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https://www.teacherspayteachers.com/Browse/Search:force%20and%20motion%20assessment | 1,540,266,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00419.warc.gz | 1,103,667,231 | 55,695 | showing 1-24 of 1,398 results
Force and motion can be a difficult concept for students. However, learning more about how forces all around us affect how we and other objects move can be a lot of fun! Concepts covered: force, friction, gravity, balanced forces, unbalanced forces, contact forces, non-contact forces, mass vs. weig
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Also included in: Force and Motion: Complete Set Bundle
\$4.00
61 Ratings
4.0
PDF (11.27 MB)
This is an assessment that covers force and motion essential standards 3.P.1. It includes multiple choice and short answer questions. Check out my TPT store for force and motion lesson activities. Border graphics purchased from Etsy. All purchases intended for single classroom use only. Thanks a
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25 Ratings
4.0
PDF (539.18 KB)
Important terms: Direction, gravity, mass, speed, motion, position, force, velocity, balanced forces and unbalanced forces. AKS/STANDARDS – Approximately 2.5-3 weeks of instruction • 2. obtain, evaluate, and communicate information about the relationship between balanced and unbalanced forces 2a.
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\$3.00
16 Ratings
3.9
PDF (701.69 KB)
This 6 page document has three parts; a 2 page assessment for students, a 2 page answer key for teachers, and a 2 page test item analysis. It's specifically aligned with the Essential Standards curriculum and corresponds with the Forces and Motion Student Note Taking Booklet for Essential Standards
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\$2.00
11 Ratings
4.0
PDF (207.18 KB)
This resource includes a short answer assessment with an answer key, and a multiple choice assessment with an answer key. The two assessments are about forces and motion including the direction and effect of gravity, applying force to objects and their reactions, and speed/time relationship. These f
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\$1.50
4 Ratings
4.0
PDF (1.01 MB)
These assessments verify that the State of Virginia Standards of Learning for First Grade Science 1.2, Force and Motion, objectives have been met through your instruction of said topic. These objectives required that “The student [will] investigate and understand that moving objects exhibit differe
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\$6.21
5 Ratings
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4th Grade Force and Motion Assessment. Multiple Choice, cloze portion, open-ended responses, measurement lab questions.
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\$2.00
4 Ratings
4.0
DOCX (127.16 KB)
A quick and easy check for understanding that allows students to match definitions and illustrate their understandings of forces.
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2 Ratings
4.0
PDF (154.64 KB)
This Force and Motion test is a great addition to your unit on Force and Motion. This quiz can be used as a formative or summative assessment. This resource makes a great end of unit assessment, but could also be used as a review activity for a future test.This test includes matching, true false, an
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1 Rating
4.0
PDF (280.28 KB)
This purchase includes a forces and motion quiz, forces and motion test review, and forces and motion test. The quiz and the test both have answer keys included. Both assessments and the review go along with my Forces and Motion Unit PowerPoint. I use this with an advanced 5th grade class, but it
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Hello Fellow Teacher! The following product can be used to assess student knowledge of force and motion. It is composed of 7 multiple choice, 5 true/false, 3 short answer, and 1 TEI (hot spot). The 5 true/false questions require the student to rewrite any false statements as a true statement. Than
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FREE
5 Ratings
3.9
PDF (267.73 KB)
The following Science assessment covers the following: magnets, force, motion, friction, gravity.
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A short 1/2 page assessment about force and speed.
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This forces and motion science assessment is based on NC Science Essential Standard 5.P.1.
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FREE
3 Ratings
4.0
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This hands-on force and motion unit is packed with science experiments, learning stations, games, printables, poems, a force and motion assessment, 12 colorful vocabulary posters, coordinating vocabulary cards, printables, and more! It contains a unique student Motion Mission notebook with six exper
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Also included in: Science Unit MEGA BUNDLE!
\$8.00
3,715 Ratings
4.0
PDF (7.63 MB)
Force and Motion and Simple Machines - Complete Unit This full unit has everything you need to teach your students about forces and motion as well as simple machines. The unit is hands-on, very detailed, and engaging for students! The concepts covered in this unit are: 1) What is a force (push vs
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\$19.99
563 Ratings
4.0
ZIP (67.6 MB)
Force and Motion 5E Lesson Plans Bundle - Everything you need to teach a unit on Force and Motion. Each of the 7 lesson plans follows the 5E model and provides you with the exact tools to teach the topics. All of the guesswork has been removed with these NO PREP lessons. ***PLEASE READ FIRST***
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\$54.00
211 Ratings
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Force, Motion, and Simple Machines - This document contains posters, task cards, and activities you can use to enhance your unit on Force, Motion, and Simple Machines. Pages included: READING COMPREHENSION PASSAGES Motion Reading Passage and Comprehension Page First Law of Motion Reading Passa
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626 Ratings
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PDF (47.55 MB)
Force and Motion The NEW Kindergarten Science standards are here! Are you prepared to teach Force and Motion? This bundle includes nine activities to make teaching this new topic a breeze for you and fun for the kids! >>>You might also like: Force and Motion Interactive Notebook.<<
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\$6.75
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ZIP (6.37 MB)
Force and Motion Labs - Students will love you for providing them with engaging labs that allow them to learn in a style best suited for their needs. They will also save you a ton of prep time. The Force and Motion station lab Bundle is a series of 8 is a plug and play station labs designed to acc
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\$31.50
310 Ratings
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Force and motion can be a difficult concept for students. However, learning more about how forces all around us affect how we and other objects move can be a lot of fun!Concepts covered: force, friction, gravity, balanced forces, unbalanced forces, contact forces, non-contact forces, mass vs. weight
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\$16.00
\$12.80
369 Ratings
4.0
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\$102.60
\$52.56
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Check out this physics science unit on motion, friction, gravity, inertia, and simple machine investigations geared to the primary elementary levels of 1st, 2nd, 3rd, and 4th grades, which includes teacher lesson plans, experiments for kids, full worksheets, science vocabulary flash cards, and more!
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Your students will love these creative activities and flip book to learn all about Force and Motion! This Force and Motion Unit includes 11 reading passages, organizers related to each concept, comprehension & connections activities, vocabulary posters, unit test, and flip book. Plus there are t
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Also included in: Force and Motion AND Simple Machines {ALL IN ONE BUNDLE!}
\$12.00
\$10.00
290 Ratings
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PDF (88.25 MB)
showing 1-24 of 1,398 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,843 | 7,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.897771 |
https://www.clutchprep.com/chemistry/practice-problems/143301/lanthanum-metal-forms-cations-with-a-charge-of-3-consider-the-following-observat | 1,638,503,742,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00229.warc.gz | 764,379,295 | 29,955 | Molecular Equations Video Lessons
Concept:
# Problem: Lanthanum metal forms cations with a charge of 3+. Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound A) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound B) is formed. Both A and B dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound C is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms.Propose identities for the substances A, B, C, and D.
###### FREE Expert Solution
compound A: lanthanum metal is exposed to air
4 La (s) + 3 O2 (g) → 2 La2O3 (s)
compound B: lanthanum metal is added to water
2 La (s) + 6 H2O (l) → 2 La(OH)3 (s) + 3 H(g)
96% (474 ratings)
###### Problem Details
Lanthanum metal forms cations with a charge of 3+. Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound A) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound B) is formed. Both A and B dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound C is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms.
Propose identities for the substances A, B, C, and D. | 417 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-49 | latest | en | 0.896465 |
http://creativefamilyfun.net/20-fun-christmas-math-activities/ | 1,500,936,135,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424931.1/warc/CC-MAIN-20170724222306-20170725002306-00608.warc.gz | 73,307,190 | 15,669 | # 20 Fun Christmas Math Activities
I don’t know about you, but I love to get a little festive this time of year with everyday things – like math practice.
So let’s practice counting, counting, patterns, symmetry, addition and more while getting in the Christmas spirit. We’ll use candy canes, reindeer, and make Christmas trees.
You’ll love all of these Christmas Math Activities. They’re so much fun and perfect for learning during the holiday season.
Disclosure: This post contains Amazon Affiliate links. Please see my Disclosure Page for more details.
### Preschool Christmas Math Activities
Get a little fine motor practice while using a hole punch to do this simple Christmas Tree counting activity from Sugar Aunts.
Practice counting with this Gingerbread Cookies Counting Tray from Teach Me Mommy.
Explore patterns with this simple Candy Cane Pattern activity from This Reading Mama (comes with a free printable).
Have some sweet counting fun with this simple M&M Christmas tree counting activity from Umbrella Tree Cafe.
Practice counting and ordering with this Evergreen Tree Math Game from Mom Inspired Life.
Count and have fun with this two-player Jingle Bell Math Game from Fun-A-Day.
Do some simple adding and counting while you build a snowman with this fun activity from And Next Comes L.
Use Christmas bows to practice sorting by size with this simple measurement activity from Learning 4 Kids.
Practice number matching and counting with this Christmas Tree counting activity (that would make a great busy bag) from Mom Inspired Life.
Play with a fun Christmas tree geoboard. Fun-A-Day shows you how to make and use this simple foam tree geoboard.
### Elementary Christmas Math Activities
Use candy canes to practice symmetry with this fun activity from Fantastic Fun and Learning.
Have you heard of Pascal’s Triangle? Explore this pattern activity with this fun Christmas tree activity from Teach Beside Me.
Create a Christmas tree out of Sierpinski fractal triangles with this STEAM activity at What Do We Do All Day.
Explore tally marks with this fun Candy Cane math game from JDaniel4’s Mom.
Practice your estimation skills with this simple LEGO math activity from Little Bins for Little Hands.
Explore tangrams with this fun (and yummy) Christmas Cookie Tangram activity from Left Brain Craft Brain.
Learn about fractions with this Snowman Fractions game from Teach Beside Me.
Explore a hundred chart and practice simple addition while playing Jingle All the Way to 100 from Simply Kinder.
Practice circular measurement, pattern making, and more with this beautiful STEAM Math Stars activity at Nurture Store.
Race your reindeer along a number line with this fun and simple Racing Reindeer game from Mom Inspired Life.
Hope you have a festive and fun time with all these Christmas math activities!
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#### Terri Thompson
Terri is a writer and mom of two elementary-aged girls. She has a passion for learning and is always looking for ways to make learning fun. You can find her at Better Than Homework where she shares fun learning activities or Creative Family Fun where she shares art, craft, and family fun ideas. | 636 | 3,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-30 | latest | en | 0.874228 |
https://farfromhomemovie.com/how-do-you-describe-variables-in-research/ | 1,718,260,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00509.warc.gz | 225,547,440 | 12,233 | 2021-07-15
How do you describe variables in research?
A variable in research simply refers to a person, place, thing, or phenomenon that you are trying to measure in some way. The best way to understand the difference between a dependent and independent variable is that the meaning of each is implied by what the words tell us about the variable you are using.
How do you identify variables in a research paper?
You can use this typical form to determine the independent and dependent variables from the title of the study. If the study title is in the form “The effects of X on Y in Z”. X is the independent variable and Y is the dependent variable – the outcome, and Z is the type of subjects represented.
What are the 3 types of variables in research?
An experiment usually has three kinds of variables: independent, dependent, and controlled.
What are sub variables in research?
A sub-variable is a variable (question or loop) which is indented to another parent variable. Furthermore, a sub-variable is subordinate to its parent variable, meaning that if, for example, a do not ask routing targets a parent variable, its sub-variable will also not be asked.
Why do we use variables in research?
While the variation of an independent variable will influence the variation of dependent variable, both variables give the study a focus. Dependent and independent variables are also important because they determine the cause and effects in research.
What are the types of variables in research?
There are different types of variables and having their influence differently in a study viz. Independent & dependent variables, Active and attribute variables, Continuous, discrete and categorical variable, Extraneous variables and Demographic variables.
Why is it important to define your variables?
Whenever you are working with data, it is important to make sure the variables in the data are defined so that you (and anyone else who works with the data) can tell exactly what was measured, and how. …
What is the role of variables in a quantitative research?
In conclusion, variables are important because they help to measure concepts in a study. Because quantitative studies focus on measuring and explaining variables, choosing the right variables is important. The first step is to identify the correct variables to measure a property.
How important is it for the researcher to identify the type of variables?
How important is it for the researcher to identify the type of variables used in the study It helps the researchers to know what type of results did they get in their studies. The research intends to achieve goals.
What are the kinds of variables in quantitative research?
Understanding Quantitative VariablesDemographic Variables. Social workers are often interested in what we call demographic variables. Independent and Dependent Variables. Categorical Variables. Ordinal Variables. Interval Variables. The Special Case of Income. The Special Case of Age. Alphanumeric Variables.
What is a dependent and independent variable in research?
The independent variable is the cause. Its value is independent of other variables in your study. The dependent variable is the effect. Its value depends on changes in the independent variable.
What is dependent variable in Research example?
The dependent variable is the variable that is being measured or tested in an experiment. For example, in a study looking at how tutoring impacts test scores, the dependent variable would be the participants’ test scores, since that is what is being measured.
Why is it important to identify the independent and dependent variable?
Determining cause and effect is one of the most important parts of scientific research. It’s essential to know which is the cause – the independent variable – and which is the effect – the dependent variable.
Why are independent variables important?
The importance of an independent variable is a measure of how much the network’s model-predicted value changes for different values of the independent variable. Normalized importance is simply the importance values divided by the largest importance values and expressed as percentages.
What is sample variable in quantitative research?
When there is no measure of a characteristic of the participants, the characteristic is called a “sample characteristic.” When the characteristic must be measured, it is called a “sample variable.”
What are the two types of variables?
Types of variables.Independent vs dependent.Confounding variables. | 841 | 4,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.942725 |
https://www.aqua-calc.com/calculate/materials-price | 1,712,970,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00094.warc.gz | 599,942,422 | 6,111 | # Convert prices and calculate cost of materials and compounds
## compounds price conversions and cost calculator
Price per
units of weight
< 0.01carat
0.02gram
1.96100 grams
4.89250 grams
7.82400 grams
9.78500 grams
19.56kilogram
9.781/2 kilogram
0.55ounce
4.448 ounces
8.87pound
4.441/2 pound
#### Entered price
The entered price per 9 ounces is equal to 4.99.
• The compounds and materials price calculator performs conversions between prices for different weights and volumes. Selecting a unit of weight or volume from a single drop-down list, allows to indicate a price per entered quantity of the selected unit.
#### Foods, Nutrients and Calories
Progresso Reduced Sodium Savory Chicken and Wild Rice Soup, UPC: 00041196740745 weigh(s) 261 grams per metric cup or 8.7 ounces per US cup [ weight to volume | volume to weight | price | density ]
15996 foods that contain Vitamin B-12. List of these foods starting with the highest contents of Vitamin B-12 and the lowest contents of Vitamin B-12, and Recommended Dietary Allowances (RDAs) for Vitamin B12
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Amazon weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Potassa [KOH or HKO] weighs 2 044 kg/m³ (127.60275 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-401B, liquid (R401B) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
The short ton per meter is a measurement unit of linear or linear mass density
The density of a material or substance is defined as its mass per unit of volume.
kGy to µGy conversion table, kGy to µGy unit converter or convert between all units of radiation absorbed dose measurement.
#### Calculators
Annulus and sector of an annulus calculator | 560 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.772904 |
https://electronics.stackexchange.com/questions/337118/2kv-ac-step-down-to-5v-without-using-transformer | 1,560,681,816,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00432.warc.gz | 434,056,227 | 34,498 | # 2kV AC step down to 5v without using transformer [closed]
I'm doing an industrial project, i need to step down the high voltage of 2kV to 5V, is it possible to step down to around 5V without transformer, is there any other alternatives? please suggest me :)
## closed as unclear what you're asking by Olin Lathrop, laptop2d, DoxyLover, Bimpelrekkie, jonkNov 1 '17 at 20:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Resistor divider + bridge rectifier. You didn't say anything about how much power you needed, or isolation ... – Bryan Boettcher Oct 30 '17 at 14:43
• Why would you not use a transformer? – Transistor Oct 30 '17 at 14:44
• Better tell us what you are trying to do. – Eugene Sh. Oct 30 '17 at 14:57
• "My device would be bulky so i need to design without transformer." Safety first. Size later. Put all the information in the question not scattered through the comments. – Transistor Oct 30 '17 at 15:25
• There is way too little information here. How much power do you need at 5 V? What's wrong with the obvious answer of using a transformer? What frequency is this AC? Is the 2 kV ground-referenced or isolated? What about the 5 V? What will be connected to the 5 V? This smells like a X-Y problem. Pop up two levels and explain what you are really trying to accomplish. Starting the close as unclear process, so answer soon. Also, answer all the questions, not just what you think matters. – Olin Lathrop Oct 30 '17 at 16:43
If all you need is to get a sample the voltage with an ADC (I saw your comment in Peter Green answer), just make a voltage divider.
Hook some 1/2W high ohm resistors in series between 2kV and ground, trying to make each one see at most 500V, and a last resistor proportional to the 5V you want. Something like this:
Vcc --- 20MΩ --- 20MΩ --- 20MΩ --- 20MΩ --x-- 200kΩ --x-- GND
The first four resistors are of high value because too low resistors will load your 2kV source, drawing a big current which may cause voltage drop and overheating on the resistors. And you want to use several of them (four), instead of just a big one, because each resistor has a voltage limit. Above that, voltage may arc from a lead to another, creating unwanted fireworks.
This network will draw a low current: about 25µA. The last resistor is where you will hook your ADC (note the x), and it´s calculated this way: 5 * 80000000 / 2000 = 200kΩ. This resistor will have about 5V of the total 2kV. The remaining 1995V will be distributed between the 20MΩ resistors. If you want to be on the safe side, decrease the 200kΩ a little, to have less voltage on it, preventing accidents by overvoltages on your ADC.
• All this assumes that the other phase is grounded. Even if it is a direct connection between a 20 kV circuit and a 5 V micro is crazy. – Transistor Oct 30 '17 at 17:26
• Some MCUs have a separate analog GND pin, that may help when the 2kV is not referenced to the ground, but I'm not sure. About direct connection, I don't see a way to isolate it and still have a compact design as the OP asked. Personally, I don't think it's crazy. People tend to be afraid and over concerned about high voltage, but the danger is not so high, if proper isolation is provided. And finally, it's 2kV, not 20kV. – Marcovecchio Oct 30 '17 at 17:42
You need to answer a few questions.
1. Do you need isolation?
2. How much power do you need?
3. What efficiency and/or power factor do you need?
If you need isolation that pretty much pushes you into using a transformer somewhere in the design though it may make more sense for it to be a high-frequency transformer in some kind of switched-mode converter.
If you don't need isolation then other approaches become possible. For example a buck converter or for small ammounts of power a capacitive dropper.
• With 2kV then isolation should be standard... – Solar Mike Oct 30 '17 at 16:08
• I'm making a device so that it should measure voltage, current and power factor... for Voltage of around 2kV i need to step down to 5v and then to adc and micro controller. But using transformer my product would be bulky, i thought of voltage divider first which is kinda very dangerous. So i thought of any other methods i can get – Dexter Oct 30 '17 at 16:19
• My device should be a size of a mobile charger, is it possible to do it for such a high voltage? – Dexter Oct 30 '17 at 16:20
2kV components are bulky. That's because of the insulation needed, but more important, because no one bothers with 2kV when he doesn't need 2MW. Even 200kW stuff would be done at 690V, to allow 1kV insulation and a wider and cheaper selection of components.
So … if you want something small at 2kV, you have to build most of the components yourself. Sometimes a series connection of low-voltage components will help (e.g. 1N4007 goes up to 1kV) but most times, you are out of luck.
So winding a 2kV->690V transformer yourself will solve a lot of headaches. You will still hardly come down to the size of a cell phone charger. These are incredibly tiny.
• You can't possibly say how small the result may be because we have no indication of the power requirement. – Olin Lathrop Oct 30 '17 at 16:46
• That's what I tried to explain. The power requirement of the piece in question doesn't count. You simply cannot buy low-power 2kV components. Because it makes no sense in most cases. – Janka Oct 30 '17 at 16:50
• But he could want 500 W, in which case it would certainly be larger than a cell phone charger, whatever that actually means. – Olin Lathrop Oct 30 '17 at 16:55
• I've asked the industry guy for the power range, if the requirement for high power then can it be done? – Dexter Oct 30 '17 at 17:05
• High power means big size. So, no. – Janka Oct 30 '17 at 17:16 | 1,584 | 6,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-26 | latest | en | 0.936085 |
https://herbie.uwplse.org/demo/7b3dca17693105ec7839df72e12ddc5d79dedb08.cf8a50cd49460eed2f57a21b8de7a625bbcdb510/graph.html | 1,590,450,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390437.8/warc/CC-MAIN-20200525223929-20200526013929-00357.warc.gz | 386,199,086 | 3,532 | Average Error: 33.4 → 10.0
Time: 25.6s
Precision: 64
$\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}$
$\begin{array}{l} \mathbf{if}\;b \le -9.497374990683389 \cdot 10^{+62}:\\ \;\;\;\;-\frac{c}{b}\\ \mathbf{elif}\;b \le -9.280942802423785 \cdot 10^{+21}:\\ \;\;\;\;\frac{\left(-4 \cdot \left(c \cdot a\right)\right) \cdot \frac{\frac{-1}{2}}{a}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}\\ \mathbf{elif}\;b \le -2.6813599241700416 \cdot 10^{-50}:\\ \;\;\;\;-\frac{c}{b}\\ \mathbf{elif}\;b \le 1.451866749331113 \cdot 10^{+79}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{b + \sqrt{b \cdot b - \left(c \cdot a\right) \cdot 4}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-b}{a}\\ \end{array}$
\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}
\begin{array}{l}
\mathbf{if}\;b \le -9.497374990683389 \cdot 10^{+62}:\\
\;\;\;\;-\frac{c}{b}\\
\mathbf{elif}\;b \le -9.280942802423785 \cdot 10^{+21}:\\
\;\;\;\;\frac{\left(-4 \cdot \left(c \cdot a\right)\right) \cdot \frac{\frac{-1}{2}}{a}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}\\
\mathbf{elif}\;b \le -2.6813599241700416 \cdot 10^{-50}:\\
\;\;\;\;-\frac{c}{b}\\
\mathbf{elif}\;b \le 1.451866749331113 \cdot 10^{+79}:\\
\;\;\;\;\frac{-1}{2} \cdot \frac{b + \sqrt{b \cdot b - \left(c \cdot a\right) \cdot 4}}{a}\\
\mathbf{else}:\\
\;\;\;\;\frac{-b}{a}\\
\end{array}
double f(double b, double a, double c) {
double r40829021 = b;
double r40829022 = -r40829021;
double r40829023 = r40829021 * r40829021;
double r40829024 = 4.0;
double r40829025 = a;
double r40829026 = r40829024 * r40829025;
double r40829027 = c;
double r40829028 = r40829026 * r40829027;
double r40829029 = r40829023 - r40829028;
double r40829030 = sqrt(r40829029);
double r40829031 = r40829022 - r40829030;
double r40829032 = 2.0;
double r40829033 = r40829032 * r40829025;
double r40829034 = r40829031 / r40829033;
return r40829034;
}
double f(double b, double a, double c) {
double r40829035 = b;
double r40829036 = -9.497374990683389e+62;
bool r40829037 = r40829035 <= r40829036;
double r40829038 = c;
double r40829039 = r40829038 / r40829035;
double r40829040 = -r40829039;
double r40829041 = -9.280942802423785e+21;
bool r40829042 = r40829035 <= r40829041;
double r40829043 = -4.0;
double r40829044 = a;
double r40829045 = r40829038 * r40829044;
double r40829046 = r40829043 * r40829045;
double r40829047 = -0.5;
double r40829048 = r40829047 / r40829044;
double r40829049 = r40829046 * r40829048;
double r40829050 = -r40829035;
double r40829051 = r40829035 * r40829035;
double r40829052 = 4.0;
double r40829053 = r40829052 * r40829044;
double r40829054 = r40829053 * r40829038;
double r40829055 = r40829051 - r40829054;
double r40829056 = sqrt(r40829055);
double r40829057 = r40829050 + r40829056;
double r40829058 = r40829049 / r40829057;
double r40829059 = -2.6813599241700416e-50;
bool r40829060 = r40829035 <= r40829059;
double r40829061 = 1.451866749331113e+79;
bool r40829062 = r40829035 <= r40829061;
double r40829063 = r40829045 * r40829052;
double r40829064 = r40829051 - r40829063;
double r40829065 = sqrt(r40829064);
double r40829066 = r40829035 + r40829065;
double r40829067 = r40829066 / r40829044;
double r40829068 = r40829047 * r40829067;
double r40829069 = r40829050 / r40829044;
double r40829070 = r40829062 ? r40829068 : r40829069;
double r40829071 = r40829060 ? r40829040 : r40829070;
double r40829072 = r40829042 ? r40829058 : r40829071;
double r40829073 = r40829037 ? r40829040 : r40829072;
return r40829073;
}
# Try it out
Results
In Out
Enter valid numbers for all inputs
# Derivation
1. Split input into 4 regimes
2. ## if b < -9.497374990683389e+62 or -9.280942802423785e+21 < b < -2.6813599241700416e-50
1. Initial program 53.9
$\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}$
2. Taylor expanded around -inf 7.0
$\leadsto \color{blue}{-1 \cdot \frac{c}{b}}$
3. Simplified7.0
$\leadsto \color{blue}{-\frac{c}{b}}$
## if -9.497374990683389e+62 < b < -9.280942802423785e+21
1. Initial program 46.9
$\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}$
2. Using strategy rm
3. Applied clear-num46.9
$\leadsto \color{blue}{\frac{1}{\frac{2 \cdot a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}$
4. Using strategy rm
5. Applied flip--46.9
$\leadsto \frac{1}{\frac{2 \cdot a}{\color{blue}{\frac{\left(-b\right) \cdot \left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c} \cdot \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}}$
6. Applied associate-/r/47.0
$\leadsto \frac{1}{\color{blue}{\frac{2 \cdot a}{\left(-b\right) \cdot \left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c} \cdot \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}} \cdot \left(\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}\right)}}$
7. Applied associate-/r*47.0
$\leadsto \color{blue}{\frac{\frac{1}{\frac{2 \cdot a}{\left(-b\right) \cdot \left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c} \cdot \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}$
8. Simplified11.6
$\leadsto \frac{\color{blue}{\frac{\frac{1}{2}}{a} \cdot \left(0 - \left(a \cdot c\right) \cdot -4\right)}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}$
## if -2.6813599241700416e-50 < b < 1.451866749331113e+79
1. Initial program 14.0
$\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}$
2. Using strategy rm
3. Applied clear-num14.1
$\leadsto \color{blue}{\frac{1}{\frac{2 \cdot a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}$
4. Using strategy rm
5. Applied *-un-lft-identity14.1
$\leadsto \frac{1}{\frac{2 \cdot a}{\color{blue}{1 \cdot \left(\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}\right)}}}$
6. Applied times-frac14.1
$\leadsto \frac{1}{\color{blue}{\frac{2}{1} \cdot \frac{a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}$
7. Applied *-un-lft-identity14.1
$\leadsto \frac{\color{blue}{1 \cdot 1}}{\frac{2}{1} \cdot \frac{a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}$
8. Applied times-frac14.2
$\leadsto \color{blue}{\frac{1}{\frac{2}{1}} \cdot \frac{1}{\frac{a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}$
9. Simplified14.2
$\leadsto \color{blue}{\frac{1}{2}} \cdot \frac{1}{\frac{a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}$
10. Simplified14.1
$\leadsto \frac{1}{2} \cdot \color{blue}{\left(-\frac{\sqrt{b \cdot b - 4 \cdot \left(a \cdot c\right)} + b}{a}\right)}$
## if 1.451866749331113e+79 < b
1. Initial program 40.4
$\frac{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}$
2. Using strategy rm
3. Applied clear-num40.5
$\leadsto \color{blue}{\frac{1}{\frac{2 \cdot a}{\left(-b\right) - \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}}}$
4. Taylor expanded around 0 5.3
$\leadsto \color{blue}{-1 \cdot \frac{b}{a}}$
5. Simplified5.3
$\leadsto \color{blue}{-\frac{b}{a}}$
3. Recombined 4 regimes into one program.
4. Final simplification10.0
$\leadsto \begin{array}{l} \mathbf{if}\;b \le -9.497374990683389 \cdot 10^{+62}:\\ \;\;\;\;-\frac{c}{b}\\ \mathbf{elif}\;b \le -9.280942802423785 \cdot 10^{+21}:\\ \;\;\;\;\frac{\left(-4 \cdot \left(c \cdot a\right)\right) \cdot \frac{\frac{-1}{2}}{a}}{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}\\ \mathbf{elif}\;b \le -2.6813599241700416 \cdot 10^{-50}:\\ \;\;\;\;-\frac{c}{b}\\ \mathbf{elif}\;b \le 1.451866749331113 \cdot 10^{+79}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{b + \sqrt{b \cdot b - \left(c \cdot a\right) \cdot 4}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-b}{a}\\ \end{array}$
# Reproduce
herbie shell --seed 1
(FPCore (b a c)
:name "(-b - sqrt(b*b - 4*a*c))/(2 * a)"
(/ (- (- b) (sqrt (- (* b b) (* (* 4 a) c)))) (* 2 a))) | 3,397 | 8,111 | {"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} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.192074 |
https://www.lmfdb.org/L/2/17/17.8/c9/0/12 | 1,638,081,722,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00550.warc.gz | 961,924,295 | 7,506 | # Properties
Label 2-17-17.8-c9-0-12 Degree $2$ Conductor $17$ Sign $-0.964 + 0.263i$ Analytic cond. $8.75560$ Root an. cond. $2.95898$ Motivic weight $9$ Arithmetic yes Rational no Primitive yes Self-dual no Analytic rank $0$
# Learn more
## Dirichlet series
L(s) = 1 + (16.9 − 16.9i)2-s + (74.3 − 179. i)3-s − 64.3i·4-s + (−2.35e3 − 973. i)5-s + (−1.78e3 − 4.30e3i)6-s + (1.81e3 − 751. i)7-s + (7.59e3 + 7.59e3i)8-s + (−1.27e4 − 1.27e4i)9-s + (−5.64e4 + 2.33e4i)10-s + (6.97e3 + 1.68e4i)11-s + (−1.15e4 − 4.78e3i)12-s − 1.65e5i·13-s + (1.80e4 − 4.35e4i)14-s + (−3.49e5 + 3.49e5i)15-s + 2.90e5·16-s + (−6.54e4 − 3.38e5i)17-s + ⋯
L(s) = 1 + (0.750 − 0.750i)2-s + (0.529 − 1.27i)3-s − 0.125i·4-s + (−1.68 − 0.696i)5-s + (−0.562 − 1.35i)6-s + (0.285 − 0.118i)7-s + (0.655 + 0.655i)8-s + (−0.648 − 0.648i)9-s + (−1.78 + 0.739i)10-s + (0.143 + 0.346i)11-s + (−0.160 − 0.0665i)12-s − 1.60i·13-s + (0.125 − 0.302i)14-s + (−1.78 + 1.78i)15-s + 1.10·16-s + (−0.189 − 0.981i)17-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut & 17 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & (-0.964 + 0.263i)\, \overline{\Lambda}(10-s) \end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut & 17 ^{s/2} \, \Gamma_{\C}(s+9/2) \, L(s)\cr =\mathstrut & (-0.964 + 0.263i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
Degree: $$2$$ Conductor: $$17$$ Sign: $-0.964 + 0.263i$ Analytic conductor: $$8.75560$$ Root analytic conductor: $$2.95898$$ Motivic weight: $$9$$ Rational: no Arithmetic: yes Character: $\chi_{17} (8, \cdot )$ Primitive: yes Self-dual: no Analytic rank: $$0$$ Selberg data: $$(2,\ 17,\ (\ :9/2),\ -0.964 + 0.263i)$$
## Particular Values
$$L(5)$$ $$\approx$$ $$0.287953 - 2.14363i$$ $$L(\frac12)$$ $$\approx$$ $$0.287953 - 2.14363i$$ $$L(\frac{11}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$F_p(T)$
bad17 $$1 + (6.54e4 + 3.38e5i)T$$
good2 $$1 + (-16.9 + 16.9i)T - 512iT^{2}$$
3 $$1 + (-74.3 + 179. i)T + (-1.39e4 - 1.39e4i)T^{2}$$
5 $$1 + (2.35e3 + 973. i)T + (1.38e6 + 1.38e6i)T^{2}$$
7 $$1 + (-1.81e3 + 751. i)T + (2.85e7 - 2.85e7i)T^{2}$$
11 $$1 + (-6.97e3 - 1.68e4i)T + (-1.66e9 + 1.66e9i)T^{2}$$
13 $$1 + 1.65e5iT - 1.06e10T^{2}$$
19 $$1 + (9.72e4 - 9.72e4i)T - 3.22e11iT^{2}$$
23 $$1 + (2.66e5 + 6.42e5i)T + (-1.27e12 + 1.27e12i)T^{2}$$
29 $$1 + (-3.90e6 - 1.61e6i)T + (1.02e13 + 1.02e13i)T^{2}$$
31 $$1 + (-1.44e6 + 3.50e6i)T + (-1.86e13 - 1.86e13i)T^{2}$$
37 $$1 + (-4.54e6 + 1.09e7i)T + (-9.18e13 - 9.18e13i)T^{2}$$
41 $$1 + (9.08e6 - 3.76e6i)T + (2.31e14 - 2.31e14i)T^{2}$$
43 $$1 + (-2.50e7 - 2.50e7i)T + 5.02e14iT^{2}$$
47 $$1 - 5.08e7iT - 1.11e15T^{2}$$
53 $$1 + (1.09e7 - 1.09e7i)T - 3.29e15iT^{2}$$
59 $$1 + (6.57e7 + 6.57e7i)T + 8.66e15iT^{2}$$
61 $$1 + (-1.88e8 + 7.81e7i)T + (8.26e15 - 8.26e15i)T^{2}$$
67 $$1 + 5.35e7T + 2.72e16T^{2}$$
71 $$1 + (5.11e7 - 1.23e8i)T + (-3.24e16 - 3.24e16i)T^{2}$$
73 $$1 + (-4.04e7 - 1.67e7i)T + (4.16e16 + 4.16e16i)T^{2}$$
79 $$1 + (5.24e7 + 1.26e8i)T + (-8.47e16 + 8.47e16i)T^{2}$$
83 $$1 + (2.90e8 - 2.90e8i)T - 1.86e17iT^{2}$$
89 $$1 + 2.89e8iT - 3.50e17T^{2}$$
97 $$1 + (-2.61e8 - 1.08e8i)T + (5.37e17 + 5.37e17i)T^{2}$$
show more
show less
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$
## Imaginary part of the first few zeros on the critical line
−15.93057303684148063986171423221, −14.36914964146034846710376232214, −12.85460873432139384452542579637, −12.42375905284696191099501901275, −11.22753936854724217803514745590, −8.204007792835576401374598022231, −7.58359373353851393849781788077, −4.56040358040895556196564079798, −2.90523523835195394168116930723, −0.865662797836297744395558625475, 3.71190566556529794932326163624, 4.49787069465201684461842647350, 6.81494326405852293520630822376, 8.500775357410407481909227166811, 10.40163861459722009397813064815, 11.72480289657472913222075801096, 14.05783037491342855900326042171, 14.95475060654385426128675603168, 15.60407251209329335536819402896, 16.47428015261393061355685060853 | 2,193 | 4,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.388256 |
http://www.singingturtle.com/category/uncategorized/number-sense/find-the-lcm/ | 1,516,573,690,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00157.warc.gz | 560,925,443 | 9,716 | # Journey to the LCM … and Beyond!
