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http://www.jiskha.com/search/index.cgi?query=Stats | 1,454,720,614,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701145578.23/warc/CC-MAIN-20160205193905-00306-ip-10-236-182-209.ec2.internal.warc.gz | 474,790,825 | 11,029 | Friday
February 5, 2016
# Search: Stats
Number of results: 1,224
stats
is there a good, relevant sites where i can get stats extra help?
October 11, 2007 by Keirson
C++ Programming
Design and create a class named Stats that has an array of 12 doubles as one of its member variables. The values in the array should be set by making 12 calls to a public member function named setValue that accepts two arguments, an integer indicating which value is being ...
June 8, 2010 by kyuu09
Stats
So i have a test in stats tomorrow and im trying to work on my study guide but im stuck at this question : Construct a confidence interval for the proportion for each case 90 percent interval 861 out of 2500. please help.
October 19, 2014 by Jessica
Introduction programing visual basic
break even analysis. suppose a certain product sells for a dollars per unit. then the revenue from selling x units of the product is ax dollars if the cost of producting each unit of the product is b dollars and the company has overhead cost of c dollars then the total cost of...
March 4, 2015 by Caitlin
Introduction programing visual basic
break even analysis. suppose a certain product sells for a dollars per unit. then the revenue from selling x units of the product is ax dollars if the cost of producting each unit of the product is b dollars and the company has overhead cost of c dollars then the total cost of...
March 4, 2015 by Caitlin
stats
Use this z-score formula for this problem: z = (x - mean)/(sd/√n) x = 1.25, 1.50 mean = 1.35 sd = 0.25 n = 40 Calculate two z-scores, then use a z-table to determine probability between the two scores. I hope this will help get you started. •stats. - Christina, Monday, ...
stats
If SS = 50 and s^2 = 10, what is n?
October 2, 2012 by eric
Stats
A certian company enjoyed 3.5% profits in year 1 over its revenue in Year 0. In Year 2, it enjoyed 5.75% profits over Year 1. In Year 3, it posted 1.25% profits over Year 2. What single, constant profit over this period of years would have produced the same profit? The teacher...
May 4, 2011 by Anonymous
College Stats
Is .09 less than .05?
May 8, 2011 by Carol
stats
given n=74 an x=32.find ^p and ^q
April 4, 2013 by Chris
stats
if we reject the null hypothesis, we conclude that:
April 1, 2010 by Anonymous
stats
Compute the following and show your steps 7! /5!.
August 7, 2010 by renae
stats
what does it mean to say alpha is set at 0.5
August 18, 2011 by nicole
math stats
how do i draw a histogram for the data (8,0,7,5,5)
January 24, 2012 by juan777
Stats
What is the spread? How can it be found from a set of numbers?
March 6, 2012 by Jill
math stats
Calculate p ̅. x1 20 n1 50 x2 55 n2 100
June 22, 2013 by jerry
Stats
Which of the following numbers could be a proabilty of an event 0, 1.15, 0.03, -0.43, 1, 0.34
October 7, 2014 by Sarah
Stats
How do you find SSA, SSB & SSE??
November 18, 2014 by Morgan
stats
A sample has a mean of M = 73 and a standard deviation of s = 5
June 18, 2015 by dj
stats
how does one make a residual plot please?
November 16, 2007 by Anonymous
Stats
What are variance and covariance? What do they mean? What are their differences/relationship?
September 22, 2008 by Blake
geography
what are some tricks that will; help me remember the stats and the capitles
February 23, 2010 by briana
stats
Let z be a random variable with a standard normal distribution, P(-1.9<=z<=2.1)
July 24, 2010 by Clay
stats
Mean is 60 and standard deviation is 7. what percentage is 73 and below.
October 12, 2010 by Christine
Stats
If a confidence interval for the mean is 24.8 < ě < 29.6. What is the margin of error?
November 1, 2010 by Erica
maths stats
Please unscramlbe aaeebllnn which is a name. Thanks
May 5, 2011 by orla
Stats
what is the range of data with a total of 1000 values?
September 4, 2011 by Ann james
bussi./stats
what is the probability that a loaf of bread is between 22.75 and 23.00?
September 6, 2011 by tim
STATS
If Ho: μ = 50 and Ha: mu. = 59 and s = 10 what is αlpha if x exceeds 56for n=25
March 15, 2013 by KP
stats 250
A z-score of z = +3.00 indicates a location that is the following
June 6, 2013 by Anonymous
stats
what is the lower and upper limit for -645.321?
February 6, 2015 by gisela
Stats
Explain what it means to have a deviation score of -18
September 8, 2015 by Matt
stats
how do you construct a relative frequency table based on an ogive?
September 6, 2008 by arielle
stats
How would r change if researchers had recordered the proportions?
February 6, 2009 by kate
stats
if you know the mean and standard deviation, how do you determine the proportion that will and will not be included
November 4, 2012 by mary
stats
what is the distinction between repeated measurements and matched items?
November 27, 2012 by jason
STATS
What percentage of the area of the standard normal distribution is between z = -1.00 and z = +1.00? How do you know this?
October 11, 2014 by Hernandez
health/stats
can cholesterol follow a normal distribution exactly? why or why not?
October 17, 2014 by rebekah
Math/Statistics
does anyone know anyting about standard deviation... I have no idea what it is! http://en.wikipedia.org/wiki/Standard_deviation http://www.robertniles.com/stats/stdev.shtml
August 5, 2007 by Kaz
stats
x and y independent exponential r.v with respctive parameters 2 and 3. find the cdf and density of z = x/y.
November 8, 2006 by ripa
stats
find standard deviation for this problem daily precip- frequency-31,1,0,2,0,1 0.oo-0.49 0.50-0.99 1.00-1.49 1.50-1.99 2.00-2.29 2.50-2.99
May 8, 2008 by ebony
Stats
If X is a normal random variable with mean 60, and a standard deviation of 2, find: A: P (X<58) B: P(X>63) C: P(57<X<64)
March 26, 2009 by April
psy Stats.
how do you Calculate percentage when given standard deviation and mean
May 10, 2011 by Kailey
Stats
Can you tell me what ∑xy means? Like is in the sum of all x times y or what?
July 10, 2011 by Carol
Stats
Find the mean and standard deviation of the following probability distribution: x 1 2 3 P(x) 0.3 0.5 0.2
October 25, 2011 by Jen
stats
if income in the us was normally distributed, what z-score would represent the top 1%?
October 31, 2011 by stacey
stats
using a calculator need p value on a r tailed test if the n=9 and the t is 3.204
December 2, 2011 by James
stats & prob.
Given the values of n=5, p= 0.15, and (1-p)=0.85, calculate the standard deviation of the data.
November 27, 2012 by Georgia
stats
Given a standard normal variable. What is the probability that z lies between 1.48 and 2.13
April 2, 2013 by chris
stats
What does the best estimate mean after doing the liner regression equation
May 6, 2013 by Anonymous
Stats help
What does the best estimate mean after doing the liner regression equation
May 7, 2013 by Anonymous
stats
how to find the are for a standard normal curve that cuts off the upper 1.1%
November 17, 2013 by Anonymous
Stats
Need help finding the area under the normal curve between z = -1.0 and z = -2.0?
April 16, 2014 by Betty
stats
what is the probability that a 6 digit phone number contains at least one 8?
July 19, 2014 by michelle
AP Stats
transformation necessary to linearize the data. Be sure to convince me that you understand.
August 10, 2014 by Hep
Stats
What is the best measure to describe the total variability in a population or sample?
March 11, 2015 by Dany
statistics
what does this mean i have to do this in a spreadsheet Descriptive stats for each numeric variable
September 13, 2015 by Anoymous
Stats
Where can I find fuel efficient car sales statistic for recent years?
October 25, 2007 by eason
college Stats
If Y has a geometric distribution with success probability p, show that P(Y= an odd integer)= p/(1-(q^2))
October 20, 2009 by james
stats help!
What 2 numbers is the emperical rule of 95% between if the P of no shows is .10 and the average party size is 10
January 13, 2011 by Lauren
stats
What are the steps to solve P(x>k)=.025 when the mean is 500, standard deviation is 100
March 13, 2011 by Lillie
stats
Find the probability that the company will meet its goal on a particular 50 miles of line
October 17, 2011 by Anonymous
What data and stats do you need to know in order to identify profitable nations in which to export.
October 5, 2012 by Lisa
Stats.
Find the standard normal area for each of the following, showing your reasoning clearly. a. P(1.22<z<2.15) b. P(2.00<z<3.00) c. P(-2.00<z<2.00) d. p(z=0.50)
April 11, 2014 by Kelly
stats
If a measured distribution has an average of 65 and a variance of 16 and the standard deviation of the random error is 4
September 9, 2014 by sas
stats
(a) Suppose n = 26 and p = 0.30. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)
October 6, 2014 by neyrod
stats
7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
February 15, 2015 by Anonymous
Stats
The heights of women in the US approximately follow a bell shaped curve. What do you think that means?
November 11, 2009 by Sam
STATS
A student's grade on an examination was transformed to a z value of 0.67. Therefore, we know that she scored approximately in the top
December 8, 2009 by Anonymous
Stats
Six letters are picked. Find the chance that they can be arranged to form the word RANDOM.
April 15, 2014 by Mario007
stats
what are the mean and standard error when using central limit theorem for dichotomous variable
August 28, 2014 by mary
stats
The probability that an observation, following a normal distribution, will lie within μ ± 1.3σ is a. 9.5 % b. 19.0% c. 40.3% d. 50.0% e. 80.6%
May 11, 2015 by meryy
Stats
Find the score that separates the top 75% from the bottom 25% if the Mean is 100 and the Standard deviation is 15
June 11, 2015 by cindy
Stats
Why is it important to know the possible errors we might make when rejecting or failing to reject the null hypothesis?
May 20, 2008 by Christine
FST (functions, stats, and trigonometry)
what is the equation for the axis of symmetry for the function f(x)=IxI?(absolute value)
October 7, 2009 by Jared
Stats
A testing service has 1000 raw scores. It wants to transform the distribution so that the mean =10 and the standard deviation = 1. To do so ?
February 18, 2010 by Z
Algebra 2: Prob and Stats
Find the sample size that produces the margin of error +-4.0% a) 325 b) 25 c) 16 d) 625
May 24, 2010 by Skye
Stats
In a distribution of scores with a mean of 234 and a standard deviation of 47, what is in the 43rd percentile ? and what formula do i use ?
February 26, 2011 by Ruben
Stats
In a distribution of scores with a mean of 234 and a standard deviation of 47, what is in the 43rd percentile ? and what formula do i use ?
February 26, 2011 by Ruben
stats
women age 20-29 is 36ml staandard deviation 7ml. What is probability greater than 45ml and between 30 and 62 ml
August 27, 2011 by anna
Stats
Five measurements are taken and are 59.5", 61.33", 60.55", 61.00", 59.8". Find the likely size of the chance error in a single measurement.
July 17, 2012 by Anonymous
STATS
The following confidence interval is obtained for a population proportion, p:0.556 < p < 0.584. What is the point estimate, p?
September 30, 2012 by 1234
stats
1. Find the sample variance and standard deviation 6, 53,11,49,35,26,28,28,31,31 s squared = (round to the nearest hundredth as needed)
October 2, 2012 by pat
stats
if x is a normally distributed random variable with amean of 8.00 if the proability for x is less than 9.54 is 0.67, then find the standard deviation of x
October 21, 2012 by tamara
Stats
Suppose X is a uniform random variable with C = 20 and d = 90. Find the probability that a randomly selected observation is between 23 and 85.
November 17, 2012 by Marie
stats
X is normally dsitributed with mean 250 and standard deviation 40. what value of X does only top 15% exceed?
March 24, 2014 by Anonymous
Stats
for a sample of n=20 how large a Pearson correlation is necessary to be statistically significant for a two tailed test
December 4, 2014 by Ivan
stats
Assume that aset of test scores is normally distributed with a mean of 100 and a standard deviation of 20 use the 68-95-99?
July 4, 2010 by Anonymous
buss/stats
what is it called when all the items in the population have the same chance of being selected for the sample?
October 8, 2011 by Tim
Stats
a die is rolled three times. find the chance that not all the rolls show 3 or more spots
December 4, 2011 by Fay
stats
. Find the mean and standard deviation of the following probability distribution: x 1 2 3 P(x) 0.4 0.25 0.35 Please show all of your work.
October 7, 2012 by Anonymous
Stats
If X and Y are uncorrelated in the population, the expected value of the estimated linear regression coefficient (slope) is 0. True False
November 5, 2012 by Mickie
stats
if i have a sample size of 200, 16% are poor, what is my mean and standard deviation. my answers do not make any sense
January 1, 2013 by gary
stats
explain and give example for each of the following types of variables: (a) equal-interval, (b) rank-over, and (c) nominal
September 2, 2013 by lucy
stats
If the effect size of a hypothesized phenomenon is .8, then how many subjects are needed to achieve .5 power? alpha = .05; two-tailed
April 23, 2015 by Anonymous
college stats
Again sorry for all the questions. please help, I am not good at stats and find these questions very difficult. I would appreciate an explanation if possible. 1.What is the sample average? 2.What is the SD of the box? 3.What is the number of draws from the box (the sample size...
July 12, 2013 by penelope
Prob and Stats
In a multiple regression with 5 predictors in a sample of 56 U.S. cities, what would be the critical value for an F test of overall significance at a= .05? A. 2.45 B. 2.37 C. 2.40 D. 2.56
Stats
The scores on a mathematics exam have a mean of 65 and a standard deviation of 7. Find the x-value that corresponds to the z-score 1.645.
April 21, 2010 by Isa
stats
three children in a class of 16 are being given awards. how many possible ways are there to pick winners if each prize is different.
September 25, 2010 by cb
What type of data is the projected return on an investment? Qualitative , continous, attribute,discrete or none of the above
August 31, 2011 by Tim
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Sylvan Learning Center Math help and tutoring, tailored to meet academic goals and needs. Sylvan's teachers create custom learning plans based on assessment, with ongoing evaluations and program updates to continue challenging students. Search by geography to find a center ...more>> Symbolab - EqsQuest This semantic search engine for math and science considers the contextual meaning of equations and expressions. Query a mix of text and symbols with Symbolab's pad, which toggles among Greek letters, operators, differentials, and "accents" (vectors, hats, ...more>> SymmeToy - Hank Hufnagel A Windows program for creating paint patterns, symmetry roses, tessellated art, and symmetrically decorated 3D polyhedron models. The program is easy to learn, fun to use and includes extensive online help. Appropriate for all ages, but especially for ...more>> Symmetric Patterns at the Alhambra, with Math Problems - Susan Addington, David Marshall The Alhambra is a walled city and fortress in Granada, Spain, built during the last Islamic sultanate on the Iberian peninsula, the Nasrid Dynasty (1238-1492). The palace is lavishly decorated with stone and wood carvings and tile patterns on most of ...more>> Symmetrier, monstre & tesselationer med The Geometer's Sketchpad - Allan Bergmann Jensen, Kalundborg, Denmark GSP Symmetries, Patterns and Tessellations. Contains a gallery of wallpaper designs constructed using the Geometer's Sketchpad with JavaSketchpad versions to explore, an illustrated paper on "The Math About Symmetries" (in Danish), a GSP guide, a links ...more>> Symmetries and the Millennium (Math Chat) - Frank Morgan; Christian Science Monitor What are the symmetries of a cube in space? What about a hypercube in 4- dimensional space? ...more>> Symmetry and Group Theory (The Geometry Junkyard) - David Eppstein, Theory Group, ICS, UC Irvine Computational and recreational geometry pointers to sites with information, problems, and lessons having to do with symmetry and group theory (tessellations, kaleidoscopes, origami, etc.). ...more>> Symmetry and Orbifolds (Geometry and the Imagination) - Conway, Doyle, Gilman, Thurston; The Geometry Center Given a symmetric pattern, what happens when you identify equivalent points? You get an object with interesting topological and geometrical properties, called an orbifold... hearts, paper dolls, wave patterns, quotient orbifolds, Moebius bands. ...more>> Symmetry and Pattern: The Art of Oriental Carpets - The Textile Museum/Math Forum In this online exhibit, the study of symmetry is used to analyze patterns in Oriental carpets. A joint project of The Textile Museum and The Math Forum. ...more>> Symmetry - Rick Engel A lesson on the symmetry of the Platonic and Archimedean solids using Polymorf, a math manipulative product. Demonstrates inscription, truncation, sectioning, and space filling with polyhedra. ...more>> Systematic Mathematics - Paul Ziegler This video-based home school math curriculum promises to "repair [the] essential foundation for your more experienced student, or create a solid math foundation for younger students." Each module consists of dozens of short lessons, delivered via DVD, ...more>> System Dynamics in Education Project (SDEP) System dynamics is a method for studying the world around us. It deals with understanding how complex systems change over time. Internal feedback loops within the structure of the system influence the entire system behavior. Math materials are available ...more>> syzygy-matrix: solutions for small problems in the classroom - Jeff LeMieux Generate cryptograms, division worksheets, fraction worksheets, and proportion problems. Select the length of dividend, denominators specified by the Washington State Standards for Grade 7, number of problems, and more. The worksheet maker prints solutions. ...more>> Syzygy Shareware - Thomas C. Bretl Freeware to "explore, see in new ways, discover patterns, and solve problems." Bretl offers simulations and visualizations for investigating probability, calculus, Mandelbrot Sets, and Sierpinski's Triangle and similar fractals by recursion; GeoFit (create ...more>> T3 Conferences - Texas Instruments, Inc. The T³ program hosts the annual T³ International Conference each spring and two-day Regional Conferences throughout the year. T³ Virtual Conferences contain video recorded sessions from T³ events. T³ conferences are appropriate for K-college, mathematics ...more>> Tables of Roman Measurement - Mike Neill Tables of Roman measurement on this site (Calender, Capacity, Length, Money, Surface, Weight) are taken from A Smaller Latin-English Dictionary, by William Smith LLD, a Classical Examiner in the University of London, John Murray, Albemarle Street, 1865. ...more>> Table-Top Earthquakes - John C. Lahr A demonstration of seismology for middle-school teachers and students that can be used to augment lessons in earth science (faulting, elastic rebound, plate motions), physics (forms of energy, elasticity, friction, magnetism, waves), math (graphing, logarithms, ...more>> TagMath - Cody Taggart Blog by a Wisconsinite now teaching at the United Nations International School in Hanoi (Vietnam). Weekly updates, 48-hour challenges, and other posts addressed to Taggart's students date back to February, 2012; and have included "Dublox," "Reviewing ...more>> Take a Little Piece of PI with You Wherever You Go - Adam Zampino Proudly display PI to 1000 decimal places in an attractive design, in various types of apparel and housewares (mugs, clocks, etc.). PI has fascinated people for centuries, and now you can have your own piece of PI to enjoy. This makes a great gift for ...more>> Taking 20 Percent Off - NASA Lewis Learning Technologies Project If a \$900 computer is on sale for 20 percent off, what is the sale price? A detailed answer is provided. From NASA's 9th Grade Math Proficiency Test. ...more>> Taking America's Measure - National Institute of Standards and Technology (NIST) Resources on this site for children include: Metric Pyramid (a pattern that can be printed, cut out, folded, and taped in the shape of a pyramid); Metric Fast Facts for Sports; Metric in the Kitchen; All You Will Need to Know About Metric; Metric Conversion ...more>> Taking a Workshop or Course - Annenberg Media Workshops and courses appropriate for preservice and inservice teachers, administrators, reformers, and policymakers. Supplemental video programs called teaching practices libraries are also available--they give a "fly-on-the-wall" view of real teachers ...more>> Talking Math Ed with the Males - Josh and Lorraine Males This blog "dedicated to all things math ed" includes lesson ideas, policies, curriculum, and research. Posts "for people to have a conversation about mathematics education," which date back to August, 2011, have included "CCSS from the Curriculum Writers ...more>> Tangoes - Rex Games, Inc. Based on the ancient Chinese Tangram puzzle, Tangoes is a learning tool that combines artistic and mathematical elements to enhance visual perception ability, develop problem solving skills, and encourage creative thinking capacity and teamwork. Tangoes ...more>> Tangram Page - S. T. Han All about tangrams, an ancient Oriental toy that has 7 pieces: 5 triangles in different sizes, 1 square, and 1 parallelogram. The object of the game is to form a given shape using all 7 pieces. Introduction, Puzzle of the Month, PC/Mac Software, Articles ...more>> Tangrams: History, Puzzles, Make It, and Links - Randy Crawford Follow the "history" link to learn about tangrams through the ages; click on "puzzles" for over 100 silhouettes to reconstruct, with solutions. "Make it" provides instructions for building your own tangram set. With references to more sites, software, ...more>> Tangrams - J. L. Read; Enchanted Mind A tangram applet that allows you to rearrange the seven forms to exactly reproduce the given image. To move the pieces, drag them with the mouse. To rotate pieces, click with the right mouse button or (on the Mac with a one-button mouse) click while holding ...more>> Tangrams - Tom Scavo Tangrams, a puzzle that helps develop spatial-visualization skills, may also be used to introduce or reinforce geometric concepts such as congruence, similarity, symmetry, etc. This unit for grades 4 through 6 uses tangrams to compute the area of polygons ...more>> Taniyama-Shimura Conjecture Proved (Math Chat) - Frank Morgan, MAA Online A proof of the full Taniyama-Shimura conjecture, partly included in Wiles's 1994 proof of Fermat's Last Theorem, was announced last week [June, 1999] at a conference in Park City, Utah, by Christophe Breuil, Brian Conrad, Fred Diamond, and Richard Taylor, ...more>> Tanzzle™ - Franco Cocchini Software for exploring tangrams. View published patterns and/or submit one of your own. Request a free license for schools and educational programs. Free 30-day trial; free viewer download. Choose from four different sets of pieces: Tangram (traditional ...more>> Tap Into Teen Minds - Kyle Pearce "Transforming mathematics education in an iPad paperless classroom." Posts, which date back to September, 2011, have included "Apple TV In The Classroom – The New Smart Board," "OAME Tap Into Teen Minds Educational Technology Presentation," "Problem Solving ...more>> TareasPlus Videos de lecciones a frente de una pizarra digital, narrado en español. Variando en duración, los asuntos de matemática incluyen aritmética, álgebra, trigonometría, precálculo, cálculo diferencial, ...more>> The Tau Manifesto - Michael Hartl "This manifesto is dedicated to one of the most important numbers in mathematics, perhaps the most important: the circle constant relating the circumference of a circle to its linear dimension... The traditional choice of circle constant is π -- but, ...more>> TCWORLD: Online Resource for Technology Coordinators A monthly publication of articles on exceptional uses of technology in schools, software reviews, technology policies, computer set-up tips and techniques, security methods, opinions on integration and access, purchasing tips, a newsletter, and more. ...more>> teachade - AP Ed Ventures An online community for teachers to discuss issues relevant to teaching, share and author curriculum, access resources that support their professional development, and have an impact on commercial providers of educational services and products. Draft ...more>> Teach California Charter Schools A consortium of charter schools and networks throughout California, organized to recruit and select outstanding teachers for positions in the Bay Area, Los Angeles, and the Sacramento and Central Valley areas of California. ...more>> TeachEngineering Digital Collection The TeachEngineering digital library provides teacher-tested, standards-based engineering content for K-12 teachers to use in science and mathematics classrooms. Engineering lessons connect real-world experiences with curricular content already taught ...more>> Teacher2Teacher - Math Forum A question-and-answer service for teachers and parents who have questions about teaching math. Staffed by a panel of teaching professionals, Presidential Awardees for Excellence in Mathematics Teaching and Teacher2Teacher Associates, T2T offers an archive ...more>> Teacher2Teacher (T2T) FAQ - Math Forum Answers to frequently asked questions from the archives of the Math Forum's question and answer service for teachers, with links to T2T discussions, Ask Dr. Math and the Ask Dr. Math FAQ, and relevant resources on the Web. ...more>> The Teacher as Facilitator - Classroom Compass The spring 1994 issue of Classroom Compass, Working for Reform, looks at the idea of teacher as facilitator - a basic part of the reformed classroom. Two classroom activities (one for young children, Mud Slide, the other for later grades, Stream Works) ...more>> Teacher Created Materials (TCM) A publisher of supplemental resource materials for early childhood, elementary, and middle school teachers. Lists of products, including TechWorks, a program for Grades K-8 for integrating technology skills into the curriculum; and TechKNOWLEDGEy books ...more>> Teacher Education and Professional Development - Mathematics Education at the Math Forum New ideas and pedagogy: articles, conference sessions, centers, and courses on the Web for pre-service education, inservice education, and the college and university level. ...more>> Teacher Education Materials Project (TE-MAT) - Horizon Research, Inc. (HRI) A site for K-12 professional development providers, with a conceptual framework "Designing Effective Professional Development" (short essays on: Understanding the Needs of the Target Audience; Deciding on the Purposes of the Professional Development; ...more>> Teacher Exchange - the Math Forum Web units, lessons, and activities organized by NCTM Standards and grade band level. Created by Math Forum staff members, educators, and collaborations between the two. Teachers may contribute their own lessons and ideas as well. ...more>> Teacher Learning That Supports Student Learning - Linda Darling-Hammond "What do teachers need to know to teach all students according to today's standards? What kinds of preservice training and ongoing professional development will make teacher success more likely?" Article from Educational Leadership, February 1998, Volume ...more>> TeacherNet - Highlights for Children A network of K-12 educational materials vendors and teachers: Teachers' Lounge (with email list and online bulletin board); Teaching K-8 (a monthly publication with articles available online); Zaner-Bloser (handwriting, spelling, reading and vocabulary, ...more>> Teacher Preparation and Professional Development, 2000 - Basmat Parsad, Laurie Lewis, and Elizabeth Farris Executive summary of the NCES report. The full report is also available as a PDF file. ...more>> Teacher Quality and Student Achievement: A Review of State Policy Evidence - Linda Darling-Hammond From Education Policy Analysis Archives, Vol. 8, No. 1, January, 2000. Abstract: Using data from a 50-state survey of policies, state case study analyses, the 1993-94 Schools and Staffing Surveys (SASS), and the National Assessment of Educational Progress ...more>> Teachers Becoming Self-Directed Learners: A Work in Progress - Horwitz, Hallam; Eisenhower National Clearinghouse How the Annenberg Tuning Protocol is used by the Penn-Merck Collaborative for Science Education to foster a constructivist view of learning with reflective, internally motivated practitioners. The program involves a 13-month graduate course designed to ...more>> Teachers' Domain - WGBH Educational Foundation A digital library of multimedia resources for classroom use and independent study. 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http://mathforum.org/ | 3,311 | 15,465 | {"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} | 2.71875 | 3 | CC-MAIN-2017-17 | longest | en | 0.847773 |
https://news.tradimo.com/100-return-in-3-years/ | 1,719,270,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00199.warc.gz | 351,247,264 | 30,779 | # 100% return in 3 Years
0
You’ve probably read that “2% a day” or “10% a week” are realistic goals. That is simply false. In the long run, these returns are difficult to implement. 2% per day means 14,000% per year.
So why is 100% in 3 years realistic?
100% in a single year is 26%. So to double your money in 3 years, you have to make a return of 26% a year.
Now let’s calculate the average monthly return. The result is 2%. So to achieve a return of 26% per year, you need a quite achievable 2% per month. | 149 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.933861 |
http://www.princeton.edu/%7Eachaney/tmve/wiki100k/docs/Maunder_Minimum.html | 1,469,811,412,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00295-ip-10-185-27-174.ec2.internal.warc.gz | 668,607,362 | 4,849 | # Maunder Minimum
related topics {math, energy, light} {rate, high, increase} {work, book, publish} {day, year, event} {area, part, region} {borough, population, unit_pref}
The Maunder Minimum (also known as the prolonged sunspot minimum) is the name used for the period roughly spanning 1645 to 1715 when sunspots became exceedingly rare, as noted by solar observers of the time.
The concept became notable after John A. Eddy published a landmark 1976 paper in Science titled "The Maunder Minimum".[1] Astronomers before Eddy had also named the period after the solar astronomer Edward W. Maunder (1851-1928) who studied how sunspot latitudes changed with time.[2] The periods he examined included the second half of the 17th century. Edward Maunder published two papers in 1890 and 1894, and he cited earlier papers written by Gustav Spörer.
Like the Dalton Minimum and Spörer Minimum, the Maunder Minimum coincided with a period of lower-than-average global temperatures.
During one 30-year period within the Maunder Minimum, astronomers observed only about 50 sunspots, as opposed to a more typical 40,000-50,000 spots in modern times.[citation needed]
## Contents
### Sunspot observations
The Maunder Minimum occurred between 1645 and 1715 when very few sunspots were observed. This was not due to a lack of observations; during the 17th century, Giovanni Domenico Cassini carried out a systematic program of solar observations at the Observatoire de Paris, thanks to the astronomers Jean Picard and Philippe de La Hire. Johannes Hevelius also performed observations on his own. The total numbers of sunspots (but not Wolf numbers) in different years were as follows:
During the Maunder Minimum enough sunspots were sighted so that 11-year cycles could be extrapolated from the count. The maxima occurred in 1676, 1684, 1695, 1705 and 1716.
The sunspot activity was then concentrated in the southern hemisphere of the Sun, except for the last cycle when the sunspots appeared in the northern hemisphere, too.
According to Spörer's law, at the start of a cycle, spots appear at ever lower latitudes until they average at about lat. 15° at solar maximum. The average then continues to drift lower to about 7° and after that, while spots of the old cycle fade, new cycle spots start appearing again at high latitudes.
The visibility of these spots is also affected by the velocity of the sun's rotation at various latitudes:
Visibility is somewhat affected by observations being done from the ecliptic. The ecliptic is inclined 7° from the plane of the Sun's equator (latitude 0°).
### Little Ice Age
The Maunder Minimum coincided with the middle - and coldest part - of the Little Ice Age, during which Europe and North America were subjected to bitterly cold winters. Whether there is a causal connection between low sunspot activity and cold winters has not been proven, however lower earth temperatures have been observed during low sunspot activity.[3] The winter of 1708-09 was extremely cold.[4]
Full article ▸
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https://espanol.libretexts.org/Educacion_Basica/Libro%3A_Fisica_-_De_Stargazers_a_Starships_(CK-12)/18%3A_Parallax/18.01%3A_Pre-Trigonometr%C3%ADa | 1,722,933,330,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00159.warc.gz | 184,422,366 | 31,516 | Saltar al contenido principal
# 18.1: Pre-Trigonometría
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La página en http://www.phy6.org/stargaze/Strig1.htm describe el problema básico de la trigonometría (dibujo anterior): encontrar la distancia a algún punto lejano C, dadas las direcciones en las que C aparece desde los dos extremos de una línea base medida AB. Este problema se vuelve algo más sencillo si:
1. La línea base es perpendicular a la línea desde su centro hasta el objeto, por lo que$$\Delta ABC$$ es simétrica. Vamos a denotar sus lados iguales como$$AB=BC=r$$.
2. La longitud$$c$$ de la línea base$$AB$$ es mucho menor que$$r$$. Esto significa que el ángulo$$\alpha$$ entre$$AC$$ y$$BC$$ es pequeño; ese ángulo se conoce como el paralaje de$$C$$, visto desde$$AB$$.
3. No pedimos gran precisión, sino que estamos satisfechos con un valor aproximado de la distancia —digamos, dentro del 1%.
El método aquí presentado ya fue utilizado por los antiguos griegos hace más de 2000 años. Sabían que la longitud de un círculo de radio$$r$$ era$$2\pi r$$, donde$$\pi$$ (una notación moderna, no uno de los griegos, aunque$$\pi$$ sea parte de su alfabeto) representa un número un poco mayor que 3; aproximadamente
$$\pi\approx\,3.14159\ldots$$
Nota
El matemático griego Arquímedes derivó$$\pi$$ con una precisión de aproximadamente 4 cifras, aunque la expresó de manera diferente, ya que las fracciones decimales aparecieron en Europa solo unos 1000 años después.
En este caso (ver Figura anterior), podemos aproximarnos$$\Delta ABC$$ como una 'rebanada' de un círculo mucho más grande; en este caso, la longitud de la línea base es aproximadamente igual a la longitud del arco correspondiente:
$c\,\approx\,c′$
Hay 360 grados en un círculo y$$\alpha$$ grados en este arco en particular; dado que 360 grados corresponden a una circunferencia de longitud de arco ($$2\pi r$$), los$$\alpha$$ grados corresponderán a una longitud de arco de
$c′\,=\,\frac{\alpha}{360^o}\times 2\pi r$
Resolviendo$$r$$ y enchufando$$c\,\approx\,c′$$, encontramos
$r\,=\,\frac{360^o}{2\pi\alpha}\times c$
Hemos resuelto para$$r$$ en cuanto a$$c$$. Por ejemplo, si sabemos eso$$\alpha\approx 6^o$$ (veremos por qué esto es relevante más adelante),$$2\pi\alpha\,=\,36^o$$ y obtenemos:
$r\,=\,10c′$
This page titled 18.1: Pre-Trigonometría is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. | 2,501 | 6,828 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.190198 |
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An anthropologist discovers an isolated tribe whose written alphabet contains only six letters (call the letters A, B, C, D, E, and F). The tribe has a taboo against using the same letter twice in the same word. It's never done. If each different sequence of letters constitutes a different word in the language, what is the maximum number of six-letter words the language can employ?
Please explain the method and solution.
Submitted: 11 years ago.
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Expert: jacket replied 11 years ago.
6*5*4*3*2*1 = 720 words.
The first letter can be any of the 6 letters, and 2nd letter then is any of the 5 left over letter, next any of the 4 left over letters, and so on, and so on. SO, that's how I came up with solution.
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triangle properties
Special O
er
60% o!
\$750
SCREEN WAREHOUSE
Only slightly damaged, these screens still present a
crystal clear image but with uneveness in the viewable
range to each side.
28 31
30
or the
sp
ecia
l offer
screens
Will the special offer screen still do the job?
The screen is suitable if the viewable range (r) reaches
all the way to the back—but you only have the sides
given on the specification in the ad.
25
28 31
30
r ?
you are here 4 193
the screens viewable distance, r?
Alti
tude
a
2
+ x
2
= 28
2
And this here
a
2
+ y
2
= 31
2
a new tool for your geometry toolbox
the screens viewable distance, r?
The Pythagorean Theorem
It’s the right answ
er, but i
t’s a
lso
a w
orld of pain…so hold tight f
or
a new t
ool for y
our t
oolbox!
The screen viewing area is a scalene triangle
This means that when you add an altitude, it doesn’t bisect
the base. Instead of creating two nice, neat congruent right
triangles, you get two different right triangles. And you don’t
know the base of either of them!
So w
e’d need to
sol
v
e this her
e
28
31
28
31
a a
All three sides
are different
lengths.
x y
30
The two right triangles still work with the Pythagorean
Theorem, and we know that x and y together make 30 (but
aren’t equal to each other) so we could figure this out with a set
of three simultaneous algebra equations. If that sounds bad,
it kinda is. Don’t go there. But in case you’re tempted, here’s
how it starts out:
Plus, x and y ar
en’t jus
t ha
lf of 30!
a
2
+ x
2
= 28
2
a
2
+ y
2
= 31
2
x + y = 30
(a
2
+ x
2
= 28
2
)
- (a
2
+ y
2
= 31
2
)
3 equations
DANGER
do not cross this line
by subtraction
x
2
- y
2
= 28
2
- 31
2
x + y = 30, so x = (30 - y) (30 - y)
2
- y
2
= 28
2
-31
2
by substitution
Now…expand this out....
194 Chapter 4
triangle properties
Wouldn’t it be dreamy if there was a
way to find the height of a scalene triangle
without simultaneous equations? But I know
it's just a fantasy....
you are here 4 195
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a. 1
b. 2
c. 3
d. 4
8. The variable (A) in the utility formula represents the:
a. investor’s return requirement.
b. investor’s aversion to risk.
c. certainty equivalent rate of the portfolio.
d. preference for one unit of return per four units of risk.
Use the following expectations on Stocks X and Y to answer questions 9 through 12 (round to
the nearest percent).
Bodie−Kane−Marcus: II. Portfolio Theory 5. Risk and Return: Past © The McGraw−Hill
Essentials of Investments, and Prologue Companies, 2003
Fifth Edition
161
5 Risk and Return: Past and Prologue
Bear Market Normal Market Bull Market
Probability 0.2 0.5 0.3
Stock X 20% 18% 50%
Stock Y 15% 20% 10%
9. What are the expected returns for Stocks X and Y?
Stock X Stock Y
a. 18% 5%
b. 18% 12%
c. 20% 11%
d. 20% 10%
10. What are the standard deviations of returns on Stocks X and Y?
Stock X Stock Y
a. 15% 26%
b. 20% 4%
c. 24% 13%
d. 28% 8%
11. Assume that of your \$10,000 portfolio, you invest \$9,000 in Stock X and \$1,000 in
Stock Y. What is the expected return on your portfolio?
a. 18%
b. 19%
c. 20%
d. 23%
12. Probabilities for three states of the economy, and probabilities for the returns on a
particular stock in each state are shown in the table below.
Probability of
Stock Performance
Probability of Stock in Given
State of Economy Economic State Performance Economic State
Good .3 Good .6
www.mhhe.com/bkm
Neutral .3
Poor .1
Neutral .5 Good .4
Neutral .3
Poor .3
Poor .2 Good .2
Neutral .3
Poor .5
Bodie−Kane−Marcus: II. Portfolio Theory 5. Risk and Return: Past © The McGraw−Hill
Essentials of Investments, and Prologue Companies, 2003
Fifth Edition
162 Part TWO Portfolio Theory
The probability that the economy will be neutral and the stock will experience poor
performance is
a. .06 c. .50
b. .15 d. .80
13. An analyst estimates that a stock has the following probabilities of return depending on
the state of the economy:
State of Economy Probability Return
Good .1 15%
Normal .6 13
Poor .3 7
The expected return of the stock is:
a. 7.8%
b. 11.4%
c. 11.7%
d. 13.0%
14. XYZ stock price and dividend history are as follows:
Year Beginning-of-Year Price Dividend Paid at Year-End
1999 \$100 \$4
2000 \$110 \$4
2001 \$ 90 \$4
2002 \$ 95 \$4
An investor buys three shares of XYZ at the beginning of 1999 buys another two shares
at the beginning of 2000, sells one share at the beginning of 2001, and sells all four
remaining shares at the beginning of 2002.
a. What are the arithmetic and geometric average time-weighted rates of return for the
investor?
b. What is the dollar-weighted rate of return. Hint: Carefully prepare a chart of cash
flows for the four dates corresponding to the turns of the year for January 1, 1999, to
January 1, 2002. If your calculator cannot calculate internal rate of return, you will
have to use trial and error.
15. a. Suppose you forecast that the standard deviation of the market return will be 20% in
the coming year. If the measure of risk aversion in equation 5.6 is A 4, what would
be a reasonable guess for the expected market risk premium?
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b. What value of A is consistent with a risk premium of 9%?
c. What will happen to the risk premium if investors become more risk tolerant?
annual HPR on the S&P 500 stock portfolio if the current risk-free interest rate is 5%?
17. What has been the historical average real rate of return on stocks, Treasury bonds, and
Treasury notes?
18. Consider a risky portfolio. The end-of-year cash flow derived from the portfolio will be
either \$50,000 or \$150,000, with equal probabilities of 0.5. The alternative riskless
investment in T-bills pays 5%.
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Essentials of Investments, and Prologue Companies, 2003
Fifth Edition
163
5 Risk and Return: Past and Prologue
a. If you require a risk premium of 10%, how much will you be willing to pay for the
portfolio?
b. Suppose the portfolio can be purchased for the amount you found in (a). What will
the expected rate of return on the portfolio be?
c. Now suppose you require a risk premium of 15%. What is the price you will be
willing to pay now?
between the required risk premium on a portfolio and the price at which the portfolio
will sell?
For problems 19–23, assume that you manage a risky portfolio with an expected rate of re-
turn of 17% and a standard deviation of 27%. The T-bill rate is 7%.
19. a. Your client chooses to invest 70% of a portfolio in your fund and 30% in a T-bill
money market fund. What is the expected return and standard deviation of your
client’s portfolio?
b. Suppose your risky portfolio includes the following investments in the given
proportions:
Stock A 27%
Stock B 33%
Stock C 40%
What are the investment proportions of your client’s overall portfolio, including the
position in T-bills?
c. What is the reward-to-variability ratio (S) of your risky portfolio and your client’s
overall portfolio?
d. Draw the CAL of your portfolio on an expected return/standard deviation diagram.
What is the slope of the CAL? Show the position of your client on your fund’s CAL.
20. Suppose the same client in problem 19 decides to invest in your risky portfolio a
proportion (y) of his total investment budget so that his overall portfolio will have an
expected rate of return of 15%.
a. What is the proportion y?
b. What are your client’s investment proportions in your three stocks and the T-bill
fund?
c. What is the standard deviation of the rate of return on your client’s portfolio?
21. Suppose the same client in problem 19 prefers to invest in your portfolio a proportion
(y) that maximizes the expected return on the overall portfolio subject to the constraint
that the overall portfolio’s standard deviation will not exceed 20%.
a. What is the investment proportion, y?
b. What is the expected rate of return on the overall portfolio?
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22. You estimate that a passive portfolio invested to mimic the S&P 500 stock index yields
an expected rate of return of 13% with a standard deviation of 25%. Draw the CML and
your fund’s CAL on an expected return/standard deviation diagram.
a. What is the slope of the CML?
b. Characterize in one short paragraph the advantage of your fund over the passive
fund.
23. Your client (see problem 19) wonders whether to switch the 70% that is invested in your
fund to the passive portfolio.
Bodie−Kane−Marcus: II. Portfolio Theory 5. Risk and Return: Past © The McGraw−Hill
Essentials of Investments, and Prologue Companies, 2003
Fifth Edition
164 Part TWO Portfolio Theory
b. Show your client the maximum fee you could charge (as a percent of the
investment in your fund deducted at the end of the year) that would still leave him
at least as well off investing in your fund as in the passive one. (Hint: The fee
will lower the slope of your client’s CAL by reducing the expected return net of
the fee.)
24. What do you think would happen to the expected return on stocks if investors perceived
an increase in the volatility of stocks?
25. The change from a straight to a kinked capital allocation line is a result of the:
a. Reward-to-variability ratio increasing.
b. Borrowing rate exceeding the lending rate.
c. Investor’s risk tolerance decreasing.
d. Increase in the portfolio proportion of the risk-free asset.
26. You manage an equity fund with an expected risk premium of 10% and an expected
standard deviation of 14%. The rate on Treasury bills is 6%. Your client chooses to
invest \$60,000 of her portfolio in your equity fund and \$40,000 in a T-bill money
market fund. What is the expected return and standard deviation of return on your
client’s portfolio?
Expected Return Standard Deviation of Return
a. 8.4% 8.4%
b. 8.4 14.0
c. 12.0 8.4
d. 12.0 14.0
27. What is the reward-to-variability ratio for the equity fund in problem 26?
a. .71
b. 1.00
c. 1.19
d. 1.91
For problems 28–30, download Table 5.3: Rates of return, 1926–2001, from www.mhhe.com/
blkm.
28. Calculate the same subperiod means and standard deviations for small stocks as Table
5.5 of the text provides for large stocks.
a. Do small stocks provide better reward-to-variability ratios than large stocks?
b. Do small stocks show a similar declining trend in standard deviation as Table 5.5
documents for large stocks?
29. Convert the nominal returns on both large and small stocks to real rates. Reproduce
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Table 5.5 using real rates instead of excess returns. Compare the results to those of
Table 5.5.
30. Repeat problem 29 for small stocks and compare with the results for nominal rates.
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Essentials of Investments, and Prologue Companies, 2003
Fifth Edition
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5 Risk and Return: Past and Prologue
<< Ïðåäûäóùàÿ ñòð. 40(èç 193 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >> | 2,499 | 8,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-39 | longest | en | 0.757746 |
https://homeofengineers.com/suppose-the-lifespan-of-a-wombat-is-normally-distributed-you-believe-that-the-mean-is-greater-than-16-you-collect-a-sample-of-size-n-43-wombats/ | 1,702,197,457,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00778.warc.gz | 335,473,462 | 19,048 | Suppose the lifespan of a wombat is normally distributed. You believe that the mean is greater than 16. You collect a sample of size n = 43 wombats and compute the average to be 17.1 years. It is known that the population standard deviation is σ = 6 years. What is the P-value and what decision should he make at a 5% significance level? | 82 | 337 | {"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-2023-50 | longest | en | 0.9692 |
https://encyclopediaofmath.org/index.php?title=Semi-group_of_holomorphic_mappings&direction=prev&oldid=51280 | 1,627,897,555,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00557.warc.gz | 253,452,944 | 13,316 | # Semi-group of holomorphic mappings
Non-linear semi-group theory is not only of intrinsic interest, but is also important in the study of evolution problems (cf. also Evolution equation). In recent years (as of 2000) many developments have occurred, in particular in the area of non-expansive semi-groups in Banach spaces. As a rule, such semi-groups are generated by accretive operators (cf. also Accretive mapping) and can be viewed as non-linear analogues of the classical linear contraction semi-groups (cf. also Contraction semi-group). Another class of non-linear semi-groups consists of the semi-groups generated by holomorphic mappings. Such semi-groups appear in several diverse fields, including, for example, the theory of Markov stochastic branching processes, Krein spaces, the geometry of complex Banach spaces, control theory, and optimization. These semi-groups can be considered natural non-linear analogues of semi-groups generated by bounded linear operators (cf. also Semi-group of operators). These two distinct classes of non-linear semi-groups are related by the fact that holomorphic self-mappings are non-expansive with respect to Schwarz–Pick pseudo-metrics (see, for example, [a8], [a7], [a6]).
Recall that a function $h$ defined on a domain (open connected subset) $\textbf{D}$ in a complex Banach space $X$ and with values in a complex Banach space is said to be holomorphic in $\textbf{D}$ if for each $x\in\textbf{D}$ the Fréchet derivative of $h$ at $x$ (denoted by $Dh(x)$ or $h'(x)$) exists as a bounded complex-linear mapping of $X$ into the Banach space containing the values of $h$. (Cf. also Banach space of analytic functions with infinite-dimensional domains.)
If $\textbf{D}$ and $\Omega$ are domains in complex Banach spaces $X$ and $Y$, respectively, then the set of holomorphic mappings from $\textbf{D}$ into $\Omega$ is denoted by $\text{Hol}(\textbf{D},\Omega)$. The notation $\text{Hol}(\textbf{D})$ is used to denote the set $\text{Hol}(\textbf{D},\textbf{D})$ of holomorphic self-mappings of $\textbf{D}$.
A family $\{S(t)\subset\text{Hol}(\textbf{D}\}$, where $t\in(0,T)$, $T>0$, is called a (one-parameter) continuous semi-group if
$$S(s+t)=S(s)\circ S(t),s,t,s+t\in(0,T),$$
and
$$\lim_{t\to 0^{+}}S(t)(x)=x,x\in D,$$
where the limit is taken with respect to the strong topology of $X$.
## Differentiability of semi-groups with respect to the parameter.
Let $\{S(t)\}$, $t\in(0,T)$, be a continuous semi-group defined on $\textbf{D}$. If the strong limit
$$g(x)=\lim_{t\to 0^{+}}\frac{1}{t}(x-S(t)(x))$$
exists for each $x\in \textbf{D}$, then $g\in\text{Hol}(\textbf{D},X)$ is called the (infinitesimal) generator of the semi-group $\{S(t)\}$. In this case the semi-group $\{S(t)\}$, $t\in(0,T)$, is said to be differentiable (or generated).
