog or strategizing for educational equity, Anika can be found bingeing Netflix, searching Spotify for gems for her workout playlist, or obsessively reading the news. LinkedIn Magoosh blog comment policy: To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!	1,361	5,892	{"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}	4.25	4	CC-MAIN-2020-24	longest	en	0.945155
https://ataiva.com/longest-vowel-chain-in-java/	1,701,611,727,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00166.warc.gz	146,359,024	8,868	## The challenge The vowel substrings in the word `codewarriors` are `o,e,a,io`. The longest of these has a length of 2. Given a lowercase string that has alphabetic characters only (both vowels and consonants) and no spaces, return the length of the longest vowel substring. Vowels are any of `aeiou`. ## The solution in Java code Option 1: ``````1 2 3 4 5 6 7 `````` ``````import static java.util.stream.Stream.of; interface Solution { static int solve(String s) { return of(s.split("[^aeiou]")).mapToInt(String::length).max().orElse(0); } } `````` Option 2: `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 `````` ``````import java.util.ArrayList; class Solution{ public static int solve(String s){ var vowels = new ArrayList(); vowels.add('a'); vowels.add('e'); vowels.add('i'); vowels.add('o'); vowels.add('u'); var arr = s.toCharArray(); int max = 0; int counter = 0; for (int i = 0; i < arr.length ; i++) { if (vowels.contains(arr[i])){ counter++; }else{ if (counter>max){ max = counter; } counter = 0; } } if (counter>max){ max = counter; } return max; } } `````` Option 3: `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````class Solution{ public static int solve(String s){ //.. String[] vowelsTable = s.split("[bcdfghjklmnpqrstvwxyz]"); int longuestVowelSubString = 0; for (String mySubString: vowelsTable){ if (mySubString.length()>longuestVowelSubString){ longuestVowelSubString = mySubString.length(); } } return longuestVowelSubString; } } `````` ## Test cases to validate our solution `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 `````` ``````import org.junit.Test; import static org.junit.Assert.assertEquals; import org.junit.runners.JUnit4; public class SolutionTest{ @Test public void basicTests(){ assertEquals(3,Solution.solve("ultrarevolutionariees")); assertEquals(2,Solution.solve("codewarriors")); assertEquals(3,Solution.solve("suoidea")); assertEquals(1,Solution.solve("strengthlessnesses")); assertEquals(11,Solution.solve("mnopqriouaeiopqrstuvwxyuaeiouaeiou")); } } ``````	651	2,046	{"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-2023-50	latest	en	0.386595
http://stackoverflow.com/questions/7283273/swapping-sections-of-an-array-with-each-other-in-c	1,394,245,981,000,000,000	text/html	crawl-data/CC-MAIN-2014-10/segments/1393999652621/warc/CC-MAIN-20140305060732-00018-ip-10-183-142-35.ec2.internal.warc.gz	178,270,504	16,239	Swapping sections of an array with each other in C++ Is there an easy method to swap sections (chunks) of arrays with each other? That is, I have an array: ``````array[0] = 1; array[1] = 2; array[2] = 3; array[3] = 4; array[4] = 5; array[5] = 6; array[6] = 7; array[7] = 8; `````` and a function called `swapSections(startX, endX, startY, endY)` which pretty much given these values swaps out the range of values determined by `endX - startX` with the range of values from `StartY` to `endY`, so from my example... if `x range = 2` and `startX = 0` and `y range = 3` and `startY = 5`, it would place array[0] and array[1] to where array[5] and array[6] are, and then place array[7] after array[6], pushing everything else down one. I am not sure how to go about this, and I was physically copying the memory across to a temp array, but I think there is a better way to do this. (btw, the end result from my example would be): ``````array[0] = 6; array[1] = 7; array[2] = 8; array[3] = 3; array[4] = 4; array[5] = 5; array[6] = 1; array[7] = 2; `````` - Why not do your own homework and then you may learn something? That is the point. – Ed Heal Sep 2 '11 at 12:22 @Ed I have tried to do my homework but with no success, hence why I post. – nyaan Sep 2 '11 at 12:24 @nyaan Please post what you have tried so far. – balki Sep 2 '11 at 14:08 You might want to look at valarray and it's slices. - i have to use `int * array = new int[]; ` – nyaan Sep 2 '11 at 12:22 @downvoter - reason? – Kornel Kisielewicz Sep 2 '11 at 12:25 One approach you can take in your example is swapping the minimum range and then "bubbling up" the last part. So: ``````array[0] = 1; --> 6 array[1] = 2; --> 7 array[2] = 3; array[3] = 4; array[4] = 5; array[5] = 6; --> 1 array[6] = 7; --> 2 array[7] = 8; `````` Then you bubble up the 8 by swapping array[7], with array[6], then array[6] with array[5], etc. until you put the 8 in the right place. Hopefully that gets you started. - The easiest way to swap sections of an array in terms of readability and effort is to use the standard C++ function `swap_ranges()` ``````#include <iostream> #include <algorithm> int main() { int a[8] = {1,2,3,4,5,6,7,8}; std::cout << "Beforeswap: "; for(int i=0; i<8; ++i) std::cout << a[i] << ' '; std::cout << '\n'; std::swap_ranges(a+0, a+2, a+5); std::cout << "After swap: "; for(int i=0; i<8; ++i) std::cout << a[i] << ' '; std::cout << '\n'; } `````` ...but it will only swap subranges of equal length, not the unequal length as in your test case. Your case is actually a combination of `swap_ranges()` and `rotate()`. - Assume you have (or can create) a function to reverse a range of elements in place, something like `reverse(array, start, end)`. You can then perform this task in four steps: ``````// (1) reverse the first range array[0] = 2; array[1] = 1; // (2) reverse the elements between the ranges array[2] = 5; array[3] = 4; array[4] = 3; // (3) reverse the second range array[5] = 8; array[6] = 7; array[7] = 6; // (4) finally, reverse the entire array array[0] = 6; array[1] = 7; array[2] = 8; array[3] = 3; array[4] = 4; array[5] = 5; array[6] = 1; array[7] = 2; `````` If you are allowed to use the Standard Library there is a `std::reverse` function in `<algorithm>` that makes this trivial.	1,141	3,308	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2014-10	latest	en	0.862065
https://wiki.seg.org/wiki/Translations:Sinusoidal_motion/13/en	1,720,995,439,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00402.warc.gz	566,559,736	25,516	Translations:Sinusoidal motion/13/en At time n = 0, the vector lies in the positive direction along the horizontal coordinate axis. Then at some arbitrary time index n, the vector will make an angle of ${\displaystyle \omega n\Delta t}$ radians with the horizontal axis. The projections of this vector on the x- and y-axes give the horizontal and vertical components, respectively: ${\displaystyle \left({\rm {\ cos\ }}\omega n\Delta t{\rm {,\ \ sin\ }}\omega n\Delta t\right)}$. Let the vertical axis be the imaginary axis (i.e., a unit distance on the vertical axis is ${\displaystyle i{\rm {=}}{\sqrt {-{\rm {1}}}}}$). We can represent the vector at time ${\displaystyle n\Delta t}$ in terms of its components ${\displaystyle {\rm {\ cos\ }}\omega n\Delta t}$ and ${\displaystyle {\rm {\ sin\ }}\omega n\Delta t}$ as follows:	222	829	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.0625	3	CC-MAIN-2024-30	latest	en	0.69673
https://quicksilverforums.com/get-most-out-of-15grams-to-ounces/	1,713,301,043,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00841.warc.gz	427,087,234	17,950	Get most out of 15grams to ounces Are 15grams to ounces you ready to unleash the magic of conversion and elevate your culinary creations? In this blog post, we dive into the world of grams to ounces, unlocking the secrets to precise measurements and innovative cooking techniques. Whether you’re a seasoned chef or a baking enthusiast, mastering this conversion is key to achieving perfection in the kitchen. Get ready to explore how 15 grams can make a world of difference in your dishes! Understanding the Conversion: Why it Matters Have you ever wondered why understanding the conversion from grams to ounces is crucial in the culinary world? Well, let’s break it down. Grams and ounces are both units of weight measurement, but they come from different systems – metric and imperial. Being able to convert between the two allows you to follow recipes accurately, ensuring your dishes turn out as intended. Precision is key when it comes to cooking and baking. A slight miscalculation in measurements can make a big difference in the final outcome of your dish. Whether you’re working with delicate pastries or savory sauces, getting the right balance of ingredients is essential for flavor and texture. By mastering the conversion between grams and ounces, you open up a world of culinary possibilities. You’ll have more flexibility in trying out new recipes and experimenting with different flavor combinations. It’s a skill that every aspiring chef should have in their toolkit! How to Convert Grams to Ounces Let’s dive into the world of converting grams to ounces. It might sound tricky at first, but once you grasp the concept, it becomes second nature. To convert grams to ounces, you need to know that 1 ounce is equivalent to approximately 28.35 grams. So, all you have to do is divide the number of grams by 28.35 to get your answer in ounces. For example, if you have 100 grams and want to know how many ounces that is, simply divide 100 by 28.35. The result will be around 3.53 ounces. This conversion comes in handy when following recipes from different parts of the world or when dealing with ingredients measured in grams rather than ounces. Understanding this simple conversion can make a big difference in your cooking and baking adventures! Common Uses for Grams and Ounces Measurements When it comes to common uses for grams and ounces measurements, the kitchen is where these units shine. Whether you’re following a recipe or measuring out ingredients for a culinary creation, having a good grasp of these measurements is essential. In baking, precise measurements are crucial for achieving that perfect cake or batch of cookies. 15 grams of vanilla extract can elevate your dessert from ordinary to extraordinary with just a small amount. For those who love to cook savory dishes, grams and ounces come in handy when portioning out spices or seasonings. Adding just the right amount of salt (around 2.5-3 grams) can make all the difference in bringing out the flavors in your dish. Even in everyday tasks like meal prepping or portion control, understanding grams and ounces can help you maintain a balanced diet without overindulging. Weighing out ingredients like nuts or grains ensures you’re getting the right serving size without going overboard. So next time you’re in the kitchen, remember the versatility of grams and ounces – they might just be the key to unlocking your inner chef! Tips for Accurate Conversions When it comes to converting grams to ounces, precision is key. To ensure accurate conversions every time, start by using a reliable conversion chart or calculator. Double-check your calculations to avoid any errors that could impact your recipe. It’s important to pay attention to the unit of measurement used in the original recipe and make sure you convert all ingredients consistently. Remember that slight variations in measurements can affect the outcome of your dish. For more complex recipes, consider investing in a digital kitchen scale for precise measurements. This will help you accurately portion out ingredients like flour or sugar without relying solely on volume measurements. If you’re still unsure about conversions, don’t hesitate to reach out for help from online cooking communities or forums. Experienced cooks may offer valuable tips and tricks for converting between grams and ounces seamlessly. By following these tips and staying diligent with your conversions, you’ll be able to confidently tackle any recipe that calls for precise measurements in grams or ounces. Tools and Resources for Easy Conversions When it comes to converting grams to ounces, having the right tools and resources can make the process much easier. One handy tool is an online conversion calculator that 15grams to ounces allows you to quickly input the amount in grams and get the equivalent in ounces instantly. These calculators are convenient and accurate, saving you time and effort. Another useful resource is a conversion chart that lists common measurements for both grams and ounces. Having this chart on hand can help you easily reference different quantities without needing to do mental math each time. It’s a practical solution for those who prefer a visual aid when converting measurements. Additionally, there are smartphone apps available that specialize in unit conversions, including grams to ounces. These apps provide quick access to conversions on-the-go, 15grams to ounces making them perfect for busy cooks or bakers who need instant results while working in the kitchen. Creative Ways to Use 15 grams in Cooking and Baking Looking to add a touch of flavor to your dishes with just 15 grams? Get ready to elevate your cooking and baking game with these creative ideas! In baking, 15 grams of cocoa powder can take your chocolate cake from good to exceptional. Sprinkle some powdered sugar over the top for that perfect finish. For savory dishes, try using 15 grams of fresh herbs like basil or cilantro as a garnish. The aroma and freshness will enhance the overall taste of your dish. Experiment with spices by mixing 15 grams of paprika, cumin, 15grams to ounces and garlic powder for a homemade taco seasoning blend. Your tacos will thank you later! When it comes to desserts, infuse 15 grams of vanilla extract into whipped cream for a deliciously fragrant topping on pies or fruit salads. Get creative in the kitchen and let those 15 grams work their magic! Conclusion Whether you’re a seasoned chef or just starting out in the kitchen, understanding how to convert measurements from grams to ounces can make 15grams to ounces a world of difference in your cooking and baking endeavors. By mastering this simple conversion, you open up a whole new realm of culinary possibilities. From precise ingredient measurements to experimenting with new recipes, knowing how to accurately convert between grams and ounces is an essential skill for any home cook. So next time you come across a recipe that calls for 15 grams of an ingredient, don’t fret – armed with the knowledge from this article, you’ll be able to confidently measure it out in ounces and create delicious dishes with ease. Happy cooking!	1,390	7,209	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2024-18	longest	en	0.922019
https://nl.mathworks.com/matlabcentral/cody/problems/1837-find-the-index-of-the-largest-value-in-any-vector-x-4-3-4-5-9-12-0-4-5/solutions/996194	1,581,988,813,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00249.warc.gz	512,279,177	15,302	Cody # Problem 1837. Find the index of the largest value in any vector X=[4,3,4,5,9,12,0,4.....5] Solution 996194 Submitted on 5 Oct 2016 by Massimo Zanetti This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass x = [2 5 2 5 12 2 4 13]; y_correct = 8; assert(isequal(maxindex(x),y_correct)) 2 Pass x = [2 4 3 5 5 3]; y_correct = 4; assert(isequal(maxindex(x),y_correct))	170	489	{"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-2020-10	latest	en	0.666767
http://mathhelpforum.com/advanced-algebra/index16.html	1,498,396,635,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00063.warc.gz	260,364,499	15,140	Advanced Algebra Help Forum: Linear and Abstract Algebra 1. ### Make sure your post belongs here • Replies: 7 • Views: 4,747 Nov 25th 2012, 02:03 PM 1. ### Lie Algebra Bracket Operator • Replies: 0 • Views: 216 Oct 19th 2015, 07:39 AM 2. ### About the Jordan-Holder theorem as stated in a book by Rotman. • Replies: 6 • Views: 358 Oct 18th 2015, 08:59 PM 3. ### Composition series for direct product (groups). • Replies: 0 • Views: 328 Oct 18th 2015, 03:34 PM 4. ### they called a few buyers • Replies: 0 • Views: 117 Oct 17th 2015, 05:59 AM 5. ### Projection Transformation on x Axis Parallel to y=2x • Replies: 2 • Views: 299 Oct 17th 2015, 01:06 AM 6. ### Difficulty in proof of Zassenhaus lemma. • Replies: 0 • Views: 327 Oct 16th 2015, 03:30 PM 7. ### A finite group with exactly two conjugacy classes has order 2. • Replies: 6 • Views: 558 Oct 16th 2015, 08:38 AM 8. ### Limit at negative infinity? • Replies: 3 • Views: 285 Oct 16th 2015, 03:50 AM 9. ### these harsher realities for the first time • Replies: 0 • Views: 158 Oct 15th 2015, 04:19 PM 10. ### Linear or Exponential • Replies: 3 • Views: 260 Oct 15th 2015, 03:25 AM 11. ### Further Applications of Linear Equations • Replies: 1 • Views: 266 Oct 12th 2015, 09:03 PM 12. ### complement and null space • Replies: 2 • Views: 271 Oct 12th 2015, 07:10 PM 13. ### Find all subgroups • Replies: 4 • Views: 278 Oct 12th 2015, 04:30 PM 14. ### Cayley table question • Replies: 11 • Views: 948 Oct 10th 2015, 04:49 AM 15. ### Finding roots of a function using (Fixed Point Iteration) Method • Replies: 1 • Views: 269 Oct 9th 2015, 07:24 AM • Replies: 3 • Views: 216 Oct 8th 2015, 07:39 AM 17. ### Proof involving the trace on linear operators • Replies: 2 • Views: 221 Oct 8th 2015, 07:19 AM 18. ### Algorithm for eigenvalues and eigenvectors • Replies: 4 • Views: 271 Oct 4th 2015, 11:38 AM 19. ### Explicitly Proving Vector Spaces • Replies: 1 • Views: 185 Sep 29th 2015, 06:15 PM 20. ### Show Isomorphism Between Set Real Numbers • Replies: 6 • Views: 356 Sep 28th 2015, 04:47 AM 21. ### Intoduction and Boolean Algebra • Replies: 2 • Views: 253 Sep 27th 2015, 01:05 AM 22. ### Determinant always equal to zero? • Replies: 3 • Views: 294 Sep 25th 2015, 05:48 PM 23. ### Need help • Replies: 4 • Views: 2,916 Sep 23rd 2015, 09:02 PM 24. ### Application of linear programming • Replies: 4 • Views: 466 Sep 23rd 2015, 07:44 PM 25. ### Polynomial of the 4th degree. • Replies: 5 • Views: 319 Sep 23rd 2015, 06:20 PM • • • • • • , , , ### svmsung.j7.1218 , Click on a term to search for related topics. Use this control to limit the display of threads to those newer than the specified time frame. Allows you to choose the data by which the thread list will be sorted.	1,026	2,762	{"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-2017-26	longest	en	0.744737
https://rationalistramble.wordpress.com/2015/12/07/stopping-rules-p-values-and-the-likelihood-principle/	1,544,829,946,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376826530.72/warc/CC-MAIN-20181214232243-20181215014243-00224.warc.gz	717,863,507	33,239	## Stopping rules, p-values, and the likelihood principle A few months ago, someone who used to be called perversesheaf came to tumblr to bash LessWrong there. Now, while there is a very large number of criticisms that can be aimed at it, both as a website and community, this person decided to bash Bayesianism, which, as readers of this blog might have noticed, is something I personally believe is probably Correct. That person has since deleted their blog, but Scott’s tumblr still has the original post, which I’m going to reproduce here, in a fashion. But first, the basics. There is a principle in statistics called the Likelihood Principle whose basic content is that, “given a statistical model, all of the evidence in a sample relevant to model parameters is contained in the likelihood function.” You will remember that if we’re trying to estimate parametres $\boldsymbol \theta$ after observing some data $D$ then the likelihood function is the following quantity: $p(D \| \boldsymbol \theta, X)$ when seen as a function of $\boldsymbol \theta$. And in particular, any function that has the form $\mathcal L(\boldsymbol \theta; D) = y(D) p(D \| \boldsymbol \theta, X)$ where $y(D)$ is any function of the data alone and not of the parametres can be seen as a likelihood function. More intuitively, we say that a function is a probability in a case where we’re wondering about the outcome when the parametres are held fixed: “If a coin was tossed ten times (experiment) and it is fair (parametre), what’s the probability that it lands heads every time (outcome)?”; and it’s a likelihood in a case where we’re wondering about the parametres when the outcome is held fixed: “If a coin was tossed ten times (experiment) and it landed heads every time (outcome), what’s the likelihood that it is fair (parametre)?” So the likelihood principle, then, says that all the information a sample can give about a parametre is contained in the function $p(D\|\boldsymbol\theta, X)$. And if one believes that Bayes’ Theorem is the correct way to deal with uncertainty, then one must, necessarily, believe that the likelihood principle is true, since: $p(\boldsymbol\theta \| D, X) \propto p(D \| \boldsymbol\theta, X) p(\boldsymbol \theta \| X)$ That is, our posterior beliefs equals our prior beliefs times a likelihood function times a normalisation constant. Let’s look at the W’s practical example. Suppose I run two experiments: 1. A (possibly biased) coin was tossed $12$ times and came out heads $h$ times; 2. Another (possibly biased) coin was tossed until it came out heads $3$ times, and was tossed a total of $n$ times. I want to know how much information each of those experiments give me about the probabilities $\theta_i$ that each of those coins will come out heads. In both cases, by Bayes’ Theorem: $p(\theta_i \| D_i, X) = \frac {p(D_i \| \theta_i, X) p(\theta_i\|X)} {\int_0^1 p(D_i \| \theta_i, X) p(\theta_i\|X) d\theta_i}$ The two experiments are not the same, though, so even if their prior distributions $p(\theta_i \| X)$ are equal, their posteriors may not be. Let’s see what the likelihood function for either experiment is. The first experiment is just a series of $12$ independent Bernoulli trials, so our likelihood function is: $p(h \| \theta , X) = \binom {12} {h} \theta^{h}(1-\theta)^{12 - h}$ In the second experiment, we toss the second coin until we have observed $3$ heads. Therefore, our likelihood function is: $p(n \| \theta , X) = \binom {n-1} {2} \theta^3 (1 - \theta)^{n-3}$ Now, suppose I ran experiment 1 and observed $3$ heads. Then I ran experiment 2 and had to toss the coin $12$ times. The likelihood functions are respectively: \begin{aligned} p(h = 3\|\theta, X) &= \binom {12} {3} \theta^3(1-\theta)^{12-3} \\ &= 220 \theta^3 (1-\theta)^9 \\ \\ p(n = 12 \| \theta, X) &= \binom {12 - 1} {2} \theta^3 (1-\theta)^{12 - 3} \\ &= 55 \theta^3 (1-\theta)^9 \end{aligned} The two likelihoods are proportional, so the likelihood principle says we should draw the same conclusions from them. Now, if $p(\theta_1 \| X) \neq p(\theta_2 \| X)$, our posteriors for both cases won’t necessarily be the same, but the point is that the impact of both experiments in our beliefs about their respective $\theta_i$ is exactly the same in both cases. It intuitively makes sense, though. The difference between both of those experiments are in when we decided to stop, in advance. The actual results of both experiments were the same: $12$ tosses, $3$ heads. The fact that, in my mind, I had decided in advance I’d stop when this or that happened, has no bearing on our actual observed results, and thus should have no impact on our actual beliefs about the actual parametre $\theta_i$. Why did I write that last paragraph? Because in frequentist p-value analyses, those two outcomes are different. Consider the case where we’re trying to decide whether the null hypothesis that the coin is fair, $H_0: \theta = 0.5$, is true, as opposed to $H_1: \theta < 0.5$. A p-value analysis goes: 1. Set a significance level in advance of the experiment. Typical values are 0.05 and 0.01; 2. Calculate the probability that an outcome at least as extreme as the observed one would have happened if the null hypothesis was true; 3. If that probability is less than the significance level, reject the hypothesis; otherwise, nothing can be said. What are the p-values of both experiments? Inserting our values in the likelihood functions, we have: \begin{aligned} p(h \leq 3 \| \theta = 0.5, X) &= \sum\limits_{i=0}^3 \binom {12} {i} * 0.5^i * 0.5^{12-i} \\ &= 0.0730 \\ \\ p(n \geq 12 \| \theta = 0.5, X) &= \sum\limits_{i=12}^{+\infty} \binom {i - 1} {2} * 0.5^3 * 0.5^{i-3} \\ &= 0.0327\end{aligned} We calculated these probabilities because “at least as extreme” varies depending on which experiment we’re running, and which hypotheses we’re testing against each other. In the first experiment, observing $3$ or less heads after tossing the coin $12$ times is an extreme result for $H_0$ compared to $H_1$; in the second, having to toss the coin $12$ or more times when we’re just waiting for $3$ heads is the extreme result instead. Based on that, the first experiment doesn’t reject the null hypothesis but the second one does, at the 5% significance level. Even though they had the exact same outcome. Using this test, the prior, completely mental choice we made about which experiment to run, completely unrelated to our beliefs about $\theta$, would somehow influence our beliefs about whether $H_0$ or $H_1$ is true. I’ve spoken before about how Bayesianism can only compare hypotheses to each other. They don’t exist in a vacuum, and a Bayesian procedure would never outright tell you to “reject a hypothesis.” It can tell you which hypotheses are more or less likely, when you give it a set of possibilities, a prior distribution over these possibilities, and a likelihood. That’s all it can do. But frequentist statistics aims to do that, and for that it takes into account stuff that to a Bayesian has nothing to do with the experiment. It takes your mind states, prior to running the experiment, as part of the experiment. It takes possible distributions of data that haven’t happened. And then they criticise Bayesianism for being too subjective. Okay, I guess. We have to remember that frequentism and Bayesianism are different things, that answer different questions, whose basic object of study just happens to have the same name – probability – but is not the same thing at all. To a frequentist, it’s a limiting frequency; to a Bayesian, it’s a measure of uncertainty. They agree a lot, but sometimes they don’t. Now let’s see perversesheaf’s post, entitled “A problem with the likelihood principle”: As someone interested in both philosophy and mathematics, I am disturbed by the rise of what I call “radical Bayesian,” or the belief that the standard probability axioms and Bayes’ rule together give us a complete description of how we ought to reason about the world (at least in principle). I obviously blame Yudkowsky and LessWrong for the bulk of this, though similar lines of thought have been advocated by certain philosophers and seem to have trickled down to undergraduate philosophy majors who’ve never read LW. (I’d just like to pause this to snicker a bit at the fact that perversesheaf thinks LW is as big as that. Like, forreal, radical Bayesianism has been around for much longer, and the LW community is wayyyy too tiny to be responsible for even a percent of that. And in fact, as far as I know, Yudkowsky himself was influenced in no small part by other AI researchers, since Bayesianism seems to be the rule in the AI and Machine Learning community [EDIT: This might not be true].) In my view, the basic error here is not new and is also made by, for example, consequentialists and some of the more radical libertarians. First, take a complicated human endeavor (epistemology, morality, politics) and create a seductive, simple model that works on a variety of easy test cases. Second, wildly overgeneralize and claim this simple model constitutes a complete theory of the endeavor in question. Third, when presented with counterexamples that evoke strong intuitions that disagree with the results your model gives, don’t admit the model is incomplete or explain away those intuitions. Instead, dogmatically stick to your guns and claim those intuitions are normatively wrong in virtue of the fact that they disagree with your model (e.g., consequentialism and the dust speck thought experiment). (I’d like to point a few things out. First, the dust speck thought experiment is about utilitarianism, not consequentialism. The former is a (tiny, tiny) subset of the latter. Second, obviously for any consistent normative moral theory, there are going to be counterintuitive thought experiments, because intuitive moral theory is inconsistent. Third, this is true of every other attempt to create a consistent (even if not necessarily simple) model of normative anything, because human intuition is inconsistent in general, and morality and inference are just two parts of it where this is glaring.) It’s not that I don’t think Bayesian thinking (or consequentialist or libertarian thinking) has [anything] to offer. It’s incredibly useful, both practically and theoretically. Rather, I don’t think it gives a complete picture of how we ought to reason. There are good reasons to believe it doesn’t. But everyone likes simple, clear-cut answers. We desperately want to resolve the deep epistemological questions that weigh on our minds, and I think this need for resolution blinds us and allows us to be duped by purported grand unified theories of rationality (or morality, or politics, etc). So instead of what I regard as the sensible, moderate position regarding Bayesian methods — they’re often very useful but sometimes inappropriate — many people end up being radical Bayesians and believe that Bayes’ rule presents The Final Answer to the question of how we should reason about uncertainty. It is the latter position I disagree with. I think su3su2u1 (and nostalgebraist and a few others) have done an admirable job of pushing back against the extreme view. In particular, su3su2u1 often brings up difficulties with prior choice and how these are problematic for Bayesianism as a complete theory of reasoning. In this post, I want to raise a different problem that I haven’t seen on tumblr (at least among the blogs I read). (The discussion with su3su2u1 is part of what inspired me to write this post. Basically, my response to this is that Bayesianism is a philosophical position, but in full generality it is literally uncomputable, actually not even approximable, and to people who think that the correct way to Bayes is using it in all practical cases (both critics and adherents) I say: ha.) The problem has two parts. First, Bayesians are committed to the (strong) likelihood principle. Second, the (strong) likelihood principle is inadequate for reasoning about hypothesis testing situations where a stopping rule is involved. Deborah Mayo presents this argument in her book Error and the Growth of Experimental Knowledge (though it is not original to her — see the book for complete bibliographic information). I will present my own take on it, along with some mathematical justification she omitted. By the likelihood principle, I mean the claim that when performing inference about an unknown parameter, only the observed data matters, and all the relevant information is contained in the likelihood function. Regarding my first claim, any Bayesian is, by definition, committed to this principle. After all, their rule for inference is “posterior = prior x likelihood.” The data affect the inference only through the likelihood function. It’s worth comparing the likelihood principle to frequentist hypothesis testing to draw out the difference. Suppose I am working for a pharmaceutical company and asked to examine data from a drug trial to determine if the drug is effective. I am handed a notebook with 97 data points. If I wanted to do Bayesian hypothesis testing, my task is relatively straightforward. I draw up a model, find a prior, compute the likelihood, compute the posterior, then find (for example) a 95% percent highest posterior density interval. If this interval is entirely positive and doesn’t contain zero, then I conclude I should my belief that the drug is effective at (at least) 95% confidence. Bam, done. Computing a frequentist p-value is a little trickier because, in addition to setting up a model, I need to concern myself with not just the data, but various counterfactual possibilities: ways the data could have been. A p-value is, by definition, the probability of the observed data given that the null hypothesis is true. The null hypothesis in this case includes the testing procedure. So I need to know, for example, whether the trial was originally planned to have 97 samples, or if it was stopped after 97 samples because the company thought the data observed so far was sufficient to show the drug was effective. Each of these cases constitutes a different null hypothesis and gives rise to a different p-value. A concrete example of how these stopping rules matter, worked out in detail and contrasted to the Bayesian approach, is given on this Wikipedia page: [link]. For a Bayesian, how the decision was made to stop data collection is irrelevant. All that matters is the observed data. My goal is to show that this ignorance of the experimental design can lead to ridiculous, obviously unwarranted inferences. Suppose the certain drug company is unethical and wants to trick the FDA into thinking a placebo pill is actually an effective way to make people happier. (Suppose the placebo is given to subjects without any indication of what the drug is for, so that we may ignore the placebo effect.) To do this, the company hires a radical Bayesian statistician (who, by definition, will not concern themselves with stopping rules) and orders a trial of the drug to proceed as follows: the drug will be tested on subjects one by one, in order, and the experiment will be stopped only when the statistician has a 95% Bayesian credence that the drug is effective. (This number can be made arbitrarily close to 100.) (I wonder if it can really be made arbitrarily close to 100? I think so, but not totally sure.) You might wonder if this is always possible. After all, doesn’t the law of large numbers (or the central limit theorem, etc) show that, over time, the data should reflect the fact that the placebo has no effect? Yes, but under certain weak assumptions, it turns out there is enough variance in the experiment to make this unscrupulous stopping rule work every time, given an unlimited number of subjects. Let’s suppose that, because of natural variation in happiness, some subjects become happier after the drug is administered, and an equal number become less happy. In fact, let’s assume that this natural variation is normally distributed with mean 0 and variance 1 on some suitable quantitative scale of happiness. (This set-up obviously relies on some dubious assumptions about how happiness works, but you can ignore the storytelling if you want and just concentrate on the fact we’re doing inference on i.i.d. normally distributed random variables with unknown mean.) Let’s also assume that the Bayesian statistician models the drug’s effectiveness as a normal distribution with known variance 1 and unknown mean and starts with a non-informative flat (improper) prior on the mean. I want to pause here to emphasize that these assumptions on the model and prior are not necessary to make the argument work and only serve to make the math easier. The model could be unknown mean and unknown variance, for instance. And, if you don’t like improper priors, we could use a proper prior with really large variance to approximate an improper prior. Or we could use any normal prior, since various asymptotic results show the prior will “wash out” and become irrelevant when the sample size is large. We could even use certain more sophisticated models. But the essential point is the same. Returning to the simplified set up, let’s suppose we are midway through the experiment and have taken $n$ samples with mean happiness change $\overline{y}$. The statistician’s posterior distribution is then (cf. the third edition of Bayesian Data Analysis by Gelman et al., page 52, though the computation is straightforward and I’d be happy to provide the details if asked): $N\left(\overline{y}, \frac 1 n \right)$ A 95% Bayesian credence interval is then given by $[\overline y - 2/\sqrt{n}, \overline y + 2/\sqrt{n}]$. (I recently explained the difference between confidence and credible intervals, but one of the things mentioned there is that frequentist confidence and Bayesian credible intervals agree exactly in the above case.) It remains to show that there is always an $n$ such that $\overline{y} \ge \frac{2}{\sqrt{n}}$ If we define $S_n$ as the sum of the first $n$ random variables in an infinite sequence of independent, identically distributed random variables with distribution $N(0,1)$, this is equivalent to asking if there is always $n$ such that $\frac{S_n}{\sqrt{n}} \ge 2$ This in turn is equivalent to asking if the event $\limsup_n \frac{S_n}{\sqrt{n}} \ge 2$ has probability one. An argument using Kolmogorov’s zero-one law shows that it does. See here for details: [link]. So the evil drug company can always arrange the trial to trick the Bayesian statistician into inferring that the placebo has a positive effect. The drug company could even tell the statistician they will stop the trial only when they get favorable results, as this won’t affect the Bayesian’s inference. (Of course, if they let on that the drug is known to be a placebo and hence affect the statistician’s priors, they’re screwed.) (I will come back to this parenthesis later.) The upshot is that Bayesian reasoning can result in horrifically poor inferences when stopping rules are involved. I want to be explicit that my position is not “Bayesianism is flawed, therefore we should all be frequentists.” While it does accommodate stopping rules well, frequentism has problems of its own. Rather, my suggestion is that we should recognize that Bayesian reasoning, while useful, has certain shortcomings. We should not treat it as The Final Answer to statistical inference. That is, we should reject radical Bayesianism. Epistemology is hard. There aren’t simple answers. I think perversesheaf is wrong. We had a ~2 month discussion about it, and in the end he disappeared. I hope he’s okay. Anyway, let’s see what can be done here. His derivation is sound. That event does in fact happen with probability $1$. However, we need to flesh this out a bit more fully before we can deal with it properly. What does it mean to say that “the Bayesian was fooled”? I can think of two answers. One, “the Bayesian believes the drug has a positive effect when it does not with 95% confidence.” Two, “the Bayesian believes the drug’s effect is higher than it actually is with 95% confidence.” The former case can only be used if the drug really really has no effect at all, or has a negative effect, whereas the latter can be used at any point and it encompasses the former. In perversesheaf’s thought experiment, the drug’s effect was zero, but for now I’ll pretend I don’t actually know what the effect is. raginrayguns wrote a derivation for the upper bound of the probability that the Bayesian was fooled in the first sense. First, we note that: \begin{aligned} P\left(\lim \sup _n \frac {S_n} {\sqrt n} \geq 2 \right) &\leq P \left( \exists n : \frac {S_n} {\sqrt n} \geq 2 \right) \\ &= P \left( \bigvee\limits_{n=1}^{+\infty} \frac {S_n} {\sqrt n} \geq 2 \right) \\ & \leq \sum\limits_{n = 1}^{+\infty} P \left( \frac {S_n} {\sqrt n} \geq 2 \right) \end{aligned} Then we calculate that sum on the right-hand side at the bottom for various possible values of the true effect $\mu \leq 0$, and its plot is given by: So the upper bound drops below $1$ somewhere between $\mu = -0.07$ and $\mu-0.08$, and for $\mu < -0.6$ the effect becomes negligible. Now I’m going to derive an upper bound to the probability that the Bayesian will be fooled in the second sense for positive true effect $\mu$. Let $S_n(\mu)$ be the sum of the $n$ first samples from a $\mathcal N(\mu, 1)$ distribution. Perversesheaf’s proof is equivalent to: $P\left(\exists n: \frac {S_n(\mu)} {\sqrt n} > 2 \| \mu = 0\right) = 1$ Reggie’s proof is equivalent to: $P\left(\exists n: \frac { S_n(\mu)} {\sqrt n} > 2 \| \mu < -0.6\right) \approx 0$ I want to derive: $P\left(\frac{S_n(\mu)} {\sqrt n} > 2 + \mu \sqrt n \| \frac {S_n(\mu)} {\sqrt n} > 2, \mu \right)$ In other words, I want to derive the probability that the Bayesian is fooled in the second sense conditional on this stopping rule having actually worked, as a function of the true value $\mu$. In specific, I’ll focus on the positive case, since Reggie took care of the negative one for me. For ease of notation, let’s call our propositions: \begin{aligned} F_{\mu} : & \frac{S_n(\mu)} {\sqrt n} > 2 + \mu \sqrt n \\ F_0 : & \frac{S_n(\mu)} {\sqrt n} > 2 \end{aligned} Where $F_{\mu}$ is “the Bayesian is fooled when the true effect is $\mu$” and $F_0$ is “the Bayesian is fooled when the true effect is $0$. Using Bayes’ Theorem, we want to calculate: $P(F_{\mu} \| F_0, \mu) = \frac {P(F_0 \| F_{\mu}, \mu) P(F_{\mu} \| \mu)} {P(F_0 \| \mu)}$ Since I’m only interested in the case where $\mu > 0$, $F_{\mu} \rightarrow F_0$ and so $P(F_0\|F_{\mu}, \mu) = 1$. Furthermore, it can be shown that $\frac {S_n(\mu)} {\sqrt n} \sim \mathcal N \left( \mu\sqrt n, 1 \right)$. Therefore: \begin{aligned} P(F_{\mu} \| F_0, \mu) &= \frac {P(F_\mu \| \mu)} {P(F_0 \| \mu)} \\ & = \frac {P\left( \mathcal N(\mu \sqrt n, 1) > 2 + \mu \sqrt n\right)} {P\left( \mathcal N(\mu \sqrt n, 1) > 2 \right)} \\ &= \frac {P\left( \mathcal N(0, 1) > 2\right)} {P\left( \mathcal N(\mu \sqrt n, 1) > 2\right)} \end{aligned} The numerator is independent of $n$, but the denominator is not. This can be fixed by noting that, since $n > 0$, $P(\mathcal N(\mu \sqrt n, 1) > 2) \geq P(\mathcal N(\mu, 1) > 2)$, whence: $P(F_{\mu} \| F_0, \mu) \leq \frac {P(\mathcal N(0, 1) > 2)} {P(\mathcal N(\mu, 1) > 2)} \approx \frac {0.0455} {\text{erf} \left(\frac{x - 2} {\sqrt 2} \right) + 1}$ And the right-hand side of the above is plotted as a function of $\mu$ below: At first, the upper bound is not terribly tight, but eventually it approaches a number we can calculate. Intuitively, for large enough $\mu$, a single sample is enough for perversesheaf’s stopping rule to apply, so a single sample will be collected and the Bayesian’s 95% credible interval will be greater than $0$. But in that case, the only way the Bayesian can be fooled (in the second sense) is if that single sample is greater than $\mu + 2$, and we know the probability for that: $P(\mathcal N(\mu, 1) > \mu + 2) = P(\mathcal N(0, 1) > 2) \approx 0.02275$. Thus, for a large enough true effect, the Bayesian will be fooled only 2.3% of the time. With my upper bound, “large enough” is $2$ or greater; however, like I said, my bound is not very tight, so it’s probably quite a bit less than that. Okay, that’s all fine and dandy, I’ve proven that for high enough values of $\mu$ the Bayesian won’t be fooled, and Reggie’s proven that for low enough values of $\mu$ the Bayesian won’t be fooled. But that’s just dodging the problem! From its description, we know that $\mu = 0$, so the Bayesian is fooled anyway! Not quite. See, the point of those derivations wasn’t to just “neener neener,” there was a purpose there. I commented, above, that I’d come back to a certain parenthesis perversesheaf made. You can go back and remind yourself of what it was, I’ll wait. Done? Okay. As perversesheaf said, if the evil drug company lets on to the Bayesian that they know the drug to be a placebo, that’ll affect the Bayesian’s priors, and then they’re screwed. Yet he won’t let the knowledge of the stopping rule itself affect the Bayesian’s priors! The likelihood principle works because, normally, $p(\text{Outcome} \| \text{Parametres}, \text{Stopping Rule})$ $= P(\text{Outcome} \| \text{Parametres})$. In other words, knowledge of a stopping rule does not typically affect the likelihood of an experiment, as I’ve shown in the coin tosses example. And indeed, it does not directly affect it, but there’s a way it can affect exactly the thing perversesheaf was afraid of affecting: the priors. In the little Bayesian Network with labelled arrows I drew above, we can see a way in which knowledge of the stopping rule can affect knowledge of $\mu$: if the choice of the stopping rule was influenced by knowledge of the true parametre, then knowledge of the stopping rule affects our priors for the parametre! And while in a case such as the coin tossing one it might be reasonable to suppose that no such knowledge exists, in this case we might very well suppose that it does! Why? Because of the whole analysis Reggie and I did up there. This stopping rule only has a reasonable chance of fooling the Bayesian, in the case where the true effect is zero or near zero. And even more than that, just a prior belief that this stopping rule is more likely to be chosen when someone wants to fool us than otherwise is enough to change our prior and screw things up. Sanity check: Suppose we purposefully sever the “Knowledge” arrow in the above network; that is, suppose we change the story a bit so that it was the Bayesian themself who chose that stopping rule, arbitrarily. So there’s no evil drug company, just you, a bunch of data, and you decide to use this stopping rule just for the lulz, without any prior knowledge of the true effect. Then the Bayesian’s posterior belief is clearly no longer “foolish”! They don’t know what the true effect is, they don’t have any prior beliefs that would justify that true effect being this or that value, so their posterior really does reflect the appropriate conclusions they should draw. Actually, the fact that the stopping rule worked at all basically rules out any value for $\mu$ below $-0.6$, so by renormalisation alone that would be enough for your point estimate to be greater than $0$ and maybe your 95% credible interval as well. And this also reflects out realistic intuitions very well. If I looked at the stopping rule someone used and did not suspect it was meant to tamper with my confidence, I would carry on as usual; if I did suspect it was meant to tamper with my confidence, that’d be extra information about the estimated parametre that I can use as part of my prior. Another interesting thing is that, “natively,” frequentist methods don’t know how to deal with stopping rules either; you need to change them appropriately, let them take the stopping rule into account, so it’s only fair that you let the Bayesian do the same. Final conjecture: This is always the case. Whenever we intuitively think that a stopping rule will fool the Bayesian, either it will affect the Bayesian’s prior and they won’t be fooled, or what we’re calling “being fooled” is not actually being fooled at all. If you allow your Bayesian statistician to be at least as smart as yourself, they can predict everything you can. ### 3 Responses to Stopping rules, p-values, and the likelihood principle 1. Professor Frink says: Bayesianism methods are definitely not the rule in practical machine learning or AI. It’s impractical for many, perhaps most, use cases. And there is a long history of literature advocating against the radical bayesian position. Besides optional stopping, Stein’s counterexample is famous. Diaconis and Freedman have a famous paper where they Bayesian parameter estimation can be ill-behaved, even with innocuous, commonly used priors like Dirichlet. It’s also easy to show that if your model space is missing the correct model, learning new information can make your overall estimates worse (increasing confidence in a wrong model), which might not push against the radical bayesian perspective but it does limit it’s practicality. • pedromvilar says: As for Bayesianism not being the rule in AI and Machine Learning: really? Okay, guess I was mistaken. I’ll edit the post to reflect this. But it was definitely the impression I got from Russell and Norvig, and it’s currently the impression I’m getting from Bishop. Long history of literature advocating against the position: Yeah, I’m aware of this, which is why I find perversesheaf’s position that this is Yudkowsky and LW’s fault ludicrous. Radical Bayesianism has been around and spreading for a looooong time. Stein, Diaconis and Freedman: Can you link me those? I’d love to read them. Model space missing the correct model: Yes, I’m also aware of this, which is why I mention in this post that “[Bayesianism] can tell you which hypotheses are more or less likely, when you give it a set of possibilities, a prior distribution over these possibilities, and a likelihood. That’s all it can do.” It’s also what I wrote in my Truth, Probability, and Unachievable Consistency post, and I even linked that post in the part where I mention su3su2u1 and say “Basically, my response to this is that Bayesianism is a philosophical position, but in full generality it is literally uncomputable, actually not even approximable, and to people who think that the correct way to Bayes is using it in all practical cases (both critics and adherents) I say: ha.” • Not sure about Stein’s counterexample, but the Diaconis and Freedman example referred to is probably the one in “On Inconsistent Bayes Estimates of Location”. Open access here: dx.doi.org/10.1214/aos/1176349843	7,513	30,937	{"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": 96, "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}	3.59375	4	CC-MAIN-2018-51	latest	en	0.925922