In this blog/video I want to take you a little bit deeper into the world of … the LCM! Yes, that fascinating little mathematical entity is beckoning us to explore it further. Several readers wrote to me after I posted my LCM “trick,” saying they want to see me demonstrate WHY this “shortcut” works. One […]
# Find the LCM – FAST!
Here’s a short video on how to find the LCM. It uses a trick that I have not seen anywhere else, and the approach is quite fast. The information in this video dovetails with the info in this post. I hope that you find this video helpful.
# How to find the LCM — Intuitive Method
Recently I’ve been interested in discovering a cool, new way to get the LCM for a pair of numbers. Criteria: Method that is short and sweet. Even more important, a method that makes sense INTUITIVELY. I can’t speak for any of you , but I’ve always felt that the standard techniques for finding the LCM […]
# Find the LCM (aka LCD) in Two Easy Steps
This is really the “Week of the LCM” for me. Just as I was finishing my last post, on a new way to find the LCM for a pair of numbers, I discovered another way to do the same thing. I was looking at the problems at the end of my last post, these problems: […] | 307 | 1,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.95123 |
https://rdrr.io/cran/fda.usc/src/R/FDR.R | 1,726,410,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00305.warc.gz | 445,174,804 | 8,630 | # R/FDR.R In fda.usc: Functional Data Analysis and Utilities for Statistical Computing
#### Documented in FDRpvalue.FDR
#' @title False Discorvery Rate (FDR)
#'
#' @description Compute the False Discovery Rate for a vector of p-values and alpha value.
#'
#' @details \code{FDR} method is used for multiple hypothesis testing to correct
#' problems of multiple contrasts.\cr If \code{dep = 1}, the tests are
#' positively correlated, for example when many tests are the same contrast.
#' \cr If \code{dep < 1} the tests are negatively correlated.
#'
#' @aliases FDR pvalue.FDR
#'
#' @param pvalues Vector of p-values
#' @param alpha Alpha value (level of significance).
#' @param dep Parameter dependence test. By default \code{dep = 1}, direct
#' dependence between tests.
#' @return Return:
#' \itemize{
#' \item \code{out.FDR}{ \code{=TRUE}. If there are significative
#' differences.}
#' \item \code{pv.FDR}{ p-value for False Discovery Rate test.}
#' }
#' @author Febrero-Bande, M. and Oviedo de la Fuente, M.
#' @seealso Function used in \code{\link{fanova.RPm}}
#' @references Benjamini, Y., Yekutieli, D. (2001). \emph{The control of the
#' false discovery rate in multiple testing under dependency}. Annals of
#' Statistics. 29 (4): 1165-1188. DOI:10.1214/aos/1013699998.
#' @examples
#' p=seq(1:50)/1000
#' FDR(p)
#' pvalue.FDR(p)
#' FDR(p,alpha=0.9999)
#' FDR(p,alpha=0.9)
#' FDR(p,alpha=0.9,dep=-1)
#' @name FDR
#' @rdname FDR
#' @export
FDR=function(pvalues=NULL,alpha=0.95,dep=1){
if (is.null(pvalues)) stop("No p-values entered")
m=length(pvalues)
if (dep<0) {const.m=sum(1/(1:m))} else {const.m=1}
spvalues=sort(pvalues)
FDR=(1-alpha)*(1:m)/(m*const.m)
return(any(spvalues<FDR))
}
#' @rdname FDR
#' @export
pvalue.FDR=function(pvalues=NULL,dep=1){
if (is.null(pvalues)) stop("No p-values entered")
m=length(pvalues)
if (dep<0) {const.m=sum(1/(1:m))} else {const.m=1}
spvalues=sort(pvalues)
pv.FDR=min(spvalues*(m*const.m)/(1:m))
return(pv.FDR)
}
###
## Try the fda.usc package in your browser
Any scripts or data that you put into this service are public.
fda.usc documentation built on Oct. 17, 2022, 9:06 a.m. | 684 | 2,140 | {"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} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.238935 |
http://www.top-law-schools.com/forums/viewtopic.php?f=6&t=102782 | 1,529,525,357,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863834.46/warc/CC-MAIN-20180620182802-20180620202802-00366.warc.gz | 523,639,302 | 12,515 | ## Does the LGB over classify?
Prepare for the LSAT or discuss it with others in this forum.
poo
Posts: 90
Joined: Wed May 13, 2009 3:12 am
### Does the LGB over classify?
I am just beginning to study and have heard from some that the LGB over classifies the games (eg there is not a big difference between the different type of basic linear games). I am not sure if it is solid advice to categorize broader so I just wanted to get some advice from people who have already taken the test.
Also, if the best way to study is to exactly follow the LGB, does anyone have a list of the games categorized by all the types in the LGB (instead of by date)?
Thanks!
skip james
Posts: 262
Joined: Sat Sep 19, 2009 2:53 am
### Re: Does the LGB over classify?
yup. and i also only liked their basic linear games strategy, nothing else.
jpSartre
Posts: 326
Joined: Sun Jan 03, 2010 11:05 am
### Re: Does the LGB over classify?
I think grouping games are well classified. What kinda strategy do you use for those james?
lawduder
Posts: 483
Joined: Sun Jun 28, 2009 10:56 am
### Re: Does the LGB over classify?
poo wrote:I am just beginning to study and have heard from some that the LGB over classifies the games (eg there is not a big difference between the different type of basic linear games). I am not sure if it is solid advice to categorize broader so I just wanted to get some advice from people who have already taken the test.
Also, if the best way to study is to exactly follow the LGB, does anyone have a list of the games categorized by all the types in the LGB (instead of by date)?
Thanks!
This is not the best way to study. The LGB is an excellent guide when starting out logic games, but you'll need to adapt it to your own method as well. Last couple of tests have had a game that read like it was meant to intentionally mess up people who adhere strictly to the LGB. Once you're completely comfortable with games, you won't need to 'classify' a game before you start, you'll just know what to do.
UTexas
Posts: 68
Joined: Mon Dec 28, 2009 10:20 pm
### Re: Does the LGB over classify?
Yes.
cubswin
Posts: 617
Joined: Mon May 25, 2009 4:40 pm
### Re: Does the LGB over classify?
I would agree with you. Over-classifying probably makes it easier to organize the info and sell books. It also makes their books substantially longer than their rivals (at least TPR and Kaplan) and gives you the impresison that the material is exhaustive.
pattymac
Posts: 210
Joined: Wed Nov 04, 2009 7:44 pm
### Re: Does the LGB over classify?
I'm going through it now, I agree though, I thought the "linear/grouping" combo was nothing more than an overloaded advanced linear game.
Cambridge LSAT
Posts: 257
Joined: Mon Aug 24, 2009 3:26 pm
### Re: Does the LGB over classify?
We're in the midst of categorizing games in a more intuitive way. We'd love to hear suggestions from you guys regarding how you'd like them classified. Thanks.
ZombiesAhead
Posts: 72
Joined: Thu Sep 24, 2009 8:47 pm
### Re: Does the LGB over classify?
If you think LGB overclassifies, take a look at lsatblog.blogspot.com (http://lsatblog.blogspot.com/2009/10/ls ... -list.html). This system actually seems a little better than PS.
I don't think LGB overclassifies but I don't like the system very well. The major categories of "linear" and "grouping" are pretty good but beyond that it doesn't do much for me.
chewdak
Posts: 106
Joined: Fri Apr 03, 2009 5:54 pm
### Re: Does the LGB over classify?
People do get obsessed with certain classification schemes. Not per se the games, but people have insisted that a particular credited answer is wrong because the question falls within a specific rubric, and as such the answer makes no sense.
febstriver
Posts: 29
Joined: Mon Dec 21, 2009 2:13 am
### Re: Does the LGB over classify?
lawduder wrote:
poo wrote:I am just beginning to study and have heard from some that the LGB over classifies the games (eg there is not a big difference between the different type of basic linear games). I am not sure if it is solid advice to categorize broader so I just wanted to get some advice from people who have already taken the test.
Also, if the best way to study is to exactly follow the LGB, does anyone have a list of the games categorized by all the types in the LGB (instead of by date)?
Thanks!
This is not the best way to study. The LGB is an excellent guide when starting out logic games, but you'll need to adapt it to your own method as well. Last couple of tests have had a game that read like it was meant to intentionally mess up people who adhere strictly to the LGB. Once you're completely comfortable with games, you won't need to 'classify' a game before you start, you'll just know what to do.
titcr.
JazzOne
Posts: 2980
Joined: Tue Sep 09, 2008 11:04 am
### Re: Does the LGB over classify?
The LGB doesn't classify enough.
abbas123
Posts: 129
Joined: Sun Jun 07, 2009 11:01 am
### Re: Does the LGB over classify?
JazzOne wrote:The LGB doesn't classify enough.
lol what do you think it should classify more?!
noelleF
Posts: 29
Joined: Wed Jan 13, 2010 5:25 pm
### Re: Does the LGB over classify?
i think so. there are wayyy too many grouping classifications...do not get those at all. They make me even more confused than not classifying grouping games at all! i like the classification here: http://lsatblog.blogspot.com/2009/07/lo ... types.html
Return to “LSAT Prep and Discussion Forum�
### Who is online
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https://kr.mathworks.com/matlabcentral/cody/problems/25-remove-any-row-in-which-a-nan-appears/solutions/123213 | 1,596,541,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00214.warc.gz | 386,157,785 | 15,654 | Cody
# Problem 25. Remove any row in which a NaN appears
Solution 123213
Submitted on 3 Aug 2012 by Falko Schindler
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% A = [ 1 5 8 -3 NaN 14 0 6 NaN ]; B_correct = [ 1 5 8 ]; assert(isequal(remove_nan_rows(A),B_correct))
2 Pass
%% A = 1:10; B_correct = A; assert(isequal(remove_nan_rows(A),B_correct))
3 Pass
%% A = [ 1 5 8 -3 NaN 14 0 6 6]; B_correct = [1 5 8; 0 6 6]; assert(isequal(remove_nan_rows(A),B_correct))
4 Pass
%% A = [ 1 3 6 NaN 3 NaN]'; B_correct = [1 3 6 3]'; assert(isequal(remove_nan_rows(A),B_correct))
5 Pass
%% A = [ 1 3 6 NaN; 3 4 2 1]; B_correct = [3 4 2 1]; assert(isequal(remove_nan_rows(A),B_correct)) | 298 | 809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.477236 |
https://www.projectpro.io/recipes/create-strides-of-length-n-from-given-1d-array | 1,719,225,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00014.warc.gz | 831,593,039 | 29,545 | # How to create strides of length n from a given 1D array in numpy
This recipe helps you create strides of length n from a given 1D array in numpy
## Recipe Objective
How to create strides of length n from a given 1D array?
Strides It is nothing but the tuple of integer values, in which the bytes of particular dimension is indicated by each one of in it. To tell how many bytes to jump in the data buffer Numpy uses strides. Stride will indicate the number of bytes to jump in the order for reaching to the next value in the given dimension which also known as axis of travel. It is always constant for given axis.
## Step 1 - Import libraries
``` import numpy as np from numpy.lib.stride_tricks import as_strided ```
## Step 2 - Take Sample data
``` Sample_data = np.array([12,13,14,15,16,17,18,19], dtype = "int32") print("This is a Sample 1D array:", Sample_data) ```
`This is a Sample 1D array: [12 13 14 15 16 17 18 19]`
## Step 3 - Create Stride
``` Result = np.lib.stride_tricks.as_strided(Sample_data,((8-2)//3+1,2),(3*4,4)) print("This is the Following Result","\n",Result, "\n") print("This is the shape of original array which is an 1D array:","\n",Sample_data.shape,"\n") print("This is the shape of our Result which is an 2D array:","\n",Result.shape) ```
```This is the Following Result
[[12 13]
[15 16]
[18 19]]
This is the shape of original array which is an 1D array:
(8,)
This is the shape of our Result which is an 2D array:
(3, 2)```
In the above we have used "np.lib.stride_tricks.as_strided(array, new_array_shape, Stride_steps)" syntax for Stride in which there various functions ar working lets understand them: array - This is nothing but the original array that we want to stride. In our case we have taken 1D array of name "Sample_data". new array shape - This is nothing but the shape of our output array, in our case the resulted array is 2D so the shape will be (3,2) which means 3 rows and 2 columns. Stride steps - It is nothing but the stride that which is measured in bytes. In our case it is (12,4) because we want to jump over 3 indices in the array in which each of them is an integer i.e 4 bytes, So therefore 3*4 = 12 for row stride step and for column is 4 because the next integer is 4 bytes away.
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In this machine learning project, you will learn to implement Regression Discontinuity Design Example in Python to determine the effect of age on Mortality Rate in Python. | 1,046 | 4,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.864848 |
http://amsterjob.com/the-13-british-colonies-map/ | 1,534,915,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219495.97/warc/CC-MAIN-20180822045838-20180822065838-00649.warc.gz | 25,798,475 | 14,312 | # The 13 British Colonies Map
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5 / 5 ( 1votes )
At the time you think of The 13 British Colonies Map, exactly what you thinking of? In essence a map is a representation of a topology or function. To get example a formula such as X=2Y maps a worth of Y to each value of Back button. Of course we all believe that mathematicians are weird and sometimes hard to understand but they have you ever seen a schematic map of a subway (underground railway) system? Perhaps you have ever seen the same network of rails specified on a more "normal" The 13 British Colonies Map of the town in which it is located? Different The 13 British Colonies Map of the extremely same thing can look quite different.
At the time you make a The 13 British Colonies Map of the flat area - a "plan" or "elevation" - things are quite simple, but when you make an effort to map a larger area, like the surface of an whole planet, things can get quite complicated if you would like your map to be level. It can be all very well to make an earth, but try turning the area of that globe into a set The 13 British Colonies Map! Yikes!
However you begin it, you conclusion plan edge-effects. As I write this information I am actually engaged in programming map-generating programs designed to generate maps of fictional landscapes. I happen to be examining the map-generators that are included in the free, open-source (GNU GPL licensed) strategy game, FreeCiv. Edge results are incredibly apparent in such maps. The The 13 British Colonies Map are basically rectangular, but you can choose to obtain them act like cylinders by "wrapping" left to right or top to lower part, or you may even have "wrap" in both guidelines. Most often people make a decision on "wrap" only kept to right, and stop the best and bottom with "polar regions". Such basic "wrapping" makes for quite extreme distortion though if you try it out with a real The 13 British Colonies Map on the planet! | 444 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-34 | latest | en | 0.939357 |
http://mathhelpforum.com/differential-equations/168254-understanding-definition-solution-ode-print.html | 1,506,335,459,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690591.29/warc/CC-MAIN-20170925092813-20170925112813-00166.warc.gz | 220,214,937 | 4,739 | # Understanding the Definition of a "Solution" of an ODE
• Jan 13th 2011, 11:45 AM
VonNemo19
Understanding the Definition of a "Solution" of an ODE
Hi.
I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:
Definition 1.1.2: Solution of an ODE
Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.
Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.
P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.
Thank You,
VN19.
• Jan 13th 2011, 11:46 AM
dwsmith
Quote:
Originally Posted by VonNemo19
Hi.
I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:
Definition 1.1.2: Solution of an ODE
Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.
Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.
P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.
Thank You,
VN19.
What page? I have this book.
• Jan 13th 2011, 11:58 AM
VonNemo19
Quote:
Originally Posted by dwsmith
What page? I have this book.
pg. 5 in the ninth edition.
• Jan 13th 2011, 12:11 PM
dwsmith
This is best seen looking at an example. Work out Example 1 b on pg 5 as well.
$\displaystyle\frac{dy}{dx}=xy^{1/2} \ \ \ y=\frac{1}{16}x^4$
$\displaystyle\frac{dy}{dx}=\frac{1}{4}x^3$
Plug in.
$\displaystyle\frac{x^3}{4}=x\left(\frac{x^4}{16}\r ight)^2\Rightarrow\frac{x^3}{4}=x\left(\frac {x^2}{4}\right)\Rightarrow\frac{x^3}{4}=\frac{x^3} {4}$
Whenever you solve a DE, you can take your y you obtain and differentiate the amount of times need to check your solution.
Example
Say you have
$y''-5y'+2y+10=0$
When you find y, you can take the second derivative, minus 5 times the first, plus 2 times y, plus 10. The nth derivatives plugged into the DE should equal the RHS.
• Jan 13th 2011, 12:32 PM
VonNemo19
Quote:
Originally Posted by dwsmith
.... The nth derivatives plugged into the DE should equal the RHS.
I don't think you quite understand what it is that I am asking. Thanks for trying anyway. I know how to plug in a solution and see if the equation reduces to an identity
I'm looking for an understanding of how to find the interval of definition of a solution.
• Jan 13th 2011, 12:35 PM
dwsmith
Quote:
Originally Posted by VonNemo19
I don't think you quite understand what it is that I am asking. Thanks for trying anyway.
I'm looking for an understanding of how to find the interval of definition. I know how to plug in a solution and see if the equation reduces to an identity
Domain of the solution.
• Jan 13th 2011, 12:50 PM
HallsofIvy
I don't know what you mean by the "domain of the differential equation". I presume you mean the domain of the function F(x,t) where the differential equation is x'= F(x, t). No, the "domain of definition" is not the intersection of the domain of F with the domain of $\phi$. For one thing, F(x, t) is a function of two variables so its domain is in $R^2$ while the domain of the function $\phi$ is in $R$. For another, it makes no sense to say the domain of definition is domain of $\phi$ since $\phi$ is the solution. Of course, its domain is the "domain of definition" of the solution.
What is true of the "domain of definition", and probably what you intended to say, is that it must be a subset of projection of the domain of F(x,t) onto x-coordinate. Exactly what subset can be very difficult to determine.
• Jan 13th 2011, 01:19 PM
VonNemo19
Quote:
Originally Posted by HallsofIvy
...
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.
So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.
Here we go:
Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.
BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy
• Jan 13th 2011, 01:55 PM
dwsmith
Quote:
Originally Posted by VonNemo19
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.
So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.
Here we go:
Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.
BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy
$x\in[-5,5] \ \ \ y\in[-5,5]$
• Jan 13th 2011, 01:58 PM
VonNemo19
Quote:
Originally Posted by dwsmith
$x\in[-5,5] \ \ \ y\in[-5,5]$
So, basically, the interval of definition is simply the domain of the solution? Is this ALWAYS the case?
• Jan 13th 2011, 01:59 PM
dwsmith
Yes, one of the its names is "domain of solution" | 1,828 | 6,717 | {"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": 36, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-39 | longest | en | 0.928578 |
https://www.3dcadforums.com/threads/law-without-streight-line-for-reference-reverse-law-parallel-line-process.8489/ | 1,611,213,841,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524270.28/warc/CC-MAIN-20210121070324-20210121100324-00424.warc.gz | 640,114,186 | 14,139 | # law without streight line for reference - reverse law+ parallel line process
#### Jazzuo
##### New member
hello
in the example i used normal law and parallel process
made laws out of white lines
and made vlue lines and green lines
but what if i had only green lines and only the blue middle line to start with, and wanted to make the top and botomn blue line.
how do i make law out of the green lines since it has no straigh one?
thank you
Jazzuo
#### MrCATIA
##### Super Moderator
I remember back in CATIA V4, Laws could be made based on Reference geometry of a line, curve, or even a surface. But it seems that with V5, the reference can only be a line. The online Help describes only using a line for the reference.
So, I don't think there is a direct way to make a Law based on the distance between two curves, such as the two green curves in your example.
Last edited:
#### MrCATIA
##### Super Moderator
One method that I'd try: develop flat patterns (unfold) of the two green curves and add a straight reference line. The distance between the two green curves can be determined by the difference of the distances between the reference and the unfolded curves.
1. make a flat surface between the two green curves
2. unfold the surface into a plane, and unfold both green curves
3. draw a line in the plane somewhat parallel to the unfolded lines (this will be the reference line)
4. make two Laws: Law1(D2) between the reference line and the first unfolded line, Law2(D1) between the reference line and second unfolded line
5. add a parallel curve from the blue line using Law1
6. add a second parallel curve, in the opposite direction, from the first parallel line using Law2. This should be the curve you're looking for.
Last edited:
#### Jazzuo
##### New member
Hi i understand what you are trying to say, but i am unable to do it in catia
by step num 2 i just get a surface turned 90 degree not a straigthen out worm
J | 454 | 1,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | latest | en | 0.942523 |
https://www.vedantu.com/maths/spheroid | 1,620,394,276,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988793.99/warc/CC-MAIN-20210507120655-20210507150655-00630.warc.gz | 1,134,205,567 | 280,940 | ×
# Spheroid
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FAQ
## Spheroid Meaning
View Notes
In-plane geometry, a spheroid shape, or ellipsoid of revolution, refers to a quadric surface obtained by rotating an ellipse about one of its principal axes. This is to say, a spheroid is an ellipsoid with 2 equal semi-diameters with a circular symmetry.
If the ellipse is revolved about its major axis, the outcome is a prolate (elongated) spheroid, shaped like a rugby ball. An oblate or flattened spheroid is an outcome of a rotating ellipse around its minor axis. If the producing ellipse is a circle, the outcome is a sphere.
### Applications of Spheroid
The neutrons and protons of an atomic nucleus are mostly found in the spherical, oblate, or prolate spherical shapes along with having a spinning axis. Deformed nuclear shapes typically form as an outcome of the competition between electromagnetic repulsion between protons, quantum shell effects, and surface tension.
### Oblate Spheroids Meaning
A ''squashed'' Spheroid of which the equatorial radius ‘a’ is larger than the polar radius ‘c’, such that a > c. To a first approximation, the shape assumed by a rotating fluid (inclusive of the Earth, that is fluid'' over astronomical time scales) is known as an oblate spheroid. The oblate spheroid can be identified parametrically by the general Spheroid equations (for a Spheroid having z-Axis as the symmetry axis).
### Oblate Spheroid Shape
The oblate spheroid shape can be defined as an assumed shape for various celestial bodies and planets that are rotating rapidly at their own axis. For example, the Earth, Saturn, and Jupiter. The English mathematician, Enlightenment scientist, and astronomer –Isaac Newton reasoned that Earth and Jupiter are oblate spheroids owing to their centrifugal force. Earth's diverse geodetic and cartographic systems are formed on reference ellipsoids, all of which are oblate.
### Example of Oblate Spheroids
An example of an oblate spheroid is the planet Jupiter with a flattening spheroid of 0.06487. Another science-fiction of an extremely oblate spheroid earth is Mesklin from Hal Clement's novel Mission of Gravity.
### Prolate Spheroids
The prolate spheroid is an estimated shape of the ball in various sports, such as rugby football. The word is also used to depict the shape of some nebulae like the Crab Nebula.
Several moons of the Solar System estimate prolate spheroids in shape, although they are really triaxial ellipsoids.
On contrary to being distorted into oblate spheroids through quick rotation, celestial objects distort little into prolate spheroids through tidal forces when they orbit a massive body in a close orbit. An utmost example is Jupiter's moon Io, which becomes slightly more or less prolate in its orbit because of little eccentricity, inducing intense volcanism. However, the major axis of a prolate spheroid does not run across the satellite's poles, but through the two points on its equator straightaway facing toward and away from the primary.
### Example of a Prolate Spheroids
• An Australian rules football is an example of a prolate spheroid.
• Many submarines have a shape that can be depicted as a prolate spheroid.
• Another example includes the atomic nucleus of the element belonging to the lanthanide and actinide groups.
• In anatomy, near-spheroid organs such as testis.
• Fresnel zones used to evaluate wave propagation and interference in space.
Other Examples of prolates include Saturn's satellites Enceladus, Mimas, and Uranus' and Tethys satellite Miranda.
### Geoid and Spheroid
The geoid is essentially described as the surface of the earth's gravity field that approximates mean sea level. The geoid is perpendicular to the direction of the force of gravity. Because the mass of the Earth is not uniform in nature at all points, the magnitude of gravity thus differs, and the shape of the geoid is irregular.
### Fun Facts
• Saturn is considered to be the most oblate planet in the Solar System. It has a flattening of 0.09796.
Q1. Why Actually is a Spheroid?
Answer: The term spheroid originally meant "an estimated spherical body", disclosing irregularities even beyond the bi- or tri-axial ellipsoidal shape, and that is how the word is frequently being used in some older publishing on geodesy (for example, referring to truncated spherical harmonic diversifications of the Earth.
Q2. Why is Planet Earth Considered an Oblate Sphere?
Answer: owing to the combined effects of rotation and gravity, the figure of the Earth (and matter of fact of all planets) is not quite a sphere, but instead is bit flattened in the direction of its axis of rotation. For that reason, in geodesy and cartography the Earth is often estimated by an oblate spheroid shape, referred to as the reference ellipsoid, instead of a sphere. The current World Geodetic System model makes use of a spheroid of which the radius is 3,963.191 mi (6,378.137 km) at the Equator and 3,949.903 mi (6,356.752 km) at the poles.
Q3. What is an Ellipsoid?
Answer: Ellipsoid is a closed surface whose all plane cross-sections are either circles or ellipses. An ellipsoid is symmetrical in shape about three mutually perpendicular axes that bisect at the centre.
If m, n, and o are the principal semiaxes, a basic equation of such an ellipsoid is x2/m2 + y2/n2 + z2/o2 = 1. A unique case occurs when m = n= o: then the surface is a sphere, and the bisection with any plane crossing through it is a circle. If two axes are equivalent, say m = n, and different from the 3rd, o, then the ellipsoid is an ellipsoid of revolution, or spheroid (a shape formed by revolving an ellipse about one of its axes). If n is smaller than both m and o, then the spheroid is said to be an oblate; but it is said to be a prolate spheroid if m and n are lesser than o. | 1,372 | 5,820 | {"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} | 3.078125 | 3 | CC-MAIN-2021-21 | longest | en | 0.915498 |
http://us.metamath.org/mpeuni/usgredg2v.html | 1,638,815,136,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00266.warc.gz | 85,928,408 | 10,522 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > usgredg2v Structured version Visualization version GIF version
Theorem usgredg2v 26025
Description: In a simple graph, the mapping of edges having a fixed endpoint to the other vertex of the edge is a one-to-one function into the set of vertices. (Contributed by Alexander van der Vekens, 4-Jan-2018.) (Revised by AV, 18-Oct-2020.)