For the finite-dimensional case, M. Abate proved in [a2] that each continuous semi-group of holomorphic mappings is everywhere differentiable with respect to its parameter, i.e., it is generated by a holomorphic mapping. In addition, he established a criterion for a holomorphic mapping to be a generator of a one-parameter semi-group. Earlier, for the one-dimensional case, similar facts were presented by E. Berkson and H. Porta in their study [a4] of linear $C_0$-semi-groups of composition operators on Hardy spaces. E. Vesentini investigated semi-groups of fractional-linear transformations that are isometries with respect to the infinitesimal hyperbolic metric on the unit ball of a Banach space [a18]. He used this approach to study several important problems in the theory of linear operators on indefinite metric spaces. Note that, generally speaking, such semi-groups are not everywhere differentiable in the infinite-dimensional case.
In fact, it can be shown (see for example, [a14]), that a continuous semi-group $S(t)$ of holomorphic self-mappings of a domain $\textbf{D}$ in $X$ is generated if and only if the convergence in (2) is locally uniform on $\textbf{D}$ (cf. also Uniform convergence).
Moreover, if $\textbf{D}$ is hyperbolic (in particular, bounded; cf. also Hyperbolic metric), then $S(t)$ can be continuously extended to all of $\textbf{R}^+=[0,\infty)$ as the solution of the Cauchy problem
$$\begin{cases}\frac{\partial u(t,x)}{\partial t}+g(u(t,x))=0,\\u(0,x)=x,\end{cases}$$
where $x\in \textbf{D}$ and $t\in \textbf{R}^+$, i.e.,
$$u(t,x)=S(t)x,x\in \textbf{D},t\in[0,\infty)$$
(see, for example, [a12], [a13]). Thus, there is a one-to-one correspondence between locally uniformly continuous semi-groups and their generators. If $S(t)$ has a continuous extension to all of $\textbf{R}=(-\infty,\infty)$, then it is actually a one-parameter group of automorphisms of $\textbf{D}$.
## Exponential and product formulas.
A holomorphic vector field
$$T_g=g(x)\frac{\partial}{\partial x}$$
on a domain $\textbf{D}$ is determined by a holomorphic mapping $g\in\text{Hol}(\textbf{D},X)$ and can be regarded as a linear operator $T$ mapping $\text{Hol}(\textbf{D},X)$ into itself, where $T_gf\in\text{Hol}(\textbf{D},X)$ is defined by
$$(T_gf)(x)=Df(x)g(x),x\in\textbf{D}$$
The set of all holomorphic vector fields on $\textbf{D}$ is a Lie algebra under the commutator bracket
$$[T_g,T_h]=\bigg[g(x)\frac{\partial}{\partial x}, h(x)\frac{\partial}{\partial x}\bigg]:=\\=(Dg(x)h(x)-Dh(x)g(x))\frac{\partial}{\partial x}$$
(see, for example, [a9], [a15], [a6]).
Furthermore, each vector field (6) is locally integrable in the following sense: for each $x\in\textbf{D}$ there exist a neighbourhood $\Omega$ of $x$ and a $\delta>0$ such that the Cauchy problem (4) has a unique solution $\{u(t,x)\}\subset\textbf{D}$ defined on the set $\{|t|<\delta\}\times\Omega\subset\textbf{R}\times\textbf{D}$.
A holomorphic vector field $T_g$ defined by (6) and (7) is said to be (right) semi-complete (respectively, complete) on $\textbf{D}$ if the solution of the Cauchy problem (4) is well-defined on all of $\textbf{R}^+\times\textbf{D}$ (respectively, $\textbf{R}\times\textbf{D}$), where $\textbf{R}^+=[0,\infty)$ (respectively, $\textbf{R}=(-\infty,\infty)$).
Thus, if $\textbf{D}$ is hyperbolic, then $T_g$ is semi-complete (respectively, complete) if and only if $g$ is the generator of a one-parameter continuous semi-group (respectively, group).
On the other hand, if $\textbf{D}$ is bounded and $\widetilde{\text{Hol}}(\textbf{D},X)$ is the subspace of $\text{Hol}(\textbf{D},X)$ consisting of all $f\in\text{Hol}(\textbf{D},X)$ that are bounded on each ball strictly inside $\textbf{D}$, then a semi-group (group) $\{S(t)\}$, $t\in \textbf{R}^+$ (respectively, $t\in\textbf{R}$), induces a linear semi-group (group) $\{\mathcal{L}(t)\}$ of linear mappings $\mathcal{L}(t):\widetilde{\text{Hol}}(\textbf{D},X)\to\widetilde{\text{Hol}}(\textbf{D},X)$, defined by
$$(\mathcal{L}(t)f)(x):=f(S(t)x),$$
where $t\in\textbf{R}^+$ ($t\in\textbf{R}$) and $x\in\textbf{D}$.
This semi-group is called the semi-group of composition operators on $\widetilde{\text{Hol}}(\textbf{D},X)$. If $\{S(t)\}$, $t\in\textbf{R}^+$ ($t\in\textbf{R}$), is $T$-continuous, (that is, differentiable), then $g\in\widetilde{\text{Hol}}(\textbf{D},X)$, $\{\mathcal{L}(t)\}$, $t\in\textbf{R}^+$ ($t\in\textbf{R}$), is also differentiable and
$$\begin{cases}\frac{\partial\mathcal{L}(t)f}{\partial t}+T_g(\mathcal{L}(t)f)=0,\\\mathcal{L}(0)f=f,\end{cases}$$
for all $f\in\widetilde{\text{Hol}}(\textbf{D},X)$, where $g=-dS(t)/dt|_{t=0}$.
In other words, a holomorphic vector field $T_g$, defined by (6) and (7), and considered as a linear operator on $\widetilde{\text{Hol}}(\textbf{D},X)$, is the infinitesimal generator of the semi-group $\{\mathcal{L}(t)\}$. It is sometimes called the Lie generator. Thus, a holomorphic vector-field $T_g$ is semi-complete (respectively, complete) if and only if it is the Lie generator of a linear semi-group (respectively, group) of composition operators on $\widetilde{\text{Hol}}(\textbf{D},X)$. This follows from the observation that
$$\mathcal{L}(t)I_{\textbf{D}}=S(t)$$
and
$$T_gI_{\textbf{D}}=g,$$
where $I_D$ is the restriction of the identity operator to $\textbf{D}$.
Moreover, using the exponential formula representation for the linear semi-group,
$$\mathcal{L}(t)f=\sum^{\infty}_{k=0}\frac{(-1)^kt^k}{k!}T^k_gf=\text{exp}[-tT_g]f$$
(see, for example, [a19], [a9], [a12]), one also has
$$S(t)=\sum^{\infty}_{k=0}\frac{(-1)^kt^k}{k!}T^k_gI_{\textbf{D}}=\text{exp}[-tT_g]I_{\textbf{D}}.$$
So, a locally uniformly continuous semi-group of holomorphic self-mappings can be represented in exponential form by the holomorphic vector field induced by its generator.
Another exponential representation on a hyperbolic convex domain can be given by using the so-called non-linear resolvent of $g$.
More precisely, let $\textbf{D}$ be a bounded (or, more generally, hyperbolic) convex domain. Then it was shown in [a12] and [a13] that $g\in\widetilde{\text{Hol}}(\textbf{D},X)$ is a generator if and only if for each $r\geq0$ the mapping $(I+rg)^{-1}$ is a well-defined holomorphic self-mapping of $\textbf{D}$.
Furthermore, if $\{G(r)\}$, $r\geq0$, is any continuous family of holomorphic self-mappings of $\textbf{D}$ such that the limit
$$g(x)=\lim_{r\to0^{+}}\frac{1}{r}(x-G(r)x)$$
exists, then $g$ is a generator and the semi-group generated by $g$ can be defined by the product formula
$$S(t)=\lim_{n\to\infty}G^n\bigg(\frac{t}{n}\bigg).$$
In particular,
$$S(t)=\lim_{n\to\infty}\bigg(I+\frac{t}{n^g}\bigg)^{-n}$$
(exponential formula), where the limits in (17) and (18) are taken with respect to the locally uniform topology on $\text{Hol}(\textbf{D},X)$.
## Flow-invariance conditions.
Let $\textbf{D}$ be a convex subset of a Banach space $X$ and let $g:\overline{\textbf{D}}\to X$ be a continuous mapping on $\overline{\textbf{D}}$, the closure of $D$. Then the following tangency condition of flow invariance
$$\lim_{h\to0^{+}}\text{dist}\frac{(x-hg(x),\overline{\textbf{D}})}{h}=0,x\in\overline{\textbf{D}},$$
is a necessary condition for the solvability of the evolution equation (4). A result of R.H. Martin [a11] shows that if $g:\overline{\textbf{D}}\to X$ is a continuous accretive mapping on $\overline{\textbf{D}}$, then (18) is also sufficient for the existence of solutions to the Cauchy problems (4). These solutions yield a continuous semi-group of contraction mappings on $\overline{\textbf{D}}$.
For the class of holomorphic mappings, an analogue of Martin's theorem was given in [a3]; namely, if $g\in\text{Hol}(\textbf{D},X)$ has a uniformly continuous extension to $\overline{\textbf{D}}$, then it is a semi-complete vector field if and only if it satisfies the boundary flow invariance condition (18).
However, there are many examples of semi-complete vector fields that have no continuous extension to $\overline{\textbf{D}}$. In particular, if $F\in\text{Hol}(\textbf{D})$, then $g=I-F$ is semi-complete (see [a12]).
For absolutely convex domains, interior flow invariance conditions can be given in terms of their support functionals.
Let $X'$ be the dual of $X$ (cf. also Duality; Adjoint space). For $x\in X$ and $x'\in X;$, the pairing $(x,x')$ will denote $x'(x)$. The duality mapping $J:X\to2^{X'}$ is defined by
$$J(x):=\{x'\in X': \text{Re}\langle x,x'\rangle=||x||^2=||x'||^2\}$$
for each $x\in X$.
If $\textbf{D}$ is the open unit ball in $X$ and $g$ maps $\overline{\textbf{D}}$ into $X$, then (18) is equivalent to the condition
$$\inf_{x'\in J(x)}\text{Re}\langle g(x),x'\rangle\geq0,x\in\partial\textbf{D}$$
For the Euclidean ball $\textbf{D}$ in $X=\textbf{C}^n$, a certain condition in this direction was established by Abate [a2]. Namely, he proved that $g\in\text{Hol}(\textbf{D},\textbf{C}^n)$ is a semi-complete vector field if and only if it satisfies the estimate
$$2\big[||r(x)||^2-|\langle r(x),x\rangle|^2\big]\text{Re}\langle r(x),x\rangle+\\+(1-||x||^2)^2\text{Re}\langle g'(x)g(x),r(x)\rangle\geq0,$$
where
$$r(x)=(1-||x||^2)g(x)+\langle g(x),x\rangle x.$$
For $n=1$ this condition becomes
$$\text{Re}\:g(z)\overline{z}\geq-\frac{1}{2}\text{Re}\:g'(z)(1-|z|^2),$$
where $z\in\Delta$, the open unit disc in the complex plane $\textbf{C}$, and $g\in\text{Hol}(\Delta,\textbf{C})$.
Despite the usefulness and simplicity of condition (23) it is not clear how (21) can be derived from (23) when $g$ has a continuous extension to $\overline{\Delta}$.
Note also that in the one-dimensional case it follows from the maximum principle for harmonic functions that (21) implies the following interior condition:
$$\text{Re}\:g(z)\overline{z}\geq\text{Re}\:g(0)\overline{z}(1-|z|^2),z\in\Delta.$$
Conversely, it is clear that (21) does result from (24) if $g$ has a continuous extension to all of $\overline{\Delta}$. It turns out that an analogue of (24) is a necessary and sufficient condition for $g$ to be semi-complete [a1]: Let $\textbf{D}$ be the open unit ball in a complex Banach space $X$. Then $g\in\text{Hol}(\textbf{D},X)$ is a semi-complete vector field on $\textbf{D}$ if and only if it is bounded on each subset strictly inside $\textbf{D}$ and one of the following conditions holds:
a) For each $x\in\textbf{D}$ there exists an $x'\in J(x)$ such that
$$\text{Re}\big{\langle} g(x)-g(0)(1-||x||^2),x'\big{\rangle}\geq0;$$
b) $\inf_{x'\in J\langle x\rangle}\text{Re}\langle2||x||^2g(x)+(1-||x||^2)g'(x)x,x'\rangle\geq0$, $x\in\textbf{D}$;
c) For each $x\in\textbf{D}$ and for each $x'\in J(x)$,
$$\text{Re}\bigg{\langle}\frac{1-||x||}{1+||x||}g'(0)x+(1-||x||^2)g(0),x'\bigg{\rangle}\leq\\\leq\text{Re}\langle g(x),x'\rangle\leq\\\leq\text{Re}\bigg{\langle}\frac{1+||x||}{1-||x||}g'(0)x+(1-||x||^2)g(0),x'\bigg{\rangle}.$$
Furthermore, equality in one of the conditions a), b) or c) holds if and only if it holds in the other conditions and $f$ is complete.
Also, if $g\in\text{Hol}(\textbf{D},X)$ is a semi-complete vector field, then $g$ is, in fact, complete if and only if its derivative at zero, $g'(0)$ is a conservative linear operator, i.e.,
$$\text{Re}\langle g'(0)x,x'\rangle=0$$
for all $x\in X$ and $x'\in J(x)$ (see [a10], [a17]).
## Parametric representations of generators.
It is well-known that a complete vector field $g$ on the open unit ball $\textbf{D}$ in a Banach space $X$ is a polynomial of degree at most $2$ (see, for example, [a15], [a6]). More precisely, $g$ has the form
$$g(x)=\alpha+Ax+P_{\alpha}(x),$$
where $\alpha$ is an element of $X$, $A$ is a conservative operator on $X$ and $P_{\alpha}$ is a homogeneous form of the second degree such that $P_{i\alpha}=iP_{\alpha}$.
Suppose now that a complex Banach space $X$ is a so-called $\text{JB^{*}}$ triple system. This is equivalent to saying that its open unit ball $\textbf{D}$ is a homogeneous domain, i.e., for each pair $x,y\in\textbf{D}$ there exists a holomorphic automorphism $F$ of $\textbf{D}$ such that $F(x)=y$ (see, for example, [a15], [a6]). Then it is well-known that for each $\alpha\in X$ there exists a homogeneous polynomial $P_{\alpha}(x)$ such that $P_{i\alpha}=iP_{\alpha}$ and the mapping $g:\textbf{D}\to X$ defined by
$$g(x)-\alpha-P_{\alpha}(x)$$
is a complete vector field on $\textbf{D}$, which is called a transvection of $\textbf{D}$ (cf. also Transvection).
The cone $\mathcal{G}$ of semi-complete vector fields on $\textbf{D}$ admits the decomposition
$$\mathcal{G}=\mathcal{G}_0\bigoplus\mathcal{G}_+,$$
where is the real Banach subspace of consisting of transvections and is the subcone of such that for each ,
In other words, admits a unique representation
(a25)
where is complete, and .
The natural examples of triple systems are a complex Hilbert space , the space of bounded linear operators on , and its subspaces such that if and only if (such subspaces are usually called -algebras). In the latter case the general form of transvections on is , where and is its conjugate. Thus, each semi-complete vector field on the open unit ball of a -algebra has the form
(a26)
where and .
In particular, when is the complex plane and , the open unit disc in , (a26) becomes
(a27)
where and
(a28)
In 1978 E. Berkson and H. Porta [a4], solving an entirely different problem, gave a parametric representation of generators on the unit disc in the complex plane. More precisely, if and only if for some , has the representation
with everywhere. This point is exactly the limit point of the semi-group generated by (that is, its Denjoy–Wolff point, cf. Denjoy–Wolff theorem). The Berkson–Porta formula has also been successfully exploited in other fields; for example, in the classical functional equations of E. Schröder and N.H. Abel (see [a5] and Functional equation; Schröder functional equation).
Let be a complex Hilbert space with inner product and let be its open unit ball. Let denote the Poincaré hyperbolic metric on [a7] (cf. also Poincaré model). A mapping is said to be -monotone if for each pair and positive the following condition holds: whenever and belong to . It was shown in [a13] that if is separable and is a bounded continuous mapping, then is -monotone if and only if it generates a semi-group of -non-expansive self-mappings of . Note also that -monotonicity can be equivalently described as follows:
For a bounded holomorphic mapping and for an arbitrary the latter condition is a criterion for to be semi-complete. For the one-dimensional case, if , then this condition becomes the Berkson–Porta representation of semi-complete vector fields.
#### References
[a1] D. Aharonov, S. Reich, D. Shoikhet, "Flow invariance conditions for holomorphic mappings in Banach spaces" Math. Proc. Royal Irish Acad. , 99A (1999) pp. 93–104 [a2] M. Abate, "The infinitesimal generators of semi-groups of holomorphic maps" Ann. Mat. Pura Appl. , 161 (1992) pp. 167–180 [a3] L. Aizenberg, S. Reich, D. Shoikhet, "One-sided estimates for the existence of null points of holomorphic mappings in Banach spaces" J. Math. Anal. Appl. , 203 (1996) pp. 38–54 [a4] E. Berkson, H. Porta, "Semi-groups of analytic functions and composition operators" Michigan Math. J. , 25 (1978) pp. 101–115 [a5] C.C. Cowen, B.D. MacCluer, "Composition operators on spaces of analytic functions" , CRC (1995) [a6] S. Dineen, "The Schwartz lemma" , Clarendon Press (1989) [a7] K. Goebel, S. Reich, "Uniform convexity, hyperbolic geometry and nonexpansive mappings" , M. Dekker (1984) [a8] L.A. Harris, "Schwarz–Pick systems of pseudometrics for domains in normed linear spaces" , Advances in Holomorphy , North-Holland (1979) pp. 345–406 [a9] J.M. Isidro, L.L. Stacho, "Holomorphic automorphism groups in Banach spaces: An elementary introduction" , North-Holland (1984) [a10] S.G. Krein, "Linear differential equations in Banach spaces" , Amer. Math. Soc. (1971) [a11] R.H. Martin Jr., "Differential equations on closed subsets of a Banach space" Trans. Amer. Math. Soc. , 179 (1973) pp. 399–414 [a12] S. Reich, D. Shoikhet, "Generation theory for semi-groups of holomorphic mappings in Banach spaces" Abstr. Appl. Anal. , 1 (1996) pp. 1–44 [a13] S. Reich, D. Shoikhet, "Semi-groups and generators on convex domains with the hyperbolic metric" Atti Accad. Naz. Lincei , 8 (1997) pp. 231–250 [a14] S. Reich, D. Shoikhet, "Metric domains, holomorphic mappings and nonlinear semi-groups" Abstr. Appl. Anal. , 3 (1998) pp. 203–228 [a15] H. Upmeier, "Jordan algebras in analysis, operator theory and quantum mechanics" , CBMS-NSF Reg. Conf. Ser. in Math. , 67 , Amer. Math. Soc. (1987) [a16] E. Vesentini, "semi-groups of holomorphic isometries" Adv. Math. , 65 (1987) pp. 272–306 [a17] E. Vesentini, "Krein spaces and holomorphic isometries of Cartan domains" S. Coen (ed.) , Geometry and Complex Variables , M. Dekker (1991) pp. 409–413 [a18] E. Vesentini, "Semi-groups of holomorphic isometries" S. Coen (ed.) , Complex Potential Theory , Kluwer Acad. Publ. (1994) pp. 475–548 [a19] K. Yosida, "Functional analysis" , Springer (1968)
How to Cite This Entry:
Semi-group of holomorphic mappings. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Semi-group_of_holomorphic_mappings&oldid=51280
This article was adapted from an original article by Simeon ReichDavid Shoikhet (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 6,363 | 20,294 | {"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": 30, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.881824 |
https://static.nationmaster.com/country-info/profiles/Belarus/Transport/Railways | 1,590,900,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410745.37/warc/CC-MAIN-20200531023023-20200531053023-00361.warc.gz | 568,515,397 | 18,776 | # Belarusian Railways Stats
#### Definitions
• Broad gauge: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other.
• Goods transported > Million ton-km: Goods transported by railway are the volume of goods transported by railway, measured in metric tons times kilometers traveled."
• Passengers carried > Million passenger-km: Passengers carried by railway are the number of passengers transported by rail times kilometers traveled.
• Rail lines > Total route-km: Rail lines (total route-km). Rail lines are the length of railway route available for train service, irrespective of the number of parallel tracks.
• Rail lines > Total route-km per million: Rail lines (total route-km). Rail lines are the length of railway route available for train service, irrespective of the number of parallel tracks. Figures expressed per million population for the same year.
• Railways, goods transported > Million ton-km: Railways, goods transported (million ton-km). Goods transported by railway are the volume of goods transported by railway, measured in metric tons times kilometers traveled.
• Railways, goods transported > Million ton-km per 1000: Railways, goods transported (million ton-km). Goods transported by railway are the volume of goods transported by railway, measured in metric tons times kilometers traveled. Figures expressed per thousand population for the same year.
• Railways, passengers carried > Million passenger-km: Railways, passengers carried (million passenger-km). Passengers carried by railway are the number of passengers transported by rail times kilometers traveled.
• Railways, passengers carried > Million passenger-km per million: Railways, passengers carried (million passenger-km). Passengers carried by railway are the number of passengers transported by rail times kilometers traveled. Figures expressed per million population for the same year.
• Standard gauge: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other.
Standard gauge=1.435-m gauge
• Standard gauge per million: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other.
Standard gauge=1.435-m gauge. Figures expressed per million population for the same year.
• Total: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other.
• Total > Per \$ GDP: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other. Per \$ GDP figures expressed per 1 billion \$ gross domestic product.
• Total > Per capita: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other. Per capita figures expressed per 1,000 population.
• Total per million: This entry states the total route length of the railway network and of its component parts by gauge: broad, dual, narrow, standard, and other. Figures expressed per million population for the same year.
STAT AMOUNT DATE RANK HISTORY
Broad gauge 2013 8th out of 25
Goods transported > Million ton-km 47,933 2008 13th out of 89
Passengers carried > Million passenger-km 8,188 2008 21st out of 86
Rail lines > Total route-km 5,503 2011 29th out of 85
Rail lines > Total route-km per million 580.91 2011 21st out of 85
Railways, goods transported > Million ton-km 49,406 2011 13th out of 80
Railways, goods transported > Million ton-km per 1000 5.22 2011 7th out of 80
Railways, passengers carried > Million passenger-km 7,941 2011 25th out of 79
Railways, passengers carried > Million passenger-km per million 838.28 2011 13th out of 79
Standard gauge 25 km 2010 54th out of 57
Standard gauge per million 2.62 km 2008 46th out of 50
Total 5,537 km 2013 32nd out of 134
Total > Per \$ GDP 149.19 km per \$1 billion of GD 2006 27th out of 118
Total > Per capita 0.572 km per 1,000 people 2008 21st out of 104
Total per million 581.13 km 2008 20th out of 104
SOURCES: CIA World Factbooks 18 December 2003 to 28 March 2011; World Bank, Transportation, Water, and Urban Development Department, Transport Division.; World Bank, Transportation, Water, and Information and Communications Technologies Department, Transport Division.; World Bank, Transportation, Water, and Information and Communications Technologies Department, Transport Division. Population figures from World Bank: (1) United Nations Population Division. World Population Prospects, (2) United Nations Statistical Division. Population and Vital Statistics Report (various years), (3) Census reports and other statistical publications from national statistical offices, (4) Eurostat: Demographic Statistics, (5) Secretariat of the Pacific Community: Statistics and Demography Programme, and (6) U.S. Census Bureau: International Database.; CIA World Factbooks 18 December 2003 to 28 March 2011. Population figures from World Bank: (1) United Nations Population Division. World Population Prospects, (2) United Nations Statistical Division. Population and Vital Statistics Report (various years), (3) Census reports and other statistical publications from national statistical offices, (4) Eurostat: Demographic Statistics, (5) Secretariat of the Pacific Community: Statistics and Demography Programme, and (6) U.S. Census Bureau: International Database.
## Belarus categories
Agriculture Environment Military Background Geography People Conflict Government Religion Cost of living Health Sports Crime Import Terrorism Culture Industry Transport Disasters Labor Travel Economy Language Weather Education Lifestyle Energy Media | 1,304 | 5,831 | {"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-2020-24 | latest | en | 0.91439 |
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Motions in GeometryDrawing Escher-Type Tessellations
# Math Investigations
Laboratory Investigation 11.2Tessellations The following steps illustrate a method of altering the sides of an equilateral triangle to obtain a non polygonal figure that will tessellate. These steps can be carried out with pencil and paper or by computer software programs. Step 1 Draw a curve from A to B. (2.0K) Step 2 Rotate the curve about point B so that A maps to C. (2.0K) Step 3 Label the midpoint of as D, and draw a curve from D to C. Rotate this curve about D so that C maps to A. (2.0K) Once the lines of the original figure have been erased, the figure that remains will tessellate. Starting Points for InvestigationsIf the preceding figure is used to form a tessellation, which of the transformations (rotation, translation, reflection) will map the preceding tessellation onto itself? Step 3 produces a curve on one side of the triangle that is said to have point symmetry because it can be rotated onto itself by a 180˚ rotation. Suppose Step 3 is used to produce a curve with point symmetry on all three sides of a triangle. Will the resulting figure tessellate? Suppose Step 3 is used to create a curve with point symmetry on each of the six sides of a regular hexagon. Will the resulting figure tessellate? Tessellations Investigation (Chapter 11, Section 2)Read Me - Tessellations Instructions (Word Format) (35.0K) Click on Geometer's Sketchpad in the left menu for informationGSP file--Investigation 11.2: Tessellations (15.0K)
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https://www.elitedigitalstudy.com/11953/a-ray-of-light-passing-through-the-point-1-2-reflects-on-the-x-axis-at-point-a-and-the-reflected-ray | 1,680,073,922,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00301.warc.gz | 840,957,231 | 10,829 | A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Asked by Abhisek | 1 year ago | 177
##### Solution :-
Consider the coordinates of point A as (a, 0)
Construct a line (AL) which is perpendicular to the x-axis
Here the angle of incidence is equal to angle of reflection
∠BAL = ∠CAL = Φ
∠CAX = θ
It can be written as
∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]
On further calculation
= 180° – θ – 180° + 2θ
= θ
So we get
∠BAX = 180° – θ
Slope of line = $$\dfrac{3-0}{5-a}$$
tanθ = $$\dfrac{3}{5-a}$$ ..............(1)
Slope of line AB = $$\dfrac{2-0}{1-a}$$
tanθ =(180° – θ) = $$\dfrac{2}{1-a}$$
- tanθ = $$\dfrac{2}{1-a}$$
- tanθ = $$\dfrac{2}{a-1}$$
From equations (1) and (2) ,
$$\dfrac{3}{5-a}$$$$\dfrac{2}{a-1}$$
By cross multiplication,
3a -3 =10-2a
$$\dfrac{13}{5}$$
Therefore, the coordinates of point A are $$( \dfrac{13}{5},0)$$
Answered by Pragya Singh | 1 year ago
### Related Questions
#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices
Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0. | 628 | 1,637 | {"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} | 4.46875 | 4 | CC-MAIN-2023-14 | latest | en | 0.808757 |
http://www.redszone.com/forums/showthread.php?51530-It-s-a-modern-miracle-this-team-is-mathematically-IN-it-at-all&s=2c98d70ef9505383dd3e5c61a253250d&p=1168034 | 1,395,091,053,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678706211/warc/CC-MAIN-20140313024506-00045-ip-10-183-142-35.ec2.internal.warc.gz | 477,081,210 | 12,737 | # Thread: It's a modern miracle this team is mathematically IN it at all?
1. ## It's a modern miracle this team is mathematically IN it at all?
How amazing? Forget the fact that the Reds are ONE UNDER .500, forget the fact that they were declared dead. But get this - The Reds run differntial is 731 run scored and 781 runs given UP!!! They are FIFTY runs below their opponents on the year! And earlier this year I heard no team has made the playoffs with a NEGATIVE run differential! Not only try negative .. but 50 under!
2. ## Re: It's a modern miracle this team is mathematically IN it at all?
If a miracle happened and they made the playoffs you sure won't find me apologizing for it. I'll just be rooting for them like crazy EVERY game they play!!
3. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by GOREDSGO32
How amazing? Forget the fact that the Reds are ONE UNDER .500, forget the fact that they were declared dead. But get this - The Reds run differntial is 731 run scored and 781 runs given UP!!! They are FIFTY runs below their opponents on the year! And earlier this year I heard no team has made the playoffs with a NEGATIVE run differential! Not only try negative .. but 50 under!
And we are just one of four teams still in it for a division title - Astros, Reds, Dodgers, and Twins. Not too shabby for a team picked to finish last or near thereto. We are also one of only three teams still not eliminated in the Wild Card race (us, Astros, and Phillies). In the Wild Card, our Eliminiation Number is 1 and the Astros is 2.
4. ## Re: It's a modern miracle this team is mathematically IN it at all?
This team reminds me of the 2003 Bengals. So close, but still with a few holes to fill before they become a serious contender. They're an improvement over previous years and if they somehow manage to get a few of those missing pieces in place they could be something special in a couple years. I hope I'm right. I'm certainly more optimistic than I was a year ago.
5. ## Re: It's a modern miracle this team is mathematically IN it at all?
You can thank the Cardinals for this "modern miracle"
6. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by BamaRed
You can thank the Cardinals for this "modern miracle"
True, but the Reds have won games lately to make up ground.
7. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by EKURed
True, but the Reds have won games lately to make up ground.
True, you can't make up games unless YOU win.
8. ## Re: It's a modern miracle this team is mathematically IN it at all?
Well, of course I'm rooting for them, but they aren't exactly "hot." They're 5-5 over the last 10.
9. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by terminator
Well, of course I'm rooting for them, but they aren't exactly "hot." They're 5-5 over the last 10.
Of course, that's hotter than 0-7.
10. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by redsmetz
Of course, that's hotter than 0-7.
I can imagine that the Houston Astros forum is also going crazy right now.
11. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by BamaRed
You can thank the Cardinals for this "modern miracle"
And the Cardinals can thank the Reds for the West Coast disaster, it all has a way of evening out.
12. ## Re: It's a modern miracle this team is mathematically IN it at all?
It's ludicrous for a sub-.500 team to be in contention for a playoff spot. It doesn't feel right at all.
13. ## Re: It's a modern miracle this team is mathematically IN it at all?
I believe 5 teams in the history of baseball have made the playoffs with a negative run differential and one, the 1987 Minnesota Twins, has won it all.
14. ## Re: It's a modern miracle this team is mathematically IN it at all?
Originally Posted by Johnny Footstool
It's ludicrous for a sub-.500 team to be in contention for a playoff spot. It doesn't feel right at all.
It just seems so strange that whoever wins the division will likely do it with a record that wouldn't give them 2nd place in ANY other division. And would put them in 4th place in a couple of them, maybe even all 3 AL divisions.
15. ## Re: It's a modern miracle this team is mathematically IN it at all?
The Cardinals may tie a record tonight. The record for a losing streak in September by a first place team is 8. In 1995 the Angels were in first lost 8 in a row and it knocked them out of the playoffs. The 1980 Royals also lost 8 in a row but that was after they had clinched a playoff spot.
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• | 1,199 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-10 | latest | en | 0.965091 |
https://homework.cpm.org/category/CON_FOUND/textbook/gc/chapter/10/lesson/10.2.1/problem/10-62 | 1,726,383,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00724.warc.gz | 262,666,458 | 15,423 | ### Home > GC > Chapter 10 > Lesson 10.2.1 > Problem10-62
10-62.
A graph of an inequality is shown at right. Decide if each of the points $\left(x,y\right)$ listed below would make the inequality true or not. For each point, explain how you know.
If a point makes the inequality true, it must be in the shaded region.
1. $\left(1,1\right)$
True
1. $\left(-3,2\right)$
False
1. $\left(-2,0\right)$
False
1. $\left(0,-2\right)$
True | 140 | 441 | {"found_math": true, "script_math_tex": 5, "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-2024-38 | latest | en | 0.741161 |
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# LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
User ID: 78331155
07/03/2021 06:29 PM
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LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Last Edited by JADR+ on 07/10/2021 08:53 PM
I'm a J & proud zio.
FE Challenge: Provide a formula which calculates the exact distance between 2 GPS coordinates that does not use the Earth's radius of 6,371 km in it's assumptions
Anonymous Coward
User ID: 78990415
United States
07/03/2021 06:30 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Hooooolllly shiiiiiitt....
AKAllie
User ID: 75566719
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07/03/2021 06:33 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Was wondering if that would happen, and low and behold yep!
Anonymous Coward
User ID: 77618993
United Kingdom
07/03/2021 06:34 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Bioluminescence Luciferase of course it would illuminate... What type of Frankenstein's have we created..
Anonymous Coward
User ID: 78990415
United States
07/03/2021 06:34 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Need more confirmation videos.
User ID: 80072785
07/03/2021 06:36 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Bump for info
Anyone got a black light and a family member that was vaxxed ???
Go now and report back
Anonymous Coward
User ID: 77618993
United Kingdom
07/03/2021 06:37 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Would a blue light that they use to find spunk on sheets In those hotel nightmares TV shows have the same effect I wonder?
User ID: 78331155
07/03/2021 06:38 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Would a blue light that they use to find spunk on sheets In those hotel nightmares TV shows have the same effect I wonder?
Quoting: Anonymous Coward 77618993
Yes
I'm a J & proud zio.
FE Challenge: Provide a formula which calculates the exact distance between 2 GPS coordinates that does not use the Earth's radius of 6,371 km in it's assumptions
Anonymous Coward
User ID: 79112896
United States
07/03/2021 06:41 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Bump for info
Anyone got a black light and a family member that was vaxxed ???
Go now and report back
365 nm black light, recent J&J shot.
No glow.
Anonymous Coward
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United States
07/03/2021 06:41 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
It's the SM-102 that causes Luciferase I expression which is detectable under UV light. Maybe will make it easier for drones to detect us.
Anonymous Coward
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07/03/2021 06:41 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
just a normal side effect of the vaccine, it means it's working as intended.
ChefElvis
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07/03/2021 06:43 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
A legend in my own mind
User ID: 78331155
07/03/2021 06:56 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
DIY IPhone Blacklight
I'm a J & proud zio.
FE Challenge: Provide a formula which calculates the exact distance between 2 GPS coordinates that does not use the Earth's radius of 6,371 km in it's assumptions
Anonymous Coward
User ID: 76802664
France
07/03/2021 06:59 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Freemasonry is strong on this guy.. Look at all those dark occult tats. We have to be very skeptical of everything we are being fed.
1. Why are #magnetchallenge videos being left up on YouTube & TikTok after the factcheckers, media & social media sites have come out strongly as them being disinformation???
2. There are 6 million Freemasons worldwide & a whole lot of street level low tier Illuminati. They "could" be setting us God & Christ faithfull antivaxxers up for death by variants targeted at the unvaxxed.. with the true mark of the beast still forth coming with the one world cryptocurrency that the entire world would know up front you can't buy, trade or sell without it.. fulfilling Bible prophecy. They need us to be informed of what the mark really is so that when we use our conscious free will to take it.. we default our souls to Satan.
HollyWho
User ID: 77513256
United States
07/03/2021 06:59 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Anything with the name Lucifer in it can't be a good thing
Anonymous Coward
User ID: 79021294
United Kingdom
07/03/2021 07:01 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Anonymous Coward
User ID: 79021294
United Kingdom
07/03/2021 07:02 PM
Report Abusive Post
Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Freemasonry is strong on this guy.. Look at all those dark occult tats. We have to be very skeptical of everything we are being fed.
1. Why are #magnetchallenge videos being left up on YouTube & TikTok after the factcheckers, media & social media sites have come out strongly as them being disinformation???
2. There are 6 million Freemasons worldwide & a whole lot of street level low tier Illuminati. They "could" be setting us God & Christ faithfull antivaxxers up for death by variants targeted at the unvaxxed.. with the true mark of the beast still forth coming with the one world cryptocurrency that the entire world would know up front you can't buy, trade or sell without it.. fulfilling Bible prophecy. They need us to be informed of what the mark really is so that when we use our conscious free will to take it.. we default our souls to Satan.
Quoting: Anonymous Coward 76802664
DuckNCover
User ID: 79893001
United States
07/03/2021 07:04 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
So is it just in the Moderna Jab...
Looks like J&J shot doesn't have Luciferase, but it will give you blood clots...
Anonymous Coward
User ID: 79112896
United States
07/03/2021 07:06 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
So is it just in the Moderna Jab...
Looks like J&J shot doesn't have Luciferase, but it will give you blood clots...
Quoting: DuckNCover
All three of the US-available ones have the clotting issue.
Anonymous Coward
User ID: 78990415
United States
07/03/2021 07:09 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Bump for info
Anyone got a black light and a family member that was vaxxed ???
Go now and report back
365 nm black light, recent J&J shot.
No glow.
Quoting: Anonymous Coward 79112896
Words are useless in a world of shills... post video or stfu.
26Degrees
User ID: 80325889
07/03/2021 07:10 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
In the right hand. Who would have imagined the deceit.
Give the shot containing the mark and it migrates to the right hand. He should scan his forehead.
This is horrible
Anonymous Coward
User ID: 80104741
07/03/2021 07:12 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Holy mother of God!!
cellular implosion
User ID: 80547695
Australia
07/03/2021 07:12 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Stay away from strong magnetic fields and gamma ray bursts.
Anonymous Coward
User ID: 79112896
United States
07/03/2021 07:14 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Stay away from strong magnetic fields and gamma ray bursts.
Quoting: cellular implosion 80547695
Welp, that might be a challenge:
CME IMPACT POSSIBLE THIS WEEKEND: A minor coronal mass ejection (CME) is expected to hit Earth this weekend. It left the sun on June 29th, propelled toward us by a B7-class solar flare.The impact will probably be too weak to cause a geomagnetic storm, but high-latitude auroras are possible on July 3rd or 4th. Aurora alerts: SMS Text.
FIRST X-FLARE IN 4 YEARS: A new sunspot emerged during the early hours of July 3rd and promptly exploded, producing the first X-class solar flare since Sept. 2017. NASA's Solar Dynamics Observatory recorded the extreme ultraviolet flash:
Anonymous Coward
User ID: 79021294
United Kingdom
07/03/2021 07:17 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Would a blue light that they use to find spunk on sheets In those hotel nightmares TV shows have the same effect I wonder?
Quoting: Anonymous Coward 77618993
yes and they show up blood also!! why is the light blue on the vien?? it is impossible!!!
Anonymous Coward
User ID: 72272362
United Kingdom
07/03/2021 07:27 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Clicks on this thread goes to make a cup of tea turns kitchen light off after finishing and lightening strikes what timing!
Anonymous Coward
User ID: 78227579
07/03/2021 07:28 PM
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Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Fake.
Dudes made other vids (lightbulb one) and is debunked.
1) Luciferase is not new...it’s been around since the 1800s and was eventually purified for cancer tracing in clinical studies. They’ve talked about using it for a vaccine identifier, but currently no vaccines have it as an ingredient.
2) if you think this amazing, look up medical vein finders, it’s not demonic magic. Also goto hot topic and get some black light paint/makeup...you can make your own glowing Covid video
G3
User ID: 80251587
United States
07/03/2021 07:31 PM
Report Abusive Post
Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
DIY IPhone Blacklight
Jadr, that video was terrible and it doesn't work
Thanks though
Anonymous Coward
User ID: 79112896
United States
07/03/2021 07:43 PM
Report Abusive Post
Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
Fake.
Dudes made other vids (lightbulb one) and is debunked.
1) Luciferase is not new...it’s been around since the 1800s and was eventually purified for cancer tracing in clinical studies. They’ve talked about using it for a vaccine identifier, but currently no vaccines have it as an ingredient.
2) if you think this amazing, look up medical vein finders, it’s not demonic magic. Also goto hot topic and get some black light paint/makeup...you can make your own glowing Covid video
Quoting: Anonymous Coward 78227579
Can't speak to point 1, but towards the 2nd ...
No need for the trip to store.
Laundry detergent liquid with 'brighteners' will glow that color.
Anonymous Coward
User ID: 79627804
07/03/2021 07:45 PM
Report Abusive Post
Re: LUCIFERASE BEING PICKED UP BY VACCINE VICTIMS USING A BLACK LIGHT? (UPDATE)
just a normal side effect of the vaccine, it means it's working as intended.
Quoting: Anonymous Coward 51393839
Lol
MISTER-BEE
User ID: 44047335
United States
07/03/2021 07:54 PM
Report Abusive Post | 3,433 | 11,892 | {"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} | 2.515625 | 3 | CC-MAIN-2023-23 | latest | en | 0.715793 |
https://www.doubtnut.com/question-answer/show-that-the-point-a4-12-b-351-c234-and-d166-are-coplanar-also-find-the-equation-of-the-plane-passi-31347869 | 1,632,188,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00605.warc.gz | 799,104,898 | 70,047 | Home
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Show that the point A(4,-1,2),...
# Show that the point A(4,-1,2), B(-3,5,1), C(2,3,4) and D(1,6,6) are coplanar. Also find the equation of the plane passing through these points.
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 19-9-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
14.6 K+
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x+y-z=1
Solution :
N//A
Image Solution
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5:27 | 441 | 1,045 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.49874 |
https://books.google.no/books?id=IooAAAAAMAAJ&dq=editions:UOM39015067252117&hl=no&output=html_text&lr= | 1,632,397,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057421.82/warc/CC-MAIN-20210923104706-20210923134706-00368.warc.gz | 198,740,311 | 11,198 | # Dictionary of the Mathematical and Physical Sciences, According to the Latest Improvements and Discoveries
James Mitchell
Sir Richard Phillips, 1823 - 576 sider
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Populære avsnitt
Side 438 - A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within called the center.
Side 508 - In higher works on trigonometry, it has been demonstrated that, in any triangle, the sines of the angles are proportional to the lengths of the sides opposite to them. In other words, sin A : sin B :: BC : AC; or, sin A : sin C:: BC : AB, and sin B : sin C::AC : A B. Hence, we have sin 44° 40' : sin 56° 20
Side 292 - A law presupposes an agent ; for it is only the mode, according to which an agent proceeds : it implies a power ; for it is the order, according to which that power acts. Without this agent, without this power, which are both distinct from itself, the law does nothing ; is nothing. The expression, ' the law of metallic nature...
Side 203 - ... winch, with as little labour as it takes to wind up a jack, though the weight of the iron, tin, and wooden circle, is about 1000 pounds.
Side 76 - In foul weather, when the mercury rises much and high, and so continues for two or three days before the foul weather is quite over, then expect a continuance of fair weather to follow.
Side 209 - Specific Gravity of a body is the relation of its weight, compared with the weight of some other body of the same magnitude. A body immersed in a fluid will sink if its specific gravity be greater than that of the fluid; but if it be less, the body will rise to the top, and will be only partly uncovered.
Side 476 - ... the object he views. There is no small speculum, but the magnifiers are applied immediately to the first focal image. From the opening of the telescope, near the place of the eye glass, a speaking-pipe runs down to the bottom of the tube, where it...
Side 396 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Side 457 - And in measuring any of these station-distances, mark accurately where these lines meet with any hedges, ditches, roads, lanes, paths, rivulets, &c ; and where any remarkable object is placed, by measuring its distance from the station-line ; and where a perpendicular From it cuts that line. And thus as you go along any main...
Side 13 - ... of the motion seemed to be from the upper part downwards. It appears also that they were in some danger of having the balloon burnt altogether; as the Marquis observed several round holes made by the fire in the lower part of it, which alarmed him considerably, and, indeed, not without reason. | 661 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.898497 |
supari.in | 1,709,088,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474690.22/warc/CC-MAIN-20240228012542-20240228042542-00650.warc.gz | 541,806,610 | 27,742 | # Golden Ratio and the Grander Scheme of Things
The golden ratio or the divine proportion/ phi, is a contemporary challenge posed to us by a renaissance genius. While some are still sceptical about its existence, others clearly see the world through the lens of the famous golden ratio. Lauded as one of Da Vinci’s most intricate illustrations, the divine proportion is prominent in all his works.
If you think that the mathematical and artistic proportions have lost essence over the stretch of time, it isn’t just prevalent in art, and architecture but the nature that shapes the very us!
## 1. Gothic and Modern Architecture
Divine proportion phi makes for the mammoth buildings and stunning architecture we see today. It is the most widely used mathematical proportion in architecture, better known as φ. Its use dates back to the Great Pyramid of Giza, built around 2560 BC, as one of the earliest examples of the use of the golden ratio. The impressions of it can now be seen in both gothic and modern architecture as well.
## 2. In the Very Core of Nature
The striking resemblances to how divine proportion carves out its application in some of the very small as well as big things in our world, is uncanny. From flower seeds to ginormous galaxies, the grand design is subtly crafted to fit the golden ratio of Da Vinci.
## 3. The Human Anatomy
The human body is based on patterns of 5, which is the basis of phi as well
5 appendages to the torso, in the arms, legs and head
5 appendages on each of these, in fingers and toes
5 openings on the face
5 sense organs for sight, sound, touch, taste and smell
The golden ratio in turn is also based on 5 as the number phi, or 1.6180339 is computed using 5 such as
5^.5*.5+.5 = Phi
The human figure too, represents and exhibits perfect proportionality in terms of the golden ratio. Such as the height of a person divided by the distance between their belly button and the ground. The Vitruvian man is a classic example of an ideal human figure as per golden ratio.
## 4. Music
Fabionni sequence is the series of numbers that have extraordinary mathematical properties. But while we look for ways we can see it, its influence is also in the sound fields we hear. The notes, chords, phrases, harmony and dynamics are all mere coincidence or natural order of the universe? – People believe that certain sounds that are “meant” to work together, actually work due to the perfect golden ratio.
## 5. Spirituality
The “Divine” proportion is quite fitting when talking about spirituality as it provides a deeper understanding of the world around us. It brings out the connectedness and hidden harmony in whatever we see. The world of mathematics (practical) is highly differentiated from the spiritual (transcendental) world, but the golden ratio seems bring them both together.
The Kaaba, one of the holiest places of worship, is said to be formed through the golden ratio. We also see a classic example of geometry channeling divine energies in the form of mandalas too.
Also strangely enough, the symbol for Phi can be thought to represent a zero, divided by 1 or Unity to create structures and things juxtaposing God’s creation the universe from nothing. There is a hidden connection to golden ratio and the anatomy of the world. Do you see more similarities too? Tell us if you view the world as a composition of the ratio?