http://mathhelpforum.com/differential-equations/195280-variation-parameters-print.html	1,511,260,075,000,000,000	text/html	crawl-data/CC-MAIN-2017-47/segments/1510934806338.36/warc/CC-MAIN-20171121094039-20171121114039-00074.warc.gz	182,883,583	3,804	# Variation of Parameters • Jan 14th 2012, 12:15 PM MuhTheKuh Variation of Parameters I have a problem understanding a certain step using the Variation of Parameters Method. take for example y''+y=sec(x)tan(x) Auxiliary Equation: mē+1=0 thus m= +/- 1 CF: Y1= Acos(x)+Bsin(x) then I let y=A(x)cos(x)+B(x)sin(x) and differentiate to y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later) differentiating again y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x) Substituting the yields now: -A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step) so now I have to solve for A' and/or B' I have no idea how to do this and when I checked the answer booklet it explained the next step by: * cos(x) - ** sin(x): A'=tanē(x) = 1 - sec(x) A= x - tan(x) + C and * sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D I fully understand how to get up to the point were it says * cos(x) - ** sin(x) and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked () differentiations to A'= or B'= I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet. Thank you guys for your time, I appreciate it. • Jan 14th 2012, 12:47 PM alexmahone Re: Variation of Parameters Quote: Originally Posted by MuhTheKuh I have a problem understanding a certain step using the Variation of Parameters Method. take for example y''+y=sec(x)tan(x) Auxiliary Equation: mē+1=0 thus m= +/- 1 CF: Y1= Acos(x)+Bsin(x) then I let y=A(x)cos(x)+B(x)sin(x) and differentiate to y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 (remember this for later) differentiating again y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x) Substituting the yields now: -A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step) so now I have to solve for A' and/or B' I have no idea how to do this and when I checked the answer booklet it explained the next step by: * cos(x) - ** sin(x): A'=tanē(x) = 1 - sec(x) A= x - tan(x) + C and * sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D I fully understand how to get up to the point were it says * cos(x) - ** sin(x) and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked () differentiations to A'= or B'= I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet. Thank you guys for your time, I appreciate it. $A'\cos x+B'\sin x=0$ $-A'\sin x+B'\cos x=\sec x\tan x$ ** In order to eliminate $B$, we multiply the * equation by $\cos x$, the ** equation by $\sin x$ and subtract to get $A'(\cos^2x+\sin^2x)=-\sec x\tan x\sin x$ $A'=-\frac{\sin^2x}{\cos^2x}$ $=-\tan^2x$ $=1-\sec^2x$ In order to eliminate $A$, we multiply the * equation by $\sin x$, the ** equation by $\cos x$ and add to get $B'(\sin^2x+\cos^2x)=\tan x$ $B'=\tan x$ • Jan 14th 2012, 12:57 PM MuhTheKuh Re: Variation of Parameters Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sinē(x)/cosē(x) or that sec(x)tan(x)cos(x) = tan(x) I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc? or how someone would put it if i had to take tan(x)cos(x)? Thanks a lot! • Jan 14th 2012, 01:01 PM alexmahone Re: Variation of Parameters Quote: Originally Posted by MuhTheKuh Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sinē(x)/cosē(x) or that sec(x)tan(x)cos(x) = tan(x) I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc? or how someone would put it if i had to take tan(x)cos(x)? Thanks a lot! Just rewrite everything in terms of sines and cosines. $-\sec x\tan x\sin x=-\frac{1}{\cos x}\frac{\sin x}{\cos x}\sin x=-\frac{\sin^2x}{\cos^2x}$ $\sec x\tan x\cos x=\frac{1}{\cos x}\tan x\cos x=\tan x$	1,351	4,019	{"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": 16, "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	4	CC-MAIN-2017-47	longest	en	0.855094
https://boards.straightdope.com/t/musical-scales/54177?page=2	1,610,820,449,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00334.warc.gz	260,285,374	13,820	# Musical Scales <-----------Massaging Screechy’s tension away using Solfege-like hand gestures Wife’s entire family are classical musicians. She, and her brothers and parents, have perfect pitch. So, to answer the question way up there, most things in the world make some sound, many are a definable note or notes. They used to play “car horn” when they were kids. You know, who gets to nail the note first. My father in law gave us an IMMENSE air conditioner. It was brand new, but the compressor vibrated at a pitch that drove him insane ( it was in his violin studio ). Interesting people… Cartooniverse Mmmmmmmmmmmmm. Kodaly. Mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm. Then again, the Guidonian hand is pretty, um, complicated… Perfect pitch? Eh, that would drive me nuts. A friend was telling me about how his professor would torment the people with perfect pitch in sightsinging class: He (the professor) would play a pitch on the piano and say, “That’s G sharp.” People with perfect pitch in the class would argue “No no no that’s a D.” “For the purpose of this exercise, this note <repeats note> is a G sharp. Now transcribe this” and he would play the melody. All the people with perfect pitch would struggle to write “G sharp” when they knew it was really “D”, as well as transpose the rest of the pitches. Cruel but fun pranks to play on the gifted. In the Washington Post dated 12-08-99 there is an excellent article on the origin of the musical scale. It also gives a good explanation of the basics of music theory. I think you can view or download articles from the Post going back 2 years. See Washingtonpost.com. The article is in Section H, pages 1 and 4. Hi! This is my first response to anything here, so please be kind… OK, about scales. First of all, sound frequency is a continuum, and the pitches we pick out are discrete points in that continuum. How many tones there are in an octave will depend on how closely or widely spaced those pitches are. So when we talk about a pentatonic scale we mean one with five pitches to the octave. The sixth pitch would be an octave above the first one. To change a five-note scale to a seven-note one you have to divide the octave differently. To hear the pentatonic scale, just play only the black keys on a piano. Several people have mentioned that Chinese music uses this scale, but it has also been used all over the world, even in Western music. Lots of hymns and folktunes are pentatonic: e.g. Amazing Grace. There is real physics behind the choice of pitches for a scale, which is why the same scales show up in different cultures. Different pitches have different frequencies of sound (i.e. how fast or slowly whatever is producing the sound is vibrating), and these frequencies are related to one another. In the middle ages, Music was taught in the universities as a part of mathematics – it was the science of Ratio rather than what we would consider music class to be today. In the middle ages in France they would demonstrate the ratios with strings (it still works, btw – I’ve tried it in class). Take a string and divide it in half (producing the ratio 2:1). Half the string will vibrate exactly twice as fast as the whole string, and produce a pitch exactly one octave above that of the whole string. You can generate all the pitches in the scale this way, dividing the string into three and then four parts: 3:2 = the interval of a fifth (e.g. C-G on a piano) 4:3 = the interval of a fourth (e.g. C-F or G-C etc) Dividing the string into 9 parts, you can generate a single tone (e.g. from C to D on the piano), with the sound ratio of 9:8. The smaller the numbers in a ratio, the better the two pitches will sound to us (music rather than noise). Using different pitches as the basis for generating new pitches, you can find all seven pitches of Western scales this way. The tradition is that the philosopher Pythagoras (6th century BC) discovered this method of determining the pitches of a scale – the kind of tuning this method produces is called “pythagorean” to this day. And this is why Western scales have seven pitches, with the 8th an octave above the first: we’ve been teaching ourselves to divide the octave this way for over two thousand years. Someone has already mentioned the American composer Harry Partch. He was fascinated with the history of tuning and intonation. His book Genesis of a Music (1949, 2nd ed. Da Capo Press, 1974) has a chapter on the history of tuning systems that is the best and most consise discussion of this topic that I know. One last thing: nobody asked it, but I can tell you where the “do re mi” syllables come from. Guido of Arezzo (an Italian musician/teacher who lived c.1000-1050) taught his students to sing a major scale using a gregorian chant hymn to St. John the Baptist called “Ut queant laxis”. Guido had noticed that each phrase of the hymn began on a different scale pitch, starting on C and moving up. He used the syllables that began each phrase as the name for that pitch: UT queant laxix REsonare fibris MIra gestorum FAmuli tuorum, SOLve polluti LAbii reatum, Sancte Joannes. [“That thy servants may freely proclaim the wonders of thy deeds, absolve the sins of their unclean lips, O holy John”] “Ut” (pronounced Oot) was changed to “Do” (pronounced Doh) in the 20th century. But I have no idea who did it, or why. OK. All the modes had names in the Greek system. They were actually described in Plato’s Republic, book 3, one of the earliest such catalogues to have survived. Here’s the complete list: Scales played by striking the white keys starting with C (the major scale): Ionian. D: Dorian. (the scale Plato eventually favours.) E: Phrygian. F (the minor scale): Lydian. G: Mixolydian. A: Aeolian (possibly what Plato refers to as “Syntonolydian”). And they are not named after bodily humours, but after regions of the Greek empire. This is the scale used on the bagpipes, with repetition, to nine notes in total. That’s why most Scots music from the pipes can be played only on the black keys. From what I’ve read, the pentatonic scale is so common in folk music because there is not much scope for variation in a simple pipe instrument. Most primitive pipes are about the same size and have the finger holes spaced about the same distance apart. (For example, I was playing my chanter for my brother-in-law the other night, and he said it sounded like a snake charmer’s pipe.) The confusing thing about reading pipe music is that by convention, the notes are written down by the usual Western European notation, rather than writing out the notes of the actual pentatonic scale, given above. Playing from G to G[sup]1[/sup] on the chanter sounds nothing like the scale of G major on the piano - the intervals are wrong. Thank you for the clarification, matt_mcl. I knew they were named after different regions and influenced the bodily humors, but was twiching too much at that point to correct it in the re-reading. jti - My sourcebook indicated Scottish music (as well at least a dozen others), but wasn’t quite sure which aspect. Pipes, duh! :sound of one hand slapping one head: On a guitar you can divide the string as Speranza described by touching (but not fretting) the string at certain points, they are called ‘touch harmonics’. If you touch at the 12th fret (exactly between the nut and the bridge) and pick it, the string will vibrate twice as fast as if you play it open, if you look at it under the right light you will see that it is two waves, the string barely moving in the middle. If you touch at the 7th fret, it will divide into three waves, and it will play a note an octave and a fifth above the open note. If you touch it at the 5th fret (which is exactly halfway between the nut and the twelth fret), it will play a note two octaves higher…the fourth fret is (I believe) two octaves and a fifth…or is it a third… Higher up, the touch points don’t correspond with the frets exactly, the next big one is about 3/4 of the way from the second to the third fret, the one that will get you three octaves higher is one fifth of the way from the second to the third. You can barely hear those last two because the string is divided up into so many waves that it’s not moving very far on each one, but on an electric with compression and maybe some distortion you can hear it. You can play the first few notes of the Star Trek theme using only touch harmonics on the A and E strings, and you can play Taps with touch harmonics on just one string. The sound effect the transporters used in some of the Star Trek movies (2 and 3 I think) incorporated the sound you get if you rapidly pick the string while sliding your finger along the string, getting all the touch harmonics along the way. A lot of modern electric guitar music also uses what’s called ‘pinch harmonics’. This is where you extend the end of your thumb a bit past the pick and touch the string as you pick it, if you do it in the right place the guitar will squeal several octaves above the note you fretted (because your hand is so close to the bridge). ZZ Top, Pantera, and Randy Rhodes-era Ozzy used it a lot. When you hear a guitar playing fairly low notes and suddenly it squeals much higher than the note before, it’s probably pinch harmonics. Before I learned that, figuring out how the pitch could jump so high so suddenly drove me nuts. The violin has the touch harmonic phenomenon as well. I can only do it with the octave, but if I noodle around some more I should be able to get the same effects. As for the Greek scales, the previous posters have forgotten the Locrian scale, and example of which would be playing the C major scale starting from B. screech-owl: Those notebooks could be a very valuable commodity. They should be protected from self-teaching theory students who might want an unethical leg up via someone else’s hard-won studies. evil grin This is because those harmonics don’t apply only to strings. All brass instruments use harmonics to produce their range of notes (ever notice there’s only three (or four) keys? That’s only eight different notes.) In the case of Taps, specifically, a bugle, which has no keys, uses harmonics to create all of its sounds. The harmonics are formed on the column of air, rather than a string. If you look, you’ll find all bugle music is written with four or five notes, total, all simple harmonics. (I’m a violinist, and I hate when the parts call for harmonics. I personally think they don’t sound very good. OTOH, a similar concept is useful for producing vibrato while playing an open string). LL LOCRIAN!!! Dammit, that’s it! Kept me awake half the night trying to remember my modes! Although one of my theory teachers gave us the impression that Locrian was a synthetic mode, ‘made up so [b-b[sup]1[/sup]] wouldn’t be lonely’, and was not an actual mode in use with the Greeks. Can anyone verify or refute that? White key scales Ionian - c-c[sup]1[/sup] Dorian - d-d[sup]1[/sup] Phrygian - e-e[sup]1[/sup] Lydian - f-f[sup]1[/sup] Myxolydian - g-g[sup]1[/sup] Aeolian - a-a[sup]1[/sup] Locrian - b-b[sup]1[/sup] [singing] And that brings us back to IO[sub]nian[/sub]. [/singing] If Doris Plays Lydian, Matt Locrian. And Olentzero, I’d be happy to share my notebooks (and subsequent knowledge) with you. Can’t guarantee everything is correct or legible, but ask away. [QUOTE] Originally posted by screech-owl LOCRIAN!!! Dammit, that’s it! Kept me awake half the night trying to remember my modes! Although one of my theory teachers gave us the impression that Locrian was a synthetic mode, ‘made up so [b-b[sup]1[/sup]] wouldn’t be lonely’, and was not an actual mode in use with the Greeks. Can anyone verify or refute that? Actually, what WE call Locrian probably was used by the Greeks, only they called it Mixolydian! It’s a common misconception that the modes used by Medieval theorists were the same as those referred to by the Greeks. In fact, when “our” modal system was developed around the 10th century CE, the monks got all the names mixed up. Just a sample of final ("tonics), our name, Greek name: B - Greeks - Mixolydian; Us - Locrian C - Greeks - Lydian; Us - Ionian D - Greeks - Phrygian; Us - Dorian E - Greeks - Dorian; Us - Phrygian You are right, though, that when the modal system was re-invented by Western European musicians, (what we call) the Locrian mode was thrown in just so we’d have a mode built on B. I’m pleasantly surprised that a thread on modal theory is getting so much interest! Medievalists of the world unite! D18 Pardon the newby screwing up the bolding. Just to get the typography right, that should be: Actually, what WE call Locrian probably was used by the Greeks, only they called it Mixolydian! It’s a common misconception that the modes used by Medieval theorists were the same as those referred to by the Greeks. In fact, when “our” modal system was developed around the 10th century CE, the monks got all the names mixed up. Just a sample of final ("tonics), our name, Greek name: B - Greeks - Mixolydian; Us - Locrian C - Greeks - Lydian; Us - Ionian D - Greeks - Phrygian; Us - Dorian E - Greeks - Dorian; Us - Phrygian You are right, though, that when the modal system was re-invented by Western European musicians, (what we call) the Locrian mode was thrown in just so we’d have a mode built on B. I’m pleasantly surprised that a thread on modal theory is getting so much interest! Medievalists of the world unite! D18 I love thinking about modes in music, and as a guitarist, it was an epiphany when I happened upon this concept and how it relates to what chords are chosen in a song, as well as what notes to choose when soloing. Thinking modally allows you to abandon the concept of “key”, making each key equally understandable as another. Wanna play in in G# minor instead of Bb minor? No problem! C (the major scale): Ionian. Much music is in this mode. It is the happy mode. D: Dorian. (the scale Plato eventually favours.) This is the Allman Brothers/Santana mode. With the sixth not flatted, it is a sad, but not as sad the aeolian. Think “Whipping Post”, or maybe “So What” by Miles Davis. E: Phrygian. Similar to Locrian. Odd and Eastern or maybe spanish sounding. Think “White Rabbit” by Jefferson Airplane or maybe “Syeeda’s Song Flute” by Coltrane, or maybe flamenco music. F (the is NOT the minor scale): Lydian. Not heard too much except perhaps in jazz or modern art music. G: Mixolydian: Contemplative and moody with it’s flatted 7th note, but with a major 3rd, not as sad. This is the Grateful Dead key, also heard a lot in rock & jazz. A: Aeolian: Actually THIS is the minor scale. The sad one. Think Niel Young or Pink Floyd. Think funeral dirge. B: Locrian. ** Also odd-ball sounding, but not as minor as Phrygian. Maybe think klezmer. Also a not on the pentatonic blues [minor] scale: Originally, the notes were 1st, halfway between minor third and third, fourth, halfway between flatted fifth and fifth, and halfway between dominant and major 7th. Those halfway notes are sometimes referred to as “blue” notes. When you can play these notes, like with voice, guitar or sax, it can really sound bluesy. Obviously, these “halfway” notes cannot be produced accurately on the piano, so some folks just mash both keys to simulate those blue notes. The pentatonic scale, both minor[blues& rock] and major[country & rock], is the salvation of the amatuer guitar soloist as well as Frank Zappa. I do it 8 days a week. Wait, that’s a regular minor, isn’t it? (There’s probably a better adjective than “regular”.) I thought the harmonic minor was: c – d – e-flat – f – g – a-flat – b – c[sup]1[/sup] If that’s not a harmonic minor, I’m going to claim it as Greg’s minor, because I’ve always liked it. Dammit, now I’m all confused. And I thought I had a headache last night while deciphering my notes from college [sub]mumblednumber[/sub] years ago. Let me look that up and get back to you, unless someone wants to correct me on this first. (What’s my batting average about now? .00007? Well, at least I am keeping you all on your toes and I know someone is actually reading my posts - 'cause from the looks of it, it sure ain’t me!) There’s harmonic minor, melodic minor (differs ascending and descending) and ‘pure’ minor (if I took notes correctly). I’m stopping at the library after work anyway: I’ll look this up. <SNIP> Greg is right on this one, screech. There’s a bazillian more minor scales, too, based on the modes from the harmonic and melodic minors. Here are a fe oddballs: Locrian natural 6th (1 2 b3 #4 5 b6 b7 8) Dorian augmented 4th (1 2 b3 #4 5 b6 b7 8) Altered Dominant bb7 (1 b2 b3 b4 b5 b6 bb7 8) Just don’t make me solo in them. There are three forms of the minor key in Western music theory. They are the melodic, the harmonic, and the natural. The melodic has only one flat in the octave, namely a minor 3rd: C D Eb F G A B C The harmonic has two flats, namely a minor third and a minor 6th: C D Eb F G Ab B C The natural has three flats, namely a minor third, a minor sixth, and a minor seventh: C D Eb F G Ab Bb C The only application I’m aware of (since I haven’t studied but a small amount of theory) is in the construction of four-part chords when using the minor keys. Just to further confuse the issue, this is true only of the ascending form. Descending it’s C B-flat A-flat G F E-flat D C. It has to do with the development of keys from modes. Originally you had the Aeolian mode: A B C D E F G A But in actual practice, the F and the G tended to be raised in ascending passages of music, and lowered back down on descending passages. Thus, the minor scale. That’s actually how the major scale came to be too. On the one hand you had the Mixolydian mode: G A B C D E F G But in actual practice, the F tended to be raised to F-sharp most of the time. Similarly you had the Lydian mode: F G A B C D E F But the tritone between the F and the B sounded “evil” - and was called “the devil in music” - so in actual fact the B tended to be lowered to a B-flat in actual music. Yes, there is an Ionian mode: C D E F G A B C But it was not set down in theory books until about the 16th century, IIRC. At any rate, not till long after the original four modes. About the same time, the Aeolian mode was codified as well. Modes and keys were codified well after the fact - as a way of explaining musical intuitions, so a lot of theory has to be massaged to fit the musical practice. D18	4,609	18,584	{"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-2021-04	latest	en	0.969269
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/08%3A_Atomic_Structure/8.0A%3A_8.A%3A_Atomic_Structure_(Answers)	1,611,262,850,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00750.warc.gz	504,685,838	26,586	$$\require{cancel}$$ 8.1. No. The quantum number $$\displaystyle m=−l,−l+1,…,0,…,l−1,l$$. Thus, the magnitude of $$\displaystyle L_z$$ is always less than L because $$\displaystyle <\sqrt{l(l+1)}$$ 8.2. $$\displaystyle s=3/2<$$ ## Conceptual Questions 1. n (principal quantum number) → total energy $$\displaystyle l$$ (orbital angular quantum number) → total absolute magnitude of the orbital angular momentum $$\displaystyle m$$ (orbital angular projection quantum number) → z-component of the orbital angular momentum 3. The Bohr model describes the electron as a particle that moves around the proton in well-defined orbits. Schrödinger’s model describes the electron as a wave, and knowledge about the position of the electron is restricted to probability statements. The total energy of the electron in the ground state (and all excited states) is the same for both models. However, the orbital angular momentum of the ground state is different for these models. In Bohr’s model, $$\displaystyle L(ground state)=1$$, and in Schrödinger’s model, $$\displaystyle L(ground state)=0$$. 5. a, c, d; The total energy is changed (Zeeman splitting). The work done on the hydrogen atom rotates the atom, so the z-component of angular momentum and polar angle are affected. However, the angular momentum is not affected. 7. Even in the ground state $$\displaystyle (l=0)$$, a hydrogen atom has magnetic properties due the intrinsic (internal) electron spin. The magnetic moment of an electron is proportional to its spin. 9. For all electrons, $$\displaystyle s=½$$ and $$\displaystyle m_s=±½$$. As we will see, not all particles have the same spin quantum number. For example, a photon as a spin 1 ($$\displaystyle s=1$$), and a Higgs boson has spin 0 ($$\displaystyle s=0$$). 11. An electron has a magnetic moment associated with its intrinsic (internal) spin. Spin-orbit coupling occurs when this interacts with the magnetic field produced by the orbital angular momentum of the electron. 13. Elements that belong in the same column in the periodic table of elements have the same fillings of their outer shells, and therefore the same number of valence electrons. For example: Li: $$\displaystyle 1s^22s^1$$ (one valence electron in the $$\displaystyle n=2$$ shell) Na: $$\displaystyle 1s^22s2p^63s^1$$ (one valence electron in the $$\displaystyle n=2$$ shell) Both, Li and Na belong to first column. 15. Atomic and molecular spectra are said to be “discrete,” because only certain spectral lines are observed. In contrast, spectra from a white light source (consisting of many photon frequencies) are continuous because a continuous “rainbow” of colors is observed. 17. UV light consists of relatively high frequency (short wavelength) photons. So the energy of the absorbed photon and the energy transition ($$\displaystyle ΔE$$) in the atom is relatively large. In comparison, visible light consists of relatively lower-frequency photons. Therefore, the energy transition in the atom and the energy of the emitted photon is relatively small. 19. For macroscopic systems, the quantum numbers are very large, so the energy difference ($$\displaystyle ΔE$$) between adjacent energy levels (orbits) is very small. The energy released in transitions between these closely space energy levels is much too small to be detected. 21. Laser light relies on the process of stimulated emission. In this process, electrons must be prepared in an excited (upper) metastable state such that the passage of light through the system produces de-excitations and, therefore, additional light. 23. A Blu-Ray player uses blue laser light to probe the bumps and pits of the disc and a CD player uses red laser light. The relatively short-wavelength blue light is necessary to probe the smaller pits and bumps on a Blu-ray disc; smaller pits and bumps correspond to higher storage densities. ## Problems 25. $$\displaystyle (r,θ,ϕ)=(\sqrt{6,}66°,27°)$$. 27. $$\displaystyle ±3,±2,±1,0$$ are possible 29. 18 31. $$\displaystyle F=−k\frac{Qq}{r^2}$$ 33. (1, 1, 1) 35. For the orbital angular momentum quantum number, l, the allowed values of: $$\displaystyle m=−l,−l+1,...0,...l−1,l$$. With the exception of $$\displaystyle m=0$$, the total number is just 2l because the number of states on either side of $$\displaystyle m=0$$ is just l. Including $$\displaystyle m=0$$, the total number of orbital angular momentum states for the orbital angular momentum quantum number, l, is: $$\displaystyle 2l+1$$. Later, when we consider electron spin, the total number of angular momentum states will be found to twice this value because each orbital angular momentum states is associated with two states of electron spin: spin up and spin down). 37. The probability that the 1s electron of a hydrogen atom is found outside of the Bohr radius is $$\displaystyle ∫^∞_{a_0}P(r)dr≈0.68$$ 39. For $$\displaystyle n=2, l=0$$ (1 state), and $$\displaystyle l=1$$ (3 states). The total is 4. 41. The 3p state corresponds to $$\displaystyle n=3, l=2$$. Therefore, $$\displaystyle μ=μ_B\sqrt{6}$$ 43. The ratio of their masses is 1/207, so the ratio of their magnetic moments is 207. The electron’s magnetic moment is more than 200 times larger than the muon. 45. a. The 3d state corresponds to $$\displaystyle n=3, l=2$$. So, $$\displaystyle I=4.43×10^{−7}A$$. b. The maximum torque occurs when the magnetic moment and external magnetic field vectors are at right angles $$\displaystyle (sinθ=1)$$. In this case: $$\displaystyle \|\vec{τ}\|=μB.$$ $$\displaystyle τ=5.70×10^{−26}N⋅m$$.. 47. A 3p electron is in the state $$\displaystyle n=3$$ and $$\displaystyle l=1$$. The minimum torque magnitude occurs when the magnetic moment and external magnetic field vectors are most parallel (antiparallel). This occurs when $$\displaystyle m=±1$$. The torque magnitude is given by $$\displaystyle \|\vec{τ}\|=μBsinθ$$, Where $$\displaystyle μ=(1.31×10^{−24}J/T)$$. For $$\displaystyle m=±1$$, we have: $$\displaystyle \|\vec{τ⃗}\|=2.32×10^{21}N⋅m$$. 49. An infinitesimal work dW done by a magnetic torque $$\displaystyle τ$$ to rotate the magnetic moment through an angle $$\displaystyle −dθ$$: $$\displaystyle dW=τ(−dθ)$$, where $$\displaystyle τ=\|\vec{μ}×\vec{B}∣$$. Work done is interpreted as a drop in potential energy U, so $$\displaystyle dW=−dU.$$ The total energy change is determined by summing over infinitesimal changes in the potential energy: $$\displaystyle U=−μBcosθ$$ $$\displaystyle U=−\vec{μ}⋅\vec{B}$$. 51. Spin up (relative to positive z-axis): $$\displaystyle θ=55°$$. Spin down (relative to positive z-axis): $$\displaystyle θ=cos^{−1}(\frac{S_z}{S})=cos^{−1}(\frac{−\frac{1}{2}}{\frac{\sqrt{3}}{2}})=cos^{−1}(\frac{−1}{\sqrt{3}})=125°.$$ 53. The spin projection quantum number is $$\displaystyle m_s=±½$$, so the z-component of the magnetic moment is $$\displaystyle μ_z=±μ_B$$. The potential energy associated with the interaction between the electron and the external magnetic field is $$\displaystyle U=∓μ_BB$$. The energy difference between these states is $$\displaystyle ΔE=2μ_BB$$, so the wavelength of light produced is $$\displaystyle λ=5.36×10^{−5}m≈53.6μm$$ 55. It is increased by a factor of 2. 57. a. 32; b. (2ℓ+1) 0 s 2(0+1) =2 1 p 2(2+1) =6 2 d 2(4+1) =10 3 f 2(6+1) =14 32 59. a. and e. are allowed; the others are not allowed. b. $$\displaystyle l=3$$ not allowed for $$\displaystyle n=1,l≤(n−1)$$. c. Cannot have three electrons in s subshell because $$\displaystyle 3>2(2l+1)=2$$. d. Cannot have seven electrons in p subshell (max of 6) $$\displaystyle 2(2l+1)=2(2+1)=6$$. 61. $$\displaystyle [Ar]4s^23d^6$$ 63. a. The minimum value of $$\displaystyle ℓ$$ is $$\displaystyle l=2$$ to have nine electrons in it. b. $$\displaystyle 3d^9$$. 65. $$\displaystyle [He]2s^22p^2$$ 67. For $$\displaystyle He^+$$, one electron “orbits” a nucleus with two protons and two neutrons ($$\displaystyle Z=2$$). Ionization energy refers to the energy required to remove the electron from the atom. The energy needed to remove the electron in the ground state of He+He+ ion to infinity is negative the value of the ground state energy, written: $$\displaystyle E=−54.4eV$$. Thus, the energy to ionize the electron is $$\displaystyle +54.4eV$$. Similarly, the energy needed to remove an electron in the first excited state of $$\displaystyle Li^{2+}$$ ion to infinity is negative the value of the first excited state energy, written: $$\displaystyle E=−30.6eV$$. The energy to ionize the electron is 30.6 eV. 69. The wavelength of the laser is given by: $$\displaystyle λ=\frac{hc}{−ΔE}$$, where $$\displaystyle E_γ$$ is the energy of the photon and $$\displaystyle ΔE$$ is the magnitude of the energy difference. Solving for the latter, we get: $$\displaystyle ΔE=−2.795eV$$. The negative sign indicates that the electron lost energy in the transition. 71. $$\displaystyle ΔE_{L→K}≈(Z−1)^2(10.2eV)=3.68×10^3eV$$. 73. According to the conservation of the energy, the potential energy of the electron is converted completely into kinetic energy. The initial kinetic energy of the electron is zero (the electron begins at rest). So, the kinetic energy of the electron just before it strikes the target is: $$\displaystyle K=eΔV$$. If all of this energy is converted into braking radiation, the frequency of the emitted radiation is a maximum, therefore: $$\displaystyle f_{max}=\frac{eΔV}{h}$$. When the emitted frequency is a maximum, then the emitted wavelength is a minimum, so: $$\displaystyle λ_{min}=0.1293nm$$. 75. A muon is 200 times heavier than an electron, but the minimum wavelength does not depend on mass, so the result is unchanged. 77. $$\displaystyle 4.13×10^{−11}m$$ 79. 72.5 keV 81. The atomic numbers for Cu and Au are $$\displaystyle Z=29$$ and 79, respectively. The X-ray photon frequency for gold is greater than copper by a factor: $$\displaystyle (\frac{f_{Au}}{f_{Cu}})^2=(\frac{79−1}{29−1})^2≈8$$. Therefore, the X-ray wavelength of Au is about eight times shorter than for copper. 83. a. If flesh has the same density as water, then we used $$\displaystyle 1.34×10^{23}$$ photons. b. 2.52 MW 85. The smallest angle corresponds to $$\displaystyle l=n−1$$ and $$\displaystyle m=l=n−1$$. Therefore $$\displaystyle θ=cos^{−1}(\sqrt{n−1}{n}$$). 87. a. According to Equation 8.1, when $$\displaystyle r=0, U(r)=−∞$$, and when $$\displaystyle r=+∞,U(r)=0$$. The former result suggests that the electron can have an infinite negative potential energy. The quantum model of the hydrogen atom avoids this possibility because the probability density at $$\displaystyle r=0$$ is zero. 89. A formal solution using sums is somewhat complicated. However, the answer easily found by studying the mathematical pattern between the principal quantum number and the total number of orbital angular momentum states. For $$\displaystyle n=1$$, the total number of orbital angular momentum states is 1; for $$\displaystyle n=2$$, the total number is 4; and, when $$\displaystyle n=3$$, the total number is 9, and so on. The pattern suggests the total number of orbital angular momentum states for the nth shell is $$\displaystyle n^2$$. (Later, when we consider electron spin, the total number of angular momentum states will be found to be $$\displaystyle 2n^2$$, because each orbital angular momentum states is associated with two states of electron spin; spin up and spin down). 91. 50 93. The maximum number of orbital angular momentum electron states in the nth shell of an atom is $$\displaystyle n^2$$. Each of these states can be filled by a spin up and spin down electron. So, the maximum number of electron states in the nth shell is $$\displaystyle 2n^2$$. 95. a., c., and e. are allowed; the others are not allowed. b. $$\displaystyle l>n$$ is not allowed. d. $$\displaystyle 7>2(2l+1)$$ 97. $$\displaystyle f=1.8×10^9Hz$$ 99. The atomic numbers for Cu and Ag are $$\displaystyle Z=29$$ and 47, respectively. The X-ray photon frequency for silver is greater than copper by the following factor: $$\displaystyle (\frac{f_{Ag}}{f_{Cu}})^2=2.7$$. Therefore, the X-ray wavelength of Ag is about three times shorter than for copper. 101. a. 3.24; b. $$\displaystyle n_i$$ is not an integer. c. The wavelength must not be correct. Because $$\displaystyle n_i>2$$, the assumption that the line was from the Balmer series is possible, but the wavelength of the light did not produce an integer value for $$\displaystyle n_i$$. If the wavelength is correct, then the assumption that the gas is hydrogen is not correct; it might be sodium instead.	3,517	12,644	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.234375	3	CC-MAIN-2021-04	latest	en	0.807677
http://wordpress.discretization.de/vismathws12/author/dadaforpeace/	1,723,411,519,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00648.warc.gz	36,409,548	10,647	Lecture Notes 5.11. and 6.11.2012 The euclidean plane isometry consists of 5 motions: • Identity • Reflection • Rotation • Translation • Glide Reflection Suppose $G<E_2$ is a discrete group. First consider the translation subgroup of $G$. It holds $G\cap \mathbb{R}^2 < E_2$. We have • $E_2 = \mathbb{R}^2 \rtimes O_2$. • $E_2 \overset{\varphi}{\longrightarrow} O_2$. • $ker(\varphi) = \mathbb{R}^2 =$Translations. • $ker(\varphi\|_G)= G \cap \mathbb{R}^2$. The discrete translation group forms a lattice of dimension $k= 0,1,2$. $k=1:$ There are no translations in $G$. This implies that $G$ has a fixed point ($G$ has no glides). Thus $G<O_2$ is a finite subgroup $G=C_n$ or $D_n$ for some $n=1,2,3$. As groups of symmetries in $\mathbb{E}^2$ these are called the rosette groups or point groups since they leave at least one point fixed. $G=C_n$ are called the cyclic group. The cup (first object) is $C_2$ since it has a $180^°$ symmetry. The recycling logo (2nd object) is $C_3$ and the star (3rd object) $C_5$ since the they are $120^°$ and $72^°$ symmetric. The group $C_1$ has no symmetry and can be represented by an or an R. $G=D_n$ are called the dihedral group. Starting with the upper left the square is $D_4$, the rhombus is $D_2$ the triangle $D_1$. The dihedral group is the group of regular polygons, so an n-gon is $D_n$. $k=1:$ These groups are called the frieze groups. They have the following definition: $G\cap \mathbb{R}^2 = \{ \tau_{nv} : v\neq 0 \in \mathbb{R}^2, n\in\mathbb{Z} \} = \langle \tau_v\rangle\cong \mathbb{Z}$. $k=2:$ These groups are called the wallpaper groups. They have the following definition: $G\cap \mathbb{R}^2 = \langle \tau_v, \tau_w\rangle \{ \tau_{nv+mw} : v,w\neq 0 \in \mathbb{R}^2, n,m\in\mathbb{Z} \} \cong \mathbb{Z}^2$. There are seven distinct subgroups (up to scaling and shifting of patterns) in the discrete frieze group generated by a translation, reflection (along the same axis) and a 180° rotation. An overview gives the following picture The Orbifold notation was the name we used in the lecture. Take a width-$n$ strip $[0,n]\times\mathbb{R}\subset\mathbb{R}$ and glue its edges to give a cylinder. Any of our seven groups will give a group symmetry of the cylinder with exactly $n$-fold rotation around vertical axis. This gives $7$ families of symmetry groups, indexed by $n$. These are finite subgroups of $O_3$. They are exactly the ones that preserve the equator (or equivalent preserve the poles). These are called the axial groups: • $nn$, rotations (abstr: $\mathbb{Z}_n$) • $nx$, rotary directions (abstr: $\mathbb{Z}_{2n}$) • $n$, (abstr: $\mathbb{Z}_n \times \mathbb{Z}_2$) • $nn$, (abstr: $D_n$) • $22n$, (abstr: $D_n$) • $2n$, (abstr: $D_{2n}$) • $22n$, ($D_n \times \mathbb{Z}_2$) Special cases for $n=2$: • $22$, a 2-fold rotation (abstr: $\mathbb{Z}_2$) • $2X$, a 4-fold rotary reflection (abstr: $\mathbb{Z}_4$) • $2$, rotation, mirror, and antipodal map (abstr: $\mathbb{Z}_2^2$) • $22$, 2 mirrors which yield a rotation (abstr: $\mathbb{Z}_2^2$) • $222$, rotations (abstr: $\mathbb{Z}_2^2$), the Klein-4-group • $22$, rotation and mirror, (abstr: $D_4$) • $222$ only reflections, (abstr: $\mathbb{Z}_2^3$) Note that the abstract groups sometimes look different as, e.g., for the group $222$, which should have $D_2$ as its abstract group instead of $\mathbb{Z}_2$. These are equivalent because $D_{2} \cong D_1 \times \mathbb{Z}_2 \cong \mathbb{Z}_2$. In general, for odd $m \in \mathbb{N}$, it holds $D_{2m} \cong D_{m} \times \mathbb{Z}_2$, and $\mathbb{Z}_{2m} \cong \mathbb{Z}_m \times \mathbb{Z}_2$. For $n=1$, only three different groups remain: • $11 = 1$, the trivial group • $1x = x$, identity and antipodal map (abstr: $\mathbb{Z}_2$) • $1* = * = 11$, a single mirror, (abstr: $\mathbb{Z}_2$) Of course, there are more discrete subgroups of $O_3$. Most of them correspond to the symmetry groups of the platonic solids. First, we give the theorem: Theorem: The finite subgroups of $O_3$ are the 7 families of axial groups and the 7 platonic symmetry groups $235$, $235$ (symmetries of icosahedron and its dual, the dodecahedron), $234$, $234$ (symmetries of the cube and the dodecahedron), $233$, $233$ (tetrahedron), and $32$ (the so-called pyritohedral symmetry). The figure on the left shows the icosahedron inside of a dodecahedron. Recall that the dual or reciprocal of a solid evolves from faces becoming vertices, roughly speaking. A symmetry of one solid is also a symmetry of its dual. The symmetry group is $235$, which has the rotation part $235$ as subgroup. (http://apollonius.math.nthu.edu.tw/d1/dg-07-exe/943251/dynamic/duality.htm) The tetrahedron has the symmetry group $233$, with rotation subgroup $233$. You can easily make a tetrahedron yourself by using the figure to the right, and search for the symmetry groups yourself. Note that the tetrahedron is dual to itself. The cube and its dual counterpart, the octahedron, have the symmetry group $234$, and $234$. The last symmetry group, the pyritohedral symmetry group, comes as the intersection of the octahedral and the dodecahedral symmetry groups, $32 = 235 \cap *234$. In the figure below, you see a pyritohedron, which is not platonic (note that the faces are irregular pentagons). Wallpaper groups Back to $E_2$ (the euclidean motions in the plane), the discrete subgroups which have a translational part of dimension 2 are called wallpaper groups. In the figure below, you find an exhaustive list of the 17 wallpaper groups, each with its fundamental domain, and generating symmetries. The 17 wallpaper groups	1,845	5,648	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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-2024-33	latest	en	0.755596
https://www.jiskha.com/questions/1782358/1-an-expression-that-describes-the-price-of-each-table-based-on-the-number-of	1,591,026,459,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00074.warc.gz	774,199,066	6,250	# MATH 1. An expression that describes the price of each table, based on the number of reductions, is 95 –2x, where x represents the number of \$2 reductions in price. Write a similar expression to describe the number of tables the carpenter can expect to sell, based on the number of reductions. B. Write an equation to describe the carpenter’s income as a product of the price of each table and the expected number of tables sold. C. The carpenter hopes to earn \$3600 to pay for his time, materials, and sales booth, as well as make a small profit. Determine the number of \$2 reductions he can apply to the price of the tables to make \$3600. D. How many reductions will produce maximum income? Indicate the number of tables sold and the price of each one when maximum occurs. 1. 👍 0 2. 👎 0 3. 👁 198 We're not fooled by your different names on the seven posts. 1. 👍 2 2. 👎 1 ## Similar Questions 1. ### math The wholesale cost of a video game is \$459.45. The original markup was 51% based on selling price. Find the final sale price after the following series of price changes: a markup of 24%, a markup of 17%, and a markdown of 10%. asked by bunny on March 24, 2014 2. ### Algebra 1- Urgent The price of products may increase due to inflation and decrease due to depreciation. Derek is studying the change in the price of two products, A and B, over time. The price f(x), in dollars, of product A after x years is asked by Anon on December 17, 2016 3. ### Math 1.Mia charges \$2.25 per hour when she babysits, plus \$5.00 for transportation expenses.Which function rule represents the amount y mia charges to babysit for x hours? A)Y = 2.25x + 5.00*** B)X = 5.00y + 2.25 C)Y = 5.00x + 2.25 D)Y asked by Reeses on May 20, 2017 4. ### Math 2.The price of products may increase due to inflation and decrease due to depreciation. Derek is studying the change in the price of two products, A and B, over time. The price f(x), in dollars, of product A after x years is asked by John on December 19, 2015 5. ### Algebra ARE THESE CORRECT? Over the yrs, the O'Connell Company has been switching from typewriters to computers. At once time the company had 815 typewriters which were wearing out at a rate of about 60 per year. At the same time, they asked by River on March 19, 2007 1. ### Social studies Which of the following accurately describes a surplus? Consumer demand for a certain car is below the number of cars that are produced. The production costs for a certain car are below the sale price of that car. A car company asked by Jen on March 13, 2012 2. ### Accounting Go to Table 10-1 which is based on bonds paying 10 percent interest for 20 years. Assume interest rates in the market (yield to maturity) decline from 11 percent to 8 percent: A. What is the bond price at 11%? B. What is the bond asked by Beth on January 10, 2010 3. ### statistics Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.42. The table on the right describes the probability of having x number of girls. asked by slomomo on February 20, 2015 4. ### Engineering Statistics "A study has shown that 20% of all college textbooks have a price of \$90 or higher. It is known that the standard deviation of the prices of all college textbooks is \$9.50. Suppose the prices of all college textbooks have a normal asked by Christian on October 11, 2011 5. ### math Number of Tutus Price 12 \$420 14 \$462 16 \$496 18 \$522 20 \$540 A dance studio is purchasing tutus for an annual performance. The table represents the number of tutus available for different prices. What is the difference between asked by Anonymous on September 17, 2019	982	3,738	{"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}	3.421875	3	CC-MAIN-2020-24	latest	en	0.945063