Hypotheses
Ref Expression
usgredg2v.v 𝑉 = (Vtx‘𝐺)
usgredg2v.e 𝐸 = (iEdg‘𝐺)
usgredg2v.a 𝐴 = {𝑥 ∈ dom 𝐸𝑁 ∈ (𝐸𝑥)}
usgredg2v.f 𝐹 = (𝑦𝐴 ↦ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}))
Assertion
Ref Expression
usgredg2v ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → 𝐹:𝐴1-1𝑉)
Distinct variable groups: 𝑥,𝐸,𝑧 𝑧,𝐺 𝑥,𝑁,𝑧 𝑧,𝑉 𝑦,𝐴 𝑦,𝐸,𝑥,𝑧 𝑦,𝐺 𝑦,𝑁 𝑦,𝑉
Allowed substitution hints: 𝐴(𝑥,𝑧) 𝐹(𝑥,𝑦,𝑧) 𝐺(𝑥) 𝑉(𝑥)
Proof of Theorem usgredg2v
Dummy variables 𝑤 𝑢 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 usgredg2v.v . . . . 5 𝑉 = (Vtx‘𝐺)
2 usgredg2v.e . . . . 5 𝐸 = (iEdg‘𝐺)
3 usgredg2v.a . . . . 5 𝐴 = {𝑥 ∈ dom 𝐸𝑁 ∈ (𝐸𝑥)}
41, 2, 3usgredg2vlem1 26023 . . . 4 ((𝐺 ∈ USGraph ∧ 𝑦𝐴) → (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) ∈ 𝑉)
54ralrimiva 2961 . . 3 (𝐺 ∈ USGraph → ∀𝑦𝐴 (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) ∈ 𝑉)
65adantr 481 . 2 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → ∀𝑦𝐴 (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) ∈ 𝑉)
72usgrf1 25973 . . . . . . . . 9 (𝐺 ∈ USGraph → 𝐸:dom 𝐸1-1→ran 𝐸)
87adantr 481 . . . . . . . 8 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → 𝐸:dom 𝐸1-1→ran 𝐸)
9 elrabi 3346 . . . . . . . . . 10 (𝑦 ∈ {𝑥 ∈ dom 𝐸𝑁 ∈ (𝐸𝑥)} → 𝑦 ∈ dom 𝐸)
109, 3eleq2s 2716 . . . . . . . . 9 (𝑦𝐴𝑦 ∈ dom 𝐸)
11 elrabi 3346 . . . . . . . . . 10 (𝑤 ∈ {𝑥 ∈ dom 𝐸𝑁 ∈ (𝐸𝑥)} → 𝑤 ∈ dom 𝐸)
1211, 3eleq2s 2716 . . . . . . . . 9 (𝑤𝐴𝑤 ∈ dom 𝐸)
1310, 12anim12i 589 . . . . . . . 8 ((𝑦𝐴𝑤𝐴) → (𝑦 ∈ dom 𝐸𝑤 ∈ dom 𝐸))
14 f1fveq 6479 . . . . . . . 8 ((𝐸:dom 𝐸1-1→ran 𝐸 ∧ (𝑦 ∈ dom 𝐸𝑤 ∈ dom 𝐸)) → ((𝐸𝑦) = (𝐸𝑤) ↔ 𝑦 = 𝑤))
158, 13, 14syl2an 494 . . . . . . 7 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → ((𝐸𝑦) = (𝐸𝑤) ↔ 𝑦 = 𝑤))
1615bicomd 213 . . . . . 6 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (𝑦 = 𝑤 ↔ (𝐸𝑦) = (𝐸𝑤)))
1716notbid 308 . . . . 5 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (¬ 𝑦 = 𝑤 ↔ ¬ (𝐸𝑦) = (𝐸𝑤)))
18 simpl 473 . . . . . . . . . 10 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → 𝐺 ∈ USGraph )
19 simpl 473 . . . . . . . . . 10 ((𝑦𝐴𝑤𝐴) → 𝑦𝐴)
2018, 19anim12i 589 . . . . . . . . 9 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (𝐺 ∈ USGraph ∧ 𝑦𝐴))
21 preq1 4243 . . . . . . . . . . 11 (𝑢 = 𝑧 → {𝑢, 𝑁} = {𝑧, 𝑁})
2221eqeq2d 2631 . . . . . . . . . 10 (𝑢 = 𝑧 → ((𝐸𝑦) = {𝑢, 𝑁} ↔ (𝐸𝑦) = {𝑧, 𝑁}))
2322cbvriotav 6582 . . . . . . . . 9 (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁})
241, 2, 3usgredg2vlem2 26024 . . . . . . . . 9 ((𝐺 ∈ USGraph ∧ 𝑦𝐴) → ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) → (𝐸𝑦) = {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁}))
2520, 23, 24mpisyl 21 . . . . . . . 8 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (𝐸𝑦) = {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁})
26 simpr 477 . . . . . . . . . 10 ((𝑦𝐴𝑤𝐴) → 𝑤𝐴)
2718, 26anim12i 589 . . . . . . . . 9 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (𝐺 ∈ USGraph ∧ 𝑤𝐴))
2821eqeq2d 2631 . . . . . . . . . 10 (𝑢 = 𝑧 → ((𝐸𝑤) = {𝑢, 𝑁} ↔ (𝐸𝑤) = {𝑧, 𝑁}))
2928cbvriotav 6582 . . . . . . . . 9 (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})
301, 2, 3usgredg2vlem2 26024 . . . . . . . . 9 ((𝐺 ∈ USGraph ∧ 𝑤𝐴) → ((𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}) → (𝐸𝑤) = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁}))
3127, 29, 30mpisyl 21 . . . . . . . 8 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (𝐸𝑤) = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁})
3225, 31eqeq12d 2636 . . . . . . 7 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → ((𝐸𝑦) = (𝐸𝑤) ↔ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁}))
3332notbid 308 . . . . . 6 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (¬ (𝐸𝑦) = (𝐸𝑤) ↔ ¬ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁}))
34 riotaex 6575 . . . . . . . . . . . 12 (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) ∈ V
3534a1i 11 . . . . . . . . . . 11 (𝑁𝑉 → (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) ∈ V)
36 id 22 . . . . . . . . . . 11 (𝑁𝑉𝑁𝑉)
37 riotaex 6575 . . . . . . . . . . . 12 (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∈ V
3837a1i 11 . . . . . . . . . . 11 (𝑁𝑉 → (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∈ V)
39 preq12bg 4359 . . . . . . . . . . 11 ((((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) ∈ V ∧ 𝑁𝑉) ∧ ((𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∈ V ∧ 𝑁𝑉)) → ({(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} ↔ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁})))))
4035, 36, 38, 36, 39syl22anc 1324 . . . . . . . . . 10 (𝑁𝑉 → ({(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} ↔ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁})))))
4140notbid 308 . . . . . . . . 9 (𝑁𝑉 → (¬ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} ↔ ¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁})))))
4241adantl 482 . . . . . . . 8 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → (¬ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} ↔ ¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁})))))
43 ioran 511 . . . . . . . . . . 11 (¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))) ↔ (¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∧ ¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))))
44 ianor 509 . . . . . . . . . . . . 13 (¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ↔ (¬ (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∨ ¬ 𝑁 = 𝑁))
4523, 29eqeq12i 2635 . . . . . . . . . . . . . . . . 17 ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ↔ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}))
4645notbii 310 . . . . . . . . . . . . . . . 16 (¬ (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ↔ ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}))
4746biimpi 206 . . . . . . . . . . . . . . 15 (¬ (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}))
4847a1d 25 . . . . . . . . . . . . . 14 (¬ (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
49 eqid 2621 . . . . . . . . . . . . . . 15 𝑁 = 𝑁
5049pm2.24i 146 . . . . . . . . . . . . . 14 𝑁 = 𝑁 → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5148, 50jaoi 394 . . . . . . . . . . . . 13 ((¬ (𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∨ ¬ 𝑁 = 𝑁) → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5244, 51sylbi 207 . . . . . . . . . . . 12 (¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5352adantr 481 . . . . . . . . . . 11 ((¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∧ ¬ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))) → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5443, 53sylbi 207 . . . . . . . . . 10 (¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))) → (𝐺 ∈ USGraph → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5554com12 32 . . . . . . . . 9 (𝐺 ∈ USGraph → (¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))) → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5655adantr 481 . . . . . . . 8 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → (¬ (((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}) ∧ 𝑁 = 𝑁) ∨ ((𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}) = 𝑁𝑁 = (𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}))) → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5742, 56sylbid 230 . . . . . . 7 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → (¬ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5857adantr 481 . . . . . 6 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (¬ {(𝑢𝑉 (𝐸𝑦) = {𝑢, 𝑁}), 𝑁} = {(𝑢𝑉 (𝐸𝑤) = {𝑢, 𝑁}), 𝑁} → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
5933, 58sylbid 230 . . . . 5 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (¬ (𝐸𝑦) = (𝐸𝑤) → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
6017, 59sylbid 230 . . . 4 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → (¬ 𝑦 = 𝑤 → ¬ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁})))
6160con4d 114 . . 3 (((𝐺 ∈ USGraph ∧ 𝑁𝑉) ∧ (𝑦𝐴𝑤𝐴)) → ((𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}) → 𝑦 = 𝑤))
6261ralrimivva 2966 . 2 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → ∀𝑦𝐴𝑤𝐴 ((𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}) → 𝑦 = 𝑤))
63 usgredg2v.f . . 3 𝐹 = (𝑦𝐴 ↦ (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}))
64 fveq2 6153 . . . . 5 (𝑦 = 𝑤 → (𝐸𝑦) = (𝐸𝑤))
6564eqeq1d 2623 . . . 4 (𝑦 = 𝑤 → ((𝐸𝑦) = {𝑧, 𝑁} ↔ (𝐸𝑤) = {𝑧, 𝑁}))
6665riotabidv 6573 . . 3 (𝑦 = 𝑤 → (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}))
6763, 66f1mpt 6478 . 2 (𝐹:𝐴1-1𝑉 ↔ (∀𝑦𝐴 (𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) ∈ 𝑉 ∧ ∀𝑦𝐴𝑤𝐴 ((𝑧𝑉 (𝐸𝑦) = {𝑧, 𝑁}) = (𝑧𝑉 (𝐸𝑤) = {𝑧, 𝑁}) → 𝑦 = 𝑤)))
686, 62, 67sylanbrc 697 1 ((𝐺 ∈ USGraph ∧ 𝑁𝑉) → 𝐹:𝐴1-1𝑉)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∨ wo 383 ∧ wa 384 = wceq 1480 ∈ wcel 1987 ∀wral 2907 {crab 2911 Vcvv 3189 {cpr 4155 ↦ cmpt 4678 dom cdm 5079 ran crn 5080 –1-1→wf1 5849 ‘cfv 5852 ℩crio 6570 Vtxcvtx 25787 iEdgciedg 25788 USGraph cusgr 25950 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-rep 4736 ax-sep 4746 ax-nul 4754 ax-pow 4808 ax-pr 4872 ax-un 6909 ax-cnex 9943 ax-resscn 9944 ax-1cn 9945 ax-icn 9946 ax-addcl 9947 ax-addrcl 9948 ax-mulcl 9949 ax-mulrcl 9950 ax-mulcom 9951 ax-addass 9952 ax-mulass 9953 ax-distr 9954 ax-i2m1 9955 ax-1ne0 9956 ax-1rid 9957 ax-rnegex 9958 ax-rrecex 9959 ax-cnre 9960 ax-pre-lttri 9961 ax-pre-lttrn 9962 ax-pre-ltadd 9963 ax-pre-mulgt0 9964 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-nel 2894 df-ral 2912 df-rex 2913 df-reu 2914 df-rmo 2915 df-rab 2916 df-v 3191 df-sbc 3422 df-csb 3519 df-dif 3562 df-un 3564 df-in 3566 df-ss 3573 df-pss 3575 df-nul 3897 df-if 4064 df-pw 4137 df-sn 4154 df-pr 4156 df-tp 4158 df-op 4160 df-uni 4408 df-int 4446 df-iun 4492 df-br 4619 df-opab 4679 df-mpt 4680 df-tr 4718 df-eprel 4990 df-id 4994 df-po 5000 df-so 5001 df-fr 5038 df-we 5040 df-xp 5085 df-rel 5086 df-cnv 5087 df-co 5088 df-dm 5089 df-rn 5090 df-res 5091 df-ima 5092 df-pred 5644 df-ord 5690 df-on 5691 df-lim 5692 df-suc 5693 df-iota 5815 df-fun 5854 df-fn 5855 df-f 5856 df-f1 5857 df-fo 5858 df-f1o 5859 df-fv 5860 df-riota 6571 df-ov 6613 df-oprab 6614 df-mpt2 6615 df-om 7020 df-1st 7120 df-2nd 7121 df-wrecs 7359 df-recs 7420 df-rdg 7458 df-1o 7512 df-2o 7513 df-oadd 7516 df-er 7694 df-en 7907 df-dom 7908 df-sdom 7909 df-fin 7910 df-card 8716 df-cda 8941 df-pnf 10027 df-mnf 10028 df-xr 10029 df-ltxr 10030 df-le 10031 df-sub 10219 df-neg 10220 df-nn 10972 df-2 11030 df-n0 11244 df-z 11329 df-uz 11639 df-fz 12276 df-hash 13065 df-edg 25853 df-umgr 25887 df-usgr 25952 This theorem is referenced by: usgriedgleord 26026
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https://solvedlib.com/n/quopnis-aeods-4-wnisstats-lab-8-2confidence-interval-place,18499182 | 1,696,475,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00111.warc.gz | 602,679,341 | 27,759 | # Quopnis Aeods 4 @@wnisSTATS LAB 8.2Confidence Interval (Place of Birth) Class Time:Names: Student Learning OutcomesThe student will calculate the 90%
###### Question:
Quopnis Aeods 4 @@wnis STATS LAB 8.2 Confidence Interval (Place of Birth) Class Time: Names: Student Learning Outcomes The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval: Collect the Data Survey the students in your class, asking them if they were born in this state: Let X = the number that were born in this state_ "T40Studextz g4lyo? 0.85 n = 40 24 in Caltopmg X= define the random variable P _ The randan varable PiS tbe qymtrcf;, In words State the estimated distribution to use-hormal dictbutic) 8irss_ dmpee X Sam; Find the Confidence Interval and Error Bound 5izq Calculate the confidence interval and the error bound. Confidence Interval: Error Bound: How much area is in both tails (combined)? & How much area is in each tail? 2 on the graph with the area in each section_ Then; fill in the number Iine Fill in the blanks with the upper and lower limits of the confidence interval and the sample proportion: CL 90 0.5 2 '-0,5 2 Fiaite R.7
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## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books
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...taken together) therefore .all the angles of a right-lined figure, together with four right angles, are equal to twice as many right angles as the figure has fides. And taking away four right angles from each, there will remain all the angles of the figure...
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...ft, I. 32. Euclid. All the anterior angles of any reoilineal figure, together with four right angles, are equal to twice as many right angles as the figure has fides. Hence the following rule. RULÉ. From double thé number of fides f übt vail: 4, and the remainder...
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# Until 2010, a state tax regulation known as the 80-20 rule
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Re: #Top150 SC: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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26 Sep 2015, 07:08
1
Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
Regulation required that X receive
>80% of income from tenants and
<20% of income from other sources.
there is no comparison here so like cannot be used for giving examples only such as needs to be used.
So A and E are out for the same reason.
A. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
B. Until 2010, a state tax regulation known as the “80-20 rule” requiring that condominium associations receive
at least 80 percent of their gross income from their tenant-shareholders, and
have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
requiring changes the meaning and structure of the sentence.
C. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive
at least 80 percent of their gross income from their tenant-shareholders, and
have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
D. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive
at least 80 percent of their gross income from their tenant-shareholders, and
have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
E. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and to have no more than 20 percent from other sources, like ground-floor rent for restaurants.
I chose D over C and others.
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#Top150 SC: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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05 Oct 2015, 05:26
[quote="souvik101990"]Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
A. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
like cannot be used to represent examples.
B. Until 2010, a state tax regulation known as the “80-20 rule” requiring that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
"Requiring that" is not correct form to modify the prev clause.
C. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
Required has to be follow subjunctive. Hence "That" is missing[color=#0000ff]
D. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
Subjunctive form is correctly used in both cases - "receive and have" , "Such as" is correctly used representing examples. Hence this is the correct answer.
E. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and to have no more than 20 percent from other sources, like ground-floor rent for restaurants.
Required has to be follow subjunctive. Hence "That" is missing. Also "like" cannot be used to represent examples..
Therefore, i feel Official answer is incorrect. souvik101990 please confirm.
Kudos if you feel i am correct.
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#Top150 SC: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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05 Oct 2015, 12:19
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2
Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
A. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and no more than 20 percent from other sources, like ground-floor rent for restaurants.
B. Until 2010, a state tax regulation known as the “80-20 rule” requiring that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
C. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
D. Until 2010, a state tax regulation known as the “80-20 rule” required that condominium associations receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
E. Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and to have no more than 20 percent from other sources, like ground-floor rent for restaurants.
Required X to Y >>> Condition, position, person, animal plant forces you to act in certain manner or to take certain action
Required that >>> Generally rules, regulations, laws, legislations enforce that certain steps need to be taken or that certain actions not to be taken
In this case, "that" must be used. Thus eliminate C & E.
Also, "like" is used for comparison and precedent only a noun or pronoun. "Such as" introduces examples. In this question, we understand that "ground floor rent" is given for comparison, so eliminate B & D.
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Re: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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28 Nov 2016, 10:12
please guys revise the source of this question. OA is A but many people disagree with like usage.
if it used for comparison, it should be : ' , income like ground-floor rent for restaurants.'
Gmat doesn't input ambiguous statements in SC question as It does in Math questions since Gmat verbal calls for simplicity and clarity of sentences not ambiguity.
Any way, I doubt this question and am quite confused.
please verbal experts, your inputs here will be highly appreciated.
thanks
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Re: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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28 Nov 2016, 10:40
hatemnag wrote:
please guys revise the source of this question. OA is A but many people disagree with like usage.
if it used for comparison, it should be : ' , income like ground-floor rent for restaurants.'
Gmat doesn't input ambiguous statements in SC question as It does in Math questions since Gmat verbal calls for simplicity and clarity of sentences not ambiguity.
Any way, I doubt this question and am quite confused.
please verbal experts, your inputs here will be highly appreciated.
thanks
It depends on whether the meaning intended is introducing example or comparing. Here introducing example is a better fit. Hence "such as" should have been used. Such a question is probably not expected in the real test.
(What is the problem with the source - is it not Veritas prep?)
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Re: #Top150 SC: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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30 Nov 2016, 02:00
1
I don't buy "like" as introducing a comparison here. Remember that "like" only compares stated nouns, so it would have to be comparing ground-floor rent to "other sources." We can't compare them to the method of collection, since that noun doesn't appear. (In any case, those wouldn't really be parallel.)
So we're left using "like" to introduce an example. While "such as" is preferred, the GMAT has used "like" on occasion. We can't always get consistency on these issues! Perhaps the GMAT is bowing to popular usage on this one, since the use of "like" to introduce examples is routine in spoken English.
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Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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26 Jun 2018, 01:11
abhishekmeister wrote:
Why is option C wrong ?
Hi abhishekmeister,
Here is choice C
Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
This sentence is in command subjunctive .
Before we go in that particular moods of the sentence, let us just start with how many types of moods are there in English language.
They are : Indicative, Imperative or command, Interrogative, Conditional, and Subjunctive mood.
What we have here in the original sentence is command subjunctive. In command subjunctive we have bossy verb and its proper order .
Bossy verb + that +Subject + bare infinitive form of verb .
Now " that " is missing in C therefore C is incorrect .
Link for further information
https://en.oxforddictionaries.com/grammar/moods
Hope it helps.
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Re: Until 2010, a state tax regulation known as the 80-20 rule [#permalink]
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13 Jul 2018, 20:06
arvind910619 wrote:
abhishekmeister wrote:
Why is option C wrong ?
Hi abhishekmeister,
Here is choice C
Until 2010, a state tax regulation known as the “80-20 rule” required condominium associations to receive at least 80 percent of their gross income from their tenant-shareholders, and have no more than 20 percent from other sources, such as ground-floor rent for restaurants.
This sentence is in command subjunctive .
What we have here in the original sentence is command subjunctive. In command subjunctive we have bossy verb and its proper order .
Bossy verb + that +Subject + bare infinitive form of verb .
Now " that " is missing in C therefore C is incorrect .
Option C is 100% grammatically correct, but option D is preferred because of the context.
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Re: Until 2010, a state tax regulation known as the 80-20 rule &nbs [#permalink] 13 Jul 2018, 20:06
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# Until 2010, a state tax regulation known as the 80-20 rule
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## Camera view matrix issue
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### #1BlackJoker Members
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Posted 16 August 2014 - 12:16 PM
Hello.
I faced with the issue when creating my camera in SharpDX.
Whaen I create camera instance I could set any camera lookAt vector and it works, but when I rebuild view matrix with other VectorUP (form Y to Z for ex) camera starts to look at the center of coordinates from the same position, but it must only change its orientation and not more.
public void SetUp(Vector3 _up)
{
Matrix tmpMatrix = ViewMatrix;
Vector3 currentLookAt = new Vector3(tmpMatrix.M13, tmpMatrix.M23, tmpMatrix.M33);
ViewMatrix = Matrix.LookAtLH(position, currentLookAt, _up);
Vector3 scale;
Quaternion rotation;
Vector3 translation;
ViewMatrix.Decompose(out scale, out rotation, out translation);
rotation = Quaternion.Normalize(rotation);
ViewMatrix = Matrix.Multiply(Matrix.Translation(Vector3.Negate(position)), Matrix.RotationQuaternion(rotation));
xAxis = new Vector3(ViewMatrix.M11, ViewMatrix.M21, ViewMatrix.M31);
yAxis = new Vector3(ViewMatrix.M12, ViewMatrix.M22, ViewMatrix.M32);
zAxis = new Vector3(ViewMatrix.M13, ViewMatrix.M23, ViewMatrix.M33);
}
Can someone say what I am doing wrong here?
### #2BlackJoker Members
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Posted 20 August 2014 - 12:40 AM
This issue is not actual already.
### #3haegarr Members
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Posted 20 August 2014 - 01:29 AM
A transformation matrix expresses a position and an orientation (letting scaling aside here). The function Matrix.LookAtLH(...) expects the camera position, the target position, and the up vector as arguments. Your routine SetUp(...) is written so that it tries to read the target position (named currentLookAt therein) from the view's transformation matrix. But there is nothing within the matrix that stores those position, and there is not enough information available to reconstruct it. The only position within that matrix is the own (i.e. view's) position.
In other words, this line
ViewMatrix = Matrix.LookAtLH(position, currentLookAt, _up);
doesn't give you what you expect.
So the routine SetUp(...) need to be revised. I suggest to do something like the following:
1.) Get the scaling values from the matrix.
2.) Get the forward vector from the matrix.
3.) Compute the new side vector by using the cross-product of forward and new up vectors.
4.) Scale the three direction vectors accordingly to the values calculated in 1.)
5.) Insert the scaled up and new side vectors into the matrix.
Edited by haegarr, 20 August 2014 - 01:37 AM.
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## Discussion of Problem 1297. Palindrome
I am getting WA at #1
Posted by Atindra Das 26 Apr 2013 01:44
I have tested for almost all test cases intentioned in discussions.
what might be the problem ?? ple help
code :
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package timusonlinejudge;
import java.util.Scanner;
/**
*
* @author dell
*/
public class N1297 {
public String getPal(String s, int i, int j,String result){
while((i>=0)&&(j<=s.length()-1)){
if((""+s.charAt(i)).equalsIgnoreCase(""+s.charAt(j))){
result = s.charAt(i)+result+s.charAt(j);
i--;j++;
}
else{
break;
}
}
return result;
}
public String longestPal(String s){
// System.out.println("---");
if(s.isEmpty()){
return "";
}
String result = s.substring(0,1);
//System.out.println("---"+result);
for(int i = 1; i<s.length();i++){
if((""+s.charAt(i-1)).equalsIgnoreCase(""+s.charAt(i))){
String temp = getPal(s,i-1,i,"");
//System.out.println(temp);
if(temp.length()>result.length()){
result = new String(temp);
//System.out.println(temp);
}
}
String temp = getPal(s,i-1,i+1,""+s.charAt(i));
//System.out.println(temp);
if(temp.length()>result.length()){
result = new String(temp);
//System.out.println(temp);
}
}
return result;
}
public static void main(String[] args) {
N1297 main = new N1297();
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
//System.out.println(input);
System.out.println(main.longestPal(input));
// TODO code application logic here
}
} | 444 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.380908 |
http://bestbikerbars.com/6xm79/e1b483-evaluate-the-expression-calculator-fractions | 1,627,495,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00589.warc.gz | 5,909,791 | 9,486 | It also shows detailed step-by-step information about the fraction calculation procedure. Simplifying an Expression With a Fraction Bar. Simplify numerical expression calculator that shows work to solve the numerical expressions having array of operations like addition, subtraction, multiplication and division altogether or any of the combination in an expression. у Y² - 4y - 5 5 (Simplify Your Answer. Use Integers Or Fractions For Any Numbers In The Expression.) Khan Academy is a 501(c)(3) nonprofit organization. The Modulo Calculator is used to perform the modulo operation on numbers. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. 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Will show you how to evaluate when = try this before resorting to the assistance from instructor. | 4,177 | 19,832 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 3.3125 | 3 | CC-MAIN-2021-31 | latest | en | 0.825995 |
https://en.wikipedia.org/wiki/Osipkov-Merritt_model | 1,544,718,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00557.warc.gz | 598,565,543 | 14,789 | # Osipkov–Merritt model
(Redirected from Osipkov-Merritt model)
Osipkov-Merritt distribution functions, derived from galaxy models obeying Jaffe's law in the density. The isotropic model, ${\displaystyle f=f(E)}$, is plotted with the heavy line.
Osipkov–Merritt models (named for Leonid Osipkov and David Merritt) are mathematical representations of spherical stellar systems (galaxies, star clusters, globular clusters etc.). The Osipkov–Merritt formula generates a one-parameter family of phase-space distribution functions that reproduce a specified density profile (representing stars) in a specified gravitational potential (in which the stars move). The density and potential need not be self-consistently related. A free parameter adjusts the degree of velocity anisotropy, from isotropic to completely radial motions. The method is a generalization of Eddington's formula[1] for constructing isotropic spherical models.
The method was derived independently by its two eponymous discoverers.[2][3] The latter derivation includes two additional families of models (Type IIa, b) with tangentially anisotropic motions.
## Derivation
According to Jeans's theorem, the phase-space density of stars f must be expressible in terms of the isolating integrals of motion, which in a spherical stellar system are the energy E and the angular momentum J. The Osipkov-Merritt ansatz is
${\displaystyle f=f(Q)=f(E+J^{2}/2r_{a}^{2})}$
where ra, the "anisotropy radius", is a free parameter. This ansatz implies that f is constant on spheroids in velocity space since
${\displaystyle 2Q=v_{r}^{2}+(1+r^{2}/r_{a}^{2})v_{t}^{2}+2\Phi (r)}$
where vr, vt are velocity components parallel and perpendicular to the radius vector r and Φ(r) is the gravitational potential.
The density ρ is the integral over velocities of f:
${\displaystyle \rho (r)=2\pi \int \int f(E,J)v_{t}dv_{t}dv_{r}}$
which can be written
${\displaystyle \rho (r)={2\pi \over r^{2}}\int _{\Phi }^{0}dQf(Q)\int _{0}^{2r^{2}(Q-\Phi )/(1+r^{2}/r_{a}^{2})}dJ^{2}\left[2(Q-\Phi )-(J^{2}/r^{2})(1+r^{2}/r_{a}^{2})\right]^{-1/2}}$
or
${\displaystyle \rho (r)={4\pi \over 1+r^{2}/r_{a}^{2}}\int _{\Phi }^{0}dQ{\sqrt {2(Q-\Phi )}}f(Q).}$
This equation has the form of an Abel integral equation and can be inverted to give f in terms of ρ:
${\displaystyle f(Q)={{\sqrt {2}} \over 4\pi ^{2}}{d \over dQ}\int _{Q}^{0}{d\Phi \over {\sqrt {\Phi -Q}}}{d\rho ^{'} \over d\Phi },\ \ \ \ \ \rho ^{'}(\Phi )=\left[1+r(\Phi )^{2}/r_{a}^{2}\right]\rho \left[r(\Phi )\right].}$
## Properties
Following a derivation similar to the one above, the velocity dispersions in an Osipkov–Merritt model satisfy
${\displaystyle {\sigma _{r}^{2} \over \sigma _{t}^{2}}=1+{r^{2} \over r_{a}^{2}}.}$
The motions are nearly radial (${\displaystyle \sigma _{r}\gg \sigma _{t}}$) for ${\displaystyle r\gg r_{a}}$ and nearly isotropic (${\displaystyle \sigma _{r}\approx \sigma _{t}}$) for ${\displaystyle r\ll r_{a}}$. This is a desirable feature, since stellar systems that form via gravitational collapse have isotropic cores and radially-anisotropic envelopes.[4]
If ra is assigned too small a value, f may be negative for some Q. This is a consequence of the fact that spherical mass models can not always be reproduced by purely radial orbits. Since the number of stars on an orbit can not be negative, values of ra that generate negative f's are unphysical. This result can be used to constrain the maximum degree of anisotropy of spherical galaxy models.[3]
In his 1985 paper, Merritt defined two additional families of models ("Type II") that have isotropic cores and tangentially anisotropic envelopes. Both families assume
${\displaystyle f=f(E-J^{2}/2r_{a}^{2})}$.
In Type IIa models, the orbits become completely circular at r=ra and remain so at all larger radii. In Type IIb models, stars beyond ra move on orbits of various eccentricities, although the motion is always biased toward circular. In both families, the tangential velocity dispersion undergoes a jump as r increases past ra.
C. M. Carollo et al. (1995)[5] derive many observable properties of Type I Osipkov–Merritt models.
## Applications
Typical applications of Osipkov–Merritt models include:
## References
1. ^ Eddington, A. (1916), The distribution of stars in globular clusters, Mon. Not. R. Astron. Soc., 76, 572
2. ^ Osipkov, L. P. (1979), Spherical systems of gravitating bodies with an ellipsoidal velocity distribution, Pis'ma v Astron. Zhur., 5, 77
3. ^ a b Merritt, D. (1985), Spherical stellar systems with spheroidal velocity distributions, Astron. J., 90, 1027
4. ^ van Albada, T. (1983), Dissipationless galaxy formation and the R to the 1/4-power law, Mon. Not. R. Astron. Soc., 201, 939
5. ^ Carollo, C. M. et al. (1995), Velocity profiles of Osipkov-Merritt models, Mon. Not. R. Astron. Soc., 276, 1131
6. ^ Lupton, R. et al. (1989), The internal velocity dispersions of three young star clusters in the Large Magellanic Cloud, Astrophys. J., 347, 201
7. ^ Nolthenius, R. and Ford, H. (1987), The mass and halo dispersion profile of M32, Astrophys. J., 305, 600
8. ^ Sotnikova, N. Ya. and Rodionov, S. A. (2008), Anisotropic Models of Dark Halos, Astron. Lett., 34, 664-674
9. ^ Lokas, E. and Mamon, G. (2001), Properties of spherical galaxies and clusters with an NFW density profile, Mon. Not. R. Astron. Soc., 321, 155
10. ^ May, A. and Binney, J. (1986), Testing the stability of stellar systems, Mon. Not. R. Astron. Soc., 221, 13
11. ^ Saha, P. (1991), Unstable modes of a spherical stellar system, Mon. Not. R. Astron. Soc., 248, 494 | 1,667 | 5,587 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.844934 |
http://forum.arduino.cc/index.php?topic=164986.msg1246819 | 1,524,778,111,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948549.21/warc/CC-MAIN-20180426203132-20180426223132-00454.warc.gz | 125,188,721 | 9,209 | Go Down
### Topic: Help with getting started with an LED driver (AS1100PL) (Read 2255 times)previous topic - next topic
#### Nicknml
##### May 06, 2013, 08:28 pm
Has anybody played with one? I've tried to find some examples of using the chip with an Arduino with no success (yea I probably should have seen what models were popular before getting them, at least they were free.) Any sample sketches/diagrams would be appreciated.
#### DVDdoug
#1
##### May 07, 2013, 12:42 amLast Edit: May 07, 2013, 12:45 am by DVDdoug Reason: 1
That is sort-of an advanced chip... I've never used one. Troubleshooting could be tricky when the display-chip acts differently than what you expect...
You probably should start by experimenting with a simpler shift register. There is an example project here. (I have used something similar to that...) If you can write a hex value to the 74HC595 chip, and see that value displayed as an 8-bit binary "number" on 8 LEDs, you'll probably feel better about using the AS1100 chip.
You'll need to read the datasheet several times!!! Make sure you understand what all the pins are for (especially the input/control pins), how the serial communication works, and what the various commands (hex codes) do.
With this chip, you write data to an internal register by sending a 16-bit word that contains a register address (4-bits) and the data (8-bits) to be sent to the register. The 4 most significant bits are "don't care", but something needs to be sent so you'll normally send zeros.
The Windows calculator in the scientific mode can convert between hexadecimal and binary, or you can memorize the 16 conversions (0-F) and you'll be able to convert numbers of any size between hex and binary in your head!!! It's much easier than converting between decimal and binary, because each nybble (group of 4 bits) converts to exactly one hex digit. (You already know 1 and 0, so that's only 14 more to learn, and some are easy to remember like F, 5, & A.)
The datasheet shows hex and binary. I suggest you use hexadecimal in your sketch (put 0x in front of the number to indicate hex). You can use decimal if you really want to, the compiler doesn't care since everything is binary inside the microprocessor anyway. (On the AS1100 datasheet, an uppercase X means "don't care" and a zero followed by a lower case x means the following digits are hex.)
-----------
The serial data communication concept is fairly simple... once you "get it". You "clock-in" data one bit at a time, and then after you've sent-in 8 or 16-bits, you "latch" that data to the parallel outputs.
You need to assign 3 Arduino-output pins (data, clock, latch). The actual signal names might be different on various chips, but the function is the same.
1. Write to the data pin (one bit high or low).
2. After the data-pin is stable, write to the clock pin (With the AS1100 chip, data is read on the clock's rising-edge).
3. Write to the clock again to reset (Write low, to get ready for another rising edge.)
4. Write the next bit to the data pin (bit high or low).
... Repeat until all bits have been written (16 bits for the AS1100).
5. When all bits have been written, Write to the latch pin. (With the AS1100, data is latched-in when on a rising edge.)
6. Reset the load pin to get ready for the next rising edge.
Usually, the exact timing is not that critical. The important thing is that the data is stable for a short period of time, before the clock-edge comes along.
#### Nicknml
#2
##### May 07, 2013, 01:18 am
Thanks for the advice, I'll grab a 74HC595. The joys of being a noob
#### Nicknml
#3
##### May 20, 2013, 08:07 pm
My 74HC595 chips arrived today. On the guide I noticed that the circuit diagrams (not the actual breadboard pictures) on the tutorial page appear to be incorrect http://arduino.cc/en/Tutorial/ShiftOut. The pin labeled as Q0 is actually Q1, pin Q7 is GND to name a few.
#### Nicknml
#4
##### May 20, 2013, 09:44 pmLast Edit: May 20, 2013, 10:57 pm by Nicknml Reason: 1
Another question, could I safely hook up a small microcontroller that uses 3.3 volts to operate the 74HC595 and supply the 74HC595 with an external 5 volt supply?
By the way, has anybody tried manually clocking in the data using switches instead of a microcontroller? I might try that tomorrow.
#### Nicknml
#5
##### May 21, 2013, 06:39 pmLast Edit: May 21, 2013, 09:42 pm by Nicknml Reason: 1
I'm now looking at the AS1113, it appears that you can drive it similarly to the 74HC595 (each bit controls the pin state of the pin that corresponds to that pin.) It also looks like I would only have to use one resistor to limit the current for all LEDs by placing it between the REXT pin and ground. So to wire that up I would just need to hook up the clock(CLK), latch(LD), data (SDI), vcc(VDD), ground, REXT pins (and of course the LED pins that I plan to use)?
UPDATE: Nevermind, apparently it's not available in a DIP package.
#### didiy
#6
##### May 22, 2013, 04:26 am
It might work, but if they don't read the other sources I suspect they won't read a sticky. Just an opinion (that I'm willing to change).
I've had beginners argue with me about the subject, they want to treat them like light bulbs and that is all there is to it.
#### Nicknml
#7
##### May 22, 2013, 10:58 pm
Ok I back to playing with the AS1100 chip. I managed to get it in test mode. Now the problem is that only DIG0 of the DIG 0-DIG7 pins is on.
My sketch code below:
Code: [Select]
`int latchPin = 8;int clockPin = 12;int dataPin = 11;int speed1 = 50;void setup() { //set pins to output so you can control the shift register pinMode(latchPin, OUTPUT); pinMode(clockPin, OUTPUT); pinMode(dataPin, OUTPUT);twelvebit(1,1,1,0,0,0,0,0,0,0,0,1); //sets clock to externaltwelvebit(1,1,1,1,0,0,0,0,0,0,0,1); //turns on display test}void loop() { }void clockpin(){ digitalWrite(clockPin, HIGH); digitalWrite(clockPin, LOW);}void twelvebit(byte b1, byte b2, byte b3, byte b4, byte b5, byte b6, byte b7, byte b8, byte b9, byte b10, byte b11, byte b12){digitalWrite(latchPin, LOW); digitalWrite(dataPin, LOW); clockpin(); digitalWrite(dataPin, LOW); clockpin(); digitalWrite(dataPin, LOW); clockpin(); digitalWrite(dataPin, LOW); clockpin(); digitalWrite(dataPin, b1); clockpin(); digitalWrite(dataPin, b2); clockpin(); digitalWrite(dataPin, b3); clockpin(); digitalWrite(dataPin, b4); clockpin(); digitalWrite(dataPin, b5); clockpin(); digitalWrite(dataPin, b6); clockpin(); digitalWrite(dataPin, b7); clockpin(); digitalWrite(dataPin, b8); clockpin(); digitalWrite(dataPin, b9); clockpin(); digitalWrite(dataPin, b10); clockpin(); digitalWrite(dataPin, b11); clockpin(); digitalWrite(dataPin, b12); clockpin(); digitalWrite(latchPin,HIGH); } `
#### Nicknml
#8
##### May 24, 2013, 02:57 am
After some more research it appears that it's almost identical to the MAX7219CNG. I've been playing around with the example code, but all I could get was all LEDs flashing momentarily. http://playground.arduino.cc/LEDMatrix/Max7219
Go Up | 1,974 | 7,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.932723 |
https://pp.ipd.kit.edu/git/libfirm/tree/ir/ana/domfront.c?id=6dfeab4016d89fb2546d7772d52b84facdcb0f7b | 1,670,058,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00841.warc.gz | 485,438,702 | 4,352 | summaryrefslogtreecommitdiffhomepage log msg author committer range
path: root/ir/ana/domfront.c
blob: 66e99e77dfa9c483b067172d60c835b0dc829ca1 (plain)
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 ``` ``````/* * This file is part of libFirm. * Copyright (C) 2012 University of Karlsruhe. */ /** * @file * @brief Algorithms for computing dominance frontiers. * @author Sebastian Hack, Daniel Grund * @date 04.05.2005 */ #include "array.h" #include "irdom.h" #include "iredges_t.h" #include "obst.h" #include "pmap.h" /** * A wrapper for get_Block_idom. * This function returns the block itself, if the block is the start * block. Returning NULL would make any != comparison true which * suggests, that the start block is dominated by some other node. * @param bl The block. * @return The immediate dominator of the block. */ static inline ir_node *get_idom(ir_node *bl) { ir_node *idom = get_Block_idom(bl); return idom == NULL ? bl : idom; } /** * Compute the dominance frontier for a given block. * * @param blk the block where the calculation starts * * @return the list of all blocks in the dominance frontier of blk */ static ir_node **compute_df(ir_node *blk, ir_dom_front_info_t *info) { ir_node **df_list = NEW_ARR_F(ir_node *, 0); /* Add local dominance frontiers */ foreach_block_succ(blk, edge) { ir_node *y = get_edge_src_irn(edge); if (get_idom(y) != blk) { ARR_APP1(ir_node *, df_list, y); } } /* * Go recursively down the dominance tree and add all blocks * into the dominance frontiers of the children, which are not * dominated by the given block. */ for (ir_node *c = get_Block_dominated_first(blk); c != NULL; c = get_Block_dominated_next(c)) { ir_node **df_c_list = compute_df(c, info); for (size_t i = ARR_LEN(df_c_list); i-- > 0;) { ir_node *w = df_c_list[i]; if (get_idom(w) != blk) ARR_APP1(ir_node *, df_list, w); } } /* now copy the flexible array to the obstack */ ir_node **const df = DUP_ARR_D(ir_node*, &info->obst, df_list); DEL_ARR_F(df_list); pmap_insert(info->df_map, blk, df); return df; } void ir_compute_dominance_frontiers(ir_graph *irg) { ir_dom_front_info_t *info = &irg->domfront; assure_irg_properties(irg, IR_GRAPH_PROPERTY_CONSISTENT_OUT_EDGES | IR_GRAPH_PROPERTY_CONSISTENT_DOMINANCE); obstack_init(&info->obst); info->df_map = pmap_create(); compute_df(get_irg_start_block(irg), info); add_irg_properties(irg, IR_GRAPH_PROPERTY_CONSISTENT_DOMINANCE_FRONTIERS); } void ir_free_dominance_frontiers(ir_graph *irg) { clear_irg_properties(irg, IR_GRAPH_PROPERTY_CONSISTENT_DOMINANCE_FRONTIERS); ir_dom_front_info_t *info = &irg->domfront; if (info->df_map == NULL) return; obstack_free(&info->obst, NULL); pmap_destroy(info->df_map); info->df_map = NULL; } /* Get the dominance frontier of a block. */ ir_node **ir_get_dominance_frontier(const ir_node *block) { ir_graph *irg = get_irn_irg(block); ir_dom_front_info_t *info = &irg->domfront; return pmap_get(ir_node*, info->df_map, block); } `````` | 1,054 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.519089 |
https://mw.lojban.org/index.php?title=BPFK_Section:_Subordinators_as_of_25_Dec_2004&oldid=111462 | 1,656,787,194,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00459.warc.gz | 457,273,480 | 15,630 | # BPFK Section: Subordinators as of 25 Dec 2004
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## Proposed Definitions And Examples
### Proposed Definition of noi
noi (NOI)
Incidental (non-restrictive) relative clause marker.