## Parent Teacher Cooperation for The Success of Online Education
As the world is finally coming to terms with the current global situation brought about by the novel…
## Origin of the Butterfly Effect: Busting the Modern Fictional Myth
The concept of the butterfly effect definitely rings a bell in the minds of people who have had…
We all know that Education is the spinal cord of our Nation. No one can grab the highest…
## Let’s talk Brainwaves: How to Enter the Theta State of Mind?
The human brain is an electrochemical organ with different frequencies of waves. The five main categories of brainwaves… | 858 | 4,002 | {"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} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.951125 |
https://www.civilconcept.com/top-50-qna-of-rcc-to-crack-mcq-for-design-of-reinforced-concrete/?v=aae084230844 | 1,653,770,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00692.warc.gz | 787,273,491 | 39,167 | # Top 151 QnA of RCC to Crack | MCQ for design of reinforced Concrete
1. The direction of shear stress is
a) horizontal b) vertical
c) both (a) and (b) above d) None of the above
2. The direction of normal stress
a) parallel to failure plane b) perpendicular to failure plane
c) both (a) and (b) above d) None of the above
3. The direction of shear stress is
a) parallel to failure plane b) perpendicular to failure plane
c) both (a) and (b) above d) None of the above
4. The actual numerical value of tensile stress is………..than the theoretical stress (engineering stress)
a) less b) more
c) equal d) any of the above
5. Allowable stress in steel is less than
a) permissible stress b) working stress
c) yield stress d) none of the above
6. According to Hook’s law, which one is wrong?
a) σαΕ b) σ/E =constant
c) MoE x Strain = Stress d) MoE x Stress = Strain
7. Stress is directly proportional to strain within the elastic limit is given by
a) Archimedes b) Hooke’s
c) Newton d) Galileo
8. Elasticity of a body is defined
a) the property by which a body returns to its original shape after removal of load
b) the property by which a body doesn’t return to its original shape after removal of load
c) the property by which a body undergoes deformation even after removal of load
d) all of the above
9. Elasticity of a body is
a) the ratio of stress to strain b) the resistance to the force acting
c) large deformability d) None of the above
10. Permanent set is
a) the force which acts permanently on the body
b) irrecoverable deformation in the body
c) the shape of the member just after completion of construction
d) ratio of Poisson’s ratio to young’s modulus
11. Stress in a member subjected to a force is
a) the ratio of modulus of elasticity to strain
b) the product of modulus of elasticity and strain
c) the resistance to the force acting
d) sum of strain and modulus of elasticity
12. Young’s modulus is the ratio of the normal stress to the
a) normal strain within the elastic limit
b) normal strain within the proportional limit
c) reciprocal of normal strain within the elastic limit
d) normal strain at the yield point
13. Modulus of rigidity is the ratio of
a) normal stress to normal strain
b) shear stress to shear strain
c) lateral stress to lateral strain
d) normal strain to normal stress
14. In any member of the structure system, the strain does not affect the size of the member
a) lateral strain b) volumetric strain
c) shear strain d) longitudinal strain
15. Plasticity of a body is
a) the ratio of stress to strain b) same as the property of elasticity
c) opposite property of elasticity d) None of the above
16. The ratio of lateral strain to longitudinal strain is called
a) Poisson’s ratio b) stress
c) Hook’s law d) None of the above
17. The product El is called
a) flexural rigidity b) torsional rigidity
c) the second moment of area d) None of the above
18. The value of Poisson’s ratio always remains
a) greater than one b) less than one
c) equal to one d) None of the above
19. The percentage elongation of material from a direct tensile test indicates
a) ductility b) strength
c) yield stress d) brittleness
20. The percentage reduction of material from a direct compressive test indicates
a) ductility b) strength
c) yield stress d) brittleness
21. The safe value of stress below which the material will not fail when subjected to reversal of stress is known as
a) fatigue stress b) endurance limit
c) elastic limit d) tolerance limit
22. The property by which the metals deforms continuously at a slow rate without further increase in stress is called
a) creep b) fatigue
c) resilience d) plasticity
23. Toughness is
a) ability to absorb energy during plastic deformation
b) higher ultimate strength
c) stress at field
d) strain energy at the field
24. The impact tests are used to determine
a) ultimate crushing strength b) toughness
c) ductility d) tenancy
a) 1 b) 2
c) 3 d) 4
26. What is tenancy?
a) ultimate strength in tension
b) ultimate strength in compression
c) ultimate shear strength
d) ultimate impact strength
27. At ordinary temperatures, how is the yield point affected with
a) not greatly influenced
b) very greatly influenced
c) not influenced at all
d) difficult to tell
28. Simple bending equation is
a)
b)
c)
d) all of the above
29. Stress in a beam due to simple bending is
a) directly proportional b) inversely proportional
c) curvilinearly related d) all of the above
30. In a composite bar the load distribution among different materials of which it is made is based on the assumption the all the materials will have
a) equal areas b) same Young’s modulus
c) same strain d)same stress
31. The stress due to temperature change in a member depends on
a) supporting conditions of ends b) length of member
c) area of cross section d) none of the above
32. Strain energy of a member may be defined as work done on it
a) to deform it b) to resist elongation
c) to resist shortening d) all of the above
33. Partial derivative of strain energy gives
a) deflection b) slop
c) moment d) redundant force
34. The strain energy stored in a specimen when strained within the elastic limit is known as
a) resilience b) proof resilience
c) strain energy d) modulus of resilience
35. The maximum strain energy stored in a specimen when strained within the elastic limit is known as
a) resilience b) proof resilience
c) strain energy d) modulus of resilience
36. Find the elongation of a rod if a load of p is applied on a rod having length, I c/s area, A and modulus of elasticity, E
a) pI/AE b) pE/AI
c) AI/pE d) pA/IE
37. Find the modulus elasticity of bar material (E) if e/s area of bar is A, length L, applied load P and elongation is A
a) pI/AΔ b) pΔ/IA
c) AΔ/pI d) pA/Δ
38. The amount of stress developed due to increase in temperature in case of SSB if 1-10m, AT-10°C, A=5cm²
a) IN/mm² b) 100 N/mm²
c) 500 N/mm² d) none of the above
39. A member having cross section 5 cmx5 cm is subjected to a tensile stress of 100 kg/cm², then the applied load should be
a) 2500 kg b) 2000 kg
c) 1500kg d) 1000 kg
40. Torsion is produced in a beam due to
b) eccentricity of load from centroidal axis
c) types of beam
d) none of the above
41. Due to torsion moment beam get
a) bent b) twisted
c) uplifted d) none of the above
43. The amount of temperature stress is in fixed beam if ……………. compared with propped cantilever.
a) more
b) less
c) equal
d) any of the above
45. Stress concentration near the void is…………. as compared to solid portion.
a) less b) more
c) equal d) none of the above
46. Hoop stress is…………..
c) circumferential tensile stress d) none of the above
47. The moment of inertia of rectangular section is.
a) BD3/12 b) a4/12
c)π/64 x BD3 d) bh3/36
48. Ratio of moment of inertia (MOI) of a rectangular object about its x-axis to its base is
a) 1/12 b) 1/8
c) 1/6 d) ¼
49. M.I of circular section
a) πD4/64 b) πD4/48
c) πD4/56 d) πD4/78
50. The moment of inertia of the square section is
a) a4/12 b) BD3/12
c) π/64 x BD3 d) bh3/36
51. The moment of inertia of the elliptical section is
a) π/64 x BD3 b) BD3/12
c) a4/12 d) bh3/36
52. The moment of inertia of triangular section about its e.g.
a) bh3/36 b) BD3/12
c) a4/12 d) π/64 x BD3
53. Along the neutral axis of a simply supported beam
a) fibres do not undergo strain
b) fibres undergo minimum strain
c) fibres undergo maximum strain
d) none of the above
54. The ratio of the moment of inertia of a circular plate and a square plate for equal depth is
a) slightly less than one
b) equal to one
c) more than one
d) equal to 3π/16
55. The section modulus of a rectangular section is proportional to
a) area of the section
b) square of the area of the section
c) product of the area and depth
d) product of the area and width
56. The algebraic sum of the vertical forces on either side of the section of a loaded beam is known as
a) shear force b) bending moment
c) deformed force d) all of the above
57. Find the shear force produced in the member BC due to load
shown in fig.
a) 4 kN b) 50 kN
c) 25 kN d) 75 kN
58. Find the bending moment at fixed end for the structure as shown in fig.
a) 0 Tm b) 10Tm
c) 20Tm d) 30Tm
59. Find the bending moment at fixed end for the structure as shown in fig.
a) 0 b) 10 Tm
c) 20 Tm d) 30 Tm
60. Find the bending moment at fixed end of cantilever as shown
in fig.
a) 0 b) 10 Tm
c) 20 Tm d) 30 Tm
61. Find the bending moment at fixed end of cantilever as shown
in fig.
a) 0 b) 10 Tm
c) 20 Tm d) 30 Tm
62. A rectangular beam carries a maximum bending moment of M. If its depth is doubled, its moment carrying capacity will be
a) M b) 2M
c) 3M d) 4M
63. The algebraic sum of the moments of the forces on either side of the section of a loaded beam is known as
a) shear force b) bending moment
c) deformed force d) all of the above
64. The shear force and bending moment are related by
a) b)
c) d)
65. The shear force on a beam and the displacement are related by
a) b)
c) d) none of the above
66. Which one is correct if the deflection of a beam is y
a)
67. Calculate the maximum BM introduced due to a udl of 4KN/m, if the span of the cantilever is 1.8m
a) 2.16 KN-M b) 6.48 KN-M
c) 1.08 KN-M d) 5.48 KN-M
68. Find the bending moment at %4 th of the span having simply supported beam with following data: point load = 200 kg at Acentre, span = 20m
a) 250 kg-m b) 500 kg-m
c) 50 kg-m d) none of the above
69. The maximum deflection in the cantilever beam subjected to an udl of W/m throughout the span is
a) b)
c) d) none of the above
70. A simply supported beam of length 1 carries a load varying uniformly from zero at left end to maximum at right end maximum bending moment occurs at a distance of
a) from left end b) from left end
c) from right end d) from right end
71. Load carrying capacity of fixed beam is
a) lesser than simply supported beam
b) lesser than cantilever beam
c) greater than simply supported beam and cantilever beam
d) all of the above
72. The difference in placing the end of a beam simply over a support and the supporting end through a hinge on rollers is that the roller support
a) can offer reaction in the plane of rollers
b) can offer moment reaction
c) will not allowed the end to lift up the deflection
d) will not offer reaction normal to the plane of rollers
73. The number of reaction components possible at a hinged end for a general loading is
a) 0 b) 1
c) 2 d) 3
74. At either ends of plane frame, maximum number of possible transverse shear forces, are
a) one b) two
c) three d) four
75. At either ends of plane frame, maximum number of possible bending moment, are
a) one b) two
c) three d) zero
76. A simply supported beam of span L carries a uniformly distributed load, w. The maximum shear force, V is
a) b)
c) d)
77. A simply supported beam of span L carries a uniformly distributed load, w. The maximum bending moment, M is
a) b)
c) d)
78. A simply supported beam of span L carries two equal concentrated loads, W at a distance of L/3 from either support. The maximum bending moment, M is
a) b)
c) d)
79. A beam is hinged at both ends and loaded with a triangular load having maximum intensity (W) at centre. Find the maximum bending moment & shear force
a) b)
c) d) all of the above
80. The shape of bending moment diagram over the length of a beam, having no external load is
a) linear b) parabolic
c) cubical d) circular
81. The shape of bending moment diagram over the length of a beam, carrying a uniformly distributed load is
a) linear b) parabolic
c) cubical d) circular
82. In any beam, there is acting uniformly moment then its curve is
a) circular arch b) parabolic arch
c) both (a) and (b) of above d) none of the above
83. The shape of bending moment diagram over the length of a beam, carrying a uniformly increasing load is
a) linear b) parabolic
c) cubical d) circular
84. The maximum bending moment due to moving load on a simply supported beam occurs
a) at the mid span b) at the supports
c) under the load d) none of the above
85. For a simply supported beam with a central load, the bending moment is
a) least at the centre b) least at the supports
c) maximum at the supports d) maximum at the center
86. For a cantilever with a uniformly distributed load, w over its entire length L, the maximum bending moment is
a) b)
c) d)
87. The maximum deflection in the cantilever beam carrying a point load (W) at end is
a) b)
c) d)
88. In a cantilever beam a point load is applied at the free end. Which one is correct?
a) SF is uniform throughout the beam
b) SF varies linearly from free end to fixed end
c) BM is uniform throughout the beam
d) none of the above
89. Which of the following end conditions permits the displacement in any direction and also rotation
a) fixed end b) hinged end
c) roller end d) free end
90. Which of the following section is the most efficient in carrying bending moments ?
a) I-section b) rectangular-section
c) T-section d) circular-section
91. In an I-section almost all the shear force is taken by
a) top flange b) web
c) bottom flange d) none of the above
92. The bending moment is maximum on a section where shear force
a) is maximum b) is minimum
c) is equal d) changes sign
93. In a continuous bending moment curve the point where it changes sign is called
a) point of inflexion
b) point of contraflexture
c) point of the virtual hinge
d) all of the above
94. For any type of loading, the no. of point of contra flexure in a simply supported beam is
a) one b) two
c) zero d) three
95. The moment diagram for a cantilever carrying the concentrated load at its free end will be
a) triangle b) rectangle
c) parabola d) cubic parabola.
96. Shear force for a cantilever carrying a uniformly distributed load over its whole length is
a) triangle b) rectangle
c) parabola d) cubic parabola.
97. When a rectangular beam is loaded longitudinally, shear develops
a) bottom fibre b) top fibre
c) neutral axis d) every horizontal plane
98. When a rectangular beam is loaded transversely, the maximum compressive force develops on
a) bottom fibre b) top fibre
c) neutral axis d) every horizontal plane
99. If the shear force along a section of a beam is zero, the bending moment at the section is
a) zero b)minimum
c) maximum d) all of the above
100. The moments in the arch will be zero, if
a) ends are hinged
b) ends are fixed
c) the arch axis coincides with the line of thrust
d) the arch axis is parallel to the line of thrust
101. The arch meant for supporting uniformly distributed loads, to avoid any bending moment must be
a) circular b) elliptical
c) parabolic d) none of the above
102. In any determinate structure, vertical load fall on the
a) hinge b) rollar
c) fixed d) all of the above
103. Three hinged arch is a …..structure.
a) indeterminate
b) determinate
c) both (a) and (b) of above
d) none of the above
104. Three hinged arch will have three hinges
a) two at the two ends and one any where in between the two
ends
b) two at the two ends and one at the crown only
c) one hinge at the crown essentially and the other two any
where
d) none of the above
105. The value of horizontal thrust produced in a arch is
a) b)
c) d)
106. The amount of bending moment induced in a cable is
a) unity b) zero
c) two d) all of the above | 4,742 | 22,479 | {"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-2022-21 | longest | en | 0.594328 |
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# What is more 2^600 or 6^200
Author Message
TAGS:
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6
Kudos [?]: 76 [0], given: 0
What is more 2^600 or 6^200 [#permalink] 30 May 2003, 02:17
00:00
Difficulty:
(N/A)
Question Stats:
33% (01:52) correct 67% (00:43) wrong based on 7 sessions
What is more 2^600 or 6^200?
Manager
Joined: 25 May 2003
Posts: 54
Followers: 1
Kudos [?]: 3 [3] , given: 0
3
KUDOS
2^600 is greater
2^600=2^(6*100)=64^100
6^200=6^(2*100)=36^100
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6
Kudos [?]: 76 [0], given: 0
skoper wrote:
2^600 is greater
2^600=2^(6*100)=64^100
6^200=6^(2*100)=36^100
Manager
Joined: 08 Apr 2003
Posts: 149
Followers: 1
Kudos [?]: 8 [0], given: 0
its comparison between
2^400 and 3^200
So the ealier one is greater...
Manager
Joined: 08 Apr 2003
Posts: 149
Followers: 1
Kudos [?]: 8 [0], given: 0
Hey skoper, that's a neat way... greatoo
Intern
Joined: 20 May 2009
Posts: 39
Followers: 0
Kudos [?]: 3 [0], given: 5
Re: 2^600 [#permalink] 26 Jul 2009, 13:34
2^600
6 = roughly 2^2.5
so 2.5 * 200 = 500
500 < 600
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5
Kudos [?]: 115 [0], given: 39
Re: 2^600 [#permalink] 26 Jul 2009, 14:37
stolyar wrote:
What is more 2^600 or 6^200?
2^600 as compared to 2^200*3^200
compare
2^400,3^200,
(2^4)^100 to (3^2)^100
16^100 to 9^100 surely 16^100 is bigger
Director
Joined: 05 Jun 2009
Posts: 852
WE 1: 7years (Financial Services - Consultant, BA)
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Kudos [?]: 207 [1] , given: 106
Re: 2^600 [#permalink] 27 Jul 2009, 00:22
1
KUDOS
I think more easier comparison will be
2^600 = (2^3)^200 = 8^200 > 6^200
_________________
Consider kudos for the good post ...
My debrief : journey-670-to-720-q50-v36-long-85083.html
Manager
Joined: 03 Jul 2009
Posts: 106
Location: Brazil
Followers: 3
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Re: 2^600 [#permalink] 28 Jul 2009, 13:07
Yes sudeep, in my opinion it is faster, +1
Re: 2^600 [#permalink] 28 Jul 2009, 13:07
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Display posts from previous: Sort by | 1,166 | 3,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2015-18 | longest | en | 0.825465 |
http://www.webdeveloper.com/forum/printthread.php?t=284081&pp=15&page=1 | 1,521,512,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647251.74/warc/CC-MAIN-20180320013620-20180320033620-00576.warc.gz | 482,208,792 | 4,858 | # [RESOLVED] New to JavaScript totally stumped
• 09-26-2013, 08:33 AM
newcodette
[RESOLVED] New to JavaScript totally stumped
Hi everyone :)
I am trying to build a JavaScript form using radio buttons and trying to achieve a
watering time calculator using 2 sets of information.
1. 4 different systems
2. 3 different types
All systems + type1 = the variable totals represented below
All systems + type2 = are multiplied by 2
All systems + type 3 = are multiplied by 3.5
If the 'largeplants' check box is selected the total needs to be multiplied by 1.5
The calculateTime function needs to calculate the watering time on click & display an alert box stating the watering time
& The reset button needs to reset the radio buttons.
<!doctype html>
<html>
<meta charset="utf-8">
<title>Perfect Watering Results</title>
<script>
//Calculate Watering Time for Plants
function calculateTime (){
var system1 = 6; //watering time for sand with jetspray
var system2 = 11;//watering time for sand with sprinkler
var system3 = 17;//watering time for sand wih minisprinker
var system4 = 50; //watering time for sand with dripper
var largeplants = * 1.5; //multiply by 1.5 if checkbox is selected
var type1 //watering time for sand with all systems
var type2 = * 2; //watering time for loam with jetspray/sprinkler/minisprinker/dripper
var type3 = * 3.5;//watering time for clay with jetspray/sprinkler/minisprinker/dripper
// Get the number of the selected answer (1, 2, 3 or 4)
for (var i = 1; i <= 4; i++) {
// Get the number of the selected answer (1, 2, or 3)
for (var i = 1; i <= 3; i++) {
}
</script>
<body>
<h1>Home Garden Watering Guide</h1>
<p>Complete the form below to calculate the recommended watering time for your home garden.<br>
<br>
<br>
<form>
<p>Watering System</p>
<label for="jet">Jet Spray</label><br>
<label for="sprink">Sprinkler</label><br>
<label for="mini">Mini Sprinkler</label>
<br>
<label for="drip">Dripper</label>
</p>
</p>
</form>
<form>
<input type="checkbox" id="largeplants" name="plants"
value="largeplants" onClick=" * 1.5 ">
</form>
<form>
<p>Soil Type</p>
<label for="sand">Sand</label><br>
<label for="loam">Loam</label><br>
<label for="clay">Clay</label></form>
<br><br>
<input type="submit" value="Calculate" id="calculateTime"
onClick="calculateTime" alert("The recommended watering time is "" minutes"> <input type="reset" value="Reset">
</body>
</html>
I know I haven't got through much so any and all help and hits to achieving this will be greatly appreciated.
Also please clarify if I do need to use a loop in this instance :)
Thanks heaps
• 09-26-2013, 08:44 AM
vwphillips
## X vBulletin 4.2.2 Debug Information
• Page Generation 0.08764 seconds
• Memory Usage 2,364KB
• Queries Executed 11 (?)
Template Usage (19):
• (1)footer
• (1)gobutton
• (1)navbar_moderation
• (1)navbar_noticebit
• (2)option
• (1)spacer_close
• (1)spacer_open
Phrase Groups Available (3):
• global
• postbit
Included Files (19):
• ./global.php
• ./includes/class_bootstrap.php
• ./includes/init.php
• ./includes/class_core.php
• ./includes/config.php
• ./includes/functions.php
• ./includes/class_friendly_url.php
• ./includes/class_hook.php
• ./includes/class_bootstrap_framework.php
• ./vb/vb.php
• ./vb/phrase.php
• ./includes/functions_calendar.php
• ./includes/class_bbcode_alt.php
• ./includes/class_bbcode.php
• ./includes/functions_bigthree.php
• ./includes/functions_notice.php
Hooks Called (41):
• init_startup
• init_startup_session_setup_start
• database_pre_fetch_array
• database_post_fetch_array
• init_startup_session_setup_complete
• global_bootstrap_init_start
• global_bootstrap_init_complete
• cache_permissions
• fetch_foruminfo
• global_state_check
• global_bootstrap_complete
• global_start
• style_fetch
• global_setup_complete
• bbcode_fetch_tags
• bbcode_create
• bbcode_parse_start
• bbcode_parse_complete_precache
• bbcode_parse_complete
• cache_templates
• cache_templates_process
• template_register_var
• template_render_output
• fetch_template_start
• fetch_template_complete
• parse_templates | 1,121 | 4,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-13 | latest | en | 0.62001 |
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Question: 1
Light year is the unit of
(A) density
(B) time
(C) distance
(D) both length and time
Ans: C
distance
Question: 2
Which of the following is a derived unit?
(A) area
(B) mass
(C) length
(D) time
Ans: A
area
Question: 3
Two spheres have equal mass and volume in the ratio 2:1. The ratio of their density is
(A) 1:2
(B) 1:4
(C) 2:1
(D) 4:1
Ans: C
2:1
Question: 4
SI unit of density is
(A) kg/m
(B) g/m3
(C) kg/m2
(D) kg/m3
Ans: D
kg/m3
Question: 5
Which of the following is correct?
(A) 1 L = 1cc
(B) 1L = 10cc
(C) 1L = 100cc
(D) 1L = 1000cc
Ans: D
1L = 1000cc
Question: 6
The amount of space occupied by a three dimensional object is known as its
(A) area
(B) mass
(C) volume
(D) density
Ans: C
volume
Question: 7
The largest distance between the earth and the sun is called as _____ position.
(A) aphelion
(B) normal
(C) perihelion
(D) none of the above
Ans: A
aphelion
Question: 8
1 litre = ____ cc.
(A) 0.1
(B) 10
(C) 100
(D) 1000
Ans: D
1000
Question: 9
The unit of volume is
(A) km
(B) cm3
(C) m2
(D) m3
Ans: D
m3
Question: 10
The shortest distance between the earth and the sun is called as _____ position.
(A) aphelion
(B) perihelion
(C) light year
(D) normal
Ans: B
perihelion | 482 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.888123 |
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# How many time does 43 go into 100?
Wiki User
2017-01-05 19:36:43
Twice... with 14 left over !
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100 ÷ 43 = 2 remainder 14 | 140 | 307 | {"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-2021-49 | latest | en | 0.743443 |
https://tradingwuul.web.app/altmiller61494tub/how-do-you-calculate-annual-occupancy-rate-hypy.html | 1,642,918,957,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00694.warc.gz | 640,986,590 | 5,656 | ## How do you calculate annual occupancy rate
Formula: Vn = (Q/Hd) x 100 where Q is the monthly (yearly) sum of occupied rooms Occupancy rates for houses, chalets, etc., can be calculated like rooms. Occupancy rate; Revenue per available room. Now, let's learn about them! Key metric 1: Average room rate / average daily rate. The average The occupancy rate is calculated as the number of beds effectively occupied The Annual Return of Hospitals Database was transferred from Statistics Canada
Occupancy rate; Revenue per available room. Now, let's learn about them! Key metric 1: Average room rate / average daily rate. The average The occupancy rate is calculated as the number of beds effectively occupied The Annual Return of Hospitals Database was transferred from Statistics Canada I am trying to figure out how ALOS, BOR & TOI should be calculated and Bed occupancy rate (BOR): The occupancy rate is a measure of utilization of the 19 Oct 2015 For Hotels, the average annual occupancy should be based on the number of rooms filled. At this time there is no adjustment for the occupancy
## Formula: Vn = (Q/Hd) x 100 where Q is the monthly (yearly) sum of occupied rooms Occupancy rates for houses, chalets, etc., can be calculated like rooms.
5 Jan 2016 For example, if a lender was underwriting this property and calculated a 98.9% breakeven occupancy, it would then be compared to the market 31 Jan 2017 RevPAR = Average Daily Room Rate x Occupancy Rate you can turn your focus to them and figure out a way to boost their profitability. 25 Oct 2013 Occupancy levels of the shopping centers were determined using a variety of methods, in how occupancies were calculated for several shopping City have been higher overall it is still down from when this annual study 1 Dec 2016 On the other end of that equation, hoteliers must factor in costs, such as how brands “calculate their reward program stay incentives based on the 30 Mar 2019 Calculating hourly occupancy of a medical clinic (based on a start and end and end time, step 3 and 4 are the problem which I cant figure out.
### Gross occupancy rate is calculated by dividing the number of rooms occupied in a given month by the total number of rooms, irrespective whether the rooms are
How do you calculate Occupancy? Formula: Occupancy = Rooms Sold / Room Available. Occupancy. ADR. ADR stands for: Average Daily Rate. It is Otherwise, I'll probably just think you're from a neighboring property trying to figure out comp-set rates (Competitive set rates local to that hotel). 24 views. We have created Post Occupancy Rate Calculator with predefined formulas. Enter 2 details and it will automatically calculate the Occupancy Rate for you. 14 Nov 2018 Vacancy rate is calculated by multiplying the number of vacant units by rate factors into the NOI, which is used to calculate the annual return. 8 Oct 2019 This is typically calculated on an annual basis and is shown as a percentage. There are three types of vacancy rates that are commonly used Formula: Vn = (Q/Hd) x 100 where Q is the monthly (yearly) sum of occupied rooms Occupancy rates for houses, chalets, etc., can be calculated like rooms. Occupancy rate; Revenue per available room. Now, let's learn about them! Key metric 1: Average room rate / average daily rate. The average
### We have created Post Occupancy Rate Calculator with predefined formulas. Enter 2 details and it will automatically calculate the Occupancy Rate for you.
occupancy levels above 85% can expect to have regular bed shortages, periodic calculated. Mortality - 90% and over versus <80%. 719. (1 study) in-hospital more than 100,000 admissions annually to the 3 hospitals of which 93,190 were.
## Your property occupancy rate is one of the most important indicators of success. It is calculated by dividing the total number of rooms occupied by the total number
We have created Post Occupancy Rate Calculator with predefined formulas. Enter 2 details and it will automatically calculate the Occupancy Rate for you.
How do you calculate Occupancy? Formula: Occupancy = Rooms Sold / Room Available. Occupancy. ADR. ADR stands for: Average Daily Rate. It is Otherwise, I'll probably just think you're from a neighboring property trying to figure out comp-set rates (Competitive set rates local to that hotel). 24 views. We have created Post Occupancy Rate Calculator with predefined formulas. Enter 2 details and it will automatically calculate the Occupancy Rate for you. 14 Nov 2018 Vacancy rate is calculated by multiplying the number of vacant units by rate factors into the NOI, which is used to calculate the annual return. 8 Oct 2019 This is typically calculated on an annual basis and is shown as a percentage. There are three types of vacancy rates that are commonly used Formula: Vn = (Q/Hd) x 100 where Q is the monthly (yearly) sum of occupied rooms Occupancy rates for houses, chalets, etc., can be calculated like rooms. | 1,102 | 4,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-05 | latest | en | 0.956587 |
http://oeis.org/A139137 | 1,604,109,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00439.warc.gz | 72,389,676 | 4,518 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A139137 Expansion of phi(q) / phi(q^3) in powers of q where phi() is a Ramanujan theta function. 8
1, 2, 0, -2, -2, 0, 4, 4, 0, -6, -8, 0, 10, 12, 0, -16, -18, 0, 24, 28, 0, -36, -40, 0, 52, 58, 0, -74, -84, 0, 104, 116, 0, -144, -160, 0, 198, 220, 0, -268, -296, 0, 360, 396, 0, -480, -528, 0, 634, 694, 0, -832, -908, 0, 1084, 1184, 0, -1404, -1528, 0, 1808, 1964, 0, -2316, -2514, 0, 2952, 3196 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700). LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Eric Weisstein's World of Mathematics, Ramanujan Theta Functions FORMULA Expansion of f(q, -q^2) / f(-q, q^2) in powers of q where f(,) is Ramanujan's two-variable theta function. - Michael Somos, Apr 04 2015 Expansion of eta(q^2)^5 * eta(q^3)^2 * eta(q^12)^2 / (eta(q)^2 * eta(q^4)^2 * eta(q^6)^5) in powers of q. Euler transform of period 12 sequence [ 2, -3, 0, -1, 2, 0, 2, -1, 0, -3, 2, 0, ...]. G.f. is a period 1 Fourier series which satisfies f(-1 / (12 t)) = 3^(1/2) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A132002. - Michael Somos, Apr 04 2015 G.f.: (Sum_{k in Z} x^k^2) / (Sum_{k in Z} x^(3*k^2)). G.f.: Product_{k>0} P(12, x^k)^2 / (P(3, x^k) * P(6, x^k)^3) where P(n, x) is n-th cyclotomic polynomial. Convolution inverse of A132002. - Michael Somos, Apr 04 2015 a(n) = (-1)^n * A252706(n). - Michael Somos, Apr 04 2015 a(3*n + 2) = 0. a(3*n) = A132002(n). a(3*n + 1) = 2 * A139135(n). EXAMPLE G.f. = 1 + 2*q - 2*q^3 - 2*q^4 + 4*q^6 + 4*q^7 - 6*q^9 - 8*q^10 + 10*q^12 + ... MATHEMATICA a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, q] / EllipticTheta[ 3, 0, q^3], {q, 0, n}]; (* Michael Somos, Apr 04 2015 *) PROG (PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^5 * eta(x^3 + A)^2 * eta(x^12 + A)^2 / (eta(x + A)^2 * eta(x^4 + A)^2 * eta(x^6 + A)^5), n))}; CROSSREFS Cf. A132002, A139135, A252706. Sequence in context: A138021 A166065 A252706 * A138231 A155100 A076880 Adjacent sequences: A139134 A139135 A139136 * A139138 A139139 A139140 KEYWORD sign AUTHOR Michael Somos, Apr 10 2008 STATUS approved
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Last modified October 30 20:11 EDT 2020. Contains 338090 sequences. (Running on oeis4.) | 1,142 | 2,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.543173 |
https://www.numbersaplenty.com/813491367 | 1,680,106,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00750.warc.gz | 1,004,850,243 | 3,140 | Search a number
813491367 = 31331672863
BaseRepresentation
bin110000011111001…
…110010010100111
32002200201122100220
4300133032102213
53131223210432
6212415535423
726105363643
oct6037162247
92080648326
10813491367
11388216117
121a852a573
13cc6c7950
147a07bc23
154b63e72c
hex307ce4a7
813491367 has 16 divisors (see below), whose sum is σ = 1205772288. Its totient is φ = 484460640.
The previous prime is 813491359. The next prime is 813491387. The reversal of 813491367 is 763194318.
It is not a de Polignac number, because 813491367 - 23 = 813491359 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (813491387) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 335223 + ... + 337640.
It is an arithmetic number, because the mean of its divisors is an integer number (75360768).
Almost surely, 2813491367 is an apocalyptic number.
813491367 is a deficient number, since it is larger than the sum of its proper divisors (392280921).
813491367 is a wasteful number, since it uses less digits than its factorization.
813491367 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 672910.
The product of its digits is 108864, while the sum is 42.
The square root of 813491367 is about 28521.7700537677. The cubic root of 813491367 is about 933.5071517116.
The spelling of 813491367 in words is "eight hundred thirteen million, four hundred ninety-one thousand, three hundred sixty-seven". | 470 | 1,579 | {"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-2023-14 | latest | en | 0.824183 |
https://mathoverflow.net/questions/385984/degenerations-of-hyperelliptic-coverings | 1,653,035,934,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00712.warc.gz | 452,502,393 | 24,567 | # Degenerations of hyperelliptic coverings
Take six distinct points $$p_1,\dots,p_6\in\mathbb{P}^1$$ and consider the double covering $$f:C\rightarrow \mathbb{P}^1$$ ramified over $$p_1,\dots,p_6\in\mathbb{P}^1$$. Then $$C$$ is a smooth curve of genus two.
Can we degenerate $$C$$ to a singular rational curve or to a union of smooth rational curves by collapsing some of the $$p_i$$ together?
## 1 Answer
If $$p_1 = p_2 \ne p_3 = p_4 \ne p_5 \ne p_6$$ then the normalization of the double cover branched at the divisor $$D = \sum_{i=1}^6 p_i$$ is a smooth irreducible rational curve. If also $$p_5$$ and $$p_6$$ collide, the normalization of the double cover is the union of two smooth rational curves.
• Thank you very much. By embedding the curve in $\mathbb{P}(1,1,3)$ I can see that in the second case the curve degenerates to a union of two rational curves and that in the first case we get an irreducible curve with two singular points. But why is this last curve rational?
– user125056
Mar 9, 2021 at 20:05
• The double covering branched along the zero locus of a polinomial $f(x)$ is given by the equation $y^2 = f(x)$. If $f(x)$ has a root of multiplicity 2 (say at $x = 0$) then (etale) locally the double covering is given by $y^2 = x^2$. So, locally, it has two branches and each maps isomorphically to the base. Therefore, after normalization the morphism is not ramified over this point. Thus, when $f$ has degree 6 with two roots of multiplicity 2 and two roots of multiplicity 1, the normalization is a double covering branched only at two points, hence rational. Mar 10, 2021 at 6:42 | 475 | 1,606 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-21 | latest | en | 0.892959 |
https://www.trustudies.com/question/1194/a-child-has-a-die-whose-six-faces-sho/ | 1,679,419,894,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00333.warc.gz | 1,149,586,008 | 9,524 | 3 Tutor System
Starting just at 265/hour
# A child has a die whose six faces show the letters as given below: The die is thrown once. What is the probability of getting (i) A? (ii) D?
The total number of possible outcomes (or events) = 6
$$P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes}$$
(i) The total number of faces having A on it = 2
P (getting A) $$= \ \frac{2}{6} \ = \ \frac{1}{3} \ = \ 0.33$$
(ii) The total number of faces having D on it = 1
P (getting D) $$= \ \frac{1}{6} \ = \ 0.166$$ | 180 | 542 | {"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} | 4.0625 | 4 | CC-MAIN-2023-14 | longest | en | 0.854419 |
https://accrabookfest.com/top/how-many-words-is-the-first-harry-potter-book.html | 1,638,628,634,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00016.warc.gz | 156,919,160 | 19,883 | # How many words is the first harry potter book
1,084,170 words
## How long is the first Harry Potter book?
Harry Potter and the Philosopher’s StoneCover for one of the earliest UK editionsAuthorJ. K. RowlingPages223 (UK Edition) 332 (2014 UK Edition) 309 (US Edition) 336 (2013 US Edition) 256 (Illustrated Edition)ISBN0-7475-3269-9Followed byHarry Potter and the Chamber of Secrets
## Is 50000 words enough for a novel?
NOVEL: A manuscript over 40,000 words is considered to be a novel. However, very few novels these days are as short as that. Generally a 50,000-word novel would be the minimum word count. Most novels are between 60,000 and 100,000 words.
## How many pages is 70 000 words in a book?
It depends on the font you are using, of course, but in general, 250-300 words per page. Therefore, a 55,000 word book should be about 200 manuscript pages. A 100,000 word book would be about 400. Editors like 12 point font.
## What is the longest book ever written?
A la recherche du temps perdu
## How many unique words are in Harry Potter?
VocabularyBook, author, yearUnique wordsWordsThe Adventures of Tom Sawyer by Mark Twain (1876)7,89671,122The Hobbit by JRR Tolkien (1937)6,91196,072The Lion, The Witch, and The Wardrobe by C.S. Lewis (1950)3,52039,166Harry Potter and The Sorcerer’s Stone by J.K. Rowling (1998)6,18577,883
## How did Hermione die?
On April 16, Riddle sets a mountain troll made immune to sunlight on Hermione to kill her. Harry and Fred and George Weasley come help her.
## Is Hermione Voldemort’s daughter?
Hermione isn’t Voldemort’s daughter. She is the daughter of “Wendell Wilkins” and “Monica Wilkins” (after she modified their memories). Delphini Riddle is Voldemort’s and Bellatrix Lestrange’s daughter.
You might be interested: Question: When to start taking birth control?
## How is Snape the Half Blood Prince?
Snape was born to Eileen Prince, a witch, and Tobias Snape, a Muggle, making him a half-blood (hence the name, “Half-Blood Prince”). This is rare for a Death Eater, as remarked in the last book, though Voldemort himself also had a Muggle father. … Snape was very eager to leave his home to go to Hogwarts.
## How short can a novel be?
A. There are general guidelines for each literary category: Short stories range anywhere from 1,500 to 30,000 words; Novellas run from 30,000 to 50,000; Novels range from 55,000 to 300,000 words, but I wouldn’t recommend aiming for the high end, as books the length of War & Peace aren’t exactly the easiest to sell.
## Is 60000 words too short for a novel?
For most publishers, a book is “novel-length” when it’s between 50,000 and 110,000 words. At a writers conference I recently attended, publishing veteran Jane Friedman said 80,000 words is good for most fiction, below 60,000 isn’t novel length territory, and above 120,000 is likely too much.
## How many pages is 50000 words?
50,000 words is 100 pages single spaced, 200 pages double spaced. 60,000 words is 120 pages single spaced, 240 pages double spaced. 70,000 words is 140 pages single spaced, 280 pages double spaced.
## How many pages is 90000 words?
Here’s how 90,000 words (which is 360 pages at 250 words per page) breaks down by act: Act One: 90 pages (22,500 words)7 мая 2018 г.
## How many book pages is 60000 words?
In the days of the typewriter, a double-spaced page with 1-inch margins would hold an average of 250 words. So you could assume that since 4 pages = 1000 words, 240 pages = 60,000 words, which was the typical length for most mainstream and mystery novels. | 945 | 3,560 | {"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-2021-49 | latest | en | 0.902545 |
https://articlebiz.com/article/1051622646-using-an-apr-calculator-for-auto-purchase-decisions | 1,708,536,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00343.warc.gz | 115,342,642 | 9,221 | ### Using an APR Calculator for Auto Purchase Decisions
• Author Laura Ginn
• Published January 17, 2014
• Word count 729
Buying a new car can be exciting. However, the value of a car depreciates beginning with the moment you drive it off the lot. As a result, if you purchase a car with unfavorable financing terms, you could be paying for a car long after its resale value has decreased below what you are paying for the car. Other factors may contribute to making a car a better or worse overall bargain. Using an annual percentage rate (APR) calculator to compute your monthly payments can help you determine whether the financing terms that you are being offered are actually favorable, or if the purchase itself is a good idea.
Elements of an Auto Loan
When you purchase a new car, you must look beyond the sticker price to determine the actual cost that you will pay. For instance, the elements of a typical auto loan include not only the sticker price (principle) but also the APR, the length of the loan, the value of any down payment or trade-in and the amount of the balloon payment due at the end of the loan term, if any. Altering the amounts of any of the elements included in the loan will have an affect on the other elements. Altering certain elements of an auto loan can also determine the total price that you pay for your car.
Bursting the Balloon (Payment)
For instance, many car loans include a large balloon payment at the end of the loan term. The inclusion of a balloon payment reduces the size of regular monthly payments that you must make. The principle behind including a balloon payment is that you will have more time to accumulate the funds necessary to satisfy that obligation and therefore own the car free and clear. However, without proper planning, you may find yourself at the end of your loan term with a seriously devalued car and facing a payment that you cannot afford. By using estimates that include results generated by an APR calculator, you can compute how much you should increase your down payment or monthly installment payments to reduce the looming balloon payment to something closer to a regular monthly installment payment.
Using an APR Calculator to Computer a Faster Payoff
Making a larger down payment, or receiving a larger credit for the trade-in for your old car can also allow you to pay off your car loan faster, and may reduce the total amount that you pay for the car. You can use an APR calculator along with a loan calculator to plug in different down payment and monthly payment amounts to compare how of an affect each change will make. You may find that the additional amount you would have to offer as a down payment or make in monthly installments is more than you can afford in relationship with the rest of your budget. On the other hand, you may be pleasantly surprised to learn that with modest adjustments in your down payment or monthly installments that you can pay off your car completely several months earlier than the stated terms of your car loan.
Other Auto-Related Costs
In determining the true cost of your car, you must account for related costs that are not included in the terms of your loan. For instance, license tags, insurance, maintenance, petrol and parking fees all add to the cost of owning and operating a car, and yet none are typically included in either the sticker price or the terms of the car loan. However, you can account for at least some of these expenses in a loan calculator along with an APR calculator to determine the true monthly cost you will incur for a particular car.
To construct an expanded loan calculator along with an APR calculator, determine the yearly insurance fee for covering the car you want to buy, the average petrol mileage plus a rough estimate of your monthly driving mileage, average estimated costs for regular maintenance such as oil changes and tune-ups and the average yearly cost for any parking or garage fees. You can even make side-by-side comparisons of the ownership costs associated with two or more different cars. The results of your comparison may reveal that despite its lower sticker price, the petrol-guzzling used car you were considering may actually be more expensive in the long run than a fuel-efficient new car.
Chris Blank appreciates how useful an APR calculator can be when it comes to working out how much you can afford to pay for your loan. Visit uSwitch.com/credit-cards/guides/what-is-apr/ to find out more about how to find an affordable loan.
Article source: https://articlebiz.com | 912 | 4,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.957799 |
https://shikshika.org/solutions/ncert-12-maths-solutions-of-relations-and-functions-ex-1-4/ | 1,670,291,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00680.warc.gz | 548,763,191 | 43,671 | # NCERT 12 Maths Solutions of Relations and Functions – Chapter 1 Ex 1.4
NCERT 12 Maths Solutions of Relations and Functions – Chapter 1 Ex 1.4
Q 1. Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
i. On Z+, defined * by a* b = a- b
ii. On Z+, defined * by a* b = ab
iii. On R, defined * by a* b = ab2
iv. On Z+, defined a* b = |a-b|
v. On Z+, defined * by a* b = a
Sol. (i) On Z+, * defined * by a* b = a- b
It is not a binary operation as the image of (1, 2) under * is 1*2 = 1-2 = -1∉ Z+
(ii) On Z+, * is defined by a*b = ab
It is seen that for each a, b ϵ Z+, there is a unique element ab in Z+. This means that * carries each pair (a, b) to a unique element
a*b = ab in Z+
Therefore, * is a binary operation.
(iii) On R, * is defined by a*b = ab2
It is seen that for each a, b ϵ R, there is a unique element ab2 in R.
This means that * carries each pair (a, b) to a unique element a*b = ab2 in R. Therefore, * is binary operation.
(iv) On Z+, * is defined by a*b = |a-b|
It is seen that for each a, b ϵ Z+ there is a unique element |a-b| in Z+
This means that * carries each pair (a, b) to a unique element a*b = |a-b| in Z+. Therefore, * is a binary operation.
(v) On Z+, * is defined by a*b = a
It is seen that for each a, b ϵ Z+, there is a unique element a ϵ Z+
This means that * carries each pair (a, b) to a unique element
a*b = a in Z+
Therefore, * is a binary operation
Q 2. For each binary operation * defined below, determine whether * is binary, commutative or associative.
i. On Z, define a * b = a- b
ii. On Q, define a * b = ab + 1
iii. On Q, define a * b = ab/2
iv. On Z+, define a * b = 2ab
iv. On Z+, define a * b = ab
v. On R – {-1}, define a * b = a/b+1
Sol. (i) On Z, * is defined by a*b = a-b
a-b ϵ Z, so the operation * is binary
It can be observed that 1*2 = 1-2 = -1 and 2*1 = 2-1 = 1
Therefore, 1*2 ≠ 2*1, where 1, 2 ϵ Z
Hence, the operation * is not commutative
Also, we have (1*2)*3 = (1-2)*3 = -1*3 = -1-3 = -4
1*(2*3) = 1*(2-3) = 1* -1 = 1-(-1) = 2
Therefore, (1*2)*3≠ 1*(2*3), where 1, 2, 3 ϵ Z
Hence, the operation * is not associative
(ii) On Q, * is defined by a*b = ab+1
ab+1 ϵ Q, so operation * is binary
It is known that
ab = ba for a, b ϵ Q
Therefore, ab + 1 = ba + 1 for a, b ϵ Q
Therefore, a*b = a*b for a, b ϵ Q
Therefore, the operation * is commutative. It can be observed that
(1*2)*3 = (1×2+1)*3 = 3*3 = 3×3 +1 = 10
1*(2*3) = 1*(2×3+1) = 1*7+1 = 8
Therefore, (1*2)*3 ≠1*(2*3), where 1, 2, 3 ϵ Q
Therefore, the operation * is not associative
(iii) On Q, * is defined by a* b = ab/2
ab/2 ϵ Q, so the operation * is binary. It is known that
ab= ba for a, bϵ Q
Therefore, ab/2 = ba/2 for a, b ϵ Q
Therefore, a*b = b*a for a, bϵ Q
Therefore, the operation * is commutative
For all a, b, c ϵ Q, we have
Therefore, (a*b)*c = a*(b*c)
Therefore, the operation * is associative
(iv) On Z+, * is defined by a*b = 2ab
2abϵ Z+, so the operation * is binary operation
It is known that
ab ba for a, b Z
Therefore, 2ab = 2ab for a, b ϵ Z+
Therefore, a*b = b*a for a, bϵ Z+
Therefore, the operation * is commutative. It can be observed that
(1*2)*3 = 2(1×2)*3 = 4*3 = 24×3 = 212
1*(2*3) = 1*22×3 =1×26= 1*64 = 264
Therefore, (1*2)*3≠ 1*(2*3), where 1, 2, 3 ϵ Z+
Therefore, the operation * is not associative.
(v) On Z+, * is defined by a*b = ab
ab ϵ Z+, so the operation * is binary operation
It can be observed that 1*2 = 12 = 1 and 2*1 = 21 = 2
Therefore, 1*2 ≠ 2*1, where 1, 2,ϵ Z+
Therefore, the operation * is not commutative
It can be observed that
(2*3)*4 = 23*4 = 8*4 = 84 = (3)4 = 212
2*(3*4) = 2*34 = 2*81 = 281
Therefore, (2*3)*4≠2*(3*4) ; where 2, 3, 4 ϵ Z+
Therefore, the operation * is not associative
(vi) On R-(-1), * is defined by a*b = a/b+1
a/b+1ϵ R for b ≠-1, so that operation * is binary
It can be observed that
1*2 = 1/2+1=1/3 and 2*1 = 2/1+1=2/2=1
Therefore, 1*2 ≠ 2*1 where 1, 2 ϵ R-{-1}
Therefore, the operation * is not commutative
It can also be observed that
Therefore, (1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ R-{-1}
Therefore, the operation * is not associative
Q 3. Consider the binary operation ^ on the set {1, 2, 3, 4, 5} defined by a^b = min{a, b}. Write the multiplication table of the operation.
Sol. The binary operation ^ on the set {1, 2, 3, 4, 5} in defined as
a^b = min{a, b} for a, b ϵ {1, 2, 3, 4, 5}
Thus, the operation table for the given operation ^ can be given as
^ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5
Q 4. Consider a binary operation * on set {1, 2, 3, 4, 5} given by the following multiplication table:
i .Compute (2*3)*4 and 2*(3*4)
ii. Is * commutative?
iii. Compute (2*3)*(4*5)
* 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5
Sol. (i) We have (2*3)*4 = (1)*4=1
And 2*(3*4) = 2*1 = 1
(ii) For every a, b ϵ {1, 2, 3, 4, 5}, we have a*b=b*a
Therefore, the operation * is commutative
(iii) We have (2*3) = 1 and (4*5) = 1
Therefore, (2*3)*(4*5) = 1*1=1
Q 5. Let * be the binary operation of the set {1, 2, 3, 4, 5} defined by a*b = HCF of a and b. Is the operation * is the same as the operation * defined in Q.4 above? Justify your answer.