https://stackoverflow.com/questions/52550/what-does-the-comma-operator-do	1,702,015,429,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00205.warc.gz	580,050,045	47,242	# What does the comma operator , do? What does the `,` operator do in C? • possible duplicate of What is the proper use of the comma operator? Aug 28, 2013 at 8:00 • As I note in my answer, there is a sequence point after the evaluation of the left operand. This is unlike the comma in a function call which is just grammatical. Jun 25, 2014 at 11:46 • @SergeyK. — Given that this was asked and answered years before the other, it is more likely that the other is a duplicate of this question. However, the other is also dual-tagged with both c and c++, which is a nuisance. This is a C-only Q&A, with decent answers. May 18, 2019 at 0:18 • Does this answer your question? How does the comma operator work, and what precedence does it have? Sep 27 at 19:21 The expression: ``````(expression1, expression2) `````` First `expression1` is evaluated, then `expression2` is evaluated, and the value of `expression2` is returned for the whole expression. • then if I write i = (5,4,3,2,1,0) then ideally it should return 0, correct? but i is being assigned a value of 5? Can you please help me understand where am I going wrong? Nov 13, 2010 at 6:55 • @James: The value of a comma operation will always be the value of the last expression. At no point will `i` have the values 5, 4, 3, 2 or 1. It is simply 0. It's practically useless unless the expressions have side effects. Nov 13, 2010 at 7:06 • Note that there is a full sequence point between the evaluation of the LHS of the comma expression and the evaluation of the RHS (see Shafik Yaghmour's answer for a quote from the C99 standard). This is an important property of the comma operator. Jun 20, 2014 at 20:47 • `i = b, c;` is equivalent to `(i = b), c` because because assignment `=` has higher precedence than the comma operator `,`. The comma operator has the lowest precedence of all. Feb 2, 2019 at 6:10 • I worry that the parentheses are misleading on two counts: (1) they're not necessary — the comma operator doesn't have to be surrounded by parentheses; and (2) they could be confused with the parentheses around the argument list of a function call — but the comma in the argument list is not the comma operator. However, fixing it is not entirely trivial. Maybe: In the statement: `expression1, expression2;` first `expression1` is evaluated, presumably for its side-effects (such as calling a function), then there is a sequence point, then `expression2` is evaluated and the value returned… May 18, 2019 at 0:22 I've seen used most in `while` loops: ``````string s; { //do something } `````` It will do the operation, then do a test based on a side-effect. The other way would be to do it like this: ``````string s; while(s.len() > 5) { //do something } `````` • Hey, that's nifty! I've often had to do unorthodox things in a loop to fix that problem. Sep 4, 2009 at 1:47 • Although it'd probably be less obscure and more readable if you did something like: `while (read_string(s) && s.len() > 5)`. Obviously that wouldn't work if `read_string` doesn't have a return value (or doesn't have a meaningful one). (Edit: Sorry, didn't notice how old this post was.) Mar 25, 2010 at 8:05 • @staticsan Don't be afraid to use `while (1)` with a `break;` statement in the body. Trying to force the break-out part of the code up into the while test or down into the do-while test, is often a waste of energy and makes the code harder to understand. Sep 27, 2012 at 6:20 • @jamesdlin ... and people still read it. If you have something useful to say, then say it. Forums have problems with resurrected threads because threads are usually sorted by date of last post. StackOverflow doesn't have such problems. Nov 29, 2012 at 13:36 • @potrzebie I like the comma approach much better than `while(1)` and `break`; Mar 11, 2016 at 14:19 The comma operator will evaluate the left operand, discard the result and then evaluate the right operand and that will be the result. The idiomatic use as noted in the link is when initializing the variables used in a `for` loop, and it gives the following example: ``````void rev(char s, size_t len) { char first; for ( first = s, s += len - 1; s >= first; --s) /^^^^^^^^^^^^^^^^^^^^^^^/ putchar(s); } `````` Otherwise there are not many great uses of the comma operator, although it is easy to abuse to generate code that is hard to read and maintain. From the draft C99 standard the grammar is as follows: ``````expression: assignment-expression expression , assignment-expression `````` and paragraph 2 says: The left operand of a comma operator is evaluated as a void expression; there is a sequence point after its evaluation. Then the right operand is evaluated; the result has its type and value. 97) If an attempt is made to modify the result of a comma operator or to access it after the next sequence point, the behavior is undefined. Footnote 97 says: A comma operator does not yield an lvalue. which means you can not assign to the result of the comma operator. It is important to note that the comma operator has the lowest precedence and therefore there are cases where using `()` can make a big difference, for example: ``````#include <stdio.h> int main() { int x, y ; x = 1, 2 ; y = (3,4) ; printf( "%d %d\n", x, y ) ; } `````` will have the following output: ``````1 4 `````` The comma operator combines the two expressions either side of it into one, evaluating them both in left-to-right order. The value of the right-hand side is returned as the value of the whole expression. `(expr1, expr2)` is like `{ expr1; expr2; }` but you can use the result of `expr2` in a function call or assignment. It is often seen in `for` loops to initialise or maintain multiple variables like this: ``````for (low = 0, high = MAXSIZE; low < high; low = newlow, high = newhigh) { / do something with low and high and put new values in newlow and newhigh / } `````` Apart from this, I've only used it "in anger" in one other case, when wrapping up two operations that should always go together in a macro. We had code that copied various binary values into a byte buffer for sending on a network, and a pointer maintained where we had got up to: ``````unsigned char outbuff[BUFFSIZE]; unsigned char ptr = outbuff; ptr++ = first_byte_value; ptr++ = second_byte_value; send_buff(outbuff, (int)(ptr - outbuff)); `````` Where the values were `short`s or `int`s we did this: ``````((short )ptr)++ = short_value; ((int )ptr)++ = int_value; `````` Later we read that this was not really valid C, because `(short )ptr` is no longer an l-value and can't be incremented, although our compiler at the time didn't mind. To fix this, we split the expression in two: ``````(short )ptr = short_value; ptr += sizeof(short); `````` However, this approach relied on all developers remembering to put both statements in all the time. We wanted a function where you could pass in the output pointer, the value and and the value's type. This being C, not C++ with templates, we couldn't have a function take an arbitrary type, so we settled on a macro: ``````#define ASSIGN_INCR(p, val, type) ((((type) *)(p) = (val)), (p) += sizeof(type)) `````` By using the comma operator we were able to use this in expressions or as statements as we wished: ``````if (need_to_output_short) ASSIGN_INCR(ptr, short_value, short); latest_pos = ASSIGN_INCR(ptr, int_value, int); send_buff(outbuff, (int)(ASSIGN_INCR(ptr, last_value, int) - outbuff)); `````` I'm not suggesting any of these examples are good style! Indeed, I seem to remember Steve McConnell's Code Complete advising against even using comma operators in a `for` loop: for readability and maintainability, the loop should be controlled by only one variable, and the expressions in the `for` line itself should only contain loop-control code, not other extra bits of initialisation or loop maintenance. It causes the evaluation of multiple statements, but uses only the last one as a resulting value (rvalue, I think). So... ``````int f() { return 7; } int g() { return 8; } int x = (printf("assigning x"), f(), g() ); `````` should result in x being set to 8. • It does. And it is set to 11 if you leave out the outer braces. Quite interesting and definitely worth a compiler warning for some cases. Oct 12, 2020 at 10:57 As earlier answers have stated it evaluates all statements but uses the last one as the value of the expression. Personally I've only found it useful in loop expressions: ``````for (tmp=0, i = MAX; i > 0; i--) `````` The only place I've seen it being useful is when you write a funky loop where you want to do multiple things in one of the expressions (probably the init expression or loop expression. Something like: ``````bool arraysAreMirrored(int a1[], int a2[], size_t size) { size_t i1, i2; for(i1 = 0, i2 = size - 1; i1 < size; i1++, i2--) { if(a1[i1] != a2[i2]) { return false; } } return true; } `````` Pardon me if there are any syntax errors or if I mixed in anything that's not strict C. I'm not arguing that the , operator is good form, but that's what you could use it for. In the case above I'd probably use a `while` loop instead so the multiple expressions on init and loop would be more obvious. (And I'd initialize i1 and i2 inline instead of declaring and then initializing.... blah blah blah.) • I presume you mean i1=0, i2 = size -1 Aug 13, 2009 at 14:05 I'm reviving this simply to address questions from @Rajesh and @JeffMercado which i think are very important since this is one of the top search engine hits. Take the following snippet of code for example ``````int i = (5,4,3,2,1); int j; j = 5,4,3,2,1; printf("%d %d\n", i , j); `````` It will print ``````1 5 `````` The `i` case is handled as explained by most answers. All expressions are evaluated in left-to-right order but only the last one is assigned to `i`. The result of the `(` expression )`is`1`. The `j` case follows different precedence rules since `,` has the lowest operator precedence. Because of those rules, the compiler sees assignment-expression, constant, constant .... The expressions are again evaluated in left-to-right order and their side-effects stay visible, therefore, `j` is `5` as a result of `j = 5`. Interstingly, `int j = 5,4,3,2,1;` is not allowed by the language spec. An initializer expects an assignment-expression so a direct `,` operator is not allowed. Hope this helps.	2,740	10,418	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2023-50	latest	en	0.947099
http://www.studymode.com/essays/Standard-Deviation-And-Deviation-x-Suppose-1577378.html	1,529,716,474,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00041.warc.gz	498,666,929	25,127	# Standard Deviation and Deviation X. Suppose Pages: 6 (1259 words) Published: April 8, 2013 A survey is conducted in Chicago (population 2,800,000) using random-digit-dialing equipment which place calls at random to residential phones, both mobile and landline. The purpose of the survey is to determine the percentage of adults who would favor a half-cent increase in the sales tax to help fund public transportation. Four hundred adults are interviewed and 36% of them favor the proposal. Answer the next two questions. 1. The sample size for this sample survey appears to be a) 400 b) 2,800,000 c) 144 d) 1,008,000 2. The 36% is a a) Parameter b) Margin of error c) Chance of 144 people agreeing to the statement d) Statistic 3. Event A occurs with probability 0.05. Event B occurs with probability 0.75. If A and B are disjoint, which statement is true? a) P(A and B) = 0 b) P(A or B) = 0.80 c) P(A and B) = 0.0375 d) Both (a) and (b) are true. 4. Event A occurs with probability 0.05. Event B occurs with probability 0.75. If A and B are independent, which statement is true. e) P(A and B) = 0 a) P(A or B) = 0.80 b) P(A and B) = 0.0375 c) Both (a) and (b) are true. A marketing research firm wishes to determine if the adult men in Laramie, Wyoming would be interested in a new upscale men's clothing store. From a list of all residential addresses in Laramie, the firm selects a simple random sample of 100 and mails a brief questionnaire to each. Use this information to answer the next three questions. 5. The population of interest is a) all adult men in Laramie, Wyoming. b) all residential addresses in Laramie, Wyoming. c) the members of the marketing firm that actually conducted the survey. d) the 100 addresses to which the survey was mailed. 6. The sample in this survey is a) all adult men in Laramie, Wyoming. b) all residential addresses in Laramie, Wyoming. c) the members of the marketing firm that actually conducted the survey. d) the 100 addresses to which the survey was mailed. 7. The chance that all 100 homes in a particular neighborhood in Laramie end up being the sample of residential addresses selected is a) the same as for any other set of 100 residential addresses. b) exactly 0. Simple random samples will spread out the addresses selected. c) reasonably large due to the “cluster” effect. d) 100 divided by the size of the population of Laramie. Costs for standard veterinary services at a local animal hospital follow a Normal distribution with a mean of \$80 and a standard deviation of \$20. Answer the next three questions. 8. Give the sample space for the costs of standard veterinary services. a) {X ≥ 0} b) { 0 ≤ X ≤ 80} c) {0 ≤ X ≤ 160} d) None of these. 9. What is the probability that one bill for veterinary services costs less than \$95? a) 0.75 a) 0.7734 b) 0.2266 c) 0.15 10. What is the probability that one bill for veterinary services costs between \$75 and \$105? a) 1 a) 0.25 b) 0.4013 c) 0.4931 11. In an instant lottery, your chances of winning are 0.2. If you play the lottery five times and outcomes are independent, what is the probability that you win at least once? a. 0.2 a) 0.08192 b) 0.32768 c) 0.67232 A commuter must pass through 4 traffic lights on her way to work, and will have to stop at each one that is red. Let the random variable be X = number of red lights. The following table is the probability distribution for X. Answer the next four questions. \|X \|P(X = x) \| \|0 \|0.05 \| \|1 \|0.25 \| \|2 \|0.15 \| \|3 \|0.40 \| \|4 \|0.15 \| 12. Give the sample space S for the number of red lights that the commuter stops at. a) {1, 2, 3, 4} b) {0, 1, 2, 3, 4} c) {0.05, 0.25, 0.15, 0.40, 0.15} d) {0, 3, 2, 3, 3, 3, 2, 4, 3, 3, 3, 2, 0, 3 } 13. What is the probability that the commuter... Please join StudyMode to read the full document	1,129	3,815	{"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}	3.671875	4	CC-MAIN-2018-26	latest	en	0.927228
https://www.scribd.com/presentation/1461130/Unit-One1	1,529,318,307,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00268.warc.gz	891,139,134	26,616	# Unit One What you need to know at first! Observations Interaction of one or more of your senses with the environment or your surroundings. 1. 2. 3. 4. 5. Sight Hearing Touch Taste Smell Observations Continued Senses are limited so we use instruments to help improve our powers of observation. Examples:    Telescopes Scales Rulers Inferences Interpretation of one or more observations  Includes proposing explanations or reaching conclusions, like taking a guess  Example: The scratches on the bed rock were caused by a glacier Prediction Inference based on observations that indicate what will happen in the future  Example: Weather predictions (the only job that you can be wrong more then right and not get fired!) Classification The grouping of objects together based on common characteristics (observable properties)  Example: Shape Color Measurement A way of describing with greater accuracy, observations using numbers A direct comparison with a known standard Contain at least one of three basic dimensional quantities    Time Length Mass Must contain correct units ex: 5 cm Forms of Measurement Metric (SI) the one that everyone else uses but Americans. English: what we use Unit Time Length Mass Metric Second, minute, hour, day ,year Millimeter, centimeter, meter, kilometer gram, kilogram English Same as metric Inch, foot, mile Ounce, pound Time Instant in which something happens or period during which change occurs “time of day” which deals with the apparent motion of the sun in the sky Length Measurement of the distance between 2 points Mass Amount of matter in an object, mass never changes MASS DOES NOT EQUAL WEIGHT Weight is a measure of the pull of the earth’s gravity on a quantity of matter (body) If you travel into space and escape Earth’s gravity, you would become “weightless” Properties of matter that use some mathematical combination of basic dimensional quantities 1. 2. 3. 4. Volume Density Pressure Speed Volume Amount of space an object takes up  Ex. V(of a rectangle) = length x width x height  V= L x W x H cm3  Can also find volume by seeing how much water an object displaces in a graduated cylinder. Density Property of matter that combines mass and volume  D = mass/volume  D = m/v (g/cm3) Pressure Measure of force, or weight on a given area Example: newton’s/meter² or lb./in² Speed Measure of rate of motion Example: meter/sec. or miles/hour Percent Deviation (Error) Equation located on front cover of E.S.R.T. Measures how wrong a measurement is. Caused by faulty instruments, careless observations % Error Example The weather report said the air temperature was 35° F and a student thought the air temperature was 25° F. What is your % error (deviation)? % deviation = 35° F – 25° F x 100 35° F = 10 X 100 35 = .2857 X 100 = 28.57 Percent deviation basically tells you how much you measurement is off, in percent, from what you should have gotten. Density The quantity of material contained in a certain space Something densely packed has a large quantity of material crowded into a small amount of space Density continued The greater the mass of a substance, the greater it’s volume will be (direct relationship). Thus, when graphing substances with different densities, the more dense the substance, the steeper the slope. Phases of Matter As substances undergo a change of phase, it’s density changes  Density increases as it changes from gas to liquid to solid (water is the exception to the rule) Objects denser then water will sink, and Density Changes Density CANNOT be changed by:   Cutting the object into pieces Changing the shape (molding Clay) Density CAN be changed by:   Adding or removing heat or temperature Pressure More Density and Phases As material cools, it contracts. It becomes denser because the volume decreases but the mass stays the same. This is true for water except as it reaches 4o Celsius. At 4o, water expands until it reaches 0o. Gases are affected more by pressure and temperature then solids and liquids because their particles are more spread out. Air is a mixture of gases When air is heated it expands, creating a larger volume and smaller density. Because cooler air is more dense, it goes down, as less dense warmer air rises Effect of Temperature and Pressure on the Density of Gases Air Pressure Air Pressure is a measure of the force or weight of the atmosphere pushing down on the earths surface. The denser the air, the greater the pressure. Cold air gives a greater pressure then warm air. Remove Heat -Molecules slow down -Molecules contract -Volume decreases Remove Press -Molecules expand -Volume increases All Matter Solid, Liquid, Gass -molecules speed up -molecules expand -volume increases -Molecules compact -Volume Decreases Density Examples Decrease Increase s s -Hot air -Rings Balloon -Jar lid (run under hot water when lid is stuck come off easier -Fingers shrink Increases Decrease s -Tightly -Ears pop packed snowball when flying or going up mountainrelease of pressure Graphing Relationships As one value goes up on a graph, what happens to the other? Direct Relationship As one value goes up or down, the other value does the same thing. Examples:  As you heat something up, volume increases  As the mass of a substance increases, so does it’s volume (density remains the same)  As pressure goes up so does the density (packing a snowball) Indirect/Inverse Relationship As one value goes up, the other value does the opposite.  As volume increases pressure decreases  When density increases volume decreases Cyclic Relationship As time progress, one goes up and the other goes up and then back down again.  Example: day light, seasons, tides Constant Relationship As one value goes up, the other stays the same.	1,329	5,798	{"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}	3.765625	4	CC-MAIN-2018-26	latest	en	0.88881
http://en.allexperts.com/q/Electronic-Components-2406/2013/1/escalator-design.htm	1,490,725,207,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218189802.72/warc/CC-MAIN-20170322212949-00393-ip-10-233-31-227.ec2.internal.warc.gz	115,140,280	8,309	# Electronic Components/Escalator Design. Question Escalator Escalator QUESTION: Dear Deirdre http://en.wikipedia.org/wiki/Escalator http://science.howstuffworks.com/transport/engines-equipment/escalator.htm http://www.reedconstructiondata.com/documents/FS/specdata/143100_10001243_01_SD_ http://www.au.schindler.com/aus_index/aus-newpage-16/aus-newpage-61.htm http://www.kone.com/countries/en_US/Escalators/Escalator-Design-Tools/Pages/defa Escalators, like moving walkways, are powered by constant-speed alternating current motors and move at approximately 1–2 feet (0.30–0.61 m) per second. The typical angle of inclination of an escalator to the horizontal floor level is 30 degrees with a standard rise up to about 60 feet (18 m). 1. Is the Angle of inclination of 30 degree standard to the Horizontal floor and standard rise of about 60 feet standardized in all escalator designs while manufacturing ?. i.e. Can the Angle of inclination differ to 45 degree or standard rise to say 80 meters or speed vary i.e. 2-3 feet (0.61 m - 0.91 m) per second ?. 2. Is the Load carrying capacity (weight of the Riders) in a escalator set to a maximum limit ?. i.e. Every Escalator designed has a maximum load carrying capacity similar to a Elevator device ? In this case, how safety of the passengers is ensured if overload happens ?. what is the overload protection mechanism inbuilt within the escalator ?. As soon as there is a overload of extra weight, the escalator device stops moving upwards (Step Up Motion) or downwards (Step Down Motion) ?. 3. In events of electric power breakdowns where escalator comes to a standstill, how passengers safety is ensured as they is a chance of losing balance while moving up or down ?. Is the escalator machine battery powered up in case of mains power failure ?. Thanks & Regards, Prashant S Akerkar ANSWER: Question 1. An escalator is designed individually for each installation. So the length and angle can vary depending on the needs of a particular building. Question 2. Escalators can be self-limiting. It is a fairly simple calculation to determine how many people will fit on the escalator, the average weight of people, and to design the escalator to support that number of people, as well as some safety margin. That is why you never see a "maximum load" placard on an escalator - it will easily handle the number of people who fit on it. The major malfunctions with escalators are usually some type of control issue - including changes in speed or direction. These aren't typically caused by overload, but by control system failure. Question 3. It's fairly easy to design a gear-based system that will simply lock when power is removed. A worm type gear-reduction system will automatically lock when power is removed, and I expect this is the sort of mechanism used in escalators. While I'm not certain, this is how I expect they are designed. The same arrangement is used for tuning machines on musical instruments - such as guitars - precisely because they lock in position when the source of power is removed from the driving gear. http://en.wikipedia.org/wiki/Worm_drive ---------- FOLLOW-UP ---------- QUESTION: Dear Deirdre Thank you. 1. if a Comparison between Escalator and Elevator Systems is done in terms of Manufacturing, Installation, Maintenance Costs, Escalators will be more expensive (i.e. Cost incurred for Manufacturing, Installing and Maintenance) than Elevators ? 2. Can we implement Escalators in already built structures viz Shopping Malls, Departmental stores, Buildings, Airports, Metro Rails etc. Thanks & Regards, Prashant S Akerkar Escalators and elevators have different purposes. Escalators take up more floor space, but allow larger numbers of people to move between floors more quickly. There are dedicated up and down escalators. A separate escalator needs to be installed between each floor. An elevator permits fewer people to select between multiple floors. To handle a greater number of people, multiple elevators may be installed. In case of a power failure, an escalator is typically safer - they can still be used as stairs if the power fails. Questioner's Rating Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Dear Deirdre Thank you. Thanks & Regards, Prashant S Akerkar Electronic Components Volunteer #### Deirdre Hebert ##### Expertise I can answer most questions regarding electronic components, what they are and how to use them. ##### Experience I worked for a number of years in the electronic component testing industry, designing and building automated test equipment for the electronic manufacturing industry. Publications Numerous technical manuals for the equipment we built. Education/Credentials UNH, CCAF and others Past/Present Clients General Electric, Motorola, Ford, Sensonor and others.	1,099	4,871	{"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}	2.796875	3	CC-MAIN-2017-13	latest	en	0.820333
http://article.sciencepublishinggroup.com/html/10.11648.j.ijsd.20150101.11.html	1,603,361,277,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00157.warc.gz	8,649,168	12,036	International Journal of Statistical Distributions and Applications Volume 1, Issue 1, September 2015, Pages: 1-4 Characterizations and Infinite Divisibility of Extended COM-Poisson Distribution Huiming Zhang School of Mathematics and Statistics, Central China Normal University, Wuhan, China Huiming Zhang. Characterizations and Infinite Divisibility of Extended COM-Poisson Distribution. International Journal of Statistical Distributions and Applications. Vol. 1, No. 1, 2015, pp. 1-5. doi: 10.11648/j.ijsd.20150101.11 Abstract: This article provides some characterizations of extended COM-Poisson distribution: conditional distribution given the sum, functional operator characterization (Stein identity). We also give some conditions such that the extended COM-Poisson distribution is infinitely divisible, hence some subclass of extended COM-Poisson distributions are discrete compound Poisson distribution. Keywords: Conway-Maxwell-Poisson distribution, conditional distribution, discrete compound Poisson distribution, infinitely divisible, Stein identity 1. Extended COM-Poisson Distribution The Conway-Maxwell-Poisson distribution (COM-Poisson distribution) was firstly briefly introduced by Conway and Maxwell (1962) for modeling of queuing systems with state-dependent service times, see Shmueli et al. (2005) for details. The probability mass function (p.m.f.) is given by where and . Chakraborty(2015) introduced the extended COM-Poisson distribution from Imoto (2014), Chakraborty and Ong (2014): Definition A: A r.v. X is said to follow the extended COM-Poisson distribution with parameters [ECOMP ] if its p.m.f. is given by where the parameter space is . The extended COM-Poisson includes many COM type distributions. For example, COM-Poisson, COM-negative binomial etc., see Chakraborty(2015). It is easy to obtain the p.m.f. of recurrence relation of the ECOMP (1) with . The recurrence relation (1) will be useful for arriving some characterizations and properties of the extended COM-Poisson distribution. There are many statistical models pertain to COM-Poisson for modeling many types of count data, such as COM-Poisson regression (see Guikema and Goffelt (2008)), COM-Poisson in survival analysis (see Rodrigues et al. (2009), Sellers and Shmueli (2010)), COM-Poisson INGARCH time series models (see Zhu (2012)), COM-Poisson distribution chart in statistical process control (see Saghir et al. (2009)), zero-inflated COM-Poisson distribution (see Barriga and Phillips (2014)), fitting data from actuarial science (see Khan and Khan (2010)). Undoubtedly, extended COM-Poisson distribution with 4 parameters are more flexible than COM-Poisson distribution, negative binomial distribution and so on. It contains a wide range of statistical properties (log-concave and log-convex, under- and over dispersed). Especially, for the phenomenon of over-/equi-/underdispersion, there are many applications in applied statistics, including public health, medicine, and epidemiology, see the literatures review by Kokonendji (2014). So we can consider statistical models above in terms of extended COM-Poisson distribution in the future. 1.1. Functional Operator Characterization We know the p.m.f. of ECOMP has a simple recurrence relation given by (1). Brown and Xia (2001) studied a very large class of stationary distribution of birth-death process with arrival rate and service rate by the recursive formula: . Thus we can construct that arrival rate and service rate to characterize the ECOMP, , see also Daly and Gaunt (2015). This will be useful for functional characterization in Stein’s method. The following theorem presents a characterization for the ECOMP distributions. Theorem A: (Stein identity) Let , the r.v. X has distribution ECOMP iff the equation holds for any bounded function . The Theorem A can be directly obtained from the work of Brown and Xia (2001). On the one hand, we just check the expectation is 0, and then the sufficiency is true. On the other hand, to show the necessary, just take function g to be the indicator of {k}, hence we have the recurrence relation (1). 1.2. Conditional Distribution Characterization For the two independent r.v.’s ECOMP and ECOMP, the distribution of sum is So, the conditional distribution of is We naturally define the extended negative hypergeometric distribution with p.m.f. (2) where is the normalization constant. And we denote (2) as . Next, we cite a general result by Patil and Seshadri (1964) for characterizing a large class of discrete distributions, see also Kagan et al. (1973). And we obtain the conditional distribution characterization of ECOMP. Lemma A: Let X and Y be independent discrete r.v.’s and two dimensional function be . If where is a nonnegative function, then where Theorem B: Let X and Y be the independent discrete r.v. with and . If the conditional distribution of is extended negative hypergeometric distribution, then ECOMP and ECOMP. Proof: By using Lemma A and the definition of , we have . Then So, we have and . Let, then Hence we get the result. 2. Log-Convex and Infinitely Divisible For a discrete r.v. X with p.g.f. We say that X is infinitely divisible if where is the p.g.f. of a certain discrete r.v. . Feller's characterization of the infinite divisible discrete r.v. (see Feller (1971)) shows that a discrete r.v. is infinitely divisible iff its distribution is a discrete compound Poisson distribution with following p.g.f. where is the Lévy measure(or parameters) of a infinitely divisible distribution. It satisfies . Actually, the discrete compound Poisson distribution is the probability distribution of the sum of a number of iid non-negative integer-valued r.v.’s, where the sum is Poisson-distributed r.v.. On the one hand, the discrete compound Poisson(DCP) distributed r.v. can be represented as the sum of n i.i.d. r.v.’s , where and are independent with . On the other hand, X can be decomposed as sum of weighted Poisson: where are independently Poisson distributed with parameter . For the detailed theoretical and applied treatment of discrete infinitely divisible and discrete compound Poisson, we refer readers to section 2 of Steutel and van Harn (2003), section 9.3 of Johnson et al. (2005), Zhang et al. (2014), Zhang et al. (2013). A discrete r.v. X with has log-concave (log-convex) p.m.f. if Steutel (1970) proved that all log-convex discrete distributions are infinitely divisible, see also Steutel and van Harn (2003). So, it is easy to get infinite divisibility of ECOMP when satisfy some conditions. Theorem C: The ECOMP infinitely divisible distribution when satisfy the following conditions: . Proof: From (1), we have Then, the equality holds if for Notice that goes to its minimum as . Hence, we need to make sure the infinite divisibility of ECOMP. Applying the recurrence relation (Lévy-Adelson-Panjer recursion) of p.m.f. of DCP distribution, , see Buchmann (2003) and the Remark 1 in Zhang et al. (2014), therefore DCP case of ECOMP has the alternative recurrence relation. The parameters and of DCP case of ECOMP are determined by the following systems of equation: where . The is ECOMP log-concave distribution if satisfy the inequality for Note that ensures validity of the inequality above. For more theoretical results about general log-concavity of discrete distributions, see Balabdaoui et al. (2013), Saumard and Wellner (2014). Acknowledgments In the author’s future articles, more new characterizations and properties related to this notes in aspects of COM-Poisson distribution will appear, and Extended COM-Poisson distribution in aspects of related statistical models are in progress. If you are interested it, please contact me without hesitation. References 1. Balabdaoui, F., Jankowski, H., Rufibach, K., Pavlides, M. (2013). Asymptotics of the discrete log-concave maximum likelihood estimator and related applications. Journal of the Royal Statistical Society: Series B (StatisticalMethodology), 75(4), 769-790. 2. Barriga, G. D., Louzada, F. (2014). The zero-inated Conway-Maxwell-Poisson distribution: Bayesian inference, regression modeling and inuence diagnostic. Statistical Methodology, 21, 23-34. 3. Brown, T. C., Xia, A. (2001). Stein's method and birth-death processes. Annals of probability, 1373-1403. 4. Buchmann, B., Grbel, R. (2003). Decompounding: an estimation problem for Poisson random sums. The Annals of Statistics,31(4), 1054-1074. 5. Chakraborty, S. (2015). A new extension of Conway-Maxwell-Poisson distribution and its properties. arXiv preprintarXiv:1503.04443. 6. Chakraborty, S., Ong, S. H. (2014), A COM-type Generalization of the Negative Binomial Distribution, Accepted in April 2014, to appear in Communications in Statistics-Theory and Methods 7. Conway, R. W., Maxwell, W. L. (1962). A queuing model with state dependent service rates. Journal of Industrial Engineering, 12(2), 132-136. 8. Daly, F., Gaunt, R. E. (2015). The Conway-Maxwell-Poisson distribution: distributional theory and approximation.arXiv preprint arXiv:1503.07012. 9. Feller, W. (1971). An introduction to probability theory and its applications,Vol. I. 3rd., Wiley, New York. 10. Guikema, S. D.,Goffelt, J. P. (2008). A exible count data regression model for risk analysis. Risk analysis, 28(1), 213-223. 11. Imoto, T. (2014). A generalized Conway–Maxwell–Poisson distribution which includes the negative binomial distribution. Applied Mathematics and Computation, 247, 824-834. 12. Kagan, A. M., Linnik, Y. V., Rao, C. R. (1973). Characterization problems in mathematical statistics, Wiley. 13. Khan, N. M., Khan, M. H. (2010). Model for Analyzing Counts with Over-, Equi-and Under-Dispersion in Actuarial Statistics. Journal of Mathematics and Statistics, 6(2), 92-95. 14. Kokonendji, C. C. (2014). Over-and Underdispersion Models. Methods and Applications of Statistics in Clinical Trials: Planning, Analysis, and Inferential Methods, Volume 2, 506-526. 15. Patil, G. P., Seshadri, V. (1964). Characterization theorems for some univariate probability distributions. Journal of the Royal Statistical Society. Series B (Methodological), 286-292. 16. Sellers, K. F., Shmueli, G. (2010). Predicting censored count data with COM-Poisson regression. Robert H. Smith School Research Paper No. RHS-06-129. 17. Rodrigues, J., de Castro, M., Cancho, V. G., Balakrishnan, N. (2009). COM-Poisson cure rate survival models and an application to a cutaneous melanoma data. Journal of Statistical Planning and Inference, 139(10), 3605-3611. 18. Saghir, A., Lin, Z., Abbasi, S. A., Ahmad, S. (2013). The Use of Probability Limits of COM-Poisson Charts and their Applications. Quality and Reliability Engineering International, 29(5), 759-770. 19. Saumard, A., Wellner, J. A. (2014). Log-concavity and strong log-concavity: a review. Statistics Surveys, 8, 45-114. 20. Sellers, K. F., Borle, S., Shmueli, G. (2012). The COM-Poisson model for count data: a survey of methods and applications. Applied Stochastic Models in Business and Industry, 28(2), 104-116. 21. Shmueli, G., Minka, T. P., Kadane, J. B., Borle, S., Boatwright, P. (2005). A useful distribution for ftting discrete data: revival of the Conway-Maxwell-Poisson distribution. Journal of the Royal Statistical Society: Series C (Applied Statistics), 54(1),127-142. 22. Steutel, F. W. (1970). Preservation of infinite divisibility under mixing and related topics. MC Tracts, 33, 1-99. 23. Steutel, F. W., Van Harn, K. (2003). Infinite divisibility of probability distributions on the real line. CRC Press, New York. 24. Zhang, H., He J., Huang H. (2013).On nonnegative integer-valued Lévy processes and applications in probabilistic number theory and inventory policies American Journal of Theoretical and Applied Statistics,2 (5), 110-121. 25. Zhang, H., Liu, Y., Li, B. (2014). Notes on discrete compound Poisson model with applications to risk theory. Insurance: Mathematics and Economics, 59, 325-336. 26. Zhu, F. (2012). Modeling time series of counts with COM-Poisson INGARCH models. Mathematical and Computer Modelling, 56(9), 191-203. Contents 1. 1.1. 1.2. 2. Article Tools	3,147	12,225	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2020-45	latest	en	0.780024
https://codedump.io/share/ZnYuD5klSObw/1/performance-of-dickey-fuller-test-on-nearly-non-stationary-series	1,529,856,106,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267866965.84/warc/CC-MAIN-20180624141349-20180624161349-00523.warc.gz	574,437,590	9,521	Ira Saktor - 1 year ago 95 R Question # Performance of DIckey Fuller test on nearly non-stationary series when applying an (augmented) dickey fuller test on nearly non-stationary series (i.e. y_t = 0.97y_{t-1} + e_t), the test should - from what I have read - perform quite poorly. Yet when I apply this test to sufficiently long series (1000 time periods), i get very accurate results. When I perform the test on shorter series (100 time periods) the results are indeed poor. Is there some way how to quantify the impact of sample size on DF test? Or is my code wrong and the test should indeed perform poorly? For T = 100 indeed poor performance: ``````# set seed for reproducibility set.seed(3) # number of repetitions repetitions <- 100 # length of time series time <- 100 # generate matrices of 100 nearly non stationary processes of length 100 unitRootMatrix <- NULL nearlyNonStationaryMatrix <- NULL for (i in 1:repetitions) { # create vector of iid random errors errors2 <- rnorm(time, mean = 0, sd = 2) # create placeholders for processes temp2 <- rep(0, time + 1) for (j in 2:time) { temp2[j] = 0.95temp2[j-1] + errors2[j - 1] } # bind recent process to previous processes nearlyNonStationaryMatrix <- cbind(nearlyNonStationaryMatrix, temp2) } # Augmented Dickey Fuller test library(tseries) temp <- NULL pvals <- NULL for (i in 1:ncol(nearlyNonStationaryMatrix)) { pvals <- c(pvals, temp) } sum(pvals > 0.1) ## [1] 83 sum(pvals > 0.05) ## [1] 90 sum(pvals > 0.01) ## [1] 98 `````` for T = 1000 not so much: ``````# set seed for reproducibility set.seed(3) # number of repetitions repetitions <- 100 # length of time series time <- 1000 # generate matrices of 100 nearly non stationary processes of length 1000 unitRootMatrix <- NULL nearlyNonStationaryMatrix <- NULL for (i in 1:repetitions) { # create vector of iid random errors errors2 <- rnorm(time, mean = 0, sd = 2) # create placeholders for processes temp2 <- rep(0, time + 1) for (j in 2:time) { temp2[j] = 0.95*temp2[j-1] + errors2[j - 1] } # bind recent process to previous processes nearlyNonStationaryMatrix <- cbind(nearlyNonStationaryMatrix, temp2) } # Augmented Dickey Fuller test library(tseries) temp <- NULL pvals <- NULL for (i in 1:ncol(nearlyNonStationaryMatrix)) { pvals <- c(pvals, temp) } sum(pvals > 0.1) ## [1] 0 sum(pvals > 0.05) ## [1] 0 sum(pvals > 0.01) ## [1] 5 `````` Any thoughts, comments will be much appreciated! :) ADF is known to perform poorly against near-unit-root alternatives, it is, nevertheless, consistent. ``````set.seed(3) library(tseries) repetitions <- 10000 sim <- function(time){ x <- replicate(repetitions , arima.sim(list(ar=0.95), time, innov=rnorm(time, mean=0, sd=2))) #near unit root pvals <- rep(0, repetitions) for (i in 1:ncol(x)) {	863	2,782	{"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}	2.65625	3	CC-MAIN-2018-26	latest	en	0.777981
https://de.mathworks.com/matlabcentral/cody/problems/21-return-the-3n-1-sequence-for-n/solutions/224674	1,579,941,318,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00229.warc.gz	399,938,089	15,661	Cody # Problem 21. Return the 3n+1 sequence for n Solution 224674 Submitted on 28 Mar 2013 by Rodrigo Portugal 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 = 1; c_correct = 1; assert(isequal(collatz(n),c_correct)) 2 Pass %% n = 2; c_correct = [2 1]; assert(isequal(collatz(n),c_correct)) 3 Pass %% n = 5; c_correct = [5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct)) 4 Pass %% n = 22; c_correct = [22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct))	221	624	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2020-05	latest	en	0.466608
https://www.coin-or.org/CppAD/Doc/lu_ratio.cpp.htm	1,516,398,128,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00382.warc.gz	880,809,129	3,792	Prev Next Index-> contents reference index search external Up-> CppAD Appendix numeric_ad LuRatio lu_ratio.cpp Appendix-> Faq directory Theory glossary Bib wish_list whats_new deprecated compare_c numeric_ad addon License numeric_ad-> BenderQuad opt_val_hes LuRatio LuRatio-> lu_ratio.cpp lu_ratio.cpp Headings $\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }$ LuRatio: Example and Test # include <cstdlib> // for rand function # include <cassert> # include <cppad/cppad.hpp> namespace { // Begin empty namespace CppAD::ADFun<double> NewFactor( size_t n , const CPPAD_TESTVECTOR(double) &x , bool &ok , CPPAD_TESTVECTOR(size_t) &ip , CPPAD_TESTVECTOR(size_t) &jp ) { using CppAD::AD; using CppAD::ADFun; size_t i, j, k; // values for independent and dependent variables CPPAD_TESTVECTOR(AD<double>) Y(nn+1), X(nn); // values for the LU factor CPPAD_TESTVECTOR(AD<double>) LU(nn); // record the LU factorization corresponding to this value of x AD<double> Ratio; for(k = 0; k < nn; k++) X[k] = x[k]; Independent(X); for(k = 0; k < nn; k++) LU[k] = X[k]; CppAD::LuRatio(ip, jp, LU, Ratio); for(k = 0; k < nn; k++) Y[k] = LU[k]; Y[nn] = Ratio; // use a function pointer so can return ADFun object ADFun<double> FunPtr = new ADFun<double>(X, Y); // check value of ratio during recording ok &= (Ratio == 1.); // check that ip and jp are permutations of the indices 0, ... , n-1 for(i = 0; i < n; i++) { ok &= (ip[i] < n); ok &= (jp[i] < n); for(j = 0; j < n; j++) { if( i != j ) { ok &= (ip[i] != ip[j]); ok &= (jp[i] != jp[j]); } } } return FunPtr; } bool CheckLuFactor( size_t n , const CPPAD_TESTVECTOR(double) &x , const CPPAD_TESTVECTOR(double) &y , const CPPAD_TESTVECTOR(size_t) &ip , const CPPAD_TESTVECTOR(size_t) &jp ) { bool ok = true; using CppAD::NearEqual; double eps99 = 99.0 std::numeric_limits<double>::epsilon(); double sum; // element of L * U double pij; // element of permuted x size_t i, j, k; // temporary indices // L and U factors CPPAD_TESTVECTOR(double) L(nn), U(nn); // Extract L from LU factorization for(i = 0; i < n; i++) { // elements along and below the diagonal for(j = 0; j <= i; j++) L[i * n + j] = y[ ip[i] * n + jp[j] ]; // elements above the diagonal for(j = i+1; j < n; j++) L[i * n + j] = 0.; } // Extract U from LU factorization for(i = 0; i < n; i++) { // elements below the diagonal for(j = 0; j < i; j++) U[i * n + j] = 0.; // elements along the diagonal U[i * n + i] = 1.; // elements above the diagonal for(j = i+1; j < n; j++) U[i * n + j] = y[ ip[i] * n + jp[j] ]; } // Compute L * U for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { // compute element (i,j) entry in L * U sum = 0.; for(k = 0; k < n; k++) sum += L[i * n + k] * U[k * n + j]; // element (i,j) in permuted version of A pij = x[ ip[i] * n + jp[j] ]; // compare ok &= NearEqual(pij, sum, eps99, eps99); } } return ok; } } // end Empty namespace bool LuRatio(void) { bool ok = true; size_t n = 2; // number rows in A double ratio; // values for independent and dependent variables CPPAD_TESTVECTOR(double) x(nn), y(nn+1); // pivot vectors CPPAD_TESTVECTOR(size_t) ip(n), jp(n); // set x equal to the identity matrix x[0] = 1.; x[1] = 0; x[2] = 0.; x[3] = 1.; // create a fnction object corresponding to this value of x CppAD::ADFun<double> FunPtr = NewFactor(n, x, ok, ip, jp); // use function object to factor matrix y = FunPtr->Forward(0, x); ratio = y[nn]; ok &= (ratio == 1.); ok &= CheckLuFactor(n, x, y, ip, jp); // set x so that the pivot ratio will be infinite x[0] = 0. ; x[1] = 1.; x[2] = 1. ; x[3] = 0.; // try to use old function pointer to factor matrix y = FunPtr->Forward(0, x); ratio = y[nn]; // check to see if we need to refactor matrix ok &= (ratio > 10.); if( ratio > 10. ) { delete FunPtr; // to avoid a memory leak FunPtr = NewFactor(n, x, ok, ip, jp); } // now we can use the function object to factor matrix y = FunPtr->Forward(0, x); ratio = y[nn]; ok &= (ratio == 1.); ok &= CheckLuFactor(n, x, y, ip, jp); delete FunPtr; // avoid memory leak return ok; } Input File: example/general/lu_ratio.cpp	1,461	4,351	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2018-05	latest	en	0.352226
http://www.stata.com/statalist/archive/2005-08/msg00925.html	1,432,283,681,000,000,000	text/html	crawl-data/CC-MAIN-2015-22/segments/1432207924799.9/warc/CC-MAIN-20150521113204-00162-ip-10-180-206-219.ec2.internal.warc.gz	737,220,027	3,014	Re: st: Indexing Question From "kelly johnson" To statalist@hsphsun2.harvard.edu Subject Re: st: Indexing Question Date Mon, 29 Aug 2005 12:10:28 -0400 What if I want to do the same calculation, but in a specific time period. Eg. in my data set, I want to calculate teh percentage change in X between 7/2/05-7/5/05 (for all observations with data in that time period). Thank you! Kelly From: Ulrich Kohler <kohler@wz-berlin.de> To: statalist@hsphsun2.harvard.edu Subject: Re: st: Indexing Question Date: Mon, 29 Aug 2005 18:00:20 +0200 . by id (date), sort: gen change = (X[_N]-X[1])/X[1] * 100 Uli kelly johnson wrote: > I have a question regading calculations and indexing. Below I have an > except of a larrge data set. How can I calculate the percentage change in > the variable X from the earliest date to the last date (e.g. in the case > where id==1, the percentage change in X from 7/1/05-7/6/05). I want to do > this by id for all my observations (i have thousands of obsevations and the > date periods are of varying lengths...). I know it's of the 'by id:' type > of statements, but I just can't get it right. Thank you. > > Kelly > > > date id X > 7/1/05 1 .2 > 7/2/05 1 .3 > 7/3/05 1 .2 > 7/4/05 1 .1 > 7/5/05 1 .5 > 7/6/05 1 .6 > 6/16/05 2 .4 > 6/17/05 2 .4 > 6/18/05 2 .4 > 6/19/05 2 .2 > 6/20/05 2 .1 > 6/21/05 2 .3 > 6/22/05 2 .5 > 4/11/05 3 .2 > 4/12/05 3 .8 > 4/13/05 3 .7 > 4/14/05 3 .8 > 4/15/05 3 .2 > 4/16/05 3 .2 > 4/17/05 3 .3 > 4/18/05 3 .1 > 4/16/05 3 .5 > ... > > _________________________________________________________________ > Express yourself instantly with MSN Messenger! Download today - it's FREE! > http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/ > > * > * For searches and help try: > * http://www.stata.com/support/faqs/res/findit.html > * http://www.stata.com/support/statalist/faq > * http://www.ats.ucla.edu/stat/stata/ -- kohler@wz-berlin.de +49 (030) 25491-361 * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ _________________________________________________________________ Is your PC infected? Get a FREE online computer virus scan from McAfeeŽ Security. http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963 * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/	898	2,470	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2015-22	longest	en	0.708793
isb.creanet-64.fr	1,642,589,357,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00017.warc.gz	36,887,048	12,321	# KS3 Maths Mathematics are the stem of all other scientific subjects. Numerical and computational fluency is increasingly important in the digital era. The study of this subject is also a great opportunity for students to get a first approach of logic and to develop rigorous reasoning skills. Students learn mainly about four areas of mathematics: numbers, geometry, algebra and probabilities & statistics. They will develop a core knowledge of these fields and start building basic tools to understand the world they live in. # KS4 GCSE Maths Foundation This GCSE syllabus will reinforce and extend the topics approached in KS3. It covers: • Further fractions and powers study • Further work on geometrical measures such as length, area and volume, including trigonometric ratios • Extended study of linear equations, inequalities and graphs • An approach of formulae and more complex graphs • A variety of ways to represent data and to use probabilities and statistics • An introduction to vectors. The Edexcel GCSE is graded using the 9 – 1 system, where 9 is the highest grade and 1 the lowest. Grades 9 – 4 are considered ‘pass’ grades. At Foundation level, the highest grade that it is possible for a student to achieve is a 5. # KS4 GCSE Mathematics Higher This GCSE syllabus covers all the topics listed above under Foundation, in addition, it also covers in-depth: • Quadratic, cubic and exponential functions • Trigonometry in all kind of triangles and extended to all types of angles • The circle theorems • Interpretation of equations and graphs and the way they transform. KS5 A Level Mathematics The KS5 mathematics syllabus covers a range of pure mathematics applied mathematics topics, allowing students to get a feel for the mathematics involved in different fields of engineering. The pure mathematics syllabus allows students to achieve mastery of the GCSE most advanced algebra topics and take them to an extended study of calculus. Some of these mathematics are applied to the study of motion and forces and to advanced probabilities and statistical enquiries.	425	2,096	{"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-2022-05	latest	en	0.911603
http://www.148apps.com/app/579232274/	1,524,546,105,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00421.warc.gz	340,619,858	11,821	### What's New - Stores values like tax that probably remain the same - Added Simple Amortisation calculator to loans - Added separators to long results values e.g. 2,400 - Enlarged keypad buttons for easier entry and readability - Added Tip Amount value to the tip calculator ### App Description We’ve created Calcul8 to be a perfect all-in-one calculator for all those everyday scenarios that can get us scratching our heads. Simply choose a calculator, plug in the numbers and Calcul8 will give you the answers as if by magic. Calcul8 really is an everyday calculator wizard! We've taken 6 really useful calculators and lavished them with beautiful pixels and animation to make Calcul8 a joy to use. Here's what's in the box: ✪ Percentage Calculator ✪ Working out percentage calculations by hand can be extremely confusing. Calcul8 can give you answers to 8 popular calculations: ∙Percentage (%) of a Number - What is the % (a) of a value (b) ∙Percentage (%) Changed - How much value (a) has changed from value (b) ∙Percentage (%) Difference - What is the % difference between value (a) and (b) ∙Original Number - The value (a) is (b)% of what number ∙Percentage (%) Increase - What is value (a) increased by % (b) ∙Percentage (%) Decrease - What is value (b) decreased by % (b) ∙Before Percentage (%) Decrease - What was value (a) before a (b)% decrease ∙Before Percentage (%) Increase - What was value (a) before a (b)% increase ✪ Sale Price Calculator ✪ Sale shopping? Yes! But how much is 35% off? Calcul8 will tell you how much you will pay and how much you will save. ✪ Date Calculator ✪ Performing calculations with dates and times is almost impossible to work out in your head. Calcul8 can help you with 4 popular calculations: ∙Age - A person’s age and time until their next birthday ∙Difference - How much time there is between two dates ∙What Day - What day a particular date falls on, and which week of the year it is in ∙Arithmetic - Add or subtract time from a date ✪ Gratuity Calculator ✪ You’ve all had a great meal and maybe a few too many glasses of wine. But now it’s time to work out how to split the bill. Easy... let Calcul8 work it out for you! ✪ Loan Calculator ✪ Loans. A necessity for a lot of people these days. Understand exactly what a loan will cost you with these 5 useful calculations: ∙Simple Amortisation - Amortised interest mortgage calculation ∙Simple Interest - Simple interest calculation ∙Compound Daily - Interest calculated daily ∙Compound Monthly - Interest calculated monthly ∙Compound Annually - Interest calculated annually ✪ Pocket Calculator ✪ A simple everyday pocket calculator ### iPhone Screenshots (click to enlarge) ### App Changes • November 23, 2012 Initial release • December 04, 2012 Price increase: \$0.99 -> \$1.99 • December 04, 2012 Price decrease: \$1.99 -> \$0.99 • January 08, 2013 Price increase: \$0.99 -> \$1.99 • February 22, 2013 New version 1.1 • June 23, 2013 Price decrease: \$1.99 -> \$0.99 • June 28, 2013 Price increase: \$0.99 -> \$1.99 • October 30, 2013 Price decrease: \$1.99 -> FREE! • November 01, 2013 Price increase: FREE! -> \$1.99 • November 30, 2013 Price decrease: \$1.99 -> FREE! • December 02, 2013 Price increase: FREE! -> \$1.99	850	3,237	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.34375	3	CC-MAIN-2018-17	latest	en	0.919007