• The "relative" part means that it relates a clause to a sumti. It attaches to a sumti to provide additional information about the referents of that sumti.
• The "clause" part means that it is followed by a full bridi (which means it sometimes must be terminated with ku'o, the NOI selma'o terminator, or vau, the general bridi terminator, particularily if one wishes to add another sumti to the outer bridi).
• The "non-restrictive" part means that the information in the noi clause is not used to restrict the set of things that the sumti noi is attached to refers to. The noi bridi gives additional information about the referents of the sumti noi is attached to.
• Inside a noi clause, ke'a indicates the precise place of the bridi that the sumti is intended to fill.
• For logical scoping purposes, the scope of a noi clause is entirely outside the scope of the statement in which it is contained; its scope occurs at the point immediately after the scope in which it was contained ends. The noi clause should be considered, for scoping purposes, as occuring in its own virtual sentence (technically, its own "statement" production in the formal grammar) after both the one in which it is contained and all further statements that are logically connected to the one in which it was contained.
• noi immediately follows a simple sumti. With description sumti, the relative clause can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a clause attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku) the clause applies to all the referents of the sumti, whether there is an outer quantifier or not.
• For la description sumti, any noi clause that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a noi clause is considered to be inside the terminator ku.
### Examples of noi Usage
le gerku noi blanu cu barda
The dog, which incidentally is white, is big.
la alis noi ru'i senci cu sezysku
Alice, who was continuously sneezing, said to herself.
ro bifce noi pendo mi cu ji'a xebni lo sigja
All bees, which are friends of mine, also hate cigars.
This sentence was taken from the #lojban IRC channel, and asserts that every bee (or wasp or hornet) that exists (ignoring metaphysical issues) is both a friend of the speaker and hates cigars. This is probably not what the author intended.
la fengu lo smacu noi fy ke'a cpacu cu penmi le zdani
Fury met a mouse, F (Fury) got it (the mouse), in the house.
Had to re-order the translation a bit to make the English work; in the Lojban the "met" part comes after the comma-delimited clause.
### Proposed Definition of poi
poi (NOI)
Restrictive relative clause marker.
• The "relative" part means that it relates a clause to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti.
• The "clause" part means that it is followed by a full bridi (which means it sometimes must be terminated with ku'o, the NOI elma'o terminator, or vau, the general bridi terminator, particularily if one wishes to add another sumti to the outer bridi).
• The "restrictive" part means that the information in the poi clause is used to restrict the set of things that the sumti poi is attached to refers to. In other words, out of the referents of the sumti that poi is attached to (which, for example, in the case of lo dacti can be a great many things indeed) the sumti is actually intended by the speaker to refer only to those things for which the bridi in the poi clause is also true.
• Inside a poi clause, ke'a indicates the precise place of the bridi that the sumti is intended to fill.
• For unquantified sumti, the clause selects from all the referents of the sumti just those that satisfy it; when an inner quantifier is present it indicates how many those referents are. For quantified sumti, the quantification is over just those referents of the sumti that satisfy the clause.
• poi immediately follows a simple sumti. With description sumti, the relative clause can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a clause attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the clause even if there is an outer quantifier.
• For la description sumti, any poi clause that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a poi clause is considered to be inside the terminator ku.
### Examples of poi Usage
mi djica lo skami tanxe poi cmalu
I want a computer box which is small.
pau re'i pat ta poi zvati le canko cu mo
Question to Pat: that which at the window is what?
Pat: What is that at the window?
abu tavla le mensi ro le cizra se lifri poi do ke'a puzi ca'o tcidu
A (for Alice) talked to the (her) sister about all the strange experiences which you about them have just been reading.
da poi gerku zo'u da vasxu
There exists at least one thing which is a dog; this thing breathes.
### Proposed Definition Of voi
voi (NOI)
Descriptive (non-veridical) restrictive relative clause marker.
• The "relative" part means that it relates a clause to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti.
• The "clause" part means that it is followed by a full bridi (which means it sometimes must be terminated with ku'o, the NOI
selma'o terminator, or vau, the general bridi terminator, particularily if one wishes to add another sumti to the outer bridi).
• The "restrictive" part means that the information in the voi clause is used to restrict the set of things that the sumti voi is attached to refers to. In other words, out of all the possible things the sumti that voi is attached to could refer to (which, for example, in the case of lo dacti is a great many things indeed) the sumti is actually intended by the speaker to refer only to those things that the sumti could refer to for which the bridi in the voi clause is also true.
• The non-veridical part means that the speaker is making no claime that the bridi in the voi clause actually matches objective reality.
• Inside a voi clause, ke'a indicates the precise place of the bridi that the sumti is intended to fill.
• For unquantified sumti, the clause selects from all the referents of the sumti just those that satisfy it; when an inner quantifier is present it indicates how many those referents are. For quantified sumti, the quantification is over just those referents of the sumti that satisfy the clause.
• voi immediately follows a simple sumti. With description sumti, the relative clause can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a clause attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the clause even if there is an outer quantifier.
• For la description sumti, any voi clause that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a voi clause is considered to be inside the terminator ku.
### Examples of voi Usage
ti voi nanmu cu ninmu
This which is (non-veridically) a man is a woman.
The classic example of voi usage, presumably referring to a case of mistaken identity or a transvestite or transgendered individual.
mi xagji .iku'i .oisai ponse no da voi cidja
I am hungry. However, horrors!, posses no thing which is food.
This is presumably intended to handle under-exaggeration; the speaker probably has something that could be eaten. This is a quote from the #lojban IRC channel.
mrilu su'o ciki'o da voi festi mi
Mail, three thousand of them which are waste products of me.
This case, also from the #lojban IRC channel, is probably intended to deal with very loose usage of festi, and is probably intended to mean "Three thousand pieces of spam e-mail".
ganse vasxu le nicte vacri voi ranti
Breathing the night air, which is soft.
Presumably, voi is being used to deal with the fact that ranti probably does not literally apply to air.
ku'i ro da voi vi selsnu zo'u .ai skicu da fo ledu'u xukau catni
However, all things which are here-at subjects of conversation: it is intended that they will be described as to whether they are official.
This is from the Lojban translation of the lojban.org web site, and the voi is being used to deal with the usage of vi to translate the English "here on this site", when a web site has no physical location. The original had zu'o instead of zo'u, which I have corrected.
### Proposed Definition for ne
ne (GOI)
Incidental (non-restrictive) relative phrase marker.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide additional information about the referents of that sumti.
• The "phrase" part means that it is followed by another sumti.
• The meaning of ne is that the attached sumti is relevant to or associated with the first sumti in some way.
• The "non-restrictive" part means that the information in the ne phrase is not used to restrict the set of things that the sumti ne is attached to refers to. The attached phrase gives additional information about the sumti no'u is attached to.
• ne is often used for the loosest form of possession, such as that between a person and the chair they happen to be sitting in at the moment, but can be used for any form of association at all. It is essentially equivalent to noi srana.
• Instead of attaching a sumti, ne can be used to attach a sumtcita clause, in which case the sumtcita clause is said to apply only to the sumti it is attached to.
• ne immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (beforep the terminator ku) the claupse applies to all the referents of the sumti, whether there is an outer quantifier or not.
• For la description sumti, any ne phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a ne phrase is considered to be inside the terminator ku.
### Examples of ne Usage
le blabi gerku ne mi cu batci do
The white dog, which incidentally is associated with me, bit you.
This need not be "my" dog in the English sense, but could be the dog I'm walking for a friend, or the dog closest to me, or whatever.
le nuntra be la uiliam ca le nu cfari cu fadni i ku'i le nuntolclite be lai norman ne ubu
The behaviour of William during the beginning is ordinary. However the rudeness of Norman, which incidentally is associated with him (William).
ti'e ko'a ne li 2.6 cu mutce sutra
I hear that it, which has something to do with the number 2.6, is very fast.
xu naku me le cnano pe le tai tcima ne vi do
Is it not the case that those among the norm which is associated with the form of the weather, which is near you?
Don't you normally have such weather?
### Proposed Definition for pe
pe (GOI)
Restrictive phrase. pe is one of Lojban's restrictive relative phrase markers.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti.
• The "phrase" part means that it is followed by another sumti.
• The meaning of pe is that the attached sumti is relevant to or associated with the first sumti in some way.
• The "restrictive" part means that the information in the pe phrase is used to restrict the set of things that the sumti pe is attached to refers to. In other words, out of all the possible things the sumti that pe is attached to could refer to (which, for example, in the case of lo dacti is a great many things indeed) the sumti is actually intended by the speaker to refer only to those things that the sumti could refer to which are associated with the sumti marked by pe.
• pe is often used for the loosest form of possession, such as that between a person and the chair they happen to be sitting in at the moment, but can be used for any form of association at all. It is essentially equivalent to poi srana.
• Instead of attaching a sumti, pe can be used to attach a sumtcita clause, in which case the sumtcita clause is said to apply only to the sumti it is attached to.
• pe immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the phrase even if there is an outer quantifier.
• For la description sumti, any pe phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a pe phrase is considered to be inside the terminator ku.
### Examples of pe Usage
le kabri pe le mi pendo cu cmalu
My friend's cup is small.
Presumably the friend does not own the cup, but is merely holding it, or something similar.
le mi pendo pe le kabri cu cmalu
The cup's my friend is small.
My friend which is associated with the cup is small.
This illustrates one of the non-English-like ways pe can operate.
le lisri pe mi cu clani je se badri
The story associated with me is long and sad.
le gerku pe le mi pendu cu prami lo cakla
The dog of my fiend loves chocolate.
It is likely that po, which indicates stronger possession, was wanted here.
mi viska le cripu pe vi le rirxe
I see the bridge which is near the river.
### Proposed definition for no'u
no'u (GOI)
Incidental identity. no'u is Lojban's non-restrictive appositive phrase marker.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide additional information about the referents of that sumti.
• The "phrase" part means that it is followed by another sumti.
• The meaning of no'u is that the attached sumti has exactly the same refferents as the first sumti, which is what the "appositive" part means.
• The "non-restrictive" part means that the information in the no'u clause is not used to restrict the set of things that the sumti no'u is attached to refers to. The attached phrase gives additional information about the sumti no'u is attached to.
• It exactly equates two sumti referents, and is often used to mention names of things in passing. It is essentially equivalent to noi du".
• no'u immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (beforep the terminator ku) the claupse applies to all the referents of the sumti, whether there is an outer quantifier or not.
• For la description sumti, any no'u phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a no'u phrase is considered to be inside the terminator ku.
### Examples of no'u Usage
le nanmu no'u la djim. cu terpemci
The man, who incidentially is Jim, is a poet.
.au mi ka'e jarco le mi'a mlatu no'u la dinas do
Desire. I could show you our cat, Dina.
mi ba stidi so'u cnino gismu no'u zo nagra e zo narga e zo ranga e zo ragna
I will suggest a few new gismu: nagra, narga, ranga, and ragna.
### Proposed definition for po'u
po'u (GOI)
Restrictive identity. po'u is Lojban's restrictive appositive phrase marker.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti. po'u immediately follows a simple sumti; for descriptions sumti it can appear in a variety of places, the semantics of which are beyond the scope of this definition.
• The "phrase" part means that it is followed by another sumti.
• The meaning of po'u is that the referents of the sumti that the po'u clause is attached to are limited to those that are identical to the referents of the po'u phrase, which is what the "appositive" part means.
• The "restrictive" part means that the information in the po'u phrase not used to restrict the set of things that the sumti po'u is attached to refers to.
• It exactly equates two sumti referents, and is often used to mention names of things in passing. It is essentially equivalent to poi du".
• po'u immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the phrase even if there is an outer quantifier.
• For la description sumti, any po'u phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a po'u phrase is considered to be inside the terminator ku.
### Examples of po'u Usage
le gerku po'u le mi pendo cu cinba mi
The dog which is my friend kisses me.
This is actually a somewhat strange use of po'u; poi se pendo mi is much more common for this sort of thing.
la blabi ractu sepi'o le kinli je cmalu voksa cu bacru tcidu le cmene po'u zo alis
The white rabbit in a sharp, small voice loudly read the name, "Alice".
mi cusku za'u pa le jufra .i do tugni ma po'u le mi jufra
I said more than one sentence. You agree with what which is a sentence of mine?
Note that the first le should really be lo. za'u jufra would be even better.
### Proposed definition for po
po (GOI)
Is specific to. po is one of Lojban's restrictive relative phrase markers.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti.
• The "phrase" part means that it is followed by another sumti.
• The meaning of po is that the attached sumti is specifically relevant to or associated with the first sumti in some way.
• The "restrictive" part means that the information in the po phrase is used to restrict the set of things that the sumti po is attached to refers to. In other words, out of all the possible things the sumti that po is attached to could refer to (which, for example, in the case of lo dacti is a great many things indeed) the sumti is actually intended by the speaker to refer only to those things that the sumti could refer to which are associated with the sumti marked by po.
• po is often used for the standard sense of possession, i.e. physical or legal ownership. It is essentially equivalent to poi traji lo ka ce'u ckini ke'a vau fa.
• po immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the phrase even if there is an outer quantifier.
• For la description sumti, any po phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a po phrase is considered to be inside the terminator ku.
### Examples of po Usage
le botpi po mi cu spofu
My bottle is broken.
mi troci le nu jimpe le krinu po le do cortu
I try to understand the reason / justification associated with your pain.
Probably rinka be was meant, rather than krinu po.
xu do tcidu le samselmri po la noras
### Proposed definition for po'e
po'e (GOI)
Which belongs to. po'e is one of Lojban's restrictive relative phrase markers.
• The "relative" part means that it relates a phrase to a sumti. It attaches to a sumti to provide specifying information about the referents of that sumti.
• The "phrase" part means that it is followed by another sumti.
• The meaning of po'e is that the attached sumti is inalienably associated with the first sumti in some way. An inalienable association is one that cannot be taken away, such as the association between you and your arm.
• The "restrictive" part means that the information in the po'e phrase is used to restrict the set of things that the sumti po'e is attached to refers to. In other words, out of all the possible things the sumti that po'e is attached to could refer to (which, for example, in the case of lo dacti is a great many things indeed) the sumti is actually intended by the speaker to refer only to those things that the sumti could refer to which are associated with the sumti marked by po'e.
• po'e is used for things like people's limbs or parental relationships or other inalienable things. It is essentially equivalent to poi jinzi ke se steci srana.
• po'e immediately follows a simple sumti. With description sumti, the relative phrase can also be attached inside the sumti, before or after the selbri; when attached before the selbri (right after the gadri) it is equivalent to a phrase attached after the terminated sumti (after the terminator ku); when attached after the selbri (before the terminator ku), the inner quantifier indicates the number of referents that satisfy the phrase even if there is an outer quantifier.
• For la description sumti, any po'e phrase that occurs before the terminator ku is considered part of the name.
• When attached to a description sumti with no explicit terminators, a po'e phrase is considered to be inside the terminator ku.
### Examples of po'e Usage
le birka po'e mi cu spofu
My arm is broken.
mi kucli le du'u ki'u makau le xriso cu krici le du'u le malcevni cu djica le pruxi po'e mi
I am curious as to why Christians believe that the devil wants my soul.
le galxe po'e mi cu kanro
My throat is healthy.
### Proposed Definition for vu'o
vu'o (VUhO)
Long scope relative clause/phrase marker. Normally, a relative clause or phrase sumti binds to the last sumti to its immediate left, regardless of sumti connectors. To have a relative clause or phrase bind to every member of a connected group of sumti, place vu'o after the sumti and before the relative clause or phrase cmavo.
### Examples of vu'o Usage
la frank. .e la djordj. vu'o noi nanmu cu klama le zdani
Frank and George, who are both men, go to the house.
la eduin joi la morkar vu'ono'u lei turni be la mersias fa'u la nortumbrias
Edwin along with Morkar (?), who are the rulers of Mersia (?) and Northumbria, respectively
mi na jimpe le du'u makau nilbra be le plini .e le gerku vu'o noi se tavla do
I don't understand how big the planet and the dog, which you are talking about, are.
### Proposed Definition for zi'e
zi'e (ZIhE)
Relative clause/phrase joiner. Normally, a relative clause or phrase sumti binds to the last sumti to its immediate left, which means that it is impossible to apply more than one relative marker to the same simple sumti (description sumti are a bit more complicated, and the ways in which one can attach a relative clause or phrase to one are beyond the scope of this definition). To attach another relative marker to the same sumti, place zi'e after the end of the first relative marker's data (which normally needn't be terminated) and place the next relative marker immediately after the zi'e. Using zi'e to mix poi and noi clauses (or pe and ne, and so on) is, for very subtle reasons, not well defined.
### Examples of zi'e Usage
le botpi po mi zi'e poi blanu cu spofu
My blue bottle is broken.
do ponse ma poi drata zi'epe ne'i le do daski
You have what else that has to do with the inside of your pocket?
je'u ro lo prenu poi mi xamgu djuno zi'e poi ponse le slanu kerfa zi'e poi ponse lo blabi skapi cu pilno ry. vau .u'isai
Truthfully, all the people which I know well and who have dreadlocks and are white use bicycles Amusement.
The author has special insight into this sentence, which was found on the #lojban IRC channel, having wrote it. ry refers to lo relxilma'e.
### Proposed Definition for ge'u
ge'u (GEhU)
End relative phrase. ge'u is an elidable terminator that indicates the end of relative or sumtcita phrases. It is usually elidable in simple situations. It (or possibly another teminator, such as ku) is required, however, when one wishes to apply a logical connective to a sumti that has a relative or modal phrase attached (otherwise you are simply added an additional part to the phrase).
### Examples of ge'u Usage
li'o pensi le cmalu no'u la alis ge'u e ro le abu se manci se lifri
... thinking about the small thing which is Alice, and all of her wondrous experiences.
pilno le vlaturge'a pe la guaspis ge'u .e le lojbo gerna .ui
Use the word structure of Gua\spi and the grammar of Lojban. Happiness
Note that the original did not have the ge'u, which made it mean something like "Use the word structure of both ( Gua\spi and Lojban grammar)", which is basically nonsense.
### Proposed Definition for ku'o
ku'o (KUhO)
End relative clause. ku'o is an elidable terminator that indicates the end of NOI relative clauses. It can always be replaced by some other combination of terminators (ku, vau, and kei in particular are often relevant), but its use is preferred in complex clauses, where it can often replace several other terminators.
### Examples of ku'o Usage
la djan. cu lafti da poi grana ku'o gi'e desygau da
John lifts at least one thing which is a pole and shakes it.
so'o jinru minji noi nenri le xamsi ku'o e so'o verba noi kakpa lo canre le loldi lo mudri canpa ku'o e li'o
Several wet machines, which are in the sea, and several children, which dig in sand out of the floor with wooden shovels, and ...
pe'i do poi merko ku'o sipna
I opine that those of you who are American are sleeping.
### Proposed Definition for goi
goi (GOI)
Pro-sumti assign. goi is used to define or assign ko'a and fo'a series pro-sumti. The two sumti before and after goi are intended to refer to the same thing; in computer science terminology, goi unifies its arguments if either one is undefined. If both arguments are already defined, it is assumed that a re-definition is taking place, and the right-hand argument takes on the value of the left-hand argument (essentially matching the English ", which we'll call "). For explicit re-definition, use da'o to the right of one sumti mark the one that should be re-defined, or da'o nai to mark the one that should be left alone. goi is approximately equivalent to the Latin word "sive".
### Examples of goi Usage
la .alis. goi ko'a klama le zarci .i ko'a cu blanu
Alice, who we'll call "she", goes to the store. She is blue.
Note that the second sentence refers to the literal color blue.
ma goi ko'a
What is it that ko'a refers to?
xy. goi le xrabo bangu
X, which I am using to indicate the Arabic language
## Formal Definitions
(AKA conversion formulas)
|| noi | sumti noi ke'a broda | sumti to ri xi rau broda toi
poi + PA sumti | PA sumti poi broda | PA da poi ge me sumti gi broda
poi + sumti (no PA) | sumti poi broda | lo me sumti je broda
poi + ro da | ro da poi broda cu brode | ro da zo'u ganai da broda gi da brode
poi + su'o da | su'o da poi broda cu brode | su'o da zo'u ge da broda gi da brode
voi | voi ke'a broda | poi skicu ke'a fo lo ka ce'u broda
ne | ne sumti | noi ke'a srana sumti
pe | pe sumti | poi ke'a srana sumti
no'u | no'u sumti | noi ke'a du sumti
po'u | po'u sumti | poi ke'a du sumti
po | po sumti | poi sumti cu traji lo ka ce'u ckini ke'a
po'e | po'e sumti | poi ke'a jinzi ke se steci srana sumti
vu'o | sumti vu'o relative-clauses | lo me sumti me'u ku relative-clauses
goi, ko'a unassigned | ko'a goi sumti / sumti goi ko'a | sumti noi ca'e ko'a du ke'a
goi, both unassigned | ko'a goi ko'e | sei ca'e zo ko'a zo ko'e co'a snidu'i se'u
goi, both assigned | ko'a goi ko'e | sei ca'e zo ko'e co'a sinxa ko'a se'u
zi'e + noi | sumti noi subsentence1 zi'e noi subsentence2 | sumti noi ge subsentence1 gi subsentence2
zi'e + poi | sumti poi subsentence1 zi'e poi subsentence2 | sumti poi ge subsentence1 gi subsentence2
zi'e + PA poi + noi | PA sumti poi ke'a broda zi'e noi ke'a brode, PA lo me sumti je broda ku to ri xi rau brode toi
zi'e + noi + poi | sumti noi ke'a broda zi'e poi ke'a brode | lo me sumti to ri broda toi je brode
zi'e + PA noi + poi | PA sumti noi ke'a broda zi'e poi ke'a brode | undefined
||
• It is possible to build conversion formulas for "PA da poi", for each PA, but many of those formula will be different from each other. The two given here are representative, and are the two important ones. Others should be handled on a case-by-case basis, or just considered irreducable, as necessary.
• Making the zi'e work requires doing the other expansions first. The zi'e expansions are not generalized to more than two links, but it shouldn't be hard to figure out.
• The "ri xi rau" in "zi'e + PA poi + noi" is intended to count back to the outer "lo me".
• The "ri xi rau" in "noi" is intended to count back to the preceding sumti.
• The definitions with "me" in them rely on CLL-style "me" being selected by the BPFK (in particular, over ma'oste-style "me"). Any other choice will require revisiting of those definitions.
• The "both unassigned" and "both assigned" "goi" forms are rather wierd and to be avoided in normal speech (although they may make more sense in programmatic or mathematical contexts).
## Notes
• Many of these definitions are substantially similar. Please bear in mind that they will normally be used in isolation.
• Example of the difference between noi and poi: "ro prenu noi xamgu cu klama" is a monumentally different claim from "ro prenu poi xamgu cu klama". The former is "All people are good, and they go." The latter is "All the good people go". Formally, the former is "ro prenu ge xamgu gi klama", and the latter is "ro da poi prenu zo'u da ga nai brode gi brodi". With su'o, however, they are equivalent.
• voi is pretty amazingly useless. This is borne out by the amount of usage it has seen: I have collected above, as far as I can tell, more than 80% of the extant usages of it. ne is, if anything, worse (i.e. less used). po'e is not much used either, but that's hardly a surprise. vu'o seems to be underutilized as well, but I suspect that's more in error than anything else. The winner of the no-usage prize, however, seems to be ge'u. This, however, seems to have been an error in many cases: {mi po do ge'u .e da} means something completely different than {mi po do .e da}.
• I don't know who cares that goi is Latin 'sive', but I'm sure it's not me.
• I have no idea why only pe and ne can be used with sumtcita clauses.
## Impact
• WRT goi, the right-leaning specification in the case of both arguments being defined is wholly new.
• po in CLL was stated to mean "se steci srana", but no-one really understood what that meant. I believe that the change (to traji lo ka ce'u ckini) covers the CLL meaning better and changes no usage.
• I am aware of no other substantial impacts of this proposal.
{MODULE(module=>BPFK_PollSubordinators)}BPFK Poll: Subordinators{MODULE} | 7,994 | 32,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.927448 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/26-determine-if-input-is-odd/solutions/1076603 | 1,606,373,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00395.warc.gz | 561,143,208 | 18,812 | Cody
# Problem 26. Determine if input is odd
Solution 1076603
Submitted on 7 Dec 2016 by Anthony Muscarello
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
2 Pass
n = 2; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
3 Pass
n = 28; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
4 Pass
n = 453; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
5 Pass
n = 17; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
6 Pass
n = 16; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
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Start Hunting! | 249 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.507757 |
http://www.ques10.com/p/28595/graph-theory-and-its-applications-question-paper-1/ | 1,560,757,252,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998440.47/warc/CC-MAIN-20190617063049-20190617085049-00266.warc.gz | 297,997,132 | 8,722 | Question Paper: Graph Theory and its Applications Question Paper - May 2016 - Computer Science (Semester 4) - Visveswaraya Technological University (VTU)
0
## Graph Theory and its Applications - May 2016
### VTU Computer Science (Semester 4)
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary
1(a) Define : i) Complete Bipartite Graph ii) Regular Graph and iii) Induced Subgraph. Give one example for each. 6 marks
1(b) For a graph with n vertices and m edges, if δ is minimum and Δ is maximum of the degrees of vertices show that δ ≤ (2m/n) ≤ Δ. 6 marks
1(c) Show that the following graph is isomorphic to Kuratowski's second Graph (K3.3)
6 marks
1(d) Write a note on Konigsberg's seven Bridge problems. 6 marks
2(a) Prove that a connected planer graph G with n vertices and m edges has exactly (m - n + 2) regions in everyone one of its diagrams. 6 marks
2(b) State and explain Kuratowski's theorem. Show that the graphs K5 and K303 are non-planer by re-drawing them. 6 marks
2(c) Find the chromatic polynomial for the following graph. If 5 colors are available, in how many ways can the vertices of this graph be properly colored?
6 marks
3(a) Define Trees, Prove that a tree with n-vertices has n-1 edges. 6 marks
3(b) Define: i) Spanning Tree ii) Rooted Tree and iii) Full Binary Tree. Give on example for each. 6 marks
3(c) Explain prefix codes. Obtain an optimal prefix code for the message MISSION SUCCESSFUL. Indicate the code for the message. 6 marks
4(a) Explain the steps in Dijkstra's shortest path algorithm. 6 marks
4(b) Using Prim's algorithm, find a minimal spanning tree for the weighted graph shown below.
6 marks
4(c) Three boys B1, B2, B3 and four girls G1, G2, G3, G4 are such that i) B1 is a cousin of G1, G3, G4 , ii) B2 is a cousin of G2 and G4 , iii) B3 is a cousin of G2 and G3. If boys must marry a cousin girl, find the possible sets of such couples. 6 marks
5(a)(i) How many arrangements are there for all letters in the word SOCIOLOGICAL? 6 marks
5(a)(ii) In how many of these arrangements
• A and G are adjacent
• All the vowels are adjacent
• 6 marks
5(b) Find the co-efficient of
i) x9y3 in the expansion of (2x - 3y)12.
ii) a2b3d5 in the expansion of (a + 2b - 3c + 2d + 5)16
6 marks
5(c) In how many ways can one distribute eight identical balls into four distinct containers so that i) no container is left empty? ii) the fourth container gets an odd number of balls? 6 marks
6(a) State and prove the principle of Inclusion - Exclusion for n sets. 6 marks
6(b) In how many ways can the 26 letters of the English alphabet be permuted so that none of the patterns CAR, DOG, PIN, and BYTE occurs? 6 marks
6(c) In how many ways can be integers 1, 2, 3, ..... 10 be arranged in a line so that no even integer is in its natural place? 6 marks
6(d) An apple, a banana, a mango, and an orange are to be distributed to four boys B1, B2, B3, B4. The boys B1 and B2 do not wish to have apple, the boy B3 does not want banana or mango, and B4 refluses orange. In how many ways the distribution can be made so that no boy is displeased? 6 marks
7(a) Using the generating functions, find the number of i) non negative, and ii) positive, integer solutions of the equation x1 + x2 + x3 + x4 = 25. 6 marks
7(b) A bag contains a large number of red, green, white and black marbles, with atleast 24 of each colour. In how many ways can one select 24 og these marbles, so that there are even number of white marbles and atleast six black marbles. 6 marks
7(c) Using exponential generating function, find the number of ways in which four of the letters in the word ENGINE be arranged 6 marks
8(a) Solve the recurrence relation.
an = 2ah/2 + (n-1) for n = 2k, k ≥ 1, given a1 = 0.
6 marks
8(b) The number of virus affected files in a system is 1000 (to start with) and this increase 250% every two hours. Using a recurrence relation determine the number of virus affected files in the system after one day. 6 marks
8(c) Find and solve a recurrence relation for the number of binary sequences of length n ≥ 1 that have no consecutive O's. 6 marks | 1,184 | 4,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-26 | latest | en | 0.827942 |
https://yourgametips.com/word-games/what-is-the-daily-double-in-jeopardy/ | 1,716,132,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00213.warc.gz | 975,959,284 | 39,800 | # What is the daily double in jeopardy?
## What is the daily double in jeopardy?
DAILY DOUBLE: When a Contestant chooses a clue that is a Daily Double, before the clue is read, the Contestant must decide how much money they are going to wager (you cannot wager more than you have, and it must be in units of \$200.) If the response is correct, the Contestant is awarded the amount wagered.
## How do you do the daily double on Jeopardy?
Before the clue is revealed, the contestant who selects the Daily Double must declare a wager, from a minimum of \$5 to a maximum of his/her entire score (known as a “true Daily Double”) or the highest clue value available in the round, whichever is greater.
Are Daily Double questions harder?
Unfortunately my dataset only indicates a question is a Daily Double but not where on the board it was found so I couldn’t verify this. At 39% correct Final Jeopardy falls in between the second hardest (42% for \$1600) and hardest (35% for \$2000) questions in the Double Jeopardy round.
How many questions are in a game of jeopardy?
61 questions
### What age did Alex Trebek die?
80 years (1940–2020)
### How did Alex Trebek die?
game show host dies with cancer aged 80. Alex Trebek, the long-time host of American television quiz show Jeopardy!, has died at the age of 80. Mr Trebek announced he had been diagnosed with stage-four pancreatic cancer in March 2019.
How long will Alex Trebek live?
Trebek died at his home in Los Angeles on Novem, at the age of 80, after more than 18 months fighting pancreatic cancer. His remains were cremated, and given to his wife.
Who is taking over jeopardy for Alex Trebek?
Producers announced Monday that Jennings, who won 74 games in a row and claimed the show’s “Greatest of All Time” title in a competition last year, will host episodes that will air beginning Jan. 11. A long-term replacement for Trebek, who died of cancer on Nov. 8, will be named later.
## Is Alex Trebek dying?
Los Angeles, California, United States
## Is Alex from jeopardy dying?
What happen to Alex Trebek?
Novem: Alex Trebek passes away The news of Trebek’s death was confirmed on the game show’s Twitter page Sunday. “Jeopardy! is saddened to share that Alex Trebek passed away peacefully at home early this morning, surrounded by family and friends.
Who will replace Alex Trebek when he dies?