Sol. The binary operation *, on the set {1, 2, 3, 4, 5} is defined as a*b = HCF of a and b, the operation table for the operation* can be given as
* 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5
We observe that the operation table for the operation * and operation * in Q.4 are the same. Thus, the operation * is the same as the operation *.
Q 6. Let * be the binary operation on N given by a*b = LCM of a and b.
i. Find 5*7, 20*16
ii. Is * commutative?
iii. Is * associative?
iv. Find the identity of * in N
v. Which elements of N are invertible for the operation *?
Sol. The binary operation * on N is defined as a*b = LCM of a and b
i. We have 5*7 = LCM of 5 and 7 = 35
and 20*16 = LCM of 20 and 16 = 80
ii. It is known that
LCM of a and b = LCM of b and a for a, b ϵ N
Therefore, a*b = b*a. Thus, the operation * is commutative
iii. For a, b, c ϵ N, we have
(a*b)*c = (LCM of a and b)*c = LCM of a, b, and c
a*(b*c) = a*(LCM of b and c) = LCM of a, b, and c
Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative.
iv. It is known that
LCM of a and 1 = a = LCM of 1 and a, a ϵ N
a*1 = a = 1*a, a ϵ N
Thus, 1 is the identity of * in N
v. An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a*b = e = b*a
Here, e=1. This means that
LCM of a and b = 1 = LCM of b and a
This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.
Q 7. Is * defined on the set A = {1, 2, 3, 4, 5} by a*b = LCM of a and b, a binary operation? Justify your answer.
Sol. The operation * on the set A = {1, 2, 3, 4, 5} is defined as a*b = LCM of a and b. Now, 2*3 = LCM of 2 and 3 = 6. But 6 does not belong to the given set. Hence, the given operation * is not a binary operation.
Q 8. Let * be the binary operation on N defined by a*b = HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Sol. The binary operation * on N is defined as a*b = HCF of a and b.
It is known that HCF of a and b = HCF of b and a for a, b ϵ N.
Therefore, a*b = b*a. Thus, the operation * is commutative.
For a, b, c ϵ N, we have (a*b)*c = (HCF of a and b)* c = HCF of a, b and c
a*(b*c) = a*(HCF of b and c) = HCF of a, b and c
Therefore, (a*b)*c =a*(b*c)
Thus, the operation * is associative
Now, an element e ϵ N will be the identity for the operation * if
a*e = a = e*a, for ∀ a ϵ N.
But this relation is not true for any a ϵ N
Thus, the operation * does not have identity in N.
Q 9. Let * be a binary operation on the set Q of rational number as follows:
i. a*b = a-b
ii. a*b = a2+b2
iii. a*b = a + ab
iv. a*b = (a-b)2
v. a*b = ab/4
vi. a*b = ab2
Find which of the binary operation are commutative and which are associative?
Sol. (i) On Q, the operation * is defined as a*b = a-b
It can be observed that for 2, 3, 4 ϵ Q, we have
2*3 = 2-3 = -1 and 3*2 = 3-2 = 1
2*3 ≠ 3*2
Thus, the operation * is not commutative
It can also be observed that
(2*3)*4 = (-1)*4 = -1-4 = -5
And 2*(3*4) = 2*(-1) = 2-(-1) = 3
2*(3*4) ≠2*(3*4)
Thus, the operation * is not associative
(ii) On Q, the operation * is defined as a*b =
For a, b ϵ Q, we have
a*b = = = b*a
Therefore, a*b = b*a
Thus, the operation * is commutative
It can be observed that
(1*2)*3 = ()*3 = (1+4)*4 = 5*4 = = 41
1*(2*3) = 1*() = 1*(4+9) = 1*13 = = 170
(1*2)*3 ≠1*(2*3) where 1, 2, 3 ϵ Q
Thus, the operation * is not associative
(iii) On Q, the operation * is defined as a*b = a+ ab
It can be observed that
1*2 = 1+1 x 2 = 1+2 = 3, 2*1 = 2 + 2 x 1 = 2+2 = 4
1*2 ≠ 2*1, where 1, 2ϵQ
Thus, the operation * is not commutative
It can be observed that
(1*2)*3 = (1+1×2)*3 = 3*3 = 3+3×3 = 3+9=12
1*(2*3) = 1*(2+2×3)= 1*8 =1+1×8 = 9
(1*2)*3 ≠1*(2*3) where 1, 2, 3 ϵ Q
Thus, the operation * is not associative
(iv) On Q, the operation * is defined by a*b = (a-b)2
For a, b ϵ Q, we have
a*b = (a-b)2 and b*a = (b-a)2 = [-(a-b)]2 = (a-b)2
Therefore, a*b = b*a
Thus, the operation * is commutative. It can be observed that
(1*2)*3 = (1-2)2*3 = (-1)2*3 = 1*3 = (1-3)2 = (-2)2= 4
1*(2*3) = 1*(2-3)2 = 1*(-1)2 = 1*1 = (1-1)2 = 0
(1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ Q
Thus, the operation * is not associative
(v) On Q, the operation * is defined as a*b = ab/4
For a, b ϵ Q, we have a*b = ab/4 = ba/4 = b*a
Therefore, a*b = b*a
Thus, the operation * is commutative
For a, b, c ϵ Q, we have
Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative.
(vi) On Q, the operation * is defined as a*b = ab2
It can be observed that for 2, 3 ϵ Q
2*3 = 2.32 = 18 and 3*2 = 3.22 = 12
Hence, 2*3 3*2
Also,
Thus, the operation * is not commutative.
It can also be observed that 1, 2, 3 ϵ Q
(1*2)*3 = (1.22)*3 = 4*3 = 4.32 = 36
1*(2*3) = 1*(2.32) = 1*18 = 1.182 = 324
(1*2)*3 1*(2*3)
Also,
Thus, the operation * is not associative
Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative.
Q 10. Show that none of the operation given in Q.9 has identity.
Sol. An element e ϵ Q will be the identity element for the operation * if
a*e = a = e*a, a ϵ Q
i. a*b = a-b
If a*e=a, a≠ 0 → a-e = a, a ≠0 → e = 0
Also, e*a =a →e-a =a →e=2a
e=0=2a, a≠0
But the identity is unique. Hence this operation has no identity.
ii. a*b = a2 +b2
If a*e = a, then a2 + e2 = a
For a = -2, (-2)4 +e2 = 4+e2 ≠-2
Hence, there is no identity element.
iii. a*b = a+ab
If a*e =a → a+ae = a→ ae=0 →e=0, a≠
Also if
→ e*a=a → e+ea =a → e = , a≠0
But the identity is unique. Hence, this operation has no identity.
iv. a*b = (a-b)2
If a*e = a, then (a-e)2 = a. A square is always positive, so for
a = -2, (-2, -e)2 ≠ -2
Hence, there is no identity element
v. a*b = ab/4
If a*e = a, then ae/4 = a. Hence, e=4 is the identity element
a*4 = 4*a = 4a/4 = a
vi. a*b = ab2
If a*e = a → ae2 = a
→ e2 =1 → e±1
But identity is unique. Hence this operation has no identity.
Therefore, only part (v) has an identity element.
Q 11. Let A = N x N and * be the binary operation on A defined by (a, b)* (c, d) = (a+c, b+d). Show that * is commutative and associative. Find the identity element for * on A, if any.
Sol. Given that A = N x N
And * is a binary operation on A and is defined by
(a, b)*(c, d) = (a+c, b+d)
Let (a, b), (c, d)ϵ A
Then, a, b, c, d ϵ N
We have (a, b)*(c, d) = (a+c, b+d)
And (c, d)*(a, b) = (c+a, d+b) = (a+c, b+d)
[Addition is commutative in the set of natural numbers]
Therefore, (a, b)*(c, d) = (c, d)*(a, b)
Therefore, the operation * is commutative.
Now, let (a, b)(c, d), (e, f) ϵ A
Then, a, b, c, d, e, f ϵ N
We have, ((a, b)*(c, d))*(e, f) = (a+c, b+d)*(e, f) = (a+c+e, b+d+f)
And (a, b)*((c, d)*(e, f)) = (a, b)*(c+e, d+f) = (a+c+e, b+d+f)
Therefore, ((a, b)*(c, d))*(e, f) = (a, b)*((c, d)*(e, f))
Therefore, the operation * is associative.
An element e = (e1, e2) ϵ A will be an identity element for thye operation * if
a*e = a= e*a ∀ a =(a1, a2) i.e.,
(a1+e1, a2+e2)= (a1, a2) = (e1+a1, e2+a2)
Which is not true for any element in A.
Note that a+e = a for e = 0 but 0 does not belong to N.
Therefore, the operation * does not have any identity element.
Q 12. State whether the following statements are true or false? Justify
i. For an arbitrary binary operation * on a set N, a * a = a ∀ a ϵ
ii. If * is a commutative binary operation on N, then a*(b*c) = (c*b)*a
Sol. i. Define an operation * on N as a*b = a + b ∀ a, b ϵ N
Then, in particular, for b=a=3, we have 3*3 = 3+3 = 6≠3
Therefore, statement (i) is false
(ii) RHS = (c*b)*a = (b*c)*a [* is commutative]
= a*(b*c) [Again, as * is commutative]
=LHS
Therefore, a*(b*c) = (c*b)*a
Therefore, statement (ii) is true.
Q 13. Consider a binary operation * on N defined as a*b = a3+b3. Choose the correct answer.
i. * is both associative and commutative
ii. * is commutative but not associative
iii. * is associative but not commutative
vi. * is neither commutative nor associative
Sol. On N, the operation * is defined as a*b = a3+b3
For a, b ϵ N, we have
a*b = b3 +b3 = b3 a3 = b*a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that
(1*2)*3 = (13+23)*3 = 9*3 = 93+33 = 729 + 27 = 756
(1*(2*3) = 1*(23+33)= 1*(8+27) = 1*35
= 13+353 = 1+(35)3 = 1 + 42875 = 42876
Therefore, (1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ N
Therefore, the operation * is not associative
Hence, the operation * is commutative but not associative
Thus, the correct answer is (b).
Updated: November 17, 2022 — 2:31 pm | 5,637 | 13,988 | {"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} | 4.84375 | 5 | CC-MAIN-2022-49 | latest | en | 0.90515 |
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8th Grade Math App (iOS & Android) | 552 | 2,082 | {"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-2024-38 | latest | en | 0.859957 |
http://www.physicsforums.com/showthread.php?p=3384501 | 1,394,233,093,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999651631/warc/CC-MAIN-20140305060731-00002-ip-10-183-142-35.ec2.internal.warc.gz | 498,125,688 | 7,080 | # A but linear algebra
by Grieverheart
Tags: maximal, rank, string, supersymmetry, theory
P: 31 1. The problem statement, all variables and given/known data Ok, it seems I need to refresh my linear algebra a bit. In the string theory exams, we had a part about space-time supersymmetry of the superparticle. On of the questions was this: Argue that the matrix $\Gamma^{\mu}p_{\mu}$ can have at most half maximal rank upon imposing the equations of motion. (Recall that any matrix can be brought to Jordan normal form, meaning that it is diagonal with at most 1s in places right above the diagonal entries. 2. Relevant equations Equations of Motion (the relevant ones): $\Gamma^{\mu} p_{\mu} \dot{\theta}=0$, where $\dot{\theta}(\tau)$ is an anti-commuting spinor in space-time and has 32 components (we're working in 10 dimensions. 3. The attempt at a solution I write $\Gamma^{\mu}p_{\mu} \dot{\theta}=0$ as (I use 3x3 for simplicity) $\begin{pmatrix} a & 1 & 0\\ 0 & b & 1\\ 0 & 0 & c \end{pmatrix} \begin{pmatrix} \dot{\theta_0}\\ \dot{\theta_1}\\ \dot{\theta_2} \end{pmatrix}=0$ From this I get 3 equations, but I'm not really sure how they affect the rank.
Related Discussions Calculus & Beyond Homework 5 Calculus & Beyond Homework 12 Calculus & Beyond Homework 7 Calculus & Beyond Homework 8 Calculus & Beyond Homework 6 | 380 | 1,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-10 | longest | en | 0.857249 |
https://www.assignmentguruh.com/need-3-pages-paper-apa-format-7-formatted-paper/ | 1,623,887,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626122.27/warc/CC-MAIN-20210616220531-20210617010531-00195.warc.gz | 593,240,177 | 16,636 | # Need 3 pages paper APA format 7 formatted paper
Subject: Applied Statistical Methodology for Business Research Design
Develop and write a 3-5 page APA 7 th ed. Formatted paper:
Using your textbook, LIRN-based research, and the Internet, provide a complete solution to the Case Problem (Specialty Toys) presented at the end of Chapter 6 in the textbook. Include in-text citations.
.
Prepare a managerial report that addresses the following issues and recommends an order quantity for the weather Teddy product. In your report, answer the following five questions:
Use the sales forecaster’s prediction to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation.
Compute the probability of a stockout for the order quantities suggested by members of the management team.
Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000 units, and best case in which sales = 30,000 units.
One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stockouts. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios?
Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation.
Please be sure to include in-text citations and peer reviewed references in APA format in your discussion post.
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 643 | 3,070 | {"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-2021-25 | latest | en | 0.913901 |
http://www.crcpress.com/product/isbn/9780849373794 | 1,369,288,734,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702900179/warc/CC-MAIN-20130516111500-00070-ip-10-60-113-184.ec2.internal.warc.gz | 405,896,740 | 12,557 | ## Numerical Solutions for Partial Differential Equations: Problem Solving Using Mathematica
Series:
Published:
Author(s):
Hardback
\$97.95
ISBN 9780849373794
Cat# 7379
### Features
• Comprehensive - covers basic principles, modern methods, and program implementation
• Self-contained - explains all Mathematica functions used in the text
• Complete - includes the meaning of each line of each program
• Versatile - can be used by researchers, practitioners, and students
• ### Summary
Partial differential equations (PDEs) play an important role in the natural sciences and technology, because they describe the way systems (natural and other) behave. The inherent suitability of PDEs to characterizing the nature, motion, and evolution of systems, has led to their wide-ranging use in numerical models that are developed in order to analyze systems that are not otherwise easily studied. Numerical Solutions for Partial Differential Equations contains all the details necessary for the reader to understand the principles and applications of advanced numerical methods for solving PDEs. In addition, it shows how the modern computer system algebra Mathematica® can be used for the analytic investigation of such numerical properties as stability, approximation, and dispersion.
Introduction to Mathematica
Symbolic Computations with Mathematica
Numerical Computations with Mathematica
Finite Difference Methods for Hyperbolic PDEs
Construction of Difference Schemes for the Advection Equation
The Notion of Approximation
Fourier Stability Analysis
Elementary Second-Order Schemes
Algorithm for Automatic Determination of Approximation Order of Scalar Difference Schemes
Monotonicity Property of Difference Schemes
TVD Schemes
The Construction of Difference Schemes for Systems of PDEs
Implicit Difference Schemes
Von Neumann Stability Analysis in the Case of Systems of Difference Equations
Difference Initial- and Boundary-Value Problems
Construction of Difference Schemes for Multidimensional Hyperbolic Problems
Determination of Planar Stability Regions
Curvilinear Spatial Grids
Finite Difference Methods for Parabolic PDEs
Basic Types of Boundary Conditions for Parabolic PDEs
Simple Schemes for the One-Dimensional Heat Equation
Runge-Kutta Methods
Finite Volume Method
Approximate Factorization Scheme
Dispersion
Numerical Methods for Elliptic PDEs
Boundary-Value Problems for Elliptic PDEs
A Simple Elliptic Solver
The Finite Element Method
Numerical Grid Generation
Local Approximation Study of Finite Volume Operators on Arbitrary Grids
Local Approximation Study of Difference Schemes on Logically Rectangular Grids
Appendix Glossary of Programs
Index
Each Chapter also includes a list of references.
### Editorial Reviews
:"...A completely new edition designed for an era of calculators and handheld computers."
@:- James A. Cargal, Troy State University, Montgomery, Alabama, in The Journal of Undergraduate Mathematics and Its Applications, 1996
@
Resource OS Platform Updated Description Instructions
7379.zip All Windows Version September 14, 2006 | 630 | 3,073 | {"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} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.844786 |
https://www.kylesconverter.com/density/ounces-per-cubic-inch-to-tonnes-per-thousand-cubic-feet | 1,726,230,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00034.warc.gz | 781,682,631 | 5,810 | # Convert Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet
### Kyle's Converter > Density > Ounces Per Cubic Inch > Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet
Ounces Per Cubic Inch (oz/in3) Tonnes Per Thousand Cubic Feet (t/MCF) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Tonnes Per Thousand Cubic Feet to Ounces Per Cubic Inch
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Ounce (avoirdupois) per Cubic Inch:
oz/in3
1 Tonne per Thousand Cubic Feet:
Mass of 1 tonne per volume of a thousand cubic feet. Metric tonne of exactly 1 000 kilograms. International foot length of 0.3048 meters. Approximate density of 35.3146667214886 kilograms per cubic meter. 1 t/MCF ≈ 35.3146667214886 kg/m3.
Link to Your Exact Conversion
Conversions Table
1 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 48.98870 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 3429.1583
2 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 97.97680 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 3919.0381
3 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 146.963990 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 4408.9178
4 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 195.9519100 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 4898.7976
5 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 244.9399200 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 9797.5952
6 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 293.9279300 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 14696.3928
7 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 342.9158400 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 19595.1904
8 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 391.9038500 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 24493.988
9 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 440.8918600 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 29392.7856
10 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 489.8798800 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 39190.3808
20 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 979.7595900 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 44089.1784
30 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 1469.63931,000 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 48987.9759
40 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 1959.51910,000 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 489879.7595
50 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 2449.3988100,000 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 4898797.5949
60 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 2939.27861,000,000 Ounces Per Cubic Inch to Tonnes Per Thousand Cubic Feet = 48987975.949 | 953 | 2,976 | {"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-2024-38 | latest | en | 0.455148 |
https://tutorialprime.com/570916242/ | 1,604,044,739,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00506.warc.gz | 567,382,548 | 11,932 | "> (Answered) You are in charge of obtaining \$250,000 worth of funding for a new project your company is considering. You can issue common equity in any of the... - Tutorials Prime
## (Answered) You are in charge of obtaining \$250,000 worth of funding for a new project your company is considering. You can issue common equity in any of the...
You are in charge of obtaining \$250,000 worth of funding for a new project your company is considering. You can issue common equity in any of the...
You are in charge of obtaining \$250,000 worth of funding for a new project your company is considering. You can issue common equity in any of the following amounts:Option 1: 3000 shares at \$60 per shareOption 2: 2000 shares at \$70 per shareOption 3: 1500 shares at \$80 per shareIn any case, you will finance the rest with private debt. The cost of the private debt is a function of the amount that you borrow. You are assured in any of the scenarios of borrowing at least \$70,000, and at that level, you will have a cost of debt of 6%. For every \$10,000 additional debt beyond that, the cost increases by .5%. The cost of the common stock 9.1% if you borrow \$70,000 (i.e., Option 1), but increases by 1% for Option 2, and another 1% for Option 3. The company’s tax rate is 35%. A) Given this, which of the options above provides the best scenario?WACC: Option 1-WACC: Option 2-WACC: Option 3-B) Suppose the project is expected to generate NCFs of \$75,000 during the first four years and then \$50,000 for the fifth and final year. Given this, what is that Net Present Value of the project? Â
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WE GUARANTEE, THAT YOUR PAPER WILL BE WRITTEN FROM SCRATCH AND WITHIN A DEADLINE. | 674 | 2,733 | {"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} | 2.609375 | 3 | CC-MAIN-2020-45 | longest | en | 0.933823 |
https://www.financialexpress.com/archive/the-basics-of-option-premium-determination/197990/ | 1,585,554,855,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00137.warc.gz | 944,139,758 | 7,457 | # The basics of option premium determination
How are option premiums (prices) determined
While supply and demand ultimately determines the price of options, several factors have a significant impact on option premiums, which are
Spot Price of the underlying assets
The exercise price of the option
The volatility in the underlying markets.
Time remaining to expiration
Risk free rate of interest
Dividend (only for option on equity)
Define breakeven points
Breakeven point is the point at which there is no net loss or gain, one has just broken even. The maximum amount the option buyer can lose is the premium that he originally paid. It is that point where the payoff of the buyer is exactly equal to the amount of premium paid. To calculate the breakeven point on options, one uses the strike price and the premium.
Breakeven point Calculation
Call Option = Strike Price + Premium amount.
Put Option = Strike Price - Premium amount.
What is put/call ratio and how is it used
The put-call ratio is simply the number of puts traded divided by the number of calls traded. It can be computed daily, weekly, or over any time period. It can be computed for stock options, index options, or options on futures. Some market technicians believe that a high volume of puts relative to calls indicates investors are bearish, whereas a high ratio of calls to puts shows bullishness.
What does the term delta mean
Delta measures the rate of change of an option premium with respect to a price change in the underlying asset. Delta is a measure of price sensitivity at any given moment. Not all options move point for point with their underlying asset.
If an underlying asset moves 0.50 points and the option only moves 0.25 points, its delta is 50%; that is, the option is only 50% as sensitive to the movement of the underlying asset.
The delta is between 0 and +1 for calls and between 0 and -1 for puts (thus a call option with a delta of 0.5 will increase in price by 1 tick for every 2 tick increase in the underlying).
How does deep-in-the-money differ from just In-the-money
When someone refers to a deep-in-the-money option they are referring to a call or a put with a delta close to 1.00 (or -1.00 for puts). This option moves very close to a one for one ratio with underlying asset movement up and down, and is often viewed as the equivalent for long or short underlying asset.
Courtesy MCX Training | 520 | 2,421 | {"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-2020-16 | latest | en | 0.926705 |
https://www.slideserve.com/channer/simple-functions | 1,516,482,034,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00189.warc.gz | 985,800,554 | 17,570 | Simple Functions
1 / 41
# Simple Functions - PowerPoint PPT Presentation
Simple Functions. Writing Reuseable Formulas. In Math. Suppose f (x) = 2 x 2 +5 f(5)=? f(5) = 2*5 2 +5 =55 f(7) = 2*7 2 +5 =103. 135. 165. Problem.
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### Simple Functions
Writing Reuseable Formulas
In Math
• Suppose f (x) = 2 x2+5
• f(5)=?
• f(5) = 2*52+5 =55
• f(7) = 2*72+5 =103
135
165
Problem
Using OCD, design and implement a program that computes the area and circumference of an Australian Rules Football field, which is an ellipse that is (ideally) 165m x 135m.
a
b
b
a
Problem Generalization
Using OCD, design and implement a program that computes the area and circumference of an ellipse.
Behavior
Our program should display on the screen its purpose and prompt the user for the length and width of the ellipse. It should then read the length and width from the keyboard. It should then compute and display the ellipse’s area and circumference, along with a descriptive label.
Objects
Description Type Kind Name
screen ostream varying cout
purpose string constant --
prompt string constant --
length double variable length
width double variable width
keyboard istream varying cin
circumference double variable circumference
area double variable area
p double constant PI
a double variable halfLength
b double variable halfWidth
Operations
Description Predefined? Library? Name
display strings yes iostream <<
compute circumf. no -- --
- multiply doubles yes built-in *
- add doubles yes built-in +
- divide doubles yes built-in /
- exponentiation yes cmath pow()
- square root yes cmath sqrt()
compute area no -- --
- multiply doubles yes built-in *
display doubles yes iostream <<
Algorithm
0. Display via cout the purpose of the program plus a prompt for the length and width of an ellipse.
1. Read length, width from cin.
2. Compute halfLength = length/2.0; halfWidth = width/2.0.
3. Compute area = PI * halfLength * halfWidth.
4. Compute circumference = 2.0 * PI * sqrt((halfLength2 + halfWidth2)/2.0).
5. Display area and circumference with descriptive labels.
Coding
#include <iostream> // cin, cout, <<, >>, ...
#include <cmath> // sqrt(), pow(), ...
using namespace std;
int main()
{
const double PI = 3.14159;
cout << "\nTo compute the area and circumference"
<< "\n of an ellipse, enter its length and width: ";
double length, width;
cin >> length >> width;
double halfLength = length/2.0,
halfWidth = width/2.0;
double area = PI * halfLength * halfWidth;
double circumference = 2.0 * PI *
sqrt((pow(halfLength, 2.0) + pow(halfWidth, 2.0))/2.0);
cout << "\nThe area is " << area
<< "\n and the circumference is " << circumference
<< endl;
}
Testing
To compute the area and circumference
of an ellipse, enter its length and width: 165 135
The area is 17494.7
and the circumference is 473.589
Problem
We’ve done a fair amount of work in creating these formulas, but have no way to directly reuse that work if we ever need to use these ellipse-related formulas again...
Solution: Rewrite those formulas as functions.
Using Functions (i)
#include <iostream> // cin, cout, <<, >>, ...
#include <cmath> // sqrt(), pow(), ...
using namespace std;
double EllipseArea(double length, double width);
double EllipseCircumference(double length, double width);
int main()
{
cout << "\nTo compute the area and circumference"
<< "\n of an ellipse, enter its length and width: ";
double length, width;
cin >> length >> width;
double area = EllipseArea(length, width);
double circumference = EllipseCircumference(length,width);
cout << "\nThe area is " << area
<< "\n and the circumference is " << circumference
<< endl;
}
Using Functions (ii)
const PI = 3.14159;
double EllipseArea(double length, double width);
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return PI * halfLength * halfWidth;
}
double EllipseCircumference(double length, double width);
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return 2.0 * PI *
sqrt((pow(halfLength, 2.0) + pow(halfWidth, 2.0))/2.0);
}
Simplify for math functions
const PI = 3.14159;
double f(double x, double y);
{
double a = x/2.0,
b = y/2.0;
return PI * a * b;
}
// f(x,y)= p(x/2)(y/2)
double g(double x, double y);
{
double a = l/2.0,
b = w/2.0;
return 2.0 * PI *
sqrt((pow(a, 2.0) + pow(b, 2.0))/2.0);
}
//g(x,y) = 2 p
Function Prototypes
A function prototype acts as a declaration of the function, allowing it to be called.
A function prototype must precede any call or definition of a function, or a compiler error will occur.
Pattern:
ReturnTypeName (ParameterDeclarations);
Example Prototypes
#include <iostream> // cin, cout, <<, >>, ...
#include <cmath> // sqrt(), pow(), ...
using namespace std;
double EllipseArea(double length, double width);
double EllipseCircumference(double length, double width);
int main()
{
cout << "\nTo compute the area and circumference"
<< "\n of an ellipse, enter its length and width: ";
double length, width;
cin >> length >> width;
double area = EllipseArea(length, width);
double circumference = EllipseCircumference(length,width);
cout << "\nThe area is " << area
<< "\n and the circumference is " << circumference
<< endl;
}
Function Definitions
A function definition contains statements that dictate its behavior when it is called.
A function must be defined in order to be called, or else a linker error will occur.
Pattern:
ReturnTypeName (ParameterDeclarations) { StatementList }
Example Definitions
const double PI = 3.14159;
double EllipseArea(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return PI * halfLength * halfWidth;
}
double EllipseCircumference(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return 2.0 * PI *
sqrt((pow(halfLength, 2.0) + pow(halfWidth, 2.0))/2.0);
}
Parameters
Parameters are function variables for which the caller can specify values.
Parameters are defined between the parentheses of a function’s definition.
double EllipseArea(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return PI * halfLength * halfWidth;
}
165
135
double EllipseArea(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return PI * halfLength * halfWidth;
}
Arguments
When a function is called, its caller can pass it values called arguments which are stored in the function’s parameters.
double area = EllipseArea(165, 135);
The function then runs using its parameter values.
Program Structure
C++ programs typically follow this pattern:
#include directives
Function prototypes
Main function
Function definitions
OCD with Functions
1. Specify the desired behavior of the program.
2. Identify the objects needed.
3. Identify the operations.
a. If any operation is not predefined:
Write a function to perform it.
4. Organize objects and operations into an algorithm.
Problem
Suppose that in order to solve a different problem, a different program needs to compute the area of an ellipse?
Options:
• Copy-and-paste EllipseArea() from our previous program into the new program...
• Store EllipseArea() and EllipseCircumference() in a library module, so that programs can share them.
Library Modules
A library module consists of three files:
• A header file (usually with a .h suffix) containing shareable function prototypes.
• An implementation file (with suffix .cpp) containing shareable function definitions.
• A documentation file (with suffix .doc) containing documentation for the library.
Example
Since we are creating a library to share functions that describe an ellipse, we name our library ellipse, with
• implementation file ellipse.cpp, and
• documentation file ellipse.doc.
ellipse.h
In ellipse.h, we place the function prototypes:
/* ellipse.h
* Date: January 1998.
* Purpose: Functions for computing ellipse attributes.
*/
double EllipseArea(double length, double width);
double EllipseCircumference(double length, double width);
// ... plus any others we want to provide while we’re at it
To use these functions, a program must contain
#include ”ellipse.h”
ellipse.cpp
Their definitions are placed in ellipse.cpp:
/* ellipse.cpp
* Date: January 1998.
* Purpose: Functions for computing ellipse attributes.
*/
#include "ellipse.h"
const double PI = 3.14159;
double EllipseArea(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return PI * halfLength * halfWidth;
}
...
ellipse.cpp (cnt’d)
...
double EllipseCircumference(double length, double width)
{
double halfLength = length/2.0,
halfWidth = width/2.0;
return 2.0 * PI *
sqrt((pow(halfLength, 2.0) + pow(halfWidth, 2.0))/2.0);
}
This file can be compiled separately from any program that uses it.
ellipse.doc
The documentation file is a copy of the header file, plus function specifications:
/* ellipse.doc
* Date: January 1998.
* Purpose: Functions for computing ellipse attributes.
*/
/*********************************************************
* Compute the area of an ellipse. *
* Receive: length, width, two double values. *
* Return: the area of the corresponding ellipse. *
*********************************************************/
double EllipseArea(double length, double width);
ellipse.doc (cnt’d)
// ... first part of ellipse.doc
/*********************************************************
* Compute the circumference of an ellipse. *
* Receive: length, width, two double values. *
* Return: the circumference of ellipse defined by *
* length and width. *
*********************************************************/
double EllipseCircumference(double length, double width);
// ... plus any others we want to provide while we’re at it
By storing the documentation in a separate file, we provide information on how to use the library without cluttering its header file.
#include
When the C++ compiler processes a #include directive, it
• opens the file named in the directive; and
• replaces the #include directive with the contents of that file.
When the file contains function prototypes, the effect of the #include directive is to insert those prototypes into the program.
Translation
Translating a program into machine language consists of two steps:
1. Compilation, in which the program is converted into the computer’s machine language; and
2. Linking, in which any calls to functions defined outside of that program are bound to function definitions.
int main()
{
// ...
}
C++
Compiler
00100111010
10110001001
...
11101100100
file.cpp
file.obj
Translation (cnt’d)
Compilation creates a binaryobject file (usually with a .o or .obj suffix).
Linking binds multiple object files into a single binary executable file that can be run.
00100111010
10110001001
...
11101100100
00100111010
10110001001
...
11101100100
11100101011
10010001000
...
10101101101
01101101011
11010101001
...
00101100100
file1.obj
11100101011
10010001000
...
10101101101
C++
file2.obj
01101101011
11010101001
...
00101100100
fileN.obj
file.exe
Using a Library
To use the library, a program must
• #include its header file (usually above the main function), inserting the function prototypes so that they can be called.
• be linked to its implementation file.
Failure to do either of these produces an error.
Example
#include <iostream> // cin, cout, <<, >>, ...
using namespace std;
#include "ellipse.h" // insert ellipse prototypes
int main()
{
cout << "\nTo compute the area and circumference"
<< "\n of an ellipse, enter its length and width: ";
double length, width;
cin >> length >> width;
double area = EllipseArea(length, width);
double circumference = EllipseCircumference(length,width);
cout << "\nThe area is " << area
<< "\n and the circumference is " << circumference
<< endl;
}
Compiler Errors
A compiler error occurs if a program calls a function for which no prototype is given.
This can occur if you call a library function and neglect to #include the header file containing its prototype.
- You’ll be trying to use a function that has not been declared.
A linker error occurs if a program calls a function for which the linker is unable to find a definition.
This can occur if a program calls a function but the linker is not told to use the library implementation file or object file containing that function’s definition.
How this is done varies from platform to platform, but often involves a project file.
OCD with Libraries
1. Specify the desired behavior of the program.
2. Identify the objects needed.
3. Identify the operations.
a. If an operation is not predefined:
Write a function to perform it.
b. If an operation is likely to be reuseable someday:
Store its function in a library and access it from there.
4. Organize objects and operations into an algorithm.
Summary
Functions provide a programmer with a means of extending C++ with new operations that are not predefined in the language.
A function prototype must precede its call, or a compiler error will occur.
A function definition must be provided, or a linker error will occur.
Libraries provide a means of storing functions so that multiple programs can use them.
Summary (Cnt’d)
• Libraries consist of three files: for function prototypes, definitions, and documentation.
To use a library, a program must
• #include its header file, or a compiler error will occur.
• Be linked to its implementation file (or the object file produced by compiling the implementation file), or a linker error will occur. | 3,390 | 14,335 | {"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} | 3.53125 | 4 | CC-MAIN-2018-05 | latest | en | 0.783145 |
http://www.chegg.com/homework-help/water-15-c-drained-large-reservoir-using-two-horizontal-plas-chapter-14-problem-75p-solution-9780073327488-exc | 1,418,858,064,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802765002.8/warc/CC-MAIN-20141217075245-00150-ip-10-231-17-201.ec2.internal.warc.gz | 526,597,453 | 12,011 | # CHEGG TEXTBOOK SOLUTIONS FOR Fundamentals of Thermal-Fluid Sciences 3rd Edition
## Chapter Problem
PROBLEM
Chapter: Problem:
2 Stars(4 users)
Water at 15°C is drained from a large reservoir using two horizontal plastic pipes connected in series. The first pipe is 20 m long and has a 10-cm diameter, while the second pipe is 35 in long and has a 4-cm diameter. The water level in the reservoir is 18 m above the centerline of the pipe. The pipe entrance is sharp-edged, and the contraction between the two Pipes is sudden. Neglecting the effect of the kinetic energy correction factor, determine the discharge rate of water from reservoir.
STEP-BY-STEP SOLUTION
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Chegg is one of the leading providers of homework help for college and high school students. Get homework help and answers to your toughest questions in math, calculus, algebra, physics, chemistry, science, accounting, English and more. Master your homework assignments with our step-by-step solutions to more than 3000 textbooks. If we don't support your textbook, don't worry! You can ask a homework question and get an answer in as little as two hours. With Chegg, homework help is just a few clicks away. | 447 | 1,895 | {"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-2014-52 | latest | en | 0.903543 |
https://chemistry.stackexchange.com/questions/108505/how-to-distinguish-two-molecules-with-same-formula-but-different-geometry-with-v | 1,713,733,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00286.warc.gz | 141,747,808 | 40,712 | # How to distinguish two molecules with same formula but different geometry with vibrational spectroscopy
There is a sample that we know is all made up by only one of these two molecule:
how you can see they have the same formula but two different geometries. The question is if it is possible to know by which of these two molecules the sample is made with an experiment involving vibrational spectroscopy (IR or Raman). The following is a quite wide answer I gave to myself but I'm not sure it is right, so please correct me or confirm what I have written.
We know that if we are considering small oscillations, the vibrational part of the wave function of a molecule with $$N$$ atoms is the product of $$3N-6$$ quantum armonic oscillators eigenfunctions, and we also know that in a classic way to proceed the nuclei can move in $$3N-6$$ differents normal modes. Fixed one of these normal modes, if it is for example IR active, we know that all the $$3N-6$$ oscillators moving in that normal mode can absorb photons from the external radiation. So if the external radiation has a frequency band enough large to contain all the proper frequencies of the oscillators, we'll notice $$3N-6$$ absorbtion lines. While if we are considering a not IR active normal mode of course we won't see any absorbtion lines. In general the nuclei of a molecule move on a linear combination of all the normal modes. So we can think that each IR active normal mode of the combination gives the same $$3N-6$$ lines while each not IR active mode gives zero lines. So I would expect to observe for my real sample 12 absorbtion lines (I am not considering overtones and the noise of the media) which correspond to the proper frequencies of the oscillators. So since the two molecule have exactly the same oscillators (it only changes their position), I should observe the same 12 lines on both cases, so I'd say that it is not possible to distinguish between the two. The same argument can be used for the Raman. However I'm almost sure that I'm wrong and a way there must exists.
• – MaxW
Jan 24, 2019 at 20:58
• Basically your question is how to identify wich modes of a molecules are IR or Raman active. It seems you are aware. The point is that you wrote "have the same armonic oscillators" which is not true even at a first glance as you have already realised if you have followed the link given above. After that you see how to discern between the two in finer details. Jan 25, 2019 at 9:18
## 1 Answer
The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $$x, \,y, \,z$$ and $$xy, \,xz$$ etc operators. The $$x, \,y, \,z$$ operators transform as does a dipole (so IR transitions) and act as an alias for the symmetry species.
For example in the $$C_{2v}$$ point group each IR transition must be one of $$A_1,\,B_1 ,\, B_2$$ species and as there are 12 transitions there must be several of the same symmetry but different frequency.
The Raman transitions correspond to product terms , $$x^2,\, xy$$ etc. (because Raman spectroscopy depends of the polarisability which is proportional volume and so area in cross-section ) and these have also $$A_1,\,B_1 ,\, B_2$$ symmetry species as well as $$A_2$$ species (in $$C_{2v}$$) so we do expect to see some different transition frequencies between IR and Raman spectra. We do not necessarily know what the symmetry species are for these transitions, just that there is a difference.
In the case of $$D_{2h}$$ point group as there is a centre of inversion, there are no common transitions between IR and Raman spectra. Look at the point group and you will see this.
In more detailed analysis the polarisation of Raman signals can be used to determine which transition is $$A_1$$ etc. Spectroscopy books give details of this.
look at Understanding group theory easily and quickly for an explanation of point groups. | 955 | 4,068 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.936874 |
https://www.newsday.com/sports/baseball/mets/bay-moves-up-in-mets-run-factor-1.3082313 | 1,571,139,286,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00152.warc.gz | 1,036,258,035 | 21,447 | 46° Good Morning
46° Good Morning
# Bay moves up in Mets Run Factor
New York Mets left fielder Jason Bay reacts after hitting a walk off RBI single in the bottom of the tenth inning to defeat the New York Yankees 3-2 at CitiField. (July 3, 2011) Photo Credit: Christopher Pasatieri
It's an understatement to say Jason Bay has struggled all year, but he did make a slight jump in this week's Mets Fun Factor. Since Aug. 1, Bay has raised his average 11 points to .249 with two home runs and four RBIs. He also has a mild nine game hitting streak through Aug. 7.
PLAYER G PA R RBI HR MRF/G MRF/PA
Jose Reyes 99 462 80 37 5 1.143 .242
Carlos Beltran 98 419 61 66 15 1.143 .267
David Wright 54 240 34 33 8 1.093 .246
Ike Davis 36 149 20 25 7 1.056 .255
Angel Pagan 82 354 44 39 4 0.963 .223
Justin Turner 80 346 36 42 4 0.925 .214
Jason Bay 85 360 41 40 8 0.859 .203
Daniel Murphy 109 423 49 49 6 0.844 .217
Ruben Tejada 52 195 13 19 0 0.615 .164
Lucas Duda 58 170 16 22 3 0.603 .206
Mike Nickeas 9 26 3 3 1 0.556 .192
Josh Thole 79 269 17 25 2 0.506 .149
Ronny Paulino 56 183 15 13 1 0.482 .148
Nick Evans 19 41 7 3 1 0.474 .220
Scott Hairston 68 121 17 22 7 0.471 .264
Jason Pridie 67 159 17 16 3 0.448 .189
Willie Harris 84 180 25 12 1 0.429 .200
Fernando Martinez 11 23 3 2 1 0.364 .174
Brad Emaus 14 42 2 1 0 0.214 .071
Chin-Lung Hu 22 23 2 1 0 0.136 .130
What is the Mets Run Factor?
The Mets Run Factor is a fairly simple statistical metric. It takes the "Runs produced" sabermetric created by Bill James and divides it in two different ways.
Equation 1
The first equation is R + RBI - HR / G = Runs produced per game, as indicated in the chart above as MRF/G.
Equation 2
The second equation is R + RBI - HR / PA = Runs produced per plate appearance, as indicated in the chart above as MRF/PA.
The Mets Run Factor is updated every Monday. See past Mets Run Factor reports. | 699 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-43 | latest | en | 0.783724 |
https://www.coursehero.com/file/8991048/hwset2solution/ | 1,529,882,567,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00400.warc.gz | 783,381,911 | 68,159 | hwset2solution
# hwset2solution - ECE 453 Homework Assignment 2 Solution...
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ECE 453: Homework Assignment 2 Solution 1-11 Solution a. s ( t ) = 10 cos[2 π 10 8 t 2 cos(2 π 10 4 t + π / 4)] . This signal is in the form A cos( ω c t + θ ( t )) , where θ ( t ) = 2 cos(2 π 10 4 t + π / 4) . Hence, the signal is purely angle modulated. The message signal consists of a single tone with frequency f m = 10 4 Hz = 10 kHz. The peak phase deviation is 2 radians. The instantaneous frequency deviation ω ( t ) = d dt θ ( t ) = 4 π 10 4 sin(2 π 10 4 t + π / 4) . The peak frequency deviation, f max , is 20 kHz, so using Carson’s rule, the bandwidth is BW = 2( f m + f max ) = 2(10 + 20) kHz = 60 kHz. b. s ( t ) = 50 cos[2 π 10 8 t ] cos[2 π 10 4 t + π / 4] . This signal is in the form Am ( t ) cos[2 π 10 8 t ] , where m ( t ) = cos(2 π 10 4 t + π / 4) . Hence, the signal is amplitude modulated using DSB-SC. As in part a., the message signal, m ( t ) , is a tone with frequency f m = 10 kHz. Hence, the bandwidth occupied by s ( t ) is BW = 2 f m = 20 kHz. 1-12 Solution s ( t ) = 50 cos( ω c t + 5 sin(2 π 1000 t )) . This signal is in the form A cos( ω c t + θ ( t )) , where θ ( t ) = 5 sin(2 π 1000 t ) . Hence, the signal is purely angle modulated. The message signal consists of a single tone with frequency f m = 1000 Hz = 1 kHz. a. The peak phase deviation is 5 radians. b. The instantaneous frequency deviation ω ( t ) = d dt θ ( t ) = 10 4 π cos(2 π 1000 t ) s 1 .
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Jill Tulane University ‘16, Course Hero Intern | 786 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-26 | latest | en | 0.835406 |
www.pledgesports.org | 1,670,032,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00566.warc.gz | 1,000,494,406 | 18,396 | # Everything You Need to Know About the Variety of Betting Odds
When you decide to engage in online sports betting, you are sure to find out that the notion of odds can be tricky. Firstly, because there are three kinds of those. Let’s get to know more about this classification of common odds among different bookmakers.
## Decimal Odds
Looking at this type of odds, it’s rather easy to count how much you can win if your bet is correct. Your payout will be equal to the sum of your stake multiplied by the odds. This way you’ll find the potential sum of money you can receive. However, you’ll have to subtract the stake for net profit. Let’s take an example to understand the term better:
You wish to bet \$20 on Manchester United to beat Chelsea at 1.5. So, all you need to do is to multiply 20 by 1.5 (the decimal odds). If your prediction comes to life, you’ll get \$30. Subtract the stake, and you’ll see that your profit is \$10.
## Fractional odds
This kind of odds is even simpler and very widespread in the UK. When you look at it, the left number is the return, while the right one is the input. Let’s imagine another situation as an example. The odds are 3/1 that Liverpool will keep a clean sheet in the UEFA Super Cup. If you bet \$10 and the prediction is correct, you get \$40. Subtract the initial stake of \$10, and you’ll get \$30 of profit.
This kind of odds may also be inverted. It happens when the bet is very likely to win. For instance, when it’s 3/10 and you bet the same amount of \$10, you’ll get \$13, i.e. \$3 of pure profit.
## Vegas odds
This kind of odds is also called money-line betting. The main peculiarity is that the odds are based on winning precisely \$100. When you look at these odds, you’ll see positive numbers that represent favorites or negative numbers for the underdogs.
You can observe a similar technique in the online casino games odds.
Let’s take a look at the following example to understand this notion better:
If the odds are -400, you’ll need to stake \$400 to win \$500.
In case the odds are +400, you can bet \$100 and get a total of \$500.
Create a Campaign Get in touch
Looking for more information? Why not send us a quick message below and we'll get back to you shortly. | 525 | 2,247 | {"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-2022-49 | latest | en | 0.928457 |
https://percent.info/minus/87/how-to-calculate-9310-minus-87-percent.html | 1,638,380,849,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00548.warc.gz | 516,605,861 | 2,719 | 9310 minus 87 percent
This is where you will learn how to calculate nine thousand three hundred ten minus eighty-seven percent (9310 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9310 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 9310 of something.
9310
(100%)
87 percent means 87 per hundred, so for each hundred in 9310, you want to subtract 87. Thus, you divide 9310 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(9310 ÷ 100) × 87
= 8099.7
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 9310:
The dark blue not covered up by the pink is 9310 minus 87 percent. Thus, we simply subtract the 8099.7 from 9310 to get the answer:
9310 - 8099.7
= 1210.3
The explanation and illustrations above are the educational way of calculating 9310 minus 87 percent. You can also, of course, use formulas to calculate 9310 minus 87%.
Below we show you two formulas that you can use to calculate 9310 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
9310 - ((9310 × 87/100))
9310 - 8099.7
= 1210.3
Formula 2
Number × (1 - (Percent/100))
9310 × (1 - (87/100))
9310 × 0.13
= 1210.3
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
9320 minus 87 percent
Here is the next percent tutorial on our list that may be of interest. | 450 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.870393 |
https://www.fmaths.com/interesting-about-math/quick-answer-what-is-term-in-math.html | 1,632,183,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00195.warc.gz | 801,153,339 | 11,011 | # Quick Answer: What Is Term In Math?
## What is a term in math with example?
In algebra, terms are the values on which the mathematical operations take place in an expression. A term can be a constant or a variable or both in an expression. In the expression, 3a + 8, 3a and 8 are terms. Here is another example, in which 5x and 7 are terms that form the expression 5x + 7.
## Whats a term mean?
(Entry 1 of 2) 1: a word or expression that has an exact meaning in some uses or is limited to a subject or field legal terms. 2: a period of time fixed especially by law or custom a school term. 3 terms plural: conditions that limit the nature and scope of something (as a treaty or a will) the terms of a contract.
## How do you identify a term in math?
A term can be a signed number, a variable, or a constant multiplied by a variable or variables. Each term in an algebraic expression is separated by a + sign or J sign. In, the terms are: 5x, 3y, and 8. When a term is made up of a constant multiplied by a variable or variables, that constant is called a coefficient.
## What is a variable term in math?
Algebra – Basic Definitions A Variable is a symbol for a number we don’t know yet. It is usually a letter like x or y. A number on its own is called a Constant. A Coefficient is a number used to multiply a variable (4x means 4 times x, so 4 is a coefficient)
You might be interested: Often asked: What Is Convert In Math?
## What is a term in school?
An academic term (or simply term ) is a portion of an academic year, the time during which an educational institution holds classes. In Northern Hemisphere countries, this means that the academic year lasts from August, September, or October to May, June, or July.
## What is a term number?
more In Algebra a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs, or sometimes by divide.