https://moam.info/approximate-method-for-calculating-a-thickwalled-cylinder-with-wwwresearchgatene_606f0810097c47e5558b456f.html	1,618,838,154,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00071.warc.gz	502,884,135	12,123	Approximate method for calculating a thickwalled cylinder with ...www.researchgate.net › publication › fulltext › Approxim Approximate method for calculating a thickwalled cylinder with ... › publication › fulltext › Approxim... › publication › fulltext › Approxim...by V Andreev · ‎2018 · ‎Cited by 1 · ‎Related articlesThis paper considers an approximate method for solving th Approximate method for calculating a thickwalled cylinder with rigidly clamped ends Vladimir Andreev1,* 1 Moscow State University of Civil Engineering, Yaroslavskoe shosse, 26, Moscow, 129337, Russia Abstract. Numerous papers dealing with the calculations of cylindrical bodies [1 -8 and others] have shown that analytic and numerical-analytical solutions in both homogeneous and inhomogeneous thick-walled shells can be obtained quite simply, using expansions in Fourier series on trigonometric functions, if the ends are hinged movable (sliding support). It is much more difficult to solve the problem of calculating shells with builtin ends. 1 Introduction As was shown in [8], the natural restriction of the method of separation of variables in the calculation of radially inhomogeneous cylinders is the impossibility of satisfying arbitrary boundary conditions at the ends. This paper considers an approximate method for solving the axisymmetric problem for a thick-walled cylinder with rigid fixed ends. The problem is solved in displacements. 2 State of the problem In solving the axisymmetric problem will be sought displacements in the form:  u (r , , z )  v(r , , z )   umn (r )  cos m cos kn z; m n 0  0     vmn (r )  sin m cos kn z; m 1 n 0  w(r , , z ) (1)   wmn (r )  cos m sin kn z , m n 1  0 From (1), the boundary conditions automatically follow * Corresponding author: [email protected] E3S Web of Conferences 33, 02026 (2018) https://doi.org/10.1051/e3sconf/20183302026 HRC 2017 rz  z  0 z  0, H ; w  (2) which correspond to the hinged (sliding) support (Н – cylinder height). Below we consider an approximate calculation method cylinder with rigid fixing ends. In this case, the boundary conditions at the ends are written in the form: z  0, H ; u w 0 . (3) Fig. 1 shows half the cross-section of a cylindrical shell for two ways of fixing the ends. Fig 1. Thick-walled cylinder, a – hinged movable support, b – built-in (rigid fixing) end. 3 Approximate method for calculating The proposed method is based on the algorithm for calculating cylinders for the action of surface loads, which use an expanded area, compensating loads and the collocation method. We consider the problem when the external loads are symmetric with respect to the plane z1=H/2 (Fig. 2). Let us increase the height of the cylinder in both directions by an amount  and introduce the coordinate z=z1+  We apply to the lateral surfaces of the cylinder in the interval [0,] and [H+, H+2] such compensating loads pac, pbc, qac,qbc, that at the given points r=r1 and r=r2 at z= and z= +H (Fig. 3) the following conditions are satisfied: r r1, r2 ; u w 0 . (4) Fig. 3. Compensating loads in expanded area. Fig. 2. Calculation scheme for a cylinder with an expanded area. If equalities (4) are considered as conditions of real fixation, then it should be added that at the other points of the layers z   and z    H the displacements can be either positive or negative. In this case, it is necessary to accept the hypothesis of an ideal contact between the ends of the cylinder and the support surfaces, as is done in most contact 2 E3S Web of Conferences 33, 02026 (2018) https://doi.org/10.1051/e3sconf/20183302026 HRC 2017 problems, for example, in the calculation of beams on an elastic foundation. Let us consider the nature of the compensating loads. If we imagine that these loads are the reaction of a semi-infinite array to the walls of a cylinder embedded in it, then as the distance from the layer z   to the layer z  0 , the tangential and normal loads should tend to zero. For a description of such loads can be approximately selected functions of the form:   z  f0     f   0   0  z  ;  z (5) H  . 2 Such functions well approximate normal loads and worse – tangents, which near the layer have a peak character and when z   they are equal to zero. Nevertheless, if we bear in mind that when the compensating loads are decomposed into Fourier series, the peak z  nature of the loads qac and qbc will also correspond quite accurately to their representations in the form of a series (Fig. 4). Fig. 4. Representation of function f by the Fourier series : - - - - formula (5), (1 - N  24 ; 2 - N  48 ). Fourier series The essence of this method is as follows. Representing compensating loads pac, pbc, qa ,qbc, in the form (5) and varying the parameters  and  , it is necessary to achieve the conditions (4) at the points A and B (see Fig. 3). It should be added that when using the collocation method, large deficiency of conditions (4) are possible at the remaining points of the interval a, b on the layer z   . Therefore, the choice of the above parameters of the c functions (5) must also be accompanied by an analysis of the indicated deficiency and its minimization. In this	1,519	5,293	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2021-17	latest	en	0.844745
https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1202/Unit_8%3A_Gases/8.8%3A_Non-ideal_(Real)_Gases	1,638,276,551,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00262.warc.gz	236,870,969	28,483	# 8.8: Non-ideal (Real) Gases Learning Objectives • To recognize the differences between the behavior of an ideal gas and a real gas. • To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law. The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. ## Pressure, Volume, and Temperature Relationships in Real Gases For an ideal gas, a plot of $$PV/nRT$$ versus $$P$$ gives a horizontal line with an intercept of 1 on the $$PV/nRT$$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in Figure $$\PageIndex{1}$$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in Figure $$\PageIndex{1}$$). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $$\PageIndex{2}$$ for $$N_2$$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $$\PageIndex{3}$$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected. Thus as shown in Figure $$\PageIndex{2}$$, at low temperatures, the ratio of $$PV/nRT$$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $$PV/nRT$$ versus $$P$$ plot for many gases. Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). ## The van der Waals Equation The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation, $\left(P + \dfrac{an^2}{V^2}\right) (V − nb)=nRT \label{6.9.1}$ a and b are empirical constants that are different for each gas. The values of $$a$$ and $$b$$ are listed in Table $$\PageIndex{1}$$ for several common gases. Table $$\PageIndex{1}$$: van der Waals Constants for Some Common Gases (see Table A8 for more complete list) Gas a ( (L2·atm)/mol2) b (L/mol) He 0.03410 0.0238 Ne 0.205 0.0167 Ar 1.337 0.032 H2 0.2420 0.0265 N2 1.352 0.0387 O2 1.364 0.0319 Cl2 6.260 0.0542 NH3 4.170 0.0371 CH4 2.273 0.0430 CO2 3.610 0.0429 The pressure term in Equation $$\ref{6.9.1}$$ —$$P + (an^2/V^2$$)—corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $$n^2/V^2$$ represents the concentration of the gas ($$n/V$$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $$\PageIndex{4}$$. The volume term—$$V − nb$$—corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the an2/V2 term must be added to the measured pressure to correct for these effects. Example $$\PageIndex{1}$$ You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? Given: volume of cylinder, mass of compound, pressure, and temperature Strategy: A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. B Obtain a and b values for Cl2 from Table $$\PageIndex{1}$$. Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. Solution: A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): $n=\dfrac{m}{M}=\rm\dfrac{500\;g}{70.906\;g/mol}=7.052\;mol$ Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure: $P=\dfrac{nRT}{V}=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L}=43.1\;atm$ If chlorine behaves like an ideal gas, you have a real problem! B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $$\PageIndex{1}$$. Solving for P gives $\begin{split}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\\&=\rm28.2\;atm\end{split}$ This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. Exercise $$\PageIndex{1}$$ A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the 1. ideal gas law. 2. van der Waals equation. Answer: a. 77 atm; b. 67 atm ## Liquefaction of Gases Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $$a$$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. A large value of a indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $$\PageIndex{5}$$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. ## Summary No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids.	2,932	12,816	{"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}	2.859375	3	CC-MAIN-2021-49	latest	en	0.8983
https://www.czesciogrodowe.pl/24819/1+carbon+atom+and+1+oxygen+atom+in+korea.html	1,638,304,882,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00076.warc.gz	812,627,195	5,710	# 1 carbon atom and 1 oxygen atom in korea ### 1 If 16g of oxygen contains 1 mole of oxygen atoms, … Dear Student, 1. The mass of 1 atom = Gram atomic mass of atom/ 6.022 × 10 23 So, the mass of 1 atom of oxygen = Gram atomic mass of oxygen atom/ 6.022 × 10 23 = 16g/ 6.022 × 10 23 = 2.66 × 10-23 g Due to time constraint, it would not be possible for us to provide the answers to all your queries. ### Structure & Reactivity: Atoms: Protons, Neutrons, Electrons If carbon in nature is about 99% 12 C and 1% 13 C, then what is the average weight of a carbon atom? Problem AT2.4. Note that 14 C is even rarer than 13 C, because 14 C is converted into 14 N via radioactive decay. ### carbon \| Facts, Uses, & Properties \| Britannica Carbon (C), nonmetallic chemical element in Group 14 (IVa) of the periodic table.Although widely distributed in nature, carbon is not particularly plentiful—it makes up only about 0.025 percent of Earth’s crust—yet it forms more compounds than all the other elements … ### Trending Articles about how many carbon atoms are in … 2020/8/21· Your source for the latest how many carbon atoms are in carbon dioxide articles. Follow how many carbon atoms are in carbon dioxide trends, innovations and developments on echemi. Fluid, Electrolyte, and Acid-base Balance ### Home Learning with BBC Bitesize - KS3 Secondary … Carbon dioxide contains one carbon atom and two oxygen atoms Salt contains one atom of sodium and one atom of chlorine We represent compounds using the syols for each of the elements in it. ### What Is An Atom? - 2012/11/13· Science expert Emerald Robinson explains what an atom is. To view over 15,000 other how-to, DIY, and advice videos on any topic, visit /p> ### The distance between the center of carbon and oxygen … Mass of carbon atom (m 1) = 12 Mass of oxygen atom ( m 2 ) = 16 Consider, the situation when C-atom is loed at origin, it is assumed this way because, center of mass has to be calculated with respect to C-atom. ### Carbon Monoxide—Oxygen Atom Reaction: The Journal … The kinetics and light emission of the carbon monoxide—oxygen atom reaction have been investigated. At pressures above 0.2 mm, carbon dioxide is formed by a bimolecular association reaction which appears to violate the electron spin conservation rule. The ### Structure of Hydrogen, Carbon, Oxygen and Nitrogen An oxygen atom has six electrons in its outer shell and a carbon atom has four electrons on its outer shell. The two oxygen atoms and 2 carbon atoms will share two electrons to form four covalent bonds, by doing this it will make a carbon dioxide molecule. ### 1 Lecture 3 Other Atoms in Organic Chemistry – nitrogen, oxygen … 1 Lecture 3 Other Atoms in Organic Chemistry – nitrogen, oxygen and halogens Valence Possibilities for Neutral Atoms Atom sp3 sp2 sp sp # bonds # lone pairs Zeff ∆χ C C C C N O F N O N carbon nitrogen oxygen ### Can you form a molecule by bonding 2 oxygen atoms … Covalent bonds hold the two oxygen atoms and one hydrogen atom together. The substance is the "protonated form of superoxide". It may also be formed by removing a hydrogen atom … ### Formal Charge - uni-haurg.de Figure 2 extends the procedure demonstrated in Figure 1 to methanol, CH 3 OH, where the oxygen atom has two lone pairs of electrons. Figure 2 Assigning Electrons II Since the two lone pairs on the oxygen are not shared with any other atoms, they are assigned to ### bonding in carbonyl compounds - the carbon oxygen … 2020/8/17· The oxygen atom Oxygen''s electronic structure is 1s 2 2s 2 2p x 2 2p y 1 2p z 1. The 1s electrons are too deep inside the atom to be concerned with the bonding and so we''ll ignore them from now on. Hybridisation occurs in the oxygen as well. It is easier to 2 ### Sigma Bonds with sp3 Hybrid Orbitals Sigma Bonds with sp 3 Hybrid Orbitals Atoms that have 4 bonds, 3 bonds and 1 lone pair, 2 bonds and 2 lone pairs, or 1 bond and 3 lone pairs need four hybrid orbitals 109 degrees apart. Coining an s orbital, a p x orbital, a p y orbital, and a p z orbital makes four, sp 3 orbitals in a tetrahedral array. ### The Elements Water is a compound with the formula H 2 O, meaning that one molecule (or formula unit) of water contains two hydrogen atoms and one oxygen atom. The compound sodium hydrogen carbonate has the formula NaHCO 3 , meaning that a single formula unit of this compound contains one atom of sodium, one atom of hydrogen, one atom of carbon, and three atoms of oxygen. ### How many electrons in ONE MOLE of carbon dioxide? \| … 2017/4/8· Therefore, you can say that every atom of carbon will contain #6# electrons and every atom of oxygen will contain #8# electrons. This means that you will have #"total no. of e"^(-) = overbrace(6 * 1.37 * 10^(24))^(color(blue)("coming from C atoms")) + overbrace(8 ### Which of the following statements is true? Boiling and … 2020/8/14· Which of the following statements is true? Boiling and melting points are chemical properties. A molecule of oxygen can have atoms of carbon in it. Elements are substances made of only one type of atom. The nuer of electron shells in an atom is called the ### Carbon Atom Images, Stock Photos & Vectors \| … Find carbon atom stock images in HD and millions of other royalty-free stock photos, illustrations and vectors in the Shutterstock collection. Thousands of new, high-quality pictures added every day. 33,883 carbon atom stock photos, vectors, and illustrations are ### what is co6 what is radicle what is define of veleoncy … what is co6 what is radicle what is define of veleoncy what is the charge of potassium charcolate how atom and molecules differ difference between ele - Chemistry ### Hidden Oxygen for Android (2020) - MobyGames Hidden Oxygen is an atom arranger puzzle game. The player''s goal is to figure out the correct layout of oxygen atoms on a square board. By understanding of the board, the next, correct move can be deduced.There are a nuer of rules. Oxygen atoms cannot be ### what is difference between atom and element? atom is a single unit like C (carbon atom) O(oxygen atom) which do not exist independently except Nobel gases (He,Ne, etc.) but element may be a single atom C ( carbon element / carbon atom) or a coination of two or similar atoms O2 (oxygen element) S8(sulphur element) ### Organic Nomenclature An oxygen atom between two sp 3-hybridized carbons. Epoxides A three atom ring containing one oxygen atom and two carbon atoms in which all atoms are sp 3-hybridized. Aldehydes A carbonyl group in which the carbon is bound to a hydrogen as well as ### Chem4Kids: Carbon: Orbital and Bonding Info Carbon Dioxide This is a carbon dioxide molecule. When you breathe out, you usually breathe out carbon dioxide. With the formula CO 2 that means there are two oxygen (O) atoms and one carbon (C) atom. If you look closely at the dot structure, you''ll see that ### Stoichiometry of Elements and Compounds A.1 INTRODUCTION In this appendix, we will look at some of the quantitative relationships associated with the mass of elements and compounds. This subset of chemistry is called stoichiometry, a word derived from the Greek word “stoikheion”, meaning element. ### Week 2, Lab 1: 9/15: Atoms - University of Michigan 28) Find the syol for carbon on the periodic table. Compare the nuer given just below the syol for carbon with the average mass (amu) of one carbon atom and with the mass (grams) of 6.022 x1023 carbon atoms. 29) Does any carbon atom have a	1,853	7,516	{"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-2021-49	latest	en	0.831164
https://forum.lazarus.freepascal.org/index.php/topic,62081.msg468991/topicseen.html?PHPSESSID=k86o4la3ppia1r13fh4dma80a5	1,680,326,512,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00439.warc.gz	315,033,792	9,944	### Bookstore Computer Math and Games in Pascal (preview) Lazarus Handbook ### Author Topic: [Solved] Query an image in a TPopupMenu (Read 371 times) #### petevick • Full Member • Posts: 197 ##### [Solved] Query an image in a TPopupMenu « on: January 30, 2023, 06:57:54 pm » I know that you'd add an image to a TPopupMenu by this code snippet.... Code: Pascal [Select][+][-] 2. bmp:= TBitmap.Create; 4. try 5. imgList.GetBitmap(idx, bmp); 6. bmp.Canvas.Brush.Color := clBlue; 7. bmp.Canvas.Brush.Style := bsSolid; 8. bmp.Canvas.Rectangle(0, 0, bmp.Width, bmp.Height); 9. bmp.TransparentColor := clWhite; 10. bmp.TransparentMode := tmFixed; 11. bmp.Transparent := true; 13. finally 14. bmp.Free; 16. end; ....but how would you query what the Canvas.Brush.Color is for any particular image in the TPopupMenu ? I was originally thinking it'd probably be a similar code structure to the above, but now I'm not so sure It might not even be possible of course « Last Edit: January 30, 2023, 08:51:57 pm by petevick » Pete Vickerstaff Linux Mint 20.3 Cinnamon, Lazarus 2.2.0, FPC 3.2.2 #### GetMem • Hero Member • Posts: 4031 ##### Re: Query an image in a TPopupMenu « Reply #1 on: January 30, 2023, 08:17:54 pm » Since Idx represents a unique image in the ImageList, save the Brush.Color to a dynamic array. Retrieve with FBrushColor(idx). #### petevick • Full Member • Posts: 197 ##### Re: Query an image in a TPopupMenu « Reply #2 on: January 30, 2023, 08:51:26 pm » Since Idx represents a unique image in the ImageList, save the Brush.Color to a dynamic array. Retrieve with FBrushColor(idx). Thanks for the reply GetMem. I get an Quote	507	1,729	{"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}	2.703125	3	CC-MAIN-2023-14	latest	en	0.617387
https://mapleprimes.com/products/Maple?page=13	1,601,087,119,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00126.warc.gz	493,215,423	40,472	## How to change the format of returned entities in a... Hi, I have created a Maple Workbook in which two worksheets are included. One worksheet does all the work and is password protected, while the other allows the user to insert input parameters and to view results. In the password protected worksheet, I return the required results using the following statement: > return matrix(G),matrix(A),matrix(DR),figplot where the last returned item, figplot, is a figure (display(..., size = [600,600])). When the results are requested in the other worksheet, Maple stacks the results on the same line, and when the width is not enough, the results overflow into the line next to it. I have two questions: 1- How can I print the returned entities (matrix(G), matrix(A), matrix(DR), figplot) on seperate lines instead of having them stacked in the same line? 2- Although I am specifying the size of the plot, when returned, the plot is very small. I tried to to return the output in a column matrix as follows: > return matrix(3,1,[matrix(G), [matrix(A),matrix(DR)], figplot]) and it worked but the column matrix brackets encompass the results (looks ugly) and the figure shown is still very small and not according to the required size. Thank you very much for your help ## Solve a PDE with a known solution... Can anyone help me to find a solution to for the attached partial differential equation ? I want to find a general solution to a partial differential equation by assuming that I know one solution, called , and trying to find another solution by assuming that the general solution in the form of . I want to restrict the second solution to be in the form of so that it satisfies the PDE, and is a function of times . The latter makes error as the maple identifies that the function depends on only one variable . Could you please help me to find a solution for in the form ? Also, I have trouble with defining the operator Do in the attached file. When it operates on , maple gives D. It is not clear for me that whether this derivative is with respect to or . I need is to define Do in a way so that the derivatives are correctly taken with respect to different separate variables. ## Performance discriminant procedure... Hi all, we know Maple provided discrim method to find discriminant of a polynomial I want to write a similar method with independent variable is ,... my code is below Some examples Maple already similar method? If not have, we can improve performance it? Thank you very much. ## Solve a system of partial differential equations w... Can anyone help me to solve the attached system of PDEs with a given expression for the HINT such as I am not able to set such an arbitray HINT function for system of PDEs. SysPDE_HINT.mw Thank you, ## Inverse Laplace Transform sort of works and sort o... Here are 4 statements that attempt to use invlaplace on the exponential function. Two work, two don't. Does anyone know why the two that don't work do that? Thank you. __________________________ with(inttrans); [addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace, invmellin, laplace, mellin, savetable] invlaplace(exp(-s),s,t); Dirac(t - 1) invlaplace(exp(s),s,t); invlaplace(exp(s), s, t) invlaplace(exp(s),s,t) assuming s<0,s::real; invlaplace(exp(s), s, t) invlaplace(exp(-s),s,t) assuming s<0,s::real; Dirac(t - 1) What is going on here? ## How to numerically solve nonlinear ODE... Hello! How do I solve ODE numerically. I tried, but I fall into a singularity when t is greater than 0.01. ## 3d plot question... I am a new user of Maple 2020. I have found that when I create a 3D plot, the point probe does not return the coordinates of the point that I click on. I would appreciate any help in using the point probe in a 3d plot. Thanks. ## How can I apply simplify command in parallel?... Dear All, I want to apply the ‘simplify’ command, in parallel, for the simplification of some parameters. Both Grid:-Map and Grid:-Run commands are tested. There is no error in both, whereas no simplification is implemented. It seems that the ‘simplify’ command correctly works on only ‘Master’ node, namely anywhere we are typing. Can anyone help me to simplify in parallel. I examined two following codes. 1) with (Grid); for i from 1 to nops(dummy_UU1) do freenode:=WaitForFirst(): Run(freenode,simplify,[dummy_UU1[i]],assignto='dummy_UU2'[i]): end do: Wait(); 2) The following code is correctly executed and resulted in the simplification of dummy_UU1 components in serial. for i from 1 to nops(dummy_UU1) do dummy_UU2[i]:=simplify(dummy_UU1[i]): end do: ## How can I extract a best-fit function for a list o... Good day to all. I have a data set consisting of 12 points in the x-y plane. These points form a periodic pattern and I was wondering if it is possible to obtain a curve that is a best-fit for these points. I have reason to expect that this pattern will repeat for successive points (i.e. points 13, 14, ..., 30). Does anyone know of a way to obtain a function that can represent this behavior? MaplePrimes_Best_Fit.mw ## plot3d of stable values... Hi, I have this values I want to plot as three dimension how can I plot it? ApproximateSol := proc (w, t) options operator, arrow; w^2+t+rtable(1 .. 1, 1 .. 1, {(1, 1) = (.2960214364piecewise(0 <= w and w < 1, 1)+.1939557186piecewise(0 <= w and w < 1, 4w-2)+0.2635024425e-1piecewise(0 <= w and w < 1, 16w^2-16w+3))(0.7620592980e-10piecewise(0 <= t and t < 1, 1)+0.1532687805e-9piecewise(0 <= t and t < 1, 4t-2)+0.8496500063e-10piecewise(0 <= t and t < 1, 16t^2-16t+3))+(-.2380729574piecewise(0 <= w and w < 1, 1)-0.9481780593e-1piecewise(0 <= w and w < 1, 4w-2)+0.442773709e-1piecewise(0 <= w and w < 1, 16w^2-16w+3))(-0.2362176714e-9piecewise(0 <= t and t < 1, 1)+0.6793447600e-10piecewise(0 <= t and t < 1, 4t-2)+0.2039654777e-9piecewise(0 <= t and t < 1, 16t^2-16t+3))+(0.7931761947e-1piecewise(0 <= w and w < 1, 1)+0.967928372e-2piecewise(0 <= w and w < 1, 4w-2)-0.2747829289e-1piecewise(0 <= w and w < 1, 16w^2-16w+3))(-0.1118492678e-9piecewise(0 <= t and t < 1, 1)-0.5202991920e-10piecewise(0 <= t and t < 1, 4t-2)+0.1128321001e-9piecewise(0 <= t and t < 1, 16t^2-16*t+3))}, datatype = anything, subtype = Matrix, storage = rectangular, order = Fortran_order) end proc ## Differential Transform Method... Could anyone help me out to convert the equation into differential transform method ## Import of Equation into word... Hello Everyone, can anyone explain how to import a mathematical equation from maple to word directly? ## The strandbeest Maple 2020 The strandbeest is a walking machine developed by Theo Jansen. Its cleverly designed legs consist of single-degree-of-freedom linkage mechanisms, actuated by the turning of a wind-powered crankshaft. His working models are generally large - something of the order of the size of a bus. Look for videos on YouTube. Commercially made small toy models are also available. This one sells for under \$10 and it's fun to assemble and works quite well. Beware that the kit consists of over 100 tiny pieces - so assembling it is not for the impatient type. Here is a Maple worksheet that produces an animated strandbeest. Link lengths are taken from Theo Jansen's video (go to his site above and click on Explains) where he explains that he calculated the optimal link lengths by applying a genetic algorithm. Here is a Maple animation of a single leg. The yellow disk represents the crankshaft. And here are two legs working in tandem: Here is the complete beest, running on six legs. The crankshaft turns at a constant angular velocity. The toy model noted above runs on twelve legs for greater stability. ## Write a program to display all different relation... Write a program to display all different relations on a set with 4 elements.	2,217	7,949	{"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-2020-40	latest	en	0.888869
https://mgtoml.com/tag/1-mg-to-ml/	1,713,982,250,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00868.warc.gz	361,926,907	48,089	1 MG To ML 1 milligram of liquid equals 0.001 milliliter The formula of 1 mg to ml: MG = 1 ml / (1000 * D) MG = milligrams ML = milliliters D = Density(g/mL) Method to Convert 1 mg to ml Conversion of 1 milligram to milliliters depends on the density of the substance, which varies from one … Read more	88	303	{"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-2024-18	latest	en	0.750951
https://www.adda247.com/school/bernoullis-theorem/	1,722,926,650,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00536.warc.gz	497,758,288	121,561	Home » Physics » Bernoulli's Theorem # Bernoulli’s Theorem- Statement, Equation, Derivation, Applications Bernoulli’s Theorem: Bernoulli’s Theorem, also known as Bernoulli’s principle, is an important theorem related to the movement of fluids. The formula is named after the famous Swiss Mathematician and Physicist Daniel Bernoulli. He postulated this theorem in 1738. This theorem provides a deep insight into fluid dynamics and explains the nature of the fluid while moving. It is applicable for a streamlined or steady flow. In this article, we will go through various aspects of this principle and will cover its formula, derivation, and applications. ## Bernoulli’s Theorem Bernoulli’s theorem is an important theorem for moving fluids that relates their pressure and velocity. The theorem states that the total amount of pressure energy, kinetic energy, and potential energy per unit volume is constant along any streamline flow of fluid. In other words, the combined mechanical energy of a moving fluid, which consists of the gravitational potential energy of elevation, the energy related to fluid pressure, and the kinetic energy of fluid motion, remains constant. This theorem is basically based on the conservation of energy principle. It must be noted that fluids comprise both liquids and gases. Using this theorem, it can be said that the pressure of a fluid is inversely related to its velocity, i.e., P ∝ 1/V where P = pressure of the fluid V = velocity of the fluid ## Bernoulli’s Theorem Equation The statement of the Bernoulli’s theorem stated above can be converted into mathematical expression. This is also known as the Bernoulli’s Theorem equation or Bernoulli’s Principle Formula. The equation of Bernoulli’s Theorem is given by: P + ρgh + (1/2)ρv² = constant where, P = pressure ρ = density of the fluid v = velocity of the fluid h = height of the fluid container g = acceleration due to gravity It is also known as Bernoulli’s equation. ## Bernoulli’s Theorem Derivation We can find out the formula written above for the Bernoulli’s principle. Let us derive this formula using a diagram. In the above figure we can see that a fluid is flowing from Side A to side B The height as side A is h1 while the height of the fluid at Side B is h2 The cross section area at A = a1 Cross section area at B = a2 The pressure of fluid at side A = P1 pressure of the fluid at side B = P2 Velocity of the fluid at side A = V1 Velocity of the fluid at side B = V2 As we know that fluid flows from high pressure to low pressure, so P1 > P2 Let us assume that during a certain period of time, the fluid displaced at side A be dX1 while the displacement of the fluid be dX2 at side B so that the same amount of fluid volume is passed. Then a small amount of work done (dw) on the fluid will be given by = F1.dX1 – F2.dX2 where F1 is the force applied at side A and F2 is the force applied at side B as F = P.A where P = pressure and A = area so, dw = P1.a1.dX1 – P2.a2.dX2 as area and length = volume, so, a.dX = dv dw = P1.dv – P2.dv or, dw = dv (P1-P2) The total energy of the given fluid system will be equal to the change in the potential energy and kinetic energy of the fluid. i.e., dw = du + dk (because we assume that no viscous force is present) —–(i) where du = change in potential energy and dk = change in kinetic energy dk = (1/2)m1.(V1)² – (1/2)m2.(V2)² as we go with the assumption that the density of the fluid remains constant at all the points as mass = density x volume so, dk = (1/2)ρ.dv.(V1)² – (1/2)ρ.dv.(V2)² dk = (1/2)ρ.dv(V1² – V2²) du = m2.g.h2 – m1.g.h1 du = ρ.dv.g.h2 – ρ.dv.g.h1 or, du = ρ.dv.g(h2-h1) putting the values of dw, du, and dk in equation (i), we get dv (P1-P2) = ρ.dv.g(h2-h1) + (1/2)ρ.dv(V1² – V2²) cancelling out dv from each term, we get (P1-P2) = ρ.g(h2-h1) + (1/2)ρ(V1² – V2²) on rearranging the above terms, we get P1 + ρ.gh1 + (1/2)ρV1² = P2 + ρ.gh2 + (1/2)ρV2² as we can see the total energy is conserved at both points Hence, P + ρgh + (1/2)ρv² = constant ## Bernoulli’s Theorem and Conservation of Energy It is important to note that the Bernoulli’s theorem obeys the principle of conservation of energy. In other words, we can use the principle of energy conservation to obtain the equation for the Bernoulli’s equation. As we can observe from the above derivation that the net work done by the fluid results from its change of potential energy (due to height difference) and change in kinetic energy (due to velocity). If we take the viscosity of the fluid into account, then, the equation will also have to include the thermal energy into account. Simply put, Bernoulli’s theorem obeys the energy conservation rule, i.e., total energy of a fluid = constant at all points in its flow ## Bernoulli’s Theorem Equation PDF As this theorem is one of the most asked topics in examination, so it is important for students to learn this concept thoroughly. Below we have provided the PDF for the Bernoulli’s principle equations in different cases. These equations will help students in solving different types of problems. Bernoulli’s Theorem Equations ## Bernoulli’s Theorem Applications There are many applications of the Bernoulli’s theorem in the real world. This theorem is widely used in many disciplines, including cardiovascular physiology and aviation. Some of the applications of Bernoulli’s theorem are stated below. • Finding Pressure Drops in Pipes: The pressure drop via a pipe can be calculated using this theorem. The friction between the fluid and the pipe walls is what results in the pressure drop. • Aerodynamics: The lift produced by airplane wings is explained by Bernoulli’s theorem. The design of the wings makes the upper surface of the wing move air more quickly than the bottom surface. As a result, there is a pressure difference between the wing’s top and bottom surfaces, which produces lift. • Cardiovascular System: Hemodynamics uses this Theorem to analyze blood flow in the circulatory system. It adds to the explanation of events like the Bernoulli effect, in which decreased pressure causes heart valves to close. • Venturi meter: It is a tool that gauges how quickly liquid moves through pipes and is based on Bernoulli’s principle. • Hydraulic Engineering: Engineers employ Bernoulli’s principle to create effective water purification systems, pipelines, and dams. The theorem helps these systems’ fluid flow and pressure distribution be optimized. ## Bernoulli’s Theorem Examples Some of the solved questions related to Bernoulli’s principle is given below so that students can have a proper understanding of this topic. Example 1: A horizontal pipe filled with water flowing from point A to B. Find the pressure at a different place when the pressure at point A is 500 Pa, given the velocity of water is 2 m/s at A and 1m/s at B, and the height difference is insignificant. Solution: As the height is constant, so the potential energy component will be nil Hence, the Bernoulli’s theorem equation for such case becomes P1 + (1/2)ρV1² = P2 + (1/2)ρV2² given, P1 = 500 Pa V1 = 2m/s V2 = 1 m/s as ρ (density of water) = 1000 kg/m³ putting these values in the standard equation 500 + (1/2) x 1000 x 2² = P2 + (1/2) x 1000 x 1² 500 + 2000 = P2 + 500 P2 = 2000 Pa Example 2: What will be the pressure exerted by a static fluid at a point 10 meter above the ground if the pressure at at point 5 m above the ground is 25 Pa. Take the density of the fluid to be 500 kg/m³ and g as 10. Solution: As the fluid is static so the velocity will be 0 Hence, the kinetic energy component will be nil Using the modified equation of Bernoulli’s principle P1 + ρ.gh1 = P2 + ρ.gh2 given, P1 = 25 Pa ρ = 500 kg/m³ h1 = 5m h2 = 10m g = 10 substituting these values in the standard equation 25 + 500 x 10 x 5 = P2 + 500 x 10 x 10 25 + 25000 = P2 + 5000 P2 = 2025 Pa Example 3: What will be the total energy of an ideal fluid at the height 100 meter, if its total energy is 500 Joule at the height 50 meter? Solution: As we have been given total energy at height 50 m = 500 Joule As the Bernoulli’s theorem is based on the conservation of energy Hence, total energy = constant in an ideal fluid Hence the total energy will remain the same So, total energy at the height 100 m = 500 Joule Sharing is caring! ## FAQs ### What is Bernoulli's Principle? Bernoulli's Principle states that the total amount of pressure energy, kinetic energy, and potential energy per unit volume is constant along any streamline in a steady flow of an incompressible, inviscid fluid. ### Give the formula for Bernoulli's Theorem. The formula of Bernoulli's Theorem is given by: P + ρgh + (1/2)ρv² = constant where, P = pressure ρ = density of the fluid v = velocity of the fluid h = height of the fluid container g = acceleration due to gravity ### What is meant by an Ideal fluid? An ideal fluid is a non-viscous fluid that has no palpable forces between the layers of liquids that are moving relative to one another	2,408	9,080	{"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}	4.25	4	CC-MAIN-2024-33	latest	en	0.944697
http://mbe.modelica.university/behavior/functions/interpolation/	1,558,333,204,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232255773.51/warc/CC-MAIN-20190520061847-20190520083847-00428.warc.gz	131,048,996	32,858	Interpolation # Interpolation¶ In this chapter, we will example different ways of implementing a simple one dimension interpolation scheme. We’ll start with an approach that is written completely in Modelica and then show an alternative implementation that combines Modelica with C. The advantages and disadvantages of each approach will then be discussed. ## Modelica Implementation¶ ### Function Definition¶ For this example, we assume that we interpolate data in the following form: Independent Variable, x Dependent Variable, y x_1 y_1 x_2 y_2 x_3 y_3 … … x_n y_n where we assume that . Given this data and the value for the independent variable that we are interested in, our function should return an interpolated value for . Such a function could be implemented in Modelica as follows: function InterpolateVector "Interpolate a function defined by a vector" input Real x "Independent variable"; input Real ybar[:,2] "Interpolation data"; output Real y "Dependent variable"; protected Integer i; Integer n = size(ybar,1) "Number of interpolation points"; Real p; algorithm assert(x>=ybar[1,1], "Independent variable must be greater than or equal to "+String(ybar[1,1])); assert(x<=ybar[n,1], "Independent variable must be less than or equal to "+String(ybar[n,1])); i := 1; while x>=ybar[i+1,1] loop i := i + 1; end while; p := (x-ybar[i,1])/(ybar[i+1,1]-ybar[i,1]); y := pybar[i+1,2]+(1-p)ybar[i,2]; end InterpolateVector; Let’s go through this function piece by piece to understand what is going on. We’ll start with the argument declarations: input Real x "Independent variable"; input Real ybar[:,2] "Interpolation data"; output Real y "Dependent variable"; The input argument x represents the value of the independent variable we wish to use for interpolating our function, the input argument ybar represents the interpolation data and the output argument y represents the interpolated value. The next part of the function contains: protected Integer i; Integer n = size(ybar,1) "Number of interpolation points"; Real p; The part of our function includes the declaration of various protected variables. As we saw in our Polynomial Evaluation example, these are effectively intermediate variables used internally by the function. In this case, i is going to be used as an index variable, n is the number of data points in our interpolation data and p represents a weight used in our interpolation scheme. With all the variable declarations out of the way, we can now implement the algorithm section of our function: algorithm assert(x>=ybar[1,1], "Independent variable must be greater than or equal to "+String(ybar[1,1])); assert(x<=ybar[n,1], "Independent variable must be less than or equal to "+String(ybar[n,1])); i := 1; while x>=ybar[i+1,1] loop i := i + 1; end while; p := (x-ybar[i,1])/(ybar[i+1,1]-ybar[i,1]); y := pybar[i+1,2]+(1-p)ybar[i,2]; The first two statements are assert statements that verify that the value of x is within the interval . If not, an error message will be generated explaining why the interpolation failed. The rest of the function searches for the value of i such that . Once that value of i has been identified, the interpolated value is computed as simply: where ### Test Case¶ Now, let’s test this function by using it from within a model. As a simple test case, let’s integrate the value returned by the interpolation function. We’ll use the following data as the basis for our function: x y 0 0 2 0 4 2 6 0 8 0 If we plot this data, we see that our interpolated function looks like this: In the following model, the independent variable x is set equal to time. The sample data is then used to interpolate a value for the variable y. The value of y is then integrated to compute z. model IntegrateInterpolatedVector "Exercises the InterpolateVector" Real x; Real y; Real z; equation x = time; y = InterpolateVector(x, [0.0, 0.0; 2.0, 0.0; 4.0, 2.0; 6.0, 0.0; 8.0, 0.0]); der(z) = y; annotation (experiment(StopTime=6)); end IntegrateInterpolatedVector; We can see the simulated results from this model in the following plot: There are a couple of drawbacks to this approach. The first is that the data needs to be passed around anywhere the function is used. Also, for higher dimensional interpolation schemes, the data required can be both complex (for irregular grids) and large. So it is not necessarily convenient to store the data in the Modelica source code. For example, it might be preferable to store the data in an external file. However, to populate the interpolation data from a source other than the Modelica source code, we will need to use an ExternalObject. ## Using an ExternalObject¶ The ExternalObject type is a special type used to refer to information that is not (necessarily) represented in the Modelica source code. The main use of the ExternalObject type is to represent data or state that is maintained outside the Modelica source code. This might be interpolation data, as we will see in a moment, or it might represent some other software system that maintains its own state. ### Test Case¶ For this example, we will flip things around and start with the test case. This will provide some useful context about how an ExternalObject is used. The Modelica source code for our test case is: model IntegrateInterpolatedExternalVector "Exercises the InterpolateExternalVector" parameter VectorTable vector = VectorTable(ybar=[0.0, 0.0; 2.0, 0.0; 4.0, 2.0; 6.0, 0.0; 8.0, 0.0]); Real x; Real y; Real z; equation x = time; y = InterpolateExternalVector(x, vector); der(z) = y; annotation (experiment(StopTime=6)); end IntegrateInterpolatedExternalVector; Here the main difference between this and our previous test case is the fact that we don’t pass our data directly into the interpolation function. Instead, we create a special variable vector whose type is VectorTable. We’ll discuss exactly what a VectorTable is in a moment. But for now think of it as something that represents (only) our interpolation data. Other than the creation of the vector object, the rest of the model is virtually identical to the previous case except that we use the InterpolateExternalVector function to perform our interpolation and we pass the vector variable into that function in place of our raw interpolation data. Simulating this model, we see that the results are exactly what we would expect when compared to our previous test case: ### Defining an ExternalObject¶ To see how this most recent test case is implemented, we’ll first look at how the VectorTable type is implemented. As mentioned previously, the VectorTable is an ExternalObject type. This is a special type in Modelica that is used to represent what is often called an “opaque” pointer. This means that the ExternalObject represents some data that is not directly accessible (from Modelica). In our case, we implement our VectorTable type as: type VectorTable "A vector table implemented as an ExternalObject" extends ExternalObject; function constructor input Real ybar[:,2]; output VectorTable table; external "C" table=createVectorTable(ybar, size(ybar,1)) annotation(IncludeDirectory="modelica://ModelicaByExample.Functions.Interpolation/source", Include="#include \"VectorTable.c\""); end constructor; function destructor "Release storage" input VectorTable table; external "C" destroyVectorTable(table) annotation(IncludeDirectory="modelica://ModelicaByExample.Functions.Interpolation/source", Include="#include \"VectorTable.c\""); end destructor; end VectorTable; Note that the VectorTable inherits from the ExternalObject type. An ExternalObject can have two special functions implemented inside its definition, the constructor function and the destructor function. Both of these functions are seen here. #### Constructor¶ The constructor function is invoked when an instance of a VectorTable is created (e.g., the declaration of the vector variable in our test case). This constructor function is used to initialize our opaque pointer. Whatever data is required as part of that initialization process should be passed as argument to the constructor function. That same data should be present during instantiation (.e.g, the data argument in our declaration of the vector variable). The definition of the constructor function is unusual because, unlike our previous examples, it does not include an algorithm section. The algorithm section is normally used to compute the return value of the function. Instead, the constructor function has an external clause. This indicates that the function is implemented in some other language besides Modelica. In this case, that other language is C (as indicated by the "C" following the external keyword). This tells use that the table variable (which is the output of this function and represents the opaque pointer) is returned by a C function named createVectorTable which is passed the contents and size of the ybar variable. Following the call to createVectorTable is an annotation. This annotation tells the Modelica compiler where to find the source code for this external C function. The essential point here is that from the point of view of the Modelica compiler, a VectorTable is just an opaque pointer returned by createVectorTable. It is not possible to access the data behind this pointer from Modelica. But this pointer can be passed to other functions, as we shall see in a minute, that are also implemented in C and which do know how to access the data represented by the VectorTable. #### Destructor¶ The destructor function is invoked whenever the ExternalObject is no longer needed. This allows the Modelica runtime to clean up any memory consumed by the ExternalObject. An ExternalObject instantiated in a model will generally persist until the end of the simulation. But an ExternalObject declared as a protected variable in a function, for example, may be created and destroyed in the course of a single expression evaluation. For that reason, it is important to make sure that any memory allocated by the ExternalObject is released. In general, the destructor function is also implemented as an external function. In this case, calling the destructor function from Modelica invokes the C function destroyVectorTable which is passed a VectorTable instance as an argument. Any memory associated with that VectorTable instance should be freed by the call to destructor. Again, we see the same types of annotations used to inform the Modelica compiler where to find the source code for the destoryVectorTable function. #### External C Code¶ These external C functions are implemented as follows: #ifndef _VECTOR_TABLE_C_ #define _VECTOR_TABLE_C_ #include <stdlib.h> #include "ModelicaUtilities.h" /* Here we define the structure associated with our ExternalObject type 'VectorTable' / typedef struct { double x; /* Independent variable values / double y; /* Dependent variable values / size_t npoints; / Number of points in this data / size_t lastIndex; / Cached value of last index / } VectorTable; void createVectorTable(double data, size_t np) { VectorTable table = (VectorTable) malloc(sizeof(VectorTable)); if (table) { / Allocate memory for data / table->x = (double) malloc(sizeof(double)np); if (table->x) { table->y = (double) malloc(sizeof(double)np); if (table->y) { / Copy data into our local array / size_t i; for(i=0;i<np;i++) { table->x[i] = data[2i]; table->y[i] = data[2i+1]; } / Initialize the rest of the table object / table->npoints = np; table->lastIndex = 0; } else { free(table->x); free(table); table = NULL; ModelicaError("Memory allocation error\n"); } } else { free(table); table = NULL; ModelicaError("Memory allocation error\n"); } } else { ModelicaError("Memory allocation error\n"); } return table; } void destroyVectorTable(void object) { VectorTable table = (VectorTable )object; if (table==NULL) return; free(table->x); free(table->y); free(table); } double interpolateVectorTable(void object, double x) { VectorTable table = (VectorTable )object; size_t i = table->lastIndex; double p; ModelicaFormatMessage("Request to compute value of y at %g\n", x); if (x<table->x[0]) ModelicaFormatError("Requested value of x=%g is below the lower bound of %g\n", x, table->x[0]); if (x>table->x[table->npoints-1]) ModelicaFormatError("Requested value of x=%g is above the upper bound of %g\n", x, table->x[table->npoints-1]); while(i<table->npoints-1&&x>table->x[i+1]) i++; while(i>0&&x<table->x[i]) i--; p = (x-table->x[i])/(table->x[i+1]-table->x[i]); table->lastIndex = i; return ptable->y[i+1]+(1-p)*table->y[i]; } #endif This is not a book on the C programming language so an exhaustive review of this code and exactly how it functions is beyond the scope of the book. But we can summarize the contents of this file as follows. First, the struct called VectorTable is the data associated wit the VectorTable type in Modelica. This includes not just the interpolation data (in the form of the x and y members), but also the number of data points, npoints, and a cached value for the last used index, lastIndex. Next, we see the createVectorTable function which allocates an instance of the VectorTable structure and initializes all the data inside it. That instance is then returned to the Modelica runtime. Following the definition of createVectorTable is the definition of destroyVectorTable which effectively undoes what was done by createVectorTable. Finally, we see the function interpolateVectorTable. This is a C function that is passed an instance of the VectorTable structure and a value for the independent variable and returns the interpolated value for the dependent variable. This function performs almost exactly the same function as the InterpolateVector function presented earlier. The Modelica runtime provides functions like ModelicaFormatError so that external C code can report errors. In the case of interpolateVectorTable, these functions are used to implement the assertions we saw previously in InterpolateVector. The lookup of i is basically the same except that instead of starting from 1 each time, it starts from the value of i found in the last call to interpolateVectorTable. #### Interpolation¶ We’ve seen how interpolateVectorTable is defined, but so far we haven’t seen where it is used. We mentioned that performs very much the same role as InterpolateVector, but using a VectorTable object to represent the interpolation data. To invoke interpolateVectorTable from Modelica, we simple need to define a Modelica function as follows: function InterpolateExternalVector "Interpolate a function defined by a vector using an ExternalObject" input Real x; input VectorTable table; output Real y; external "C" y = interpolateVectorTable(table, x) annotation(IncludeDirectory="modelica://ModelicaByExample.Functions.Interpolation/source", Include="#include \"VectorTable.c\""); end InterpolateExternalVector; We mentioned previously that VectorTable is opaque and that Modelica code cannot access the data contained in the VectorTable. The Modelica function InterpolateExternalVector invokes its C counterpart interpolateVectorTable which can access the interpolation data and, therefore, perform the interpolation. ## Discussion¶ As was discussed previously, the initial interpolation approach required us to pass around large amounts of unwieldy data. By implementing the VectorTable, we were able to represent that data by a single variable. An important thing to note about the ExternalObject approach, which isn’t adequately explored in our example, is that the initialization data can be completely external to the Modelica source code. For simplicity, the example code shown in this section initializes the VectorTable using an array of data. But it could just as easily have passed a file name to the initialization code. That file could then have been read by the createVectorTable function and the contents of the VectorTable structure could have been initialized using the data from that file. In many cases, this approach not only makes managing the data easier, but leveraging C allows more complex (new or existing) algorithms to be used. The next section includes another example of how external C code can be called from Modelica.	3,637	16,351	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.296875	3	CC-MAIN-2019-22	latest	en	0.696028