LeVar Burton Burton has become a social media favorite to take the hosting reins, and more than 9,900 people have signed a petition calling on Sony Pictures Entertainment to make him the host. The former “Star Trek” star and “Reading Rainbow” host mentioned the possibility of hosting “Jeopardy!” in September.
### Is Alex Trebek still taping Jeopardy?
Alex Trebek, the beloved host of the “Jeopardy!” game show, died on November 8 following a battle with pancreatic cancer. He was 80. It was taped at the end of October, just over a week before he died. The show announced that it will resume production at the end of November with a series of interim guest hosts.
### When was the last Jeopardy show taped 2020?
October 29
What is the latest news on Alex Trebek’s health?
March 2020: Alex Trebek gives a one-year update on his pancreatic cancer. A year to the day that Trebek revealed his pancreatic cancer diagnosis, he gave an update on his health. “The one-year survival rate for stage IV pancreatic cancer patients is 18 percent. | 787 | 3,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-22 | latest | en | 0.958802 |
https://www.radiantthinking.us/thought-experiments/14.html | 1,556,037,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578605555.73/warc/CC-MAIN-20190423154842-20190423180842-00234.warc.gz | 808,300,238 | 7,144 | ## 14
When we talk about judging, about getting real, we're usually talking about limits. In rocket science parlance, we talk about dealing with "constraints."
If Johnny can eat one apple in five minutes, how many apples can he eat in sixty minutes? If you do the math you get twelve apples, but Johnny's mother knows better because she's a realist. She knows he's going to start slowing down on the second apple and his little stomach—which is smaller than his eyes—will probably not accommodate the third apple.
Johnny's stomach has its limits. Johnny might be able to imagine eating twelve apples in one hour, and he might bet his lunch money that he's going to do it—but we know better. (Don't try this bet against Paul Newman, however. In Cool Hand Luke, he bet his prison inmates that he could eat fifty hard-boiled eggs, and won— which just proves that you shouldn't underestimate limits either.)
Knowing your limits is an important aspect of thinking about a problem. When you plan a long trip, you have to allow for a number of limits including the size of your gas tank, the speed limit, your budget, even your physical limitations.
In rocket science, these limits are extraordinarily important. The amount of onboard propellant determines how long a spacecraft can continue to operate in space. In the multi-ton (mobile home sized) communications satellites that make MTV possible, the value of one year's worth of propellant is over \$100 million. That's just for one hundred pounds of propellant. The satellite itself is worth \$1 billion—until it runs out of precious station-keeping propellant, when its value goes down to zero. Then another satellite has to be launched into a 22,000-mile-high orbit. Amazingly, several of these satellites are launched every year.
When Armstrong and Aldrin landed on the moon, they had to deal with some very serious limits. They had to land softly enough to avoid damaging their spacecraft, they had to avoid obstacles, and they had to do it quickly—before their propellant ran out—or they'd fall to their deaths on the lunar surface.
Unfortunately, the Apollo 11 lunar module missed its landing site by four miles (due to a one inch per second velocity error at the beginning of their descent orbit), and Neil and Buzz found themselves hovering over a boulder-strewn field. They had a fuel gauge that was as inaccurate as the one in your car. (We can go to the moon, but why can't we make an accurate gas gauge?) Mission Control back in Houston anxiously called out, "sixty seconds," their estimate of how much longer the fuel would last. Then, "Thirty seconds." Then they waited and listened. Buzz Aldrin read off altitudes and descent speeds, "forty feet, two and a half down. Picking up some dust."
By this time the fuel gauge was on empty. But you know when your car's gas gauge is on empty how it might still have a gallon left—or it might be bone dry? It depends on the car. The lunar module fuel gauge had an error of about 2 percent—and it was registering empty. They could run out of fuel at any second.
Finally Neil Armstrong called out, "Houston, Tranquility Base here. The Eagle has landed."
To which Mission Control replied, "Roger, Tranquility, we copy you on the ground. You've got a bunch of guys about to turn blue. We're breathing again. Thanks a lot."
So the moral of our story is that when you try to solve problems that have all sorts of limits, you are thinking like the rocket scientists and the astronauts who made the first landing on the moon happen. It is easy and fun to say, "Think outside of the box," and there is a time for that kind of thinking. But when you want to get real, you have to stay inside the constraint box—that's where the challenge is.
0 0 | 814 | 3,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | longest | en | 0.971938 |
https://saoirse-ronan.net/vertex-form-in-real-life-quiz-how-much-do-you-know-about-vertex-form-in-real-life-77381 | 1,568,888,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573476.67/warc/CC-MAIN-20190919101533-20190919123533-00134.warc.gz | 639,633,333 | 6,728 | # Vertex Form In Real Life Quiz: How Much Do You Know About Vertex Form In Real Life?
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https://sigmaxl.com/ANOMBinomPropOneWay.shtml | 1,550,680,298,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00547.warc.gz | 683,350,497 | 9,076 | Include Top
# How Do I Perform ANOM Binomial Proportions One-Way in Excel Using SigmaXL?
1. Open ANOM Examples.xlsx, click on the Test Scores Binomial Prop tab. This is standardized
math test score data from 10 elementary schools (Example 3.4 from the ANOM book used
with author permission). We are testing to see if there is a difference between schools at an
alpha = 0.01 level.
2. Click SigmaXL > Graphical Tools > Analysis of Means (ANOM) > ANOM Binomial Proportions
One-Way. Ensure that the entire data table is selected. If not, check Use Entire Data Table.
Ensure that the entire data table is selected. If not, check Use Entire Data Table.
3. Click Next. Select Proficient, click Numeric Data Variable (Y) >>; select Enrollment, click
Subgroup Column or Size >>; select School, click Optional Group Category (X) >>. Set Alpha
Level
= 0.01:
4. Click OK. The ANOM Binomial Proportions One-Way chart is shown below:
5. The resulting ANOM decision chart shows that three schools are performing at significantly low
levels and two schools are performing at significantly high levels.
6. SigmaXL automatically checks to see if the sample sizes are large enough for the normal
approximation to the Binomial to be valid, i.e., np and n(p-1) are > 5. Here we see: Warning:
Sample sizes are too small to use the normal approximation to the binomial distribution.
Sample #1 has minimum n(1-p) = 2.0. Note that the warning does not show all occurrences,
just the sample(s) with smallest failed np or n(1-p).
7. This does not mean that the chart results are invalid, an obvious out or in will not likely be
affected, but the results should be used with caution, and if possible, more data collected.
# Web Demos
Our CTO and Co-Founder, John Noguera, regularly hosts free Web Demos featuring SigmaXL and DiscoverSim | 455 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-09 | longest | en | 0.866184 |
https://goodroot.info/prime-numbers-between-50-and-60/ | 1,620,263,016,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00101.warc.gz | 299,826,567 | 11,062 | # Prime Numbers Between 50 And 60, Math Challenge
Prime Numbers Between 50 And 60. Only two prime numbers lie between 50 and 60. This prime numbers generator is used to generate the list of prime numbers from 1 to a number you specify. Use the sieve of eratosthenes. A prime number (or prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. They are 53 and 59. 53 59 there are no remaining multiples of 7. 53 55 59 remove remaining multiples of 5: The prime number which exist between 50 to 60 are 53,59. By euclid's theorem, there are an infinite number of prime numbers. 51 53 55 57 59 remove remaining multiples of 3: 57 is not a prime number cuz it is divisible by 3. Total 2 prime numbers exist. All of the numbers between 50 and 60 can be factors. A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 50 51 52 53 54 55 56 57 58 59 60 remove all multiples of 2:
Prime Numbers Between 50 And 60 Indeed lately is being sought by consumers around us, perhaps one of you. People now are accustomed to using the net in gadgets to view image and video data for inspiration, and according to the title of this post I will talk about about Prime Numbers Between 50 And 60.
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## Prime Numbers Between 50 And 60 , Https Encrypted Tbn0 Gstatic Com Images Q Tbn And9Gcrpoizzvrikhb2Coxrsv1L2Pp 4Hpfh9Tju8Q Usqp Cau
Even Numbers Definition List Of Even Numbers Up To 100. 50 51 52 53 54 55 56 57 58 59 60 remove all multiples of 2: They are 53 and 59. 53 59 there are no remaining multiples of 7. A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 53 55 59 remove remaining multiples of 5: Use the sieve of eratosthenes. Total 2 prime numbers exist. All of the numbers between 50 and 60 can be factors. By euclid's theorem, there are an infinite number of prime numbers. Only two prime numbers lie between 50 and 60. The prime number which exist between 50 to 60 are 53,59. This prime numbers generator is used to generate the list of prime numbers from 1 to a number you specify. 51 53 55 57 59 remove remaining multiples of 3: A prime number (or prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. 57 is not a prime number cuz it is divisible by 3.
How can you find prime numbers? Prime numbers between 1 and 100 are as follows: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Have only two prime numbers between 50 and 60. Average = total sum of numbers / total count of numbers input parameters & values: 53 55 59 remove remaining multiples of 5: All of the numbers between 50 and 60 can be factors.
## 51 is 17×3 (play cards with 3 people, and you'll know this!) so 57 isn't prime by inference, as every 2nd odd number is divisible by 3!
Hence, the answer is 2. By euclid's theorem, there are an infinite number of prime numbers. Are the prime numbers between 50 and 60. Is \(\dfrac{3}{4}\) a prime number? This page contains a list of the first 120 prime numbers. A prime number is a number which is only divisible by. Each composite number can be factored into prime factors and individually all of these are unique the smallest prime number defined by modern mathematicians is 2. 23 29 31 37 41 43 47. 53 and 59 as 51 is divisible by 17 and 57 is divisible by 19 and all then even integers are divisible by 2. Explore what are prime numbers with the help of examples. This prime numbers generator is used to generate the list of prime numbers from 1 to a number you specify. Prime numbers between 50 and 60. What is a prime number? A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: What are the prime numbers between. Prime numbers between 0 and 60. How many prime numbers are there between 60 and 70? To be prime, a number must be divisible only by 1 and the number itself which. They are 53 and 59. 1 is not a prime number. ∴ the correct option is 'b'. 53 59 there are no remaining multiples of 7. Read there are no remaining multiples of 7, so you are done. Click here to see all problems on numbers word problems. Hence, the answer is 2. Have only two prime numbers between 50 and 60. The first ten prime numbers are. Prime numbers between 1 and 100 are as follows: B.if other person where to be asked the same question and he/she. Only two prime numbers lie between 50 and 60. Two prime numbers are always coprime to each other. | 1,983 | 7,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.842816 |
https://www.bartleby.com/solution-answer/chapter-10-problem-7p-fundamentals-of-financial-management-mindtap-course-list-15th-edition/9781337395250/cost-of-common-equity-with-and-without-flotation-the-evanec-companys-next-expected-dividend-d1-is/6423be60-efad-11e8-9bb5-0ece094302b6 | 1,571,211,295,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00425.warc.gz | 795,036,146 | 43,226 | Chapter 10, Problem 7P
### Fundamentals of Financial Manageme...
15th Edition
Eugene F. Brigham + 1 other
ISBN: 9781337395250
Chapter
Section
### Fundamentals of Financial Manageme...
15th Edition
Eugene F. Brigham + 1 other
ISBN: 9781337395250
Textbook Problem
# COST OF COMMON EQUITY WITH AND WITHOUT FLOTATION The Evanec Company’s next expected dividend, D1 is $3.18; its growth rate is 6%; and its common stock now sells for$36.00. New stock (external equity) can be sold to net $32.40 per share. a. What is Evanec’s cost of retained earnings, rs? b. What is Evanec’s percentage flotation cost, F? c. c What is Evanec’s cost of new common stock, re? a. Summary Introduction To determine: Cost of retained earnings. Introduction: Cost of Retained Earnings: A company can raise its capital by issuing new shares or by retaining its current earnings. The retained earnings is assumed as free of cost generally but it is wrong because the cost of the retained earnings can be measured as per the opportunity cost principle. If the company has no retained earnings then the company issues new shares in which the shareholders earn the dividend. Thus, the loss of dividend can be considered as an opportunity cost of the retained earnings. Explanation Given information: Next expected dividend is$3.18.
Growth rate is 6% or 0.06.
Current price of stock is \$36.
The formula to calculate cost of retained earnings is,
r=D1P0+g
Where
• r is the cost of retained earnings.
• D1 is the next expected dividend.
• P0 is the current price of the stock
b.
Summary Introduction
To determine: The percentage of flotation cost.
Introduction:
Flotation Cost:
The cost which occurred when the new shares are issued by the company is called floatation cost. It increases the cost of newly issued shares.
c.
Summary Introduction
To determine: The cost of equity from the new stock.
Introduction:
Cost of Equity:
It is the cost of the company while raising finance by issuing equity. It is the earnings from an investment to the firm’s equity investors. It is the return to the stockholder holders’ equity investments. The issue of new stock incurs the flotation cost.
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Fundamentals of Financial Management, Concise Edition (with Thomson ONE - Business School Edition, 1 term (6 months) Printed Access Card) (MindTap Course List) | 663 | 2,922 | {"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} | 3.234375 | 3 | CC-MAIN-2019-43 | longest | en | 0.924295 |
http://stackoverflow.com/questions/1860783/strange-double-to-int-conversion-behavior-in-c | 1,455,282,219,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163729.14/warc/CC-MAIN-20160205193923-00169-ip-10-236-182-209.ec2.internal.warc.gz | 214,439,297 | 20,483 | # strange double to int conversion behavior in c++
The following program shows the weird double to int conversion behavior I'm seeing in c++:
``````#include <stdlib.h>
#include <stdio.h>
int main() {
double d = 33222.221;
printf("d = %9.9g\n",d);
d *= 1000;
int i = (int)d;
printf("d = %9.9g | i = %d\n",d,i);
return 0;
}
``````
When I compile and run the program, I see:
``````g++ test.cpp
./a.out
d = 33222.221
d = 33222221 | i = 33222220
``````
Why is i not equal to 33222221? The compiler version is GCC 4.3.0
-
Floating point representation is almost never precise (only in special cases). Every programmer should read this: What Every Computer Scientist Should Know About Floating-Point Arithmetic
In short - your number is probably 33222220.99999999999999999999999999999999999999999999999999999999999999998 (or something like that), which becomes 33222220 after truncation.
-
Okay thanks, that makes sense. Then what's the preferred method to convert to an int without having to worry about this? The round function? (but that still involves a double->int conversion..) – user924 Dec 7 '09 at 15:52
user924: int i = (int)(d>=0?d+0.5:d-0.5); fixes your problem – MaR Dec 7 '09 at 15:56
Adding 0.5 ends up with a "round to nearest". If you sort of want "round down", but with a bit of room to correct this kind of imprecision, then you can add a smaller value. For example `(d+0.001)`, or `(d + d*DBL_EPSILON)`. If you're going to be multiplying numbers by 1000 and you want correct results, it might be an idea to get a library for decimal arithmetic. – Steve Jessop Dec 7 '09 at 16:08
Or just use `i = (int)nearbyint(d)`, from the C99 math library. – Useless Dec 7 '09 at 17:05
Sure, if the rounding mode matches the way you want to round. – Steve Jessop Dec 7 '09 at 17:37
When you attach a debugger and inspect the values, you will see that the value of `d` is actually `33222220.999999996`, which is correctly truncated to 33222220 when converted to integer.
There is a finite amount of numbers that can be stored in a double variable, and 33222221 is not one of them.
-
Actually, 33222221 is one of them. The problem here is that 33222.221 is not. – Steve Jessop Dec 7 '09 at 16:02
Due to floating point approximation, 33222.221 may actually be 33222.220999999999999. Multiplied by 1000 yields 33222220.999999999999. Casting to integer ignores all decimals (round down) for a final result of 33222220.
-
If you change the "9.9g" in your `printf()` calls to 17.17 to recover all possible digits of precision with a 64-bit IEEE 754 FP number, you get 33222220.999999996 for the double value. The int conversion then makes sense.
-
I don't want to repeat the explanations of the other comments.
So, here is just an advice to avoid problems like the one described:
1. Avoid floating point arithemtics in the first place whereever possible (especially when computation is involved).
2. If floating point arithmetics is really necessary, you must not compare numbers by operator== by all means! Use your own comparison function instead (or use one supplied by some library), which does something like an "is almost equal" comparison using some kind of epsilon compare (either absolute or relative to the number's magniture).
See for example the excellent article
``````http://www.cygnus-software.com/papers/comparingfloats/comparingfloats.htm
`````` | 935 | 3,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-07 | latest | en | 0.847408 |
https://inst.eecs.berkeley.edu/~cs61a/sp13/slides/25.py | 1,660,239,043,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00170.warc.gz | 312,823,085 | 2,322 | # Trees as nested tuples t = ((1, 2), (3, 4), 5) def count_leaves(tree): """Return the number of leaves in a tree. >>> count_leaves(t) 5 """ if type(tree) != tuple: return 1 return sum(map(count_leaves, tree)) def map_tree(tree, fn): """Return a tree with fn mapped to the leaves of tree. >>> map_tree(t, lambda x: x*x) ((1, 4), (9, 16), 25) """ if type(tree) != tuple: return fn(tree) return tuple(map_tree(branch, fn) for branch in tree) # Tree class (binary trees with internal values) class Tree(object): """A tree with internal values.""" def __init__(self, entry, left=None, right=None): self.entry = entry self.left = left self.right = right def __repr__(self): args = repr(self.entry) if self.left or self.right: args += ', {0}, {1}'.format(repr(self.left), repr(self.right)) return 'Tree({0})'.format(args) def fib_tree(n): """Return a Tree that represents a recursive Fibonacci calculation. >>> fib_tree(3) Tree(1, Tree(0), Tree(1)) """ if n == 1: return Tree(0) if n == 2: return Tree(1) left = fib_tree(n - 2) right = fib_tree(n - 1) return Tree(left.entry + right.entry, left, right) def count_entries(tree): """Return the number of entries in a Tree. >>> count_entries(fib_tree(6)) 15 """ if tree is None: return 0 return 1 + count_entries(tree.left) + count_entries(tree.right) def big_tree(left, right): """Return a tree of elements between left and right. >>> big_tree(0, 12) Tree(6, Tree(2, Tree(0), Tree(4)), Tree(10, Tree(8), Tree(12))) """ if left > right: return None split = left + (right - left)//2 return Tree(split, big_tree(left, split-2), big_tree(split+2, right)) # Memoization def memo(f): cache = {} def memoized(n): if n not in cache: cache[n] = f(n) return cache[n] return memoized def fib(n): """Compute the nth Fibonacci number. >>> fib(35) 5702887 """ if n == 1: return 0 if n == 2: return 1 return fib(n - 2) + fib(n - 1) @memo def fib_memo(n): """Compute the nth Fibonacci number. >>> fib_memo(35) 5702887 >>> fib_memo(100) 218922995834555169026 """ if n == 1: return 0 if n == 2: return 1 return fib_memo(n - 2) + fib_memo(n - 1) # From lecture 23/24: Recursive lists class Rlist(object): """A recursive list consisting of a first element and the rest. >>> s = Rlist(1, Rlist(2, Rlist(3))) >>> len(s) 3 >>> s[0] 1 >>> s[1] 2 >>> s[2] 3 """ class EmptyList(object): def __len__(self): return 0 empty = EmptyList() def __init__(self, first, rest=empty): self.first = first self.rest = rest def __repr__(self): f = repr(self.first) if self.rest is Rlist.empty: return 'Rlist({0})'.format(f) else: return 'Rlist({0}, {1})'.format(f, repr(self.rest)) def __len__(self): return 1 + len(self.rest) def __getitem__(self, i): if i == 0: return self.first return self.rest[i - 1] def extend_rlist(s1, s2): """Return a list containing the elements of s1 followed by those of s2. >>> s = Rlist(1, Rlist(2, Rlist(3))) >>> extend_rlist(s.rest, s) Rlist(2, Rlist(3, Rlist(1, Rlist(2, Rlist(3))))) """ if s1 is Rlist.empty: return s2 return Rlist(s1.first, extend_rlist(s1.rest, s2)) def map_rlist(s, fn): """Return an Rlist resulting from mapping fn over the elements of s. >>> s = Rlist(1, Rlist(2, Rlist(3))) >>> map_rlist(s, lambda x: x * x) Rlist(1, Rlist(4, Rlist(9))) """ if s is Rlist.empty: return s return Rlist(fn(s.first), map_rlist(s.rest, fn)) def filter_rlist(s, fn): """Filter the elements of s by predicate fn. >>> s = Rlist(1, Rlist(2, Rlist(3))) >>> filter_rlist(s, lambda x: x % 2 == 1) Rlist(1, Rlist(3)) """ if s is Rlist.empty: return s rest = filter_rlist(s.rest, fn) if fn(s.first): return Rlist(s.first, rest) return rest # Take 1: Sets as unordered sequences s = Rlist(1, Rlist(2, Rlist(3))) # A set is an Rlist with no duplicates def empty(s): return s is Rlist.empty def set_contains(s, v): """Return true if set s contains value v as an element. >>> set_contains(s, 2) True >>> set_contains(s, 5) False """ if empty(s): return False if s.first == v: return True return set_contains(s.rest, v) def adjoin_set(s, v): """Return a set containing all elements of s and element v. >>> t = adjoin_set(s, 4) >>> t Rlist(4, Rlist(1, Rlist(2, Rlist(3)))) """ if set_contains(s, v): return s return Rlist(v, s) def intersect_set(set1, set2): """Return a set containing all elements common to set1 and set2. >>> t = adjoin_set(s, 4) >>> intersect_set(t, map_rlist(s, lambda x: x*x)) Rlist(4, Rlist(1)) """ return filter_rlist(set1, lambda v: set_contains(set2, v)) def union_set(set1, set2): """Return a set containing all elements either in set1 or set2. >>> t = adjoin_set(s, 4) >>> union_set(t, s) Rlist(4, Rlist(1, Rlist(2, Rlist(3)))) """ set1_not_set2 = filter_rlist(set1, lambda v: not set_contains(set2, v)) return extend_rlist(set1_not_set2, set2) | 1,489 | 4,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.594702 |
https://bigideasmathanswers.com/eureka-math-grade-4-module-1-lesson-11/ | 1,721,312,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00592.warc.gz | 114,864,635 | 43,707 | ## Engage NY Eureka Math 4th Grade Module 1 Lesson 11 Answer Key
### Eureka Math Grade 4 Module 1 Lesson 11 Problem Set Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,311 and 268
first adding the ones place = 1 + 8 = 9.
ten place : 1 + 6 = 7
hundreds place = 3 + 2 = 5
thousands place = 6
Sum = 6,579
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,311 and 1,268
first adding the ones place = 1 + 8 = 9.
ten place : 1 + 6 = 7
hundreds place = 3 + 2 = 5
thousands place = 6 + 1 = 7
Sum = 7,579
c.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,314 and 1,268
first adding the ones place = 4 + 8 = 12. 12 is represented as 10 ones + 2 ones that’s 10 ones make 1 ten and 2 ones. So 1 is carried to the tens place and added to the tens place values.
ten place : 1 + 6 + 1 = 8
hundreds place = 3+ 2 = 5
thousands place = 6 + 1 = 7
Sum = 7,582.
d.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,314 and 2,493
first adding the ones place = 4 + 3 = 7.
ten place : 1 + 9 = 10. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 0 is represented in tens place
hundreds place = 3+ 4 + 1 = 8
thousands place = 6 + 2 = 8
Sum = 8,807
e.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 8,314 and 2,493
first adding the ones place = 4 + 3 = 7.
ten place : 1 + 9 = 10. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 0 is represented in tens place
hundreds place = 3 + 4 + 1 = 8
thousands place = 8 + 2 = 10. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands = 1
Sum = 10,807.
f.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 12,378 and 5,463
first adding the ones place = 8 + 3 = 11. 10 ones + 1 ones = 11. 10 ones make 1 ten . So 1 is carried to the tens place and added to the tens place values and 1 is represented in ones place
ten place : 7 + 6 + 1(carry on) = 14. 10 ten and 4 tens = 14 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 3 + 4 + 1(carry on) = 8
thousands place = 2 + 5 = 7
ten thousands = 1
Sum = 17,841
g.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 52,098 and 6,048
first adding the ones place = 8 + 8 = 16. 10 ones + 6 ones = 16. 10 ones make 1 ten . So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place
ten place : 9 + 4 + 1(carry on) = 14. 10 ten and 4 tens = 14 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 0 + 0 + 1(carry on) = 1
thousands place = 2 + 6 = 8
ten thousands = 5
Sum = 58,146
h.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 34,698 and 71,840
first adding the ones place = 8 + 0 = 8.
ten place : 9 + 4 = 13. 10 tens + 3 tens = 13 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 3 is represented in tens place
hundreds place = 6 +8 + 1(carried number) = 15 = 10 hundred + 5 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 5 is represented in hundreds place
thousands place = 4 + 1 + 1(carried number) = 6.
ten thousands = 3 + 7 = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 1
Sum = 106,538.
i.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 34,698 and 71,840
first adding the ones place = 1 + 5 = 6.
ten place : 1 + 4 = 5.
hundreds place = 8 + 4 = 12 = 10 hundred + 2 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 2 is represented in hundreds place
thousands place = 4 + 6 + 1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands = 4 + 5 + 1(carried number) = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 5 + 3 + 1(carried number) = 9
Sum = 901,256.
j. 527 + 275 + 752
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 527 , 275 and 752
first adding the ones place = 7 + 5 + 2 = 14 = 10 ones + 4 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 4 is represented in ones place
ten place : 2 + 7 + 5 + 1(carried number) = 15 = 10 tens + 5 tens. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 5 is represented in tens place
hundreds place = 5 +2 + 7 + 1(carried number)= 15 = 10 hundred + 5 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 5 is represented in hundreds place
thousands place = 4 + 6 + 1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands = 4 + 5 + 1(carried number) = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 5 + 3 + 1(carried number) = 9
Sum = 1,554
k. 38,193 + 6,376 + 241,457
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 38,193 + 6,376 + 241,457
first adding the ones place = 3 + 6 + 7= 16 = 10 ones + 6 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place
ten place : 9 + 7 + 5 + 1(carried number) = 22 = 20 tens + 2 tens. 20 tens make 2 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 2 is represented in tens place
hundreds place = 1 + 3 + 4 + 1(carried number)= 10. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 8 + 6 + 1 +1(carried number) = 16 =10 thousand + 6 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 6 is represented in thousands place
ten thousands =3 + 4 + 1(carried number) = 8 .
Hundred thousands = 2
Sum = 286,026.
Draw a tape diagram to represent each problem. Use numbers to solve, and write your answer as a statement.
Question 2.
In September, Liberty Elementary School collected 32,537 cans for a fundraiser. In October, they collected 207,492 cans. How many cans were collected during September and October?
Number of cans collected by the Liberty Elementary School for a fundraiser in September are = 32,537
Number of cans collected by the Liberty Elementary School for a fundraiser in October are = 207,492
Total number of cans collected during September and October are = 32,537 + 207,492 =
Question 3.
A baseball stadium sold some burgers. 2,806 were cheeseburgers. 1,679 burgers didn’t have cheese. How many burgers did they sell in all?
Number of cheeseburgers sold at baseball stadium = 2,806
Number of burgers without cheese sold at baseball stadium are = 1,679
Total number of burgers sold in all are = 2,806 + 1,679
Question 4.
On Saturday night, 23,748 people attended the concert. On Sunday, 7,570 more people attended the concert than on Saturday. How many people attended the concert on Sunday?
Number of people attended the concert on Saturday night = 23,748
Number of people attended the concert on Sunday are = 7,570 more than on Saturday
Total people attended the concert on Sunday = 23,748 + 7, 570
### Eureka Math Grade 4 Module 1 Lesson 11 Exit Ticket Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 23,607 and 2,307
first adding the ones place = 7 + 7 = 14. 14 is represented as 10 + 4 that’s 1 ten and 4 ones. So 1 is carried to the tens place and added to the tens place values.
ten place : 0 + 0 + 1 = 1
hundreds place = 6 + 3 = 9
thousands place = 3 + 2 = 5
ten thousands place = 2
Sum = 25,914.
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 3,948 and 278
ones place = 8 + 8 = 16. 16 is represented as 10 + 6 that’s 1 ten and 6 ones. So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place.
ten place : 4 + 7 + 1(carried number) = 12 = 10 + 2 that’s 1 hundred and 2 tens. So 1 is carried to the hundreds place and added to the hundreds place values and 2 is represented in tens place.
hundreds place = 9 + 2 + 1(carried number) = 12 = 10 + 2 that’s 1 thousand and 2 hundred. So 1 is carried to the thousands place and added to the thousands place values and 2 is represented in hundreds place.
thousands place = 3 + 1(carried number) = 4
Sum = 4,226.
c. 5,983 + 2,097
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 5,983 and 2,097
ones place = 3 + 7 = 10. 10 is represented as 10 + 0 that’s 1 ten and 0 ones. So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place.
ten place : 8 + 9 + 1(carried number) = 18 = 10 + 8 that’s 1 hundred and 8 tens. So 1 is carried to the hundreds place and added to the hundreds place values and 8 is represented in tens place.
hundreds place = 9 + 0 + 1(carried number) = 10 = 10 + 0 that’s 1 thousand and 0 hundred. So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place.
thousands place = 5 + 2 + 1(carried number) = 8
Sum = 8,080
Question 2.
The office supply closet had 25,473 large paper clips, 13,648 medium paper clips, and 15,306 small paper clips. How many paper clips were in the closet?
Number of large paper clips = 25,473
Number of medium paper clips = 13,648
Number of small paper clips = 15,306
Total number of paper clips = 25,473 + 13,648 + 15,306
### Eureka Math Grade 4 Module 1 Lesson 11 Homework Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 7,909 and 1,044
first adding the ones place = 9 + 4 = 13 = 10 ones + 3 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 3 is represented in ones place
ten place : 0 + 4 + 1(carried number) = 5
hundreds place = 9 + 0 = 9.
thousands place = 7 + 1 = 8.
Sum = 8,953.
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 27,909 and 9,740
first adding the ones place = 9 + 0 = 9
ten place : 0 + 4 = 4.
hundreds place = 9 + 7 = 16. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 6 is represented in hundreds place
thousands place = 7 + 9 +1(carried number) = 17 =10 thousand + 7 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 7 is represented in thousands place.
ten thousands =2 + 1(carried number) = 3 .
Sum = 37,649.
c.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 827,909 and 42,989
first adding the ones place = 9 + 9= 18 = 10 ones + 8 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 8 is represented in ones place
ten place : 0 + 8 + 1(carried number) = 8.
hundreds place = 9 + 9 = 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 7 + 2 +1(carried number) = 10 =10 thousand + 0 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands =2+ 4 + 1(carried number) = 7 .
Hundred thousands = 8
Sum = 870,898.
d.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,205 and 11,845
first adding the ones place = 5 + 5 = 10 = 10 ones + 0 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place
ten place : 0 + 4 + 1(carried number) = 5.
hundreds place = 2 + 8 = 10. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 9 + 1 +1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands =8 + 1 + 1(carried number) =10. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 1 is represented in ten thousands place
Hundred thousands = 2 + 1(carried over number) = 3
Sum = 301,050.
e.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 547,982 and 114,849
first adding the ones place = 2 + 9 = 11 = 10 ones + 1 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 1 is represented in ones place
ten place : 8 + 4 + 1(carried number) = 13. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 3 is represented in tens place
hundreds place = 9 + 8 + 1(carried number)= 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 7 + 4 +1(carried number) = 12 =10 thousand + 2 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 2 is represented in thousands place
ten thousands =4 + 1 + 1(carried number) = 6.
Hundred thousands = 5 + 1 = 6.
Sum = 662,831.
f.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 258,983 and 121,897.
first adding the ones place = 3 + 7 = 10 = 10 ones + 0 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place
ten place : 8 + 9+ 1(carried number) = 18. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 8 is represented in tens place
hundreds place = 9 + 8 + 1(carried number)= 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 8 + 1 +1(carried number) = 10 =10 thousand + 0 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands =5 + 2 + 1(carried number) = 8.
Hundred thousands = 2 + 1 = 3.
Sum = 380,880.
g.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 83,906 and 35,808
first adding the ones place = 6 + 8 = 14 = 10 ones + 4 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 4 is represented in ones place
ten place : 0 + 0 + 1(carried number) = 1.
hundreds place = 9 + 8 = 17. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 7 is represented in hundreds place
thousands place = 3 + 5 +1(carried number) = 9.
ten thousands =8 + 3 = 11. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 1 is represented in ten thousands place
Hundred thousands = 1(carried number) = 1.
Sum = 119,714.
h.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,999 and 91,849
first adding the ones place = 9 + 9 = 18 = 10 ones + 8 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 8 is represented in ones place
ten place : 9 + 4 + 1(carried number) = 14. 10 tens make 1 hundreds . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 9 + 8 +1(carried number)= 18 . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 9 + 1 +1(carried number) = 11. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place.
ten thousands =8 + 9 +1(carried number) = 18. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 8 is represented in ten thousands place
Hundred thousands = 2 +1(carried number) = 3.
Sum = 381,848
i.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,999 and 91,849
first adding the ones place = 0 + 0 = 0.
ten place : 0 + 0 = 0.
hundreds place = 9+1= 10 . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 4 + 5 +1(carried number) = 10. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place.
ten thousands =5 + 4 +1(carried number) = 10. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 7 + 2 +1(carried number) = 1. 10 hundred thousands make 1 million . So 1 is carried to the millions place and added to the millions place values and 0 is represented in hundred thousands place.
Millions = 1 (carried over number) = 1.
Sum = 1,000,000.
Draw a tape diagram to represent each problem. Use numbers to solve, and write your answer as a statement.
Question 2.
At the zoo, Brooke learned that one of the rhinos weighs 4,897 pounds, one of the giraffes weighs 2,667 pounds, one of the African elephants weighs 12,456 pounds, and one of the Komodo dragons weighs 123 pounds.