## Can a term be one word?
In linguistic analysis, a phrase is a group of words (or possibly a single word ) that functions as a constituent in the syntax of a sentence, a single unit within a grammatical hierarchy. A phrase typically appears within a clause, but it is possible also for a phrase to be a clause or to contain a clause within it.
## Can two words be a term?
For the basic question of whether the words constitute a phrase, the answer is a Yes. Also, when the words so together serve to convey a certain definite sense, different from the meanings of the individual words, and this sense is accepted over time, they are generally called a ‘set phrase’.
## Is a Factor in math?
Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The prime factors of a number or an algebraic expression are those factors which are prime.
## What are 3 types of variables?
There are three main variables: independent variable, dependent variable and controlled variables.
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## What is variable explain with example?
In mathematics, a variable is a symbol or letter, such as “x” or “y,” that represents a value. For example, a variable of the string data type may contain a value of ” sample text” while a variable of the integer data type may contain a value of “11”.
## What are the main types of variables?
There are six common variable types:
• DEPENDENT VARIABLES.
• INDEPENDENT VARIABLES.
• INTERVENING VARIABLES.
• MODERATOR VARIABLES.
• CONTROL VARIABLES.
• EXTRANEOUS VARIABLES. | 853 | 3,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-39 | latest | en | 0.949561 |
https://wmbriggs.com/post/10071/ | 1,639,037,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363689.56/warc/CC-MAIN-20211209061259-20211209091259-00215.warc.gz | 655,428,434 | 36,872 | Statistics
# Every Family Has Children Until They Have A Boy: Probability Stumper
The Grand Old Stumper
Reader Bryan Davies quoted a poser from Math Overflow (have we heard of that?) which read:
In a country in which people only want boys every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country?
The question is ill-posed as stated because it includes several unstated assumptions. One set might be, and the one I’ll use, is that the country begins with no kiddies and with n couples, all the same age and who never die during their reproductive years and can reproduce at will and do, always on New Year’s Day, a day of celebration. Further, no babies are killed or die before they are born and none or killed or die once they escape into the wild. Babies are born once per year for every couple until success (a boy!). And no immigration nor emigration.
One last assumption is no genetic engineering or other meddling: forget the kind of things that happen in China and India. Why complicate things? Now, something causes each child to be a boy or girl, but we do not know what this something is in each case. Probability is a measure of information, not of biology. Therefore, given there are only two concrete choices, we deduce from our assumptions the probability (which I repeat measures our uncertainty, not the biology) is 1/2 for boys, same for girls.
So, Year 0, there are no boys, no girls and no ratio neither.
Year 1, the uncertainty in the number of boys will (given our assumptions) follow a binomial, characterized with p = 1/2 and n chances. Pr(0 boys | assumptions) = (1/2)n, Pr(1 boy | assumptions) = n * (1/2)n, Pr(2 boys | assumptions) = (n choose 2) * (1/2)n and so forth. The “(1/2)n” is always there because of a nifty quirk of the binomial with p = 1/2.
The proportion of boys to girls follows right from this. If there are 0 boys, the proportion is 0/n, because there must be, given our assumptions, n girls. The probability of seeing this proportion is the same as seeing 0 boys. And so on for all the other proportions, 1/(n-1), 2/(n-2), etc., except if all boys are born then the proportion is infinite (well, n/0 anyway). Lastly, assuming n is even, the most likely occurrence is (if n is even, or within rounding if not) n/2 boys, giving a proportion of 1/1. This follows because p = 1/2.
Visually (with n = 8), we might have seen this:
BBBGGGGG
With a (for boys) b1 = 3 and (for girls) n – b1 = 5 and thus a ratio of 3/5.
Year 2. Those couples who have had a boy exit the competition, the remainder have another go. The uncertainty in the number of boys in this new crop will again be characterized with a binomial with the same p but with n – b1 chances. Again, the most likely outcome, to our knowledge, is (n – b1)/2 new boys. That’s again because p = 1/2.
This year might have given, say, a b2 = 3, thus
BBBGG
The ratio counts Year one’s b1 boys and n – b1 girls plus this year’s crop, for a total of b1 + b2 = 6 boys and (n – b1) + (n – b1 – b2) = 7. The ratio is 6/7.
Year 3 is a repeat, our uncertainty another binomial but with (n – b1 – b2) = 2 chances. The most likely number of boys is 1. Suppose we see two boys:
BB
The total boys is 8, the total girls 7, for a ratio of 8/7.
If you are mathematically inclined, you will notice this ratio is not 1/1, which (I’m guessing) is the answer the examiner wants. It’s close, though.
The reason it is close is that each year the most likely occurrence, to within rounding, is half boys, half girls from the couples who are still going at it. Adding all those halfs up, as it were, gives half-and-half boys and girls as the most likely final outcome. But this isn’t necessarily the outcome.
We could figure the probability of seeing 8 total boys and 7 total girls easily but tediously enough. It involves calculating the probability of seeing 3 boys and 5 girls in Year one and 3 boys and 2 girls in Year two and 2 boys and 0 girls in Year 3. But then we’d have to figure the other ways (if any exist) to get 8b/7g in three years. Conceptually simple, because each combination follows a binomial with known parameters, but, as claimed, tedious to run through.
Large n
Now it could have been, for whatever sized n, that Year 1 saw all boys. We know the probability of this is (1/2)n, which is always greater than 0 for any finite n (which it always will be). Meaning, our knowledge does not preclude an infinite ratio: it could happen, especially with small n.
For large n, we follow the same pattern as above. But eventually two things happen: the kiddies become adults and pair off and begin producing their own children, and eventually the thus created grandpas and grandmas cease their efforts. How many child bearing years does a woman have, after all, assuming she’s pushing out a kid a year?
Ceasing to produce is easy to account for, but figuring the number of new couples is hard, because that depends on—you guessed it—the proportion of extant boys-now-men to girls-now-women. If the proportion of boys and girls is not 1/1, and while this is the most likely it is not certain, then some boys or girls will go marriageless. You then have to assume if they’re going to remain that way (easy), or if the strays can marry the strays which probably will come along the next year (hard, because how long will they remain fecund?).
Now all this is discrete and tedious, but if one has the energy and time it could all be ploughed through. We also have the sense that, because of the symmetries and clear assumptions, that the “system” will reach a limit where the proportions of boys to girls is roughly equal.
It may be equal in any year, but it’s more likely, we guess, to only “near” equal, where we can work out what it means to be near.
Categories: Statistics
### 39 replies »
1. Scotian says:
This is an old chestnut that I remember from my youth. As you note the usual solution assumes a continuum and all sorts of infinities. It would be interesting to run a computer simulation and look for attractors. The sex ratio could be varied, kin avoidance could be added, all sorts of things. I wonder if this has been done. I know that population simulations are very popular in biology.
2. Briggs says:
Scotian,
In this, like in most questions, there isn’t any problem per se introducing continuity, hence infinities. But it’s how one approaches the limit that matters. Is crucial, even. It is never a good idea to jump right to infinity: think of needle dropping problems and the like.
3. Under the (perhaps unlikely) assumption that both sexes are always equally likely (with no tendency of any parental pair to have more girls than boys or vice versa)we have the following:
The expected proportion of first-born children who will be boys is 1/2.
The expected proportion of second-born children who will be boys is 1/2.
etc.
4. More tediously, but still kind of interesting due to its display of a cute summation identity:
The expected fraction of all families that will have just one child is 1/2. Of the rest 1/2 will get a boy on the second try, and so on.
This gives the total expected number of boys equal to the total number of families times 1/2+1/4+1/8+etc=1 (which is too well known to be interesting and in any case it was obvious that,in the absence of twins, every family would have exactly one boy).
But the expected number of girls counted by family size is a bit more interesting. We get none of the one child cases, one in each two child family, two in each threesome and so on. Thus the total number of girls is equal to the total number of families times the sum from n=1 to infinity of n/(2^(n+1)).(Which is also equal to one but may be less well known)
PS There is now an apocryphal story about John von Neumann in which he instantly gave the answer 1/2 and when the student who posed the question said “Ah, so you saw how to do it by considering birth order rather than family size” vN replied “No I just used family size and summed the series”
5. Briggs says:
Ed,
Thanks. Interesting thing about “expected values” is that they’re usually not expected. Take the value on a die: “expected value” of 3.5, which is impossible to expect. EVs have some nice mathematical properties, but they’re best not to start with in unfamiliar situations. Can lead you astray. So can infinities, as I mentioned to Scotian.
Suppose, like my Uncle Pat and Aunt Patty (n = 1) you kept trying for a boy, as the rules stated, but quit after 4 kids. They had 4 girls and said enough’s enough. All couples, like I say in the text, must quit sometime. Landsburg assumes they can continue indefinitely, an assumption which would please my Uncle, but is not realistic.
Now (use Landsburg’s chart), the probability of B is 1/2, and GB is 1/4, GGB is 1/8, and GGGB is 1/16. But then the probability of GGGG is also 1/16 (we must write everything that can happen and the probabilities of everything that can happen must add to 1). That’s a B/G ratio of 0/4 = 0, which has (of course) a probability of 1/16.
Again, the right answer depends on when you’re looking. The expected value doesn’t help much unless you’re peering into infinity.
For my Aunt and Uncle, the ratio of B/G was always 0 (0/1, 0/2, 0/3, then 0/4). But then that’s because of their stopping rule (nature builds one in, even if you don’t).
For stopping rule of 0 (no kids), the B/G ratios (probabilities) are 0/0 (1).
For stopping rule of 1, the B/G ratios (probabilities) are 1/0 (1/2), 0/1 (1/2).
For stopping rule of 2, the B/G ratios (probabilities) are 1/0 (1/2), 1/1 (1/4), 0/1 (1/4).
For stopping rule of 3, the B/G ratios (probabilities) are 1/0 (1/2), 1/1 (1/4), 1/2 (1/8), 0/3 (1/8).
Etc., up until, what, 15? 20? Not much more than that, anyway. Notice that with n = 1 the most likely B/G ratio is 1/0. There can never be more than 1 boy, right? The only time the ratio is 1/1 is with stopping rules of at least 2, and each has probability 1/4.
6. Scotian says:
Alan, the von Neumann story is usually told in reference to a fly and bicycles or trains.
7. Briggs says:
Alan, Scotian,
Hadn’t seen that before. Love it!
8. Yes Scotian, I was just having a bit of fun with it (which is why I said “there is *now* an apocryphal story…”) – but thanks for posting those links to the original version.
9. I don’t think it matters what the limit on family size is. The expected proportion of boys will always be 50% (based again on the firstborn, secondborn, etc breakdown for the simplest explanation).
What does change is the expected total number of children which is twice the number of families if we can keep trying forever but a bit less than that if our maximum number of tries is limited.
[eg with a cap of 3 tries the expectation is that half of the families will get a boy on the first try, of those left half will get a boy on the second try, and half of the rest will get a boy on the third try for an expected number of 1/2+1/4+1/8=7/8 boys per family; and for girls it’ll be none in the one child families, one in each of the families that gets a boy on the second try, two in each of the families that gets a boy on the third try, and three in each of the families that never gets a boy for an expected total of 0(1/2)+1(1/4)+2(1/8)+3(1/8)=7/8 ]
10. Ed says:
Mr. Briggs, thanks for the reply, I was really looking at the post with other eyes and will look closer to what you’ve stated.
11. Scotian says:
Alan, all the von Neumann stories may be apocryphal. Thus the change in trains versus bicycles and in who asks him the puzzle. Briggs may accuse you of jumping to the continuum too quickly in your finite family size example. This is why I would like to see random walk like simulations under the conditions of a distribution of family size, fertility rates, finite initial population, a realistic boy/girl birth ratio, and so on. As happens with random walks once the initial even sex ratio deviates from half and half due to chance it might take a long time to come back to parity. This might have enormous implications for population dynamics or I might be off my rocker (even chances).
By the way there is that evil word chance again!
12. Timotheos says:
Hmmm, working on the assumption that the families will copulate onto infinity until having a boy, and that we are looking at this from a non-specified time range, I would say that the average number of girls that a family would produce would be the Sum of All Terms from 0 to infinity of the function x*((1/2)^x), which would be equal to 2.
Given that the odds of having a boy is a finite number, it makes no sense to say that there could be an infinite period of time in which only girls, and no boys, were born, since those odds would be equal to the limit of x as x approaches infinity of the function (1/2)^x, which is equal to 0. So the number of boys per family must be equal to 1.
Thus, the ratio of boys to girls is 1 to 2, which makes 2/3rds of the population girls, and 1/3rd of the population boys.
13. Casey says:
So long as the probability of a pregnancy producing either a boy or a girl is independent of previous pregnancy outcomes, the “stopping rule” does not matter, half of pregnancies result in girls and half in boys, so the population proportions will balance similarly. There are lots of descriptions of the reproductive cycle that will give this outcome.
14. David Engle says:
The “stopping rule” certainly does matter. The problem requires that all mothers which have not yet produced a son cannot stop producing girls until they do (and must stop upon producing a single son.) Clearly the population will have more girls.
expectation of girls is ( sum of n from 1 to infinity of ( (1/n)*(1/2)^n ) )
15. “Clearly the population will have more girls.” But,as so often happens, what is “clear” is often NOT true.
16. Scotian says:
Clearly you can not fool mother nature in this manner by such a passive process. My, and I admit trivial point, was to ask about the approach to equilibrium from an initial imbalance and the effect of non-passive factors. This could take just one generation. In a sense the experiment has already been done as many people have done exactly what the puzzle states to no effect. Even a minority of people trying this would be visible in the population, if it was possible. Of course given that the natural boy/girl birth ratio is about 105/100 it might be hard to detect this in practice. To those who state otherwise I will leave it up to you to find the error in your math. Be ruthless.
Of more, not puzzle, interest is the effect not of birth ratio but of reproductive age sex-ratio due to differential mortality or other means of avoiding child bearing, such as birth control, lack of partners, etc. Here one is interested in population dynamics and the conditions under which populations crash. Related to this one can ask what is the probability of a small breeding population producing all boys or all girls? Can small sub-populations evolve that show a strong sex birth selection that is parasitic on the larger population? If any biologist is reading this, do you know of any examples in the biosphere? I know that in some animals sex ratio is temperature dependent and then there are bees.
Of course none of this has anything to do with the solution to the puzzle.
17. David Engle says:
Hmmm… Having slept on it, I realize my earlier comment was wrong. The number of girls and boys in the population will be equal. The answer is not as obvious as I’d thought at first.
Perhaps the simplest argument is correspondence. Each and every mother must have exactly one son, and each and every son must have exactly one mother, therefore there must be a 1:1 correspondence between males and females in the population.
18. Speed says:
This is an example of a statistician making something more difficult and tedious than necessary. Alan Cooper’s answer (the one I came up with after a couple of minutes — I am typing this with one hand as the other is busy patting myself on the back) makes it understandable and therefore teachable to the uninitiated. It is also the type of reasoning that people need to use in daily life and scientists in thinking about and explaining their experiments.
There is still the need for a rigorous solution and clear definition of the experiment but the long explanation will be lost on most people.
19. Briggs says:
Speed,
The only problem with Alan’s answer—admittedly a small one—is that it’s strictly wrong.
The answer, as I’ve tried (perhaps unsuccessfully) to emphasize, depends on when you look. We cannot say with certainty that the answer is 1/2 boys and girls always and everywhere. Not for small n, not for infinite n, not for infinite couplings, neither.
I showed (in comments) that in the situation where n = 1 and with most stopping rules there is only a 1/4 chance of equal numbers of boys and girls.
So while Alan’s summary might guide the intuition toward the right answer (and see my text for a similar guide), it can lead you astray.
Not everything is easy.
20. Speed says:
Briggs wrote, “The only problem with Alan’s answer—admittedly a small one—is that it’s strictly wrong.”
But then there is no strictly right answer. There are a range of possible right answers — some more likely than others. Or the right answer is a range of possible answers, some more likely than others. It is nice to acknowledge this and it could become a terrific teaching moment but is far beyond the scope of the “poser” and I can’t imagine working through the equations at a cocktail party.
It is a problem that can teach middle school students (and non-technically trained adults) how to solve interesting and non-obvious puzzles.
21. Briggs says:
Speed,
“But then there is no strictly right answer.”
No, sir. That is false. There is a strictly, deducibly (yes, deducibly) right answer for every specification of n and stopping rule. See the comments above where I worked out the strictly right answer for an n = 1 and for all stopping rules.
It can also be worked out—and I gave notes how—for larger n. In fact, given our computing power, this should be easy to do. That’s homework for an ambitious reader!
Perhaps the difficulty is what you’re calling a “right answer”. I think maybe (I’m guessing) you say “right answer” means the exact proportion that must occur under certain specified circumstances. This does not exist except a stopping rule of 0. All—as in all—other situations are distributions over possibilities.
It’s too bad we can’t work this out at a cocktail party (but then you’ve never been to one of mine), but that’s life.
22. “It’s too bad we can’t work this out at a cocktail party” – I’ll drink to that! But I’m not sure what I said that you think is wrong. I tried to be sure to refer to the 50% ratio always as an expectation, but maybe I did miss one. Of course, just like the Spanish Inquisition (see http://www.youtube.com/watch?v=Tym0MObFpTI), no one expects the expectation (in fact if I suspected doctored data I might specify exactly 50 Heads in 100 coin tosses as a low p-value event on which to reject the null hypothesis of Boolean trial with 100 reps at p=0.5). But for the ultimate argument that stopping conditions have no effect I like Casey’s argument (even though he unfortunately said “half…will …” rather than “the expectation is that half … will…”). In ANY set of ova attacked by a 50-50 mix of equally powerful blue and pink sperms the expected proportion of male babies is 50%.
23. Briggs says:
Alan,
Apologies. Misspoke in a hurry. I meant “strictly wrong” in the sense there exists an exact proportion with certainly.
24. Timotheos says:
Eeek! I just realized that I set up my summation wrong. I made the subtle mistake of adding together the odds of an outcome times the number of girls it would return, not by the odds of ONLY that number of girls being returned. In effect, I counted everything twice. So my summation should have been x*((1/2)^(x+1)), which is equal to 1 and not 2. And I see that Alan Cooper already beat me to the punch, so now I’m thirsty.
Another slightly more helpful way to write summation would be to take the odds of getting a girl and then add the odds of getting an additional girl, generating the summation sequence 1/2, 1/4, 1/8..etc, which is exactly the same as the sequence for the boys. This has the intuitive advantage of lining up the same numbers for both sexes, which makes the solution a little easier to process.
Still, it may seem strange that the stopping policy had no effect on the proportion. But I think that’s because our intuition is screaming that changing a factor should have SOME effect, and not because it is telling us anything about the proportion. And while instigating the stopping policy does not change the proportion of boys to girls, it certainly does change the number or boys and girls, since we supposed that the couples could copulate ad infinitum for both boys and girls, which would allow for an infinite number of both. Adding in the stopping policy necessitates the number of boys and girls in each family is finite, so the stopping policy certainly puts a large restriction on the number, but not the proportion, of boys and girls in the population.
25. David Engle says:
The stopping policy does have an effect on the ratio.
Every son must have one mother, but not every mother has a son, and no mother can have more than one son, therefore there must be more mothers than sons.
The problem is that the wording of the puzzle implies that every woman eventually has a male child, and we work out the series summation using this assumption. In reality this isn’t true, some women will chose to have no children, some will fail to find mates, some will die before they succeed in producing a male child. Because of these cases, which aren’t accounted for in the summation, women will outnumber men.
26. Yes, David, if there is a limit on family size then the number of boys will be less than the number of mothers. BUT, even though some mothers may have more than one girl, others will have less, and in fact the expected number of girls will also be less than the number of mothers (to exactly the same extent that the expected number of boys is). The REAL reason why this must be true was provided above by Casey, and the only reason for counting by family size or whatever is in order to understand how arguments which seem to contradict him go wrong.
If you look back at my first answer you will see that it really covers all of these cases. If the maximum allowed number of children per family is N, then, for N=0 the expected numbers of boys and girls are both zero. For N=1 the only child in each family has equal chances of being a boy or girl so the expected numbers of each sex per family are both 1/2. For N=2 the second children (allowed only in those of the families that start with girls) also have equal chances of being boy or girl, so the expected family type frequencies are 1/2 Bonly, 1/4GB and 1/4 GG for an expected number of boys per family of 1/2(1)+1/4(1)=3/4 and of girls 1/4(1)+1/4(2)=3/4. I did the case N=3 in an earlier response, and the general case follows the same pattern.
27. Briggs says:
Alan,
And there’s the weakness of expected values. With n = 1, there is only a 1/4 chance of equal boys and girls for all stopping rules 2 or greater. A pretty wide disparity over the expected value guesstimate.
28. Yes, and for a single one child family there’s ZERO chance of seeing the expected numbers of boys and girls.
But then as Monty Python told us “NoOne Expects…”
29. Nullius in Verba says:
It doesn’t even need an infinite summation. All the births that actually take place are 50:50 events. So 50% of them (on average) will be boys.
The stopping rule is a complete red herring. The only way it could have an effect is if the birth probabilities for any given family were not independent – for example, if different people had a different (i.e. not 50%) boy/girl birth ratio determined by their genetics, but the mean of these different birth ratios across the population was 50%. Then waiting for a boy would selectively remove those parents with a propensity to have boys, leaving those with the propensity to have girls.
Anyway, the rest of it is just Briggs’ usual point about the expectation not being the outcome. Something made even more obvious by pointing out that if you have 5 parental pairs, the expected number of children is 2.5, which seems rather unlikely to happen. (One would hope.)
To which I think most people would reply: “You know perfectly well what I meant.” It’s not really all that complicated.
30. Nullius in Verba says:
Typo – that should be “2.5 male children”.
31. David Engle says:
“All the births that actually take place are 50:50 events. So 50% of them (on average) will be boys.”
And there is the problem, you are not considering the births that do not take place. In the case where there is no birth at all, the mother (1 girl) out numbers the son that does not exist (0 boys). This has no effect on the ratio of the (still theoretical) growth rate which is 50:50, but until someone’s actually born this population ratio is still 1:0.
The problem is the imprecise sematics of question make us cofuse the growth ratio and the population ratio, as well as the terms: family, mother and female. As is often the case in math if the original wording were more precise we wouldn’t be arguing at all.
32. David Engle says:
The stopping rule is a red herring in the sense that it does not alter the birth ratio.
but the stoping rule still has a very important effect, it imposes a hard constraint that boys can never out number girls no matter what the other conditions of the system are.
33. aGrimm says:
Matt: Not being a statistician, I’ve always wanted to know what the odds are for/against my family’s lineage. Perhaps you would be so kind?
Great-grandfather: 4 boys
Grandfather: 2 boys
Father: 9 boys (Uncle: 4 boys)
No girls until some dum-dum (me) broke the chain. Trust me, boys are easier to raise and require a lot less toilet paper.
– Thanks
34. aGrimm says:
Matt: I got to thinking that my request (previous post) is rude without at least offering something in exchange. Being retired, the only thing I can offer is some positive feedback as to the influence of your site. First some background: the one statistics course I took 40 years ago, I principally employed to count things radioactive during my career. The course did teach me to recognize, most times, when I was getting my leg pulled by statistical shenanigans. I’ve been following your site since I saw your first posting to Watts Up With That. I have learned a lot more about recognizing statistical shenanigans (Yay p value!),for which I thank you very much. But your influence does not stop with me for I am the go-to guy in my rather extensive circle of family and friends when it comes to science and math questions. I cannot glibly talk statistics, but I’m one of those who has the ability to translate science/math into laymen’s terms. Your debunking of so many things gets translated to this group and hopefully to their circle of family and friends. If I am an example, your influence goes well beyond the list of commenters here. And you should be rightfully proud of your work.
35. Nullius in Verba says:
“And there is the problem, you are not considering the births that do not take place.”
The births that don’t take place have no effect on the population statistics. By definition.
“but the stopping rule still has a very important effect, it imposes a hard constraint that boys can never out number girls no matter what the other conditions of the system are.”
Suppose in the first generation that every family has a boy and stops. What are the populations of boys and girls?
36. David Engle says:
(g+m)/(f+s)
where g is the number of girls who have no children, lets say 10 for in one possible case,
m is the number of mother who have children, lets also say 10,
f is the number of fathers, lets say 5, ( yeap, this is also possilbe )
s is the number of sons ( which your forcing to be equal to m), so also 10
notice the ratio is 20:15 in favor of girls, and every subsequent generation in this example can never have more boys than girls, ever, NO MATTER how subseqent random births happen.
37. Does no-one ever die in your world, David? It is true (and obvious without any fancy formulas)that if no mother can have more than one male child then the number of males who have ever lived must be less than the number of such females. But if every woman has one male child and then dies, then the number of males will (for one brief lifetime)certainly exceed that of females.
38. Scotian says:
Sorry David but you are wrong. The mistake that most people are making is reading too much significance into the condition that only the birth of a girl leads to another attempt. The condition of the puzzle is identical to any family being given the right to try for an additional child independent of the sex of the previous one based on some sort of lottery. For example, half of the original, then half of that half and so on. It doesn’t even have to be exactly half as it could be determined by coin toss to simulate the randomness of birth. This would not produce families of only one boy but it is equivalent as far as the sex ratio goes and you see that there is no bias for either sex and that boys could very well outnumber girls by chance. After the experiment is complete the children can easily be redistributed into “families” of one boy each. You might argue that the stopping point might leave you will an excess of girls with no boys to balance them, although strictly speaking this can only happen if there are no boys at all, but you can have an excess of boys too. The latter is forgotten by the simple fact that the boys can be used to make up single child families. Once you realize that the puzzle is only about distribution and not sex ratio the rest follows. It is sort of like topology. | 7,183 | 30,301 | {"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} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.955652 |
https://numbermatics.com/n/7381625501/ | 1,725,719,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00534.warc.gz | 414,792,959 | 6,895 | # 7381625501
## 7,381,625,501 is an odd composite number composed of three prime numbers multiplied together.
What does the number 7381625501 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 18 divisors.
7381625501 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eighteen divisors.
## Prime factorization of 7381625501:
### 832 × 101 × 1032
(83 × 83 × 101 × 103 × 103)
See below for interesting mathematical facts about the number 7381625501 from the Numbermatics database.
### Names of 7381625501
• Cardinal: 7381625501 can be written as Seven billion, three hundred eighty-one million, six hundred twenty-five thousand, five hundred one.
### Scientific notation
• Scientific notation: 7.381625501 × 109
### Factors of 7381625501
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 287
### Divisors of 7381625501
• Number of divisors d(n): 18
• Complete list of divisors:
• Sum of all divisors σ(n): 7619578398
• Sum of proper divisors (its aliquot sum) s(n): 237952897
• 7381625501 is a deficient number, because the sum of its proper divisors (237952897) is less than itself. Its deficiency is 7143672604
### Bases of 7381625501
• Binary: 1101101111111101010101010100111012
• Base-36: 3E2TUOT
### Squares and roots of 7381625501
• 7381625501 squared (73816255012) is 54488395037013501001
• 7381625501 cubed (73816255013) is 402212926313780697870270626501
• The square root of 7381625501 is 85916.3866849625
• The cube root of 7381625501 is 1947.0809272941
### Scales and comparisons
How big is 7381625501?
• 7,381,625,501 seconds is equal to 234 years, 37 weeks, 11 hours, 31 minutes, 41 seconds.
• To count from 1 to 7,381,625,501 would take you about four hundred sixty-nine years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 7381625501 cubic inches would be around 162.3 feet tall.
### Recreational maths with 7381625501
• 7381625501 backwards is 1055261837
• The number of decimal digits it has is: 10
• The sum of 7381625501's digits is 38
• More coming soon!
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The information we have on file for 7381625501 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 7381625501, math, Factors of 7381625501, curriculum, school, college, exams, university, Prime factorization of 7381625501, STEM, science, technology, engineering, physics, economics, calculator, seven billion, three hundred eighty-one million, six hundred twenty-five thousand, five hundred one.
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Some bits of this website may not work unless you switch it on. | 1,021 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.835901 |
https://ask.csdn.net/questions/1053814 | 1,585,994,541,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370521574.59/warc/CC-MAIN-20200404073139-20200404103139-00235.warc.gz | 360,478,766 | 13,801 | Play the Dice 骰子的问题
Problem Description
There is a dice with n sides, which are numbered from 1,2,...,n and have the equal possibility to show up when one rolls a dice. Each side has an integer ai on it. Now here is a game that you can roll this dice once, if the i-th side is up, you will get ai yuan. What's more, some sids of this dice are colored with a special different color. If you turn this side up, you will get once more chance to roll the dice. When you roll the dice for the second time, you still have the opportunity to win money and rolling chance. Now you need to calculate the expectations of money that we get after playing the game once.
Input
Input consists of multiple cases. Each case includes two lines.
The first line is an integer n (2<=n<=200), following with n integers ai(0<=ai<200)
The second line is an integer m (0<=m<=n), following with m integers bi(1<=bi<=n), which are the numbers of the special sides to get another more chance.
Output
Just a real number which is the expectations of the money one can get, rounded to exact two digits. If you can get unlimited money, print inf.
Sample Input
6 1 2 3 4 5 6
0
4 0 0 0 0
1 3
Sample Output
3.50
0.00 | 314 | 1,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-16 | latest | en | 0.915992 |
https://sqlpete.wordpress.com/2016/10/ | 1,498,714,339,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00305.warc.gz | 821,280,207 | 30,881 | # Archive for October, 2016
### GUIDs
GUIDs (Globally Unique Identifiers, aka UUIDs) are simply a string of 32 random hexadecimal numbers (that is: characters 0-9 and A-F), separated into five groups by dashes.
Here’s a GUID:
``C2DFE25B-6C1B-46B3-9497-DA45EF76D994``
All modern languages are able to generate them; here’s how I generated it in SQL:
``````SELECT NEWID()
GO
------------------------------------
C2DFE25B-6C1B-46B3-9497-DA45EF76D994
(1 row(s) affected)
``````
A GUID is simply a big random number, presented in a human-readable form. How big is ‘big’? With 32 hex digits, it means a GUID can take any of 16^32 = 2^128 values. (2^128 is approximately 3.4 x 10^38)
GUIDs are big. They’re so big, that you could label every atom in the universe using just 3 GUIDs. In fact, it’d be massive overkill: 3 GUIDs have a potential 2^384 values between them, which is equal to 3.940201 x 10^115; the number of atoms in the universe is estimated at 10^82, many orders of magnitude less.
Because GUIDs can take such an enormous range of values, the chances of generating a duplicate are minuscule. Quote:
“In other words, only after generating 1 billion UUIDs every second for the next 100 years, the probability of creating just one duplicate would be about 50%.” (wikipedia)
The ‘U’ in GUID basically means ‘unique for all practical purposes you’re likely to ever be involved with’ (unless you work for CERN, in which case I take it back).
So, that’s great: we have this construct that for all intents and purposes is unique (and I’ll treat it as such from here on), and we can generate one any time we want one. But how are they used?
## Usage
The most common usage of GUIDs is as keys for referring to other pieces of information, especially a block of structured information. For example, when I request a customer’s credit file, there’s a GUID, right near the top of the file. If I need to refer to that credit file again (whether inside my organisation, or with the issuing bureau), I can refer to it by the GUID, and we all know exactly which file I mean — not just the customer/address it refers to, but the data as it stood at that point.
### In databases
Now, database tables need a primary key to identify each row – and, by definition, the value of the key has to be unique. So it would seem a natural thing to want to have a GUID as a primary key. Even better: not only will we ensure that every row in our table will be unique, but every row in every table can be uniquely identified, in every database in the world! And you don’t even need to request a GUID from your database server when you create the data for a row, you can pre-generate primary keys in your C# code, and use them before they ever need to be stored on the server!
Sounds too good to be true, so what’s the catch?
### The catch
First off, most developers, analysts (and even DBAs) talk about ‘primary keys’ when they mean clustering keys – often, they’re the same piece of information, but they absolutely don’t have to be. The primary key is the piece of data that uniquely identifies a row in a table. The clustering key is the piece of data that determines the order of the data when it’s stored (on disk). More often than not, a straightforward incrementing integer (1,2,3…) can do the job of both, but it’s an informed choice that the database developer should be making.
When the clustering key is an incrementing integer, organising the data on disk is easy: the data goes in the next available slot. But when it’s (effectively) a random number, where does it go? The database has to make guesses about how much data there’s likely to be; guesses that it’ll have to re-assess every time a new row needs to be INSERTed into the database – worst case, it’s re-organising the data on disk every few INSERTs. This is really inefficient, and causes unnecessary stress on your server.
Internally in SQL Server, GUIDs take up 16 bytes of space, compared to the 4 bytes of an INT, or 8 bytes of a BIGINT. That’s not a major issue, unless you have lots of indexes on your table: indexes on tables automatically contain the clustering key, so with a GUID clustering key, every single index defined on that table will also contain the GUID. Potentially lots of valuable space used up, if you’re not careful.
• They cause inefficiencies under the hood: the server can’t make it’s usual good guesses about where to store data. NB: There is such a thing as a SEQUENTIAL GUID (Info here at MSDN), which lessens the impact – personally, I still wouldn’t bother.
• They take up four times more space than traditional INTs, which could be a problem if you have lots of indexes.
• Table JOINs are slower; SQL Server is optimised for joining tables together via simple integers.
There’s another (very important) reason not to use them that people tend to overlook: it makes debugging and tracking down errors incredibly painful! Incrementing numbers are intuitive, easy to memorise (if they’re small enough), easy to compare (“x+z” happened after “x”)… but GUIDs are just a ‘blob’ of data, there’s nothing intuitive about them.
### How to use GUIDs, pain-free
It’s simple: add a GUID as a normal column and index it!
``````ALTER TABLE dbo.Person
GO
-- ...UPDATE the table to fill PersonUID here ...
CREATE NONCLUSTERED INDEX IX_Person_PersonUID
ON dbo.Person(PersonUID)
GO
``````
That’s as complex as it needs to be.
``````00000000-0000-0000-0000-000000000001 | 1,338 | 5,465 | {"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} | 2.765625 | 3 | CC-MAIN-2017-26 | longest | en | 0.906438 |
http://www.aerospaceweb.org/question/aerodynamics/q0130.shtml | 1,508,325,388,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00592.warc.gz | 407,617,515 | 9,112 | Ground Effect and WIG Vehicles
I remember several years ago I saw an episode of "Wings" on the Discovery Channel that talked about a plane the Russians developed and tested that was meant for inter-continental travel over the ocean. The special thing about it was it flew only ten feet above the water. Do you have any information about it?
- question from Nick Lengyel
I've heard that Boeing is making a huge airplane, the Pelican, that will use "the WIG effect." Would this airplane prove to be practical? And if so, why wasn't the idea used before?
- question from Jonathan
The class of vehicles you mention are popularly known as Ground Effect Vehicles, or Wing in Ground (WIG) Effect Vehicles. As you indicate, what makes this class of craft unique is the fact that they fly at altitudes on the order of tens of feet, or a few meters. In so doing, they take advantage of a peculiar aerodynamic principle known as the ground effect.
To understand what ground effect is and how it functions, we first need to take a step back and explain some aerodynamic properties of an airplane wing. When producing lift, a wing generates strong swirling masses of air off both its wingtips. As discussed in a previous question on the creation of lift, a wing generates lift because there is a lower pressure on its upper surface than on its lower surface. This difference in pressure creates lift, but the penalty is that the higher pressure flow beneath the wing tries to flow around the wingtip to the lower pressure region above the wing. This motion creates what is called a wingtip vortex. As the wing moves forward, this vortex remains, and therefore trails behind the wing. For this reason, the vortex is usually referred to as a trailing vortex. One trailing vortex is created off each wingtip, and they spin in opposite directions as illustrated below.
Creation of trailing vortices due to a difference in pressure above and below a lifting surface
While trailing vortices are the price one must pay for generating lift, their primary effect is to deflect the flow behind the wing downward. This induced component of velocity is called downwash, and it reduces the amount of lift produced by the wing. In order to make up for that lost lift, the wing must go to a higher angle of attack, and this increase in angle of attack increases the drag generated by the wing. We call this form of drag induced drag because it is "induced" by the process of creating lift.
Effect of downwash in decreasing lift and increasing drag
That having been said, let us now explore what happens when an aircraft flies very close to the ground. The phenomenon is most often observed when an airplane is landing, and pilots often describe a feeling of "floating" or "riding on a cushion of air" that forms between the wing and the ground. The effect of this behavior is to increase the lift of the wing and make it more difficult to land.
However, there is no "cushion of air" holding the plane up and making it "float." What happens in reality is that the ground partially blocks the trailing vortices and decreases the amount of downwash generated by the wing. This reduction in downwash increases the effective angle of attack of the wing so that it creates more lift and less drag than it would otherwise. This phenomenon is what we call ground effect, as shown below.
Ground effect and its influence on trailing vortices
An additional bonus of ground effect that becomes more significant as speed increases is called ram pressure. As the distance between the wing and ground decreases, the incoming air is "rammed" in between the two surfaces and becomes more compressed. This effect increases the pressure on the lower surface of the wing to create additional lift.
As you might expect, the impact of ground effect increases the closer to the ground that a wing operates. As indicated in the plot shown below, ground effect typically does not exist when a plane operates more than one wingspan above the surface. At an altitude of 1/10 wingspan, however, induced drag is decreased by half.
Decrease in induced drag due to ground effect
When all of these benefits are taken into account, we find that a vehicle operating in ground effect has the potential to be much more efficient than an aircraft operating at high altitude. The aerodynamic efficiency of an aircraft is expressed through a quantity called the lift-to-drag ratio, or L/D. In steady, level, non-accelerating flight, a plane's lift is equal to its weight, and the amount of thrust required is equal to the drag it produces. Therefore, the L/D ratio is a measure of the weight that can be carried for a given amount of thrust. The higher the L/D, the more efficient the vehicle. Typical L/D values for conventional, subsonic aircraft are on the order of 15 to 20. By comparison, a ground effect vehicle could, in theory, achieve L/D ratios closer to 25 or 30.
Though ground effect has been known since the early days of flight, most pilots regarded it as nothing more than a nuisance that changed the flying qualities of their aircraft during takeoff and landing. Nevertheless, many researchers soon realized that this phenomenon could be exploited to produce a new class of highly efficient craft known as WIG vehicles. Most of the pioneering research into these vehicles was performed in West Germany and the Soviet Union.
Perhaps the most successful researcher in this field was Rostislav Alexeiev, head of the Central Hydrofoil Design Bureau in the Soviet Union from the 1950s through the early 1970s. Alexeiev began his career developing hydrofoils, which are boats fitted with underwater wings. When the boat moves, these wing create lift that pulls the boat hull up and out of the water and allows the craft to cruise at higher speeds. But Alexeiev quickly realized that a hydrofoil can only go so fast due to the drag created by the dense water through which it "flies." Why not instead raise the entire vehicle out of the water and cruise at even higher speeds, he reasoned. This line of thought led to a new vehicle with wings above the surface of the water, a vehicle the size of a boat and able to carry a massive payload but able to cruise at the speed of an aircraft. Another advantage of such a vehicle would be the ability to fly at very low altitudes, below the detection range of enemy radar. Alexeiev dubbed this new class of vehicle the Ekranoplan, which is Russian for "screen plane." His early designs were numbered under the SM series, an acronym standing for Samorodnaya Model, meaning "self sustained craft."
Alexeiev developed a number of sub-scale designs in experimental tests that culminated in 1965 with the completion of the KM , the largest WIG vehicle ever built. Making its first flight on 18 October 1966, the KM was powered by a whopping 10 turbojet engines and weighed up to 540 tons. Eight of the engines were mounted near the vehicle's nose so that their thrust could be deflected underneath the wing to create an initial cushion of air that raised the KM out of the water. Once the craft was traveling fast enough that the wings generated sufficient lift to keep the vehicle above the water, the thrust was redirected aft to increase velocity.
KM Ekranoplan
The KM was modified numerous times to evaluate the effects of different design elements. Among these changes included varying the wingspan from 105 ft (32 m) to 131 ft (40 m) and increasing the fuselage length from 302 ft (92 m) to 347 ft (106 m). The KM, also dubbed the Caspian Sea Monster by American observers who spotted the craft in satellite surveillance, remained in use until 1980 when it crashed during takeoff.
Although Alexeiev passed away in 1980, his design bureau continued to build and test Ekranoplans until the collapse of the Soviet Union. One of the more successful concepts was the Lun, which began trials on the Caspian Sea in 1987. The Lun was of similar design to the KM but smaller and built to carry anti-ship cruise missiles for high-speed attacks against American carrier battle groups.
Lun Ekranoplan with missile launchers mounted above the fuselage
However, the changing political tide and struggling economy in the Soviet Union caused the Ekranoplan concept to fall out of favor, and a second Lun was never completed. The Alexeiev Design Bureau remains in operation and continues to propose new designs for civilian use, but the market has not yet developed.
Nevertheless, Boeing has recently taken interest in the WIG phenomenon and proposed a concept for a massive craft to meet a US Army need for a long-range heavy transport. Called the Pelican, the 500 ft (153 m) span vehicle would carry up to 2,800,000 lb (1,270,060 kg) of cargo while cruising as low as 20 ft (6 m) over water or up to 20,000 ft (6,100 m) over land. Unlike the Soviet concepts, the Pelican would not operate from water, but from conventional runways using a series of 76 wheels as landing gear.
Boeing Pelican ground effect vehicle
The outboard wing sections, shown tilted downward in the above image to optimize efficiency in ground effect, would be swiveled upward to provide clearance for runway operations. Power for the craft would be supplied by four advanced turboprop engines. The Pelican would also have two cargo decks. The lower deck would be sized for large cargos, including up to 17 main battle tanks, while the upper deck would be used for troops or cargo pallets. With a maximum takeoff weight up to 3,000 tons, the Pelican would have a wing area of more than an acre.
The Pelican is a logical extension of Rostislav Alexeiev's work. As big as the KM prototype was, it was simply too small to take full advantage of the benefits of ground effect. Indeed, Alexeiev had proposed much larger versions of the KM and Lun able to carry up to a thousand armed troops and their ground vehicles. The principal advantage of a WIG vehicle is the ability to move very heavy loads in a craft with a relatively small wingspan and low aspect ratio wing with great aerodynamic efficiency. Exemplifying this strategy is Boeing's claim that the Pelican is capable of transporting 750 tons over 10,000 nm (18,530 km) when cruising in ground effect, but can carry the same load only 6,500 nm (12,045 km) when out of ground effect.
So the WIG concept is clearly not new, and has been implemented by several manufacturers. However, none has really caught on, and the primary reason is that ground effect craft only become truly practical for very large vehicles, even larger than the massive KM. The reasoning is as follows. As discussed above, the amount of lift a flying vehicle needs to generate is directly related to its weight. The heavier a plane is, the more lift it needs, so the larger its wings must be (for the same cruise speed). This basic relation becomes a problem when we consider very heavy aircraft. As payload weight increases, wing size increases which requires larger and heavier structures that further increase overall weight. As weight increases, additional thrust and fuel is required to push the vehicle at its desired cruise speed over the required range, and the need for larger or additional engines plus greater fuel capacity further increases the overall weight of the vehicle. This trend pushes manufacturers towards increasingly complicated and expensive design solutions that make very large and heavy aircraft unprofitable to build.
However, the beauty of ground effect is that a given amount of wing area produces more lift near the ground than it would at high altitude. Or in other words, the same payload can be transported with a much smaller wing, which translates directly into a smaller, lighter, and more fuel efficient craft. It is therefore not surprising that most advocates of WIG vehicles have focused on very large vehicles, like Boeing's Pelican. This is not to say that vehicles at the other end of the size spectrum have been ignored. Quite the contrary, far more small WIG craft have been built and flown than large craft like the KM. Manufacturers around the world, particularly in Germany, have constructed a variety of small WIG vehicles designed for a handful of passengers. One such example is the L-325, a four-passenger craft built by the American company Flarecraft.
Flarecraft L-325 wing in ground effect vehicle
So why haven't these smaller craft become a commercial success? Unfortunately, most of the improvements in efficiency that make a large WIG vehicle attractive do not translate as well across the spectrum to their smaller cousins. The same reductions in induced drag and the ability to lift a given payload with a smaller wing still apply, but the benefits are not as pronounced. While a craft operating close to the ground experiences a reduction in induced drag, other forms of drag are increased. Most importantly, WIG vehicles experience greater skin friction drag simply because the air is denser at sea level than it is at high altitude. A large vehicle can tolerate this increase because the decrease in induced drag is far more significant. But for a small vehicle, the decrease in induced drag and increase in skin friction drag are more equal, so only marginal improvements in overall efficiency are possible.
Furthermore, the increased drag created by denser air at low altitudes limits maximum speed, so a WIG craft will take more time to travel a given distance than a comparable aircraft operating at high altitude. Again, this performance penalty is less significant for a large WIG craft since they would likely be used for transporting large cargos, and speed can be sacrificed for lower cost. Commercial passengers, on the other hand, would be less likely to accept longer travel times than are currently possible with modern airliners.
For these reasons, the future of WIG craft remains uncertain. The potential benefits of ground effect are indeed attractive, but it is unclear whether those benefits are significant enough to warrant construction of large enough vehicles to take full advantage of them. The Pelican is an intriguing concept that could revolutionize the transportation of very heavy payloads, but only time will tell whether or not it is just a pipe dream.
An excellent resource for further information on the aerodynamics, design, and history of Wing in Ground Effect vehicles as well as details about specific craft and their manufacturers is The WIG Page.
- answer by Jeff Scott, 29 June 2003
Related Topics: | 2,984 | 14,503 | {"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-2017-43 | longest | en | 0.961635 |
https://www.geeksforgeeks.org/category/competitive-programming/page/28/ | 1,611,563,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00387.warc.gz | 803,319,512 | 19,499 | # Category Archives: Competitive Programming
Given two arrays A[] and B[], and an integer K, the task is to find the number of ways to select two subarrays of the… Read More
Prerequisites: MO’s algorithm, SQRT Decomposition Given an array arr[] of N elements and two integers A to B, the task is to answer Q queries… Read More
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Given two arrays arr1[] and arr2[] and an integer K, our task is to find the number elements in the first array, for an element… Read More
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Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a Perfect number.… Read More
Given a sorted array arr[] of N integers and a number K, the task is to write the C program to find the upper_bound() and… Read More
Given an array arr[] of integer elements, the task is to find the length of the largest sub-array of arr[] such that all the elements… Read More
Given a Matrix of size N*N filled with 1‘s and 0‘s, the task is to find the maximum distance from a 0-cell to its nearest… Read More
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This article focuses on how to implement your solutions and implement them fast while doing competitive programming. Setup Please refer Setting up a C++ Competitive… Read More
Given a Prufer sequence of a Tree, the task is to print the nodes with prime-degree in this tree. Examples: Input: arr[] = {4, 1,… Read More
Given two integers, X and K, the task is to find if X2 is divisible by K or not. Here, both K and X can… Read More
Given an array arr consisting of N elements and Q queries represented by L and R denoting a range, the task is to print the… Read More | 517 | 2,141 | {"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-2021-04 | latest | en | 0.855686 |
https://www.sanfoundry.com/fluid-mechanics-interview-questions-answers-freshers/ | 1,721,899,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00530.warc.gz | 850,612,670 | 22,425 | # Fluid Mechanics Questions and Answers – Thermodynamic Properties & Compressibility
This set of Fluid Mechanics Questions and Answers for freshers focuses on “Thermodynamic Properties, Compressibility and Bulk Modulus”.
1. If there is no exchange of heat between system and surrounding where system comprises of a compressible fluid but the heat is generated due to friction, the process is an adiabatic.
a) True
b) False
Explanation: For process to be adiabatic, there is no heat exchange and no heat generation within fluid.