https://www.esaral.com/q/find-the-general-solution-of-each-of-the-following-equations-45065	1,716,745,827,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00060.warc.gz	646,204,548	11,426	# Find the general solution of each of the following equations: Question: Find the general solution of each of the following equations: $\cos x-\sin x=-1$ Solution: To Find: General solution. Given: $\cos x-\sin x=1 \Longrightarrow \cos \left(x+\frac{\pi}{4}\right)=\frac{-1}{\sqrt{2}}=\cos \frac{3 \pi}{4}$ [divide $\sqrt{2}$ on both sides and $\cos (x-y)=\cos x \cos y-\sin x \sin y$ ] So $\sin x=0$ or $\cos x=0$ Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 k \pi \pm \alpha, k \in I$ $\Rightarrow x+\frac{\pi}{4}=2 k \pi \pm \frac{3 \pi}{4} \Rightarrow x=2 k \pi \pm \frac{3 \pi}{4}-\frac{\pi}{4} \Rightarrow x=2 k \pi+\frac{3 \pi}{4}-\frac{\pi}{4}$ or $\Rightarrow x=2 k \pi-\frac{3 \pi}{4}-$ $\frac{\pi}{4}$ $\Rightarrow x=2 k \pi-\pi$ or $x=2 k \pi+\frac{\pi}{2}$ So general solution is $x=2 n \pi+\frac{\pi}{2}$ or $x=(2 n-1) \pi$ where $n \in I$	332	882	{"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}	4.3125	4	CC-MAIN-2024-22	latest	en	0.394124
https://www.nagwa.com/en/videos/358121480438/	1,582,410,180,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145729.69/warc/CC-MAIN-20200222211056-20200223001056-00322.warc.gz	834,539,128	5,868	# Video: KS1-M18 • Paper 1 • Question 23 98 − ＿ = 28. 04:06 ### Video Transcript 98 take away what equals 28. Let’s draw a bar model to help us understand what this question is asking us. This bar represents the first number in the subtraction. It’s the number that we take away something from. And it’s worth 98. This bar represents part of 98. It’s the part that we take away. But we don’t know what this bar is worth. This is the value we need to find out. Our final bar represents what’s left after we’ve worked out the subtraction. This bar is worth 28. Another way to look at this problem is to start with 28 and work backwards. Because we’re working backwards, we need to use the inverse operation, which means we need to add instead of subtract. So instead of working out 98 take away something equals 28, we can start with 28 and think about what we add to it to make 98. We’re working backwards and so we add instead of subtract. 28 plus what equals 98. Let’s look at how the number 28 changes as it becomes 98. We can model 28 as two tens and eight ones. And then, we can model the number 98 as nine tens and eight ones. How does this number change to get to this number? Let’s start by looking at how the ones change. 28 has eight ones. And 98 also has eight ones. We don’t need to add any ones. So we know that the number we add must end in a zero. It has no ones. Now, let’s look at how the tens change. And they do change. We can see that they become a lot larger. 28 has two tens. And 98 has nine tens. How many tens do we add to get from two tens to nine tens? Well, two plus seven equals nine. So we must add seven tens or 70. 98 is 70 more than 28. And so we can say that 98 take away 70 equals 28. Let’s use a place value grid and some counters just to check that our answer is correct. Here’s the number 98, nine tens and eight ones. Watch how the two digits behave as we take away our seven tens or 70. The tens digit is going to get smaller or decrease. But because we’re only taking away seven tens and no ones, watch how the ones digit stays the same. Let’s take away 70 then, seven tens, and see whether we end up with 28. We’ll start with 98, 88, 78, 68. That’s three tens or 30 we’ve already taken away, 58, 48, 38. That’s now six tens or 60 that we’ve taken away. We’ve got to take away one more ten, 28. And so 98 take away 70 equals 28.	626	2,378	{"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}	4.6875	5	CC-MAIN-2020-10	latest	en	0.913384
http://hismath.blogspot.com/2008/12/jsh-breaking-galois-theory.html	1,521,637,742,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257647649.70/warc/CC-MAIN-20180321121805-20180321141805-00018.warc.gz	130,502,552	5,513	JSH: Breaking Galois Theory In the ring of algebraic integers consider the special construction: 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 with x=1+sqrt(-6) which is chosen because it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = 7. Notice that 7(175x^2 - 15x + 2) has 7(1-sqrt(-6)) as a factor and 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where a_1(x)a_2(x) has 7(1+sqrt(-6)) as a factor, but a_1(x) is coprime to a_2(x), because a_1(x) + a_2(x) = 7x-1 as I remind that the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. There are multiple apparent contradictions with this example. If you divide 7 from both sides, then 1-sqrt(-6) remains a factor of both sides, but then factors of 7 that remain for BOTH a's are coprime to it, while also being coprime to each other, but in each case with (5a_1(x) + 7) and (5a_2(x)+ 7) the 7 on the right in each the result after dividing 7 from both sides must be coprime to the 'a' opposite it, but then in at least one case what remains must have factors in common with 1-sqrt(-6). Those attacking the validity of this example as a counter-example to standard Galois Theory should give the factors of the a's in common with 7. Oh, the resolution? It's easy. There is no actual contradiction.	479	1,335	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2018-13	latest	en	0.91047
https://theorycircuit.com/ic-555-ic-741/high-intensity-led-strobe-circuit/	1,726,536,400,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00806.warc.gz	523,690,320	25,528	# High Intensity LED Strobe Circuit Last Updated on March 16, 2024 Strobe light device need to produce regular flashes of light and it can create stroboscopic effect. Here high intensity LED strobe circuit designed with timer IC 555 and few external components. Here we have used 1 watt white LED to produce high intensity light. Timer IC 555 is configured as Astable multivibrator and it will produce continuous square pulse depends on the timing resistor and timing capacitor value. Circuit Diagram ### Components Required 1. IC 555 2. 1W white LED 3. Variable Resistor 100KΩ 4. Resistor 10KΩ, 10Ω/1w each one 5. Capacitor 0.1µF, 0.01µF each one 6. Battery This 1 watt white LED will have two terminals named as Anode (+) and Cathode (-), if you want to run this circuit continuously then use a LED heat sink with 1 watt LED. ### Construction & Working Construction of this circuit begins with Timer IC 555 and timing elements Resistor R1 & RV1 then Capacitor C1. Timer IC Pin 8 and 4 connected to the positive terminal of battery and pin 1 to negative supply, pin 5 connected with negative supply through capacitor C2. Resistor R1, RV1 and Capacitor C1 connected serially and between Discharge pin 7, threshold pin 6 and trigger pin 2. Output pin 3 from timer IC 555 is connected with 1 Watt white LED through R2 Resistor. Output Time (t) = 0.693(R1 + 2RV1).C Here RV1 value is current position Resistance value (because its variable Resistor) and not Whole RV1 Resistance value. Depends on the timing elements value, ON and OFF square pulse generated by the timer IC 555 (Know more about 555 astable multivibrator) and output from pin 3 is applied to the white led, now the LED starts to blink according to the ON and OFF pulses. ## 2 thoughts on “High Intensity LED Strobe Circuit” 1. Maya says: I am using 1.5 X 2 volts alkaline AAA batteries , how to strobe 1 or 2 or 3 green lights in parallel , will NE7555 chip produce 100mA at 3V ? Should i use fets or transistors ? If I use voltage multiplier with inductor or capacitors will I get 100 mA at 3V .	541	2,073	{"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-2024-38	latest	en	0.819153
http://mathhelpforum.com/advanced-algebra/182620-prove-artinian-ring-r-division-ring.html	1,480,740,063,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00474-ip-10-31-129-80.ec2.internal.warc.gz	172,743,483	10,044	# Thread: Prove the Artinian ring R is a division ring 1. ## Prove the Artinian ring R is a division ring From a practice exam: Let $R$ be a ring with identity $1$, and assume that $R$ is right Artinian. Prove that if $R$ has no nonzero nilpotent ideals and no idempotent elements other than $0$ and $1$, then $R$ is a division ring. I thought maybe we could use Schur's lemma for this one, but I don't see how to make it work with or without that result. Any help would be much appreciated. Thanks! 2. Originally Posted by hatsoff From a practice exam: I thought maybe we could use Schur's lemma for this one, but I don't see how to make it work with or without that result. Any help would be much appreciated. Thanks! it's a trivial result of Artin-Wedderburn theorem. 3. Ah, thanks!	212	793	{"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2016-50	longest	en	0.952979
http://www.statisticool.com/education/subtractionstory.html	1,686,379,237,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00089.warc.gz	81,431,718	3,986	If you find any of this useful, please consider donating via PayPal to help keep this site going. # Subtraction Story 4/28/20 With the recent Coronavirus Disease 2019 (COVID-19), many children are at home because schools are closed. Education and fun, however, should not stop. Here is a spreadsheet I created that gets kids practicing subtraction and addition with subtraction stories. In the spreadsheet, there is a column that randomly selects from several hundred male and female names. There are columns for the numbers 0 through 20 with their English equivalents. This is because I have the spreadsheet randomly choose if it will display the actual number or the English equivalent in the subtraction stories. If you want to use numbers larger than 20, you will need to add on to these columns. Numbers are randomly selected from the range 2 to 20. I didn't allow 0 or 1 to be selected because it made the problems too easy, as well as increased the minor programming logic I'd need to add to account for singular words and other math and grammar special cases. There are columns that are used to randomly select a subject, as well as the present and past tense verbs relating to this subject. Note that I tried to include some funny, gross, or entertaining examples that kids might like. The spreadsheet actually creates three subtraction stories. In each subtraction story the student needs to solve for the unknown x. The first subtraction story is of the "10-x=3" type. The second subtraction story is of the "10-3=x" type. The third subtraction story is of the "x-4=3" type. There are probability parameters that you can set for stories 1 and 2 that will make the spreadsheet randomly select from the three subtraction stories with your specified probabilities. Note, the probability of selecting the third subtraction story is just calculated as 1-(probability story 1 + probability story 2). I set the Print Area, that is, the area that will be printed out when you click Print, to be the area with the words in the outlined box. This way, you can just open up the spreadsheet and print. You can also press F9 to refresh the random numbers and generate a new subtraction story. You are free to edit the number range, the names, subjects, verbs, and so on, as you need. If you find this spreadsheet useful, please credit Statisticool.com.	496	2,359	{"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}	3.640625	4	CC-MAIN-2023-23	latest	en	0.945652
https://www.gogeometry.com/school-college/p868-cyclic-quadrilateral-five-rectangles-centers.htm	1,721,347,731,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00352.warc.gz	684,682,625	3,061	# Geometry Problem 868: Cyclic Quadrilateral, Circle, Five Rectangles, Four Centers, Congruence. Level: High School, College, Mathematics Education The figure below shows a cyclic quadrilateral ABCD. ABEF, BCGH, CDJK, and ADLM are rectangles of centers O1, O2, O3, and O4, respectively, so that AF = CD, BH = AD, CK = AB, and DL = BC. Prove that O1O2O3O4 is a rectangle. See also: Typography of problem 868. Home \| Search \| Geometry \| Problems \| All Problems \| Open Problems \| Visual Index \| 10 Problems \| 861-870 \| Cyclic Quadrilateral \| Circle \| Rectangle \| Congruence \| Perpendicular lines, 90 degrees \| Email \| Solution / comment	176	640	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2024-30	latest	en	0.801597
https://edspi31415.blogspot.com/2022/01/12-days-of-christmas-integrals-x2-expx2.html	1,656,713,248,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00540.warc.gz	274,013,895	15,941	## Monday, January 3, 2022 ### 12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx 12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx On the Tenth day of Christmas Integrals, the integral featured today is... ∫ x^2 ∙ exp(x)^2 dx exp(x) = e^x u = exp(x) du = exp(x) dx 1 ÷ exp(x) du = dx 1 ÷ u du = dx u^2 = exp(x)^2 ln(u) = x ln(u)^2 = x^2 Then: ∫ x^2 ∙ exp(x)^2 dx = ∫ ln(u)^2 ∙ u^2 ∙ 1/u du = ∫ u ∙ ln(u)^2 du Note: We did this integral on the Seventh Day of 12 Days of Christmas integrals (December 31, 2021). Check out that post for details! = u^2/2 ∙ ln(u)^2 - u^2/2 ∙ ln(u) + u^4/4 + C Recall u = exp(x): = 1/2 ∙ (exp(x))^2 ∙ x^2 - 1/2 ∙ (exp(x))^2 ∙ x + 1/4 ∙ (exp(x))^2 + C Eddie All original content copyright, © 2011-2022. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. ### Retro Review: Texas Instruments BA-Solar Retro Review: Texas Instruments BA-Solar Finance + Solar + 1980s Quick Facts Model: BA-Solar Company: Texas Instruments Years: 1986 - 199...	430	1,234	{"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-2022-27	latest	en	0.650591
https://www.physicsforums.com/threads/if-p-is-even-and-q-is-odd-then-binom-p-q-is-even.1064363/	1,722,728,985,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00112.warc.gz	752,847,366	27,623	# If p is even and q is odd, then ##\binom{p}{q}## is even • issacnewton issacnewton Homework Statement Prove that if ## p, q \in \mathbb{N} ## and ##p## is even and ##q## is odd, then ## \binom{p}{q}## is even. Relevant Equations Mathematical Induction If ##p## is even and ##q## is odd, then ##p = 2m## and ##q = 2n - 1##, for some ##m, n \in \mathbb{N}##. Let ##P(m,n)## be the statement that ## \binom{2m}{2n-1}## is even. Now, I am going to use mathematical induction here. My plan is to prove ##P(1, n)## and ##P(m, 1)## as base cases. For the inductive case, I want to assume ##P(m-1, n)## and ##P(m, n-1)## and prove ##P(m, n)##. With this plan, I will prove ##P(m, n)## for all two dimensional grid of ##m## and ##n##. For ##P(1, n)##, letting, ##m=1##, the binomial coefficient is $$\binom{2}{2n-1}$$ So, ##0 \leqslant 2n-1 \leqslant 2##. Since ##2n-1## is odd, I must have ##2n-1 = 1##. Hence $$\binom{2}{2n-1} = \binom{2}{1} = 2$$ which is even. Hence ##P(1, n)## is true. Now, for ##P(m, 1)##, letting ##n=1##, the binomial coefficient is $$\binom{2m}{1} = 2m$$ which is even. So, ##P(m, 1)## is also true. Next, for the inductive case, I assume that ##P(m-1, n)## and ##P(m, n-1)## are true. So, $$\binom{2m - 2}{2n-1} \text{ is even}$$ $$\binom{2m}{2n-3} \text{ is even}$$ And I want to prove that $$\binom{2m}{2n-1} \text{ is even}$$ I tried using the recurrence relation here. $$\binom{2m}{2n-1} = \binom{2m-1}{2n-1} + \binom{2m-1}{2n-2 }$$ But I don't quite get how to proceed from here. This problem may be done with induction on one variable but I wanted to learn induction on two variables, which is more difficult. There are probably other ways to traverse the 2D grid of m , n. But, I wanted to start with this approach. I would try to perform an induction on ##n## for any fixed number ##p=2m## because it has the nicer formula: $$\binom{p}{q}=\dfrac{p-q+1}{q}\cdot \binom{p}{q-1}$$ For ##n=1## we get ##\displaystyle{\binom p 1 =p=2m}\;## that is even. The formula gives \begin{align} \binom p q &=\binom{p}{2n-1}=\dfrac{2m-(2n-1)+1}{2n-1}\cdot \binom{p}{q-1}\\[6pt] &=\dfrac{2m-(2n-1)+1}{2n-1}\cdot\dfrac{2m-(2n-2)+1}{2n-2}\cdot\binom{p}{q-2} \end{align} Does that help? And please check whether I had a typo in the formulas. issacnewton You need to show separately that if ##P(m-1, n)## is true, then ##P(m,n)## is true, and if ##P(m,n-1)## is true, then ##P(m,n)## is true. First, suppose ##P(m-1,n)## is true. Then, as you claimed, ##\binom{2m-2}{2n-1}## is even. Continue the recurrence: $$\binom{2m-1}{2n-1} = \binom{2m-2}{2n-1} + \binom{2m-2}{2n-2}, \quad \binom{2m-1}{2n-2} = \binom{2m-2}{2n-2} + \binom{2m-2}{2n-3}$$ Then $$\binom{2m}{2n-1} = \binom{2m-2}{2n-3} + 2\binom{2m-2}{2n-2} + \binom{2m-2}{2n-1}$$ In the right hand side of this equation, the last two term terms are even (the last term is even by the induction hypothesis). By the binomial identity ##n\binom{n-1}{k-1} = \binom{n}{k}##, we get ##(2m-1)\binom{2m-2}{2n-3} = (2n-2)\binom{2m-1}{2n-2}## is an even number, so since ##2m-1## is odd, then ##\binom{2m-2}{2n-3}## is even. Thus ##\binom{2m}{2n-1}## is the sum of three even numbers, which is even. I leave it to you to show that if ##P(m,n-1)## is true, then ##P(m,n)## is true. Just an aside -- the binomial identity ##n\binom{n-1}{k-1} = k\binom{n}{k}## gives a short proof of the statement: If ##p## is even and ##q## is odd, then the equation ##(p+1)\binom{p}{q} = (q+1) \binom{p+1}{q+1}## shows that ##(p+1)\binom{p}{q}## is even, so ##\binom{p}{q}## is even (since ##p+1## is odd). This argument was used in the induction step above. PeroK and issacnewton Euge said: You need to show separately that if ##P(m-1, n)## is true, then ##P(m,n)## is true, and if ##P(m,n-1)## is true, then ##P(m,n)## is true. This is not necessarily true. If one can show by other means that ##P(m,1)## is true for any ##m##, then it is sufficient to show that ##P(m,n-1)## implies ##P(m,n)##. Since $${2m \choose 1} = 2m,$$ ##P(m,1)## is true by construction and no inductive step on ##m## is necessary. issacnewton and docnet Euge, Once I prove ##P(m,1)## and ##P(1, n)## and prove the implication $$P(m-1, n) \wedge P(m, n-1) \Longrightarrow P(m,n)$$ I can span whole two dimensional grid of ##(m,n)##. Is that not correct approach ? This not mathematical induction but in relation with Legendre formula　,https://en.wikipedia.org/wiki/Legendre's_formula we expect $$\sum_{i=1}^\infty \lfloor \frac{p}{2^i} \rfloor > \sum_{i=1}^\infty \lfloor \frac{p-q}{2^i} \rfloor + \sum_{i=1}^\infty \lfloor \frac{q}{2^i} \rfloor$$ which might be easier to prove. issacnewton issacnewton said: Euge, Once I prove ##P(m,1)## and ##P(1, n)## and prove the implication $$P(m-1, n) \wedge P(m, n-1) \Longrightarrow P(m,n)$$ I can span whole two dimensional grid of ##(m,n)##. Is that not correct approach ? Yes, this is a correct approach. I thought you wanted to prove the original statement with standard two-variable induction, in which you induct along the horizontal and vertical directions. Sorry about that. You can also induct as @Orodruin suggested or induct on along the lines ##m + n = k##. There are many ways to do induction in two or more variables. Thanks Euge. Can you please explain how can I do induction along the lines ##m+n = k## ? I just want to learn as many ways as possible since induction in 2 variables is not covered in introductory classes. When ##m + n = 2##, ##m =1##, ##n = 1## and ##\binom{2\cdot 1}{2\cdot 1-1} = \binom{2}{1} = 2##, so ##P(m,n)## is true in this case. Now assume ##k > 2## and ##P(m,n)## is true whenever ##m + n < k##. If ##m + n = k##, then by the induction hypothesis ##\binom{2(m-1)}{2(n-1)-1} = \binom{2m-2}{2n-3}## and ##\binom{2(m-1)}{2n-1} = \binom{2m-2}{2n-1}## are even. Thus, ##\binom{2m}{2n-1} = \binom{2m-2}{2n-3} + 2\binom{2m-2}{2n-2} + \binom{2m-2}{2n-1}## is the sum of three even numbers, which is even. Hence ##P(m,n)## is true whenever ##m + n = k##. Last edited: since the number being computed is the number of subsets of (odd) size q in a set of (even) size p, just note that for every odd subset of an even set, the complementary set is also odd. hence the odd subsets occur in pairs. fresh_42 mathwonk said: since the number being computed is the number of subsets of (odd) size q in a set of (even) size p, just note that for every odd subset of an even set, the complementary set is also odd. hence the odd subsets occur in pairs. You'll need to elaborate on that. Take ##p = 4, q = 3##. Using the set A, B, C, D, the subsets are: ABC, ABD, ACD, BCD How do they pair off in your scheme? @PeroK: dohhh! oops, my mistake. of course, in your example, the subsets of size 3 are paired with those of size 1, which are even, since p is. which is perhaps why induction is relevant. i'll try to think more. well p=4 is ok as above. and p=6 is ok by combining the same remark with the fact that "6 choose 3" is even since the original flawed argument works exactly when p = 2q? hmmm... ok, a set P is even iff it has an involution f:P-->P, f^2 = id, with no fixed points, i.e. iff the points can be paired up, (x, f(x)) with f(x)≠x. If a set P is even with fix point free involution f:P-->P, and q is an odd number ≤ card(P) = p, then I claim f induces a fix point free involution on the set of subsets S of size q in P. I.e. since S has an odd number q of elements, the fix point free f cannot induce an involution on S itself, so for every odd subset S, the involved set f(S) is a different set, but of the same cardinality q as S. qed. In your example we can use the involution (AB), (CD), which pairs the subsets ABC, ABD, ACD, BCD, up with ABD, ABC, BCD, ACD. i.e. it pairs them as (ABC,ABD), (ACD,BCD). I believe this argument is not only shorter, easier, and more insightful than an inductive argument, but now also actually correct. thank you! Here is a version you can explain to "a person in the street": imagine a church congregation composed entirely of married couples, hence with an even number of people. Now ask how many committees one can form from this congregation having a fixed odd number q of members. These committees can also be paired up as follows: to each committee there is a "dual" committee obtained by replacing each member by that member's spouse. This dual committee cannot be the same as the original committee unless it is composed entirely of married couples, but that is impossible since the number of members q is odd. Thus the total number of such committees is even. (The first argument proved only that the total number of committees with a (variable) odd number of members is even. Notice this argument is abstractly the same, but with a different involution, one that preserves cardinality. Moral: never give up on an argument just because it is wrong.) You will probably believe me that I actually did give this explanation to the two people seated next to me at the pot-luck I attended tonight. ... (I seem to recall they got up and left soon afterward.) Last edited: issacnewton mathwonk said: Here is a version you can explain to "a person in the street": imagine a church congregation composed entirely of married couples, hence with an even number of people. Now ask how many committees one can form from this congregation having a fixed odd number q of members. These committees can also be paired up as follows: to each committee there is a "dual" committee obtained by replacing each member by that member's spouse. This dual committee cannot be the same as the original committee unless it is composed entirely of married couples, but that is impossible since the number of members q is odd. Thus the total number of such committees is even. I'm not convinced. Suppose we have 3 couples: ##A_1, A_2, B_1, B_2, C_1, C_2##. We could have the subset ##A_1, B_1, C_1## and this cannot be paired with another subset using your scheme, as far as I can see. The simplest argument that I could find is to use the basic binomial identity twice for the inductive step: $$\binom p q = \binom {p-1} {q} + \binom {p-1} {q - 1} = \binom {p -2} {q} + 2\binom {p-2} {q-1} + \binom {p-2} {q-2}$$That identity can be justified by considering the subsets that have or have not the first element in the set. Last edited: mathwonk anuttarasammyak said: This not mathematical induction but in relation with Legendre formula　,https://en.wikipedia.org/wiki/Legendre's_formula we expect $$\sum_{i=1}^\infty \lfloor \frac{p}{2^i} \rfloor > \sum_{i=1}^\infty \lfloor \frac{p-q}{2^i} \rfloor + \sum_{i=1}^\infty \lfloor \frac{q}{2^i} \rfloor$$ which might be easier to prove. Formula to be proved is $$\sum_{i=1}^\infty \lfloor \frac{2(m+n-1)}{2^i} \rfloor > \sum_{i=1}^\infty \lfloor \frac{2m-1}{2^i} \rfloor + \sum_{i=1}^\infty \lfloor \frac{2n-1}{2^i} \rfloor$$ $$LHS-RHS=(m+n-1)-(m-1)-(n-1)+\sum_{i=2}^\infty \lfloor \frac{2(m+n-1)}{2^i} \rfloor - \sum_{i=2}^\infty \lfloor \frac{2m-1}{2^i} \rfloor - \sum_{i=2}^\infty \lfloor \frac{2n-1}{2^i} \rfloor$$ $$=1+[ \sum_{i=2}^\infty \lfloor \frac{2(m+n-1)}{2^i} \rfloor - \sum_{i=2}^\infty \lfloor \frac{2m-1}{2^i} \rfloor - \sum_{i=2}^\infty \lfloor \frac{2n-1}{2^i} \rfloor ]$$ [ ] part is not negative as shown below. So LHS > RHS. x=m+a, y=n+b, 0 ##\leq## a,b < 1 $$\lfloor x+y \rfloor - \lfloor x \rfloor - \lfloor y \rfloor = \lfloor a+b \rfloor \geqq 0$$ Last edited: This was buried in my answer, but for a short, non-inductive proof of the result, observe that ##(p+1)\binom{p}{q} = (q+1)\binom{p+1}{q+1}## is even because ##q + 1## is; since ##p+1## is odd, we deduce ##\binom{p}{q}## is even. issacnewton, anuttarasammyak and PeroK @PeroK: I don't understand your objection. According to the rule for forming the dual committee, the subset A1B1C1 is dual to the subset A2B2C2, obtained by replacing each member by their spouse/partner. I.e. just change each subscript to the opposite subscript. Abstractly, just apply the global involution to each element of the subset. PeroK mathwonk said: @PeroK: I don't understand your objection. According to the rule for forming the dual committee, the subset A1B1C1 is dual to the subset A2B2C2, obtained by replacing each member by their spouse/partner. I.e. just change each subscript to the opposite subscript. Abstractly, just apply the global involution to each element of the subset. Yes, I see it now. I was thinking only one replacement for some reason. Last edited: ah yes, that's what happens in the somewhat atypical example of "4 choose 3" that I used as illustration. for clarity I should have said the dual of ABC is BAD = ABD,..... Last edited: I just wanted to present the strong induction proof as suggested by Euge in post #9. Let ##k = m+n##. Here the base case is ##k=2## and ##k=3##. For ##k=2##, I have ##m=1, n=1## and Euge has shown that ##P(1,1)## is true. For ##k=3##, I have two cases to consider. Case 1) ##m=1, n=2## $$\binom{2}{2(2)-1} = \binom{2}{3} = 0 \text{ using definition}$$ which is an even number. Case 2) ##m=2, n=1## $$\binom{2(2)}{2(1)-1} = \binom{4}{1} = 4$$ which is also an even number. So, ##P(1, 2)## is true and ##P(2, 1)## is true, which means that the statement is true for ##k=3##. Now, let ## k \geqslant 3 ## be arbitrary in ##\mathbb{N}##. Also, assume that ##P(m,n)## is true for ##2 \leqslant s \leqslant k ##. I have to show that the statement is true for ##k+1##. Since ##k+1 = m+n+1##, there are two cases to consider. Case 1) Prove that ##P(m+1, n)## is true $$\binom{2(m+1)}{2n-1} = \binom{2m+2}{2n-1}$$ Using recurrence relation, I have $$= \binom{2m+1}{2n-1} + \binom{2m+1}{2n-2}$$ $$= \binom{2m}{2n-1} + \binom{2m}{2n-2} + \binom{2m}{2n-2} + \binom{2m}{2n-3}$$ $$\binom{2(m+1)}{2n-1} = \binom{2m}{2n-1} +2 \binom{2m}{2n-2} + \binom{2m}{2n-3} \cdots\cdots (1)$$ Since statement is true for ##k=m+n## by inductive hypothesis, ##P(m,n)## is true, so $$\binom{2m}{2n-1} \text{ is even}$$ first term in eq (1) is even. second term in eq (1) is even due to the multiplying factor of 2. Now, ##2 \leqslant k ##. So, ## 2 \leqslant k-1 \leqslant k##. By inductive hypothesis, the statement is true for ##k-1 = m+(n-1)##. Hence ##P(m, n-1)## is true. $$\therefore \binom{2m}{2(n-1)-1} = \binom{2m}{2n-3} \text{ is even}$$ third term in eq (1) is even. Hence LHS in eq (1) is even $$\binom{2(m+1)}{2n-1} \text{ is even}$$ So, ##P(m+1, n)## is true. Case 2) Prove that ##P(m, n+1)## is true $$\binom{2m}{2(n+1)-1} = \binom{2m}{2n+1} = \binom{2m-1}{2n+1} + \binom{2m-1}{2n}$$ $$= \binom{2m-2}{2n+1} + \binom{2m-2}{2n} + \binom{2m-2}{2n} + \binom{2m-2}{2n-1}$$ $$= \binom{2m-2}{2n+1} + 2\binom{2m-2}{2n} + \binom{2m-2}{2n-1}$$ $$\binom{2m}{2(n+1)-1} = \binom{2(m-1)}{2(n+1)-1} + 2\binom{2m-2}{2n} + \binom{2(m-1)}{2n-1} \cdots\cdots (2)$$ second term in eq (2) is even due to multiplying factor 2. By inductive hypothesis, statement is true for ##k = m+n = (m-1) + (n+1)##. So, ##P(m-1, n+1)## is true. So, first term in eq (2) is even $$\binom{2(m-1)}{2(n+1)-1} \text{ is even}$$ Also, ##2 \leqslant k-1 \leqslant k ## and again by inductive hypothesis, statement is true for ##k-1 = (m-1) + n ##. Hence, ##P(m-1, n)## is true. So, third term in eq (2) is even $$\binom{2(m-1)}{2n-1}\text{ is even}$$ Hence LHS in eq (2) is even $$\binom{2m}{2(n+1)-1} \text{ is even}$$ So, ##P(m, n+1)## is true. So, statement is true for ##k+1 = m+n+1##. By principle of strong induction, ##P(m,n)## is true for all ##m,n \in \mathbb{N}##. Does this look like a valid strong induction proof ? Thanks Hi @issacnewton, in your inductive step you assumed ##k \ge 3## and the claim holds whenever ##2 \le m + n \le k##. So it is unnecessary to prove ##P(m,n)## is true whenever ##m + n = 3##. Also, in your inductive step, you assume ##k = m + n##, but it should be ##k + 1 = m + n##, since you are trying to prove ##P(m,n)## when ##m + n = k + 1##. Then note ##P(m-1, n)## and ##P(m-1,n-1)## hold by the induction hypothesis. Hence ##\binom{2(m-1)}{2n-1} + \binom{2(m-1)}{2(n-1)-1} + 2\binom{2m-2}{2n-2}## is even, i.e., ##\binom{2m}{2n-1}## is even. issacnewton ### Similar threads • Precalculus Mathematics Homework Help Replies 11 Views 484 • Precalculus Mathematics Homework Help Replies 10 Views 453 • Precalculus Mathematics Homework Help Replies 4 Views 4K • Precalculus Mathematics Homework Help Replies 2 Views 863 • Precalculus Mathematics Homework Help Replies 9 Views 3K • Precalculus Mathematics Homework Help Replies 12 Views 2K • Precalculus Mathematics Homework Help Replies 4 Views 2K • Precalculus Mathematics Homework Help Replies 2 Views 2K • Precalculus Mathematics Homework Help Replies 9 Views 2K • Precalculus Mathematics Homework Help Replies 17 Views 634	6,029	16,724	{"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": 2, "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-2024-33	latest	en	0.777797
https://vdocuments.net/6-tools-of-tqm.html	1,695,524,116,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00260.warc.gz	650,985,350	23,637	# 6 tools of tqm of 36 /36 TOOLS OF TQM TOOLS OF TQM Check sheets Check sheets Scatter Diagrams Scatter Diagrams Cause-and-Effect Diagram Cause-and-Effect Diagram Pareto Charts Pareto Charts Flow Charts Flow Charts Histograms Histograms Statistical Process Control Statistical Process Control (SPC) (SPC) Post on 30-Sep-2015 21 views Category: ## Documents Embed Size (px) DESCRIPTION This ppt is on tools of tqm TRANSCRIPT • TOOLS OF TQM Check sheetsScatter Diagrams Cause-and-Effect DiagramPareto ChartsFlow ChartsHistogramsStatistical Process Control (SPC) • Seven Tools for TQM • Tools of TQMTools for generating ideasCheck sheetScatter diagramCause and effect diagramTools to organize dataPareto chartsProcess charts (Flow diagrams)Tools for identifying problemsHistogramsStatistical process control chart • ChecklistSimple data check-off sheet designed to identify type of quality problems at each work station; per shift, per machine, per operator • Use check-lists for testingEarsNeckFeet • Scatter DiagramsA graph that shows how two variables are related to each otherData can be used in a regression analysis to establish equation for the relationship • Chart1 34 19 16 11 6 3 Y: No. of customer complaints per month Sheet1 X: No. of training pgms for employees per monthY: No. of customer complaints per month 034 119 216 311 46 53 Correlation Coeffcient-0.9561305767 Maruti free check-up camps per monthNo. of customers for servicing their cars per month 0139 1203 2376 3407 4575 5605 Correlation Coeffcient0.9830879702 Free Desserts to no. of customers per day per monthNo. of Customers per day per month 048 126 256 322 454 536 Correlation Coeffcient-0.0368621586 Sheet1 Y: No. of customer complaints per month Sheet2 Sheet3 • Chart2 139 203 376 407 575 605 Sheet1 X: No. of training pgms for employees per monthY: No. of customer complaints per month 034 119 216 311 46 53 Correlation Coeffcient-0.9561305767 Maruti free check-up camps per monthNo. of customers for servicing their cars per month 0139 1203 2376 3407 4575 5605 Correlation Coeffcient0.9830879702 Free Desserts to no. of customers per day per monthNo. of Customers per day per month 048 126 256 322 454 536 Correlation Coeffcient-0.0368621586 Sheet1 Y: No. of customer complaints per month Sheet2 Sheet3 • Chart3 48 26 56 22 54 36 Sheet1 X: No. of training pgms for employees per monthY: No. of customer complaints per month 034 119 216 311 46 53 Correlation Coeffcient-0.9561305767 Maruti free check-up camps per monthNo. of customers for servicing their cars per month 0139 1203 2376 3407 4575 5605 Correlation Coeffcient0.9830879702 Free Desserts to no. of customers per day per monthNo. of Customers per day per month 048 126 256 322 454 536 Correlation Coeffcient-0.0368621586 Sheet1 Y: No. of customer complaints per month Sheet2 Sheet3 • Cause and Effect DiagramUsed to find problem sources/solutionsOther namesFish-bone diagram, Ishikawa diagramStepsIdentify problem to correctDraw main causes for problem as bonesAsk What could have caused problems in these areas? Repeat for each sub-area. • Cause and Effect Diagram Example • Cause and Effect Diagram ExampleMethodManpowerMaterialMachineryToo many defectsMain CauseMain Cause • Cause and Effect Diagram ExampleMethodManpowerMaterialMachineryDrillOvertimeSteelWoodLatheToo many defectsSub-Cause • Cause and Effect Diagram ExampleMethodManpowerMaterialMachinery • Fishbone Chart Problems with Airline Customer Service • Pareto AnalysisTechnique that displays the degree of importance for each elementNamed after the 19th century Italian economistOften called the 80-20 RulePrinciple is that quality problems are the result of only a few problems e.g. 80% of the problems caused by 20% of causes • Wine Glass Defects (63 defects) • Pareto Analysis of Wine Glass Defects (63 Defects) Chart2 4266.6666666667 1082.5396825397 590.4761904762 496.8253968254 2100 No. of defects Cumulative Percentage Chart1 • Example: (Identification of Problems)Following are the problems of a failing service center :Phones are only answered after many rings. Staff seem distracted and under pressure. Engineers do not appear to be well organized. They need second visits to bring extra parts. This means that customers have to take more holiday to be there a second time. • They do not know what time they will arrive. This means that customers may have to be in all day for an engineer to visit. Staff members do not always seem to know what they are doing. Sometimes when staff members arrive, the customer finds that the problem could have been solved over the phone. • Classification of ProblemsLack of staff training: items 5 and 6: (51 complaints)Too few staff: items 1, 2 and 4: (21 complaints)Poor organization and preparation: item 3:(2 complaints) By doing the Pareto analysis above, the manager can better see that the vast majority of problems (69%) can be solved by improving staff skills. • FlowchartsUsed to document the detailed steps in a processOften the first step in Process Re-Engineering • HistogramsA chart that shows the frequency distribution of observed values of a variable like service time at a bank drive-up window Displays whether the distribution is symmetrical (normal) or skewed • Control ChartsImportant tool used in Statistical Process Control The UCL and LCL are calculated limits used to show when process is in or out of control • Patterns to Look for in Control Charts • CASE STUDY:RICOH COMPANY LIMITED SHORTENING CUSTOMERS TELEPHONE WAITING TIME • At (1): The operator receives a call but due to lack of experience or knowledge, does not know where to connect the call. At (2): The receiving party cannot answer the phone quickly, perhaps because he is unavailable, and there is no substitute. • One partner out of the office topped the list, occurring a total of 172 times. Customers who had to wait a long time averaged 29.2 daily, which accounted for 6% of the calls received everyday. • Measures and ExecutionTaking lunch in three different shifts, leaving at least two operators on the job at all timesAlso brought in a helper operator from the clerical sectionAsking all employees to leave messages when leaving their desksCompiling a directory listing of the personnel and their respective jobs • Support different situations by specific tools This slide introduces the Cause and Effect Diagram. The next several slide show the development of a simple example. If time is available, it would be helpful to ask students to develop their own examples.This slide illustrates a Cause and Effect Chart for a practical problem.This chart enables you to discuss some of the information which can be obtained from the Process Control Charts.	1,646	6,861	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2023-40	latest	en	0.612559
https://hackaday.io/project/162873-mac-mini-2018-hexa-core-cooling-analysis/log/158421-plans-a1a2-plan-b-plan-c-details	1,550,362,116,000,000,000	text/html	crawl-data/CC-MAIN-2019-09/segments/1550247481249.5/warc/CC-MAIN-20190216230700-20190217012700-00190.warc.gz	583,146,926	12,340	# Plans A1/A2, Plan B, & Plan C Details A project log for Mac mini 2018 Hexa Core Cooling Analysis Aiding the stock cooling of the CPU in a 2018 Mac mini Michael O'Brien 01/19/2019 at 09:200 Comments As I've been working on the project, I've been able to figure out what needs to be done and what I can do in order to try and achieve it. As such, I've been developing backup plans concurrently to each other. What I'll be discussing here is the figurative fleshing out of the said ideas and bringing up the rationalization of that has cause me to pump over to the "next level" of cooling, thus the next plan. Most of what I've already written thus far is regarding Plan A. What I'd like to achieve is a solution that fits inside the mini if at all possible. Note: I'm not providing many links in this log right now due to the WIP-state a lot of the research I'm conducting. With whichever plan(s) I execute, I will have full documentation. ---------- more ---------- # Plan A1/A2 Shush, it's in the title... Recap though. There are 2 heat pipes in the current solution provided by Apple. I'm betting on them being the limiting factor in cooling this SFF (small form factor) computer. There are a variety of combinations of aftermarket heat pipes that can move greater power away from the CPU. Similarly, the cold plate would need to redesigned, a little, to accommodate for this change. A common method of increasing surface area if that of creating louvers. Now why do I want more surface area? There are 2 problems when dealing with the heat coming from the i7-8700B. The first is removing it from the die in totality and then dissipating it by the fins. The other is spreading it away from the die reducing the effective heat flux so that the cold plate can do its job better. In conjunction with a sandwich plate, placing a piece of PGS film between the die and cold plate can achieve this. The current power density (heat flux) for the ~120 W peak, ~81 W sustained, over the 149.6 mm^2 die area is ~54-80 W/cm^2. With that PGS film, that can go down to ~10.2% of what it currently is. The 2 versions of this plan deal with figuring out if I want a louvered or sandwich plate. # Plan B The problem with heat pipes is that the long they get, the less power they can transmit. Currently, My best case scenario gets my Qmax up from an estimated 60 W to ~107 W. I mean, when on burst workloads, the solution already helps and does a kinda okay job with thermals not spiking after you pt on better paste, but a 78% increase in capacity might not get me as far as I need. I'm looking for a max operating temperature of 85 °C in a 35 °C environment, which means I need to drop the overall absolute thermal resistance of the cooling solution to a little less than half of what is currently is. The math isn't quite there. If though, I use 2 shorter heat pipes, like a particular 8 mm pair I'm spying, then I have the capacity of removing ~150 W right off the bat, but they are only 100 mm long. I'd make a junction piece of copper to terminate them with and then use 2-3 other heat pipes coming off the fin stack and mating with that pair in the block. The junction block would allow for direct contact, but also greater contact area and thus lower resistance to transfer the heat to the fin stack. However, there is a clearance problem, or more specifically, 3 of them that make this a daunting task. • First up is the RAM EMI cage limiting me to a pair of 8 mm heat pipes; well I could use a 10 mm further way, but that's not an option. • The second obstruction is the right-most fan screw mount on the fin stack. I have to be clever to get around it and I haven't figured out how to yet. • Finally the third is the speaker in the mini. It causes the largest opening between the cold plate and the fin stack to be 20-22 mm, I estimate. I haven't gotten to modeling it yet. Larger heat pipes means larger bend radii. It's a SFF computer so something has to give and though the speaker is crappy, I don't intend on removing anything just yet. Anyhow, deciding which Plan A to use and if I'll double down and produce an extra set of parts doubling my cost on Plan B is dependant upon one factor I mentioned before: thermal resistance. Now I won't talk about it much here because I have a log file that is forthcoming that deals with this topic explicitly but to reiterate, I'm attempting to double the performance of Apple's cooling solution. Granted, 85 °C is definitely not ideal, but it certainly is acceptable. If I cannot find sufficient information to support that modifications to the air cooling setup will net at least a 2x increase, I will have to ditch them. If the math is supportive enough but practical human implementation without specialized manufacturing equipment doesn't produced acceptable results, I will ditch them. # Plan C Exotic cooling. A category that hardly a soul touches. The simple end is water cooling, which is no longer niche. Next up you might chill your water. Then maybe dry ice. To top it off, liquid nitrogen. All of these have inherent drawbacks aside from some of them not being practical for a daily driver computer. One such item that very few people have toyed with but might actually be perfect for this application is liquid cooling, but at 0.6 W / (m * K), water isn't high on my list as a coolant. Even still, I'm confident that there if I go with liquid cooling, this is still a feasible task. I don't possess the time to establish low count contacts to have special parts machined, but I'm away that metal foams exist. Instead of dealing with having micro-fins, micro-channels, or micro-posts machined into a low-profile, direct die water block, I can solder copper foam to a plate with an o-ring and bolt it all together. Restrictive, yes, but I have only one thing in the loop! Looking around at radiators, it is apparent that a good, all copper, single 120 mm radiator can dissipate more than 100 W of heat with a 10 °C delta in water temps. Why am I considering this? Well, this mini will be paired next to a 2012 quad core and both will be mounted underneath a shelf. This shelf is being modified to have 2, 120 mm fans circulate air and their RPM will be directly controlled by its neighboring mini. The internal fan connectors are Molex's Pico-EZMate and you can pickup pre-crimped wire from Mouser along with SMD board connectors and female wire housings. I digress. On to of the shelf will be a collection of 2.5" and 3.5" drives in HDD and SSD flavors. Behind them will be enough room to squeeze in a small pump. The only fabrication problem I have as a result of this Plan is that I need to physically modify the fin stack to allow tubing to enter and exit the case. Now about that coolant... # Plan C's Coolant Stop. I know all about Aluminum and Gallium, and it's alloys. Just read the rest will ya? Anyhow, gallium-based liquid metal alloys have been explored for their potential for cooling. The trademarked Galinstan isn't specifically on the table for use and when you drop that capitalization, "galinstan" references nearly all alloys that come from the ternary mix of gallium, indium, and tin. Most who have dealt with higher end cooling are familiar with 3 common liquid metal products, Conductonaut being the most recognizable to some, that are all in this family. The thing is, not even their datasheets give much information as to the specific ratios used for the mix. Additional reading indicates doping with antimony as an antioxidant, bismuth to reduce viscosity, and zinc reduces the melting point. Despite the common "16.5 W / (m * K)" claim to Galinstan, it actually ranges from ~22-35 W / (m * K) on the usable temperatures for computer CPU cooling. What has always bugged me is that Thermal Grizzly claims a melting point near 8 °C and a thermal conductivity of ~73 W / (m * K) or more than double that of any public data I can find on galinstan. Additionally, to make your own, you need gallium, which can be a tad expensive for how much you get, to which ~60% of your traditional liquid metal would consist of this. But, I found 2 research documents that claim a 30-10-60 (% by weight), Gallium-Indium-Tin mix is still eutectic and has a melting point of 12 °C. That's half the Gallium or less of every other mix I've seen. Optimizing for buying the metals for the least dollar-per-gram cost, I can reportedly pick up all of the needed metals for ~\$170 and will be able to produce ~98 ml of liquid metal. Compare that to Conductonaut at \$15/gram, this works out to \$0.27/gram. Now, anyone who is more familiar with chemistry than me, I'm aware of a big caveat that I need to test. According to the ternary phase diagram for GaInSn alloys, nearly any reduction in Gallium increases the melting point well beyond 30 °C. The problem though is that eGaIn, which is a 75-25-0 mix, has a melting point of 16 °C, but in fact shouldn't be a liquid at that temp according to the binary phase plot. GaInSn in a 62-25-13 mix has a melting point of 5 °C. GaInSnZn in a 71-15-13-1 mix has a melting point of 3 °C. and just 4% by weight of 67-29-4 of GaInZn melts at 13 °C. There are others too, but it appears that that 1-5% of zinc by weight can alter the crystalline structures heavily. Moreover, with more tin in the 30-10-60 mix, the thermal conductivity has got to be higher, or so I assume. Add up to 3-5% by weight of a couple other metals and if the alloy behaves, I couldn't be happier. # But Pumps! Flow Rates! Water vs galinstan alloys: Water Galinstan-like alloys Units Density 1.0 ~6.4 gram / ml Thermal Conductivity ~0.6 ~24-35 (73 tops?) W / (m * K) Specific Heat 4183 ~296 J / (kg * K) Absolute Viscosity 1.0 2.4 cP Yeah, so??? Even if you pick a middle ground thermal conductivity of 30 W / (m * K), these alloys have about 5x the thermal conductivity of water and they are 6 times denser. If you account for the specific heat and take in account density, 1 ml of water will accept 41.83 J / ml for a 10 °C temperature increase and galinstan alloys will be about 18.94 J / ml. Heat transfer happens better when there is a greater temperature delta and this is achieved. Thermal conductivity is 30x better which means that you need 1/30th the fluid to conduct the same amount of heat from the system. It also means you don't need a massive flow rate to achieve the same cooling performance. It also means you don't need complicated geometry and flow patterns to maximize the surface area to achieve the same cooling performance. You just need a lot less of it and it doesn't need any maintenance except for the occasional hose swap in the pump. Pump? There are a few options that can be had. If there were enough room, I'd try for a magnetohydrodynamic pump, of MHP. Using a TEC, apparently these can be pretty low on power consumption. I don't have enough space for a 200 W TEC though. I could go for a water lubricated D5 of DDC pump, but I don't know what the life of that ceramic bearing would be, nor do I have an lubricity data on galistan. There is some friction research out there, but I doubt the properties are comparable. Next up, peristaltic pumps. If you've used a mix-you-own drink from a specific soft drink company, you've used a peristaltic pump. They are non-contact with the fluid involved and the pump heads can be separate from the motors. Even better still is that the motors are commonly stepper motors, which can be practically silent.	2,776	11,476	{"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}	2.5625	3	CC-MAIN-2019-09	latest	en	0.938595