Given
One of the rhinos weights at the zoo = 4,897 pounds.
One of the giraffes weights at the zoo = 2,667 pounds
One of the African elephants weights at the zoo = 12, 456 pounds.
One of the Komodo dragons at the zoo weighs = 123 pounds.
a. What is the combined weight of the zoo’s African elephant and the giraffe? | 6,163 | 21,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.776964 |
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# If (|p|!)^p = |p|!, which of the following could be
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If (|p|!)^p = |p|!, which of the following could be [#permalink]
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21 Feb 2012, 21:40
2
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If $$(|p|!)^p = |p|!$$, which of the following could be the value(s) of $$p$$ ?
A. -1
B. 0
C. 1
D. -1 and 1
E. -1, 0, and 1
How is OA E . factorials are not even defined for - numbers. Pls clarify.
Q 15 Diagnostic test
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Posts: 431
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
### Show Tags
04 Dec 2012, 23:47
3
3
[quote="GMATPASSION"]If $$(|p|!)^p = |p|!$$, which of the following could be the value(s) of $$p$$ ?
Test Values
A. -1 ==> $$|-1|^-1$$=$$|-||!$$ ==> $$\frac{1}{1}=1$$ ==> -1 is valid
B. 0 ==> $$|0|!^0=|0|!$$ ==> $$1^0=1$$ ==> $$1=1$$ ==> 0 is valid
C. 1 ==> $$1^1=1$$ ==> 1 is valid
D. -1 and 1
E. -1, 0, and 1
I just learned that 0! = 1. Thanks!
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Posts: 50580
Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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21 Feb 2012, 21:49
GMATPASSION wrote:
If $$(|p|!)^p = |p|!$$, which of the following could be the value(s) of $$p$$ ?
A. -1
B. 0
C. 1
D. -1 and 1
E. -1, 0, and 1
How is OA E . factorials are not even defined for - numbers. Pls clarify.
Q 15 Diagnostic test
You are right factorial of negative number is undefined. But it's not the case here since we have |p|!, so it's factorial of absolute value of p.
Hope it's clear.
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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22 Feb 2012, 00:04
Sorry missed the "mod". Silly mistake
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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22 Feb 2012, 03:16
1
I would go with D rather than E as 0^0 is not defined for GMAT
D. -1 and 1
(|p|!)^p = |p|! = |-1|!)^-1 = 1^-1 = 1
(|p|!)^p = |p|! = |1|!)^1 = 1^1 = 1
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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22 Feb 2012, 10:15
1
Isn't 0! equal to 1? I don't see where you see the 0^0.
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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22 Feb 2012, 10:24
1
You are right: 0^0 is not possible for LHS.
My copy of this question has (|p|!)^p=|p^p|!, so either it's some other question or it has already been revised. In its current form p=0 is a valid solution.
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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25 Oct 2014, 17:02
2
I haven't touched math for quite sometime since I chose to get a PhD in molecular genetics, and even though I'm having a blast on this site "learning" how to solve problems, some things just make no sense.
How can 0!=1?
who made up that rule? were they doing meth?
so if I have zero eggs then I can arrange them 0!=1 ways in my fridge
which is awesome for two main reasons.
a) I can put my zero eggs in the fridge one way without caring if I have or not space for eggs.
b) If I don't buy eggs then I'm burning one egg-arranging calorie quantum constantly so I'll be skinny in no time. Just don't buy eggs.
Man I love math, too bad I'll fail on the GMAT test.
Math Expert
Joined: 02 Sep 2009
Posts: 50580
If (|p|!)^p = |p|!, which of the following could be [#permalink]
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26 Oct 2014, 04:47
2
2
quijotes wrote:
I haven't touched math for quite sometime since I chose to get a PhD in molecular genetics, and even though I'm having a blast on this site "learning" how to solve problems, some things just make no sense.
How can 0!=1?
who made up that rule? were they doing meth?
so if I have zero eggs then I can arrange them 0!=1 ways in my fridge
which is awesome for two main reasons.
a) I can put my zero eggs in the fridge one way without caring if I have or not space for eggs.
b) If I don't buy eggs then I'm burning one egg-arranging calorie quantum constantly so I'll be skinny in no time. Just don't buy eggs.
Man I love math, too bad I'll fail on the GMAT test.
Why does 0! = 1?
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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26 Oct 2014, 05:46
The GMAT has not tested the fact 0!=1, and I highly doubt that they would test if students know 0!=1.
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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21 Mar 2016, 20:17
If (|p|!)p=|p|!(|p|!)p=|p|!, which of the following could be the value(s) of pp ?
A. -1
B. 0
C. 1
D. -1 and 1
E. -1, 0, and 1
How is OA E . factorials are not even defined for - numbers. Pls clarify.
Q 15 Diagnostic test
OA is E.
0! = 1(as studied in school books in India)
one can solve for all the values -1 , 0 , 1.
Thanks!
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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23 Mar 2016, 07:58
Here We can test to see that all the values are possible.
Also we knoew he negative factorials are not defined but due to mod all the values are positive.
hence E
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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25 Jul 2016, 02:58
1
GMATPASSION wrote:
If $$(|p|!)^p = |p|!$$, which of the following could be the value(s) of $$p$$ ?
A. -1
B. 0
C. 1
D. -1 and 1
E. -1, 0, and 1
How is OA E . factorials are not even defined for - numbers. Pls clarify.
Q 15 Diagnostic test
The only factorial that are perfect squares are 0! and 1!
-1, 0 , 1 all three are possible
since |-1|= 1
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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06 Aug 2017, 02:05
how is -1 valid? I still do not get it
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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22 Aug 2017, 06:56
1
pclawong wrote:
how is -1 valid? I still do not get it
The absolute value sign.
(|-1|!)^ (-1) = 1/(1!) = 1.
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Re: If (|p|!)^p = |p|!, which of the following could be [#permalink]
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13 Nov 2017, 12:05
0! = 1 because an empty set can only be ordered in 1 way.
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Re: If (|p|!)^p = |p|!, which of the following could be &nbs [#permalink] 13 Nov 2017, 12:05
Display posts from previous: Sort by | 3,078 | 9,216 | {"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} | 3.921875 | 4 | CC-MAIN-2018-47 | latest | en | 0.837464 |
http://questaal.org/tutorial/asa/energybands/ | 1,524,586,884,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946807.67/warc/CC-MAIN-20180424154911-20180424174911-00204.warc.gz | 265,917,524 | 8,386 | # Generating Energy Bands in the ASA
This tutorial shows how to generate energy bands for PbTe using the ASA code lm, but the --band switch applies equally to all codes that make crystal energy bands. See, for example a corresponding tutorial that generates bands with lmf.
### Introduction
You can make energy bands along symmetry lines you specify, using any code with band-generating capability (lmf, lm, tbe, and lumpy), using the --band switch, in symmetry line mode. Some information to help you construct symmetry lines for any crystal lattice can be found here.
You must have a potential which can make a single-particle hamiltonian. For the ASA, this means that the potential parameters are available. Usually they are generated as a byproduct of the self-consistency cycle), but it can be something else, e.g. potential parameters modified by the Levenberg-Marquardt fitting algorithm.
This tutorial begins assuming you have completed Introductory tutorial for the ASA, and have self-consistent ASA calculation for PbTe in your working directory.
### Tutorial
Energy bands are typically plotted along specified symmetry lines, which information you supply through a symmetry-line file.
Cut and paste the box below into file syml.pbte
41 .5 .5 .5 0 0 0 L to Gamma (Lambda)
41 0 0 0 1 0 0 Gamma to X (Delta)
21 1 0 0 1 .5 0 X to W (Z)
41 1 .5 0 0 0 0 W to Gamma (no name?)
syml.pbte has rows such as:
41 .5 .5 .5 0 0 0 L to Gamma (Lambda)
...
0 0 0 0 0 0 0
Each line of text specifies a line segment in k-space. In the first line of this file the segment runs from $k=(1/2,1/2,1/2)$ to $0$ in units of $2\pi/a\ (L\ to\ \Gamma)$. ‘41’ specifies the number of k points are generated in this segment. At present the remainder of the line is not used. The last line tells the line reader not to read any more lines.
These are typical lines for the fcc lattice. The structure of this file is described in detail on this page.
For drawing bands, use the --band switch in the symmetry line mode. You don’t have to do anything special because it is the default mode; lm will generate bands along symmetry lines specified in syml.pbte
To generate the bands, simply run
lm pbte --quit=rho
lm pbte --band:fn=syml
The –band switch tells lm to generate energy bands. They are written to file bands.pbte along the specified k-points (rather than the standard cycle of integration of states over the Brillouin zone. The structure of the bnds file is documented on this page.
Once you have this file, it can be plotted with your preferred plotting package. Here, we will use Questaal’s plbnds and fplot utilities:
echo -12,8,5,10 | plbnds -fplot -ef=0 -scl=13.6 -dat=green -lt=1,bold=3 -lbl=L,G,X,W,G - bnds.pbte
will generate a fplot command file, plot.plnds, which can be run with
fplot -f plot.plbnds
This will give you your desired postscript file fplot.ps.
Note You can combine the plbnds and fplot instructions in one command that generates fplot.ps directly. The command below also shrinks the figure by a factor 0.7.
echo -12,8,5,10 | plbnds -fplot~scl=.7~sh -ef=0 -scl=13.6 -dat=green -lt=1,bold=3 -lbl=L,G,X,W,G - bnds.pbte
ASA bands of PbTe without Spin Orbit coupling
#### Colouring Energy Bands
It is also possible to colour the energy bands. The idea behind this is to assign one or more weights to each band, which can be used to assign a color by the graphics plotting package. Weights are taken from a Mulliken decomposition of an eigenstate into orbital contributions. The sum of all orbital contributions adds to one; by selecting out a subset of orbitals you get a fractional weight which is used as the colour.
To obtain bands with one, two, or three color weights, run the –band command with the list of orbitals to include as follows:
--band~col=orbital-list~...
--band~col=orbital-list1~col2=orbital-list2~...
--band~col=orbital-list1~col2=orbital-list2~col3=orbital-list3~...
Each orbital is associated with a site and some lm character. In the ASA, there is one orbital belonging to a site and a particular angular momentum (lm quantum numbers). It is a tedious job to extract their position in the hamiltonian, but lm does this for you. Do
lm pbte --quit=ham --pr55
You should see the following:
Orbital positions in hamiltonian, resolved by l:
Site Spec Total By l ...
1 Pb 1:9 1:1(s) 2:4(p) 5:9(d)
2 Te 10:18 10:10(s) 11:13(p) 14:18(d)
3 E 19:27 19:19(s) 20:22(p) 23:27(d)
4 E 28:36 28:28(s) 29:31(p) 32:36(d)
Suppose we want to assign two color weights, the first color associated with Pb character, the second with Te character. Do
lm pbte --band~col=1:9~col2=10:18~fn=syml
Note that orbital-list1 and orbital-list2 are Questaal-style integer lists.
Inspect bnds.pbte. It has the same bands as before, but now each k-point has two additional entries for the Mulliken weights. To make a postscript file with color weights, do
echo -12,8,5,10 | plbnds -fplot~scl=.7~sh -ef=0 -scl=13.6 -dat=green -lt=1,bold=3,col=0,0,0,colw=1,0,0,colw2=0,1,0 -lbl=L,G,X,W,G - bnds.pbte
Coloured ASA bands of PbTe without Spin Orbit coupling | 1,497 | 5,293 | {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-17 | latest | en | 0.848305 |
https://www.fishpond.co.nz/Books/Simplified-LRFD-Bridge-Design-Jai-B-Kim-Edited-by-Robert-H-Kim-Edited-by/9781466566514 | 1,571,277,007,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00019.warc.gz | 896,126,139 | 14,157 | SmartSellTM - The New Way to Sell Online
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Simplified LRFD Bridge Design
By
Rating
Product Description
Product Details
LRFD Method of Bridge Design
Limit States
Strength Limit States for Superstructure Design
Resistance Factors, , for Strength Limits
Number of Design Lanes, NL
Multiple Presence Factor of Live Load, m
Load Combinations for the Strength I Limit State
Simple Beam Live Load Moments and Shears Carrying Moving Concentrated Loads per Lane
Live Load Moments and Shears for Beams (Girders)
Design Examples
Design Example 1: Reinforced Concrete T-Beam Bridge
Design Example 2: Load Rating of Reinforced Concrete T-Beam by the Load and Resistance Factor Rating (LRFR) Method
Design Example 3: Composite Steel-Concrete Bridge
Design Example 4: Longitudinal Steel Girder
Design Example 5: Reinforced Concrete Slabs
Design Example 6: Prestressed Interior Concrete Girder
Design Example 7: Flexural and Transverse Reinforcement for 50 ft Reinforced Concrete Girder
Design Example 8: Determination of Load Effects Due to Wind Loads, Braking Force, Temperature Changes, and Earthquake Loads Acting on an Abutment
Practice Problems
Practice Problem 1: Noncomposite 60 ft Steel Beam Bridge for Limit States Strength I, Fatigue II, and Service
Practice Problem 2: 161 ft Steel I-Beam Bridge with Concrete Slab
Practice Problem 3: Interior Prestressed Concrete I-Beam References
Primary References
Supplementary References
Appendix A: Distribution of Live Loads per Lane for Moment in Interior Beams (AASHTO Table 4.6.2.2.2b-1)
Appendix B: Distribution of Live Loads per Lane for Moment in Exterior Longitudinal Beams
(AASHTO Table 4.6.2.2.2d-1)
Appendix C: Distribution of Live Load per Lane for Shear in Interior Beams (AASHTO Table 4.6.2.2.3a-1)
Appendix D: Distribution of Live Load per Lane for Shear in Exterior Beams (AASHTO Table 4.6.2.2.3b-1)
Appendix E: U.S. Customary Units and Their SI Equivalents
Index
Jai B. Kim, PE, PhD, is a professor emeritus of civil and environmental engineering at Bucknell University. He was department chairman for 26 years. Also, currently he is a bridge consultant (since 1980) and president of BKLB Structural Consultants, Inc. Recently he was a structural engineer at the Federal Highway Administration (FHWA). He has been active in bridge research for over 40 years, and was a member of the Transportation Research Board Committee of Bridges and Structures. He also served on the Structural PE Exam Committee of the National Council of Examiners and Surveying (NCEES) for many years. He holds a BSCE and MSCE from Oregon State University and a PhD from University of Maryland. Robert H. Kim, PE, MSCE, is chief design engineer for BKLB Structural Consultants, Inc. He has extensive experience in the design, research, and construction of highway bridges. He has authored and presented several papers related to bridge engineering. Robert's three books, Bridge Design for the Civil and Structural Professional Exams, Second Edition; Timber Design, Seventh Edition; and Civil Discipline Specific Review for the FE/EIT Exam are well-read by both students and engineers. In 2013, he is working on a bridge rehabilitation design in Connecticut. He holds a BS from Carnegie Mellon University and a MSCE from Penn State University. Jonathan R. Eberle, BSCE is engaged in research with focus on the seismic design and analysis of structures at Virginia Polytechnic Institute and State University as a graduate student. He holds a BSCE from Bucknell University. Contributors: Dave M. Mante, MSCE performed a rigorous full-scale laboratory testing on an innovative concrete bridge deck system and he is performing research on prestressed concrete girders at Auburn University. Eric J. Weaver, PE, M.Eng, performed research on fatigue and life-cycle analysis of steel truss bridges at Lehigh University. He worked for NASA's space shuttle program and currently he is working as a structural engineer for Westinghouse Electric Co.
Reviews
"The book appears to be a simplified and organized approach to the basics of LRFD Bridge Superstructure Design and should serve as a good review and reference manual for the PE Examination."
--Stevan C. Wilver, Larson Design | 970 | 4,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-43 | latest | en | 0.725763 |
https://www.kylesconverter.com/density/centigrams-per-liter-to-tonnes-per-cubic-meter | 1,726,734,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00697.warc.gz | 787,189,212 | 5,581 | # Convert Centigrams Per Liter to Tonnes Per Cubic Meter
### Kyle's Converter > Density > Centigrams Per Liter > Centigrams Per Liter to Tonnes Per Cubic Meter
Centigrams Per Liter (cg/l) Tonnes Per Cubic Meter (t/m3) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Tonnes Per Cubic Meter to Centigrams Per Liter
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Centigram per Liter:
Mass of 1 centigram per volume of a liter. Equivalent density of 0.01 kilograms per cubic meter. 1 cg/l ≈ 0.01 kg/m3.
1 Tonne per Cubic Meter:
Mass of 1 tonne per volume of a cubic meter. Metric tonne of exactly 1 000 kilograms. Equivalent density of 1 000 kilograms per cubic meter. 1 t/m3 ≈ 1000 kg/m3.
Conversions Table
1 Centigrams Per Liter to Tonnes Per Cubic Meter = 070 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0007
2 Centigrams Per Liter to Tonnes Per Cubic Meter = 080 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0008
3 Centigrams Per Liter to Tonnes Per Cubic Meter = 090 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0009
4 Centigrams Per Liter to Tonnes Per Cubic Meter = 0100 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.001
5 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001200 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.002
6 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001300 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.003
7 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001400 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.004
8 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001500 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.005
9 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001600 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.006
10 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0001800 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.008
20 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0002900 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.009
30 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.00031,000 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.01
40 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.000410,000 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.1
50 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.0005100,000 Centigrams Per Liter to Tonnes Per Cubic Meter = 1
60 Centigrams Per Liter to Tonnes Per Cubic Meter = 0.00061,000,000 Centigrams Per Liter to Tonnes Per Cubic Meter = 10 | 762 | 2,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.420047 |
http://setiweb.org/percent-error/percent-error-with-zero-in-denominator.php | 1,516,305,780,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00191.warc.gz | 304,017,913 | 4,076 | Home > Percent Error > Percent Error With Zero In Denominator
# Percent Error With Zero In Denominator
The freshman laboratory is not the same as a research lab, but we hope that the student will become aware of some of the concerns, methods, instruments, and goals of physics You can only upload photos smaller than 5 MB. We know nothing about the reliability of a result unless we can estimate the probable sizes of the errors and uncertainties in the data which were used to obtain that result. i do understand that according to the formula one of the errors would result in a divide by 0 but by adding this Inf to all the other errors, using MAPE navigate here
The relative error in the numerator is 1.0/36 = 0.028. He claimed (again incorrectly) that it had an upper bound of 100. Prentice-Hall, 1988. The absolute value in this calculation is summed for every forecasted point in time and divided by the number of fitted pointsn. https://en.wikipedia.org/wiki/Approximation_error
B: DETERMINATE AND INDETERMINATE ERRORS Experimental errors are of two types: (1) indeterminate and (2) determinate (or systematic) errors. 1. Manufacturer's performance guarantees for laboratory instruments are often expressed this way. 2. If the formalism is applied blindly, as it often is, sophisticated precision may be claimed when it does not exist at all. share|cite|improve this answer answered Feb 18 '14 at 7:34 Claude Leibovici 75.8k94193 In my case, this shifts the problem to where Y_cal + Y_exp is near zero. (However, in
But this experimenter is still obligated to provide a reasonable estimate of the experimental error (uncertainty). are 1.4 to 1. This is standard notation: exp(x) means the same as ex. The coefficients will turn out to be positive also, so terms cannot offset each other.
of the "true" value of Q, Qtrue, is 58%, and the odds against it lying outside of one A.D.M. Please upload a file larger than 100x100 pixels We are experiencing some problems, please try again. Trending 10+10*0+10=? 22 answers What Time would be 74 Minutes before 11:20? 20 answers What is 3/4 of 48? 18 answers More questions If there are 6 apples and you take The equations in this document used the SYMBOL.TTF font.
One illustrative practical use of determinate errors is the case of correcting a result when you discover, after completing lengthy measurements and calculations, that there was a determinate error in one ISBN0-8018-5413-X. ^ Helfrick, Albert D. (2005) Modern Electronic Instrumentation and Measurement Techniques. It is always the same problem with that. APPENDIX II.
H. In the mathematical field of numerical analysis, the numerical stability of an algorithm in numerical analysis indicates how the error is propagated by the algorithm. Then, don't forget, that you are also obligated to provide an experimental error estimate, and support it. p.53.
If the error a is small relative to A, and b is small relative to B, then (ab) is certainly small relative to AB, as well as small compared to (aB) check over here EXAMPLES Example 1: A student finds the constant acceleration of a slowly moving object with a stopwatch. Thanks! By analysis of the scatter of the measurements, the uncertainty is determined to be m = 0.07 gm.
They are important when your actual(exact) value is very large. Retrieved from "https://en.wikipedia.org/w/index.php?title=Approximation_error&oldid=736758752" Categories: Numerical analysis Navigation menu Personal tools Not logged inTalkContributionsCreate accountLog in Namespaces Article Talk Variants Views Read Edit View history More Search Navigation Main pageContentsFeatured contentCurrent eventsRandom We were probably thinking that a forecast that is too large is a positive error. his comment is here In some cases you may know, from past experience, that the measurement is scale limited, that is, that its uncertainty is smaller than the smallest increment you can read on the
But, if I simply divide, either by the true signal, the approximation, or various combinations of the two, the relative error shoots to infinity near the zero-crossings. This is 0.25%. This document is © 1996, 2014 by Dr.
## etc.
The relative error in the denominator is 1.0/106 = 0.0094. Difference rule for determinate errors. t = 4.2 ± 0.2 second. This is a useful relation for converting (or comparing) A.D.M.
In it, you'll get: The week's top questions and answers Important community announcements Questions that need answers see an example newsletter By subscribing, you agree to the privacy policy and terms And I was wondering how to make it in percentage. ! Let N represent the numerator, N=G+H. http://setiweb.org/percent-error/percent-error-is.php One usually doesn't know.
PROPAGATION OF INDETERMINATE ERRORS Indeterminate errors have unknown sign. You can only upload files of type PNG, JPG, or JPEG. The relative error in the denominator is z/Z. | 1,128 | 4,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.936093 |
https://research.stlouisfed.org/fred2/series/PLOINVFIA624NUPN | 1,462,018,617,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111838.20/warc/CC-MAIN-20160428161511-00211-ip-10-239-7-51.ec2.internal.warc.gz | 989,103,128 | 19,343 | # Price Level of Investment for Finland
2010: 94.14957 PPP of Investment over Exchange Rate
Annual, Not Seasonally Adjusted, PLOINVFIA624NUPN, Updated: 2012-09-17 12:06 PM CDT
1yr | 5yr | 10yr | Max
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate.
Source Indicator: pi
Source: University of Pennsylvania
Release: Penn World Table 7.1
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(a) Price Level of Investment for Finland, PPP of Investment over Exchange Rate, Not Seasonally Adjusted (PLOINVFIA624NUPN)
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copy to all
Create your own data transformation: [+]
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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Suggested Citation
``` University of Pennsylvania, Price Level of Investment for Finland [PLOINVFIA624NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/PLOINVFIA624NUPN, April 30, 2016. ```
Retrieving data.
Graph updated. | 580 | 2,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | latest | en | 0.85418 |
https://excelmate.wordpress.com/tag/named-ranges/ | 1,600,538,500,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192783.34/warc/CC-MAIN-20200919173334-20200919203334-00107.warc.gz | 395,609,990 | 27,079 | # Blog Archives
## Excel – Creating Dynamic Named Ranges
If you use named ranges in Excel, you’ll know how useful they can be. If you are not familiar with them check my blog ( http://wp.me/p2EAVc-99 ) on setting up named ranges and how they can be used.
If you are currently using them you will also know that it can be a bit of pain having to update the range all the time as new records are added or removed from your data. If your data does change regularly, wouldn’t it be nice to have a named range that automatically adjusts to the correct number of rows or columns? Well…you can, by creating dynamic ranges.
At the heart of a dynamic range is the OFFSET function. Before we embark on creating ranges, let’s look at the syntax first as it is not that straightforward.
The official syntax you get from Microsoft when you enter the function is;
=OFFSET(reference, rows, columns, [height], [width])
…which doesn’t really tell you much if you have never used this function before.
Reference: every range has a starting point, even dynamic ones. Reference is like an anchor point from which the rest of the range is referenced.
Rows: is the number of rows away from the reference or anchor point. Positive numbers represent rows down, and negative numbers rows up.
Columns: is the number of columns away from the reference or anchor point. Positive numbers represent columns to the right, and negative numbers columns to the left.
=OFFSET(\$A\$1,0,0 would mean the range starts from A1
=OFFSET(\$A\$1,1,0 would mean the range starts from A2 (1 row below)
=OFFSET\$A\$1,0,1 would mean the range starts from B1 (1 column to the right)
…and so on and so forth.
Height: this is optional, but represents the number of rows you want to include in your range. So if we had =OFFSET(\$A\$1,0,0,5 then the range would be 5 rows starting from A1.
Width: another optional argument. This sets the number of columns that make up your range. Continuing the formula from above, if we have =OFFSET(\$A\$1,0,0,5,3) then we would set our range to 5 rows high and 3 columns wide starting at A1.
Using OFFSET to create a defined range
This of course gives us a fixed range, as we are specifically defining the number of rows and columns we want in the named range.
So how do we get it to be dynamic?
We need to build in functions that calculate the number of rows and/or columns within the OFFSET function. Generally speaking we do this within the height and width parts of the function, but you may need to alter the other arguments such as when you work with dynamic charts.
So let’s build a dynamic range;
First of all, you can’t create a dynamic named range in the same way as a normal named range i.e. select a bunch of cells and give them a name. We need to go to the NAME MANAGER and click on NEW, or go to DEFINE NAME.
In either case, give your range a name.
Decide whether the range is specific to the worksheet or can be referenced from anywhere in the workbook.
Add a comment…purely optional as a note to yourself or anyone else vaguely interested in the named range.
And then the important bit – REFERS TO: Rather than entering a basic range, this is where we have to use the OFFSET function.
Starting with a basic table, I want to create a dynamic range that automatically works out the number of rows in a fixed width table of 5 columns.
Working out what we have and what we need to create our dynamic range
So our formula will look like this;
=OFFSET(Sheet1!\$A\$1,1,0,COUNT(Sheet1!\$A:\$A),5)
In plain English…starting 1 row below A1, count the number of cells that contain numbers in column A (to work out number of rows) and set the width to 5 columns.
In this example I am excluding the heading row, but you may well need to include it, if for example, you are creating a dynamic range to use in a Pivot table.
Below, we can see the correctly selected data.
Correctly selected range based on rows found
If I add a few rows of data, the range is automatically recalculated to include them.
Recalculated range with extra rows
If I want the range to check the number of columns too then I need to modify the formula as follows;
=OFFSET(Sheet1!\$A\$1,1,0,COUNT(Sheet1!\$A:\$A),COUNTA(Sheet1!\$1:\$1))
Recalculated range with extra columns and rows
Now, not only does the formula count the number of cells containing numbers in column A, it counts the number cells containing text (hence COUNTA)across the top of my data table to give me the number of columns that make up my table.
So now you can create either a one-way or two-way dynamic range depending on your personal requirements, and no need to manually update the REFERS TO cell references.
In another blog, I will show you how you can use the OFFSET function to create dynamic charts.
## Excel – Creating & Using Named Ranges
When you first come across named ranges they don’t seem to be the most useful of things in Excel, yet the more you learn in Excel the more uses and benefits you will discover in using them.
So what is a named range?
Basically, it is a name that has been given to a single cell or group of cells in a worksheet. But why call it by something that is longer than its address?
Let’s look at an example naming a single cell.
In A1 let’s enter the UK VAT rate (sales tax) of 20%. In some cells elsewhere in the worksheet I have some values and I want to calculate the VAT on each of those.
Formula showing use of absolute reference
You can see my values in column A, the calculated VAT in column B and for ease of understanding I have displayed the formula in column C. Basic stuff you might say…a simple calculation using \$ symbols to fix the cell address of the 20% rate.
Now let’s give our 20% cell a sensible name that reflects what it contains e.g. VAT_Rate. Note the use of an underscore in the name. When creating named ranges do not use spaces…it won’t work. Also be aware that calling a cell VAT20 for instance, which may seem like a sensible name, will actually take you to cell address VAT20. Don’t forget that if you are using 2007 or later, columns go all the way to XFD so add underscores or something to the name that will not turn it into a cell reference.
To give your cell a name, the easiest way is to click in the NAME BOX.
The Name Box
Where you see the cell address, start typing the name you want to give to your cell remembering the rules about names I mentioned earlier. Make sure you press ENTER otherwise the name won’t be saved.
If you want to check that the name works, click on the arrow on the right hand side of the NAME BOX, click on the name you have just created and all being well the cell you gave a name to will be highlighted.
Name added to the Name Box
“Fantastic!” I hear you say…”but you have just spent about 30 seconds creating and testing a name that is three times the length of the actual cell address and no obvious benefits…would have been quicker typing \$A\$1!”.
And so it would seem. But let’s rewrite our formula…
Start writing your formula exactly as before but instead of typing in \$A\$1, start typing the name you assigned to the cell. As we are entering a formula, when I type the first letter of the name it comes up with a list of functions, and rather conveniently, named ranges beginning with the same letter…
Named ranges appear in the functions/names list when you type a formula
Either arrow down to the name or type the name in full and press enter as you would for any normal formula once all the cell references etc. have been entered. And because I’m using a name, it has a fixed location so no need for those \$ symbols.
Same formula as before but using the new named range
The end result in column B is no different to what we had before, but when you look at the formula, instead of some cell reference we can now see a name that gives us an indication of what is actually being calculated. So there is advantage number 1 of a NAMED RANGE. If you are used to building more complex formulas, then they will become much shorter and what they are calculating becomes much more obvious.
The second advantage is that I can now make reference to my cell anywhere in the workbook without the need to know where it is, or even ty and find it to click on it to build it into any of my formulas.
Using the named range elsewhere in the same workbook
Just type the name as part of your formula as before, select the name from the list that appears, and done! Simple, and if you look at the formula you know exactly what it is calculating – the name tells you.
If you forget a name that you have created, you can bring up a list of all named ranges in your workbook by pressing the F3 key as you build your formula. Click on the name you want to use and click on OK.
Use F3 to see a list of all named ranges in your workbook
So far I have only named a single cell, but you can create named ranges for entire blocks of cells, and you can even have overlapping named ranges.
Naming a block of cells
Here I have manually named a block of cells containing sales data for “North”. I could then repeat this for each column, however, I will show you a more efficient way of doing this a bit further on, but here it is just to show that blocks of cells can be named too.
I can now write formulas that refer to the block of named cells;
=SUM(SalesNorth)
…which will sum up the cells A2:A12. No need for me to physically select any cells, just use the name and once again when I look at my formula I know exactly what it is adding up.
Hopefully by now you can start seeing some of the advantages of using named ranges.
A bit of a random geek fact for you: if you zoom down to 39% or less, named ranges are actually displayed in the worksheet (works best with blocks of cells rather than single ones).
Zooming down to 39% (or less) to see named ranges in a worksheet
The quickest way to name a range is to simply type the name in the NAME BOX and press enter but you can also do it through the NAME MANAGER or DEFINE NAME in the FORMULAS tab.
Choosing the scope of a named range
Use the dialog box to name your range, decide on its scope i.e. is referencing the name limited to a single worksheet or can it be used across the entire workbook? You can add comments, personally I never bother but the choice is yours. And finally which cell(s) does your name refer to? Click on OK when done. The new name you have created will then appear in the drop down list in the NAME BOX.
If you go through the NAME MANAGER click on NEW in the top left hand corner to open the same dialog box as above.
The name manager
If you want to EDIT or DELETE your named ranges you need to come through the NAME MANAGER. Unfortunately, you cannot delete named ranges via the name box.
If you have dozens of named ranges, you can use the FILTER button to limit the list by filtering on different criteria.
Using the filter in the name manager
I mentioned earlier a quick way to generate range names. Previously I created named ranges by selecting the cells myself and using the NAME BOX to assign a name. That’s fine if you only have 2-3 to create but what if you have loads of columns or rows that you would like to refer to by name?
Using a basic two-way table (you can do this on any size of table by the way) highlight the whole table. On the FORMULAS tab, click on CREATE FROM SELECTION or if you like keyboard shortcuts use Ctrl + Shift + F3 and you will see this…
Using Create From Selection to create named ranges
Decide which row/column values you want to use as range names and tick/un-tick the relevant boxes in the dialog window. Click on OK. Nothing apparent happens until you check out the NAME BOX or the NAME MANAGER.
The full list of named ranges in the workbook
We now have a whole bunch of newly created named ranges using the row and column headings from the table!
Showing the automatically created named ranges
Test them out if you want by clicking on any of the names and see which cells are highlighted. You can now start writing formulas using as many names as necessary;
=SUM(Yr_2009,Yr_2011)
Now tell me that’s not better than =SUM(Sheet4!B2:B5,Sheet4!D2:D5)!
Finally, be aware that all the methods I have shown you here apply to FIXED dimension ranges i.e. the cell(s) you select and name remain unchanged unless you go into the NAME MANAGER and edit the REFERS TO information adding further rows/columns. You can create dynamic ranges but I will write about those separately as they are quite a big topic in their own right.
A named range CAN be automatically resized as long as new rows/columns are added WITHIN the boundaries of the original named range. If my named range is from A1 to D20, as long as I insert or delete rows between rows 1 and 20, or insert/delete columns between A and D, the range will include those automatically. If I add data from row 21 or column E and beyond I would need to manually change the REFERS TO references in the NAME MANAGER. | 2,937 | 12,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-40 | latest | en | 0.91021 |
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OpenStudy (anonymous):
How do you write .046 as a percent
So a percentage is basically a number over 100. We can turn a decimal into a percentage, therefore, by multiplying it by 100. Is that enough to help?
OpenStudy (anonymous):
Correct!
OpenStudy (anonymous):
Another way to think about percents is this: If you had \$0.046 dollars how many pennies would that be? Answer: 0.50 dollars --> 50 cents then 0.046 dollars --> 4.6 cents. You simply moving the decimal to the right 2 places. It is allot easier to say the '75% of dentist agree' then to have to say '0.75 dentists in our sample'.