2. For a compressible fluid, if there is no change in specific volume at constant temperature, what type of process it is?
a) Isothermal process
c) Polytropic process
d) None of the mentioned
Explanation: As, specific volume remains constant, density remains constant. Therefore for given temperature there is no change in volume. hence, the process is isothermal.
3. If the fluid is incompressible, do thermodynamic properties play an important role in its behaviour at varying temperature and pressure?
a) Yes
b) No
c) Depends on the fluid
d) None of the mentioned
Explanation: If fluid is incompressible there is not much change in observed properties with variation in temperature and pressure. Hence, no perceivable change.
4. If for same temperature and pressure change, the value of bulk modulus is compared for isothermal process and adiabatic process, which one would be higher?
a) Isothermal process
c) Value is constant for both the processes
d) None of the mentioned
Explanation: For isothermal process
K=p
K=kp
where K=Bulk modulus
k=Polytropic constant
p=Pressure.
5. The value of gas constant is same for all the gases
a) True
b) False
Explanation: The value of gas constant depends on molecular weight. As the molecular weight is different, gas constant will be different.
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6. Calculate the pressure exerted by 9 kg of air at a temperature of 20℃ if the volume is 0.8m3. Assuming ideal gas laws are applicable.
a) 946 kN/m2
b) 1892 kN/m2
c) 1419 kN/m2
d) None of the mentioned
Explanation: Ideal gas Law: PV=nRT
n=M/m
P=(9*8314*293)/28.97=946 kN/m2.
7. A gas weighs 16 N/m3 at 30℃ and at an absolute pressure of 0.35 N/mm2. Determine the gas constant.
a) 708.23
b) 354.11
c) 531.17
d) 1062.34
Explanation: R=P/(ρ*T)=3500000*9.81/16*303=708.23.
8. A cylinder of 0.8 m3 in volume contains superheated steam at 70℃ and .4 N/m2 absolute pressure. The superheated steam is compressed to .3 . Find pressure and temperature.
a) 0.74 N/m2, 422.3℃
b) 1.48 N/m2, 422.3℃
c) 0.74 N/m2, 844.6℃
d) 1.48 N/m2, 844.6℃
Explanation: For polytropic process,
P2=(v1/v2)n *P1
=(0.8/0.3)1.3 * 0.4 ……..(for superheated stream n=1.3)
=.74 N/m2
T1=P1v1/nR=422.3℃.
9. Determine the compressibility of an incompressible fluid, if the pressure of the fluid is changed from 70 N/m2 to 130 N/m2. The volume of the liquid changes by 0.15 percent.
a) 0.0025 m2/N
b) 0.0050 m2/N
c) 0.0070 m2/N
d) 0.0012 m2/N
Explanation: Compressibility=1/Bulk Modulus
=1/K
K=(dp*V/dv)
=60/0.15
=400
Compressibility=.0025.
10. What is the variation of cp, cv and k in case of gases when the temperature increases?
a) cp and cv decreases with temperature, and k increases
b) cp and cv increase with temperature, and k decreases
c) cp and cv increase with temperature, and k increases
d) cp and cv decreases with temperature, and k decreases
Explanation: cp is molar heat capacity at constant pressure. As temperature is increased, enthalpy increases, heat capacity increases.
Same is for cv, cp is molar heat capacity at constant volume.
However cp-cv=R and cp/cv = R
Hence, as cp, cv increases R decreases.
Sanfoundry Global Education & Learning Series – Fluid Mechanics.
To practice all areas of Fluid Mechanics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,129 | 3,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.878621 |
https://www.edaboard.com/showthread.php?199195-Theory-of-Motor-(EMF-theory-) | 1,571,569,300,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00510.warc.gz | 875,263,496 | 20,596 | # Theory of Motor (EMF theory???)
1. ## Theory of Motor (EMF theory???)
Hi,
can anyone tell me the back emf voltage is against the voltage supply from the motor terminal?
is this related to the inductance characteristic? can anyone give me a details explanation.
thanks.
regards,
sysysy
•
2. ## Re: Theory of Motor (EMF theory???)
the rotating device can act as a motor and a generator. When you supply to that device, then it works as a motor, if you draw voltage that is generated from the rotating device by rotating it, you call it generator. So, when you supply voltage to the rotting device, you wil have the emf, that is the voltage generated to counter the your supply voltage
3. ## Re: Theory of Motor (EMF theory???)
Hi,
1 more things, the speed of the motor is based on a average current right?
then, how to relate it when we provide more voltage to the motor, then the motor speed will run more faster?
can u tell me how this relate to the current?
because as i understand : V(supply voltage)-CEMF(counter emf)= I(current)R(resistance) and the CEMF = w(speed).k(constant of motor)
now i dunno why when we increase V, then the speed will increase as well.
thank.
regards,
sysysy
---------- Post added at 15:04 ---------- Previous post was at 14:39 ----------
I think i figured it out edi.
Thanks.
4. ## Re: Theory of Motor (EMF theory???)
the speed of the motor is based on a average current right
The shown equations clarify, that the statement is incorrect. Saying it's based on voltage would be almost correct in contrast.
The motor type hasn't been mentioned yet. To avoid misunderstandings, it should be said, that the equations are valid for a motor with constant excitation, e.g. a PMDC motor.
So what's the role of motor current? Basically it's proportional to the load torque. An unloaded motor is consuming only a small idle current, representing the torque of friction losses and additional electrical losses like armature eddy currents. They are in fact speed dependent, and in so far, the idle current is based on speed, rather than the other way round.
•
5. ## Re: Theory of Motor (EMF theory???)
Thanks for the rectification.
Yes, it is a permanent magnet DC motor
So i have tried to come out the equation w = V/k - TR/k square
so from equation above, when increase of V, then w will increase. ( there is ntg relate with I, am i correct)?
but i am not really understand with that idle current.
•
6. ## Re: Theory of Motor (EMF theory???)
From this equation, w (↑) = V (↑) /k- TR/k²
actually i wanna say the when voltage increase, so the speed also will increase,
but as i know, speed in inversely proportional to torque. but the equation seem like cant explain this...
anyone can give me the correct information?
7. ## Re: Theory of Motor (EMF theory???)
but as i know, speed in inversely proportional to torque
No, where did you this?
8. ## Re: Theory of Motor (EMF theory???)
because when we reduce the gear of motor, it will increase torque of the motor but reduce speed of motor.
anyway, from what i have analysed in the equation
w (↑) = V (↑) /k- TR/k².... i assume v,k and r is constant. so when T is large, then w will decrease, and when T is small, w will be increase...am i correct?
this is becaus v is already fix...so when v minus large T....w become small and when V minus small T, then w become larger....
9. ## Re: Theory of Motor (EMF theory???)
Inverse proportional is usually understood as f(x) ~ 1/x not f(x) ~ -x, that's why I'm asking.
I think, the discussed equation is pretty clear. Not considering generatoric operation, the motor has it's maximum speed with no load (T=0), the speed drops, when the load increases.
Some people call it a "theory", a practical man measures motor speed with different loads, put the points in a diagram, and draw a line through it.
10. ## Re: Theory of Motor (EMF theory???)
Hi, Fvm,
I think my confusion raise up after i read a explanation from this. i show u and maybe u can interpret in correct way.
The explanation say that Torque is directly proportional with speed.
hope can u point out something from there.
thanks.
•
11. ## Re: Theory of Motor (EMF theory???)
The quoted literature doesn't apply to a PMDC motor, it's about variation of the field current of a DC motor with electrical exictation (shunt wound motor).
Furthermore, the load configurations isn't specified, at least not on the shown page. Thus it's unclear what sets the torque in this experiment.
12. ## Re: Theory of Motor (EMF theory???)
Can you elaborate on torque and its relation to speed?? Because, In some motors, I have read of Four quadrant operations, where speed and torque maybe directly or indirectly proportional?? Does that relation have anything to do with constant excitation?
--[[ ]]-- | 1,149 | 4,816 | {"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} | 3.28125 | 3 | CC-MAIN-2019-43 | latest | en | 0.924757 |
https://www.liaoxuefeng.com/discuss/001409195742008d822b26cf3de46aea14f2b7378a1ba91000/0015025442910166bf0810a7d5442dc856947283f64a2b3000 | 1,508,430,314,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00106.warc.gz | 975,173,670 | 10,958 | ### 关于作者
Discuss / Python / 交作业
### 交作业
Topic source
#### weareguyshot
#1 Created at ...
def triangles(n): l=[1] num=1 while num<=n: yield l l1=[0]+l l2=l+[0] l=[l1[i]+l2[i] for i in range(len(l1))] num=num+1 for i in triangles(10): print i
• 1 | 99 | 251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-43 | latest | en | 0.412631 |
www.waxius.com | 1,606,713,219,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00413.warc.gz | 928,315,946 | 7,070 | # First Daily Coding Challenge
It’s Day 1 of the Daily Coding Problem. So here it is:
# Problem #1
`Given a list of numbers and a number `k`, return whether any two numbers from the list add up to `k`.`
For example, given `[10, 15, 3, 7]` and `k` of `17`, return true since `10 + 7` is `17`.
Bonus: Can you do this in one pass?
## Approach:
As a Mental Exercise… here’s what I’d to do approach this:
First… I’d create an array of numbers and let it remain static.
[4, 6, 13, 8, 17, 3, 9] .. I could add more numbers, but lets keep it simple with numbers under 20.
Then I’d pick a value for K that I KNOW would work.. like 20. because two of my numbers 17 and 3 will add to 20.
Second… I’d create a nested FOR-LOOP that indexes through the array and picks the first number and holds onto it in a variable x. The inner loop with go through the remaining numbers and place one at a time into variable y. It will then check to add x+y and see if it equals k. IF it does… print out x+y=k to the screen.
After it checks.. the inner loop goes to the next number until there are no more numbers left.
It would go like this:
4 + 6 = 10… 10 does not equal 20 so nothing happens…
4 + 13 = 17… 17 does not equal 20 so nothing happens…
… and so on.
After trying to add 4 to all the other numbers… the outer loop moves on to the next number 6.
6 + 13 = 19…. 19 does not equal 20 so nothing happens…
6 + 8 = 14…. 14 does not equal 20 so nothing happens…
Notice that I did not add 6 to the first array number 4. We already checked 4 + 6 so there’s no reason to check 6 + 4 now.
Eventually… the Outer loop will grab 17 for variable X and the inner loop will grab 3 for y. and we will print 17 + 3 = 20. x+y=k.
So that’s my plan. I don’t know if this is the “right” answer, but it’s what I’ve got.
## Solution:
using System;
namespace Daily_Program_Challenge_1
{
class Program
{
static void Main(string[] args)
{
int[] array = new int[] { 4, 6, 13, 8, 17, 3, 9, 12, 11};
int k = 20;
int matches = 0;
for (int i = 0; i < array.Length; i++)
{
int first = array[i];
for (int j = i + 1; j < array.Length; j++)
{
int second = array[j];
if (first + second == k)
{
Console.WriteLine(first + ” + ” + second + ” = ” + k + ” True!”);
matches++;
}
}
}
Console.WriteLine(“There were “+matches+” matches found!”);
}
}
} | 766 | 2,327 | {"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-2020-50 | latest | en | 0.688359 |
https://www.physicsforums.com/threads/values-of-a-and-b-in-pendulum-equation.806468/ | 1,531,782,753,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00520.warc.gz | 948,255,447 | 14,279 | # Homework Help: Values of a and b in pendulum equation?
1. Apr 3, 2015
### ajrvega22
Ok guys so for my lab report I am given an equation of period=2pi(length/a)^b and through this equation and the slope and y-int of my log-log graph I am suppose to solve for the values of a and b. I know that taking the log of the equation gives me log(T)=log(2pi)+blog(L/a) and this relates to the slope intercept equation of y=mx+b where b is the slope. Therefore I know log(T) is y, blog(L/a) is mx and log(2pi) is b. I just can't figure out how to solve for the values of a and b?
2. Apr 3, 2015
### ehild
Use that log(L/a)=log(L)-log(a). Write the equation for log(T) as function of log(L).
3. Apr 3, 2015
### brianeyes88677
Spread your functions into logT=b*logL+log2π-b*loga. Compare to the regression line y=m*x+n from your experiment ,where b is the slope and log2π-b*loga is the intercept. | 265 | 894 | {"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} | 3.171875 | 3 | CC-MAIN-2018-30 | latest | en | 0.897565 |
https://howtoexcel.net/average-down-calculator | 1,709,132,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00865.warc.gz | 305,395,296 | 26,226 | Average Down Calculator
Amount Invested # of Shares Average Price Current Share Price
Amount to Invest Shares Needed New Average
## What is an average down calculator?
An average down calculator can help you determine whether it's worth it to buy more shares of a stock that has fallen in value. When you buy shares at a lower price, then that lowers the cost of your holdings. That can make it easier to turn a profit on it if the stock rallies.
For instance, if you invested \$1,000 into a company and that allowed you to own 100 shares, then your average purchase price would have been \$10 per share. Now, if the stock were to go on to decline to \$5 per share, you could buy more shares to bring your average down.
Let's say in Scenario A, you invest another \$1,000. At \$5, you would be able to buy another 200 shares. At that point, you would own 300 shares while having invested a total of \$2,000. Your will now be \$6.67 per share. And the more shares you buy at the reduced price, the lower you can bring down your average. Let's take a look at another scenario.
In Scenario B, let's say you invested \$3,000. At that level of investment, you could be an additional 600 shares of the company Then, you would have 700 shares for a total investment of \$4,000 -- bringing your average down to \$5.71. That's slightly lower than the \$6.67 in Scenario A if you did the smaller investment.
## How does this average down calculator work?
Regardless of how many prior purchases you have made, all you need to do for this calculator is enter the total value of your investment in a stock, and how many shares you own of it. This calculator will then tell you what your average cost is.
Then, you can enter the current stock price. Next, what the calculator will do is tell you how low you can get your average by buying at different increments of \$1,000 (assuming you buy at the current stock price). This can help you determine if it makes sense (e.g. doesn't cost a whole lot) to get your average down. The goal here is for you to be able to input the current price as it stands today to see whether it makes sense executing on a trade or not.
However, you can also do what-if scenarios on different price points. Although you may not want to buy at today's price, you can input a lower price to see if it's worth buying the stock if it falls even further. That way, you can be ready to make a trade if it hits that price.
## When should you average down?
Whether you average down or not ultimately depends on your risk tolerance and the hopes you have for an investment. If you believe the stock is significantly undervalued and due to recover, then averaging down can make a lot of sense; you're effectively adjusting your investment's cost down so that it's cheaper and easier to turn a profit on it down the road.
However, by averaging down, you are also investing more money into a single stock. The danger here is now you become less diversified and that can put you at greater risk because you become more dependent on the performance of that individual stock.
It's important to evaluate both factors when making a decision of whether to average down or not. But if it's a quality investment and you're comfortable with the level of risk, it could be a good move. The average down calculator can help you make that determination.
If you liked this average down calculator please give this site a like on Facebook and also be sure to check out some of the many templates that we have available for download. You can also follow us on Twitter. | 777 | 3,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-10 | longest | en | 0.958214 |
https://students.flinders.edu.au/my-course/course-rules/topic/educ9766/2021 | 1,726,313,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00558.warc.gz | 516,476,354 | 17,833 | Year
2021
Units
4.5
Contact
1 x 2-hour workshop weekly
Prerequisites
1 EDES9003 - Introduction to Educational Statistics
1a EDES9003D - Introduction to Educational Statistics
1b EDUC9762 - Introduction to Statistics
Must Satisfy: ((1 or 1a or 1b))
Enrolment not permitted
1 of EDES9005, EDES9005D has been successfully completed
Assumed knowledge
Familiarity with excel and SPSS or similar programs.
Topic description
This topic is designed for students who wish to understand and use advanced statistical techniques. The topic aims to prepare students in education to select and employ appropriate analytical procedures for the examination of data collected in surveys, quasi-experimental research studies and longitudinal studies, as well as to draw appropriate conclusions and interpret the research findings from such studies. The topic concentrates on the understanding and use of the analytical procedures of linear and multiple regression, and data reduction techniques. Multilevel regression and structural equation modelling are introduced. The programs SPSS, AMOS and HLM are used. A central goal of the topic is to enable students to evaluate critically contemporary educational research and to contribute to that research.
Educational aims
Students will:
• Understand the assumptions underlying multivariate techniques and be aware of the limitations of these techniques in complex survey and longitudinal studies
• Understand multivariate and multilevel methods and know when and how to apply them
• Be able to use standard statistical software to conduct multivariate and multilevel analyses
• Be able to write interpretive reports in which the results of multivariate and multilevel analyses are accessible to non-specialists
Expected learning outcomes
On completion of this topic you will be expected to be able to:
1. Perform multivariate estimation and hypothesis testing
2. Use exploratory data reduction techniques, such as principal component analysis and factor analysis, and confirmatory factor analysis
3. Describe the assumptions and limitations of multivariate statistical techniques used to in the analysis of complex educational survey data
4. Select and justify appropriate multivariate and multilevel techniques to address an educational problem
5. Use standard statistical programs to analyse complex survey data sets to answer educational problems
6. Write interpretive reports demonstrating appropriate use, interpretation and communication of multivariate and multilevel statistical results for non-specialist audiences | 484 | 2,558 | {"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} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.866912 |
https://de.scribd.com/doc/142557976/Word-Problems-System-of-Linear-Equation-in-Two-Variables-doc | 1,621,383,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00236.warc.gz | 220,141,788 | 112,589 | Sie sind auf Seite 1von 7
# TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
1. A collection of 105 coins consists of 1.00 peso and 5.00 pesos. If the total
value is 205.00 pesos, find the number of coins of each denomination in the
collection.
2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00
for each dress and P1,000.00 for each pair of shoes. He sold the dresses at a
profit of 20% and the shoes at a profit of 50%. If his total profit was P18,000.
How many dresses and shoes did he buy?
3. The sum of two numbers is 115. The difference is -21. Find the numbers.
4. Two angles are complementary, one angle is 42 degrees more than half
the other. Find the angles.
5. A two digit number is 6 times the sum of its digits. The tens digit is 1
more than the units digit. Find the original number.
6. The sum of two numbers is 29. One number is 7 less than twice the other
number. Find the two numbers.
7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers.
Eugene bought 6 hamburgers and 4 french-fries French fries for P330.00.
Erick bought 4 hamburgers and 9 french fries for P505.00. What is the cost
of each french-fries and each hamburgers?
8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all,
how many of each denomination does she have?
9. A vending machine takes only nickels and dimes. There are 5 times as
many dimes as nickels in the machine. The face value of the coins is \$4.40.
How many of each coin are in the machine.
10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab
supply room has only 50% and 20% sulfuric acid solution on hand, how much
of each should be mixed to obtain twelve liters of a 30% solution.
1 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
SOLUTION
1. A collection of 105 coins consists of 1.00 peso and 5.00 pesos. If the total value is 205.00 pesos,
find the number of coins of each denomination in the collection.
Solution by elimination:
Let x = number of 1 peso coins
Y= number of 5 peso coins
( )
( )
'
+
+
2 205 5
1 105
equation y x
equation y x
Subtract the first equation from the second equation( equation 2 minus equation
1) , the result is
100 4 y
25 y
, substitute in the first equation
X + 25 = 105
80 x
Answer # 1: There are 80 1 peso coins and 25 5 peso coins.
2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00 for each dress and
P1,000.00 for each pair of shoes. He sold the dresses at a profit of 20% and the shoes at a profit of
50%. If his total profit was P18,000. How many dresses and shoes did he buy?
Solution by elimination method:
Let x = number of dresses
Y = number of shoes
Equation:
( )( ) ( )( )
'
+
+
18000 1000 5 . 0 400 2 . 0 : 2 equation
48000 1000 400 : 1 equation
y x
y x
Divide equation 1 by 100 the result is
480 10 4 + y x
, call this equation 3
Simplify equation 2, the result is
18000 500 80 + y x
, divide this by 20, the result is
900 25 4 + y x
, call this equation 4
2 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
Subtract equation 3 from equation 4
( )
420 15
480 10 4
900 25 4
+
+
y
y x
y x
28 y
, substitute in equation 3 and
solve for x.
480 10 4 + y x
becomes
( ) 480 28 10 4 + x
200 4 x
50 x
3. The sum of two numbers is 115. The difference is -21. Find the numbers.
Solution by elimination method: Let x and y be the numbers
'
+
21 : 2 equation
115 : equation1
y x
y x
Subtract equation 2 from equation 1, the result is
( )
136 2
21
115
+
y
y x
y x
68 y
, substitute this in equation 1 and solve for x.
115 + y x
becomes
115 68 + x
47 x
Answer #3: The numbers are 68 and 47
4. Two angles are complementary, one angle is 42 degrees more than half the other. Find the angles.
Solution by substitution method: Let x and y be the measure of the angles
'
+
+
y x
y x
2
1
42 : equation2
90 : equation1
Substitute equation 2 in equation 1, the result is
90
2
1
42 +
,
_
## + y y , multiply both sides by 2 it becomes
180 2 84 + + y y
3 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
180 3 84 + y
96 3 y
32 y
substitute this in equation 2
( ) 58 32
2
1
42 + x
Answer # 4: The measure of the angles are 32 and 58 degrees.
5. A two digit number is 6 times the sum of its digits. The tens digit is 1 more than the units digit.
Find the original number.
Solution by elimination method: Let x = units digit
Y = tens digit
The value of the number is x+10y .
( )
'
+
+ +
x y
y x y x
1 : 2 equation
6 10 : 1 equation
Simplify equation 1 it becomes,
y x y x 6 6 10 + +
0 4 5 y x
y x 4 5
( ) x x + 1 4 5 we substitute equation 2.
x x 4 4 5 +
4 x , therefore y=5.
Answer: The original number is 54.
6. The sum of two numbers is 29. One number is 7 less than twice the other number. Find the two
numbers.
Solution by substitution method: Let x and y be the numbers
'
+
7 2 : 2 equation
29 : 1 equation
y x
y x
Substitute equation 2 in equation 1, the result is
( ) 29 7 2 + y y
36 3 y
4 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
12 y
, substitute this in equation 2 to solve for x.
( ) 7 12 2 x
17 x
Answer # 6 : The numbers are 17 and 12.
7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers. Eugene bought 6
hamburgers and 4 french-fries French fries for P330.00. Erick bought 4 hamburgers and 9 frenchfriyes
for P505.00. What is the cost of each french-fries and each hamburgers?
Solution by elimination method: Let x = number of hamburgers
Y = number of French fries
'
+ +
+ +
4 equation this call , 1515 27 12 3 505 9 4 : 2 equation
3 equation this call , 660 8 12 2 330 4 6 : 1 equation
y x y x
y x y x
Subtract equation 3 from equation 4, the result is
( )
855 19
660 8 12
1515 27 12
+
+
y
y x
y x
45 y
, substitute this in equation 1 and solve for x,
( ) 330 45 4 6 + x
150 6 x
25 x
Answer #7: The cost of each hamburger is P25.00 and the cost of each French
fries is P45.00
8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all, how many of each
denomination does she have?
Solution by elimination method: Let x= number of P20 bills
Y=number of P50 bills
'
+
+
1490 50 20 : 2 equation
40 : 1 equation
y x
y x
Multiply equation 1 by 20, the result is
800 20 20 + y x
, subtract this from equation 2, the result is
5 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
( )
690 30
800 20 20
1490 50 20
+
+
y
y x
y x
23 y
, substitute this in equation 1, and solve for x.
40 23 + x
17 x
Answer #8: There are 17 P20 bills and 23 P50 bills
9. A vending machine takes only nickels and dimes. There are 5 times as many dimes as nickels in
the machine. The face value of the coins is \$4.40. How many of each coin are in the machine.
Solution by elimination method: Let x = number of nickels
Y= number of dimes
'
40 . 4 10 . 05 . : equation2
x 5 y : equation1
y x
Multiply equation 2 by 100, the result is,
440 10 5 + y x
, then use equation 1 the result is
( ) 440 5 10 5 + x x
440 55 x
8 x , substitute in equation 1 and solve for y
( ) 8 5 y
40 y
Answer #9: There are 8 nickels and 40 dimes.
10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply
room has only 50% and 20% sulfuric acid solution on hand, how much of each
should be mixed to obtain twelve liters of a 30% solution.
Solution by elimination method: Let x = amount of 50% sulfuric acid solution in
liters
Y = amount of 20% sulfuric acid solution
in liters
( ) ( ) ( )
'
+
+
12 3 . 0 2 . 0 5 . 0 : equation2
12 : equation1
y x
y x
Multiply equation 2 by 10, the result is,
36 2 5 + y x
, call this equation 4
6 To suggest a topic e-mail reviewermath@yahoo.com. thanks
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION
IN TWO VARIABLES)
Multiply equation 1 by 2, the result is
24 2 2 + y x
, call this equation 5
Subtract equation 5 from equation 4, the result is
( )
12 3
24 2 2
36 2 5
+
+
x
y x
y x
4 x , substitute in equation 1 and solve for y, the result is
12 4 + y
8 y
Answer #10: They need 4 liters of 50% sulfuric acid and 8 liters of 20% sulfuric
acid.
7 To suggest a topic e-mail reviewermath@yahoo.com. thanks | 2,719 | 8,370 | {"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} | 3.796875 | 4 | CC-MAIN-2021-21 | latest | en | 0.911427 |
https://www.eathyreading.website/2021/02/how-to-solve-price-elasticity-of-supply.html | 1,670,358,522,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00116.warc.gz | 792,899,223 | 25,187 | # HOW TO SOLVE PRICE ELASTICITY OF SUPPLY
Price elasticity of supply is the degree of responsiveness of quantity supplied to changes in price.
Mathematically, it is expressed as
$PES=\frac{\%∆Q_s}{\%P}$
Where $\%Q_s$ is the percentage change in quantity supplied
$\%P$ is the percentage change in price
The accompanying examples will illustrate the steps in calculating price elasticity of supply
Example 1
Determine the price elasticity of supply if a $10\%$ increase in price results in an $18\%$ increase in quantity supplied.
$PES=\frac{\%∆Q_s}{\%P}$
$PES=\frac{18}{10}$
$PES=1.8$
The numerical value of price elasticity of supply is greater than $1$, hence supply is relatively elastic.
Example 2
Calculate the price elasticity of supply if a change in the price of books from €4 to €2 causes the quantity supplied to decrease from 10 pens to 5 pens.
$PES=\frac{\%∆Q_s}{\%P}$
Unlike the first example, we were not given the percentage change in price and quantity, hence we solve for percentage changes using the midpoint method.
For percentage change in quantity supplied,
$Q_s=\frac{Q_1-Q_0}{\frac{Q_1+Q_0}{2}}\times100$
$Q_1=10$, $Q_0=5$
$Q_s=\frac{5-10}{\frac{5+10}{2}}\times100$
$Q_s=\frac{-5}{7.5}\times100$
$Q_s=-66.7\%$
For percentage change in price,
$P=\frac{P_1-P_0}{\frac{P_1+P_0}{2}}\times100$
$P_1=2$, $P_0=4$
$P=\frac{2-4}{\frac{2+4}{2}}\times100$
$P=\frac{-2}{3}\times100$
$P=-66.7\%$
$PES=\frac{-66.7\%}{-66.7\%}$
$PES=1$
The numerical value of price elasticity of supply is one, hence, supply is said to be unitary elastic.
Example 3
Determine the price elasticity of supply if a rise in the price of books from €4 to €8 causes the number of books supplies to rise from 8 books to 20 books.
Percentage change in quantity supplied is
$Q_s=\frac{Q_1-Q_0}{\frac{Q_1+Q_0}{2}}\times100$
$Q_1=20$, $Q_0=8$
$Q_s=\frac{20-8}{\frac{20+8}{2}}\times100$
$Q_s=\frac{12}{14}\times100$
$Q_s=85.7\%$
percentage change in price,
$P=\frac{P_1-P_0}{\frac{P_1+P_0}{2}}\times100$
$P_1=8$, $P_0=4$
$P=\frac{8-4}{\frac{8+4}{2}}\times100$
$P=\frac{4}{6}\times100$
$P=66.7\%$
$PES=\frac{88.7\%}{66.7\%}$
$PES=1.3$
The coefficient of price elasticity of supply is greater than one, hence supply is relatively elastic.
Example 4
Determine the price elasticity of supply if the quantity of bags supplied increases from 4500 to 5500 as a result of a rise in the price of bags from €14 to €18 per bag.
Percentage change in quantity supplied is
$Q_s=\frac{Q_1-Q_0}{\frac{Q_1+Q_0}{2}}\times100$
$Q_1=5500$, $Q_0=4500$
$Q_s=\frac{5000-4500}{\frac{5500+4500}{2}}\times100$
$Q_s=\frac{1000}{5000}\times100$
$Q_s=20\%$
percentage change in price,
$P=\frac{P_1-P_0}{\frac{P_1+P_0}{2}}\times100$
$P_1=18$, $P_0=14$
$P=\frac{18-14}{\frac{18+14}{2}}\times100$
$P=\frac{4}{16}\times100$
$P=25\%$
$PES=\frac{20\%}{25\%}$
$PES=0.8$
As you can see, the numerical value of the Price elasticity of supply is 0.8, which is lesser than one but greater than 0. Therefore, supply is relatively inelastic.
Example 5
A rise in the price of shoes from €14 to €18 per shoe causes the quantity supplied to rise from 4000 to 6000 shoes. Describe the price elasticity of supply
$Q_1=6000$, $Q_0=4000$
$Q_s=\frac{6000-4000}{\frac{6000+4000}{2}}\times100$
$Q_s=\frac{2000}{5000}\times100$
$Q_s=40\%$
$P_1=18$, $P_0=14$
$P=\frac{18-14}{\frac{18+14}{2}}\times100$
$P=\frac{4}{16}\times100$
$P=25\%$
$PES=\frac{40\%}{25\%}$
$PES=1.6$
The supply can be described as relatively elastic since the numerical value of PES is greater than one.
Related post
Example 6
The price of new cars increases by $5\%$. If the price elasticity of supply for new cars is 0.6, the quantity supplied of new cars will rise by how many percent?
$PES=\frac{\%∆Q_s}{\%P}$
Here percentage change in price{5\%} is given and the numerical value of Price elasticity of supply{0.6) is given, so, we would solve it as a mathematician does.
$0.6=\frac{\%∆Q_s}{5}$
Cross multiply
$0.6\times5=\%∆Q_s$
$3=\%∆Q_s$
$\%∆Q_s=3$
So, the quantity supplied will rise by $3\%$
Help us grow our readership by sharing this post | 1,470 | 4,089 | {"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} | 4.53125 | 5 | CC-MAIN-2022-49 | latest | en | 0.715629 |
https://economics.stackexchange.com/questions/51137/existence-of-a-walrasian-equilibrium | 1,719,357,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00201.warc.gz | 191,445,143 | 37,449 | # Existence of a Walrasian Equilibrium
I was wondering how to tackle part b of this question. I've already completed part a) using the usual method of equating the MRS (kink point for the second utility function) then looking at feasibility but part b is confusing me as to where to start.
Do i look at the assumptions of the existence of a Walrasian equilibrium such as the utility functions being strictly increasing, concave and continuous and if so what do I do with the information? If anyone could give me a starting point i'd appreciate that.
Thanks
So after solving a), you have the demand functions $$x_{11}(p)$$, $$x_{21}(p)$$, $$x_{12}(p)$$, $$x_{22}(p)$$, right?
An equilibrium exists if there is such a $$p$$ for which there is no excess demand or excess supply*. In this scenario, the supply is constant, given by the endowment thus you are looking for a price $$p$$ for which $$x_{11}(p) + x_{21}(p) = \omega_{11} + \omega_{21}$$ and $$x_{12}(p) + x_{22}(p) = \omega_{12} + \omega_{22}.$$ If you have calculated the demand functions correctly it is enough to use one of these market clearing equations to determine for which parameters $$a$$,$$b$$ the equation can hold, as the other equation follows from this one and the budget constraints.
*A special case is if $$p = 0$$, in which case excess supply is possible in equilibrum. | 346 | 1,349 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.929671 |
https://gist.github.com/aammd/ba73669e90ea27b8af2f | 1,656,258,092,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103269583.13/warc/CC-MAIN-20220626131545-20220626161545-00503.warc.gz | 324,797,767 | 20,868 | {{ message }}
Instantly share code, notes, and snippets.
aammd/fizzbuzz.md
Last active Jan 25, 2017
An example of calculating "FizzBuzz" in R using dplyr and magrittr
Apparently, there is a simple problem called Fizz buzz which is sometimes used to identify competent programmers. A good opportunity to practice some `dplyr` and `magrittr` tricks.
```library(dplyr)
library(magrittr)
library(knitr)
1:100 %>%
data.frame %>%
set_names("num") %>%
tbl_df %>%
mutate(fizz=num %>% mod(3) %>% equals(0) %>% ifelse("fizz",num),
buzz=num %>% mod(5) %>% equals(0) %>% ifelse("buzz",fizz),
fizzbuzz= num %>% mod(15) %>% equals(0) %>% ifelse("fizzbuzz",buzz)
) %>%
select(num,fizzbuzz) %>%
kable("markdown")```
num fizzbuzz
1 1
2 2
3 fizz
4 4
5 buzz
6 fizz
7 7
8 8
9 fizz
10 buzz
11 11
12 fizz
13 13
14 14
15 fizzbuzz
16 16
17 17
18 fizz
19 19
20 buzz
21 fizz
22 22
23 23
24 fizz
25 buzz
26 26
27 fizz
28 28
29 29
30 fizzbuzz
31 31
32 32
33 fizz
34 34
35 buzz
36 fizz
37 37
38 38
39 fizz
40 buzz
41 41
42 fizz
43 43
44 44
45 fizzbuzz
46 46
47 47
48 fizz
49 49
50 buzz
51 fizz
52 52
53 53
54 fizz
55 buzz
56 56
57 fizz
58 58
59 59
60 fizzbuzz
61 61
62 62
63 fizz
64 64
65 buzz
66 fizz
67 67
68 68
69 fizz
70 buzz
71 71
72 fizz
73 73
74 74
75 fizzbuzz
76 76
77 77
78 fizz
79 79
80 buzz
81 fizz
82 82
83 83
84 fizz
85 buzz
86 86
87 fizz
88 88
89 89
90 fizzbuzz
91 91
92 92
93 fizz
94 94
95 buzz
96 fizz
97 97
98 98
99 fizz
100 buzz
jangorecki commented Oct 20, 2015
I hope nobody is solving this way on R interviews :)
```f = seq(3,100,3) | 621 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.474787 |
https://www.mmo-champion.com/threads/1209298-Perl-programming?p=18703476&viewfull=1 | 1,516,562,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00631.warc.gz | 969,765,186 | 16,869 | 1. ## Perl programming?
So I'm taking a beginning Perl programming class for my IA degree at my local University, and I'd like to note that my programming skills are limited to basic basic Java and HTML. Yikes right? but anyway I was given an optional assignment that I just can't figure out for the life of me. All we had to do was modify a little bit of our code to get a sum, but of course I can't figure it out.
See we started with (or at least this was the direction I went) :
(note there is code above this, but nothing that affects the loop.)
\$i = \$x;
print "Multiple of fours between \$x and \$y are:\n";
while (\$i <= \$y) {
if (\$i % 4 == 0) {
print "\$i\t";}
\$i = \$i + 1;}
And this works. The program takes \$x and \$y from <STDIN> checks to see if it is a multiple of 4 and if so prints it.
So if x and y are say 1 and 10 the output shows up as: "4 (tab space) 8" on my screen.
Then the teacher had to mess with me and write down an optional step. The step is to display the sum of all the multiples of four.
I tried:
\$i = \$x;
print "Multiple of fours between \$x and \$y are:\n";
while (\$i <= \$y) {
if (\$i % 4 == 0) {
print "\$i\t";
\$z = \$z + \$i;
print "Sum = \$z";}
\$i = \$i + 1;}
Which kinda works? If i use the same x and y values 1 and 10 my output is "4 (tab space) Sum = 48 (tab space) Sum = 12" and I want "4 (tab space) 8 (tab space) Sum = 12"
So even though this is optional, I'd really like to know why my code isn't doing what I thought it would be doing. If anyone who knows a bit more about Perl than I do or even just progamming basics in general. All input would be appreciated.
2. Hi there,
I don't know perl, but I do know some other languages (tested this with PHP)..
Here goes:
Code:
```\$i = \$x;
\$z = 0; //Don't know if this is the right way to declare a variable, but you get the idea
print "Multiple of fours between \$x and \$y are:\n";
while (\$i <= \$y)
{
if (\$i % 4 == 0)
{
print "\$i\t";
\$z = \$z + \$i;
}
\$i = \$i + 1;
}
print "Sum = \$z";```
I declared \$z before your while loop and put 'print sum' after your while loop.
You want to do it this way because the way you have it, it prints "\$i (tab) Sum = \$z" everytime the if-condition is true. That's why it prints "4 (tab space) Sum = 48 (tab space) Sum = 12" (it went through the if two times).
What you want to do is let it print the "Sum = \$z" when it's done with the while loop.
I hope you understand what I mean, kind of a vague explanation sorry :P
3. I get what you're saying about it being in the loop. Even though I'm new to programming, I feel like I should of at least caught that.
Anyways, I do appreciate the your help and and the speed of your response so thank you very much
P.S. Do you happen to know of any sites where I may be able find like practice problems or other similar things to work on improving my coding?
4. No problem really, always nice to help someone
Also, don't worry too much. Not seeing little things like that happens every now and then when the brain activates derp-mode.
I haven't programmed in Perl so I'm afraid I can't help you with finding a usefull site, maybe someone can
5. Originally Posted by Shevik
Do you happen to know of any sites where I may be able find like practice problems or other similar things to work on improving my coding?
I assume you mean in general, and not just within perl.
You should have a look at project euler. It's a great site, with many good problems, which you can solve in the language of your choice.
As for perl specific problems, you should go to your university library and borrow "programming perl" and "learning perl". Both have problems throughout the book, which you could solve.
Hope that helps
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• | 1,040 | 3,871 | {"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} | 2.9375 | 3 | CC-MAIN-2018-05 | latest | en | 0.925082 |
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# ME410_Quiz3 - ME410/610 Intro to System Control Quiz 03...
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Unformatted text preview: ME410/610 Intro to System Control Quiz 03 WRITE YOUR NAME ON THE BACK go- Answer the following questions for the transfer function: G(s) = [52 + 4s + 4]/[s3 + 3s2 + 8.55] What are the poles and zeros? /r 7 31:) :1 ; Mr 5;» :13 ”‘3 3 K ~33 4 H fl 3 if " r a t“ if? ":7: 3f ,, 7): \f {-32, Ll C5353} g Q \$ ”I -' ,M_M..._'_«,_,W.._._,._v~— “MWW 1 3 :1 . 1,, Graph the poles and zeros in the s-plane. Is the systemstable? (RX 0;; Y‘i’l CE ‘3 (j 3’, 5 g 3’ l ' Jig)? 1». T3 3 “1 I 3 1‘ _. 7/“ r‘ ti; r z. {3;}? fife/7A mm: a 703\$ Fall 2010 Sept. 24 ...
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http://mathhelpforum.com/calculus/189199-annoying-trigonometric-substitution-integral.html | 1,526,811,255,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00382.warc.gz | 190,291,680 | 10,854 | 1. Annoying trigonometric substitution integral
How do you solve this integral: From 0 to a, x^2*sqrt(a^2-x^2)
I have no idea where pi comes from, so here's what I do
Let x=a*sin(theta)
My work ends up to be (4a^4/32) - (a^4sin4a/32)
Help!
2. Re: Annoying trigonometric substitution integral
When you get your new integral in the theta domain, what is the expression for that?
3. Re: Annoying trigonometric substitution integral
Originally Posted by some_nerdy_guy
How do you solve this integral: From 0 to a, x^2*sqrt(a^2-x^2)
I have no idea where pi comes from, so here's what I do
Let x=a*sin(theta)
Unfortunately, you error occurs somewhere in here- in exactly the part you did not show. Did you change the limits of integration as you worked? If x= a sin(theta), then when x= 0, a sin(theta)= 0 so theta= 0. When x= a, a= a sin(theta) so sin(theta)= 1, theta= pi/2.
My work ends up to be (4a^4/32) - (a^4sin4a/32)
Help!
4. Re: Annoying trigonometric substitution integral
Okay, thanks guys. I've figured out the problem, which is, as HallsofIvy stated, I didn't change my limits of integration in which 'a' should have been 'pi/2.' Could anyone possibly explain why if x=a, then theta=pi/2. I don't understand the logic behind that.
5. Re: Annoying trigonometric substitution integral
Originally Posted by some_nerdy_guy
Could anyone possibly explain why if x=a, then theta=pi/2. I don't understand the logic behind that.
That's because of the substitution, the substitution was $\displaystyle x=a\sin(\theta) \Leftrightarrow \theta=\arcsin\left(\frac{x}{a}\right)$
So if $\displaystyle x=a$ that means $\displaystyle \theta=\arcsin\left(\frac{a}{a}\right)=\arcsin(1)= \frac{\pi}{2}$
6. Re: Annoying trigonometric substitution integral
Ahhh, okay. I really need to brush up on my trig.
7. Re: Annoying trigonometric substitution integral
When you do a substitution on a definite integral, you must change over three things: the limits, the integrand, and the differential. For an indefinite integral, of course, while you don't need the limits, you must still transform the integrand and the differential. | 606 | 2,131 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.832939 |
http://mathematica.stackexchange.com/questions/tagged/series-expansion+syntax | 1,409,607,189,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535920694.0/warc/CC-MAIN-20140909043344-00088-ip-10-180-136-8.ec2.internal.warc.gz | 337,548,851 | 11,635 | # Tagged Questions
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### How do I plot a Taylor polynomial of various degrees in two variables? [closed]
I want to plot the Taylor polynomials of $f(x,y)= Sin(1 + x + y^2)/(4 + x^2 + y^2)$ of degrees 4 and 7 around the point $(0,0)$ over the rectangle $[-\pi,\pi]\times [-\pi,\pi]$ I am currently using ...
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### Proper treatment of roots and powers in Series?
I have the following problem in Mathematica 9 on Linux. I let Mathematica compute the Series expansion: ... | 310 | 1,152 | {"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} | 2.59375 | 3 | CC-MAIN-2014-35 | latest | en | 0.846943 |
https://www.convertunits.com/from/alen/to/nanon | 1,675,066,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00181.warc.gz | 735,953,349 | 12,644 | ## Convert alen [Danish] to nanon
alen nanon
Did you mean to convert alen [Danish] alen [Scandinavia] alen [Swedish] to nanon
How many alen in 1 nanon? The answer is 1.5931177314004E-9.
We assume you are converting between alen [Danish] and nanon.
You can view more details on each measurement unit:
alen or nanon
The SI base unit for length is the metre.
1 metre is equal to 1.5931177314004 alen, or 1000000000 nanon.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between alen [Danish] and nanon.
Type in your own numbers in the form to convert the units!
## Quick conversion chart of alen to nanon
1 alen to nanon = 627700000 nanon
2 alen to nanon = 1255400000 nanon
3 alen to nanon = 1883100000 nanon
4 alen to nanon = 2510800000 nanon
5 alen to nanon = 3138500000 nanon
6 alen to nanon = 3766200000 nanon
7 alen to nanon = 4393900000 nanon
8 alen to nanon = 5021600000 nanon
9 alen to nanon = 5649300000 nanon
10 alen to nanon = 6277000000 nanon
## Want other units?
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## Definition: Nanon
The SI prefix "nano" represents a factor of 10-9, or in exponential notation, 1E-9.
So 1 nanon = 10-9 n.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 536 | 1,790 | {"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-2023-06 | latest | en | 0.780233 |
https://kr.mathworks.com/matlabcentral/profile/authors/19988784?detail=answers | 1,642,436,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00713.warc.gz | 409,664,948 | 21,346 | Community Profile
# Bathala Teja
Last seen: 약 1달 전 2021 이후 활성
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FFT analysis using powergui in simulink
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forming a function handles in matrix
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issue in solving set of ode's
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ODE45 is taking hours and hours to compute
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How to solve system of first order ODE's by forming matrix
Q: dy1/dt = 2*y1+5*y2+y3 dy2/dt = 7*y1+9*y2 dy3/dt = 4*y1+y2+3*y3 y1(0) = 0, y2(0)= 0, y3(0) = 0; i want to plot y1(t), y2...
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답변 | 1,576 | 4,660 | {"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} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.756791 |
http://www.expertsmind.com/questions/write-an-algorithm-to-find-outputs-number-of-cars-301103356.aspx | 1,611,113,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00046.warc.gz | 143,181,707 | 12,579 | ## Write an algorithm to find outputs number of cars, Data Structure & Algorithms
Assignment Help:
A company is carrying out a survey by observing traffic at a road junction. Every time a car, bus or lorry passed by road junction it was noted down. 10 000 vehicles were counted during survey.
Write an algorithm, using an algorithm, that:
• inputs all 10000 responses
• Outputs number of cars, buses and Lorries that passed by the junction during survey
• Outputs number of vehicles which weren't cars, buses or Lorries during survey
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Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report! | 579 | 2,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.90603 |
https://www.airmilescalculator.com/distance/ybr-to-yyd/ | 1,632,521,412,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00062.warc.gz | 657,182,070 | 40,084 | # Distance between Brandon (YBR) and Smithers (YYD)
Flight distance from Brandon to Smithers (Brandon Municipal Airport – Smithers Airport) is 1193 miles / 1920 kilometers / 1037 nautical miles. Estimated flight time is 2 hours 45 minutes.
Driving distance from Brandon (YBR) to Smithers (YYD) is 1401 miles / 2254 kilometers and travel time by car is about 26 hours 6 minutes.
## Map of flight path and driving directions from Brandon to Smithers.
Shortest flight path between Brandon Municipal Airport (YBR) and Smithers Airport (YYD).
## How far is Smithers from Brandon?
There are several ways to calculate distances between Brandon and Smithers. Here are two common methods:
Vincenty's formula (applied above)
• 1192.968 miles
• 1919.896 kilometers
• 1036.661 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
• 1189.377 miles
• 1914.117 kilometers
• 1033.541 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 Brandon Municipal Airport
City: Brandon
Country: Canada
IATA Code: YBR
ICAO Code: CYBR
Coordinates: 49°54′36″N, 99°57′6″W
B Smithers Airport
City: Smithers
Country: Canada
IATA Code: YYD
ICAO Code: CYYD
Coordinates: 54°49′28″N, 127°10′58″W
## Time difference and current local times
The time difference between Brandon and Smithers is 2 hours. Smithers is 2 hours behind Brandon.
CDT
PDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 161 kg (355 pounds).
## Frequent Flyer Miles Calculator
Brandon (YBR) → Smithers (YYD).
Distance:
1193
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1193
Round trip? | 492 | 1,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.830426 |
http://mathematica.stackexchange.com/questions/27007/find-radii-of-concentric-circles-in-image/27010 | 1,461,895,804,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110356.23/warc/CC-MAIN-20160428161510-00204-ip-10-239-7-51.ec2.internal.warc.gz | 117,348,079 | 21,851 | # Find radii of concentric circles in image
I have an image of concentric circles, I would like to find the radii of the circles (only the innermost few are important).
I've had a go using what I could find in previous posts, but am a bit confused about which method I should be using - if I need MorphologicalComponents, or whether to use SelectComponents "count" or "equivalentradius", or colornegate etc. Sometimes the circles are broken (especially when I binarize), so I need to look for incomplete circles too..