https://aramuseum.org/least-common-multiple-of-15-and-12/	1,660,437,758,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00383.warc.gz	126,548,208	5,149	LCM that 12 and 15 is the smallest number among all typical multiples that 12 and 15. The first couple of multiples that 12 and also 15 are (12, 24, 36, 48, 60, 72, 84, . . . ) and also (15, 30, 45, 60, . . . ) respectively. There space 3 typically used techniques to discover LCM the 12 and also 15 - by division method, by prime factorization, and by listing multiples. You are watching: Least common multiple of 15 and 12 1 LCM the 12 and also 15 2 List of Methods 3 Solved Examples 4 FAQs Answer: LCM the 12 and also 15 is 60. Explanation: The LCM of 2 non-zero integers, x(12) and also y(15), is the smallest hopeful integer m(60) that is divisible through both x(12) and y(15) without any remainder. The methods to uncover the LCM that 12 and also 15 are described below. By Listing MultiplesBy element Factorization MethodBy division Method ### LCM the 12 and 15 through Listing Multiples To calculation the LCM that 12 and also 15 by listing out the common multiples, we can follow the given listed below steps: Step 1: list a couple of multiples that 12 (12, 24, 36, 48, 60, 72, 84, . . . ) and 15 (15, 30, 45, 60, . . . . )Step 2: The common multiples native the multiples that 12 and also 15 are 60, 120, . . .Step 3: The smallest common multiple that 12 and also 15 is 60. ∴ The least usual multiple of 12 and 15 = 60. ### LCM the 12 and 15 by element Factorization Prime administrate of 12 and 15 is (2 × 2 × 3) = 22 × 31 and (3 × 5) = 31 × 51 respectively. LCM the 12 and also 15 have the right to be obtained by multiplying prime factors raised to their respective greatest power, i.e. 22 × 31 × 51 = 60.Hence, the LCM the 12 and 15 by element factorization is 60. ### LCM the 12 and 15 by division Method To calculate the LCM that 12 and 15 through the division method, we will divide the numbers(12, 15) by their prime determinants (preferably common). The product of these divisors gives the LCM of 12 and also 15. Step 3: proceed the steps until just 1s space left in the last row. The LCM of 12 and 15 is the product of every prime numbers on the left, i.e. LCM(12, 15) by division method = 2 × 2 × 3 × 5 = 60. ☛ likewise Check: ## FAQs ~ above LCM that 12 and also 15 ### What is the LCM that 12 and 15? The LCM that 12 and 15 is 60. To find the least usual multiple (LCM) of 12 and also 15, we require to discover the multiples that 12 and 15 (multiples that 12 = 12, 24, 36, 48 . . . . 60; multiples of 15 = 15, 30, 45, 60) and choose the the smallest multiple that is specifically divisible through 12 and 15, i.e., 60. ### Which the the following is the LCM of 12 and 15? 36, 60, 35, 11 The value of LCM the 12, 15 is the smallest usual multiple that 12 and 15. The number to solve the given condition is 60. ### If the LCM that 15 and also 12 is 60, find its GCF. LCM(15, 12) × GCF(15, 12) = 15 × 12Since the LCM of 15 and also 12 = 60⇒ 60 × GCF(15, 12) = 180Therefore, the greatest usual factor = 180/60 = 3. ### How to find the LCM the 12 and also 15 by element Factorization? To find the LCM of 12 and also 15 making use of prime factorization, us will uncover the element factors, (12 = 2 × 2 × 3) and also (15 = 3 × 5). LCM of 12 and also 15 is the product of prime determinants raised to their respective greatest exponent amongst the number 12 and also 15.⇒ LCM of 12, 15 = 22 × 31 × 51 = 60. See more: What Is The Greatest Common Factor Of 56 And 84, 112, Gcf Of 56 And 84 ### What is the Relation between GCF and also LCM the 12, 15? The adhering to equation can be supplied to express the relation between GCF and LCM the 12 and 15, i.e. GCF × LCM = 12 × 15.	1,162	3,625	{"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}	4.8125	5	CC-MAIN-2022-33	latest	en	0.932537
https://homesteady.com/how-7600613-use-basic-calculator.html	1,511,069,563,000,000,000	text/html	crawl-data/CC-MAIN-2017-47/segments/1510934805362.48/warc/CC-MAIN-20171119042717-20171119062717-00518.warc.gz	624,852,639	16,163	# How to Use a Basic Calculator Even if you know how to do basic arithmetic, a calculator saves you time and effort when you need a quick answer. ## Simple Equations A calculator is a handy tool. It is a handy tool for performing simple, everyday math--like you may need for budgeting or grocery shopping--without pencil and paper. Press the "On" button to begin. Type the first number of the equation you want to solve. Press the key for the operation you want to perform: division (÷), multiplication (X), subtraction (-), or addition (+). Type the number you are adding, subtracting, multiplying or dividing. Press the equals (=) sign key. ## Percentages Type in the number you need to know a percentage of. Press the multiplication (X) key. Type in the percentage amount you want to know. Press the percentage (%) key. ## Square Root Type the number whose square root you wish to find. Press the square root ('√') key. Press the "C" key to clear the display, if you wish to solve another equation. • Calculator ## Tips • A calculator has memory that is able to temporarily store values. To store a number in memory, press the "MS" key. To bring up the value kept in the memory, use the "MR" key. To add to the value in the memory, press the "M+" key. Once you bring up a stored number using the "MR" key you can use the regular function keys to perform whatever operation you need. To clear the memory, press the "MC" key. • You can perform longer computations as long as you do not press the "C" key. The "C" key clears all input to the calculator while the "CE" key clears just the most recent entry. Use the "CE" key if you make a mistake in a long computation--this way you won't need to begin over again.	400	1,733	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2017-47	longest	en	0.862335
https://www.experts-exchange.com/questions/10064841/Check-time-between-two-TDateTimes.html	1,524,402,289,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125945596.11/warc/CC-MAIN-20180422115536-20180422135536-00589.warc.gz	778,456,949	28,746	Improve company productivity with a Business Account.Sign Up x Solved # Check time between two TDateTimes Posted on 1998-07-19 Medium Priority 296 Views Last Modified: 2010-08-05 OK, I've started making a simple alarm program and I've run into a fairly major problem - I'm saving all my alarms OK and successfully recalling them as TDateTimes, but there's a difficulty. Are you sitting comfortably? Say I have alarms at 08:30, 11:15 and 16:30. This is fine if the app starts before the earliest alarm - it can progress through the sequence easily enough. But since the app runs on a network, it is getting started and stopped all the time, so if a user logs on and the app runs at 10:13, it needs to reorganize the order in which the alarms come. Bear in mind that this has to be flexible: a user might stay logged on for over 24 hours... I've tried doing stuff like Now-AlarmOne but mathematical operators seem to go haywire with this! Good Luck! Ed 0 Question by:edmoore 6 Comments LVL 3 Expert Comment ID: 1358201 The TDateTime format allows you to compare times. You can write a thing like this: if time1> time2 then... Do you understand? It will compare the first time and the seccond. Can you exaplin better what is your problem and what you don't understand? 0 LVL 10 Expert Comment ID: 1358202 Most mathematical operators should work fine with TDateTime. The most commonly used is - But you can also / * and + with it. (Time2-Time1)2460 is the amount of minutes passed between Time1 and Time2. A TDateTime is just a Double. The whole part are the days since 1899. The fraction is the part of the day. 1.5 = 1 day 12 hours 1 hour is 1/24 1 minute 1/(2460) 1 second 1/(246060) You can also use DecodeDate EncodeDate DecodeTime and EncodeTime. Regards Jacco 0 Expert Comment ID: 1358203 Could you define your problem more? Lets say a user logs on at 10:13, he had an alarm set at 08:30 do you wont that alarm to go off? Should the 08:30 alarm go off every day or only on the set date? Let me know and I will help you. 0 LVL 8 Expert Comment ID: 1358204 ? Time and date are stored in a float type (or is it double?). Heck doesn't mind, either way the date is defined with the numbers before the decimal separator and time is defined with these after the decimal sperator. Now when you compare two times it is fairly simple isn't it? the number later on the day will always be greater then the other.... Or am I missing the question here? Zif. 0 LVL 6 Accepted Solution Holger101497 earned 280 total points ID: 1358205 ok, several people so far have said that the date is in the integer part and the time in the frac-part. The reason I'm posting this as an answer is because I think I see your problem: These are DAILY alarms, right? 8:30 every day? You can "solve" your problems by using the frac-part of your times, thereby ignoring the date/day. IF (Frac(Now)>Frac(Alarm)) // later than alarm AND (Frac(Now)<Frac(Alarm)+5/24/60) //alarm time less than 5 minutes ago THEN SoundAlarm; // btw: "now" is a function - instead of calling it several times, store result in a variable. Also reduces strange effects if "Now" changes from line to line g* - it rarely hits a critical value, but if it only happens once in a thousand times, it's almost impossible to track down :-)) Of course you have to "remember"/store which alarms were used and which ones weren't... I don't really see a need to "sort" the alarms, but if you really want to: SecsRemaining:=Round( (Frac(Alarm)-Frac(Now)+1)246060) MOD (2460*60); // if you are using 16-bit Delphi, you might need to typecast to avoid overflows // this should return the seconds remaining before the alarm should be triggered the next time. Use this value to sort. Good luck! Let me know if that helped and ask for details before rejecting the answer. :o)) It's ( '-) :-)) 0 Author Comment ID: 1358206 OK, thanks everyone for helping me with this - I didn't realise how quickly it would fill up or I would have checked back more often. I'll have to award the points to the only actual formal answer, but all of you have been very helpful, thanks! Ed :) 0 ## Featured Post Question has a verified solution. Are you are experiencing a similar issue? Get a personalized answer when you ask a related question. Have a better answer? Share it in a comment. ## Join & Write a Comment Already a member? Login. A lot of questions regard threads in Delphi. One of the more specific questions is how to show progress of the thread. Updating a progressbar from inside a thread is a mistake. A solution to this would be to send a synchronized message to the… Have you ever had your Delphi form/application just hanging while waiting for data to load? This is the article to read if you want to learn some things about adding threads for data loading in the background. First, I'll setup a general applica… Watch the video to learn how one can deal with PST file corruption issue with an outstanding Kernel for Outlook PST Repair Tool easily. Using this tool, non-technical users can swiftly perform the repair process to restore their essential data witho… Watch the video to know how one can repair corrupt Exchange OST file effortlessly and convert OST emails to MS Outlook PST file format by using Kernel for OST to PST converter tool. It can convert OST to MSG, MBOX, EML to access them. It can migrate… ###### Suggested Courses Course of the Month8 days, 17 hours left to enroll #### 595 members asked questions and received personalized solutions in the past 7 days. Join the community of 500,000 technology professionals and ask your questions.	1,402	5,654	{"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}	3.15625	3	CC-MAIN-2018-17	latest	en	0.934352
https://de.zxc.wiki/wiki/Tafelintervall	1,680,209,895,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00077.warc.gz	240,823,629	8,129	# Panel interval In addition to the number of decimal places, the table interval of tabulated calculation aids is the most important key figure in tables, for example angle function or logarithm tables, astronomical ephemeris or auxiliary tables for the earth ellipsoid . The step size of the argument with which one enters the table is called the table interval . For example, this is the angle itself for trigonometric functions and the date for precalculated planets . ## Efficiency of use A balanced relationship between the interval and the number of digits in a table is essential for the efficiency and speed of looking up, because you have to interpolate between the tabulated values for greater accuracy . The only exception are three- or four-digit tables from which you can take each value directly . If there are more than 5 decimals, the table interval needs to be carefully considered if the numbers are not to fill several volumes. If the interval is too large, the linear interpolation between the columns is no longer sufficient , so that the user has to switch to the time-consuming quadratic interpolation . If, on the other hand, the table interval is too small, the size of the table or the volume of the book grows rapidly - up to the point of unusability or rapid wear and tear of a book that is too thick . The 7-digit Vega-Bremiker may serve as an example of a very efficient and balanced table . This logarithmic table , which was published in over 100 editions from 1795 to around 1960, logarithmic - trigonometric tables , along with other tables and formulas set up for use in mathematics, was calculated for military technology by the Slovenian-Austrian officer Baron von Vega in 1793-97 and quickly spread to the various countries Subjects and applications. ## Numerical example from the "Vega-Bremiker" The trigonometric part of the "Vega-Bremiker" (parts II and III) contains the trigonometric functions sine and tangent , namely in 2 steps: for angles from 0 ° to 5 ° in the table interval 1 " ( arcsecond ) and from 0 ° to 45 ° (as a result of the co-functions de facto up to 90 °) in the interval of 10 " . How wisely this choice was made over 200 years ago is shown by some values of the logarithms (increased by 10): ``` log sin Tafeldiff. log tan Intervall 1" und 2°00'00" 8,542 8192 (603 604) 8,543 0838 Tafeldifferenzen von 600: 2 00 01 8,542 8795 603 603 8,543 1442 zur Interpolation auf 0,01" 2 00 02 8,542 9397 602 603 8,543 2045 genügt Rechenschieber, 2 00 03 8,543 0000 603 604 8,543 2649 für 0,1" kurze Kopfrechnung. ``` ```4°00'00" 8,843 5845 (301 302) 8,844 6437 Die Tafeldifferenz ist 4 00 01 8,843 6146 301 303 8,844 6740 nur mehr halb so groß, 4 00 02 8,843 6447 301 302 8,844 7042 deshalb bei 5° Übergang 4 00 03 8,843 6748 301 303 8,844 7345 auf 10" Tafelintervall: ``` ```6°00'00" 9,019 2346 (2004 2026) 9,021 6202 Intervall 10": 6 00 10 9,019 4348 2002 2025 9,021 8227 Bis 45° sinken die Tafel- 6 00 20 9,019 6350 2002 2024 9,022 0251 differenzen auf 210 u. 420, 6 00 30 9,019 8351 2001 2023 9,022 2274 sind also noch sinnvoll. ``` Theoretically, one could make several gradations and thus reduce the volume of the book (which is 4.5 cm thick) somewhat - for example ``` 0 - 5° Tafelintervall 1" (wie oben), 5 - 10° Tafelintervall 5" (statt 10" wie oben) 10 - 25° Tafelintervall 10" 25 - 45° Tafelintervall 20". ``` The table differences between which each has to be interpolated would thereby become more uniform - e.g. B. for the sine (2-45 °) in the range 250-900 (instead of Vega-Bremiker 210-2400) .. and the book about 15% thinner. However, this small advantage would result in a large increase in calculation errors, because the manageable gradation ( 1:10 ) would be replaced by several non-round steps (1: 5: 10: 20). ## Today's meaning of table works Since the advent of the electronic pocket calculator in the 1970s, the above logarithm tables have lost much of their importance, but similar tables are important for various functions such as harmonic spherical surface functions , elliptic integrals or for solving transcendent equations . They are also of lasting importance as auxiliary tables for complex technical and structural tasks, etc. In the history of mathematics and the history of technology , the optimal choice of panel intervals has been an important task in preparing various calculations. Even today, every Astronomical Yearbook shows the experienced user whether this choice has been given sufficient attention. ## Variable speed of the planets The ephemeris (projections) of the 5 bright planets Mercury to Saturn for the year 2008 serve as an example . These freely visible planets have orbital times between 0.24 and 30 years. In the usual table interval of 10 days, Mercury moves up to 20 ° further in the starry sky, Venus and Mars by around 10 °, Jupiter and Saturn only by a maximum of 2.5 ° and 1 ° respectively. Therefore it makes sense to adapt the table intervals to this speed. The German calendar for Sternfreunde only does this for Mercury (5-day intervals), while the other four planets are tabulated at 10-day intervals. A very practical yearbook, the Austrian sky calendar , on the other hand, uses time quite useful, because you usually only have to interpolate in quarters of an interval and this is easier in your head than with tenths. ## swell • Vega-Bremiker , logarithmic-trigonometric manual . 100th edition, Weidmannsche Verlagsbuchhandlung, Berlin 1959. • Th.Neckel, O.Montenbruck, Ahnerts Astronomisches Jahrbuch 2008 . Stars and Space Publishing House, Heidelberg 2007. • H.Mucke, Himmelskalender 2008 . Austrian Astro Association, Vienna 2007.	1,626	5,822	{"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}	3.8125	4	CC-MAIN-2023-14	latest	en	0.892529
http://www.gurufocus.com/term/ROE/TIF/Return%2Bon%2BEquity/Tiffany%2B%2526amp%253B%2BCo	1,480,827,473,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698541187.54/warc/CC-MAIN-20161202170901-00110-ip-10-31-129-80.ec2.internal.warc.gz	503,810,195	33,178	Switch to: Tiffany & Co (NYSE:TIF) Return on Equity 14.44% (As of Jul. 2016) Return on equity is calculated as net income divided by its average shareholder equity. Tiffany & Co's annualized net income for the quarter that ended in Jul. 2016 was \$423 Mil. Tiffany & Co's average shareholder equity for the quarter that ended in Jul. 2016 was \$2,928 Mil. Therefore, Tiffany & Co's annualized return on equity (ROE) for the quarter that ended in Jul. 2016 was 14.44%. TIF' s Return on Equity Range Over the Past 10 Years Min: 6.82 Max: 19.41 Current: 15.43 6.82 19.41 During the past 13 years, Tiffany & Co's highest Return on Equity (ROE) was 19.41%. The lowest was 6.82%. And the median was 16.49%. TIF's Return on Equity is ranked higher than 77% of the 931 Companies in the Global Luxury Goods industry. ( Industry Median: 5.95 vs. TIF: 15.43 ) Definition Tiffany & Co's annualized Return on Equity (ROE) for the fiscal year that ended in Jan. 2016 is calculated as ROE = Net Income (A: Jan. 2016 ) / ( (Total Equity (A: Jan. 2015 ) + Total Equity (A: Jan. 2016 )) / 2 ) = 463.9 / ( (2835.1 + 2911.4) / 2 ) = 463.9 / 2873.25 = 16.15 % Tiffany & Co's annualized Return on Equity (ROE) for the quarter that ended in Jul. 2016 is calculated as ROE = Net Income (Q: Jul. 2016 ) / ( (Total Equity (Q: Apr. 2016 ) + Total Equity (Q: Jul. 2016 )) / 2 ) = 422.8 / ( (2942.9 + 2912.2) / 2 ) = 422.8 / 2927.55 = 14.44 % * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. In the calculation of annual return on equity, the net income of the last fiscal year and the average total shareholder equity over the fiscal year are used. In calculating the quarterly data, the Net Income data used here is four times the quarterly (Jul. 2016) net income data. Return on Equity is displayed in the 15-year financial page. Explanation Return on Equity (ROE) measures the rate of return on the ownership interest (shareholder's equity) of the common stock owners. It measures a firm's efficiency at generating profits from every unit of shareholders' equity (also known as net assets or assets minus liabilities). ROE shows how well a company uses investment funds to generate earnings growth. ROEs between 15% and 20% are considered desirable. The factors that affect a companys Return on Equity (ROE) can be illustrated with the Du Pont Formula: Return on Equity (ROE) (Q: Jul. 2016 ) = Net Income / Average Shareholder Equity = 422.8 / 2927.55 = (Net Income / Revenue) * (Revenue / Average Total Assets) * (Average Total Assets / Average Equity) = (422.8 / 3726.4) * (3726.4 / 5125.55) * (5125.55 / 2927.55) = Net Profit Margin * Asset Turnover * Leverage Ratio = 11.35 % * 0.727 * 1.7508 = Return on Assets * Leverage Ratio = 8.25 % * 1.7508 = 14.44 % Note: The Net Income data used here is four times the quarterly (Jul. 2016) net income data. The Revenue data used here is four times the quarterly (Jul. 2016) revenue data. * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. With this breakdown, it is clear that if a company grows its Net Profit Margin, its Asset Turnover, or its Leverage, it can grow its return on equity. Be Aware The net income used here is the net income to common shareholders. Because a company can increase its return on equity by having more financial leverage, it is important to watch the leverage ratio when investing in high ROE companies. Like ROA, ROE is calculated with only 12 months data. Fluctuations in companys earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective. Asset light businesses require very few assets to generate very high earnings. Their ROEs can be extremely high. Related Terms Historical Data * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Tiffany & Co Annual Data Jan07 Jan08 Jan09 Jan10 Jan11 Jan12 Jan13 Jan14 Jan15 Jan16 ROE 15.01 18.37 13.32 15.26 18.14 19.41 16.82 6.82 17.43 16.15 Tiffany & Co Quarterly Data Apr14 Jul14 Oct14 Jan15 Apr15 Jul15 Oct15 Jan16 Apr16 Jul16 ROE 18.05 17.19 5.26 27.40 14.71 14.60 12.70 22.65 11.96 14.44 Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names \| Earn affiliate commissions by embedding GuruFocus Charts GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)	1,273	4,515	{"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-2016-50	longest	en	0.878087
https://gmatclub.com/forum/436-amazing-gmat-data-sufficiency-questions-with-answers-143279.html?kudos=1	1,508,601,866,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00654.warc.gz	712,231,745	53,311	It is currently 21 Oct 2017, 09:04 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # 436 Amazing GMAT Data Sufficiency Questions with Answers new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Status: Married Joined: 07 Oct 2012 Posts: 15 Kudos [?]: 457 [20], given: 2 Concentration: Human Resources, Accounting GMAT Date: 01-12-2013 GPA: 3.76 WE: Consulting (Hospitality and Tourism) 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 07:29 20 KUDOS 47 This post was BOOKMARKED Last edited by problogger99 on 10 Dec 2012, 06:42, edited 4 times in total. Kudos [?]: 457 [20], given: 2 Director Affiliations: GMATQuantum Joined: 19 Apr 2009 Posts: 603 Kudos [?]: 489 [4], given: 17 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 16:56 4 KUDOS Expert's post Every single question in these two documents is from the Official GMATPrep Test database. These are clearly the best source of questions after the Official Guides. Just make sure you do these after you have taken the two Official Practice Tests in the GMATPrep software. Dabral Kudos [?]: 489 [4], given: 17 Moderator Joined: 01 Sep 2010 Posts: 3355 Kudos [?]: 9071 [0], given: 1154 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 07:30 the source ?? give me info please about these PDF Thanks _________________ Kudos [?]: 9071 [0], given: 1154 Senior Manager Joined: 22 Dec 2011 Posts: 294 Kudos [?]: 296 [0], given: 32 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 07:40 Cheers Kudos [?]: 296 [0], given: 32 Manager Joined: 21 Sep 2012 Posts: 234 Kudos [?]: 405 [0], given: 63 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 08:59 These are gmat prep questions unless he's mixed some questions from other sources. I've seen someone made a similar post before... Kudos [?]: 405 [0], given: 63 Moderator Joined: 01 Sep 2010 Posts: 3355 Kudos [?]: 9071 [0], given: 1154 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 17:30 dabral wrote: Every single question in these two documents is from the Official GMATPrep Test database. These are clearly the best source of questions after the Official Guides. Just make sure you do these after you have taken the two Official Practice Tests in the GMATPrep software. Dabral Ok but what about for instance the 5 question of the second dcument ?? Quote: 5. 354-!-item-!-187;#058&000214 If x, y, and z are integers, is x + y + 2z even? (1) x + z is even. (2) y + z is even. This doesn't make much sense ..................... _________________ Kudos [?]: 9071 [0], given: 1154 Director Affiliations: GMATQuantum Joined: 19 Apr 2009 Posts: 603 Kudos [?]: 489 [0], given: 17 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 29 Nov 2012, 17:45 carcass, Why? It is exactly the same as the question in the GMATPrep software. Is it because of the 2z there, because we know 2z is always even, so why did they put it there? The question is really equivalent to Is x+y even? Here each statement alone is insufficient. 1) x = 3, z = 3, y = 4 Answer is No x = 2, z = 2, y = 4 Answer is Yes. Insufficient. same logic for statement 2 as well. Now when we combine the two statements, do you see a clean way to do it that involves a simple step. No paper or pencil required. Cheers, Dabral Kudos [?]: 489 [0], given: 17 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16598 Kudos [?]: 273 [0], given: 0 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 02 Dec 2013, 22:54 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Manager Joined: 13 Oct 2013 Posts: 136 Kudos [?]: 57 [0], given: 129 Concentration: Strategy, Entrepreneurship Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 05 Jul 2014, 10:58 do you have solutions ? problogger99 wrote: _________________ --------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way Kudos [?]: 57 [0], given: 129 Intern Joined: 16 Jul 2014 Posts: 9 Kudos [?]: 1 [0], given: 4 GPA: 3.56 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 07 Jan 2015, 13:23 What about this? My PDF is missing the question link and shows random data for all the questions. 51. 4284-!-item-!-187;#058&003822 What is the total value of Company H's stock? (1) Investor P owns 4 1 of the shares of Company H’s total stock. (2) The total value of Investor Q’s shares of Company H’s stock is \$16,000. Kudos [?]: 1 [0], given: 4 Math Expert Joined: 02 Sep 2009 Posts: 41892 Kudos [?]: 129111 [0], given: 12194 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 07 Jan 2015, 19:43 Expert's post 1 This post was BOOKMARKED viraj1188 wrote: What about this? My PDF is missing the question link and shows random data for all the questions. 51. 4284-!-item-!-187;#058&003822 What is the total value of Company H's stock? (1) Investor P owns 4 1 of the shares of Company H’s total stock. (2) The total value of Investor Q’s shares of Company H’s stock is \$16,000. Discussed here: what-is-the-total-value-of-company-h-s-stock-119515.html _________________ Kudos [?]: 129111 [0], given: 12194 Intern Joined: 29 Aug 2015 Posts: 1 Kudos [?]: [0], given: 1 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 08 Sep 2015, 03:59 I cannot see the links for the solutions. The links are broken or weird characters. Can someone help? Kudos [?]: [0], given: 1 Math Expert Joined: 02 Sep 2009 Posts: 41892 Kudos [?]: 129111 [0], given: 12194 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 08 Sep 2015, 04:01 prachimehta wrote: I cannot see the links for the solutions. The links are broken or weird characters. Can someone help? All those questions with solutions are in our data base: viewforumtags.php Hope it helps. _________________ Kudos [?]: 129111 [0], given: 12194 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16598 Kudos [?]: 273 [0], given: 0 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 01 Oct 2016, 10:32 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16598 Kudos [?]: 273 [0], given: 0 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] ### Show Tags 03 Oct 2017, 04:14 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Re: 436 Amazing GMAT Data Sufficiency Questions with Answers [#permalink] 03 Oct 2017, 04:14 Display posts from previous: Sort by # 436 Amazing GMAT Data Sufficiency Questions with Answers new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2,646	9,217	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2017-43	latest	en	0.826792
https://www.airmilescalculator.com/distance/gov-to-dpo/	1,606,858,371,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00468.warc.gz	533,281,153	60,294	# Distance between Nhulunbuy (GOV) and Devonport (DPO) Flight distance from Nhulunbuy to Devonport (Gove Airport – Devonport Airport) is 2074 miles / 3337 kilometers / 1802 nautical miles. Estimated flight time is 4 hours 25 minutes. Driving distance from Nhulunbuy (GOV) to Devonport (DPO) is 2791 miles / 4491 kilometers and travel time by car is about 51 hours 20 minutes. ## Map of flight path and driving directions from Nhulunbuy to Devonport. Shortest flight path between Gove Airport (GOV) and Devonport Airport (DPO). ## How far is Devonport from Nhulunbuy? There are several ways to calculate distances between Nhulunbuy and Devonport. Here are two common methods: Vincenty's formula (applied above) • 2073.550 miles • 3337.055 kilometers • 1801.866 nautical miles Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth. Haversine formula • 2079.838 miles • 3347.175 kilometers • 1807.330 nautical miles The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points). ## Airport information A Gove Airport Country: Australia IATA Code: GOV ICAO Code: YPGV Coordinates: 12°16′9″S, 136°49′4″E B Devonport Airport City: Devonport Country: Australia IATA Code: DPO ICAO Code: YDPO Coordinates: 41°10′10″S, 146°25′47″E ## Time difference and current local times The time difference between Nhulunbuy and Devonport is 1 hour 30 minutes. Devonport is 1 hour 30 minutes ahead of Nhulunbuy. ACST AEDT ## Carbon dioxide emissions Estimated CO2 emissions per passenger is 226 kg (498 pounds). ## Frequent Flyer Miles Calculator Distance: 2074 Elite level bonus: 0 Booking class bonus: 0 ### In total Total frequent flyer miles: 2074 Round trip?	509	1,864	{"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-2020-50	latest	en	0.817183
https://www.cosmonomy.eu/eng/cosmonomy/structure-of-matter.htm	1,611,342,907,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703531335.42/warc/CC-MAIN-20210122175527-20210122205527-00134.warc.gz	729,196,875	16,125	# THE CIRCULAR MOTION AND FINITE FREE SPACE with a limit in distance Attention! They exist translational errors The longest distance (displacement) in the finite and isotropic space theoretically corresponds in divergence 180º. The return back (approaching) in the initial point begins in the -180º degrees. The declination of the straight line motion, in perfect circle theoretically is the longest in the 90º degree and at the return in -90º (or 270º) degrees. If we see this length of longest distance in the finite space as dynamical so as a body that is moved with constant speed. In equal times equal arcs are covered. The longest length of the first semicircle (arc of 180º degrees) is covered in the same time lapse t that is covered the other arc of 180º in the return. It is considered that acceleration exists (centripetal) in the circular motion with constant speed, because the direction of motion changes. We can a paradoxical thought observing the phenomenon of circular motion. This same observation help us for the view about a curve space, as a finite and isotropic space with a limit in the longest distance imposes for all things. We can consider a motion with slowed down, in concerning to a arbitrary point of departure. The rate of deceleration increases with the lapse of time and, in the longest distance that finds in the 180º degrees where the straight line distance stops, we can consider that the speed is zeroed for a moment. In the phase of return and approach back to the initial point, motion can be considered accelerating to the initial point, up to speed acquires the maximum value in the point of departure. In the circular motion coexist two opposite motions and equivalents two opposite time intervals and intermediary phases, where the angle of divergence of straight line increases at maximum. The observation of circular motion with constant speed and its cross-correlation with equivalent phenomenon of regular deceleration or acceleration in straight line remind us an other strange equivalence in physics. It reminds the coincidental equivalence of inertia mass and gravitational mass, that Einstein tried unsuccessfully to understands. An observer that is accelerated in a closed laboratory and pushed contrary to the direction of motion can consider, that a gravitational field attract him, if he does not find a way to search outside of his laboratory. Similarly, here we observe the case of the circular motion is described as phenomenon of straight line motion, that is altered and reversed periodically The above brief description of the circular or periodical motion such as a phenomenon of immobility or stagnant situation and such as a phenomenon of periodical change with limits and inversion in motion acquires particular theoretical interest from the application of this concept in motion of light. Do not forget, that motion of light does not only to a direction (as material bodies) except when the conditions prevent its transmission. Since we immediately consider a finite space with a limit in the longest distance and inevitable divergence from the rectilinear motion, comes in our brain the thought about propagation of bodiless light. Because the radius in a finite space (not Euclidean space, as if motion happens always on the surface of a large ball) cannot be unlimited straight line and forms a circumference with increase of distance. How does such a phenomenon of obligatory divergence of the straight line can influences motion of electromagnetic waves? Can we consider as an opposite and weak force prevent smoothly the rectilinear motion? The " spheres " of waves run along in a radius where ceases to is straight line as the increase of distance (and this declination means not concentric circles and deformity in the spherical form). An important detail exists still, when we observe this detail in the above theoretical form of a equivalent circular motion. We can detect some phenomena of high speed in periodical change. This detail results when the circular motion is not perfect circular, but is almost circular (≈360º). In this case where the circular motion (which we describe equivalent as periodical phenomenon of deceleration/acceleration), does not formed in perfect circle 360º, then the two phases of longest distance and approach to the initial point probably are not precisely equal. In other words, the rate of equivalent deceleration and the rate of equivalent acceleration and the corresponding times in that these two motions happen possibly have a small difference. They exist clues, that from a extreme small difference in these two phases of the circular motion (or in corresponding phenomena of periodical change) in microscopic creation of matter a leftover is result or an alleviation of energy or a new phenomenon... If the rate of deceleration (in a physical process) differ from the rate of acceleration (at the inversion of process), then time interval t in which the speed Vmax is acquired theoretically will differ than time interval t' in that the speed is decreased. The rate of deceleration and the rate of acceleration (in the physical changes) likely are not equal and this inequality rather is related with that motion is not interrupted perfectly and the changes does not acquire null value. This thought springs from a lot of observations. (...) They exist all clues, that multitude of particular phenomena that is observed in the structure of matter and which can be described with different terminology in science, they constitute special cases of more general phenomena that are observed in the daily experience and are analyzed in rational thought. Phenomena, such as are motion, change of speed, time interval, covered length, circular motion, frequency and rate etc. Attention! Many translational errors are exist. LIKELY LIMITS OF THE SIMPLEST RELATIONS AND OBSERVATIONS If the most minimum rate of acceleration amin is expressed by the constant G - that is fixed with masses M of one kilo where are attracted with force F when are found in distance 1m - then this rate amin in combination with the known limit of superior speed c us helps we advance in certain first calculations about the likely limits of the Universe. With the simplest script that is based on two universal constants (c and G) finds following potential sizes of time and length (length of radius, diameter or perimetric) : (The supposal of accelerating mass 1kg with a force 6,6725 ×10-11 N and with a limit in the superior speed c, was the initial thought from that began the effort to expressed with terms of physics, the philosophical interpretation about the Universe as completed and constant on contrary to the individual materially things). WE SUPPOSE that a force F size 6,6725 × 10-11 Ν accelerate a body mass =1kg 1N =1kgr • m/sec2 1ly = 9,46073 ×1015 m 1 Mpc = 106 pc ≈ 3,2615 ×106 ly × 9,46073 ×1015 m ≈ 30,856170 × 1021 m The acceleration is result general from the formula a = F/m a =F/Ma = 6,6725 × 10-11 Ν /1kg = 6,6725 × 10-11 m /sec2 In how much time T the speed of mass M =1kg will become equal with the speed of light c, when it begins from null speed, that is to say in how much time it will become Vm=c ? It's law of the speed : V=a t a=V / t → t = V / a If V=c then Tm = 2,9979245 × 108 m/sec / 6,6725 × 10-11 m/sec2 = 4,49295×1018 sec = c/G A body M=1kg and acceleration a=6,6725 × 10-11 m/s2 takes the speed c in time inerval T=4,49295 ×1018 sec, in other units 14,2372994125 ×1010 years. In this time interval T how much distance S in linear motion (straigh line) it can have covered ? This we find from the law of distance : S=1/2 a t2 Sm = 1/2 × am × tm2 (Distance S of mass m = 1kg) Sm = 1/2 × (6,6725 ×10-11 m/sec2 ) × (4,49295×1018 sec)2 = 6,73475432 ×1026 m 1pc (parsec)= 3,086333 ×1016 m 0,673475432 ×1027 m / 3,086333 ×1016 = 2,1821216 ×1010 pc Also, if we multiply time T= 4,49295 ×1018 sec multiplied by the meters where light moved per sec, so we find the meters where light will have moved in this number T, with its regular speed from the start. In time 4,49295 ×1018 sec of the body 1kg, light will have moved twofold distance Slight : S light= (4,49295×1018 sec) × (2,997924 ×108 m/sec)= 1,346952 ×1027 m = c2/G IN PARSEC: 1,346952 ×1027 m / 3,086333 × 1016 = 4,36424 × 1010 pc Finally, in time Τ=4,49295 ×1018 sec where the body M=1kg is need to arrives in speed of light c with acceleration a= 6,6725 × 10-11 m/sec2 , light in same time do double distance S Universe ( 0,673475432 ×1027 m × 2 )= 1,346952 ×1027 m (Theorem Merton about the speed until a limit in the increase). Dividing time T = Vc / amin = 4,49295×1018 sec by 2π we find : T / 2π = 0,449295 ×1019 sec / 6,2831852 = 0,07150752 × 1019 sec (0,07150752 × 1019 sec) × (31,68808781 × 10-9 ) = 2,2659365 × 1010 = 22,659365 × 109 years 1 earthly year ≈ 31,5576 × 106 sec \| 1sec = 31,688087 × 10-9 year We supposed that a mass 1kg is accelerated with regularly applied the force that results from the constant G of gravity. Actually, the free space is not neither level neither absolutely empty. SAMPLES ΤUniverse = 4,492955 ×1018 sec SUniverse = 6,734769 ×1026 m = 2,1823619 ×104 Mpc (1st scenario) SUniverse = 1,346954 ×1027 m = 4,364724 ×104 Mpc (2d scenario) SUniverse = 6,734769 ×1026 m / 2π = 1,07187183 ×1026 m = 3,473337 ×103 Mpc (3d scenario) SUniverse = 1,346954 ×1027 m / 2π = 2,14374297 ×1026 m = 6,946674 ×103 Mpc (4d scenario) Do not lose the continuity ! Useful relations [-] 1 year = 365,25 days ×24hours ×60minutes ×60sec =31,5576 ×106 sec (31 557 600 sec) 1 sec = 1/ 31,5576 ×106 = 3,168808781 ×10-8 earthly years 1 light year (ly) = 31,5576 ×106 sec × 2,9979245 ×108 m/sec = 9,46073 × 1015 m 1 parsec = 3,2615 ly = 3,0857 ×1016 m 1Mpc = 106 pc = 3,2615 ly ×106 × 9,46073 ×1015 m = 3,0857 ×1022 m pi = 3,14159265358979323846... √2 = 1,4142135 Observe that when we describe the free space as finite and with divergence from the rectilinear motion, then are presented trigonometrical relations and numbers of geometry in the circle. Angle in degreesº for arc length: S 360 / 2π r Length of arc S per degree = φº 2π r / 2 × 180º Stotal = V T = 2π r (For regular speed V) Radius r = Stotal / 2 π \| Diameter d = Stotal String of arc 180º = Diameter d String of arc 90º = √2 x r Centripetal acceleration acentr = V2 / r (Observation: The centripetal acceleration results without mass. The mass however in nature is presented as a result by the change in a speed). Acceleration (relation with length S) a = V2 / S ( a=V V / V t = V/t) → V =√ a S Acceleration (relation with time t) a = V / t → V = a t Acceleration (relation with frequency) a = V f → V=a / f Acceleration (relation with time and length) a = S / t2 The formula Vκ = 2π r / T has particular importance for the case, because it connects the speed with time and with the perfect circular motion between them and concerning the radius with the insert of relation 2pi. The formula resolved as for period T, radius r and is become: Vκ = 2π r / TT = 2π r / Vκ r = T × V / 2π2 π = Τ × Vκ / r LENGTH OF A ARC In order to we find the length of arc that corresponds in each degree on a perfect circle we use the formula: Arc length = φ° 2π r /360 (where φ° is the angle in degrees) Surface of ball: S = 4πR2 For radius: r =√S/4π MORE INVESTIGATION WITH THE FOLLOWING RELATIVE AND KNOWN TYPES : Linear speed V=S/t (length of arc where be motion / corresponding time) Angular speed ω=φ/t (angle where follows the επιβατ radius / cor. time) Relation of linear and angular speed : V=2πR/t and ω=2π/t It result: V=ω R Relation between angular speed and frequency: ω=2π f ω = V / r = 2π f = 2π / T = φ T → φ = ω / T = ω f → V = ω r = 2π f r = 2π r / T = S / t → T = 2π /ω = 2π r / V = 1 / f → f = ω / 2π = V / 2π r = 1 / T → r = V / ω = V / 2π f = V T / 2π CONCISE TABLE OF SAMPLES AROUND MAX LENGTH S and TIME T OF THE UNIVERSE Slight Universe = 1,346954 ×1027 m = c2 / G T Universe = 4,492954 ×1018 sec = c / G Smass Universe = 6,73477 ×1026 m 4,492954 ×1018 sec = 14,237312 ×1010 years R Universe = 2,14374 ×1026 m T Universe / 2π = 7,150758 × 1017 S Universe / 2 = 6,73477 ×1026 m T Universe x 2 = 8,985908 ×1018 c2 / S Universe = 6,6725 × 10-11 =G T Universe x π = 1,41150 ×1019 S Universe2 = 1,814285 × 1054 T Universe2 = 2,01866 × 1037 1 / S Universe = 0,7424158 × 10-27 =G/c2 1 / T Universe = 2,225709 × 10-19 =G/c S Universe / T Universe = 2,997925 × 108 =c R Universe2 = 4,59562 × 1052 1/ R Universe = 4,66474479 × 10-27 D Universe2 = 1,838257 × 1053 1/ D Universe = 2,33237 × 10-27 For Smass Universe = 6,73477 ×1026 m Length of arc per degree = 1,87076 ×1024 m For Slight Universe = 1,346954 ×1027 m Length of arc per degree = 3,74153 ×1024 m 1st PUBLICATION www.kosmologia.gr ©2009-10 ISBN 978-960-93-2431-1 WILL BECOME CORRECTIONS AND WILL BE CONTINUED... Now, it is not privilege of few leading physicists to speak about the Universe and its limits with terms of Science and with language of numbers. Whoever can thinks and searches theoretically with knowledge of medium education student! Because the Universe has constant min and max limits... forever and everywhere. #### THAT DIRECT RIGHTLY THE RESEARCH In physical explanation, already we have concluded the connection between of material carriers with the existence of simultaneous quantity of energy, which is presented with the form of " empty " but finite space. Already we have concluded a close relation of nuclear force with the dynamic connection that material things have in a minimal distance from the free space and immediately with the total energy of the complete Universe. Already we have comprehended how the complete Universe is permanently present and participates in the presence of individual things, even in most microscopic dimensions. The research in microscopic dimensions is research about the limits of the Universe. The existence of limits in the Universe and in certain physical activities would not be revealed by the research in microscopic dimensions, if limits did not exist. They are exist limits in the Universe, as roughly we have determined them and for this reason, we have the advantage to detect these limits through observations of the most microscopic processes. Go to Top	3,870	14,378	{"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}	2.984375	3	CC-MAIN-2021-04	latest	en	0.895625