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Minimum sum subsequence such that at least one of every four consecutive elements is picked
• Difficulty Level : Medium
• Last Updated : 26 Apr, 2021
Given an array arr[] of positive integers. The task is to find minimum sum subsequence from the array such that at least one value among all groups of four consecutive elements is picked.
Examples :
```Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 6
6 is sum of output subsequence {1, 5}
Note that we have following subarrays of four
consecutive elements
{(1, 2, 3, 4),
(2, 3, 4, 5),
(3, 4, 5, 6)
(4, 5, 6, 7)
(5, 6, 7, 8)}
Our subsequence {1, 5} has an element from
all above groups of four consecutive elements.
And this subsequence is minimum sum such
subsequence.
Input : arr[] = {1, 2, 3, 3, 4, 5, 6, 1}
Output : 4
The subsequence is {3, 1}. Here we consider
second three.
Input: arr[] = {1, 2, 3, 2, 1}
Output: 2
The subsequence can be {1, 1} or {2}
Input: arr[] = {6, 7, 8}
Output: 6
Input: arr[] = {6, 7}
Output: 6```
The idea is similar to LIS problem. We store minimum sum subsequence ending with every element of arr[]. We finally return minimum of last four values.
```dp[i] stores minimum sum subsequence (with at least
one of every four consecutive elements) of arr[0..i]
such that arr[i] is part of the solution. Note that
this may not be the best solution for subarray
arr[0..i].
We can recursively compute dp[i] using below formula
dp[i] = arr[i] + min(dp[i-1], dp[i-2], dp[i-3], dp[i-4])
Finally we return minimum of dp[n-1], dp[n-2],
dp[n-4] and dp[n-3]```
Below is the implementation of above idea.
## C++
`// C++ program to find minimum sum subsequence``// of an array such that one of every four``// consecutive elements is picked.``#include ``using` `namespace` `std;` `// Returns sum of minimum sum subsequence``// such that one of every four consecutive``// elements is picked from arr[].``int` `minSum(``int` `arr[], ``int` `n)``{`` ``// dp[i] is going to store minimum sum`` ``// subsequence of arr[0..i] such that arr[i]`` ``// is part of the solution. Note that this`` ``// may not be the best solution for subarray`` ``// arr[0..i]`` ``int` `dp[n];` ` ``// If there is single value, we get the`` ``// minimum sum equal to arr[0]`` ``if` `(n == 1)`` ``return` `arr[0];` ` ``// If there are two values, we get the`` ``// minimum sum equal to the minimum of`` ``// two values`` ``if` `(n == 2)`` ``return` `min(arr[0], arr[1]);` ` ``// If there are three values, return`` ``// minimum of the three elements of`` ``// array`` ``if` `(n == 3)`` ``return` `min(arr[0], min(arr[1], arr[2]));` ` ``// If there are four values, return minimum`` ``// of the four elements of array`` ``if` `(n == 4)`` ``return` `min(min(arr[0], arr[1]),`` ``min(arr[2], arr[3]));` ` ``dp[0] = arr[0];`` ``dp[1] = arr[1];`` ``dp[2] = arr[2];`` ``dp[3] = arr[3];` ` ``for` `(``int` `i = 4; i < n; i++)`` ``dp[i] = arr[i] + min(min(dp[i - 1], dp[i - 2]),`` ``min(dp[i - 3], dp[i - 4]));` ` ``// Return the minimum of last 4 index`` ``return` `min(min(dp[n - 1], dp[n - 2]),`` ``min(dp[n - 4], dp[n - 3]));``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 1, 2, 3, 3, 4, 5, 6, 1 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ``cout << minSum(arr, n);`` ``return` `0;``}`
## Java
`// Java program to find minimum sum subsequence``// of an array such that one of every four``// consecutive elements is picked.``import` `java.io.*;` `class` `GFG {` ` ``// Returns sum of minimum sum subsequence`` ``// such that one of every four consecutive`` ``// elements is picked from arr[].`` ``static` `int` `minSum(``int``[] arr, ``int` `n)`` ``{`` ``// dp[i] is going to store minimum sum`` ``// subsequence of arr[0..i] such that arr[i]`` ``// is part of the solution. Note that this`` ``// may not be the best solution for subarray`` ``// arr[0..i]`` ``int``[] dp = ``new` `int``[n];` ` ``// If there is single value, we get the`` ``// minimum sum equal to arr[0]`` ``if` `(n == ``1``)`` ``return` `arr[``0``];` ` ``// If there are two values, we get the`` ``// minimum sum equal to the minimum of`` ``// two values`` ``if` `(n == ``2``)`` ``return` `Math.min(arr[``0``], arr[``1``]);` ` ``// If there are three values, return`` ``// minimum of the three elements of`` ``// array`` ``if` `(n == ``3``)`` ``return` `Math.min(arr[``0``], Math.min(arr[``1``], arr[``2``]));` ` ``// If there are four values, return minimum`` ``// of the four elements of array`` ``if` `(n == ``4``)`` ``return` `Math.min(Math.min(arr[``0``], arr[``1``]),`` ``Math.min(arr[``2``], arr[``3``]));` ` ``dp[``0``] = arr[``0``];`` ``dp[``1``] = arr[``1``];`` ``dp[``2``] = arr[``2``];`` ``dp[``3``] = arr[``3``];` ` ``for` `(``int` `i = ``4``; i < n; i++)`` ``dp[i] = arr[i] + Math.min(Math.min(dp[i - ``1``], dp[i - ``2``]),`` ``Math.min(dp[i - ``3``], dp[i - ``4``]));` ` ``// Return the minimum of last 4 index`` ``return` `Math.min(Math.min(dp[n - ``1``], dp[n - ``2``]),`` ``Math.min(dp[n - ``4``], dp[n - ``3``]));`` ``}` ` ``// Driver code`` ``static` `public` `void` `main(String[] args)`` ``{`` ``int``[] arr = { ``1``, ``2``, ``3``, ``3``, ``4``, ``5``, ``6``, ``1` `};`` ``int` `n = arr.length;`` ``System.out.println(minSum(arr, n));`` ``}``}` `// This Code is contributed by vt_m.`
## Python3
`# Python 3 program to find minimum sum``# subsequence of an array such that one``# of every four consecutive elements is picked.` `# Returns sum of minimum sum subsequence``# such that one of every four consecutive``# elements is picked from arr[].``def` `minSum(arr, n):` ` ``# dp[i] is going to store minimum sum`` ``# subsequence of arr[0..i] such that`` ``# arr[i] is part of the solution. Note`` ``# that this may not be the best solution`` ``# for subarray arr[0..i]`` ``dp ``=` `[``0``] ``*` `n` ` ``# If there is single value, we get`` ``# the minimum sum equal to arr[0]`` ``if` `(n ``=``=` `1``):`` ``return` `arr[``0``]` ` ``# If there are two values, we get the`` ``# minimum sum equal to the minimum of`` ``# two values`` ``if` `(n ``=``=` `2``):`` ``return` `min``(arr[``0``], arr[``1``])` ` ``# If there are three values, return`` ``# minimum of the three elements of`` ``# array`` ``if` `(n ``=``=` `3``):`` ``return` `min``(arr[``0``],`` ``min``(arr[``1``], arr[``2``]))` ` ``# If there are four values,`` ``# return minimum of the four`` ``# elements of array`` ``if` `(n ``=``=` `4``):`` ``return` `min``(``min``(arr[``0``], arr[``1``]),`` ``min``(arr[``2``], arr[``3``]))` ` ``dp[``0``] ``=` `arr[``0``]`` ``dp[``1``] ``=` `arr[``1``]`` ``dp[``2``] ``=` `arr[``2``]`` ``dp[``3``] ``=` `arr[``3``]` ` ``for` `i ``in` `range``( ``4``, n):`` ``dp[i] ``=` `arr[i] ``+` `min``(``min``(dp[i ``-` `1``], dp[i ``-` `2``]),`` ``min``(dp[i ``-` `3``], dp[i ``-` `4``]))` ` ``# Return the minimum of last 4 index`` ``return` `min``(``min``(dp[n ``-` `1``], dp[n ``-` `2``]),`` ``min``(dp[n ``-` `4``], dp[n ``-` `3``]))` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` ` ``arr ``=` `[ ``1``, ``2``, ``3``, ``3``, ``4``, ``5``, ``6``, ``1` `]`` ``n ``=` `len``(arr)`` ``print``(minSum(arr, n))` `# This code is contributed by ita_c`
## C#
`// C# program to find minimum sum subsequence``// of an array such that one of every four``// consecutive elements is picked.``using` `System;` `class` `GFG {` ` ``// Returns sum of minimum sum subsequence`` ``// such that one of every four consecutive`` ``// elements is picked from arr[].`` ``static` `int` `minSum(``int``[] arr, ``int` `n)`` ``{`` ``// dp[i] is going to store minimum sum`` ``// subsequence of arr[0..i] such that arr[i]`` ``// is part of the solution. Note that this`` ``// may not be the best solution for subarray`` ``// arr[0..i]`` ``int``[] dp = ``new` `int``[n];` ` ``// If there is single value, we get the`` ``// minimum sum equal to arr[0]`` ``if` `(n == 1)`` ``return` `arr[0];` ` ``// If there are two values, we get the`` ``// minimum sum equal to the minimum of`` ``// two values`` ``if` `(n == 2)`` ``return` `Math.Min(arr[0], arr[1]);` ` ``// If there are three values, return`` ``// minimum of the three elements of`` ``// array`` ``if` `(n == 3)`` ``return` `Math.Min(arr[0], Math.Min(arr[1], arr[2]));` ` ``// If there are four values, return minimum`` ``// of the four elements of array`` ``if` `(n == 4)`` ``return` `Math.Min(Math.Min(arr[0], arr[1]),`` ``Math.Min(arr[2], arr[3]));` ` ``dp[0] = arr[0];`` ``dp[1] = arr[1];`` ``dp[2] = arr[2];`` ``dp[3] = arr[3];` ` ``for` `(``int` `i = 4; i < n; i++)`` ``dp[i] = arr[i] + Math.Min(Math.Min(dp[i - 1], dp[i - 2]),`` ``Math.Min(dp[i - 3], dp[i - 4]));` ` ``// Return the minimum of last 4 index`` ``return` `Math.Min(Math.Min(dp[n - 1], dp[n - 2]),`` ``Math.Min(dp[n - 4], dp[n - 3]));`` ``}` ` ``// Driver code`` ``static` `public` `void` `Main()`` ``{`` ``int``[] arr = { 1, 2, 3, 3, 4, 5, 6, 1 };`` ``int` `n = arr.Length;`` ``Console.WriteLine(minSum(arr, n));`` ``}``}` `// This code is contributed by vt_m.`
## Javascript
``
Output:
`4`
Alternate Solution :
First of all, think that we have only four elements then our result is at least four given elements. Now, in the case, if we have more than four elements then we must maintain an array sum[] where sum[i] includes the possible minimal sum up to i-th element and also i-th element must be a part of the solution.
While computing sum[i], our base condition is that arr[i] must be a part of sum[i] and then we must have an element from last four elements. So, we can recursively compute sum[i] as sum of arr[i] and minimum (sum[i-1], sum[i-2], sum[i-3], sum[i-4]). Since there are overlapping subproblems in the recursive structure of our problem, we can use Dynamic Programming to solve this problem. And for the final result we must compute minimum of last four values of sum[] array as result must contain an element from last four elements.
## C++
`// CPP program to calculate``// minimum possible sum for given constraint``#include ``typedef` `long` `long` `ll;``using` `namespace` `std;` `// function to calculate min sum using dp``int` `minSum(``int` `ar[], ``int` `n)``{`` ``// if elements are less than or equal to 4`` ``if` `(n <= 4)`` ``return` `*min_element(ar, ar + n);` ` ``// save start four element as it is`` ``int` `sum[n];`` ``sum[0] = ar[0];`` ``sum[1] = ar[1];`` ``sum[2] = ar[2];`` ``sum[3] = ar[3];` ` ``// compute sum[] for all rest elements`` ``for` `(``int` `i = 4; i < n; i++)`` ``sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));` ` ``// Since one of the last 4 elements must be`` ``// present`` ``return` `*min_element(sum + n - 4, sum + n);``}` `// driver program``int` `main()``{`` ``int` `ar[] = { 2, 4, 1, 5, 2, 3, 6, 1, 2, 4 };`` ``int` `n = ``sizeof``(ar) / ``sizeof``(ar[0]);`` ``cout << ``"Minimum sum = "` `<< minSum(ar, n);`` ``return` `0;``}`
## Java
`// Java program to calculate``// minimum possible sum for given constraint``import` `java.util.Arrays;` `class` `GFG``{` `// function to calculate min sum using dp``static` `int` `minSum(``int` `ar[], ``int` `n)``{`` ``// if elements are less than or equal to 4`` ``if` `(n <= ``4``)`` ``return` `Arrays.stream(ar).min().getAsInt();` ` ``// save start four element as it is`` ``int` `[]sum = ``new` `int``[n];`` ``sum[``0``] = ar[``0``];`` ``sum[``1``] = ar[``1``];`` ``sum[``2``] = ar[``2``];`` ``sum[``3``] = ar[``3``];` ` ``// compute sum[] for all rest elements`` ``for` `(``int` `i = ``4``; i < n; i++)`` ``//sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));`` ``sum[i] = ar[i] + Arrays.stream(Arrays.copyOfRange(`` ``sum, i - ``4``, i)).min().getAsInt();` ` ``// Since one of the last 4 elements must be`` ``// present`` ``return` `Arrays.stream(Arrays.copyOfRange(`` ``sum, n - ``4``, n)).min().getAsInt();``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ``int` `ar[] = { ``2``, ``4``, ``1``, ``5``, ``2``, ``3``, ``6``, ``1``, ``2``, ``4` `};`` ``int` `n = ar.length;`` ``System.out.println(``"Minimum sum = "` `+ minSum(ar, n));``}``}` `// This code is contributed by PrinciRaj1992`
## Python3
`# Python3 program to calculate``# minimum possible sum for given constraint` `# function to calculate min sum using dp``def` `minSum(ar, n):`` ` ` ``# if elements are less than or equal to 4`` ``if` `(n <``=` `4``):`` ``return` `min``(ar)` ` ``# save start four element as it is`` ``sum` `=` `[``0` `for` `i ``in` `range``(n)]`` ``sum``[``0``] ``=` `ar[``0``]`` ``sum``[``1``] ``=` `ar[``1``]`` ``sum``[``2``] ``=` `ar[``2``]`` ``sum``[``3``] ``=` `ar[``3``]` ` ``# compute sum[] for all rest elements`` ``for` `i ``in` `range``(``4``, n):`` ``sum``[i] ``=` `ar[i] ``+` `min``(``sum``[i ``-` `4``:i])` ` ``# Since one of the last 4 elements must be`` ``# present`` ``return` `min``(``sum``[n ``-` `4``:n])` `# Driver Code``ar ``=` `[``2``, ``4``, ``1``, ``5``, ``2``, ``3``, ``6``, ``1``, ``2``, ``4``]``n ``=` `len``(ar)``print``(``"Minimum sum = "``, minSum(ar, n))` `# This code is contributed by Mohit Kumar`
## C#
`// C# program to calculate``// minimum possible sum for given constraint``using` `System;``using` `System.Linq;` `class` `GFG``{` `// function to calculate min sum using dp``static` `int` `minSum(``int` `[]ar, ``int` `n)``{`` ``// if elements are less than or equal to 4`` ``if` `(n <= 4)`` ``return` `ar.Min();` ` ``// save start four element as it is`` ``int` `[]sum = ``new` `int``[n];`` ``sum[0] = ar[0];`` ``sum[1] = ar[1];`` ``sum[2] = ar[2];`` ``sum[3] = ar[3];`` ``int` `[]tempArr;`` ` ` ``// compute sum[] for all rest elements`` ``for` `(``int` `i = 4; i < n; i++)`` ``{`` ``//sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));`` ``tempArr = ``new` `int``[4];`` ``Array.Copy(sum, i - 4, tempArr, 0, 4);`` ``sum[i] = ar[i] + tempArr.Min();`` ``}`` ` ` ``// Since one of the last 4 elements must be`` ``// present`` ``tempArr = ``new` `int``[4];`` ``Array.Copy(sum,n-4,tempArr,0,4);`` ``return` `tempArr.Min();``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{`` ``int` `[]ar = { 2, 4, 1, 5, 2, 3, 6, 1, 2, 4 };`` ``int` `n = ar.Length;`` ``Console.WriteLine(``"Minimum sum = "` `+`` ``minSum(ar, n));``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output:
`Minimum sum = 4`
Time Complexity: O(n).
Thanks to Shivam Pradhan (anuj_charm) for providing this alternate solution.
This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 5,752 | 16,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.867565 |
https://web2.0calc.com/questions/help-plz_27 | 1,563,923,878,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529737.79/warc/CC-MAIN-20190723215340-20190724001340-00537.warc.gz | 585,714,779 | 5,852 | +0
# HelP pLz!!
0
332
1
In the diagram below, angle BAC = 24 degrees and AB = AC.
If angle ABC = y degrees, what is the value of y?
Nov 20, 2017
#1
0
Since angle BAC = 24 degrees and AB = AC, then:
180 - 24 =156 The sum of the other 2 angles.
Since angle ABC = angle ACB, then:
y = 156/2 =78 degrees.
Nov 20, 2017
#1
0
Since angle BAC = 24 degrees and AB = AC, then:
180 - 24 =156 The sum of the other 2 angles.
Since angle ABC = angle ACB, then:
y = 156/2 =78 degrees.
Guest Nov 20, 2017 | 194 | 505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-30 | latest | en | 0.565898 |
http://www.chegg.com/homework-help/exercise-appropriate-round-answers-nearest-tenth-percent-unl-chapter-11.1-problem-67e-solution-9780321501073-exc | 1,469,959,107,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828314.45/warc/CC-MAIN-20160723071028-00278-ip-10-185-27-174.ec2.internal.warc.gz | 374,786,998 | 17,869 | View more editions
# TEXTBOOK SOLUTIONS FOR A Survey of Mathematics with Applications 8th Edition
• 6720 step-by-step solutions
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Chapter: Problem:
For Exercise, where appropriate, round answers to the nearest tenth of a percent unless specified otherwise.
Reselling a Car Quincy Carter purchased a used car for \$1000. He decided to sell the car for 10% above his purchase price. Quincy could not sell the car so he reduced his asking price by 10%. If he sells the car at the reduced price, will he have a profit or a loss or will he break even? Explain how you arrived at your answer.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
A Survey of Mathematics with Applications | 8th Edition
9780321501073ISBN-13: 0321501071ISBN: Authors:
Alternate ISBN: 9780321512970, 9781256347255 | 239 | 969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-30 | latest | en | 0.866803 |
http://yourfinancebook.com/gold-karat/ | 1,481,054,158,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541995.74/warc/CC-MAIN-20161202170901-00048-ip-10-31-129-80.ec2.internal.warc.gz | 821,803,576 | 11,816 | What is Gold Karat – How it’s measured in market
While buying gold ornaments or jewellery you must have noticed the seller or representative often describing the unit in Karat.
Karat is a unit of measure to know the purity of gold. If the seller tells you a particular jewellery of 22 karat then it means, that jewellery has 91.06% (22/24) of gold and balance 8.4% of metals like silver, zinc etc.
Gold is very soft to make jewellery, for this reason seller use to mix silver, copper or zinc to it.
You should not get confused between carat and karat. Both are used to measure purity but “carat” is specifically used to measure purity of precious stones and “Karat” is used to measure purity of Gold.
Gold jeweleries are measured in different Karat and based on that, price has been calculated. Karat is calculated as 24 times of the gold mass divided by the total mass.
24 karat is fine and pure having 99.99% gold mass in it and balances other metals. Similarly in 22 karat you have 22/24 part of pure gold and balance of other metals.
Karat Gold in % Other Metal in % 24k 99.99% 0.01% 22k 91.6% (22/24) 8.4% (2/24) 18k 75% (18/24) 24% (6/24) 14k 58.5% (14/24) 41.5% (10/24) 10k 41.7% (10/24) 58.3% (14/24) 6k 25% (6/24) 75% (18/24)
For example, if you have any gold jewellery of 22k then after melting it you will get 91.6% of pure gold and balance will be other materials mixed to it. While exchanging at a jewellery shop the buyer will also pay you the value of 91.6%. In India, you will find most jewellery with 22k i.e. 91.6% of gold in it.
All most all jewellery will have a mark in it to let you know the purity of gold used in it. If you did not find it then ask the seller to show you. You can also test jewellery to identify if it’s made up of pure gold.
How to test purity of Gold
You can test purity of gold with the help of nitric acid test. You can ask your jeweler to test it on your behalf as nitric acid is very difficult to get from market. To do this test you need to keep your jewellery in a stainless steel container and then take few drops of nitric acid and put it on the jewellery. If it is made up of pure gold then you will not see any reaction otherwise the jewellery will turn green or milk-colored if it’s gold plated.
You can also request the jewelers to carry on electronic test for you. This device will test gold’s electrical conductivity to let you know if it’s actual gold and will also let you know its purity. | 658 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-50 | longest | en | 0.947363 |
https://www.digsilent.de/en/faq-powerfactory/tags/emt.html?page=2 | 1,627,629,119,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153934.85/warc/CC-MAIN-20210730060435-20210730090435-00477.warc.gz | 727,489,650 | 16,274 | ### Do you have an example of an intermittent phase-to-ground fault on a cable?
You will need to perform an EMT-type simulation for this purpose. To replicate the fault on the cable insulation, the cable must be explictly divided in two sections at the point where the faulty…
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### Would you have an example for a wind turbine driving asynchronous machine?
Refer to the attached example. It describes an induction generator with the necessary dynamic models. The reactive power is balanced using a switched capacitor. The dynamic models are a shaft model…
Tags | 597 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.858691 |
https://www.physicsforums.com/threads/lebesgue-integral-jordan-measurable.555355/ | 1,550,831,262,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247515149.92/warc/CC-MAIN-20190222094419-20190222120419-00521.warc.gz | 903,802,177 | 11,778 | # Lebesgue integral(jordan measurable)
1. Nov 30, 2011
### gotjrgkr
Hi!
I have a guess. Could you give me your opinion about my guess??
Let A be a rectifiable set(or jordan measurable set).This is defined in a book "Analysis on manifolds" by munkres. You can refer to it in p.112-113.
Now, let f be a bounded function over the set A, and suppose f is integrable over A.
Then $\int$$_{A}$f = $\int$$_{I}$g where I is a large rectangle containg the set A and g is a function with domain I whose a value at x$\in$A is f(x) and a value at x$\in$I-A is 0.
Then under this assumption, A is measurable, f is also measurable on A, and f is lebesgue integrable over A and the lebesgue integral and $\int$$_{A}$f are equal.
Is my guess true?? | 220 | 738 | {"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} | 2.84375 | 3 | CC-MAIN-2019-09 | latest | en | 0.938794 |
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Biot number (mass transfer)
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations. Gives a simple index of the ratio of the heat transfer resistances ... more
Blade root bending moment load due to yaw
The blade root bending moment due to the wind turbine yaw operation. The yaw rate can be calculated for passive yaw, or is defined by the design for active ... more
Block and tackle - efficiency approximation - with friction factor
A block and tackle is a system of two or more pulleys with a rope or cable threaded between them, usually used to lift or pull heavy loads.A more precise ... more
Block and tackle - efficiency approximation - without friction factor
f is the number of sheaves in the purchase, and there is a roughly % loss of efficiency at each sheave due to friction, then:
... more
Block and tackle - effort required to raise a given weight
A block and tackle is a system of two or more pulleys with a rope or cable threaded between them, usually used to lift or pull heavy loads.The formula used ... more
Boltzmann distribution in mechanics
In statistical mechanics and mathematics, a Boltzmann distribution (also called Gibbs distribution) is a probability distribution, probability measure, or ... more
Boltzmann factor in mechanics
In statistical mechanics and mathematics, a Boltzmann distribution (also called Gibbs distribution) is a probability distribution, probability measure, or ... more
Boundary shear stress (for natural rivers)
Assuming a single, well-mixed, homogeneous fluid and a single acceleration due to gravity (both are good assumptions in natural rivers, and the second is a ... more
Bradley model of the force applied on a contact area between two spheres
Contact mechanics is the study of the deformation of solids that touch each other at one or more points. When two solid surfaces are brought into close ... more | 397 | 1,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-05 | longest | en | 0.916167 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
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Ineq. #1:
Ineq. #2:
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Number Theory
In my opinion, number theory is an extremely fascinating subject. There is much crazy
number theory out there (ever heard of the Riemann Hypothesis?). Fortunately, the stuff in
contest math is much simpler than that. It is mostly ignored in high school curriculum, we
seem to like to do a lot of algebra and calculus. However, do not less this fool you, number
theory is not nearly as “advanced” as those subjects. At the core, number theory is very
simple. In fact, most of the time, the problems can be stated in very simple terms and at
the same time, the tools we have are much more limited than in other subjects. This is good
news and bad news. Good news is it doesn’t take long to learn the tools. Bad news is it
requires much trickiness.
Now that the little introductory blurb is done. Let us go over some common notation.
1 Notation
1. N,Z,Zp,Z/pZ,Q,Q0,R, and C We will take these to be the natural numbers, integers,
p-adic integers, integers in mod p, rational numbers, irrational numbers, reals, and
complex numbers respectively. Don’t worry about the p-adic numbers if you don’t
know what those are, that was just thrown in to warn some people of the real standard
notations.
2. The above things are all systems or playgrounds. They are where we are working.
Most of the time, in math class without even knowing it, we make the assumption
we are working in the reals. Sometimes, we venture into the complex numbers. For
example, what if I asked two simple questions:
• Solve for x: x2 = 1 Well the answer is, it depends. What if I told you that 3 is
a solution? What if I told you, that indeed there are 4 solutions: x = 1, 3, 5, 7?
You would think I was crazy until you realized that I was working in mod 8.
• Solve for x: x2 = −1 What is the answer? You might be more suspicious now.
In the reals there are no solutions! In fact, this is essentially the only polynomial
equation that has no solutions in the reals. As soon as we add i to the reals,
we can solve this equation. We also happen to now be able to solve all such
equations. We have created the complex numbers! But I claim that in fact the
solutions could be x = 2, 3 You are probably on to me now. I’m talking about
mod 5! But maybe crazier than that, x = i, j, k, are also solutions. Working with
quaternions (this is where those vectors and cross products and stuff come from),
we have these solutions. Our ARML shirt was blatantly wrong last year, since
i * j = k which is certainly not real.
I hope you are all sufficiently confused now. Apparently, an innocent polynomial
equation could have more than the degree number of solutions, no solutions, or not
even make any sense at all. Fortunately, all you have to avoid all this craziness is make
sure you specify the system you are working in. In that spirit, let us learn some more
ways to specify the system you are in. If we put a little floating plus to the right of
any of these symbols we get the positive numbers. Most of the time, we put a little
cross floating to the right of these signs we get the same system with zero removed.
For example N is essentially the same as . Also, sometimes you might see things
that look like R[x]. What this means, is we add x as an element to the reals and see
what kind of system we get. In this case, we get polynomials. Another example is that
R[i] is essentially the same as C. As you can tell all of these systems can be related to
each other. It is extremely important to understand the relationships between them.
In fact, I believe all of these systems can be based directly back to the integers, so if
ever there is some sort of contradiction in math and the whole thing collapses, we must
have gotten the integers wrong. All these systems have rules and we start calling them
groups, rings, fields, etc. As you learn more math, I’m sure you will study them if you
haven’t already. Maybe one day you’ll finally understand Druker’s shirt. ring to
rule them all: {0}.
3. In general, variable names in math are actually quite standard. For example, for some
reason w, x, y, z are useful as variables. Maybe the numbers at the beginning of the
alphabet are used for constants most often. Often times, s, t, u, v are most useful as
some sort of parameters. p is almost always a prime and in the case when we need
two primes, we either use subscripts or sometimes venture to use q. m, n are often
generic integers, with m frequently used for mods and n for numbers. i, j, k are most
useful as indices, but sometimes k gets used in division. q, r are obviously related and
commonly related to division as quotient and remainder. These are not strict rules,
but I have seen an instructor change all the ps in an equation to ms just because we
pointed out it didn’t necessarily have to be prime.
Other notation and terminology will be explained as they come up. As a note about
the general style of the lecture, it is bad. But beyond that, I am going to omit most
proofs so as to not turn this into a textbook. If you have any questions, you can always
talk to me personally.
2 Divisibility and Primes
2.1 Preliminary Notions
Most of us know what it means for something to divide something else. However, for our
purposes, it is important to define exactly what it means. For a, b ∈ Z and a ≠ 0, we say
that adividesb iff s.t. b = ka We write a|b to indicate that adividesb. a is called a
divisor of b and b is called a multiple of a. A prime p is prime iff it has exactly two distince
postive divisors. Any other integer is composite except 1 and −1.
Here are two more concepts that most everybody is already familiar with. The greatestcommondivisor
of two integers a, b(not both 0) is k = gcd(a, b) = (a, b) where s.t. d|a, d|b , k ≥ d Two
integers a and b are relativelyprime or coprime, if (a, b) = 1. The leastcommonmultiple
of integers a and b is defined as the smallest positive integral multiple of both a and b. We
write it as lcm[a, b] = [a, b].
Work out for yourself this gcd and lcm junk for those crazy numbers 0 and 1. Also, note
that these definitions can be generalized in many different ways. One way is by defining gcd
and lcm for n numbers and the other is to generalize to different systems, not the integers.
2.2 Fundamental Theorem of Arithmetic
There are several fundamental theorems that lead up to the Fundamental Theorem of Arithmetic.
We will briefly cover them and omit the proofs.
1. Division Algorithm Most of you probably know how to divide, but probably never
thought of it in this way. Thankfully given any two integers a and b with b ≠ 0, you can
divide a by b. In more precise terms: Given a, b ∈ Z, 0 ≤ r < |b|.
We call a, b, q, and r the dividend, divisor, quotient, and remainder respectively. This
is known as the division algorithm for unkown reasons.
2. Bezout’s Identity Using the division algorithm we can solve an extremely interesting
question. Given two integers a and b, the equation ax + by = 1 has integers solutions
iff (a, b) = 1. This is an extremely important idea that was on the AMC 12A as a
problem this year. Notice that this also implies that ax + by = d has solutions for all
integer d. What d can you solve for if (a, b) ≠ 1?
3. Euclid’s Lemma This can be easily proven with Bezout’s Identity. It states that if
a|bc and (a, b) = 1, then a|c This is an interesting and intuitive fact to know. However
it is most useful in proving the final result.
4. Fundamental Theorem of Arithmetic This is also known as unique factorization.
It states that given any integer n, it can be uniquely expressed as the product of distinct | 2,059 | 8,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-22 | latest | en | 0.940819 |
https://cutiumum.net/ratio-and-proportion-fractions-worksheet/ | 1,596,592,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00455.warc.gz | 232,236,199 | 32,879 | # Ratio And Proportion Fractions Worksheet
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Class 10 LAKHMIR SINGH AND MANJIT KAUR Solutions Physics Chapter 3 - Sources of Energy
Sources of Energy Exercise 121
Solution 1
Nuclear fuels (like uranium)
Solution 2
The amount of heat produced by burning 1 gram of a fuel completely is known as its calorific value.
Solution 3
Calorific value of LPG is 50kJ/gm means that if 1 gram of LPG is burnt completely, then it will produce 50kJ of heat energy.
Solution 4
LPG, due to its higher calorific value.
Sources of Energy Exercise 122
Solution 5
Ignition temperature of fuel can be defined as the minimum temperature to which a fuel must be heated so that it may catch fire and start burning.
Solution 6
If ignition temp of a fuel is 80oC, this means that minimum 80oC is required for the ignition of the fuel.
calorific
Solution 8
A source of energy is one which can provide adequate amount of energy in a convenient form over a long period of time.
Two main categories of the source of energy are:
i. Renewable source of energy
ii. Non- renewable source of energy
Solution 9
A good source of energy is one:
(i) which would do a large amount of work per unit mass,
(ii) which is cheap and easily available,
(iii) which is easy to store and transport,
(iv) which is safe to handle and use.
Solution 10
A non-renewable source of energy is defined as the source of energy which has accumulated in nature over a very, very long time and cannot be quickly replaced when exhausted.
Ex. Coal, petroleum etc.
Solution 11
A renewable source of energy is the one which is being produced continuously in nature and is inexhaustible.
Ex. wind energy, ocean thermal energy etc.
Solution 12
Renewable source of energy can be used again and again endlessly, where as non-renewable source of energy cannot be renewed once used.
Ex. Renewable sources of energy are wind energy, ocean energy.
Non-renewable sources of energy are coal, fossil fuel.
Solution 13
Fossil fuels are known as non-renewable sources of energy because fossil fuels once used cannot renewed or regained.
Solution 14
Air and water, because both air and water can be used again and again endlessly, they never get exhausted.
Solution 15
Petroleum and coal are non-renewable sources of energy because they cannot be used again once exhausted.
Solution 16
(a) Renewable source of energy - wind, tides, wood
Non- Renewable source of energy - coal, petroleum, natural gas
(b) The above classification is based on the fact that renewable sources are inexhaustible, whereas non-renewable sources are exhaustible.
Solution 17
Coal is a non-renewable source of energy because it has accumulated in the earth over a very. very long time, and if all the coal gets exhausted, it cannot be produced quickly in nature.
Wood is considered as a renewable source of energy because if trees are cut to obtain wood, then more trees will grow in due course of time.
Solution 18
(a) The material which is burnt to produce heat energy is known as a fuel.
Ex. Wood, coal, LPG, kerosene, diesel etc.
(b) Characteristics of ideal fuel:
(i) It should have high calorific value.
(ii) It should burn without giving out any smoke or harmful gases.
(iii) It should have proper ignition temperature.
(iv) It should cheap and easily available.