So far I have:
i = Import["http://i.imgur.com/oTTM9MG.jpg"];
b = Binarize[i, {0.3, 1}];
m = MorphologicalComponents[b];
c = SelectComponents[m, {"Count", "Holes"},
1000 < #1 < 20000 && #2 > 0 &] // Colorize
or, using :
i = Import["http://i.imgur.com/oTTM9MG.jpg"];
disk = ColorNegate[Binarize[i, {0.3, 1}]];
rings = ComponentMeasurements[
450 <= #1[[1]] <= 550 && 325 <= #1[[2]] <= 375 &&
10 <= #2 <= 500 &];
Show[{disk,
Graphics[{{Red, Circle[rings[[1, 2]][[1]], rings[[1, 2]][[2]]]}}]}]
Am I making it harder than it is? Could someone bump in the right direction?
both this, How to find circular objects in an image? and this, Finding the centroid of a disk in an image were helpful (but I need to expand it to multiple circles and fit partial circles).
Any help would be appreciated. Thanks!
-
This reminds me of electron diffraction rings. What is the picture of? – rcollyer Jun 14 '13 at 15:47
Its from a Fabry-Perot etalon en.wikipedia.org/wiki/Fabry%E2%80%93P%C3%A9rot_interferometer (being used to look at the anomalous Zeeman effect in mercury :) ) – Jeff Jun 14 '13 at 16:01
Looked a bit like electron or x-ray diffraction to me too but the central spot from the incident beam looked a bit too weak. – s0rce Jun 14 '13 at 18:34
(I guess I should add, the F.P. etalon is being imaged with a B&W digital camera..) – Jeff Jun 17 '13 at 15:14
What you could do is apply an edge filter and find the threshold which binarizes your image best:
i = Import["http://i.imgur.com/oTTM9MG.jpg"];
edges = LaplacianGaussianFilter[ColorNegate[i], 2];
Manipulate[Binarize[edges, t], {t, 0, .1}]
After that you could select all objects with a certain radius or you throw out all small objects with a specific "Count". You have to decide then what you prefer as radius. I thought maybe the "MeanCentroidDistance" gives a quite stable measure
circles =
SelectComponents[
MorphologicalComponents[LaplacianGaussianFilter[ColorNegate@i, 2], 0.0056],
"Count", # > 300 &];
Colorize[circles]
ComponentMeasurements[circles, "MeanCentroidDistance"]
(*
{26 -> 271.952, 33 -> 262.778, 129 -> 221.202, 157 -> 209.482,
329 -> 154.293, 398 -> 136.493}
*)
-
Thanks very much! This is along the same lines I was going but much more sophisticated (and your approach works!). I'll try this with all my data (I'm afraid the data may get a little noisy, but I'll get to that later). Thanks! – Jeff Jun 14 '13 at 16:11
Hi, I've played around with this quite abit and like the solution. However, MorphologicalComponents requires the circles to be connected (I think?) Is there a way to use only partial circles? e.g. if I use only the middle section of the previous image, i.imgur.com/JkRlEvW.png – Jeff Jun 17 '13 at 15:16
@Jeff If you have circles which are not complete, than you'll run indeed into problems. Have you tried to work with a circular Hough transform? Mathematica provides a similar approach in ImageLines for linear structures. For arbitrary circles it is computational even more expensive but if you like I can post another answer with some code. – halirutan Jun 18 '13 at 23:23
You can use the center detection from this answer:
Basically, the center you're looking for is a point in the image for which every gradient points towards or away from the center. That means we can minimize this error term:
squaredError = 1/2 ({cx - x, cy - y}.{-gy, gx})^2;
which leads to a linear equation system:
errDerivative = Expand[D[squaredError, {{cx, cy}}]];
linearSystem = {{D[errDerivative, cx],
D[errDerivative, cy]}, -errDerivative /. {cx -> 0, cy -> 0}}
Now we can simply insert the gradients from the image into this equation system:
gradientX = ImageData@GaussianFilter[img, 1, {0, 1}];
gradientY = ImageData@GaussianFilter[img, 1, {1, 0}];
ls = Total[
y -> yArr}, {-2, -1}];
center = LinearSolve @@ ls;
The center location is in indices (starting with 1) and image processing functions want coordinates (starting with 0, at the bottom left corner), so I have to convert the coordinates:
center[[1]] -= 1;
Then I can apply a polar transform
maxRadius = 250;
polar = ImageTransformation[img,
center + {Cos[#[[1]]], Sin[#[[1]]]}*#[[2]] &, {360, maxRadius},
DataRange -> Full,
PlotRange -> {{0, 360 \[Degree]}, {1, maxRadius}}]
The mean brightness for each row gives a measure of the strength of each radius:
radiusStrength = Mean /@ ImageData[polar, DataReversed -> True];
peakX = Position[
True][[All, 1]];
ListLinePlot[radiusStrength, Epilog -> {Red, Point[peaks[[-4 ;;]]]}]
The radii with the highest strengths correspond nicely with the radii you're looking for:
Show[img,
Graphics[{Red, Dashed, Circle[center, #] & /@ peaks[[-4 ;;, 1]]}]]
Note that while this is more complex than a 'ComponentMeasurements'-based solution it is much more robust. You don't have to adjust any thresholds or parameters. And it should work fine, even if parts of the circles are occluded or hardly visible.
-
+1 Great answer! - is the slight waviness in the polar` image showing the fact that the original image isn't circular or is it an optical illusion or inaccuracy? – cormullion Jun 14 '13 at 14:04
@cormullion: I'm not 100% sure. Since the center is a least squares estimate using all gradients in the image, the error could be due to gradients that don't point towards the center. Or maybe my index -> coordinate conversion is off by one? – nikie Jun 14 '13 at 14:09
The coordinate conversion was off by one. Fixed it. I love MMA for image processing, but the distinction between indices and coordinates is ugly. Especially as there is no built-in function to convert them. – nikie Jun 14 '13 at 14:19
Stuff like this is awesome. Can you recommend any resources to get started with MMA image processing? I'd like to play around with it sometime. – Reid Jun 14 '13 at 14:33
@Reid This online event should be worth catching. – cormullion Jun 14 '13 at 16:46 | 1,768 | 6,390 | {"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} | 3.25 | 3 | CC-MAIN-2016-18 | longest | en | 0.850135 |
https://socratic.org/questions/a-block-of-wood-is-floating-in-water-the-weight-of-the-part-of-the-block-above-w | 1,596,625,215,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00413.warc.gz | 511,963,607 | 5,933 | # A block of wood is floating in water. The weight of the part of the block above water is one third of the total weight of the block. What is the the part the water by the block of weight of wood, using Archimedes' principle?
Feb 15, 2016
The wood is $\frac{2}{3}$ as dense as water.
#### Explanation:
The question is not clear but I think this is likely what you are looking for.
If $\frac{2}{3}$ of the block of wood is submerged then
$\textcolor{w h i t e}{\text{XXX}}$the mass of water displaced by $\frac{2}{3}$ of the volume of the block of wood is equal to the mass of the block of wood.
Another way to say this is:
each unit volume of wood can be supported by $\frac{2}{3}$ of a unit volume of water. | 192 | 715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 3.859375 | 4 | CC-MAIN-2020-34 | latest | en | 0.941388 |
http://www.convertunits.com/from/bar/to/MN+/+(m%5E2) | 1,501,036,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425751.38/warc/CC-MAIN-20170726022311-20170726042311-00588.warc.gz | 396,091,273 | 10,867 | # ››Convert bar to meganewton/square metre
bar MN / (m^2)
Did you mean to convert bar to meganewton/square metre millinewton/square metre
How many bar in 1 MN / (m^2)? The answer is 10.
We assume you are converting between bar and meganewton/square metre.
You can view more details on each measurement unit:
bar or MN / (m^2)
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0E-5 bar, or 1.0E-6 MN / (m^2).
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between bars and meganewtons/square meter.
Type in your own numbers in the form to convert the units!
# ››Want other units?
You can do the reverse unit conversion from MN / (m^2) to bar, or enter any two units below:
# Enter two units to convert
From: To:
I'm feeling lucky, show me some random units.
# ››Definition: Bar
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
# ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 457 | 1,703 | {"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-2017-30 | latest | en | 0.863616 |
https://www.jianshu.com/p/135312d1c1aa | 1,642,907,842,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00330.warc.gz | 799,898,649 | 66,565 | # Quartz 2D编程指南之三：路径(Path)
Figure 3-1 Quartz supports path-based drawing
Figure 3-2 A path that contains two shapes, or subpaths
Figure 3-3 A clipping area constrains drawing
Figure 3-4 Multiple paths; each path contains a randomly generated circle
Figure 3-5 Defining an arc with two tangent lines and a radius
Figure 3-6 Multiple paths; each path contains a randomly generated curve
Figure 3-7 A cubic Bézier curve uses two control points
Figure 3-8 A quadratic Bézier curve uses one control point
Quartz的一些函数将路径的子路径看成是闭合的。这些函数显示地添加一条直线来闭合 子路径，如同调用了CGContextClosePath函数。
Figure 3-9 Multiple paths; each path contains a randomly generated ellipse
Figure 3-10 Multiple paths; each path contains a randomly generated rectangle
CGPathCreateMutable，取代CGContextBeginPath
CGPathMoveToPoint，取代CGContextMoveToPoint
CGPathCloseSubpath，取代CGContexClosePath
Table 3-1 Parameters that affect how Quartz strokes the current path
linewidth是线的总宽度，单位是用户空间单元。
linejoin属性指定如何绘制线段间的联接点。Quartz支持表3-2中描述的联接样式。
Table 3-2 Line join styles
linecap指定如何绘制直线的端点。Quartz支持表3-3所示的线帽类型。默认的是butt cap。
Table 3-3 Line cap styles
Linedash pattern(虚线模式)允许我们沿着描边绘制虚线。我们通过在CGContextSetLineDash结构体中指定虚线数组和虚线相位来控制虚线的大小及位置。
CGContextSetLineDash结构如下：
voidCGContextSetLineDash(
CGContextRefctx,
floatphase,
constfloatlengths[],
size_t count,
);
Figure 3-11 Examples of line dash patterns
Quartz提供了表3-4中的函数来描边当前路径。其中一些是描边矩形及椭圆的便捷函数。
Table 3-4 Functions that stroke paths
CGContextBeginPath(context);
for(k =0; k < count; k +=2)
{
}
CGContextStrokePath(context);
Figure 3-12 Concentric circles filled using different fill rules
Quartz提供了表3-5中的函数来填充当前路径。其中一些是填充矩形及椭圆的便捷函数。
Table 3-5 Functions that fill paths
result = (alpha * foreground) + (1 - alpha) *background
“颜色与颜色空间”章节里面详细讨论了颜色值的alpha组件，该组件用于指定颜色的透明度。在本章的例子中，我们可以假设颜色值是完全不透明的(alpha = 0)。对于不透明的颜色值，当我们用普通混合模式时，所有绘制于背景之上的绘图都会遮掩住背景。
Figure 3-13 The rectangles painted in the foreground
Figure 3-14 The rectangles painted in the background
Figure 3-15 Rectangles painted using normal blend mode
Figure 3-16 Rectangles painted using multiply blend mode
Figure 3-17 Rectangles painted using screen blend mode
Figure 3-18 Rectangles painted using overlay blend mode
Figure 3-19 Rectangles painted using darken blend mode
Figure 3-20 Rectangles painted using lighten blend mode
Figure 3-21 Rectangles painted using color dodge blend mode
Figure 3-22 Rectangles painted using color burn blend mode
Figure 3-23 Rectangles painted using soft light blend mode
Figure 3-24 Rectangles painted using hard light blend mode
Figure 3-25 Rectangles painted using difference blend mode
Figure 3-26 Rectangles painted using exclusion blend mode
Figure 3-27 Rectangles painted using hue blend mode
Figure 3-28 Rectangles painted using saturation blend mode
Figure 3-29 Rectangles painted using color blend mode
Figure 3-30 Rectangles painted using luminosity blend mode
Listing 3-1 Setting up a circular clip area
CGContextBeginPath(context);
CGContextAddArc(context, w/2, h/2, ((w>h) ? h : w)/2,0,2*PI,0);
CGContextClosePath(context);
CGContextClip(context);
Table 3-6 Functions that clip the graphics context | 1,002 | 3,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.306368 |
https://www.dpmms.cam.ac.uk/~twk/my-book-cor.html | 1,660,240,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00033.warc.gz | 633,028,872 | 4,848 | # Corrections to the Pleasures of Counting For First Reprint
PLEASE NOTE: Since html does not deal easily with formula those who have browsers which can deal with such things will be better off using DVI PS or LATEX
Here is a list of corrections and additions for the first reprinting. I should like to thank Robin Chapman of Exeter (first mistake spotted), Professor Cassels, Dr Garling, Dr Osborne, Professor Barenblatt, and Prof J. Taylor (all of Cambridge), Harold N. Ward (University of Virginia), Adam Atkinson, Paul Shallhorn and Pat Henry for spotting various problems and for telling me about them. I would particularly like to thank Douglas Quadling and Dr Francis Clarke (Swansea) for long lists of useful comments.
Side note on page 54
Some bats have a mechanism for making themselves temporarily deaf when emitting sound bursts. (See [260], a book evidently written by a man who has never seen a bat he did not love.) They use high frequency sound because their prey is small.
Side note on page 190
According to Arnol'd in [264], Kolmogorov started as a student of history. His first paper, written when he was seventeen, concerned the mediaeval tax records in Novgorod. After he had presented his conclusions to a seminar he asked the historian in charge whether he agreed with them. `Young man,' the professor said, `in history we need at least five proofs for any conclusion.' Next day, Kolmogorov switched to mathematics.
Side note on page 221
When we talked about oxygen absorption we assumed that the area for absorption A was related to the volume V of an animal in the same way as the surface area of a sphere is related to its volume so that A is proprtional to V to the power 2/3. But we noted on page 218 that lungs look very `fractal'. If the surface of the lung has `Richardson number' different from the surface of a sphere then we would have A is proportional to V to the power beta with beta not equal to 2/3 and the slope of the mouse-elephant curve becomes easier to explain. (See pages 342-5 of [261].)
Side note on page 392
However, most of those whose opinions I respect reject Penrose's views on the nature of human thought `not because they are crazy, but because they are not crazy enough'.
Side note on page 501
(Added in second printing.) If you wish to write complex formulae, I strongly recommend Grätzer's Math Into LaTeX [263].
Side note on page 505
Hacker lore says that two heads are faster than one and three heads slightly faster than two. Thereafter, adding manpower to a software project actually makes it slower. The only way to speed up a project is to use cleverer people. Hence the three Knuth rule. If three Knuths working together cannot do it, it cannot be done.
Page 53, line before (i) add
(The restrictions on theta-subscript1 are rather arbitrary, but some restrictions are needed.)
Page 53, near middle of written matter.
Replace angle BCX by angle ACX.
Page 53, 4 lines up.
Replace a=AX cos theta_{1} by AX=a sin theta_{1}.
Page 53, last displayed formula should read.
a sin theta_{1}=2r sin theta_{1}-theta_{2} over 2
Page 59 middle. Replace
for 1 leq i leq n\$ and guess that that a_{m+1} will be close to
T_{n}(a_{m-n+1},a_{m-n+2},dots ,a_{m})=P(n)
by for 0 leq i leq n-1 and guess that that a_{m+1} will be close to}
T_{n-1}(a_{m-n},a_{m-n+1},dots ,a_{m})=P(n)
Page 79, line 2
replace k=b=1 by k=c=1
replace b(t)=1-r(t) by b(t)=1+r(t)
Page 79, first displayed formula
replace r'(t)=-(1-r(t))r(t) by
r'(t)=-(1+r(t))r(t)
Page 79, second displayed formula replace r(t) over 1-r(t)=r(0) over 1-r(0) e^{-t} by
r(t) over 1+r(t)=r(0) over 1+r(0) e^{-t}
Page 79, end of first para replace
r(t) tends to 0 for all t>0.
by
r(t) tends to 0 as t tends to infinity.
Page 120
In the 2nd displayed formula replace tau=gl^{-1}t^{-2} by tau=gl^{-1}t^{2}. As a consequence, in the 4th, 5th and 6th formula g/l should be replaced by l/g. Finally in line -7 replace `varies inversely with' by `is proportional to".
Page 121, REPLACE side note as follows
According to materials scientists, when molten glass cools it remains a liquid but one whose coefficient of viscosity increases as the temperature decreases. Thus we can blow glass at high temperatures, mold it at lower temperatures and so on. However, the demonstration of liquid properties for glass at room temperature lies at the very edge of modern experimental technique.
Page 126, add at end of first side note
The treatment of dimension I have given follows the traditional pattern in glossing over certain points. In [261] Barenblatt takes a more modern approach and shows that, if we think a little harder, we can understand a lot more.
Page 133, bottom line
Replace `wave front' by `shock front'.
Page 143, Page 143, first equation after 7.1
Page 166. In BOTH Figure 8.2(a) and Figure 8.2(b)
Replace squareroot2 on x-axis by 1/squareroot2.
Pages 167 to 171.
A factor of 1/2 is missing in many of the formulae. The correct factor may, of course, be found by setting f=1. Alternatively: wherever ))b/N appears replace with ))b/(2N); on page 168 lines 7, 8 and 9 replace f(rb/N)b/N by f(rb/N))b/(2N); page 171, second displayed formula, replace h by h/2.
Pages 175 and 176
Replace l/g by g/l wherever it occurs.
Page 177, 3/4 down
Replace `collected paper with' by `collected papers) with' and `velocity?)' by `velocity?'.
Page 186, top third TWICE
Replace K(t) by Kt.
Page 187, one third down
Replace `not much bigger than l' by `not much bigger than l^{2}'.
Page 264, Figure (e)
Replace 3 on top path by 2. Replace 1 on almost vertical path by 0.
Page 264, Figure 11.5
Label on edge XY replace 10p+10 by p+10
Page 274, last line but one of Exercise 11.2.6
Replace 10p+10 by p+10.
Page 278 line 19
Extra space between ) [and] are
Page 290 line -5 replace
for all integer r with R geq 1.
by
for all integers r with r geq 1.
Page 291, last displayed formula
Replace pi by its square root.
Pages 331, 391, 392
The correct possesive form is Hodges'.
Page 331, 3/4 way down. Replace
A is encoded by SA, B by SB and so on.
by
A is encoded by S(A), B by S(B) and so on.
Page 364, 1/4 way down
Replace `Feynmann' by `Feynman'.
Page 403 First part of proof
There are three places where the power 1-uN should be replaced by (1-u)N. It is not made clear where the condition 0 leq u leq 1 comes in (observe that u/(1-u) leq 1). I have rewritten the proof slightly to make this clearer.
Page 419, First line Exercise 17.4 (ii)
Replace chi(0)<0 by chi(0)>0.
Page 419, Second line Exercise 17.4 (ii)
Replace chi(t)<0 by chi(t)>0.
Page 420, line 19
Replace `nth generation' by `n+1st generation'.
Page 428, last line of (i)
Replace `K=L-1/4' by `K=1/4-L'.
Page 428, first line of (ii)
Replace `K=-1/4' by `K=1/4'.
Page 428, displayed formula in (iii)
Remove the factor of 1/2.
Page 428, last displayed formula
Replace `L-1/4, by `1/4-L'.
Page 440.Replace first displayed formula in (ii) by
dI/dS=dI/dt dt/dS= (beta SI-gamma I)/-beta SI=-1+gamma/beta S}
Page 442, third displayed formula up should read
J=2S_{0}(S_{0}/(rho-1).
Page 442, second displayed formula up should read
S_{0}(S_{0}/(rho-1)=(1+nu/rho)
Page 443, line 10
Replace `Claude' by `Claud'.
Page Page 490, line 10
Page 517, line 5 ending with 37-8
Page 528
[260] J. D. Altringham. Bats, Biology and Behaviour. OUP, Oxford, 1996
[261] G. I. Barenblatt. Scaling, Self-similarity and Intermediate Asymptotics. CUP, Cambridge, 1996
[262] G. K. Batchelor. Kolmogoroff's theory of locally isotropic turbulence. Proceedings of the Cambridge Philosophical Society, 43:533-59, 1947.
[263] G. A. Grätzer. Math into LaTeX. Birkhäuser, Boston, 1996.
[264] S. H. Lui. An interview with Vladimir Arnol'd. Notices of the AMS, 44(3):432-8, April 1997.
historians, killjoy, routed, 104, 153, 190
Page 531, under Knuth K. E., add item
three Knuth rule, 505
Page 531, replace
lungs, as surfaces of nearly infinite area, 218
by
lungs, as fractals, 218, 221
Page 531, under menagerie, mathematical, add item
bats, 54
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# Arrays and quantifiers - PowerPoint PPT Presentation
Arrays and quantifiers. Programming Fundamentals 19 Feliks Klu ź niak. a : integer array of N elements Sorted : a[ 0 ] =< a[ 1 ] and a [ 1 ] =< a[ 2 ] and ... and a [ N – 2 ] =< a[ N – 1 ]. a : integer array of N elements
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### Arrays and quantifiers
Programming Fundamentals
19
Feliks Kluźniak
a : integer array of N elements
Sorted : a[ 0 ] =< a[ 1 ] and a[ 1 ] =< a[ 2 ] and
... and a[ N – 2 ] =< a[ N – 1 ]
Arrays and predicates
a : integer array of N elements
Sorted : a[ 0 ] =< a[ 1 ] and a[ 1 ] =< a[ 2 ] and
... and a[ N – 2 ] =< a[ N – 1 ]
Unsorted : a[ 0 ] > a[ 1 ] or a[ 1 ] > a[ 2 ] or
... or a[ N – 2 ] > a[ N – 1 ]
NOTE: Recall de Morgan’s laws, and not (x =< y) = (x > y) .
Arrays and predicates
a : integer array of N elements
Sorted : a[ 0 ] =< a[ 1 ] and a[ 1 ] =< a[ 2 ] and
... and a[ N – 2 ] =< a[ N – 1 ]
Unsorted : a[ 0 ] > a[ 1 ] or a[ 1 ] > a[ 2 ] or
... or a[ N – 2 ] > a[ N – 1 ]
NOTE: Recall de Morgan’s laws, and not (x =< y) = (x > y) .
This is unwieldy, and does not work very well for large N !
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (A i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
This is generalized conjunction:
for every i
such thati is positive and smaller than N – 1,
a[ i ] is not greater than a[ i + 1 ] .
i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (A i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
This is generalized conjunction:
for every i we most often say ”for all”, but mean ”for every”/”for each”
such thati is positive and smaller than N – 1,
a[ i ] is not greater than a[ i + 1 ] .
i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (Ai : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
quantifier
i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (Ai: 0 =< i< N – 1 : a[ i] =< a[ i+ 1 ])
quantifier, dummy variable
i: 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (Ai: 0 =< i< N – 1 : a[ i ] =< a[ i+ 1 ])
quantifier, dummy variable
Note: The dummy variable is bound by the quantifier: we cannot substitute anything for it, and the value of the entire formula does not depend on it.
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (Ai: 0 =< i< N – 1 : a[ i] =< a[ i+ 1 ])
quantifier, dummy variable
Note: The dummy variable is boundby the quantifier: we cannot substitute anything for it, and the value of the entire formula does not depend on it.
The variables on which the value of the formula depends are the free variables . In this case they are N and the various elements of the array a.
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (A i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
quantifier, dummy variable, range
i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (A i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
quantifier, dummy variable, range, quantified formula
i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1]
Arrays and predicates
The (bounded) universal quantifier :
Sorted : (A i : 0 =< i < N – 1 : a[ i ] =< a[ i + 1 ])
This is simply shorthand for
true and a[ 0 ] =< a[ 1 ] and a[ 1 ] =< a[ 2 ] and ...
.... and a[ N – 2 ] =< a[ N – 1 ]
Notice that when N =< 1, this expression reduces to
true !
Arrays and predicates
The (bounded) existential quantifier :
Unsorted : (E i : 0 =< i < N – 1 : a[ i ] > a[ i + 1 ])
This is generalized disjunction:
there exists ani
such thati is positive and smaller than N – 1,
and a[ i ] is not greater than a[ i + 1 ] .
i : 0 =< i < N – 1 : a[ i ] > a[ i + 1]
Arrays and predicates
The (bounded) existential quantifier :
Unsorted : (E i : 0 =< i < N – 1 : a[ i ] > a[ i + 1 ])
This is simply shorthand for
false or a[ 0 ] =< a[ 1 ] or a[ 1 ] =< a[ 2 ] or ...
.... or a[ N – 2 ] =< a[ N – 1 ]
Notice that when N =< 1, this expression reduces to
false !
Arrays and predicates
The dummy variable is bound (i.e., not free).
One must be careful to avoid a clash of names.
P( y ) : (A x : 0 < x < N : x > y)
Q : 9 < x and x < 99 and P( x )
P( x ) is not equivalent to (A x : 0 < x < N : x > x)
but to (A z : 0 < z < N : z > x)
(where z is some unused variable).
Arrays and predicates
de Morgan’s laws that:
(not (A v : r : P)) = (E v : r : not P)
(not (E v : r : P)) = (A v : r : not P)
Arrays and predicates
(A v : r : P)
is equivalent to ( A v :: r implies P )
Hence the pleasant and very important property that a universally quantified formula with an empty range is true!
Arrays and predicates
(A v : r : P)
is equivalent to ( A v :: r implies P )
Hence the pleasant and very important property that a universally quantified formula with an empty range is true!
NOTE:
”Empty range” does not mean that there is no range expression, as in
( A x :: P( x ) ). It means that the range expression is false, as in
( A i : 0 < i < 1 : P( x ) ) .
Arrays and predicates
From our generalised de Morgan’s laws it follows that an existentially quantified formula with an empty range is false.
(not (A v : r : P)) = (E v : r : not P)
Arrays and predicates
Sorted( n ) : (A i : 0 =< i < n – 1 : a[ i ] =< a[ i + 1 ])
Unsorted( n ) : (E i : 0 =< i < n – 1 : a[ i ] > a[ i + 1 ])
For an array with only one element the range is empty. So the array is sorted, and is not unsorted.
Arrays and predicates
(S i : 0 =< i < N : a[ i ])
The sum of the elements of ain the range [ 0, N ),
i.e., a[ 0 ] + a[ 1 ] + a[ 2 ] + .... + a[ N – 1 ] .
NOTE: This is not a predicate: its value is numerical.
Arrays and predicates
(S i : 0 =< i < N : a[ i ])
(P i : 0 =< i < N : a[ i ])
The product of the elements of ain the range [ 0, N ) ,
i.e., a[ 0 ] * a[ 1 ] * a[ 2 ] * .... * a[ N – 1 ].
NOTE: These are not predicates: their values are numerical.
Arrays and predicates
(S i : 0 =< i < N : a[ i ])
(P i : 0 =< i < N : a[ i ])
(N i : 0 =< i < N : a[ i ] = 0)
The number of elements of ain the range [ 0, N ) whose value is 0.
(Here, a[ i ] = 0 is just an example of a predicate and a[ i ] is just an example of an expression).
NOTE: These are not predicates: their values are numerical.
Arrays and predicates
(S i : 0 =< i < N : a[ i ])
(P i : 0 =< i < N : a[ i ])
(N i : 0 =< i < N : a[ i ] = 0)
The counting quantifier has no classical counterpart, the first two do:
N - 1
N - 1
a i
a i
i = 0
i = 0
Arrays and predicates
Let j < k, and let P be some predicate.
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(A i : 0 =< i < j : P( i ))(A i : 0 =< i < k : P( i ))
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(A i : 0 =< i < j : P( i )) is implied by
(A i : 0 =< i < k : P( i ))
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(A i : 0 =< i < j : P( i )) is implied by
(A i : 0 =< i < k : P( i ))
Because
P( 0 ) and P( 1 ) and .... and P( j )
is implied by
P( 0 ) and P( 1 ) and .... and P( j ) and ... and P( k )
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(E i : 0 =< i < j : P( i )) (E i : 0 =< i < k : P( i ))
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(E i : 0 =< i < j : P( i )) implies (E i : 0 =< i < k : P( i ))
Arrays and predicates
Let j < k, and let P be some predicate.
Which of the following two predicates is stronger?
(E i : 0 =< i < j : P( i )) implies (E i : 0 =< i < k : P( i ))
Because
P( 0 ) or P( 1 ) or ... or P( j )
implies
P( 0 ) or P( 1 ) or ... or P( j )or ... or P( k )
Arrays and predicates
So it will normally be more natural to use a universally quantified formula for expressing an invariant when we do something with an array.
Recall that our method is to find an invariant that is weaker than the desired outcome, and then to strengthen it as the program progresses:
P
P and not B
desired
• There is a whole class of programs that traverse an array sequentially. They share the following properties:
• The invariant is of the form
• ( A j : 0 =< j < k : P( j ) ) and k =< N
• The loop condition is of the form
• k != N and .... (sometimes simply k != N )
• The termination argument is
• the invariant implies N – k >= 0, and N – k strictly decreases with each iteration (becausekincreases).
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. Let us write a program that finds whether a contains the integer m .
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. Let us write a program that finds whether a contains the integer m .
Specifically, we want variable k to be the lowest value such that a[ k ] = m . If a does not contain m, we want k = N .
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. Let us write a program that finds whether a contains the integer m .
Specifically, we want variable k to be the lowest value such that a[ k ] = m . If a does not contain m, we want k = N .
This can be expressed as follows:
R: (A i : 0 =< i < k : a[ i ] != m) and (a[ k ] = m or k = N)
k
all elements are different from m
m
k
all elements are different from m
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Notice that P( k ) is a straightforward weakening of R .
k
all elements are different from m
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Clearly, P(k) and a[ k ] = m means that we found the answer.
k
all elements are different from m
m
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Clearly, P(k) and a[ k ] = m means that we found the answer, and so does P(k) and k = N .
k
all elements are different from m
m
k
all elements are different from m
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Clearly, P(k) and a[ k ] = m means that we found the answer, and so does P(k) and k = N .
P(k) is trivial to establish. How?
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Clearly, P(k) and a[ k ] = m means that we found the answer, and so does P(k) and k = N .
P(k) is trivial to establish: k := 0 (empty range!).
Arrays and predicates
R: ( sequentially. They share the following properties:A i : 0 =< i < k : a[ i ] != m) and (k = N or a[ k ] = m)
Our invariant will be P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
Clearly, P(k) and a[ k ] = m means that we found the answer, and so does P(k) and k = N .
P(k) is trivial to establish: k := 0 (empty range!). So the program writes itself:
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k) Why?
od
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k) Why?
od
P(k) will hold after the assignment k := k + 1 if P( k + 1 ) holds before the assignment.
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k) Why?
od
P(k) will hold after the assignment k := k + 1 if P( k + 1 ) holds before the assignment.
So our task is to show that
(P( k ) and k != N and a[ k ] != m) implies P( k + 1 ) . Why is this so?
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k) Why?
od
P(k) will hold after the assignment k := k + 1 if P( k + 1 ) holds before the assignment.
So our task is to show that
(P( k ) and k != N and a[ k ] != m) implies P( k + 1 ) . Why is this so?
P( k + 1 ) = (P( k ) and (a[ k ] != m) and (k + 1 =< N))
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
Let m = 4 and let the array be:
5 1 4 0
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
Let m = 4 and let the array be:
5 1 4 0
k = 0
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
Let m = 4 and let the array be:
5 1 4 0
k = 1
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
Let m = 4 and let the array be:
5 1 4 0
k = 2 BINGO!
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
If we were searching for m = 6, then we would proceed:
5 1 4 0
k = 2
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
If we were searching for m = 6, then we would proceed:
5 1 4 0
k = 3
Arrays and predicates
Our invariant will be sequentially. They share the following properties:P(k): ( A j : 0 =< j < k : a[ j ] != m ) and k =< N.
k := 0; % P(k)
while k != N and a[ k ] != m do
% P(k) and k != N and a[ k ] != m
k := k + 1 % P(k)
od
If we were searching for m = 6, then we would proceed:
5 1 4 0
k = 4
So we don’t try to access a[ 4 ]. Why?
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
which is trivially established by WHAT?
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
which is trivially established by sumsq, k := 0, 0 .
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
which is trivially established by sumsq, k := 0, 0 .
And, of course, P( k ) and k = N implies R .
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
sumsq, k := 0, 0 ;% P( k )
while k != N do
% P( k ) and k != N
sumsq := sumsq + a[ k ] * a[ k ] ; % P( k ) violated
...
od
% P( k ) and k = N, hence R
Arrays and predicates
Let sequentially. They share the following properties:a be an array with N elements. We want to establish:
R : sumsq = ( S i : 0 =< i < N : a[ i ] * a[ i ] )
Our invariant will be
P( k ) : (sumsq = ( S i : 0 =< i < k : a[ i ] * a[ i ] ) and k =< N.
sumsq, k := 0, 0 ;% P( k )
while k != N do
% P( k ) and k != N
sumsq := sumsq + a[ k ] * a[ k ] ; % P( k ) violated
k := k + 1 % progress, P( k ) restored
od
% P( k ) and k = N, hence R
Arrays and predicates
Exercise: sequentially. They share the following properties:
Let a be an array with M elements.
Write a program (together with the proof of correctness!)
that establishes:
R : nneg = ( N i : 0 <= i < M : a[ i ] < 0 )
counting
quantifier
NOTE: Don‘t do this. Do Homework 2 instead.
Arrays and predicates | 6,317 | 18,835 | {"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} | 3.84375 | 4 | CC-MAIN-2017-39 | latest | en | 0.739967 |
https://math.stackexchange.com/questions/72459/do-i-understand-complex-differentiability-correctly | 1,571,824,449,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00179.warc.gz | 590,631,309 | 31,659 | # Do I understand complex differentiability correctly?
I don't use complex analysis much in my "day job", but I'm perusing the materials for personal interest. I just wanted to check with the community about my understanding of complex analytic (holomorphic) functions, and why they are "special" compared to real-valued functions.
A far as I understood it, the definition of a complex holomorphic function is a consequence of the fact that complex numbers essentially live in $\mathbb{R}^2$ but has a funny definition of algebraic multiplication $(x_1,y_1)(x_2,y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)$ that makes it useful for defining roots of negative numbers, while at the same time still acting like real numbers when the $y$ component is zero.
Complex functions, since they can essentially be mapped to something that's $f:\mathbb R^2 \rightarrow \mathbb R^2$, have by definition derivatives that are Frechet derivatives ($2 \times 2$ linear operators in this case). However, the derivative also need to act like a single complex number, so the $2 \times 2$ linear operator needs to act like multiplication by a complex number. By the definition of complex products above, this means there are special restrictions on the matrix elements of this operator. And by extension of the fact that Frechet derivatives is just a matrix of partial derivatives, special relationships exist between the partial derivatives.
Quite a bit down the road, this essentially leads to the consequences laid out in the Cauchy Integral Theorem. Is that correct?
TLDR: the definition of complex number products, along with their $\mathbb R^2$ nature, leads to the definition of holomorphic functions and the Cauchy Integral Theorem. Correct?
• Basically correct, if stated a bit peculiarly. Your "special restrictions on the matrix elements of this operator" are called the "Cauchy-Riemann equations". – GEdgar Oct 14 '11 at 0:15
• If I am linking things correctly, the Cauchy-Reimann basically requires the Frechet derivative to be Hermitian self-adjoint (and thus behave "like a scalar"). Which nothin came first, CR or self-adjointness? – Tim Lin Oct 14 '11 at 0:24
• No, it is not Hermitian self-adjoint. Probably you need to spend some time with a complex analysis textbook and learn the more conventional terminology. After that, come back and see what it looks like as a "special" $\mathbb R^2$ function. – GEdgar Oct 14 '11 at 15:33 | 582 | 2,435 | {"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} | 3.421875 | 3 | CC-MAIN-2019-43 | latest | en | 0.914937 |
https://bramanswanderings.com/2021/06/15/a-tower-of-blocks/ | 1,680,037,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00627.warc.gz | 175,488,635 | 30,124 | ## A Tower of Blocks
Tonight on LEGO Masters they had to build a tower so my thoughts wandered to a tower that I built with Ankerstein.
This tower is not as tall as what they built tonight as it is only built with with three boxes of Ankerstein? In the picture above you see the first three boxes in the series. Number 6 is the Basic Box and 6A and 8A are Extension Boxes. The 3 boxes together make up set number 10: 6+6A+8A=10
Of course, I am sure that you wanted to see what was inside the boxes. Here are the three boxes without their covers. They are in order left to right.
You can see that each box adds new types of stones as well as simply adding more of the earlier stone types.
As you add more boxes the structures that you build become more complex.
Each set of stones comes with plan books that show you how to build a structure using the stones in the boxes that make up the set. (6+6A+8A=10)
One of the plan books shows pictures of what the completed structure should look like, and the other is a layer plan.
The structure that I built is called Lookout Tower.
Here is the completed structure. Can you imagine this structure just by looking at the stones in their boxes?
As you build a structure it slowly comes together as you add layer after layer of stones.
Here is a nice view looking down on the roof.
This building was a bit of a challenge as you had to make sure you kept everything level. Especially as the top is wider than the bottom.
You may think that it is a shaky construction since the stones are simply stacked on top of each other. However, the weight of the stones above make for a solid base.
I don’t think that it would have survived the shake table they had on Lego Masters, but it would have done pretty well considering that it is not snapped together.
I have built some taller towers, but they are usually attached to a building. This one is built with set 24.
The tallest lighthouse that I have built is over 4 feet tall, but I don’t have time to look for the pictures tonight.
Steven
A Little More Ankerstein
Anker Church Building
Anker Church Building II
This entry was posted in Ankerstein and tagged , , , , , , , , , , , . Bookmark the permalink.
### 2 Responses to A Tower of Blocks
1. Betty says:
These are beautiful! I was aware stones like this were available. I like dollhouses, so these appeal to me. Thanks for sharing!
2. Pastor Cathy says:
A Tower of Blocks
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 577 | 2,519 | {"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-2023-14 | latest | en | 0.977476 |
http://www.ehow.co.uk/how_7802497_calculate-maximum-load-crane.html | 1,529,612,607,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00338.warc.gz | 403,987,818 | 11,445 | DISCOVER
# How to calculate the maximum load for a crane
Updated March 23, 2017
A crane can only lift as high a load as its main wheel allows. This maximum wheel load depends on several factors. The rated capacity load of course matters, but an impact allowance factor determines its significance. The other factors are the weight of the crane overall and the weight of the trolley and hoist assembly. Besides the crane's load, use the maximum wheel load to choose the crane's other components, such as its beam size.
Divide the crane's weight, measured in pounds, by 2. If the crane weight is 2722 Kilogram: 6,000 / 2 = 3,000.
Add the weight of the hoist and trolley, also measured in pounds. If they weigh 11.3 Kilogram combined: 3,000 + 2,500 = 5,500.
Multiply the rated capacity load weight by the impact allowance factor. If the former is 4536 Kilogram and the latter s 1.2: 10,000 x 1.2 = 12,000.
Add the answers from steps 2 and 3: 5,500 + 12,000 = 17,500.
Divide your answer by 2: 17,500 / 2 = 3969 Kilogram. This answer is the crane's maximum load. | 295 | 1,065 | {"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} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.941143 |
https://www.shaalaa.com/question-bank-solutions/calculate-mass-charge-one-mole-electrons-subatomic-particles-discovery-of-electron_10576 | 1,718,862,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00646.warc.gz | 880,765,500 | 9,655 | # Calculate the Mass and Charge of One Mole of Electrons. - Chemistry
Calculate the mass and charge of one mole of electrons.
#### Solution
Mass of one electron = 9.10939 × 10–31 kg
Mass of one mole of electron = (6.022 × 1023) × (9.10939 ×10–31 kg)
= 5.48 × 10–7 kg
Charge on one electron = 1.6022 × 10–19 coulomb
Charge on one mole of electron = (1.6022 × 10–19 C) (6.022 × 1023)
= 9.65 × 104 C
Is there an error in this question or solution?
Chapter 2: Structure of Atom - EXERCISES [Page 69]
#### APPEARS IN
NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.1 - (ii) | Page 69
Share | 222 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.698833 |
https://www.expertsmind.com/library/determine-the-volume-of-direct-surface-runoff-in-acre-ft-5302163.aspx | 1,713,664,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00467.warc.gz | 700,308,405 | 15,260 | Determine the volume of direct surface runoff in acre-ft
Assignment Help Civil Engineering
Reference no: EM13302163
Determine the volume of direct surface runoff in acre-ft that will result from the following storm. The basin area is 1000 acres. The phi index is 0.5 in./hr. 30-minute period 1 2 3 4 Rain intensity (in./hr) 0.9 0.5 0.6 0.5
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a ship has a cross-sectional area of 4000 ft^2 at the water line when the draft ( depth below the water line of the vessel bottom) is 10 ft. How many pounds of cargo will result in an increase of draft of 2 in ( specific weight is 64 Ib/ft^3 ) | 826 | 3,387 | {"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 | latest | en | 0.915404 |
https://brainly.my/tugasan/81243 | 1,484,778,635,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00042-ip-10-171-10-70.ec2.internal.warc.gz | 794,292,907 | 10,302 | # AIGEBRA 1.(X-2)-(2-x) TULIS JAWAPAN ANDA
2
dari qasyana
## Jawapan
2015-08-11T10:52:44+08:00
(x-2) - (2-x)
= x - 2 - 2 + x
= x + x - 2 - 2
= 2x - 4
okkk
kalau nak permudahkan, = 2(x-2)
2015-08-11T14:11:29+08:00
(x-2) - (2-x)
= x - 2 - 2 + x
= x + x - 2 - 2⇒di urutkan
= 2x - 4
semoga mem bantuu yee | 181 | 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} | 3.71875 | 4 | CC-MAIN-2017-04 | latest | en | 0.142622 |
https://metanumbers.com/181466 | 1,627,379,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00321.warc.gz | 368,988,271 | 10,902 | ## 181466
181,466 (one hundred eighty-one thousand four hundred sixty-six) is an even six-digits composite number following 181465 and preceding 181467. In scientific notation, it is written as 1.81466 × 105. The sum of its digits is 26. It has a total of 3 prime factors and 8 positive divisors. There are 88,480 positive integers (up to 181466) that are relatively prime to 181466.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 26
• Digital Root 8
## Name
Short name 181 thousand 466 one hundred eighty-one thousand four hundred sixty-six
## Notation
Scientific notation 1.81466 × 105 181.466 × 103
## Prime Factorization of 181466
Prime Factorization 2 × 41 × 2213
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 181466 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 181,466 is 2 × 41 × 2213. Since it has a total of 3 prime factors, 181,466 is a composite number.
## Divisors of 181466
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 278964 Sum of all the positive divisors of n s(n) 97498 Sum of the proper positive divisors of n A(n) 34870.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 425.988 Returns the nth root of the product of n divisors H(n) 5.204 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 181,466 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 181,466) is 278,964, the average is 3,487,0.5.
## Other Arithmetic Functions (n = 181466)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 88480 Total number of positive integers not greater than n that are coprime to n λ(n) 22120 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16417 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 88,480 positive integers (less than 181,466) that are coprime with 181,466. And there are approximately 16,417 prime numbers less than or equal to 181,466.
## Divisibility of 181466
m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 5 2 8
The number 181,466 is divisible by 2.
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (181466)
Base System Value
2 Binary 101100010011011010
3 Ternary 100012220222
4 Quaternary 230103122
5 Quinary 21301331
6 Senary 3520042
8 Octal 542332
10 Decimal 181466
12 Duodecimal 89022
20 Vigesimal 12dd6
36 Base36 3w0q
## Basic calculations (n = 181466)
### Multiplication
n×i
n×2 362932 544398 725864 907330
### Division
ni
n⁄2 90733 60488.7 45366.5 36293.2
### Exponentiation
ni
n2 32929909156 5975658894902696 1084378917022412632336 196777904556389130739484576
### Nth Root
i√n
2√n 425.988 56.615 20.6395 11.2657
## 181466 as geometric shapes
### Circle
Diameter 362932 1.14018e+06 1.03452e+11
### Sphere
Volume 2.50308e+16 4.13809e+11 1.14018e+06
### Square
Length = n
Perimeter 725864 3.29299e+10 256632
### Cube
Length = n
Surface area 1.97579e+11 5.97566e+15 314308
### Equilateral Triangle
Length = n
Perimeter 544398 1.42591e+10 157154
### Triangular Pyramid
Length = n
Surface area 5.70363e+10 7.04238e+14 148166
## Cryptographic Hash Functions
md5 a0f46ea02196d8d365b531a5c0f2f620 f38dabe6b48c3a30b16dce5a88cb19a0c393152d 013d88182219a670ca9f08be8120a437247230aa56dd1e90e99014c26d23d467 4e72da44405184fa486a58c7223134d35eabbd6731778194f27810f93540938908b132a15195f756f9845c143165c76472c0fc3374455a266ca60423ffc3274b 166562cd7aa1fa275bcad5bcbe057c58817999fc | 1,426 | 4,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-31 | latest | en | 0.796695 |
https://codegolf.stackexchange.com/questions/68864/make-bracket-squares?noredirect=1 | 1,719,340,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00705.warc.gz | 148,408,277 | 47,370 | # Make Bracket Squares
Every programmer knows that brackets []{}()<> are really fun. To exacerbate this fun, groups of interwoven brackets can be transformed into cute and fuzzy diagrams.
Let's say that you have a string that contains balanced brackets, like [{][<(]})>(()). Step one is to rotate the string 45 degrees clockwise. (In Mathematica, this can be almost done with Rotate[ur_string,-pi/4]). Here is the result of the first step:
[
{
]
[
<
(
]
}
)
>
(
(
)
)
Next add a diagonal space between each character.
[
{
]
[
<
(
]
}
)
>
(
(
)
)
Next, start with the left-most bracket and draw a square between it and its partner in crime.
+---+
| |
| { |
| |
+---+
[
<
(
]
}
)
>
(
(
)
)
Repeat this process with each pair of brackets, overwriting previous characters with +s if need be.
+---+
| |
| +-+---------+
| | | |
+-+-+ |
| |
| [ |
| |
| < |
| |
| ( |
| |
| ] |
| |
+-----------+
)
>
(
(
)
)
Continue until you have made everything nice and square.
+---+
| |
| +-+---------+
| | | |
+-+-+ |
| |
| +-----+ |
| | | |
| | +---+-+---+
| | | | | |
| | | +-+-+-+ |
| | | | | | | |
| +-+-+-+ | | |
| | | | | |
+-----+-+---+ | |
| | | |
| +-----+ |
| |
+---------+
+-----+
| |
| +-+ |
| | | |
| +-+ |
| |
+-----+
Input
Input will be a single line of balanced brackets and no other characters, with each bracket being one of []{}()<>. Each type of bracket is balanced individually, though different types may overlap (this is what makes the squares look interesting). A trailing newline is optional.
Output
Output will be the interlocking square pattern generated from the bracket string. Trailing spaces and trailing newline are optional, but there mustn't be leading whitespace.
Goal
This is code-golf, fewest bytes wins.