https://www.lotterypost.com/thread/100124	1,481,395,674,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698543434.57/warc/CC-MAIN-20161202170903-00269-ip-10-31-129-80.ec2.internal.warc.gz	968,299,079	19,366	Welcome Guest You last visited December 10, 2016, 1:24 pm All times shown are Eastern Time (GMT-5:00) # GA. Lottery Another perspect. 11/12-18 Topic closed. 9 replies. Last post 12 years ago by Blackapple. Page 1 of 1 Wyncote,Pa United States Member #3206 January 3, 2004 60746 Posts Offline Posted: November 12, 2004, 3:20 am - IP Logged DYNAMIC PAIRS: 1. 10-19-80-71 2. Root sum 9 is definite 3. Regular sums 11,4 &14* 4. PICKS:801-918-711-716 5. 301-803-806-194 6. 911 7. These picks may catch the 8. 9 root sum,nevertheless 9. it should be targeted for 10. combo selection. 11. The 14 sum coincides with 12. the predicted root sum 5 13. for the nite,but there is 14. nothing hindering it from 15. DAY number action. 16. The 7 digit is highly 17. anticipated.The 3(131) in the 18. the Day number 11/11 brings 19. a 7 or 9 -Bird's 367 is one 20. number liked considering 21. 363 played 11/11 nite. Poway CA (San Diego County) United States Member #3489 January 25, 2004 14120 Posts Offline Posted: November 12, 2004, 4:02 am - IP Logged For those of you that need some help in figuring out what numbers are a root sum of 9 (that includes me, by the way!!) here they are: 009 018 027 036 045 099 117 126 135 144 189 225 234 279 288 333 369 378 459 468 477 558 567 666 999 If Blackapple is correct about the root sum of 9 being definite, then play those 25 numbers for a sure win. I'm going to play them 25 cent boxed at . That will cost me \$6.25 and a box will pay at least \$37.50, \$75 if a double and \$225 for a triple. Thanks Blackapple. I like it when you give only one sure root sum. More than that can get expensive. Wyncote,Pa United States Member #3206 January 3, 2004 60746 Posts Offline Posted: November 12, 2004, 4:24 am - IP Logged Thanks Califdude,Do you show a root nine in your computations? I know its close to showing. Poway CA (San Diego County) United States Member #3489 January 25, 2004 14120 Posts Offline Posted: November 12, 2004, 5:00 am - IP Logged Quote: Originally posted by Blackapple on November 12, 2004 Thanks Califdude,Do you show a root nine in your computations? I know its close to showing. No, I don't do root sums, only what I call sum columns. That means just the ones digit is used, so if a number has a sum of 3 or 13 or 23 it is in column 3. That gives me (for boxed numbers) 10 equal columns of 22. Root sums are only 9 columns (if you eliminate the 000) and they are either 25 or 24 numbers in the columns. Here is a part of my database: Day Number Sum Jump 11-12 mid 11-11 eve 363 12 -3 11-11 mid 131 5 -2 11-10 eve 296 17 -1 11-10 mid 242 8 -4 11-09 eve 048 12 -1 11-09 mid 904 13 -3 11-08 eve 583 16 +1 11-08 mid 843 15 5 11-07 eve 578 20 -3 11-06 eve 553 13 +4 11-06 mid 405 9 -4 11-05 eve 913 13 +3 11-05 mid 316 10 +3 11-04 eve 926 17 +1 11-04 mid 169 16 0 11-03 eve 088 16 -2 11-03 mid 170 8 +1 11-02 eve 250 7 +2 11-02 mid 339 15 -1 11-01 eve 132 6 +2 11-01 mid 004 4 +2 10-31 eve 796 22 +3 10-30 eve 298 19 +3 10-30 mid 141 6 -4 10-29 eve 118 10 -1 10-29 mid 731 11 -1 10-28 eve 633 12 -2 10-28 mid 987 24 0 10-27 eve 815 14 -3 10-27 mid 755 17 -1 10-26 eve 468 18 -3 10-26 mid 182 11 +4 10-25 eve 052 7 +3 10-25 mid 860 14 +2 10-24 eve 877 22 0 10-23 eve 408 12 -1 10-23 mid 094 13 +2 10-22 eve 506 11 -4 10-22 mid 988 25 5 10-21 eve 145 10 -2 10-21 mid 138 12 +1 The numbers in pink are doubles and the 2 that are in green are sequence numbers (3 in a row). The sum is yellow if the draw had a neighbor (2 or more numbers in sequence). The "jump" is the difference between the current sum column and the previous sum column. Here is the chart for Georgia: 5= 63 +4= 42 +3= 52 +2= 63 +1= 56 0= 46 -1= 54 -2= 48 -3= 63 -4= 61 The four numbers in yellow are the top 4 jumps. So....since the draw last night was 363 with a sum of 12, that would give a sum column of 2 (just the ones digit). If I were looking for a sum based on my jump chart, then it would be 2+5 = 7, 2+2 =4, 2-3 = 9, and 2-4 = 8. So I would be looking at numbers that had a sum with a ones digit of 7 or 4 or 9 or 8. Here is the chart of the sum columns: 0 1 2 3 4 5 6 7 8 9 000 001 002 003 004 005 006 007 008 009 019 029 011 012 013 014 015 016 017 018 028 038 039 049 022 023 024 025 026 027 037 047 048 058 059 069 033 034 035 036 046 056 057 067 068 078 079 089 044 045 055 119 066 111 077 113 088 115 099 117 118 128 129 139 112 122 114 124 116 126 127 137 138 148 149 159 123 133 125 135 136 146 147 157 158 168 169 179 134 144 145 155 156 166 167 177 178 188 189 199 226 227 228 229 239 249 222 223 224 225 235 236 237 238 248 258 259 269 233 234 244 245 246 247 257 267 268 278 279 289 299 335 255 256 266 339 277 359 288 333 334 344 336 337 338 348 349 368 369 379 389 399 345 346 347 357 358 377 378 388 479 489 444 355 356 366 367 449 459 469 488 579 499 445 446 447 448 458 468 478 569 588 589 599 455 456 457 467 477 559 578 669 679 689 699 555 466 557 558 568 668 678 688 779 789 799 556 566 567 577 677 777 778 788 888 889 899 999 666 667 If you play only the top 4 sum columns each time, you will be playing 40% of the total numbers, but you will find that the hits are above 40% (close to 50%). In the chart of the jumps for Georgia it is 45.6%. So, you can do a little bit better than "expected". Hope this helps. Poway CA (San Diego County) United States Member #3489 January 25, 2004 14120 Posts Offline Posted: November 12, 2004, 5:02 am - IP Logged I forgot to mention that the jumps that are in green are pure numbers (all even or all odd digits). Wyncote,Pa United States Member #3206 January 3, 2004 60746 Posts Offline Posted: November 12, 2004, 2:29 pm - IP Logged Thanks CalifDude I 'll study it. Douglasville, Ga United States Member #52 October 3, 2001 980 Posts Offline Posted: November 12, 2004, 3:48 pm - IP Logged like those 999 .. QP gave me 499 for evening and 258 -471 jerry Poway CA (San Diego County) United States Member #3489 January 25, 2004 14120 Posts Offline Posted: November 12, 2004, 7:02 pm - IP Logged Georgia 11/12 Midday = 479 a miss. \$\$ Notaphilist \$\$ United States Member #4290 April 9, 2004 16476 Posts Offline Posted: November 12, 2004, 7:12 pm - IP Logged WTG Blackapple, the root 9 won out! GA 11-12-04 Evening * 972 *** I am a money magnet. 12062016 Wyncote,Pa United States Member #3206 January 3, 2004 60746 Posts Offline Posted: November 12, 2004, 7:24 pm - IP Logged Yes its the root sum 9 and that 27 has not escaped. You also call it " 27 " Page 1 of 1	2,535	6,558	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2016-50	latest	en	0.75095
https://www.chegg.com/homework-help/questions-and-answers/friendly-sausage-factory-fsf-produce-hot-dogs-rate-5-000-per-day-fsf-supplies-hot-dogs-loc-q32720778	1,547,760,554,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583659340.1/warc/CC-MAIN-20190117204652-20190117230652-00083.warc.gz	737,060,864	18,050	The Friendly Sausage Factory (FSF) can produce hot dogs at a rate of 5,000 per day. FSF supplies hot dogs to local restaurants at a steady rate of 270 per day. The cost to prepare the equipment for producing hot dogs is \$66. Annual holding costs are 45 cents per hot dog. The factory operates 291 days a year. a. Find the optimal run size. (Do not round intermediate calculations. Round your answer to the nearest whole number.) b. Find the number of runs per year. (Round your answer to the nearest whole number.) Number of runs c. Find the length (in days) of a run. (Round your answer to the nearest whole number.) Run length (in days)	155	642	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2019-04	latest	en	0.90101
https://xgzav.com/online-business-secret-shopping-to-earn-money-online/	1,685,667,854,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00581.warc.gz	1,202,972,361	31,042	# Online Business – Secret Shopping To Earn Money Online Are you familiar info a lottery pool must be? Are you familiar with how is actually a a part of a one? We will take a look at this concept to listen to if it must be a part of your lottery strategy. In the best way to numbers, develop always never neglect the element of luck. Any kind of form of gambling, actually need for getting more luck that you may get. Always keep in mind that you coping more when compared with thousand possible combinations that can be called out anytime during the draw. Lets you know is also true when are playing online. May find many ways on how you can choose numbers that will make your odds of getting the jackpot prize better. In 토토사이트 , that can an simple way to get digits about the. Using the Internet, may very in order to pick out digits to one’s combinations because you will have a lot of options. Listed below some online lottery tips for picking out lucky phone numbers. The Powerball lottery calculations are based on a 1/59 for your first five white balls and 1/39 for your “red” power ball. Very first set of multipliers is 59x58x57x56x55. This group totals 600,766,320. Now divide 600,766,360 by 120 (1x2x3x4x5). Bigger in time . total is 5,006,386. Presently there a 1/39 chance to catch the “red” ball. 39 x 5,006,386 gives the real possibilities of winning the Powerball Jackpot, namely 195,249,054 to a. There are dozens not really hundreds of complicated plans and schemes out there that use slick ads and empty promises to trade the latest flavor information on the right way to win the lottery. Error many players make end up being to fall into the temptation of the slick advertising and empty promises. An individual one of followers people who always must try every new lottery “winning” system you imagine? Do you buy tickets one technique for a few weeks and then totally switch your means? online lottery website games are mostly a losing proposition. The odds are HUGE and against you. Spending time and money on various faulty systems, plans, and software diminishes your goal of actually taking home the big one. Discover one good method and keep it going for reasonable length of time. Luck plays a part, however appeared only a good part. Place make your individual luck by increasing the quantity of of games and how many of tickets you playing. How you put in place your winning lottery system and plan is more crucial in answering dilemma ‘Can I win the lottery?’. When you buy lottery tickets either in retailers or online, certainly you hope that you’ll need win the jackpots. You at least wish right now there were other ways of obtaining money within the aspect of luck instead of earning it through your day-to-day work. As an alternative to only hoping and praying that one day you is certain to get a component of luck november 23 in a lottery you join, most likely have tried many types of ways grow your opportunity. From the use of charm into the mathematical calculation,, you stick with it trying but perhaps still, you have never experienced the winning. In addition to need to use these tips below to get the best for you to get the lottery prizes before acquire hopeless in joining the lottery. If you are someone to become proficient in picking the lottery numbers instead of developing these mistakes, you do you need a proven lottery system as good as the lottery black book option.	739	3,434	{"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-2023-23	latest	en	0.933747
http://elitistjerks.com/f73/t110356-feral_mew_simulator/p5/	1,371,678,012,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368709224828/warc/CC-MAIN-20130516130024-00080-ip-10-60-113-184.ec2.internal.warc.gz	80,821,720	21,896	Elitist Jerks [Feral] Mew Simulator 03/08/11, 2:26 AM #61 a civilian Piston Honda Stenhaldi Worgen Druid Sargeras Let me clarify the math a little. When I write .15rip-f(1.1530+12)/40shred, that is my estimate for the damage gained by refreshing rip during TF when rip still has f fraction of its time remaining. You gain .15rip damage due to TF's effect, but you lose f rip, worth f30 energy, which would otherwise be spent on a TF'd shred doing 1.15shred damage for 40 energy. Thus you lose f1.1530/40shred damage. You also lose CP, but assuming you aren't refreshing rip just as savage roar is about to fall off, the worst case here is that you have to use a 1 CP roar later. This lasts a little over half as long as a full CP roar, so we can say you lose 12 energy, a little under half the cost of savage roar, in the worst case. This energy would otherwise be spent on non-TF'd shreds, so you lose 12/40shred damage. If f is small, you've probably already refreshed savage roar so this loss won't occur; if f is large, this loss will tend to approach its maximum value, since you'll be short on CP after applying the rip in the first place. Thus as a rough estimate I multiply this number by f, yielding a loss of f12/40shred damage. (This factor isn't really significant to my results in the previous post since f is close to 1 in most of the figures there. It just yields a more accurate value for the damage gain when f is small.) This yields a damage gain of .15rip-f(1.1530+12)/40shred for refreshing rip during TF. A similar calculation (but without the CP complication) yields .15rake-g1.1535/40*shred damage for refreshing rake during TF when rake still has g fraction of its time remaining. 03/08/11, 4:35 AM #62 Leafkiller Piston Honda Leafkiller Worgen Druid Stormrage Perhaps we can understand this better by looking at some of the data from the Mew output. I suspect we need to have more data, but this is a start. Using my sim script and the non-hitcap profile without the "C" statement I see the following: Number Mangles: 5.7883 Mangle Uptime (%): 98.94664 Number Shreds: 94.7792 Number Rakes: 24.8578 Rake Uptime (%): 97.75557 Number Rips: 11.4866 Rip Uptime (%): 96.62631 Number Savage Roars: 9.8142 Savage Roar Uptime (%): 92.99265 Savage Roar CP(s): 2.1991 Adding the "C" statement just before the filler shred code gives: Number Mangles: 5.7749 Mangle Uptime (%): 98.97488 Number Shreds: 93.3718 Number Rakes: 24.8272 Rake Uptime (%): 97.9173 Number Rips: 14.1193 Rip Uptime (%): 96.692 Number Savage Roars: 9.7514 Savage Roar Uptime (%): 92.89413 Savage Roar CP(s): 2.2263 There are a lot of small impacts here. Note that we are seeing an average of about 2.6 clipped Rips per run. Also the number of clipped Rakes goes slightly down. Keep in mind that some Rips are already being cast during TF, and some Rips will not be clipped because they occur close enough to the start of TF, that 5 combo points will not be reached in time to clip, and in some cases SR will be refreshed during TF using up the combo points. Here is some DPE data: Without "C": Mangle DPE: 394.34863 Shred DPE: 527.9686 Rake DPE: 1678.81946 Rip DPE: 4143.96923 Bite DPE: 685.54647 With "C": Mangle DPE: 394.91729 Shred DPE: 524.7775 Rake DPE: 1682.56594 Rip DPE: 3530.56237 Bite DPE: 685.70999 You can see the impact on the overall Rip DPE from clipping - more Rips, but around the same number of ticks. And the damage percents: Without "C": Mangle Damage (%): 1.07319 Shred Damage (%): 26.96929 Rake Damage (%): 19.67437 Rip Damage (%): 26.35181 Bite Damage (%): 2.90876 With "C": Mangle Damage (%): 1.07613 Shred Damage (%): 26.39717 Rake Damage (%): 19.698 Rip Damage (%): 27.14795 Bite Damage (%): 2.68013 The dps for the non "C" run is DPS: 22592.52336 +/- 10.45928 StdDev: 533.63694 DPS (> 25%): 21582.07916 DPS (< 25%): 25623.85593 while the dps for the "C" run is DPS: 22595.7482 +/- 10.28701 StdDev: 524.84768 DPS (> 25%): 21695.58801 DPS (< 25%): 25296.22874 I am not sure why the dps is lower in the BiTW phase in the run with the "C" statement. It is set to be 25% of the time of the fight - so the time spent in it is the same for both test runs. In the BiTW phase the code for FBing on 5 combo points occurs very early in the script - so the Rip clipping should not be happening - which means that it would have to be a condition upon entering the phase that is different. 03/08/11, 8:32 AM #64 Junlex Piston Honda Rhiannon Worgen Warrior Anachronos (EU) To try to identify some sort of pattern as to when rip would last 11 ticks vs 12 ticks I ran a short log which I have uploaded: World of Logs - Real Time Raid Analysis I ran hit/exp capped with no trinket procs, no auto-attacks, no abilities used apart from mangle, shred and rip, to try to reduce the number of potential variables but no obvious pattern emerged. From this short run of 15 rips, 13 had 12 ticks, 1 had 11 ticks (starting at 13:03:22.303) and 1 had 13 ticks (starting at 13:06:16.292). A lot more data's required though, as when I did 6 rips earlier, 2 had 12 ticks while 4 had 11 ticks. Last edited by Junlex : 03/08/11 at 8:44 AM. 03/08/11, 1:24 PM #65 Leafkiller Piston Honda Leafkiller Worgen Druid Stormrage A couple of additional thoughts. The check I added to test this does not check for the "A more powerful spell is already in effect" error. However, this should not be an issue in the current script since it is not pre-potting, and it is using the Potion of Tol'vir during Heroism which happens in the last 40 seconds of the fight where Rip is no longer being cast. The trinkets used are Unheeded Warning and Fluid Death so they are not a factor. I can't think of any other procs that affect AP outside of the potion. This is a condition that would have to be dealt with if I ever tried to implement this in Ovale though - or if I turned on pre-potting. Currently there is no call in the Mew script API to see if a Rip cannot be overwritten - but it would be very easy to add one (unfortunately I never made my Mew Eclipse environment capable of compiling the code). Coming back to the 2.6 number. The current approach I have been testing is to put all abilities outside of Shred ahead of the Rip refresh. The first test was ahead of most abilities - except for one that would impact this - the three rip shreds. I will tinker with the script to selectively prioritize the Rip refresh higher and see if I can increase the 2.6 number and along with the overall dps. Edit: testing with Rip refresh ahead of SR loses 15dps. Putting a check into the Rip Shred makes no difference at all. This code is never hit when we already have 5 combo points (not too surprising). I have been able to push the 2.6 number to over 3, but the net result was the 15dps decrease due to lower uptime on SR. Last edited by Leafkiller : 03/08/11 at 1:46 PM. 03/10/11, 7:09 PM #66 Leafkiller Piston Honda Worgen Druid Stormrage Originally Posted by Junlex To try to identify some sort of pattern as to when rip would last 11 ticks vs 12 ticks I ran a short log which I have uploaded: World of Logs - Real Time Raid Analysis I ran hit/exp capped with no trinket procs, no auto-attacks, no abilities used apart from mangle, shred and rip, to try to reduce the number of potential variables but no obvious pattern emerged. From this short run of 15 rips, 13 had 12 ticks, 1 had 11 ticks (starting at 13:03:22.303) and 1 had 13 ticks (starting at 13:06:16.292). A lot more data's required though, as when I did 6 rips earlier, 2 had 12 ticks while 4 had 11 ticks. Interesting but also frustrating data. I certainly do not see a pattern in there - other than the durations of the Rips matched the number of ticks - 22, 24 and 26 seconds. I don't know how we model that correctly. 03/11/11, 7:26 AM #67 Helistar Von Kaiser Night Elf Druid Dalaran (EU) Originally Posted by Leafkiller I am not sure why the dps is lower in the BiTW phase in the run with the "C" statement. [....] My guess is that the lack of rip refresh causes the "without C" script to enter the <25% state with 5 CPs, which means one additional FB during the <25% phase, which would match the damage% difference for FB that you report (without C: 2.9%, with C: 2.68%). You didn't paste the Number of Bites, but I guess it's higher in the "without C" case. A simple solution would be to fudge the 25% BitW threshold so that it doesn't happen always at the same time. In reality what should be done is to change the script so that if it sees that, upon approaching 25%, by not refreshing rip early it can cast one additional FB, then the early rip refresh is ignored. The calculation of the gain in Rip damage is based on the assumption that at that time CPs have no other use, which becomes false as we cross the 25% threshold. 03/21/11, 8:04 PM #68 Beanna Glass Joe Beanna Tauren Druid Conseil des Ombres (EU) Hello ! I was about to swap my profession in order to get the awesome Synapse Springs and I did some simulations on Mew just to be sure I was not mistaken in forgetting LW to become an engineer. I was rather surprised to see that engineering is about a ~50 DPS loss on my toon according to Mew and I don't find any reason to explain this DPS drop. Why Mew shows Synapse Springs as being under any +80 agi from other professions? (I correctly reduced my agility by 80 when swapping my professions on Mew.) I mean, this great activated +480 agi mini-trinket timed with TF to refresh our dots for full duration with a big AP boost and which produces a big energy pool to get the most out of it should be better than other professions, I am right? Do you think that Synapse Springs may not be modeled correctly or timed with TF in Mew's script? I dare not change my profession as Mew does not confirm to me that's a good decision. I'm not sure about my estimates on Synapse Springs. Rawr seems to consider that both professions are exactly equal, which I do not really think is correct neither... Last edited by Beanna : 03/21/11 at 8:13 PM. 03/22/11, 12:00 AM #69 tangedyn Piston Honda tangedyn Tauren Druid Thaurissan Rawr uses basic formulation, and the obvious easy way to model it would be to simply treat it as 80 agility when averaged over time. Mew Formulation does the same thing. At the moment, Mew Simulation simply uses the Springs every time it's off CD, not attempting to synch it. If you glyph TF, attempting to line synch up Synapse Springs will probably be a DPS loss. If you glyph Berserk instead, the Mew seems to indicate a small DPS increase. Not sure why it is showing up as a DPS loss with Tiger's Fury glyphed, my guess is that it's spending more time desynched that it would normally be averaged over an infinite fight duration. 03/22/11, 9:08 AM #70 Petitourson Glass Joe Petitourson Tauren Druid Conseil des Ombres (EU) I have run 20+ simulations to try and pin down what's causing the discrepancy. With the default script (using my gear for reference) Synapse Springs was always a dps loss (-11.5 < N < -2.6) until I enabled both a 10% duration randomization and the high resolution timer simultaneously (+0.35 < N < +24.1). Synapse Springs would fall behind if only one of the two options was enabled. With Leafkiller's Mew script Synapse Springs is always a dps increase, the gap widening the more advanced options I ticked on (10% duration randomization, high resolution timer) with gains comprised +4.3 < N < +22.6 All sims were run with GoBerserk since it's the only one worth glyphing as an engineer. 03/22/11, 1:22 PM #71 Beanna Glass Joe Beanna Tauren Druid Conseil des Ombres (EU) Of course I plan to use the glyph of Berserker in order to synch SS with TF! My bad, I didn't checked the boxes, especially High Resolution Timer. I've just done the simulations once more and Synapse Springs appears to be a ~+47 DPS increase for my toon so I'll finally go for engineering. Thank you very much for your advices. 04/12/11, 6:16 AM #72 Yawning Von Kaiser Yawning Night Elf Druid No WoW Account Mew-20110412 is now available. Major changes since last release:Dark Intent's periodic damage increase is now 1/2/3% per 4.1 PTR patch notes. Interrupts are reliable per 4.1 patch notes. Simulation is now the default model for all new profiles. The Simulator now allows for casting Rebirth/Tranquility durring the encounter. Furor and Predator's Swiftness are now modeled by the Simulator. Spell Vulnerability/Haste/Crit Taken are specifiable buffs/debuffs (Mostly used for Tranquility/Rebirth, also affects certain trinkets). New Simulator script calls getBerserkBaseDuration(), getEncounterDuration(), getElapsedTime(), and isAutoAttackEnabled() New Simulator script actions AUTOATTACK_START, AUTOATTACK_STOP. The Simulator now expects the script to manage the swing timer. Formulation no longer uses Ferocious Bite at all over 25% mob HP. New option that will cause Mew to model Glyph of Shred as +1/1/2 ticks. BUG: Change potion cooldown. (Was 60 sec is now 120 sec, no DPS impact for encounter durations worth modeling.) BUG: Apply the ToTT multiplier to trinket sourced damage procs. (DMC:H is still sub-par compared to other options.) BUG: Stampede buff duration is now always 8 sec regardless of the number of points in the talent. Numerous other internal changes, please see the SVN commit logs if you are curious. Note: If you have a custom script, please look at the documentation in the google code wiki or the example script to ensure that you start the swing timer. As usual questions/comments/bug reports appreciated. 04/13/11, 8:02 PM #73 Leafkiller Piston Honda Leafkiller Worgen Druid Stormrage I updated my sim script to include the updated sim script that Yawning just released along with modifications to my two scripts that incorporate changes to work with the new Mew release (the external buff code is the same in all three scripts and the interrupt and FC code is close to identical). In my so-called "universal" script, there is a variable, "rotation" that will use the default Mew script if set to "0", my script if set to "1" and an experimental script if set to "2". The experimental rotation currently emulates the Atremedes fight. The script can be found here: The Fluid Druid - View topic - Leafkiller's 4.0.6 Feral Ovale Script 04/25/11, 11:47 PM #74 tangedyn Piston Honda tangedyn Tauren Druid Thaurissan Mew-20110425 is now available. Major changes since last release: - Simulation Progress Bar updates now throttled to reduce system resource overhead - (4.1) Berserk is now off the GCD - Leafkiller's optimized Cat Strategy is now available in Samples/universal1.script - Simulations now generates a Report Output, which can be copy and pasted Notes: - We have created a Google Discussion Group at Mew Theorycraft \| Google Groups as the official feedback and support forum for Mew. We will continue to keep an eye on a few other forums, but not as frequently. - When reporting results of your own Simulation runs, please copy and paste the entire Report Output (from the Report tab) in your post. This will allow us to analyze your results better. 04/26/11, 12:21 PM #75 Leafkiller Piston Honda Worgen Druid Stormrage For anyone using my script, please keep the following in mind: The version of my script that is checked into the Mew tree is set to the Atramedes fight. There is a variable: `int rotation = 2;` that is near the top of the file - that dictates which rotation to use. There are three rotations in the file. If you set the variable to 0, it will use the default rotation (which may be out of date since I updated it on the last public release of Mew). if you set it to 1, you get my version of a "Pathwork" rotation, and if you set it to 2, you will get my version of the Atramedes fight - which is what it was set to when I added it to the Mew tree. The reason I put multiple rotations into the file was due to the slowness of the Mew file I/O. I did not want to wait for that when testing different fight scenarios. At some point I would like to have the fight scenarios added to the drop down list so they can be individually selected without having to walk through the file system. Elitist Jerks [Feral] Mew Simulator	4,327	16,234	{"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-2013-20	latest	en	0.937262
https://www.wallstreetmojo.com/profitability-index-formula/	1,642,976,758,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00156.warc.gz	1,122,188,066	48,068	# Profitability Index Formula ## What is the Profitability Index Formula? The formula for Profitability Index is simple and it is calculated by dividing the present value of all the future cash flows of the project by the initial investment in the project. Profitability Index = PV of future cash flows / Initial investment It can be further expanded as below, • Profitability Index = (Net Present value + Initial investment) / Initial investment • Profitability Index = 1 + (Net Present value / Initial investment) For eg: Source: Profitability Index Formula (wallstreetmojo.com) ### Steps to Calculate Profitability Index Below are the steps to calculate profitability index – 1. Firstly, the initial investment in a project has to be assessed based on the project requirement in terms of for machinery u0026 equipment and other expenses, which are also capital in nature. 2. Now, all the future cash flows expected from the project are required to be determined. Then the discounting factor has to be calculated based on the current expected return from an investment of similar risk. Now, using the , the present value of the future cash flows from the project can be calculated. 3. Finally, the profitability index of the project is calculated by dividing the present value of all the future value of cash flow from the project (step 2) by the initial investment in the project (step 1). ### Examples You can download this Profitability Index Formula Excel Template here – Profitability Index Formula Excel Template #### Example #1 Let us take the example of company ABC Ltd which has decided to invest in a project where they estimate the following annual cash flows: • \$5,000 in Year 1 • \$3,000 in Year 2 • \$4,000 in Year 3 At the beginning of the project, the initial investment required for the project is \$10,000, and the discounting rate is 10%. PV of cash flow in Year 1= \$5,000 / (1+10%)1 = \$4,545 PV of cash flow in Year 2 = \$3,000 / (1+10%)2 = \$2,479 PV of cash flow in Year 3 = \$4,000 / (1+10%)3 = \$3,005 So, Sum of PV of future cash flows will be: Profitability Index of the project = \$10,030 / \$10,000 As per the formula of the profitability index, it can be seen that the project will create an additional value of \$1.003 for every \$1 invested in the project. Therefore, the project is worth investing since then it is more than 1.00. #### Example #2 Let us take the example of a company A which is considering two projects: Project A Project A needs an initial investment of \$2,000,000 and a discount rate of 10% and with estimated annual cash flows of: • \$300,000 in Year 1 • \$600,000 in Year 2 • \$900,000 in Year 3 • \$700,000 in Year 4 • \$600,000 in Year 5 Initial investment = \$2,000,000 PV of cash flow in Year 1= \$300,000 / (1+10%)1 = \$272,727 PV of cash flow in Year 2 = \$600,000 / (1+10%)2 = \$495,868 PV of cash flow in Year 3 = \$900,000 / (1+10%)3 =\$676,183 PV of cash flow in Year 4 = \$700,000 / (1+10%)4 = \$478,109 PV of cash flow in Year 5 = \$600,000 / (1+10%)5 =\$372,553 So, Sum of PV of future cash flows will be: Profitability Index of Project A = \$2,295,441 / \$2,000,00 Project B The initial investment of \$3,000,000 and discount rate of 12% and with estimated annual cash flows of: • \$600,000 in Year 1 • \$800,000 in Year 2 • \$900,000 in Year 3 • \$1,000,000 in Year 4 • \$1,200,000 in Year 5 PV of cash flow in Year 1= \$600,000 / (1+12%)1 = \$535,714 PV of cash flow in Year 2 = \$800,000 / (1+12%)2 =\$637,755 PV of cash flow in Year 3 = \$900,000 / (1+12%)3 =\$640,602 PV of cash flow in Year 4 = \$1,000,000 / (1+12%)4 =\$635,518 PV of cash flow in Year 5 = \$1,200,000 / (1+12%)5 =\$680,912 So, Sum of PV of future cash flows will be: Profitability Index of Project B = \$3,130,502 / \$3,000,000 Using the formula of profitability index, it can be seen that Project A will create an additional value of \$0.15 for every \$1 invested in the project compared to Project B, which will create an additional value of \$0.04 for every \$1 invested in the project. Therefore, Company A should select Project A over Project B. ### Profitability Index Calculator You can use the following Profitability Index calculator- PV of Future Cash Flows Initial Investment Profitability Index Formula Profitability Index Formula = PV of Future Cash Flows = Initial Investment 0 = 0 0 ### Relevance and Use index formula is very important from the point of view of . It is a handy tool to use when one needs to decide whether to invest in a project or not. The index can be used for ranking project investment in terms of value created per unit of investment. • The basic idea is that – the higher the index, the more attractive the investment. • If the index is greater than equal to unity, then the project adds value to the company, or otherwise, it destroys value when the index is less than unity. ### Recommended Articles This has been a guide to Profitability Index Formula. Here we discuss how to calculate the profitability index along with practical examples, a calculator, and a downloadable excel template. You can learn more about excel modeling from the following articles –	1,373	5,207	{"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}	3.65625	4	CC-MAIN-2022-05	latest	en	0.921366
https://www.minimins.com/threads/need-new-scales.202867/	1,498,628,992,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128322870.59/warc/CC-MAIN-20170628050821-20170628070821-00182.warc.gz	911,336,804	16,693	# Need new scales... Discussion in 'Dukan Diet' started by babaloo, 30 March 2011 Social URL. 1. ### babalooMember Posts: 46 0 Start Weight: 14st7lb Current Weight: 13st4lb Goal Weight: 12st7lb Lost(%): 1st3lb(8.37%) Diet: Dukan Diet My scales are playing tricks on me...When I set it to stones and pounds I'm 13stone 11 but when I change it to pounds and kg I'm 13st7. Which one is true? So I decided to wear myself at the doctors and I was 87kg which is 13st 7. Can anyone recommend me a reliable scales that I can buy? Best wishes, L x 3. ### Maintainer Chief WITCH Posts: 11,743 101 Start Weight: 141kg Current Weight: 69.3kg Goal Weight: 65.0kg Lost(%): 71.7kg(50.85%) Diet: Self devised (By my calculations, 87K is actually 13 stone 9.5lbs...) In any event, different scales will always weigh us differently. I expect also it was a different time of the day, and you were dressed at the doctors. Can't help otherwise cos not in the right country! 4. ### babalooMember Posts: 46 0 Start Weight: 14st7lb Current Weight: 13st4lb Goal Weight: 12st7lb Lost(%): 1st3lb(8.37%) Diet: Dukan Diet Google calculator said 87 kilograms = 13.7001549 stones. Nevertheless, I weighed myself 3 times on my scales and it says different readings- was at the same time, no clothes and before food 5. ### Emmmagrammar police Posts: 1,279 11 Start Weight: 13st13lb Current Weight: 10st0lb Goal Weight: 10st0lb Lost(%): 3st13lb(28.21%) Diet: Dukan It's the .7 - that's a metric decimal, but there are 14lb in a stone. 0.7 of 14 is 9.8, or to round up 10lb. 6. ### m-mouseNot very good at this! Posts: 3,665 8 Start Weight: 15st8lb Current Weight: 10st12lb Goal Weight: 11st4lb Lost(%): 4st10lb(30.28%) Diet: dukan - conso 7. ### Maintainer Chief WITCH Posts: 11,743 101 Start Weight: 141kg Current Weight: 69.3kg Goal Weight: 65.0kg Lost(%): 71.7kg(50.85%) Diet: Self devised So you have to calculate it yourself - this is my lot each time before coming to you guys and weighing in. My scales only do kilos. 87K x 2.2 = 191.4 13 stone x 14 = 182 Therefore 191.4 = 13 stone 9.4lbs So maybe your scales are ok after all? 8. ### Maintainer Chief WITCH Posts: 11,743 101 Start Weight: 141kg Current Weight: 69.3kg Goal Weight: 65.0kg Lost(%): 71.7kg(50.85%) Diet: Self devised I'm a real cynic re body fat as you can make it say whatever you like depending on the time of day you do it, and how much you've drunk. OK they say to do it at the same time of the day, but first thing we've a totally different body fat reading to the afternoon when hydrated, so I remain sceptical. Also, I read somewhere that pear shaped people get weird readings also. That's me 9. ### babalooMember Posts: 46 0 Start Weight: 14st7lb Current Weight: 13st4lb Goal Weight: 12st7lb Lost(%): 1st3lb(8.37%) Diet: Dukan Diet Aw! Now, it's clear. Obviously Maths wasn't my strong point! 10. ### babalooMember Posts: 46 0 Start Weight: 14st7lb Current Weight: 13st4lb Goal Weight: 12st7lb Lost(%): 1st3lb(8.37%) Diet: Dukan Diet Thanks everyone, now I don't have to buy new scales! haha I just need to look through my maths GCSE text book! haha 11. ### Maintainer Chief WITCH Posts: 11,743 101 Start Weight: 141kg Current Weight: 69.3kg Goal Weight: 65.0kg Lost(%): 71.7kg(50.85%) Diet: Self devised Aren't we wonderful? We give free maths lessons, cookery, relooking (I'm currently helping a French online friend who's never ever ever blowdried her hair in her life and still has the long boring ponytail she's had for years and is considering now she's lost weight cutting it off!!)... bathroom DIYing... Mouse gives chicken training classes. Vicky list making... Scrumps chicken cooking... Popular Forums 1. MiniMins.com is a weight loss support community helping each other on their weight loss journey. We have a multitude of forums, from Slimming World and Exante, to Success Stories. Click the logo at the top right to return to the forum home page at any time. Replies: 3 Views: 1,042 2. ### Advice needed re jumping ship to slimming world !!!!!! catz_marr, in forum: Dukan Diet Replies: 5 Views: 1,099 3. ### Need help natosha120, in forum: Dukan Diet Replies: 8 Views: 1,215 4. ### Opinions needed Lindy16, in forum: Dukan Diet Replies: 6 Views: 493 Replies: 6 Views: 693	1,319	4,286	{"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-2017-26	longest	en	0.86985
https://www.aqua-calc.com/one-to-one/surface-density/kilogram-per-square-nanometer/tonne-per-square-nanometer/1000	1,581,898,676,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00147.warc.gz	658,629,040	7,645	# 1 000 kilograms per square nanometer in tonnes per square nanometer ## kg/nm² to t/nm² unit converter of surface density 1 000 kilograms per square nanometer [kg/nm²] = 1 tonne per square nanometer [t/nm²] ### kilograms per square nanometer to tonnes per square nanometer surface density conversion cards • 1 000 through 1 024 kilograms per square nanometer • 1 000 kg/nm² to t/nm² = 1 t/nm² • 1 001 kg/nm² to t/nm² = 1 t/nm² • 1 002 kg/nm² to t/nm² = 1 t/nm² • 1 003 kg/nm² to t/nm² = 1 t/nm² • 1 004 kg/nm² to t/nm² = 1 t/nm² • 1 005 kg/nm² to t/nm² = 1.01 t/nm² • 1 006 kg/nm² to t/nm² = 1.01 t/nm² • 1 007 kg/nm² to t/nm² = 1.01 t/nm² • 1 008 kg/nm² to t/nm² = 1.01 t/nm² • 1 009 kg/nm² to t/nm² = 1.01 t/nm² • 1 010 kg/nm² to t/nm² = 1.01 t/nm² • 1 011 kg/nm² to t/nm² = 1.01 t/nm² • 1 012 kg/nm² to t/nm² = 1.01 t/nm² • 1 013 kg/nm² to t/nm² = 1.01 t/nm² • 1 014 kg/nm² to t/nm² = 1.01 t/nm² • 1 015 kg/nm² to t/nm² = 1.02 t/nm² • 1 016 kg/nm² to t/nm² = 1.02 t/nm² • 1 017 kg/nm² to t/nm² = 1.02 t/nm² • 1 018 kg/nm² to t/nm² = 1.02 t/nm² • 1 019 kg/nm² to t/nm² = 1.02 t/nm² • 1 020 kg/nm² to t/nm² = 1.02 t/nm² • 1 021 kg/nm² to t/nm² = 1.02 t/nm² • 1 022 kg/nm² to t/nm² = 1.02 t/nm² • 1 023 kg/nm² to t/nm² = 1.02 t/nm² • 1 024 kg/nm² to t/nm² = 1.02 t/nm² • 1 025 through 1 049 kilograms per square nanometer • 1 025 kg/nm² to t/nm² = 1.03 t/nm² • 1 026 kg/nm² to t/nm² = 1.03 t/nm² • 1 027 kg/nm² to t/nm² = 1.03 t/nm² • 1 028 kg/nm² to t/nm² = 1.03 t/nm² • 1 029 kg/nm² to t/nm² = 1.03 t/nm² • 1 030 kg/nm² to t/nm² = 1.03 t/nm² • 1 031 kg/nm² to t/nm² = 1.03 t/nm² • 1 032 kg/nm² to t/nm² = 1.03 t/nm² • 1 033 kg/nm² to t/nm² = 1.03 t/nm² • 1 034 kg/nm² to t/nm² = 1.03 t/nm² • 1 035 kg/nm² to t/nm² = 1.04 t/nm² • 1 036 kg/nm² to t/nm² = 1.04 t/nm² • 1 037 kg/nm² to t/nm² = 1.04 t/nm² • 1 038 kg/nm² to t/nm² = 1.04 t/nm² • 1 039 kg/nm² to t/nm² = 1.04 t/nm² • 1 040 kg/nm² to t/nm² = 1.04 t/nm² • 1 041 kg/nm² to t/nm² = 1.04 t/nm² • 1 042 kg/nm² to t/nm² = 1.04 t/nm² • 1 043 kg/nm² to t/nm² = 1.04 t/nm² • 1 044 kg/nm² to t/nm² = 1.04 t/nm² • 1 045 kg/nm² to t/nm² = 1.05 t/nm² • 1 046 kg/nm² to t/nm² = 1.05 t/nm² • 1 047 kg/nm² to t/nm² = 1.05 t/nm² • 1 048 kg/nm² to t/nm² = 1.05 t/nm² • 1 049 kg/nm² to t/nm² = 1.05 t/nm² • 1 050 through 1 074 kilograms per square nanometer • 1 050 kg/nm² to t/nm² = 1.05 t/nm² • 1 051 kg/nm² to t/nm² = 1.05 t/nm² • 1 052 kg/nm² to t/nm² = 1.05 t/nm² • 1 053 kg/nm² to t/nm² = 1.05 t/nm² • 1 054 kg/nm² to t/nm² = 1.05 t/nm² • 1 055 kg/nm² to t/nm² = 1.06 t/nm² • 1 056 kg/nm² to t/nm² = 1.06 t/nm² • 1 057 kg/nm² to t/nm² = 1.06 t/nm² • 1 058 kg/nm² to t/nm² = 1.06 t/nm² • 1 059 kg/nm² to t/nm² = 1.06 t/nm² • 1 060 kg/nm² to t/nm² = 1.06 t/nm² • 1 061 kg/nm² to t/nm² = 1.06 t/nm² • 1 062 kg/nm² to t/nm² = 1.06 t/nm² • 1 063 kg/nm² to t/nm² = 1.06 t/nm² • 1 064 kg/nm² to t/nm² = 1.06 t/nm² • 1 065 kg/nm² to t/nm² = 1.07 t/nm² • 1 066 kg/nm² to t/nm² = 1.07 t/nm² • 1 067 kg/nm² to t/nm² = 1.07 t/nm² • 1 068 kg/nm² to t/nm² = 1.07 t/nm² • 1 069 kg/nm² to t/nm² = 1.07 t/nm² • 1 070 kg/nm² to t/nm² = 1.07 t/nm² • 1 071 kg/nm² to t/nm² = 1.07 t/nm² • 1 072 kg/nm² to t/nm² = 1.07 t/nm² • 1 073 kg/nm² to t/nm² = 1.07 t/nm² • 1 074 kg/nm² to t/nm² = 1.07 t/nm² • 1 075 through 1 099 kilograms per square nanometer • 1 075 kg/nm² to t/nm² = 1.08 t/nm² • 1 076 kg/nm² to t/nm² = 1.08 t/nm² • 1 077 kg/nm² to t/nm² = 1.08 t/nm² • 1 078 kg/nm² to t/nm² = 1.08 t/nm² • 1 079 kg/nm² to t/nm² = 1.08 t/nm² • 1 080 kg/nm² to t/nm² = 1.08 t/nm² • 1 081 kg/nm² to t/nm² = 1.08 t/nm² • 1 082 kg/nm² to t/nm² = 1.08 t/nm² • 1 083 kg/nm² to t/nm² = 1.08 t/nm² • 1 084 kg/nm² to t/nm² = 1.08 t/nm² • 1 085 kg/nm² to t/nm² = 1.09 t/nm² • 1 086 kg/nm² to t/nm² = 1.09 t/nm² • 1 087 kg/nm² to t/nm² = 1.09 t/nm² • 1 088 kg/nm² to t/nm² = 1.09 t/nm² • 1 089 kg/nm² to t/nm² = 1.09 t/nm² • 1 090 kg/nm² to t/nm² = 1.09 t/nm² • 1 091 kg/nm² to t/nm² = 1.09 t/nm² • 1 092 kg/nm² to t/nm² = 1.09 t/nm² • 1 093 kg/nm² to t/nm² = 1.09 t/nm² • 1 094 kg/nm² to t/nm² = 1.09 t/nm² • 1 095 kg/nm² to t/nm² = 1.1 t/nm² • 1 096 kg/nm² to t/nm² = 1.1 t/nm² • 1 097 kg/nm² to t/nm² = 1.1 t/nm² • 1 098 kg/nm² to t/nm² = 1.1 t/nm² • 1 099 kg/nm² to t/nm² = 1.1 t/nm² #### Foods, Nutrients and Calories NUNEZ DE PRADO, ORGANIC EXTRA VIRGIN OLIVE OIL, UPC: 015532100906 weigh(s) 236.7 gram per (metric cup) or 7.9 ounce per (US cup), and contain(s) 857 calories per 100 grams or ≈3.527 ounces [ weight to volume \| volume to weight \| price \| density ] ARTISAN GARLIC BREAD, UPC: 021130116317 contain(s) 300 calories per 100 grams or ≈3.527 ounces [ price ] #### Gravels, Substances and Oils CaribSea, Freshwater, African Cichlid Mix, Ivory Coast Sand weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume \| volume to weight \| price ] Chlorochromic anhydride [CrO2Cl2] weighs 1 910 kg/m³ (119.2374 lb/ft³) [ weight to volume \| volume to weight \| price \| mole to volume and weight \| density ] Volume to weightweight to volume and cost conversions for Refrigerant R-23, liquid (R23) with temperature in the range of -95.56°C (-140.008°F) to 4.45°C (40.01°F) #### Weights and Measurements Rem (Roentgen Equivalent Man) is the derived unit of any of the quantities expressed as the radiation absorbed dose (rad) equivalent [ millirem high-energy protons ] The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?" µg/US qt to oz/tsp conversion table, µg/US qt to oz/tsp unit converter or convert between all units of density measurement. #### Calculators Cube calculator. Compute the volume of a cube and its surface area	2,999	5,888	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2020-10	latest	en	0.152451
https://www.sae.org/publications/technical-papers/content/260040/?src=2003-01-0334	1,718,934,046,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00824.warc.gz	821,088,911	13,684	1926-01-01 ACTION, APPLICATION AND CONSTRUCTION OF UNIVERSAL-JOINTS 260040 Use of the universal-joint for transmitting power mechanically through an angle has been traced back to about 300 years before Cardan's period and about 400 years before a patent on a universal-joint was granted to Robert Hooke in 1664. The first reference to use of this type of joint is found in a manuscript by Wilars de Honecort, a thirteenth-century architect. A peculiarity of the universal-joint is that, as the two shafts which it unites are rotated when at an angle to each other, it imparts to the driven shaft a non-uniform rotational velocity which becomes very erratic as the angle between the shafts approaches 90 deg. This action has been analyzed by many writers by different methods, two such analyses being cited by the author of the present paper. To secure uniformity of velocity of the two shafts by the use of a pair of universal-joints, the driving and the driven shaft must form equal angles with the intermediate shaft and the axes of the joint yokes on the ends of the intermediate shaft must lie in the same plane. With such an arrangement, the universal-joint is so highly efficient at small angles that the loss of power in transmission can be detected only by delicate and accurate instruments. The theoretically correct arrangement of a double-jointed propeller-shaft in a car having a torque-arm is shown. Many typical examples of patented universal-joint design on which much inventive effort has been expended are given, together with descriptive text. Basic ideas embodied in some of these are in use in various types of universal-joints now offered in the market and that have been used for years in motor-vehicles. The more prominent of these joints are illustrated and described. The author classifies universal-joints into three general divisions: grease lubricated, oil lubricated, and non-lubricated. An ideal sought by some engineers is a joint that will perform its function satisfactorily with no lubrication or other attention, and that will have a life equal to that of the vehicle in which it is installed. A type of joint that is distinguished by a uniform driving relation between the engine and the driving wheels has a series of steel balls held in races milled in opposite faces of the jaws of the yokes. Such rolling balls add a supplemental principle to that of the universal-joint. Live-rubber balls give a cushioning effect that is highly desirable in some applications. Flexible-disc joints, which are in a class by themselves, have been greatly improved by special construction of the fabric-and-rubber discs and by the method of holding them to the spiders on the shaft ends. SAE MOBILUS We also recommend: TECHNICAL PAPER Self Adjusting Technology in a Clutch 960981 View Details TECHNICAL PAPER Rubber Wheel Abrasion Test 700687 View Details TECHNICAL PAPER Bearing, Gearing, and Lubrication Technology 780077 View Details	636	2,973	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2024-26	latest	en	0.953918
http://turing.ml/docs/quick-start/	1,544,607,870,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376823817.62/warc/CC-MAIN-20181212091014-20181212112514-00093.warc.gz	290,338,714	4,247	# Probablistic Programming in Thirty Seconds If you are already well-versed in probabalistic programming and just want to take a quick look at how Turing’s syntax works or otherwise just want a model to start with, we have provided a Bayesian coin-flipping model to play with. This example can be run on however you have Julia installed (see Getting Started), but you will need to install the packages `Turing`, `Distributions`, `MCMCChain`, and `StatPlots` if you have not done so already. This is an excerpt from a more formal example introducing probabalistic programming which can be found in Jupyter notebook form here or as part of the documentation website here. ``````# Import libraries. using Turing, StatPlots, Random # Set the true probability of heads in a coin. p_true = 0.5 # Iterate from having seen 0 observations to 100 observations. Ns = 0:100; # Draw data from a Bernoulli distribution, i.e. draw heads or tails. Random.seed!(12) data = rand(Bernoulli(p_true), last(Ns)) # Declare our Turing model. @model coinflip(y) = begin # Our prior belief about the probability of heads in a coin. p ~ Beta(1, 1) # The number of observations. N = length(y) for n in 1:N # Heads or tails of a coin are drawn from a Bernoulli distribution. y[n] ~ Bernoulli(p) end end; # Settings of the Hamiltonian Monte Carlo (HMC) sampler. iterations = 1000 ϵ = 0.05 τ = 10 # Start sampling. chain = sample(coinflip(data), HMC(iterations, ϵ, τ)); # Construct summary of the sampling process for the parameter p, i.e. the probability of heads in a coin. psummary = Chains(chain[:p]) histogram(psummary) ``````	416	1,612	{"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-2018-51	longest	en	0.848258