• Do we have to deal with nesting of the same type of bracket? e.g. for [[]] can we output two squares overlapping or do we have to output one square inside the other? Commented Jan 8, 2016 at 2:49
• One square is inside of the other. I will adjust my example. Edit: done. Commented Jan 8, 2016 at 2:53
# JavaScript (ES6), 269 274 278 296 261 bytes
Edit Saved 4 bytes thx @Neil
x=>[...x].map(c=>{g.push([],[]),z='<{[(>}])'.indexOf(c);if(z>3)for(j=a=o[z-4].pop();j<=b;j++)S(j,a,'|'),S(j,b,'|'),S(a),S(b);else o[z].push(b);b+=2},S=(y,x=j,c='-')=>g[y][x]=g[y][x]?'+':c,o=[[],[],[],[]],g=[],b=0)&&g.map(r=>[...r].map(c=>c||' ').join).join
TEST
F=x=>[...x].map(c=>{g.push([],[]),z='<{[(>}])'.indexOf(c);if(z>3)for(j=a=o[z-4].pop();j<=b;j++)S(j,a,'|'),S(j,b,'|'),S(a),S(b);else o[z].push(b);b+=2},S=(y,x=j,c='-')=>g[y][x]=g[y][x]?'+':c,o=[[],[],[],[]],g=[],b=0)&&g.map(r=>[...r].map(c=>c||' ').join).join
// Less golfed
U=x=>(
S = (y,x=j,c='-')=>g[y][x]=g[y][x]?'+':c,
o = [[],[],[],[]],
g = [],
b = 0,
[...x].map(c=>
{
g.push([],[]),
z='<{[(>}])'.indexOf(c);
if(z>3)
for(j = a =o[z-4].pop(); j <= b; j++)
S(j,a,'|'),
S(j,b,'|'),
S(a),
S(b)
else
o[z].push(b);
b += 2
}),
g.map(r=>
[...r].map(c=>c||' ').join
).join\n
)
function test() {
O.textContent=F(I.value)
}
test()
Input:<input id=I value='[{][<(]})>(())' oninput='test()'>
<pre id=O></pre>
• Why not [...r].map?
– Neil
Commented Jan 8, 2016 at 13:41
• Better still, [...r].map(c=>c||' ').
– Neil
Commented Jan 8, 2016 at 13:42
• @Neil I cannot use r.map because r is a sparse array and map skips missing elements. So I now use g, that is filled (and there are as many rows as columns in output) Commented Jan 8, 2016 at 14:08
• I didn't say r.map, I said [...r].map, and [...r] is NOT a sparse array, as you mentioned yourself in codegolf.stackexchange.com/a/61505 comment 5.
– Neil
Commented Jan 8, 2016 at 14:35
• @Neil I missed that ... it seems a nice hint, thanks Commented Jan 8, 2016 at 14:40
## Python 3, 226
n,p,*t=0,[],0,[],[],[],[]
for b in input():
r=t[ord(b)//30];r+=[n];n+=2
if b in'])}>':p+=[r[-2:]];del r[-2:]
R=range(n-1)
for y in R:print(''.join(' -|+'[sum((y in q)+2*(x in q)for q in p if x>=q[0]<=y<=q[1]>=x)]for x in R))
Example. Explanation:
n,p,*t=0,[],0,[],[],[],[] # n -> counter for input
# p -> bracket pairs
# t -> four stacks [unused, (), <>, [], {}]
for b in input(): # for each bracket b of input
r=t[ord(b)//30]; # r -> alias for b's stack
r+=[n]; # push bracket's index
n+=2 # increase counter by 2 (to add diagonal gaps)
if b in'])}>': # if b is a closing bracket
p+=[r[-2:]]; # pair the top 2 brackets on stack
del r[-2:] # pop them from stack
R=range(n-1) # n-1 is now the width/height of output
for y in R:
print(''.join(' -|+'[
sum((y in{a,b})+2*(x in{a,b})for a,b in p if x>=a<=y<=b>=x)
]for x in R))
# three nested loops:
# 1) for each line y
# 2) for each character x
# 3) for each bracket pair (a, b)
if x>=a<=y<=b>=x
# if x or y isn't in the inclusive range [a, b], we can skip it
(y in{a,b})
# if y is a or b, then the character lies on a horizontal edge of that square
# so we add 1 to the sum
2*(x in{a,b})
# if x is a or b, then the character lies on a vertical edge of that square
# so we add 2 to the sum
' -|+'[sum()]
# if it lies on a single edge, the sum will be 1 or 2 -> '-' or '|'
# if it lies on two edges, the sum will be 1 + 2 == 3 -> '+'
• You can save 2 bytes by not unpacking the bracket pairs in the last line. Commented Jan 8, 2016 at 11:50
# pb - 449 bytes
^w[B!0]{w[B=40]{b[39]}t[B]w[B!0]{w[B=T]{^b[1]}w[B=T+1]{^b[1]}w[B=T+2]{^b[2]}^[Y]^>}<[X]^w[B!1]{>}t[1]b[0]w[T!0]{>w[B=1]{t[T+1]b[0]}w[B=2]{t[T-1]b[0]}}b[3]vw[B!0]{>}^w[B!3]{b[0]<}b[0]vb[1]>[X]vv[X]b[43]t[X]^[Y]^<[X]w[B=0]{>}>[X]vv[X]b[43]w[Y!T-1]{vw[B=45]{b[43]}w[B=0]{b[124]}}vb[43]t[1]>w[B!43]{t[T+1]w[B=124]{b[43]}w[B=0]{b[45]}>}w[T!0]{^t[T-1]w[B=45]{b[43]}w[B=0]{b[124]}}w[B!43]{w[B=124]{b[43]}w[B=0]{b[45]}<}<[X]^[Y]^w[B=0]{>}b[0]>w[B=1]{b[0]>}}
I was all excited when I read this because I have a language that prints directly to a position! This challenge about positioning outputs should be easy and short!
Then I remembered pb is longwinded anyway.
With comments:
^w[B!0]{
w[B=40]{b[39]} # change ( to ' to make closing bracket calculation work
t[B]
# this used to just find the first matching bracket
# but then op clarified we had to use depth
# whoops
# <fix>
w[B!0]{
w[B=T]{^b[1]} # put a 1 above opening brackets of this type
w[B=T+1]{^b[1]} # same as before, but ugly hack to make ( work
w[B=T+2]{^b[2]} # put a 2 above closing brackets of this type
^[Y]^ # return to input line
>}
<[X]^w[B!1]{>}t[1]b[0] # set T to 1 above the opening bracket
w[T!0]{> # until T (depth) == 0:
w[B=1]{t[T+1]b[0]} # add 1 to T if 1
w[B=2]{t[T-1]b[0]} # subtract 1 from T if 2
}
b[3] # when T is 0, we've found the right one
vw[B!0]{>} # go to the end of the input
^w[B!3]{b[0]<}b[0]v # clean up the row above
# </fix>
b[1] # replace it with 1 so it's not confusing later
>[X]vv[X]b[43]t[X] # put a + at its output position and save coord
^[Y]^<[X]w[B=0]{>}>[X]vv[X]b[43] # put a + at opening bracket's output position
w[Y!T-1]{v
w[B=45]{b[43]} # replace - with +
w[B=0]{b[124]} # otherwise put |
}
vb[43] # put a + at lower left corner
t[1] # count side length + 1
>w[B!43]{
t[T+1]
w[B=124]{b[43]} # replace | with +
w[B=0]{b[45]} # otherwise put -
>}
w[T!0]{^ # create right side
t[T-1]
w[B=45]{b[43]}
w[B=0]{b[124]}
}
w[B!43]{ # create top side
w[B=124]{b[43]} # this replacement saves us from putting the last + explicitly
# which is why we counted the side length + 1, to get that
# extra char to replace
w[B=0]{b[45]}
<}
<[X]^[Y]^w[B=0]{>}b[0]>w[B=1]{b[0]>}# Go to next character (skipping 1s)
}
• I... Just... Can't... Commented Feb 10, 2016 at 2:11
# CJam, 117 bytes
q_,2*:M2#0a*\:iee{1=K/}${~)4%1>{a+aL+:L;}*}/L{2f*__~,>m*5Zb\m*{~W2$#%Mb\}%\2m*Mfb3am*}%e_2/{~2$2$=+3e<t}/" |-+"f=M/N*
Try it here.
# Ruby, 268
->z{a=(2..2*t=z.size).map{' '*t}
g=->y,x{a[y][x]=(a[y][x-1]+a[y][x+1]).sum>64??+:?|}
2.times{|h|s=z*1
t.times{|i|c=s[i]
c>?$&&(j=s.index(c.tr('[]{}()<>','][}{)(><')) (m=i*2).upto(n=j*2){|k|k%n>m ?g[k,m]+g[k,n]:h<1&&a[k][m..n]=?++?-*(n-m-1)+?+} s[i]=s[j]=?!)}} puts a} ungolfed in test program f=->z{ a=(2..2*t=z.size).map{' '*t} #make an array of strings of spaces g=->y,x{a[y][x]=(a[y][x-1]+a[y][x+1]).sum>64??+:?|} #function for printing verticals: | if adjacent cells spaces (32+32) otherwise + 2.times{|h| #run through the array twice s=z*1 #make a duplicate of the input (*1 is a dummy operation to avoid passing just pointer) t.times{|i| #for each index in input c=s[i] #take the character c>?$&&( #if ascii value higher than for \$
j=s.index(c.tr('[]{}()<>','][}{)(><')) #it must be a braket so find its match
(m=i*2).upto(n=j*2){|k| #loop through the relevant rows of array
k%n>m ?g[k,m]+g[k,n]: #if k!=n and k!m draw verticals
h<1&&a[k][m..n]=?++?-*(n-m-1)+?+ #otherwise draw horizontals (but only on 1st pass)
}
s[i]=s[j]=?!) #we are done with this set of brackets, replace them with ! so they will be ignored in next call of index
}
}
puts a
}
z=gets.chop
f[z] | 3,485 | 9,873 | {"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} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.887021 |
https://www.k5worksheets.com/tag/math/page/2/ | 1,581,965,354,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00417.warc.gz | 805,319,076 | 10,016 | #### TagMath
Understanding math is hard. Yet, is not that impossible! If you study hard, you may get a good score! What you must do is to repeat your exercise, again, and again. If you believe with your skills, and get so much confidence, we believe, that your score will be improved.
Since small, human learn to count. The skills of counting are including in the math. They tend to count since, in daily life, there are many objects and subjects which are needed to count. For example, you need to make sure the sum up of some clothes you have or the amount of money you have.
1st Grade Math Worksheets is the sheets used by the teacher in teaching the 1-grade students. Meanwhile, students may learn from the sheets about many things such as addition, multiplication and many others. Thus, to learn all these knowledge, we try to teach you with 1st-grade math worksheets. Let’s check this out:
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Teaching 6th-Grade-Students is different with teaching the students below them. However, teaching the adult and the mature one is also different isn’t it? Meanwhile, if you want to teach the adult one, you must find the best method which are very suitable for them while if you want to teach the small one, you must find the best method also. Not only the best method, but you must also find and get the right media for your learning process. Meanwhile, 6th Grade Math Worksheets is such a good media for your math class.
Simple Math Worksheets is the sheets used by the students in learning some simple math method. Meanwhile, before going to the complex math worksheets, you must teach your students with the simple math method. This is important since as a teacher, you need to explain them with the most simple first then going to more and more complex one. Teaching principal is to teach something steps by steps.
Second Grade Math Worksheets is the sheets used by the teacher in teaching their students with math, However, if you want to make your students mastered in math, you need to make sure to introduce them with some of the math skills before working with the sheets. However, as a teacher, you need also to make sure your students understand with some of the knowledge in the sheets.
Sixth Grade Math Worksheets is the sheets used by the teacher in order to teach the students with the sheets. However, if you want to use this sheets, before, you must know all the knowledge related with the sheets. You may not teach your students with the sheets until you, yourselves know what is the meaning of this sheets and how to work with this one.
Super Math Worksheets is the sheets used by the teacher in order for them to learn the students with high quality media of learning. However, if you want to look for other learning resources, we will always update you with some of the best one. Meanwhile, this math worksheets are very suitable to be used in math class. Thus, if you want to use this sheets, you must use it in math class. You can not use it in other class but math. | 757 | 3,585 | {"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-2020-10 | latest | en | 0.968767 |
https://math.stackexchange.com/questions/1088632/exact-sequence-of-finitely-generated-abelian-groups | 1,718,683,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00012.warc.gz | 325,730,473 | 35,372 | # Exact sequence of finitely generated abelian groups
Let $$0\rightarrow G_{1}\xrightarrow{f_{1}} G_{2}\xrightarrow{f_{2}} G_{3}\rightarrow 0$$ is a short exact sequence of finitely generated abelian groups. We call $\overline{G_{i}}$ the quotient of $G_{i}$ by its torsion. Then, we consider : $$0\rightarrow \bar{G_{1}}\xrightarrow{\bar{f_{1}}} \bar{G_{2}}\xrightarrow{\bar{f_{2}}} \bar{G_{3}}\rightarrow 0.$$
Prove that $\bar{f_{2}}$ is surjective and that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group.
My answer is (I'm not sure, that's why it looks like questions) : Is it true that $\bar{f_{2}}$ is surjective since $f_{2}$ is ? Is it true that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group since $Im(f_{1})=Ker(f_{2})$ ? Thank you.
• $\bar{f_2}$ is surjective because $f_2$ is surjective and a torsion element can't map to a non-torsion element. Commented Jan 2, 2015 at 16:40
Writing $G_i = \mathbb{Z}^{r_i} \oplus T_i,$ where $r_i \geq 0$ is an integer and $T_i$ is a torsion group, we have the following exact sequence: $$0 \rightarrow \mathbb{Z}^{r_1} \oplus T_1 \xrightarrow{f_1} \mathbb{Z}^{r_2} \oplus T_2 \xrightarrow{f_2} \mathbb{Z}^{r_3} \oplus T_3 \rightarrow 0.$$ From this, taking quotient by trosion subgroups, we get the following complex: $$0 \rightarrow \mathbb{Z}^{r_1} \xrightarrow{\bar{f_1}} \mathbb{Z}^{r_2} \xrightarrow{\bar{f_2}} \mathbb{Z}^{r_3} \rightarrow 0.$$ This is not exact in general. First we want to show that $\bar{f_2}$ is surjective. This follows from the fact that $f_2$ is surjective, and $f_2(t) \in T_3$ for every $t \in T_2.$ Now we want to show that $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion group. Let $x \in \text{ker}{\bar{f_2}}.$ Then $\bar{f_2}(x) = 0 \Rightarrow f_2(x) \in T_3.$ But $T_3$ is a torsion group. So there exists $n \in \mathbb{N},$ such that $f_2(nx) = nf_2(x) = 0.$ Hence there is $y \in G_1$ such that $f_1(y) = nx.$ Note that $y \notin T_1.$ So $nx = 0$ in $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}.$ This shows that, every element of $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion element. | 825 | 2,117 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.724811 |
http://www.qldscienceteachers.com/junior-science/physics/quizzes/work-and-power | 1,481,380,787,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543315.68/warc/CC-MAIN-20161202170903-00382-ip-10-31-129-80.ec2.internal.warc.gz | 658,210,730 | 4,203 | # Quiz:Work and Power
1. If a force moves an object or changes its direction, what is being done?A powerB energy in joulesC work 2. The unit used to measure Work is:A horsepowerB newtonsC joules 3. The rule for calculating Work is:A force × distanceB mass / volumeC energy / time 4. Who does more work - a man who lifts a large box from the ground up into the back of a truck, or a man who puts the same box on a trolley and wheels it up a ramp into the truck?A the man who lifts itB the man with the trolleyC they do the same work 5. A weightlifter lifts barbells of 200 newtons above his head to a height of 2 metres. How much work does he do?A 400 newtonsB 100 joulesC 400 joules 6. Energy and work use the same unit of the Joule because:A that is the unit chosen by the scientist named Pascal who studied heatB energy is required to do workC both measure the speed at which power is used 7. One kilojoule equals:A 1000 joulesB 100 joulesC 1⁄1000 joule 8. The old imperial unit for energy was the:A basal energy requirementB calorieC kilojoule 9. The rate at which work is done is called:ApowerBworkCenergy 10. Power is measured in: Ajoules Bwatts Cnewtons 11. The rule for the calculation of power is: Ajoules / seconds Benergy × time Cwork / time 12. 1 watt is equivalent to: A1 joule per second B1⁄10 newton per metre C1 newton metre 13. The old imperial unit for power was the: Akelvin Bhorsepower Ckilojoule 14. If 100 joules of work was done in 10 seconds, what power was used? A1 kilowatt B10 watts C1000 watts 15. If a weightlifter lifts 2000 newtons to a height of 2 metres in 4 seconds, how powerful is he? A1 horsepower B1 watt C1 kilowatt
Your AnswersQ 1: Q 2: Q 3: Q 4: Q 5: Q 6: Q 7: Q 8: Q 9: Q10: Q11: Q12: Q13: Q14: Q15: Correct AnswersQ 1: Q 2: Q 3: Q 4: Q 5: Q 6: Q 7: Q 8: Q 9: Q10: Q11: Q12: Q13: Q14: Q15: | 592 | 1,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2016-50 | latest | en | 0.861258 |
https://support.minitab.com/en-us/minitab/help-and-how-to/statistics/nonparametrics/how-to/mann-whitney-test/before-you-start/hypotheses/ | 1,726,457,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00545.warc.gz | 493,817,232 | 4,264 | # Hypotheses for Mann-Whitney Test
For a Mann-Whitney test, the hypotheses are as follows.
Null hypothesis
H0: η1 = η2 The median of the first population (η1) equals the median of the second population (η2).
Alternative hypothesis
Select one of the following alternative hypotheses.
H1: η1 ≠ η2 The median of the first population (η1) does not equal the median of the second population (η2). H1: η1 > η2 The median of the first population (η1) is greater than the median of the second population (η2). H1: η1 < η2 The median of the first population (η1) is less than the median of the second population (η2). | 164 | 611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.840075 |
https://www.electricalvolt.com/harmonics-and-harmonic-frequency-in-ac-circuits/ | 1,721,934,008,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00834.warc.gz | 648,477,468 | 23,502 | # Harmonics and Harmonic Frequency in AC Circuits
Harmonics refers to the additional frequencies that are generated alongside the fundamental frequency of a waveform. These frequencies are typically unwanted and can cause distortion in the wave pattern. When harmonics are superimposed onto the fundamental waveform, the resulting wave pattern becomes distorted and may lead to issues in electronic equipment.
There are two types of circuits.
1. Linear Circuit
2. Non-linear circuit
In a linear circuit, the current follows the voltage. The current increases or decreases proportionally with the increase/decrease of voltage. The V-I characteristics of the devices are linear and called linear devices. An example of a linear device is a resistor. The linear devices follow the Ohm’s Law. If the voltage applied across the resistor is sinusoidal, the current flowing through it is also sinusoidal.
In a nonlinear device, the current flowing through the device does not follow Ohm’s law. The current flowing through the nonlinear devices does not follow the voltage waveform. If the applied voltage is sinusoidal, the current flowing through the non-linear device is non-sinusoidal.
At the time of charging of the transformer, the magnetizing current drawn by the transformer does not follow the sinusoidal voltage, and the current is non sinusoidal.
According to the Fourier series, the non-sinusoidal current can be resolved into the fundamental current and the integral multiples of the fundamental frequency. The integral multiples frequency of the fundamental frequency is called Harmonics.
## What is a fundamental frequency?
The fundamental frequency of the sinusoidal waveform is the first harmonic order frequency, and it is equal to the supply frequency. The fundamental frequency is the lowest frequency f on which a complex waveform is built. In India and UK, the fundamental frequency is 50 Hz, and in the USA, the fundamental frequency of alternating voltage is 60 Hz.
A sinusoidal waveform is an alternating voltage or current which varies according to sin function.
i = Im sinwt
v = Vm Sinwt
w = 2Π f
Where,
v and i are the instantaneous values of the waveform
Vm, Im is the maximum value of the waveform
f is the lowest frequency or fundamental frequency
## What is Harmonics?
Ideally, voltage and current waveform is sinusoidal. However, because of the increased popularity of electronic power devices and other non-linear loads, the waveform quite often becomes distorted. The distorted waveform can be resolved into fundamental and integral multiples of fundamental frequency according to the Fourier series. Thus, the distorted waveform will contain harmonics.
## What is Harmonic Frequency?
If the fundamental frequency is f , the harmonic frequency will be 2f, 3f, 4f, 5f, 6f, and so on. If the frequency of the fundamental waveform is 50 Hz, the frequency of the harmonic waveform will be (2f= 2 x 50=100Hz), 3f= 3 x 50=150Hz, 200 Hz, 250Hz, 300Hz, 350 Hz, etc.
Harmonic order can be assigned according to the frequency of voltage or current waveform.
2,4,6,8 order of harmonics are called even harmonics, and the 3,5,7,9 are called the odd harmonics. 3,6,9 order harmonics are called triplen harmonics.
If the current or voltage waveform is symmetrical about the x-axis, the even number of harmonics would be zero.
The converter and inverter waveform are symmetrical about the x-axis, which is why it contains only odd-order harmonics.
The instantaneous value of the harmonic frequency can be given as;
For second-order harmonics
For a Third order harmonics
For a fourth-order harmonics
The instantaneous value of the complex waveform can be expressed as follows.
E T= Ef (fundamental) + Eh (harmonics)
Further, the harmonics order can be represented as positive and negative sequence harmonics. The harmonic frequency that has the same phase sequence as the fundamental frequency is called positive sequence harmonics. The harmonic frequency, which has an opposite phase sequence of fundamental frequency, is called negative sequence harmonics. | 861 | 4,097 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.900989 |
http://mathhelpforum.com/calculus/178349-background-infor-problem.html | 1,481,198,485,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00299-ip-10-31-129-80.ec2.internal.warc.gz | 175,987,198 | 10,270 | # Thread: Background Infor on This Problem
1. ## Background Infor on This Problem
http://www.stewartcalculus.com/data/...p_0205_stu.pdf
I’m not looking for an answer to this problem (not trying to have others do my work for me). However, I could use a little help in understanding the background information for this problem. Specifically, what particular laws or equations come into play here? What theorems and background information should I brush up in order to get started? Understanding specifically what is being asked in a word problem has always been my weak point. Thanks in advance.
2. Looks like you need to know how to calculate the derivative, how to model with polynomials, and the relationship between derivative and acceleration. More specifically, you need to know what it would mean for the function P(x) to represent height h at distance l, and height 0 at distance 0. You should also use the hint about putting a restriction on P'(x). There's a nice property that you want the graph to have near the origin and at the start of descent. Looks like you'll need to solve a system of equations at some point, too.
3. Thanks for the info. I was able to get most of the problem alright after that (we are allowed to submit two drafts of our work for professor review prior to the final being handed in) but am stumbling on the last point which, ironically enough, seems the easiest. Here is how I arrived at my answer:
If an airplane is not allowed to exceed 860 miles/h^2 and the cruising altitude is 35,000 feet and the speed is 300 miles/h… Given the formula (6hv^2)/l^2 is less than or equal to k… Find how far away the pilot should start the descent. Here is how I calculated:
h = 35,000 feet X (1 mile/5280 feet) = 6.63 miles
v = 300 miles/h
l = unknown
k = 860 miles/hour
(6hv^2)/l^2 = k
l^2 = (6hv^2)/k
l^2 = [(6)(6.63)(300^2)]/860
l^2 = 4163.02
l = 64.52
My professor says this number is too large but all my math seems right. Any idea where I screwed up? | 505 | 1,988 | {"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-2016-50 | longest | en | 0.952712 |
https://www.ablison.com/what-does-qv-mean/ | 1,718,674,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00256.warc.gz | 548,001,376 | 35,161 | # What Does QV Mean?
Understanding QV in Financial Terms===
In the field of finance, there are numerous terms that are commonly used to measure the value and performance of a company. One such term is QV, which stands for "quoted value." This metric is an important indicator of a company’s worth and is commonly used in financial analysis and valuation. In this article, we will take an in-depth look at QV, its definition, calculation, and importance in financial management.
## The Definition of QV: A Detailed Explanation
QV is a financial metric that measures the market value of a company’s outstanding shares of stock. It represents the total value of a company as determined by the current market price of its shares. Simply put, QV is the price that investors are willing to pay for a company’s shares.
In order to calculate QV, one must multiply the current market price of a company’s shares by the number of outstanding shares. This will provide an estimate of the total value of the company as determined by the market.
## How to Calculate QV in Financial Statements
Calculating QV in financial statements is relatively straightforward. First, you need to determine the current market price of a company’s shares. This can be found by looking at the stock exchange where the company’s shares are traded. Once you have the current market price, you need to multiply it by the number of outstanding shares to get the total QV.
For example, if a company has 1 million outstanding shares and the current market price is \$10 per share, the QV would be \$10 million.
## QV vs. BV: Differences and Similarities
QV is often compared to another financial metric known as BV, or "book value." While both metrics are used to measure the value of a company, they are calculated differently and serve different purposes.
BV represents the value of a company’s assets minus its liabilities and is calculated by subtracting the total liabilities from the total assets. This metric is used to determine the "book value" of a company, or the value of its assets if it were to liquidate.
QV, on the other hand, represents the market value of a company’s shares and is calculated based on the current market price of those shares. QV is used to determine the value of a company as determined by the market, rather than its book value.
## Importance of QV in Financial Analysis and Valuation
QV is a crucial metric in financial analysis and valuation. It provides investors and analysts with a sense of the market’s perception of a company’s value. A high QV indicates that investors have confidence in the company’s future growth potential, while a low QV may indicate the opposite.
In addition, QV is often used in conjunction with other financial metrics such as earnings per share (EPS) and price to earnings ratio (P/E ratio) to provide a more comprehensive picture of a company’s financial health and performance.
## Conclusion: QV as a Key Metric in Financial Management
In summary, QV is an important financial metric that measures the market value of a company’s outstanding shares. It is calculated based on the current market price of a company’s shares and is used to determine the value of a company as determined by the market. | 669 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-26 | latest | en | 0.951399 |
http://banach.millersville.edu/syllabi/math464.02_summer2005/ | 1,540,111,030,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513804.32/warc/CC-MAIN-20181021073314-20181021094814-00154.warc.gz | 34,438,920 | 4,661 | Real Analysis I
Summer Session III 2005
MATH 464.02/506.02 (3 credit), MTuWThF, 10:00A-11:30A, Wickersham 226
Prerequisites:
A grades of C- or better in MATH 310 (formerly numbered 220) Introduction to Proof, MATH 311 (formerly numbered 261) Calculus III, and MATH 322 (formerly numbered 242) Linear Algebra are the prerequisites for this course.
Instructor:
Dr. Buchanan
Office: Wickersham 218, Phone: 872-3659, FAX: 871-2320
Office Hours: 2:00P-3:00P (M-F), or by appointment
Email: Robert.Buchanan@millersville.edu
URL: http://banach.millersville.edu/~bob
Textbook:
Charles G. Denlinger, Elements of Real Analysis, July 2005.
Objectives:
Upon successful completion of this course, a student will have
• developed an ability to read and comprehend expository mathematical writing in real analysis at the upper undergraduate level,
• acquired a comprehensive understanding of the basic concepts of real analysis, including
• its axiomatic/deductive organization and flow of ideas, its foundations in the real number system (a complete ordered field),
• the fundamental notions of general topology and their uses in real analysis,
• the theory of continuous, real-valued functions of a real variable,
• the theory of differentiable, real-valued functions of a real variable,
• the theory of the Riemann integral of real-valued functions of a real variable,
• the fundamental theorem of calculus,
• sharpened his/her ability to critique mathematical ideas with constructive skepticism and in particular, will have developed an intellectual storehouse of instances in which naïve intuition would otherwise lead him/her to believe untrue statements in analysis or disbelieve true statements,
• learned that mathematical proof is the one indispensable tool for separating truth from falsehood in analysis,
• learned the essential role played by definitions in the formulation of methodology and proofs in mathematics,
• developed the specialized techniques of real analysis that allow the above objectives to be realized,
• learned significant applications of the concepts and techniques of real analysis, and
• learned the fundamental role played by real analysis in providing rigorous justification of the methods taught in elementary calculus courses.1
Course Contents:
The summer session activities may include exposure to and exploration of the following topics.
Partial Topic List:
• Real Number System ()
• Definitions and properties of fields and ordered fields
• Natural numbers in ordered fields
• Rational numbers in ordered fields
• Archimedian ordered fields
• Sequences
• Limits and convergence
• Algebra of limits
• Inequalities and limits
• Divergence to infinity
• Monotone sequences
• Subsequences
• Cauchy sequences
• Topology of the Real Number System
• Neighborhoods and open sets
• Interior, exterior, and boundary of a set
• Isolated points
• Closed sets and cluster points
• Limits of Functions
• Definition of the limit of a function
• Algebra of limits of functions
• Limits of polynomial and rational functions
• Inequalities and limits
• One-sided limits
• Continuous Functions
• Definition of continuous function
• Algebra of continuous functions
• Continuity on compact sets
• Continuity on intervals
• Applications of continuity to root finding, fixed points, etc.
• Differentiable Functions
• Definition of derivative and differentiability
• Rules for differentiation
• Relative extrema
• Monotone functions
• Rolle's Theorem and the Mean Value Theorem
• Riemann Integral
• Darboux's definition of the Riemann integral
• Integral as the limit of Riemann sums
• Algebra of integrals
• Fundamental Theorem of Calculus (first and second forms)
Tests:
There will be a take-home midterm test (given to students on Friday August 5, 2005 and due Monday, August 8, 2005) and a comprehensive final examination (given to students on Friday August 19, 2005 and due Monday, August 22, 2005). Students are not allowed to work together or discuss the midterm and final examination assignments. I will not ``curve'' midterm or exam grades.
Homework:
Students are expected to do their homework and participate in class. Students should expect to spend a minimum of three hours outside of class on homework and review for every hour spent in class. There will be four graded homework assignments during the summer session. The homework problems will be selected from the exercises in the textbook. Students should submit all homework by the date due. Late homework will not be accepted without valid excuse. It is very important that students work all assigned homework exercises, even those not designated as part of a graded assignment. Because of the pace of the summer session, a significant amount of your learning of the material will take place as you work on assigned problems. Students may work together and/or discuss any homework problems not assigned for a grade. The work you submit for graded homework exercises must represent your individual effort.
Attendance:
Students are expected to attend all class meetings. If you must be absent from class on the day an assignment is due, you must complete and hand in the assignment prior to the absence. If you know you will be absent on the day that a homework assignment or the midterm test is due, you must submit the work prior to the absence. Students who miss handing in an assignment due to an unforeseen absence should provide a valid excuse, otherwise you will not be allowed to submit the assignment.
Course grade will be calculated as follows.
Midterm test 25% Final exam 25% Homework/Projects 50%
I keep a record of students' homework and test scores. Students should also keep an individual record of graded assignments. I will not ``curve'' course grades. The course letter grades will be calculated as follows.
90-92 A 93-100 A 80-82 B 83-86 B 87-89 B 70-72 C 73-76 C 77-79 C 60-62 D 63-66 D 67-69 D 0-59 F
Course Repeat Policy
An undergraduate student may not take an undergraduate course of record more than three times. A course of record is defined as a course in which a student receives a grade of A, B, C, D, (including and ) F, U, Z or W. The academic department offering a course may drop a student from a course if the student attempts to take a course more than three times.2
Inclement Weather Policy:
If we should miss a class day due to a school closing because of weather, any activities planned for that missed day will take place the next time the class meets. For example, if a test is scheduled for a day that class is canceled on account of snow, the test will be given the next time the class meets.
Final Word:
Math is not a spectator sport. What you learn from this course and your final grade depend mainly on the amount of work you put forth. Daily contact with the material through homework assignments and review of notes taken during lectures is extremely important.
Page maintained by: Robert.Buchanan
Robert.Buchanan@millersville.edu
Last updated: | 1,523 | 6,992 | {"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-2018-43 | latest | en | 0.858603 |
https://www.advancedlearners.com/orlando/college/calculus/tutor/find.aspx | 1,695,410,267,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00084.warc.gz | 711,628,389 | 12,109 | ## Calculus Tutors
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## Davorin D
### Teaching Style
From my tutoring experience, I have noticed that students have trouble understanding the meaning of numbers and symbols on paper simply because no one has taught them how to visualize and interpret them in a real world situation. They also don't realize that they have plethora of resources and tools available to them to help them, yet they rarely utilize them. I try to give hints and clues to my students and let them obtain the right answers on their own instead of simply solving the problem for them. I believe this gives them a much better understand of the material.
### Experience Summary
Having recently graduated with a Bachelor's degree, I am continuing my education to obtain my Master's degree in mathematics. Number Theory will be my focus, as I intend to get involved in the encryption field. During my senior year as undergraduate, I have worked on campus as a teacher's assistant and at home as an online tutor for UNCG's iSchool program. I have privately tutored undergraduates needing help in pre-calculus and calculus. I enjoyed helping the students and showing them some of my own tricks and ways when it comes to solving the problems.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics with Concentration in Computer Science University of North Carolina at Greensboro BS 2007
## Jane J
### Teaching Style
I love to work with youth. I'd like to help kids to understand the course material better and achieve their academic success. I’m patient and easy going and considerate
### Experience Summary
I have been a teaching assistant for 6 years to college student. I have taught 12 courses such as linear algebra, pre-calculus, calculus and statistics. I am an experienced tutor.
### Credentials
Type Subject Issued-By Level Year
Degree Statistics and Applied mathematics University of Southern California Master 2008
## Justin J
### Teaching Style
I am a very enthusiastic tutor and, as I stated in the previous section, believe teaching in such a way that the student gains a true mastery of the given subject. In a sense, I believe in each student understanding a particular concept in their own unique way that is consistent with their unique thinking processes. I strive to do this by relating a given concept analogously to something the student already fundamentally understands. I also create new problems to solve that probe the students’ progress and comprehension. I try to make these problems as realistic as possible to make them interesting to the student.
### Experience Summary
Being a Ph.D. candidate and appointed research assistant, I very much realize the importance of true comprehension of a subject. Not only do I hold two B.S. degrees in Applied Mathematics and Chemical Engineering but I also carry credits to the equivalent of a Non-Thesis Masters in Applied Mathematics and am one course shy of a minor in Chemistry. I have tutored groups of students for Sylvan Learning Centers, conducted one-on-one sessions with Educational Enrichment and independently, and prepared and delivered lectures for advanced undergraduate chemical engineering courses. I am capable and experienced in tutoring all levels of mathematics, chemistry, chemical engineering, physics and materials science.
### Credentials
Type Subject Issued-By Level Year
Degree Chemical Engineering University of Florida Ph.D.--in progress 2010
Degree Chemical Engineering NC State University B.S. with Honors 2005
Degree Applied Mathematics NC State University B.S. with Honors 2005
## Purvesh P
### Teaching Style
I vary my teaching style depending on the student's particular disposition. I normally allow them to dictate the pace at which I tutor without digressing from our objective. Nonetheless, my goal is to be positive impact on the student. I love to help students get to a level they did not expect they could achieve, and in the same token, I am appreciative when I am humbled by challenging problems for which I have no solution. However, when faced with such a situation, I guide the student in an appropriate direction, and let them know that I will use my own resources to discover a solution and then get back to them. Without question, my style is synonymous to what computer users refer to good programs as, "user-friendly." I am patient with my tutees and steady in my demeanor so that I can gain and hold their confidence.
### Experience Summary
I have been tutoring since my sophomore year at Rutgers University. I hold the College Reading and Learning Association Level II (CRLA) certification. I tutored Mathematics (Pre-Calculus, Calculus), Chemistry, and Physics to undergraduate students at the university. I have since tutored students in these disciplines at various levels, such as middle school and high school. I have also tutored adults in the area of Information Technology, subjects such as Introduction to Programming, Web Design, and Database Design, while studying for my IT Certification. I have since attained an MS degree in Computer Science at Rensselaer Polytechnic Institute and have been working in the IT industry for the past 4 years. I continue to enjoy tutoring and teaching others; be it students, co-workers, or family members. I feel a tremendous sense of satisfaction when I see others succeed and know that I have played a part in their growth.
### Credentials
Type Subject Issued-By Level Year
Degree Computer Science Rensselaer Polytechnic Institute MS 2002
Certification Information Technology Devry University Post Baccalaureate Diploma 2001
Degree Biochemistry Rutgers University BS 1999
## Gary G
### Teaching Style
My teaching style has been, for the most part, dictated by student response. I am comfortable teaching in a traditional lecture format, in a format that uses a cooperative learning approach exclusively, or in a hybrid format. The goal is for effective learning to take place, and I believe my strongest quality is to be able to adapt in such a way that best helps students reach their academic goals.
### Experience Summary
For eight years, I taught freshman and sophomore-level mathematics courses at Arizona State University. These courses included College Algebra, Pre-Calculus, Calculus, Finite Mathematics, and Elementary Mathematics Theory. Additionally, I have tutored students in these courses both in the Mathematics Department Tutor Center, and on my own personal time.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics Southern Illinois University MS 1999
Degree Mathematics Allegheny College BS 1996
## Redha R
### Teaching Style
I teach by example and I am methodical. I show a student how to solve a problem by going very slowly and by following a sequence of steps. I make sure that the student follows understands each step before moving on to the next one. I then ask the student to solve almost the exact same problem (with a slight change in numbers for example)so that the student learns the method and how to solve the problem by herself/himself. I am very patient but expect the student to be willing to learn. I tell students that learning a subject matter is more important than getting an A in a class - this is because 1. if you learn, chances are you won't forget (at least for a long while) and 2. you will get a good grade as a result.
### Experience Summary
I have been employed by the IBM corporation for the last 27 years. I held numerous positions in hardware development, software development, telecommunication network development, project management, solution architecture, performance analysis, and others. As much as I enjoy my job, I have passion in Mathematics. I develop new ideas related to my work and expand them into U.S. patents and external technical papers. I informally tutor family and friends attending high school or college. I was a Mathematics teaching assistant at the Univ. of Pittsburgh for 2 years and at the Univ. of Michigan at Ann Arbor for 3 years. I love teaching and sharing my knowledge with others. I have published numerous technical papers and hold numerous U.S. patents as well.
### Credentials
Type Subject Issued-By Level Year
Other Arabic Native speaker Fluent current
Other French Native speaker Fluent current
Degree Electrical Engineering Univeristy of Michigan P.h.D 1990
Degree Electrical Engineering and Mathematics University of Pittsburgh M.S. 1982
Degree Computer Science and Mathematics Univerisity of Pittsburgh B.S. 1980
## Nigel N
### Teaching Style
I believe that every child is capable of learning and that every child has their own style of learning. As a teacher, I believe in differentiated instruction that allows all students an equal opportunity to learn. I believe in teaching through student exploration. Once a child is able to discover a concept on their own, they are able to better understand and remember it. My approach to teaching is to expose students to practical problems. Once these problems are understood, I introduce a mathematical concept that should allow them to discover a solution to the problem. Throughout this approach, I’m always encouraging and patient. I always stress the importance of making an attempt at solving a problem, as in my favorite quote, “Without struggle, there is no progress.”
### Experience Summary
During completion of my undergraduate degree in math, I always strived to achieve excellence in education and to pass on my love for learning. Throughout my college career, I’ve tutored at the undergraduate, the middle and the high school levels. Such exposure allowed me to determine the concepts students have the most difficulty in understanding, and I’m able to prepare accordingly. Upon completion of my degree, I accepted an analyst position with an HR Consulting firm. During these four years, I gained vital experience that allows me to transfer mathematical concepts into real world examples. Such experience has proven to be critical in helping students visualize practical examples of what they learn. I have since left the Consulting firm and entered into teaching. I’m finally doing what I enjoy the most, teaching and passing on a love for learning.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics Morehouse College BS 2001
## Darlene D
### Teaching Style
Much of my job includes mentoring new employees. Although I enjoy working with adults, I find it much more interesting to work with younger people. I would like the opportunity to tutor another student who is struggling with math. I have found that most students I have helped just needed some individual attention, and a safe place to ask questions that they would not ask in the classroom setting.
### Experience Summary
I have worked in the Computer Science field since 1985 when I first started working for the US Space Program. During that time, I earned my BS and MS degrees, and am currently working on the Doctorate in Business with Applied Computer Science. During the last four years, I have tutored a student in mathematics, ranging from basic Algebra to College Calculus. During that time, the student received all A's. Since this student is moving on to a University out of the area, I will no longer be working with him; therefore I am looking forward to helping other students.
### Credentials
Type Subject Issued-By Level Year
Degree Business with Applied Computer Sci Northcentral University Doctorate 2009
Degree Computer Science Webster University Masters 2004
Degree Business Administration Univ of Central Florida BS 1992
## Judy L.
Clermont, FL
Thought I would share Jean's grades with you. The last numbers are her grades at mid term this final nine weeks!! She has come up on five grades. I know you are a HUGE part. I want to thank you so much.
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Director of Counseling & Academic Advising, Montve
The administration and staff of Advanced Learners provide our students with effective, dedicated, well-qualified tutors. They respond promptly to our requests and their dependability has surpassed any expectation. The professional competence of Adva...
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## Othon G.
St. Augustine, FL
Things are great and I would recomend John to everyone. | 3,347 | 16,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.906352 |
http://forums.wolfram.com/mathgroup/archive/2012/Jun/msg00326.html | 1,604,077,421,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00199.warc.gz | 43,793,061 | 7,644 | Re: Replace, ReplaceAll and If time performace comparition
• To: mathgroup at smc.vnet.net
• Subject: [mg127027] Re: Replace, ReplaceAll and If time performace comparition
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Mon, 25 Jun 2012 04:01:42 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
```On 6/24/12 at 4:26 AM, rodrigomurtax at gmail.com (Murta) wrote:
>HI all I was working in some simulations with random numbers and get
>this example of performance comparition.
>randomList = RandomInteger[{-100, 100}, 10 10^6];
>(randomList /. (x_ /; x < 0 -> 0)); // AbsoluteTiming {5.747133,
>Null}
>Replace[randomList, (x_ /; x < 0 -> 0), 1]; // AbsoluteTiming
>{4.758984, Null}
>(If[# < 0, 0, #] & /@ randomList); // AbsoluteTiming {0.572200,
>Null}
>I personally prefer work with patterns because they are more compact
>and functional. Someone knows why patter is one magnitude order
>slow?? There is some trick to make it faster?
I don't know how to make the pattern matching faster. But I do
know how to create the list faster. That is:
In[1]:= randomList = RandomInteger[{-100, 100}, 10 10^6];
In[2]:= (a = If[# < 0, 0, #] & /@ randomList); // AbsoluteTiming
Out[2]= {0.597679,Null}
In[3]:= (b = Clip[randomList, {0, 100}]); // AbsoluteTiming
Out[3]= {0.086329,Null}
In[4]:= a == b
Out[4]= True
And it seems to me the syntax of Clip makes it very apparent
what the code does.
```
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` A particle moves according to the equation X =acos(pit/2) the distance covered by the particle in the time interval t=0 to t=3 is `
one year ago
Arun
23381 Points
``` A particle moves according to equation x=AcosπtThus the particle is at extreme position.Time period is = 2pi/pi = 2 seconds. it means that again when it reach the same extreme position it will do it in 2 seconds by covering 4A distance {A in each .5 second}for next 1.5 seconds it will cover 3A distance.HENce, total distance the particle will cover will be:4A + 3A = 7A.
```
one year ago
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Post Question | 354 | 1,304 | {"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-2020-05 | latest | en | 0.71039 |
https://socratic.org/questions/57f55bacb72cff58de3d640c | 1,571,053,604,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653216.3/warc/CC-MAIN-20191014101303-20191014124303-00303.warc.gz | 704,932,049 | 5,960 | # Question #d640c
Oct 5, 2016
C
#### Explanation:
Young's Modulus: $\setminus \lambda \stackrel{\setminus}{\textrm{\mathrm{de} f}} = \frac{\textrm{s t r e s s}}{\textrm{s t r a \in}} = \frac{\sigma}{\epsilon}$
$\epsilon = \frac{\Delta L}{L} , \quad \sigma = \frac{F}{A}$
A circular cross-section, but there are 2 of them so $A = \frac{\pi {D}^{2}}{2}$
$\implies \lambda = \frac{\frac{F}{A}}{\frac{\Delta L}{L}} = \frac{F L}{A \Delta L}$
$= \frac{2 F L}{\left(\pi {D}^{2}\right) \Delta L}$
$= \frac{2 \left(80 g\right) \left(10\right)}{\left(\pi {0.005}^{2}\right) \left(0.001\right)} \implies C$!! | 251 | 606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 3.640625 | 4 | CC-MAIN-2019-43 | latest | en | 0.427123 |
http://mathhelpforum.com/calculus/72270-finding-functions.html | 1,529,895,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00399.warc.gz | 214,510,831 | 10,783 | # Thread: Finding functions
1. ## Finding functions
So far, I have this:
Proving $\displaystyle |f(x)-f(y)|$ is continuous:
Need to show: $\displaystyle \forall \ \epsilon>0 \exists \delta>0 s.t \ |y-x| < \delta \ \Rightarrow |f(x)-f(y)|< \epsilon$.
$\displaystyle |y-x|< \delta \Rightarrow |y-x|^2< \delta^2$
Therefore: $\displaystyle |f(x)-f(y)| \leq (y-x)^2< \delta^2$
Let $\displaystyle \delta^2=\epsilon$
Hence $\displaystyle |f(x)-f(y)| \leq (y-x)^2< \delta^2=\epsilon$ so continuity is proved.
However, i'm sure this helps but I can't see why. I've also tried expanding the right and left hand sides but that doesn't help either.
Does anyone have any hints?
2. Hello,
You're right, the functions $\displaystyle f$ satisfying $\displaystyle |f(x)-f(y)|\leqslant (x-y)^2$ for all $\displaystyle x,y\in\mathbf{R}$ are continuous on $\displaystyle \mathbf{R}$. In fact they are even differentiable on $\displaystyle \mathbf{R}$. Can you show it ?
3. $\displaystyle |f(x)-f(y)| \leq (x-y)^2$
$\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \leq \frac{h^2}{h}=h \rightarrow 0$.
Setting $\displaystyle h=y-x$.
Therefore:
$\displaystyle |f(x)-f(y)| \leq (x-y)^2 \ \forall x,y \in \mathbb{R}$
Hence $\displaystyle |f(x)-f(y)| \leq 0$
$\displaystyle 0 \leq f(x)-f(y) \leq 0 \Rightarrow \ f(x)=f(y)$
Therefore $\displaystyle f(x)$ and $\displaystyle f(y)$ are the same number.
Therefore function f is a constant polynomial.
4. Originally Posted by Showcase_22
$\displaystyle |f(x)-f(y)| \leq (x-y)^2$
$\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)}{h} \leq \frac{h^2}{h}=h \rightarrow 0$.
You forgot one term: $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h){\color{blue}-f(x)}}{h} \leqslant \frac{h^2}{h}=h \rightarrow 0$.
Anyway this does not show that $\displaystyle f$ is differentiable at $\displaystyle x$ since it is possible for $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty$ and $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0$ to hold simultaneously.
Hint: use the same approach to show that $\displaystyle |f'(x)|\leqslant 0$.
5. Sorry about the first part, it was a typo.
Originally Posted by flyingsquirrel
You forgot one term: [tex]
Anyway this does not show that $\displaystyle f$ is differentiable at $\displaystyle x$ since it is possible for $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty$ and $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0$ to hold simultaneously.
Since we're taking the modulus of $\displaystyle f(x)-f(y)$, wouldn't this imply $\displaystyle |-\infty| \leq 0$ giving $\displaystyle +\infty \leq 0$ which is a contradiction?
6. Originally Posted by Showcase_22
Since we're taking the modulus of $\displaystyle f(x)-f(y)$, wouldn't this imply $\displaystyle |-\infty| \leq 0$ giving $\displaystyle +\infty \leq 0$ which is a contradiction?
If you are thinking about saying that $\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\leqslant 0$ implies $\displaystyle \left| \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\leqslant 0$, no, it would not lead to a contradiction because $\displaystyle a\leqslant b$ does not imply $\displaystyle |a|\leqslant |b|$. (the reason is that $\displaystyle t\mapsto |t|$ is not an increasing function) | 1,098 | 3,304 | {"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} | 4.1875 | 4 | CC-MAIN-2018-26 | latest | en | 0.560353 |
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