https://www.quizover.com/online/course/14-2-introduction-to-probability-by-openstax	1,532,096,953,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00512.warc.gz	969,744,959	20,522	14.2 Introduction to probability Page 1 / 1 This module provides an introduction which develops concepts related probability. Suppose you roll a 4-sided die , which gives equal probabilities of rolling 1, 2, 3, or 4. (Yes, these really do exist.) • A What is the probability that you will roll a 3? • B Explain, in your own words, what your answer to part (a) means. Do not use the words “probability” or “chance.” A typical 9-year-old should be able to understand your explanation. • C What is the probability that you will roll an even number? • D What is the probability that you will roll a number less than 3? • E What is the probability that you will roll a number less than 4? • F What is the probability that you will roll a number less than 7? Suppose you roll two 4-sided dice . • A Draw a tree diagram showing all the possible outcomes for both rolls. • B What is the probability that you will roll “3 on the first die, 2 on the second?” • C What is the probability that you will roll “3 on one die, 2 on the other?” • D What is the probability that the sum of both dice will be a 5? find the 15th term of the geometric sequince whose first is 18 and last term of 387 I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) hmm well what is the answer Abhi how do they get the third part x = (32)5/4 can someone help me with some logarithmic and exponential equations. 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer I don't understand what the A with approx sign and the boxed x mean it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 oops. ignore that. so you not have an equal sign anywhere in the original equation? hmm Abhi is it a question of log Abhi 🤔. Abhi I rally confuse this number And equations too I need exactly help salma But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends salma Commplementary angles hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday hi salma what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks. a perfect square v²+2v+_ kkk nice algebra 2 Inequalities:If equation 2 = 0 it is an open set? or infinite solutions? Kim The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined. Al y=10× if \|A\| not equal to 0 and order of A is n prove that adj (adj A = \|A\| rolling four fair dice and getting an even number an all four dice Kristine 22*2=8 Differences Between Laspeyres and Paasche Indices No. 7x -4y is simplified from 4x + (3y + 3x) -7y how do you translate this in Algebraic Expressions Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam Hello Uday I'm interested in Nanotube Uday this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15 Prasenjit can nanotechnology change the direction of the face of the world At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light. the Beer law works very well for dilute solutions but fails for very high concentrations. why? how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!	1,519	5,692	{"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}	4.375	4	CC-MAIN-2018-30	longest	en	0.946875
https://www.homeownershub.com/maintenance/number-of-wires-in-conduit-153520-.htm	1,548,201,384,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00351.warc.gz	827,039,761	12,612	# Number of wires in conduit Installing dishwasher. New install. Need to run electic-separate 15amp circuit. Would like to feed wires through nearby conduit run. How many conductors and neutrals can run in a 1/2 or 3/8 conduit ? <% if( /^image/.test(type) ){ %> <% } %> <%-name%> Teo2006 wrote: The limit is on the total number of wires. The capacity of those wires may need to be derated if there are more than three current carrying conductors in the conduit. Conductors that are functioning as actual neutral conductors are not counted for that purpose but grounded current carrying conductors other than true neutrals are. To answer your question about the total number of wires permitted in the conduit I would need to know the actual type and nominal size of conduit or at least it's actual internal diameter. Conduits carrying three or more wires may only be filled to 40% of their cross sectional area. -- Tom Horne "This alternating current stuff is just a fad. It is much too dangerous <% if( /^image/.test(type) ){ %> <% } %> <%-name%> <% if( /^image/.test(type) ){ %> <% } %> <%-name%> Teo2006 ( snipped-for-privacy@netscape.net) said... I created a Conduit Calculator application that is available for free at http://daxack.ca/Conduit/index.html -- Calvin Henry-Cotnam "I really think Canada should get over to Iraq as quickly as possible" <% if( /^image/.test(type) ){ %> <% } %> <%-name%> I think its safe to say you can install (3) #12 thhn wires in a 1/2 conduit. 40% fill. Assuming your dishwasher is 120v ac one hot, one neutral and one ground. On 06 Oct 2006 22:04:39 GMT, snipped-for-privacy@remove.daxack.ca.invalid (Calvin Henry-Cotnam) wrote: <% if( /^image/.test(type) ){ %> <% } %> <%-name%> ## Site Timeline • ### Kitchen faucet chatter • - next thread in Home Repair • ### Concrete step removal / replacement • - previous thread in Home Repair • ### Re: Ocasio-Cortez calls climate change our World War II, warns the world will ... • Share To HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.	575	2,223	{"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}	2.796875	3	CC-MAIN-2019-04	latest	en	0.809669
https://www.ademcetinkaya.com/2022/12/rfa-rare-foods-australia-limited.html	1,685,986,475,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00217.warc.gz	678,964,465	59,650	RARE FOODS AUSTRALIA LIMITED Research Report ## Summary With the up-gradation of technology and exploration of new machine learning models, the stock market data analysis has gained attention as these models provide a platform for businessman and traders to choose more profitable stocks. As these data are in large volumes and highly complex so a need of more efficient machine learning model for daily predictions is always looked upon. We evaluate RARE FOODS AUSTRALIA LIMITED prediction models with Supervised Machine Learning (ML) and Factor1,2,3,4 and conclude that the RFA stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy RFA stock. ## Key Points 1. Probability Distribution 2. Should I buy stocks now or wait amid such uncertainty? 3. How do you decide buy or sell a stock? ## RFA Target Price Prediction Modeling Methodology We consider RARE FOODS AUSTRALIA LIMITED Decision Process with Supervised Machine Learning (ML) where A is the set of discrete actions of RFA stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4 F(Factor)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Supervised Machine Learning (ML)) X S(n):→ (n+3 month) $∑ i = 1 n r i$ n:Time series to forecast p:Price signals of RFA stock j:Nash equilibria (Neural Network) k:Dominated move a:Best response for target price For further technical information as per how our model work we invite you to visit the article below: How do AC Investment Research machine learning (predictive) algorithms actually work? ## RFA Stock Forecast (Buy or Sell) for (n+3 month) Sample Set: Neural Network Stock/Index: RFA RARE FOODS AUSTRALIA LIMITED Time series to forecast n: 01 Dec 2022 for (n+3 month) According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy RFA stock. X axis: Likelihood% (The higher the percentage value, the more likely the event will occur.) Y axis: Potential Impact% (The higher the percentage value, the more likely the price will deviate.) Z axis (Yellow to Green): Technical Analysis% ## Adjusted IFRS Prediction Methods for RARE FOODS AUSTRALIA LIMITED 1. Paragraph 5.7.5 permits an entity to make an irrevocable election to present in other comprehensive income subsequent changes in the fair value of particular investments in equity instruments. Such an investment is not a monetary item. Accordingly, the gain or loss that is presented in other comprehensive income in accordance with paragraph 5.7.5 includes any related foreign exchange component. 2. An entity that first applies IFRS 17 as amended in June 2020 after it first applies this Standard shall apply paragraphs 7.2.39–7.2.42. The entity shall also apply the other transition requirements in this Standard necessary for applying these amendments. For that purpose, references to the date of initial application shall be read as referring to the beginning of the reporting period in which an entity first applies these amendments (date of initial application of these amendments). 3. For example, Entity A, whose functional currency is its local currency, has a firm commitment to pay FC150,000 for advertising expenses in nine months' time and a firm commitment to sell finished goods for FC150,000 in 15 months' time. Entity A enters into a foreign currency derivative that settles in nine months' time under which it receives FC100 and pays CU70. Entity A has no other exposures to FC. Entity A does not manage foreign currency risk on a net basis. Hence, Entity A cannot apply hedge accounting for a hedging relationship between the foreign currency derivative and a net position of FC100 (consisting of FC150,000 of the firm purchase commitment—ie advertising services—and FC149,900 (of the FC150,000) of the firm sale commitment) for a nine-month period. 4. At the date of initial application, an entity shall assess whether a financial asset meets the condition in paragraphs 4.1.2(a) or 4.1.2A(a) on the basis of the facts and circumstances that exist at that date. The resulting classification shall be applied retrospectively irrespective of the entity's business model in prior reporting periods. International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world. ## Conclusions RARE FOODS AUSTRALIA LIMITED assigned short-term Ba3 & long-term Ba3 forecasted stock rating. We evaluate the prediction models Supervised Machine Learning (ML) with Factor1,2,3,4 and conclude that the RFA stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy RFA stock. ### Financial State Forecast for RFA RARE FOODS AUSTRALIA LIMITED Options & Futures Rating Short-Term Long-Term Senior OutlookBa3Ba3 Operational Risk 3870 Market Risk6980 Technical Analysis8759 Fundamental Analysis3438 Risk Unsystematic9058 ### Prediction Confidence Score Trust metric by Neural Network: 79 out of 100 with 747 signals. ## References 1. Batchelor, R. P. Dua (1993), "Survey vs ARCH measures of inflation uncertainty," Oxford Bulletin of Economics Statistics, 55, 341–353. 2. Breiman L. 1996. Bagging predictors. Mach. Learn. 24:123–40 3. J. Harb and D. Precup. Investigating recurrence and eligibility traces in deep Q-networks. In Deep Reinforcement Learning Workshop, NIPS 2016, Barcelona, Spain, 2016. 4. A. Tamar, Y. Glassner, and S. Mannor. Policy gradients beyond expectations: Conditional value-at-risk. In AAAI, 2015 5. Imbens G, Wooldridge J. 2009. Recent developments in the econometrics of program evaluation. J. Econ. Lit. 47:5–86 6. Bera, A. M. L. Higgins (1997), "ARCH and bilinearity as competing models for nonlinear dependence," Journal of Business Economic Statistics, 15, 43–50. 7. K. Tuyls and G. Weiss. Multiagent learning: Basics, challenges, and prospects. AI Magazine, 33(3): 41–52, 2012 Frequently Asked QuestionsQ: What is the prediction methodology for RFA stock? A: RFA stock prediction methodology: We evaluate the prediction models Supervised Machine Learning (ML) and Factor Q: Is RFA stock a buy or sell? A: The dominant strategy among neural network is to Buy RFA Stock. Q: Is RARE FOODS AUSTRALIA LIMITED stock a good investment? A: The consensus rating for RARE FOODS AUSTRALIA LIMITED is Buy and assigned short-term Ba3 & long-term Ba3 forecasted stock rating. Q: What is the consensus rating of RFA stock? A: The consensus rating for RFA is Buy. Q: What is the prediction period for RFA stock? A: The prediction period for RFA is (n+3 month)	1,745	7,074	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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-2023-23	longest	en	0.868765
https://study.com/academy/lesson/how-to-write-1-million-in-scientific-notation.html	1,576,015,911,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575540529006.88/warc/CC-MAIN-20191210205200-20191210233200-00220.warc.gz	543,955,001	42,191	# How to Write 1 Million in Scientific Notation An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: How to Write 1000 in Scientific Notation ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:03 Steps to Solve • 2:29 Real World Application Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Laura Pennington Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions. Scientific notation is a convenient way of expressing large or small numbers. This lesson will explain the steps involved in writing 1 million using scientific notation. We will then take a look at a real-world situation in which scientific notation is used to write 1 million. ## Steps to Solve We want to write 1 million using scientific notation. Writing 1 million using scientific notation involves the same steps as writing any number using scientific notation. It's simply a matter of counting decimal places, observing which way we're moving the decimal, and then placing numbers in the right place. That doesn't sound so bad, huh? Well, it's not, so let's get started! First of all, a number written in scientific notation is a number multiplied by a power of 10. In general, the steps we use in writing a number x using scientific notation are as follows: 1. Move the decimal point in x to create a number that is between 1 and 10. To do this, we move the decimal so that there is only one digit to the left of the decimal point. That is, we move the decimal point in between the first and second digit of the number. The new number will be the number we multiply by a power of 10. 2. Count the number of places you moved the decimal point in step one. This will be the number we raise 10 to. If we moved the decimal point to the left in step one, then the number we raise 10 to will be positive, and if we moved the decimal point to the right in step one, then the number we raise 10 to will be negative. 3. Write the number using scientific notation with the information obtained in steps one and two. Alright, let's take the number 1000000 through these steps. The first step is to move the decimal point in 1000000 between the first and second digit. Therefore, we move the decimal point in between 1 and 0 in 1000000. We see that the number we will be multiplying by a power of 10 is 1. The next step is to count the number of places we moved the decimal point and to observe the fact that we are moving the decimal point to the left. We have that we moved the decimal point six places to the left. Therefore, the power we raise 10 to is 6. Great! We've made it to the third step, which is basically just filling in the blanks. We've found that in the number a x 10 b, a is 1, and b is 6, so we have that 1000000 = 1 x 10 6. See? You were right, it wasn't very hard at all! To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### 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.	850	3,721	{"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}	4.71875	5	CC-MAIN-2019-51	longest	en	0.885764
https://promovare-site.info/relationship/entity-relationship-diagram-example-for-database.php	1,566,282,064,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00286.warc.gz	616,414,799	9,648	# Entity relationship diagram example for database and templates. Use Lucidchart for all your ER diagram needs! ER Diagram Example Template · ER Diagram · University Database ER Diagram Template. An entity relationship model, also called an entity-relationship (ER) diagram, is a graphical Let's take an example of a company database. CS Home Data Modeling ERD Entities Relationships Attributes Example promovare-site.info It is a graphical representation of data requirements for a database. Some courses make up each program. Students register in a particular program and enroll in various courses. A lecturer from the specific department takes each course, and each lecturer teaches a various group of students. Relationship Relationship is nothing but an association among two or more entities. Entities take part in relationships. We can often identify relationships with verbs or verb phrases. You are attending this lecture I am giving the lecture Just loke entities, we can classify relationships according to relationship-types: A student attends a lecture A lecturer is giving a lecture. Weak Entities A weak entity is a type of entity which doesn't have its key attribute. It can be identified uniquely by considering the primary key of another entity. For that, weak entity sets need to have participation. Let's learn more about a weak entity by comparing it with a Strong Entity Strong Entity Set Strong entity set always has a primary key. It does not have enough attributes to build a primary key. It is represented by a rectangle symbol. It is represented by a double rectangle symbol. It contains a Primary key represented by the underline symbol. It contains a Partial Key which is represented by a dashed underline symbol. The member of a strong entity set is called as dominant entity set. The member of a weak entity set called as a subordinate entity set. Primary Key is one of its attributes which helps to identify its member. In a weak entity set, it is a combination of primary key and partial key of the strong entity set. In the ER diagram the relationship between two strong entity set shown by using a diamond symbol. The relationship between one strong and a weak entity set shown by using the double diamond symbol. Students have one or more given names, a surname, a student identifier, a date of birth, and the year they first enrolled. When he finishes the course, a grade such as A or B and a mark such as 60 percent are recorded. Each course in a program is sequenced into a year for example, year 1 and a semester for example, semester 1. Although it is compact, the diagram uses some advanced features, including relationships that have attributes and two many-to-many relationships. The ER diagram of the university database In our design: Each student must be enrolled in a program, so the Student entity participates totally in the many-to-one EnrollsIn relationship with Program. A program can exist without having any enrolled students, so it participates partially in this relationship. As a weak entity, Course participates totally in the many-to-one identifying relationship with its owning Program. This relationship has Year and Semester attributes that identify its sequence position. Student and Course are related through the many-to-many Attempts relationships; a course can exist without a student, and a student can be enrolled without attempting any courses, so the participation is not total. When a student attempts a course, there are attributes to capture the Year and Semester, and the Mark and Grade. For a real university, many more aspects would need to be captured by the database. The airline has one or more airplanes. An airplane has a model number, a unique registration number, and the capacity to take one or more passengers. An airplane flight has a unique flight number, a departure airport, a destination airport, a departure date and time, and an arrival date and time. ## Learning MySQL by Hugh E. Williams, Saied M.M. Tahaghoghi Each flight is carried out by a single airplane. A passenger has given names, a surname, and a unique email address. A passenger can book a seat on a flight. The ER diagram of the flight database An Airplane is uniquely identified by its RegistrationNumber, so we use this as the primary key. A Flight is uniquely identified by its FlightNumber, so we use the flight number as the primary key. The departure and destination airports are captured in the From and To attributes, and we have separate attributes for the departure and arrival date and time. Because no two passengers will share an email address, we can use the EmailAddress as the primary key for the Passenger entity. ### Entity Relationship Modeling Examples - Learning MySQL [Book] An airplane can be involved in any number of flights, while each flight uses exactly one airplane, so the Flies relationship between the Airplane and Flight relationships has cardinality 1: N; because a flight cannot exist without an airplane, the Flight entity participates totally in this relationship. A passenger can book any number of flights, while a flight can be booked by any number of passengers. N Books relationship between the Passenger and Flight relationship, but considering the issue more carefully shows that there is a hidden entity here: We capture this by creating the intermediate entity Booking and 1: N relationships between it and the Passenger and Flight entities. Identifying such entities allows us to get a better picture of the requirements. There are no requirements to capture passenger details such as age, gender, or frequent-flier number. If, instead, we assumed that the capacity is determined by the model number, we would have created a new AirplaneModel entity with the attributes ModelNumber and Capacity. The Airplane entity would then not have a Capacity attribute. Airlines typically use a flight number to identify a given flight path and schedule, and they specify the date of the flight independently of the flight number. For example, there is one IR flight on April 1, another on April 2, and so on. Different airplanes can operate on the same flight number over time; our model would need to be extended to support this. Part 2.1 - Entity relationship model diagram in dbms in hindi introduction and basics syllabus The system also assumes that each leg of a multihop flight has a different FlightNumber. Our database also has limited ability to describe airports. If we were to model the airport as a separate entity, we could use the IATA-assigned airport code as the primary key. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.	1,299	6,718	{"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}	2.625	3	CC-MAIN-2019-35	longest	en	0.934164
https://holooly.com/solutions/the-beam-is-subjected-to-the-distributed-loading-determine-the-length-b-of-the-uniform-load-and-its-position-a-on-the-beam-such-that-the-resultant-force-and-couple-moment-acting-on-the-beam-are-zero/	1,603,151,502,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00444.warc.gz	371,902,673	16,841	## Question: The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: ${ w }_{ 1 }$ = $40$ $\frac { Ib }{ ft } \quad c$ = $10$ $ft$ ${ w }_{ 2 }$ = $60$ $\frac { Ib }{ ft } \quad d$ = $6$ $ft$	107	347	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 10, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2020-45	latest	en	0.817464
http://www.jiskha.com/display.cgi?id=1167867644	1,462,553,492,000,000,000	text/html	crawl-data/CC-MAIN-2016-18/segments/1461861848830.49/warc/CC-MAIN-20160428164408-00178-ip-10-239-7-51.ec2.internal.warc.gz	612,702,122	3,786	Friday May 6, 2016 # Homework Help: Math 7th grade Posted by Ariana on Wednesday, January 3, 2007 at 6:40pm. I am very bad at fractions.Can anyone tell me what a multiplication sentance is? The book gave me a model. It is an expression in math or written form. John had 1/2 of the pizza, and he ate 3/8 of that. How much did he eat of the original pizza. 1/2 x 3/8 dont know 25.70 divided by 1/4= what	129	409	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2016-18	longest	en	0.981198
http://www.careersvalley.com/accenture-practice-probability-questions	1,369,551,750,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368706635944/warc/CC-MAIN-20130516121715-00022-ip-10-60-113-184.ec2.internal.warc.gz	382,489,401	15,525	## Accenture Practice Probability Questions In Accenture papers, you can expect few questions from probability. These would generally of conventional nature and hence you can answer these with good practice. Question 1 A leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw? a) 70/8281 b)140/20449 c ) 25/5445 d)35/5448 Solution: Total number of balls = 8 + 6 = 14 Total ways of drawing 3 balls, N(S) = 14C3 = 364 No of ways to draw 3 black balls = N(E1) for black balls = 8C3 = 56 Probability of all balls being black = P(E1) = N(E1) / N(S) = 56/ 364 = 14/91 No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20 Probability of all balls being white = P(E2) = 20 / 364 = 5/91 Probability of drawing 3 black balls in first draw and drawing 3 white balls in the second draw = P(E1) x P(E2) Therefore P(E) = 14/91 x 5/91 = 70 /8281 Question 2 Fourteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together? a) 2/9 b)1/13 c)2/13 d)1/26 Solution : In a circle of n different persons, the total number of arrangements possible = (n - 1)! Total number of arrangements = n(S) = (14 – 1)! = 13 ! Taking three persons as a unit, total persons = 12 (in 4 units) Therefore no. of ways for these 12 persons to around the circular table = (12 - 1)! = 11! In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit = n(E) = 11! X 3! Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 11! X 3 ! divided by 13! 3 x 2 divided by 13 x 12 = 1/26 Question 3 In how many different ways can the letter of the word “ECHRONICLEE” be arranged? a)1663200 b)8316000 c)3326400 d)4158000 Solution : ECHRONICLEE has 11 letters Total number of rearrangements = (Number of letters)! / (Number of 1st repetitive letter)! x (Number of 2nd repetitive letter)!.... In the word ECHRONICLEE, E occurs thrice. C occurs twice. Hence denominator of the above formula will become 3! x 2! Therefore Total number of rearrangements = 11!/ 3! 2! = 3326400 • Fresher IT Jobs	688	2,300	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.625	5	CC-MAIN-2013-20	latest	en	0.905292
http://blogs.bethel.k12.or.us/cruscher2/	1,503,413,520,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886110792.29/warc/CC-MAIN-20170822143101-20170822163101-00630.warc.gz	50,562,821	8,784	It has been a long year and we have all been working hard. Just a reminder to all students and family members, All late assignments and reproofs are due no later than Thursday, June 15th. We are going to begin our last section in Chapter 2 tomorrow. We have already taken many different assessments on various learning targets for Chapter 2. As we finish up the last section in Chapter 2, we will have time for students to retake assessments as needed. With this will come a couple days of reteaching and practice in class for the students who need it. However, many students are very eager to retake these assessments before then. If a student would like to retake any assessment, they must provide me with proof that they have practiced the skill and are ready to retake the assessment. All students have the option on completing problems from their book. They my redo problems that have already been assigned, or ask for new problems. They may chose to do test corrections on the assessment that they would like to retake. They may also complete practice on any of the numerous websites that offer videos and practice problems available online. I have posted in previous posts websites that students may use to practice solving equations, solving absolute value equations, and rewriting literal equations. Please review previous Algebra 1 posts to find these. Below you will find additional resources for solving inequalities, solving compound inequalities, and solving absolute value inequalities. For review on solving inequalities check out Solving Inequalities. For review on solving compound inequalities check out the following two websites. http://www.coolmath.com/algebra/07-solving-inequalities/06-compound-inequalities-01 http://www.coolmath.com/crunchers/algebra-problems-solving-inequalities-3 Happy practicing! Today we began Lesson 5 in Math 8. Lesson 5 is challenging, but the students are working hard to try to grasp the concepts. In this lesson, the students are learning to perform operations (addition, subtraction, multiplication, division) on number in scientific notation. We began working on addition today and will practice more tomorrow before moving to subtraction. To help your student with this learning target, please visit the website below. It contains videos and practice problems. Three day weekend! With all of the fun activities planned this weekend, don’t forget your homework. Math 8 – Complete pages 33-34 from the Lesson 4 HW packet. If you have not yet completed pages 31-32 please do so. Parents, please read the parent letter at the front of the packet. It is yours to keep. Algebra 1 – You were given a packet to practice your skills from Chapter 1. You are NOT responsible for completing the entire packet. You should do at least 10 problems from each section that you need to reproof on. Do more if you feel that you need more practice. I will list the problems that match the assessments. Remember the answer key is attached to the packet. • Solving Equations Quiz 1 and Solving Equations Procedural: #38-86 and #95-110 • Solving Equations Application: #111-113 • Literal Equations Procedural: #117-125 For a review on Solving Equations check out http://www.coolmath.com/prealgebra/16-intro-to-solving-equations. For a review on Absolute Value Equations Check out http://www.coolmath.com/algebra/18-absolute-value-equations-inequalities/02-solving-absolute-value-equations-01. For a review on Literal Equations check out https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq/alg-old-school-equations/v/solving-for-a-variable. As we get closer to Interim Progress Reports, many students and parents are concerned about the current grades. In Math 8, we have had 2 assessments so far and a 3rd one on Tuesday. Please review your child’s grades with them and help them determine what they can do to improve them. I am more than will to allow any student to retake assessments as needed. The only requirement for retaking an assessment is that they have to show me proof that they have been working on the skill either on their own or with me. As I have mentioned, 6th period flex time is limited to 23 minutes a day, 4 days a week and is shared among 4 academic classes that I teach. I encourage all students to try to use as many resources as possible for additional help such as homework club and additional help from me after school on Tuesdays. We will be practicing the learning targets continually and students will have time to retake assessments during class or flex. Below are the learning targets that we are focusing on as well as a couple of websites that may help provide additional help or practice. I hope this helps. • I can use properties of exponents. • http://www.coolmath.com/algebra/01-exponents • https://www.ixl.com/math/geometry/properties-of-exponents • I can use square root and cube root symbols. • http://www.coolmath.com/prealgebra/04-exponents/02-exponents-important-common-01 • Solving Equations Quiz: Textbook p.26 All (I will print the answer key if needed) • I can solve equations. Conceptual: Review Chapter 1 Notes and Core Concepts in Textbook • I can solve equations. Procedural: Textbook p.44-45 All • I can solve equations. Application: Textbook p.47 #12-16 Students do not have to complete all of the problems. They only need to complete enough for them to feel confident in there ability to earn a 3 or higher on the reproof. I hope this helps. Hello Parent(s)/Guardian(s), On Friday, all student received an update on their grades. Many of them are not where they would like to be in regards to meeting the standards. We will be re-teaching during class and the students will have several opportunities to retake assessments before progress reports go out. Please check with your students to see if they need any additional support from me. Many of them have asked to come in during 6th period flex time for additional support, however, the spots are limited. So, I am offering all students the opportunity to get additional help after school on Tuesdays from 3:15pm to 4:15pm. If your student is not maintaining at least a 3 on their assessments, they may need to take advantage of this opportunity. Algebra 1 parents, our textbook comes with a few helpful apps for android and apple devices. • The first app is called Big Ideas Math Videos. This app provides students with videos of extra examples being worked out for each section in the book. The videos explain step by step how to work out the various problems and can be very helpful to your student. • The second app is called Big Ideas Math Solutions. I usually assign the odd problems for homework in this class. This is because the answers to the odd problems are found in the back of the book. However, if you student is not able to figure out how the book got a particular answer, they can use this app to look at the solution as well as the steps for working out the problems. Math 8 and Algebra 1 parents, there is an app called My Dear Aunt Sally that also comes with our textbooks that can be helpful for students of all ages. This game is designed to help students practice their basic mathematics skills. The students can choose the difficulty level. The game offer practice with whole numbers for free, however to practice integers, decimals, fractions, and rational numbers, there is a small fee. I encourage all of my students to practice with at least the whole numbers part of the game to improve fluency with basic operations and the order of operations. I have noticed that for many of my Math 8 students, multiplication is a skill that needs to be practiced. We will start using multiplication charts in class to help with this. Please check my teacher webpage for additional information as it comes up. I am trying to update it as much as possible. You can find the link on the Bethel School District webpage or below. Please feel free to email if you have any questions or concerns. Candice Ruscher	1,823	8,057	{"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-2017-34	longest	en	0.959565
http://math.stackexchange.com/questions/273044/logarithm-exponent-in-chernoff-bound	1,469,330,438,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00134-ip-10-185-27-174.ec2.internal.warc.gz	165,691,372	16,262	Logarithm exponent in Chernoff bound I am applying Chernoff bound for a Poisson process with mean $\lg n$. I am putting $\delta =4$. Hence, $Pr(X<(1+4)\mu)< (\frac{e^\delta}{(1+4)^{(1+4)}})^\mu$ $= (\frac{e^\delta}{5^5})^{\lg n}=\frac{1}{c^{\lg n}}$, where $c=5^5/e^4$. Now how can I show, for some constant $k > 1$: $Pr(X<5 \lg n)<\frac{1}{n^k}$. I want to specifically know: How, $c^{\lg n} > n^k$ and the bound of $k$. Thanks. - It was a silly question. $c^{\lg n} = n^{\lg c}$. I forgot this silly relation.	204	518	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.171875	3	CC-MAIN-2016-30	latest	en	0.685946
http://excel.bigresource.com/Return-12-Months-from-Year-End-Date--AiuCRO.html	1,603,399,222,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00500.warc.gz	40,871,861	11,387	# Return 12 Months From Year End Date? Apr 17, 2014 Someone will enter their financial year end in the worksheet e.g. 31/03/2014 and I then have 12 cells below it called Month 1, Month 2 etc up to Month 12. I need the cell next to Month 1 to calculate what it would be... so for a Year End of 31/03/2014 month 1 would be April and this needs to apply to the 12 months. ## Get Month & Year X Months From Current Date May 3, 2008 I am trying to get mmm-yy date format for the next 2 consecutive months from now. I used the following code for that reason, but I am not getting. strDate = Format(DateAdd("m", 1, Date), "mmm-yy") strDate = Format(DateAdd("m", 2, Date), "mmm-yy") However if I use this, I get the current months date correctly (like May - 08) strDate = Format(DateAdd("m", 0, Date), "mmm-yy") ## Number Of Months In Year 2007 From Date Range Jan 12, 2008 I have a "start date" and an "end date". Is there any way to tell how many months are included in 2007 from those two dates using a formula? For instance, from 01-Feb-04 to 01-Feb-07, there is 1 month in 2007 (January). And likewise, 01-Jan-05 to 01-Jan-08 there is 12 months in 2007. As noted, I have the start and end dates. I don't have the "how many months in 2007". Start Date End Date How many months in 2007 01-Feb-0401-Feb-07 1 01-Jan-0301-Jan-06 0 01-Apr-0401-Apr-07 3 01-Feb-0301-Feb-06 0 01-Mar-0401-Mar-07 2 01-Feb-0401-Feb-07 1 01-Dec-0401-Dec-07 11 01-Jan-05 01-Jan-08 12 ## Look Up The Date To See Which Year It Falls In And Return The Year Oct 13, 2008 i have the following table of information Year DOB 7 01.09.96 -31.08.97 8 01.09.95 -31.08.96 9 01.09.94 -31.08.95 10 01.09.93 -31.08.94 11 01.09.92 -31.08.93 and a list of dates i need to look up the date to see which year it falls in and return the year ## Return Months Between Start Date & End Dates Jul 6, 2007 I have been trying to search for either a function or VBA that will perform the following: If I enter I want all months >=1/1/07 and <=3/31/07 it would tell me that the months in between are January, February and March and so on with all dates. ## Return A Date Based On Year / Month Jan 21, 2009 formula off here i use all the time relating to finding and sumproducts for specific months and years i.e. Jan 2008, Dec 2007.. depending on these dates excel searched through a specified range and returned me any values i wanted like No. of occurences, totals, sums etc etc it was a sumproduct formula... is there any way i could specifiy a date i.e. Jan 2009, which would search column a and return the date /and/or an account number in column b, only if the date was during jan 2009? The reason i want this is to use a lookup on the account numbers to return specific items of info, but i only need the account numbers if they occurred in specific months which i want to choose. ## User Defined Function To Return Fiscal Year Of A Date May 2, 2012 I am working with a fiscal year that starts in December and ends in November. I want to make a user defined function that will return the fiscal year of a date. I've created the below code, but it returns a zero. Function FiscalYear(DateFY) If Month(DateFY) = 12 Then Year (DateFY) + 1 Else Year (DateFY) End If End Function ## Add Months & Return Decimal As Years & Months Jul 31, 2007 I am looking for a formula that will add months and return the year. E.g. if I add 1.05 and 1.07 I should get 3.01. i.e. 3 years and 1 month. ## How To List Next 2 Months Of Year Apr 23, 2012 In cell A1 I have a drop down list with all the months of the year. What I want to do is if A1 has January selected, B1 becomes February and C1 becomes March. Basically I want the next 2 months. What happens when December is selected? How would i have January in B1 and February in C1? ## Listbox For Months Of The Year Apr 16, 2006 I am creating a budget spreadsheet. I am trying to create a list box where I choose what month I want to start with and the other months follow in each column, depending on what month picked in my list box. ## Add Up Times Worked For 12 Months Of The Year Jan 28, 2006 I need to be able to add up times worked for 12 months of the year. For example if one workd 49 hours and 23 minutes the month of Jan and the same for Feb, how do I do this and keep a running sum of them? I can't see to figure out how to format it so that it does not change to time of day since it will be over 24 hours. ## Calculate Number Of Months In A Year Feb 1, 2012 I need to automatically calculate the number of months a deal runs through 2012 dependant on the start and end date. I have attached a basic spreadsheet. Column C shows the results I would like the formula to calculate. ## Sorting Months By Calendar Year Jan 15, 2013 I have just finished sorting my anniversary list by Month (in alpha order) B C D E F 1 Hire Date MonthDay YearLast Name First Name 2 04/04/2011April 042011Emp 1 3 04/09/1996April 091996Emp 2 4 08/02/2012August022012Emp 3 5 08/09/2004August092004Emp 4 6 08/13/2001August132001Emp 5 7 08/16/2010August162010Emp 6 Is there a way to sort by Calendar year? Jan, Feb, Mar, etc. ## Loading A Combo Box With The Months Of The Year Apr 13, 2007 to populate a combo box with the months of the year, without referencing an area on a sheet. Here is the code that works for me: Private Sub UserForm_Initialize() Dim i As Long Dim varMonths As Variant varMonths = Array("Jan", "Feb", "Mar", _ "Apr", "May", "Jun", "Jul", "Aug", _ "Sep", "Oct", "Nov", "Dec") For i = LBound(varMonths) To UBound(varMonths) Next i End Sub The original question and answer can be found here: [url] I see there's a second reply which looks a little slicker, but the above code is tried and tested by me. ## Formula To Convert Fraction Of Year To Months? Jan 28, 2012 Inmates can apply for pre-release when they have half their minimum sentence served. For example 2 years 6 months. They would have to serve 1 year 3 months. I use this formula =DATE(YEAR(A1)-A5,MONTH(A1)-B5,DAY(A1)-B6) Where A1 is half the years, B5 is half the months, and B6 is half the days. The problem is when an inmate is sentenced to a minimum of 3 years and 3 months. The latter formula will not calculate 1.5 years or 1.5 months. It rounds up. What I would like to do is covert the 1.5 years to months and the 1.5 months to days. ## Formula That Averages Totals From Different Months Out Of The Year Jan 1, 2007 I need is a formula that averages totals from different months out of the year. I already have a yearly average. That was easy. But what I need is a 3 & 6 month average. I also need it to be most current, so when I am in September, it will take the 3 previous months and average them and same with the 6 month. Then when I move into October, it would take its 3 previous months, i.e. - July,August, September. ## Elapsed Months (many Months Have Gone By Since Todays Date) Jul 19, 2009 Cell A1 is a past date. In cell B1 I would like how many months have gone by since todays date. eg. Cell A1 = July 07, B1 would = 24 months. ## Calculate Number Of Months In Specific Year Between Two Dates? May 19, 2014 I am trying to calculate the number of months in a specific year between two dates. For example. Start date 01/06/2012 End Date 01/02/2013 Number of months in 2012 = 6 Number of months in 2013 = 2 How can I write a formula to give me the answer of 6 & 2 from the start and finish dates? ## Date And Month From A Column And Year Should Take Current Year May 14, 2009 I have dates in my column “A”, for example (A1 cell =22-Mar-1971), (A2 cell=30-Dec-1965). Now my requirement is in B column date and month from A column and year should take current year. Output in B column (B1 cell =22-Mar-2009), (B2 cell=30-Dec-2009) ## Convert Date From Any Year To Current Year Feb 27, 2009 DATE function won't return TODAY()'s year in the "year" slot. Is there a way convert, for example, 2/8/1963 to 2/8/2009 without using Concatenate? ## Change Total Formulas For All Tables At Once To Show Either Year-to- Date Or Total Year Oct 15, 2007 I have a sheet in my workbook with at least 180 small tables, there may be more. I woulds like to be able to change total formulas for all tables at once to show either year-to- date or total year. For example: If we have only progressed through the second period of the year, I would like to choose something to indicate period 2. At other time I may want to know the total year whether the periods are completed or not. ## Creating View By Filtering 2 Columns By Date (both Within 3 Months Of Today Date) Mar 20, 2014 Trying to do a linkback from another post located here but not having much luck doing it: [URL] I'm working with 2 date columns and trying to filter a view to only include projects with dates within 3 months of today's date. I've attached a current working file of the data and the end result i'm hoping to achieve via a macro of some sort. I've manually got it to work via formula by inserting 2 additional columns (highlighted yellow) which determine if the dates "YES" fall in this 3 month time frame of "" blank if not. create a macro which does all of this automatically without modifying any columns if this is possible ## Find Starting Date And Fill In The Following Months Until End Date Mar 4, 2013 I have collected some data on economic factors for different countries. Unfortunately, the dates when I started to calculated my economic factors are different for each country (due to the data available to me). What I would like Excel to have done is to take the date when I started to measure for e.g. country A (D3 ie 30/06/2007), copy it into column "I" (for country A, it's cell I3) and fill in the following months in the rows below (with always the date of the last day of a month) until it reaches 28th of Feb 2013. Then, it should go up to the next country (country B) take the starting date (D4, ie 31/07/2007), go to the last entry in "I" (ie I71) and paste the date in, fill in the months until 28th of Feb 2013, do the same for country C and so on. I have started to code a VBA but I am unfortunately a beginner in VBA and totally stuck at the moment. My VBA code does paste in the months but for some reason, it also changes the starting date of the first month. Moreover, I tried a workaround for the fact that Excel doesnt know when to stop; ie I introduced a "monthdiff" variable which should calculate the number of months between the starting date (which is variable and unique for each country) and the end date (which is always 28th of Feb 2013). At the moment, it only does this for country A. VB: Set rng = ActiveSheet.Range("I3" & Cells(monthdiff, "I").Address)[SIZE=4][/SIZE] I have tried to make this dynamic but have been unsuccessful so far. Spreadsheet with data&code is attached. VB: Sub Macro1() Dim mainrange As Range Dim rng As Range [Code]..... ## Filter Data By Date Within 3 Months Of Today Date? Mar 20, 2014 I'm trying to filter 2 date columns to include only data containing dates within 3 months of today's date. I see there is a data filter option for "next quarter" but not 100% sure if this covers the quarter from today's date? ## Return The Number Of Months There Are Between The Start And End Dates Nov 13, 2006 if A1 is a start date and B1 is the end date, I want C1 to return the number of months there are between the start and end dates? ## Year Month Date To Month Date Year Code Jul 28, 2009 Serial No Search E220060926320061125420060612520070824620061026720061226820061127920061226 Excel tables to the web >> Excel Jeanie HTML 4 E - Year Month Date I need F column as Month Date Year Format ## Return Remaining Time To Birthday In Months And Days Jan 27, 2014 I'm looking for a formula that will return the time left to a birthday, in Month and Days. ## Formula To Return Day Of Year Jul 15, 2008 Is there a function that will return the day of any particular year? For example, if the date 01/01/2008 is entered, it will return 1 and if 31/12/1977 is entered, it will return 365? ## Return Data Corresponding To A Certain Year Nov 22, 2007 how to lookup a bank holiday lookup sheet to drop in to another sheet in a row beneath each other using the year as a lookup only! ## Calculate Date Six Months From Cell Date Jan 25, 2010 I have a start date for a contract in cell AM1. I need AN1 show a date six months from the start date in AM1. I am fine with that part. However, I would like for AN1 to calculate not only a six-month date but also to auto-update to the next six month date from AM1 once the first six month date is about, let's say, one month past. What I am doing is calculating when a six-month inspection needs to be completed; these inspections are ongoing, so I need them to auto-update. So let's say a contract was awarded today, 25 JAN 2010. That would be value in AM1. I want AN1 to produce a date six months from 25 JAN 2010, which would be 25 JUL 2010 (yes, this way of calculating the six month date is fine). Then around, let's say, 25 AUG 2010, I want AN1 to auto-update to produce the next six month date, which would be 25 JAN 2011 (six months from 25 JUL). Obviously the function would need to relate to TODAY() in some way.	3,643	13,285	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.0625	3	CC-MAIN-2020-45	latest	en	0.851622

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