(c) Fuel A:
i. Lower calorific value of 55 kJ/g (Disadvantage)
ii. Moderate ignition temperature of 80oC (Advantage)
iii. No harmful gases produced (Advantage)
Fuel B:
i. High calorific value of 80 kJ/g (Advantage)
ii. Very low ignition temperature of 10oC (Disadvantage)
iii. Harmgul gases like CO and SO2 produced (Disadvantage)
Sources of Energy Exercise 123
Solution 30
Fuel B is the most ideal fuel because
(i) it leaves no residue on burning
(ii) it has high calorific value of 48 kJ/g
(iii) it does not burn explosively
Solution 31
Fuel Y is a better fuel because
(i) it has a moderate ignition temperature of 75oC.
(ii) it produces no harmful gases like CO on burning.
Solution 32
(i) cooking gas - D
(ii) alcohol - E
(iii) wood - B
(iv) hydrogen - C
(v) kerosene - A
Solution 33
Hydrogen gas>Methane>Petrol>Kerosene>Biogas>Wood
Solution 34
Dung cakes<Coal<Alcohol<Diesel<LPG
Hydrogen gas
Sources of Energy Exercise 130
Diesel
LPG
Kerosene
Solution 4
Coal, oil or gas.
Solution 5
Fractions obtained from petroleum:
Diesel, petrol, kerosene, petroleum gas etc.
Sources of Energy Exercise 131
Solution 6
LPG consists mainly of butane, alongwith smaller amounts of propane and ethane.
Solution 7
Compressed Natural Gas (CNG)
Solution 8
(i) LPG-Liquified Petroleum Gas
(ii) CNG-Compressed Natural Gas
Solution 9
(i) Main constituent of petroleum gas is butane.
(ii) Main constituent of natural gas is methane.
Methane
Solution 11
Uses of natural gas:
(i) As a fuel in thermal power plants
(ii) As a fuel in transport vehicles
Solution 12
CNG is used as a fuel in transport vehicles
butane
Solution 14
Natural gas is considered to be a good fuel because it has a high calorific value, burns with a smokeless flame, causes no air pollution and does not produce any poisonous gas.
Solution 15
The traditional sources of energy which are familiar to most people are called conventional source of energy.
Ex. Wood, coal etc.
Solution 16
In a thermal power plant, heat produced by burning coal is used to boil water to form steam. The steam, at high temperature and pressure, rotates the turbine and its shaft, which drives the generator to produce electricity.
Solution 17
Disadvantages of burning the fossil fuels are:
(i) The burning of fossil fuels produce acidic gases.
(ii) The burning of fossil fuels produce large amount of CO2 gas, which increases greenhouse effect.
(iii) The burning of fossil fuels produce smoke and leaves behind a lot of ash.
Solution 18
Burning of fossil fuels leads to the production of many acidic gases like sulphur-dioxide and nitrogen dioxides. These gases causes acid rain which damages trees, plants and buildings, reduces fertility of soil, and poses danger to aquatic life. The burning of fossil fuels puts a poisonous gas carbon monoxide in air. It also produces large amount of CO2 gas which damages the environment in the long run by increasing the greenhouse effect. Also, burning of fossil fuels produces smoke and a lot of ash.
Solution 19
Pollution caused by the burning of fossil fuels can be controlled by the increasing the efficiency of combustion process and by using various techniques to reduce the escape of harmful gases and ash into the surrounding air.
Solution 20
We will use LPG due to its high calorific value and smokeless flame.
Solution 21
LPG is considered as good fuel because it has a high calorific value gives a smokeless flame.
Solution 22
LPG is considered as a better fuel than coal because it has a higher calorific value, while burning it does not produce any smoke.
Solution 23
For the detection of leakage, a foul smelling substance called ethyl mercaptan is added to the LPG.
In case of LPG leakage in the kitchen, following steps must be taken:
1. The door and windows should be opened at once to allow the gas to escape.
2. The source of gas leakage should be checked and then set right with the help of a gas mechanic.
Solution 24
(a) Natural fuels formed deep under the earth from the pre-historic remains of the organisms (like plants and animals) are called fossil fuels.
Ex. Coal, petroleum and natural gas.
(b) The plants and animals which died millions of years ago and got buried deep in the earth, away from the reach of oxygen, got converted into fossil fuels due to the chemical effects of pressure, heat and bacteria.
(c) Sun is considered to be the ultimate source of fossil fuels because it was the sunlight of long ago that made plants grow and the animals which got buried in the earth also ate plants. So, plants and animals which were originally made by sun's energy only have been converted into fossil fuels.
(d) Petroleum and natural gas
(e) Coal
Sources of Energy Exercise 132
Solution 37
(a) X = Carbon
(b) Another element which is usually found in combination with carbon in fossil fuels is hydrogen.
Solution 38
Petrol is obtained from petroleum, which is a fossil fuel. Fossil fuels were originally made by sun's energy because it was the sunlight of long ago that made plants and animals grow. So, the energy in petrol originally came from the sun.
Solution 39
(a) X is ethyl mercaptan
(b) Ethyl mercaptan has a foul smell that can be detected easily.
Solution 40
(a) Catalytic converter
(b) It converts poisonous carbon monoxide into non-poisonous carbon dioxide and harmful nitrogen oxides into harmless nitrogen gas.
Sources of Energy Exercise 140
Solution 1
Potential energy into electrical energy
Kinetic energy
Solution 3
Kinetic energy of water.
Sun
15 km/hr
Solution 6
Use of wind energy
(a) in the past - in flour mills
(b) at present - for generating the electricity through wind-powered generators.
Solution 7
Copper tube of solar water heater is painted black because black colour is good absorber of heat.
Nuclear fusion
Solution 9
Solar water heater
Solar cooker
Solar cell
Solution 10
Plane mirror reflector
Solar cell
Solution 13
Solar constant is 1.4kW/m2 or 1.4 kJ/s/m2.
Solar energy received by 1 m2 area in 1 s = 1.4 kJ
Solar energy received by 1 m2 area in one hour (or 3600 s) = 1.4 x 3600 = 5040 kJ
Solution 14
sunlight; electrical
Solution 15
(a) Difference between thermal power plant and hydro power plant:
Thermal power plant uses non-renewable sources of energy like coal, oil or gas; whereas hydro power plant uses renewable source of energy i.e. water.
Thermal power plant causes pollution due to the burning of fossil fuels; whereas hydro power plant is environment friendly.
(b) Thermal power plant causes serious air pollution because it emits harmful gases and fly-ash into the air.
Sources of Energy Exercise 141
Solution 16
Sun is a renewable source of energy, where as fossil fuels are non-renewable sources of energy.
Solution 17
Concave and plane mirrors are used for making solar cooker because concave mirror converges a large amount of sun's rays at a point that is required for high heating and plane mirror reflects the rays of light in the form of a strong beam of sunlight on the top of the box that is required for moderate heating.
Solution 18
(a) Glass sheet cover
(b) A plane mirror reflector is used in a box type solar cooker so as to get a strong beam of sunlight after reflection from the mirror.
Solution 19
(a) Advantages of using solar cooker
(i) It saves precious fules like coal, kerosene etc.
(ii) It does not produce any smoke or ash.
(iii) The food cooked in a solar cooker has all its nutrients intact.
(b) Disadvantages of using solar cooker
(i) Solar cooker cannot be used during night time.
(ii) If the day-sky is covered with clouds, it will not be possible to cook the food using solar cooker.
(iii) Direction of reflector has to be changed from time to time to keep it facing the sun.
Solution 20
(a) Solar cell is a device which converts solar energy directly into electrical energy.
(b) Silicon
(c) Use of solar cell:
i. To provide electricity in artificial satellites and space probes.
ii. To provide electricity to remote areas where normal electricity transmission lines do not exist.
iii. To provide electricity to light houses
iv. To operate traffic signals, watches, calculators and toys.
Solution 21
(i) They have no moving parts and require no maintenance.
(ii) They can be set up in remote and inaccessible areas
(i) They are very expensive.
(ii) They are less efficient. They can convert only about 25 % of light falling on them into electricity.
(iii) They cannot work during night time.
Solution 22
Solar cell panel consists of a large number of solar cells joined together in a definite pattern. It is used to convert solar energy into electricity.
Advantages of solar cell panel are:-
1. It provides much more electric power than a single solar cell.
2. It is used to provide electricity in remote and inaccessible rural areas.
Solution 23
(a) The amount of solar energy received per second by one metre square area of the near earth space (exposed perpendicularly to the sun rays) at an average distance between the sun and the earth, is known as solar constant. Its value is 1.4 kJ/s/m2.
(b) Area, A=5m2; time,t=10 min=600sec; E=4200 kJ
Solar constant =E/(Axt)
=4200/(5x600)
=1.4 kJ/s/m2
Solution 24
Traditionally, the energy of flowing water has been used for rotating the water-wheels and for driving water-mills to grind wheat to make flour. The traditional use of energy of flowing water has been modified by improvements in technology and it is now used to generate electricity.
Solution 25
Traditionally, wind energy was used through windmills to pump water from a well and to grind wheat into flour. But the traditional use has been modified and now it is used for the generation of electricity.
Solution 26
(a) The electricity generated from hydro power plant is known as hydro electricity.
Water is collected in a reservoir at a height, so the water has potential energy stored in it. When the water flows down through this large height, its potential energy gets converted into kinetic energy. The fast flowing water rotates the turbine which is connected to generator through its shaft. The generator produces electricity.
(i) It does not cause any environmental pollution.
(ii) It uses the energy of flowing water, which is a renewable source of energy
(i) Large areas of land are required.
(ii) Large eco-systems get destroyed.
Solution 27
(a) Construction and working of solar cooker
A solar cooker consists of an insulated metal box or wooden box which is painted all black from inside. There is a thick glass sheet cover over the box and a plane mirror reflector is also attached to the box.
The food to be cooked is put in metal containers which are painted black from outside. These metal containers are then placed inside the solar cooker box covered with the glass
sheet. The sun's rays fall on the reflector, get reflected into the box through the upper glass and get trapped within it. | 3,283 | 14,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.942282 |
http://jsxgraph.uni-bayreuth.de/wiki/index.php?title=Extended_mean_value_theorem&oldid=6749&diff=prev | 1,600,657,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00354.warc.gz | 74,329,688 | 7,428 | # Difference between revisions of "Extended mean value theorem"
The extended mean value theorem (also called Cauchy's mean value theorem) is: For continuous functions
$f, g: [a,b] \to \mathbb{R}$
that are differentiable on the open interval <mtah>(a,b)[/itex]
### The underlying JavaScript code
var board = JXG.JSXGraph.initBoard('box', {boundingbox: [-5, 10, 7, -6], axis:true});
var p = [];
p[0] = board.create('point', [0, -2], {size:2});
p[1] = board.create('point', [-1.5, 5], {size:2});
p[2] = board.create('point', [1, 4], {size:2});
p[3] = board.create('point', [3, 3], {size:2});
// Curve
var fg = JXG.Math.Numerics.Neville(p);
var graph = board.create('curve', fg, {strokeWidth:3, strokeOpacity:0.5});
// Secant
line = board.create('line', [p[0], p[3]], {strokeColor:'#ff0000', dash:1});
var df = JXG.Math.Numerics.D(fg[0]);
var dg = JXG.Math.Numerics.D(fg[1]);
// Usually, the extended mean value theorem is formulated as
// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y())
// We can avoid division by zero with that formulation:
var quot = function(t) {
return df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X());
};
var r = board.create('glider', [
function() { return fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
function() { return fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
graph], {name: '', size: 4, fixed:true, color: 'blue'});
board.create('tangent', [r], {strokeColor:'#ff0000'}); | 531 | 1,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-40 | latest | en | 0.547789 |
https://gmatclub.com/forum/gmatprep-official-guide-11th-edition-26867.html?fl=similar | 1,511,035,614,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00670.warc.gz | 633,096,164 | 39,683 | It is currently 18 Nov 2017, 13:06
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GMATprep & Official Guide 11th Edition
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Posts: 22
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GMATprep & Official Guide 11th Edition [#permalink]
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22 Feb 2006, 11:30
Is studying this (GMATprep & Official Guide 11th Edition) alone sufficient for gmat preparation?
I have used the Kaplan resources, but think the OG and new prep software is enough study material.
Any recent gmat takers (2006) have any suggestions or advise???
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Joined: 17 Feb 2006
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22 Feb 2006, 18:55
That depends on how good you naturally are at standardized tests. If they come easy enough for you, that should be enough. If you want more prep, then get some additional materials. Princeton Review has 4 tests on the Cracking CD with access to another one online. GMATPrep doesn't come with answer explanations as far as I know (someone correct me if I'm wrong), but the PR tests do and that is extremely helpful.
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27 Feb 2006, 06:31
No question, OG11 and GMATPrep are all you need if you are comfortable with the concepts to start with.
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27 Feb 2006, 06:31
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GMATprep & Official Guide 11th Edition
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 705 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-47 | latest | en | 0.909885 |
https://www.numbersaplenty.com/4875097 | 1,716,688,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00880.warc.gz | 788,280,080 | 3,026 | Search a number
4875097 is a prime number
BaseRepresentation
bin10010100110001101011001
3100011200101011
4102212031121
52222000342
6252253521
756303053
oct22461531
910150334
104875097
11282a807
1217712a1
131018c96
1490c8d3
15664717
hex4a6359
4875097 has 2 divisors, whose sum is σ = 4875098. Its totient is φ = 4875096.
The previous prime is 4875071. The next prime is 4875109. The reversal of 4875097 is 7905784.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4756761 + 118336 = 2181^2 + 344^2 .
It is a cyclic number.
It is not a de Polignac number, because 4875097 - 211 = 4873049 is a prime.
It is not a weakly prime, because it can be changed into another prime (4875037) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2437548 + 2437549.
It is an arithmetic number, because the mean of its divisors is an integer number (2437549).
Almost surely, 24875097 is an apocalyptic number.
It is an amenable number.
4875097 is a deficient number, since it is larger than the sum of its proper divisors (1).
4875097 is an equidigital number, since it uses as much as digits as its factorization.
4875097 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 70560, while the sum is 40.
The square root of 4875097 is about 2207.9621826472. The cubic root of 4875097 is about 169.5616967501.
The spelling of 4875097 in words is "four million, eight hundred seventy-five thousand, ninety-seven". | 503 | 1,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.87815 |
https://halotools.readthedocs.io/en/latest/api/halotools.mock_observables.underdensity_prob_func.html | 1,709,353,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00863.warc.gz | 287,255,052 | 5,894 | # underdensity_prob_func¶
halotools.mock_observables.underdensity_prob_func(sample1, rbins, n_ran=None, random_sphere_centers=None, period=None, sample_volume=None, u=0.2, num_threads=1, approx_cell1_size=None, approx_cellran_size=None, seed=None)[source]
Calculate the underdensity probability function (UPF), $$P_U(r)$$.
$$P_U(r)$$ is defined as the probability that a randomly placed sphere of size $$r$$ encompases a volume with less than a specified number density.
See the Formatting your xyz coordinates for Mock Observables calculations documentation page for instructions on how to transform your coordinate position arrays into the format accepted by the sample1 argument.
Parameters:
sample1array_like
Npts1 x 3 numpy array containing 3-D positions of points. See the Formatting your xyz coordinates for Mock Observables calculations documentation page, or the Examples section below, for instructions on how to transform your coordinate position arrays into the format accepted by the sample1 and sample2 arguments. Length units are comoving and assumed to be in Mpc/h, here and throughout Halotools.
rbinsfloat
size of spheres to search for neighbors Length units are comoving and assumed to be in Mpc/h, here and throughout Halotools.
n_ranint, optional
integer number of randoms to use to search for voids. If n_ran is not passed, you must pass random_sphere_centers.
random_sphere_centersarray_like, optional
Npts x 3 array of randomly selected positions to drop down spheres to use to measure the void_prob_func. If random_sphere_centers is not passed, n_ran must be passed.
periodarray_like, optional
Length-3 sequence defining the periodic boundary conditions in each dimension. If you instead provide a single scalar, Lbox, period is assumed to be the same in all Cartesian directions. If set to None, PBCs are set to infinity, in which case sample_volume must be specified so that the global mean density can be estimated. In this case, it is still necessary to drop down randomly placed spheres in order to compute the UPF. To do so, the spheres will be dropped inside a cubical box whose sides are defined by the smallest/largest coordinate distance of the input sample1. Length units are comoving and assumed to be in Mpc/h, here and throughout Halotools.
sample_volumefloat, optional
If period is set to None, you must specify the effective volume of the sample. Length units are comoving and assumed to be in Mpc/h, here and throughout Halotools.
ufloat, optional
density threshold in units of the mean object density
number of ‘threads’ to use in the pair counting. if set to ‘max’, use all available cores. num_threads=0 is the default.
approx_cell1_sizearray_like, optional
Length-3 array serving as a guess for the optimal manner by how points will be apportioned into subvolumes of the simulation box. The optimum choice unavoidably depends on the specs of your machine. Default choice is to use max(rbins) in each dimension, which will return reasonable result performance for most use-cases. Performance can vary sensitively with this parameter, so it is highly recommended that you experiment with this parameter when carrying out performance-critical calculations.
approx_cellran_sizearray_like, optional
Analogous to approx_cell1_size, but for used for randoms. See comments for approx_cell1_size for details.
seedint, optional
Random number seed used to randomly lay down spheres, if applicable. Default is None, in which case results will be stochastic.
Returns:
upfnumpy.array
len(rbins) length array containing the underdensity probability function $$P_U(r)$$ computed for each $$r$$ defined by input rbins.
Notes
This function requires the calculation of the number of pairs per randomly placed sphere, and thus storage of an array of shape(n_ran,len(rbins)). This can be a memory intensive process as this array becomes large.
Examples
For demonstration purposes we create a randomly distributed set of points within a periodic unit cube.
>>> Npts = 10000
>>> Lbox = 1.0
>>> period = np.array([Lbox,Lbox,Lbox])
>>> x = np.random.random(Npts)
>>> y = np.random.random(Npts)
>>> z = np.random.random(Npts)
We transform our x, y, z points into the array shape used by the pair-counter by taking the transpose of the result of numpy.vstack. This boilerplate transformation is used throughout the mock_observables sub-package:
>>> coords = np.vstack((x,y,z)).T
>>> rbins = np.logspace(-2,-1,20)
>>> n_ran = 1000
>>> upf = underdensity_prob_func(coords, rbins, n_ran=n_ran, period=period, u=0.2) | 1,030 | 4,568 | {"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": 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-2024-10 | latest | en | 0.736818 |
https://careercup.com/question?id=5028453673861120 | 1,620,875,079,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00530.warc.gz | 188,482,919 | 9,010 | ## Flipkart Interview Question for Senior Software Development Engineers
Team: digital
Country: India
Interview Type: Phone Interview
Comment hidden because of low score. Click to expand.
0
of 0 vote
One way is to do in order traversal of the tree and look for count inversions, means whether the next number is smaller than the other. O(nlogn)
Comment hidden because of low score. Click to expand.
0
of 0 vote
We will do this by doing a pre order traversal by the tree and maintaining of stack of conditions at each node. i.e. if we are traversing the left tree then we add a condition (< value of node) and if right then condition (>= value of node). These are the conditions that a BST satisfies. For each node check which conditions is doesn't satisfy in the stack and print the pair for which it doesn't satisfy the condition. Running time n nodes and logn comparisons for each node no n*log(n)
Comment hidden because of low score. Click to expand.
0
Its complexity will be O(n^2), not O(n*log(n)).
Suppose, you start from root node, then to check that it follows BST property, we have to make sure all of the nodes in its left sub-tree are smaller than root. And all nodes in right sub-tree are greater or equal to it. So effectively we are comparing root's values with n-1 nodes. Therefore, its complexity would be O(n^2).
Comment hidden because of low score. Click to expand.
0
of 0 vote
You can solve in O(n*logn*logn), which is even better than O(n^2):
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;
namespace PrintWrongPairsInBST
{
class Program
{
static void Main(string[] args)
{
//Tree:
// 7
// 2 19
// 8 5 12 17
// 13 21
// 46
//Which isnt sorted
Tree t = new Tree(7);
t.left = new Tree(2);
t.right = new Tree(19);
t.left.left = new Tree(8);
t.left.right = new Tree(5);
t.right.left = new Tree(12);
t.right.right = new Tree(17);
t.left.right.left = new Tree(13);
t.right.right.left = new Tree(21);
t.right.right.left.left = new Tree(46);
Process(t,
}
static void Process(Tree t,
{
if (t == null) //Very base case
{
return;
}
if (t.left == null && t.right == null) //Leaf
{
if (value.Count > 0)
{
int[] pathValue = new int[value.Count + 1];
int index = 0;
foreach (int v in value)
{
pathValue[index++] = v;
}
pathValue[index++] = t.info;
char[] pathDirection = new char[direction.Count + 1];
index = 0;
foreach (char d in direction)
{
pathDirection[index++] = d;
}
pathDirection[index++] = 'U';
ProcessPath(pathValue,
pathDirection);
}
return;
}
//Induction
Process(t.left,
value,
direction);
value.RemoveLast();
direction.RemoveLast();
Process(t.right,
value,
direction);
value.RemoveLast();
direction.RemoveLast();
}
static void ProcessPath(int[] value,
char[] direction)
{
for (int i = value.Length - 1; i >= 0; i--)
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1 The Costs of Producing to Mass Markets Variable Costs, Fixed Costs and Minimising Them
2 Mass Market Decisions Previous lectures: negotiations for sale Mass markets: decide on many units to sell at a given posted price. Can t individually negotiate each sale. Must decide how much to produce: nature of costs and technology becomes important.
3 Can Less Be More? Akio Morita, founder of Sony, attempted to export a small transistor radio to the US (1955). He found that demand was very high (upwards of 100,000 units) but Sony did not have the capacity to produce that much.
4 A U in Your Cost Curve Morita explained: Our capacity was less than a thousand radios a month. If we got an order for one hundred thousand, we would have to hire and train new employees and expand our facilities even more. This would mean a major investment, a major expansion, and a gamble. I sat down and drew a curve that looked something like a lopsided letter U. The price for five thousand would be our regular price. That would be the beginning of the curve. For ten thousand there would be a discount, and that was the bottom of the curve. For thirty thousand the price per unit would begin to climb. For fifty thousand the price per unit would be higher than for five thousand, and for one hundred thousand units the price would have to be much more per unit than for the first five thousand
5 The Law of Diminishing Returns Morita had run into the law of diminishing returns: When you have some fixed inputs into production (such as a plant), eventually expanding output raises the marginal (or per unit) cost of production. Increasing output beyond 10,000 a year meant hiring more workers, paying overtime, crowded floor space, higher prices to suppliers etc.
6 Some Cost Concepts Total costs: all of your costs of production Fixed costs: your costs even if you do not produce any output Variable costs: your costs that are dependent on your output level
7 Average versus Marginal Cost Average Total Cost (ATC) = Total Cost/Quantity = TC/Q Marginal Cost (MC) = (Change in total cost)/(change in quantity) = TC/ Q Suppose TC(Q) = F + cq. What are ATC, AVC and MC?
8 The Shape of Typical Cost Curves MC Cost (\$ s) Quantity
9 The Shape of Typical Cost Curves MC ATC Cost (\$ s) Quantity
10 The Shape of Typical Cost Curves MC ATC Cost (\$ s) AVC Quantity
11 The Shape of Typical Cost Curves MC ATC Cost (\$ s) AVC AFC Quantity
12 The Relationship Between Marginal Cost and Average Total Cost When marginal cost is less than average total cost, average total cost is falling. MC < ATC ATC When marginal cost is greater than average total cost, average total cost is rising. MC > ATC ATC
13 The Relationship Between Marginal Cost and Average Total Cost The marginalcost curve crosses the averagetotal-cost its minimum point. Why?
14 The Relationship Between Marginal Cost and Average Total Cost MC ATC Cost (\$ s) The marginal cost curve always crosses the average total cost curve at the minimum average total cost! Quantity
15 Why the Average Total Cost Curve is U-Shaped There are two opposing forces that guarantee the short-run average total cost curve will be U-shaped: Decreasing average fixed cost Eventually increasing average variable cost caused by diminishing returns The shape of the ATC curve combines these two effects.
16 Why didn t Sony expand? Doubling capacity was a risky endeavour. Morita was uncertain that the demand for repeat business would be high enough. Expanding too quickly, Sony would have sacrificed quality and efficiency and perhaps harmed its long-run reputation.
17 Adjustment in the Long- Run Fixed inputs into production can be adjusted in the long-run. Choose optimal size or number of plants.
18 Returns to Scale A change in scale occurs when there is an equal percentage change in the use of all the firm s inputs. Suppose we double the quantities of labour and capital used in production, doubling the scale of the firm. How much will output increase? Constant returns to scale Increasing returns to scale
19 Constant Returns to Scale Constant returns to scale occur when the percentage increase in output equals the percentage increase in inputs. Doubling all inputs causes output to exactly double.
20 Increasing Returns to Scale Increasing returns to scale occur when the percentage increase in output exceeds the percentage increase in inputs. Doubling all inputs causes output to more than double.
21 Decreasing Returns to Scale Decreasing returns to scale occur when the percentage increase in output is less than the percentage increase in inputs. Doubling all inputs causes output to less than double.
22 Economies and Diseconomies of Scale Economies of scale are present when, as output increases, longrun average cost decreases. Diseconomies of scale are present when, as output increases, longrun average cost increases. Long-run average cost curves usually exhibit all three types of returns to scale.
23 The Shape of Long-Run Cost Curves U-Shaped Long-Run Average Cost: Economies of Scale: occurs when a firm has large fixed costs. ATC declines as output increases Diseconomies of Scale: occurs when some important input is limited. ATC rises as output increases
24 The Shape of Long-Run Cost Curves U-Shaped Long-Run Average Cost: The bottom of the U-Shape occurs at the quantity that minimizes average total cost This is called the Efficient Size of the firm.
25
26 Costs at the Margin Critical to any cost decision are the notions of incremental or marginal cost How much will it cost me to produce one additional unit of my product, or to supply one more unit of my service?
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Unit III: The Costs of Production & Theory of the Firm CHAPTERS 13-17 First, lets review marginal returns How many workers should you hire? Remember rationale thinkers think marginally!! Marginal = 1 additional
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CHAPTER 8 Production and Cost START UP: STREET CLEANING AROUND THE WORLD It is dawn in Shanghai, China. Already thousands of Chinese are out cleaning the city s streets. They are using brooms. On the other | 7,491 | 32,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.947212 |
https://es.mathworks.com/matlabcentral/answers/143622-if-then-with-a-range | 1,669,817,079,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00302.warc.gz | 280,543,465 | 27,463 | # If-then with a range
3 views (last 30 days)
son on 28 Jul 2014
Commented: Ben11 on 28 Jul 2014
M is from 1 to 10 ( 1,2,3....,10)
if M is odd then N = M + 1 else N = M + 2
I created this file but the answer is wrong
close all
clear all
M=1:1:10;
if (mod(M,2)==1)
N=M+1;
else
N=M+2;
end
Matlab give N = 3 4 5 6 7 8 9 10 11 12
but it should be
N = 2 4 4 6 6 8 8 10 10 12
Ben11 on 28 Jul 2014
Edited: Ben11 on 28 Jul 2014
You're almost there!
clear
clc
M = 1:10;
N = zeros(1,length(M));
for k = 1:length(M)
if mod(M(k),2) == 1
N(k) = M(k)+1;
else
N(k) = M(k)+2;
end
end
N
N =
2 4 4 6 6 8 8 10 10 12
Ben11 on 28 Jul 2014
sum(N) should do it
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Translated by | 361 | 913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.819203 |
https://noanimalpoaching.org/and-pdf/113-hyperbola-questions-and-answers-pdf-203-186.php | 1,632,123,756,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00603.warc.gz | 471,312,047 | 6,966 | # Hyperbola Questions And Answers Pdf
File Name: hyperbola questions and answers .zip
Size: 2013Kb
Published: 10.06.2021
Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:.
As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently.
## Hyperbola Problems
Conics Worksheet 3 Hyperbolas Answers. Identifying the Type of Conic from an Equation or a Picture i. Sherman Hall, B Wing, Room Worksheets have become an integral part of the education system. It is often about dealing with customers or clients who are upset, or annoyed, or downright livid.
Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:. Calculate the equation of the hyperbola with a transverse axis of 8 and a focal length of Find its equation. Calculate the equation of the hyperbola centered at 0, 0 whose focal length is 34 and the distance from one focus to the closest vertex is 2. Determine the equation of the hyperbola centered at 0, 0 that passes through the points: and. Determine the equation of the hyperbola centered at 0, 0 knowing that one focus is 2 units from one vertex and 50 from the other. A rectangular hyperbola passes through the point.
Find the coordinates of the centre, lengths of the axes, eccentricity, latus-rectum, coordinates of foci and vertices, equations of the directrices of the hyperbola. Get familiar with the important concepts of RD Sharma Class 11 Solutions - Chapter 27 Hyperbola including formula of hyperbola, equation and foci with the help free video tutorials by Doubtnut. Hyperbola is the chapter twenty-seven in RD Sharma text book for mathematics. The solutions are in video tutorial and PDF format. It will help you cover the chapter quickly and in a shorter time frame. The exercise wise solutions are further drawn together in such a way that it will make it easy for you to separately select the topic of their choice.
## Conics Worksheet 3 Hyperbolas Answers
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These solutions for Hyperbola are extremely popular among Class 11 Science students for Math Hyperbola Solutions come handy for quickly completing your homework and preparing for exams. Find the equation of the hyperbola. Draw PM perpendicular to the directrix. Find the eccentricity of the hyperbola, the length of whose conjugate axis is 3 4 of the length of transverse axis. If P is any point on the hyperbola whose axis are equal, prove that SP.
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Solo el escroto. Он с трудом сдержал улыбку. - Только лишь мошонка. Офицер гордо кивнул: - Да.
Он оглядел пустой зал.
#### Exercise 1
Зеленоватое, оно было похоже на призрак. Это было лицо демона, черты которого деформировали черные тени. Сьюзан отпрянула и попыталась бежать, но призрак схватил ее за руку. - Не двигайся! - приказал. На мгновение ей показалось, что на нее были устремлены горящие глаза Хейла, но прикосновение руки оказалось на удивление мягким. Это был Стратмор. | 1,076 | 4,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.907841 |
https://www.physicsforums.com/threads/lipschitz-functions.317079/ | 1,508,458,657,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00672.warc.gz | 962,525,513 | 16,728 | # Lipschitz functions
1. May 30, 2009
### Carl140
1. The problem statement, all variables and given/known data
1. Let 0 < a < b <= 1. Prove that the set of all Lipschitz functions of order
b is contained in the set of all Lipschitz functions of order a.
2. Is the set of all Lipschitz functions of order b a closed subspace of those
of order a?
2. Relevant equations
I know that a function f: [a,b] -> R is Lipschitz of order a if there exists a constant K
such that |f(x) - f(y)| <= K |x-y|^a and for all x,y in [a,b].
3. The attempt at a solution
Assume f is a Lipschitz function of order b then there exists some constant K such that
|f(x)-f(y)|<= K |x-y|^b. Then I need to prove that we can find some constant say C
such that |f(x) - f(y)| <= C |x-y|^a , where 0 < a < b=1.
2. May 30, 2009
### slider142
Fix a and b. Have you considered the equality K|x - y|b = K|x - y|a|x - y|b - a ?
3. May 30, 2009
### Carl140
So |f(x)-f(y)|<= K |x-y|^b implies |f(x)-f(y)|<= K |x-y|^a |x-y|^(b-a).
Therefore: |f(x) - f(y)| /|x-y|^a <= |x-y|^(b-a)
But x and y are both in [a,b] so |x-y| <= |x|+|y| = b + b = 2b.
Therefore |f(x)-f(y)|/|x-y|^a <= (2b)^(b-a).
So our constant C is then (2b)^(b-a). Is this OK?
How to show the closedness part? I know I have to take a sequence and show its closed
under the limit but really I have no clue how to proceed.
4. May 30, 2009
### slider142
What happened to K?
The first equality does not make sense, x and y are variables. Add some more rigor to your statements.
5. May 30, 2009
### Carl140
OK, thanks again.
My try:
Since f is Lipschitz of order b then there exists a constant K >0 such that
for all x, y in [a,b] we have: |f(x)-f(y)|<= K |x-y|^b.
Observe K|x-y|^b = K|x-y|^a |x-y|^(b-a).
Therefore |f(x)-f(y)| <= K |x-y|^a |x-y|^(b-a) and thus:
|f(x)-f(y)|/|x-y|^a <= K |x-y|^(b-a).
Since x, y are points in [a,b] then |x-y|^(b-a) <= (2b)^(b-a).
Therefore |f(x)-f(y)|/|x-y|^a <= K (2b)^(b-a) and hence:
|f(x)-f(y)|<= K (2b)^(b-a) |x-y|^a so f is Lipschitz of order a with constant
C = K (2b)^(b-a).
OK?
6. May 30, 2009
### HallsofIvy
Staff Emeritus
You appear to be using a and b with two different meanings here. I think more important is that since a< b< 1, 0< b-a< 1.
7. May 30, 2009
### Carl140
Halls: Sorry, I do not follow your hint/suggestion, what do you mean?
8. May 30, 2009
### slider142
Two things that Hall mentioned:
1) You are denoting the endpoints of the closed interval with the same constants you are using to denote the exponents of the Lipschitz inequality. The two are not related; if necessary use different letters, like [c, d].
2) The fact that 0 < b-a < 1 is important.
9. May 30, 2009
### Carl140
Gotcha guys.
|f(x)-f(y)| / |x-y|^a<= K |x-y|^(b-a).
Since 0 < b-a< 1 then |x-y|^(b-a) < |x-y|.
Actually I meant: then |f(x)-f(y)|/|x-y|^a <= K |x-y|^a |x-y|. But x,y are both points in [c,d] so |x-y| <= d.
Thus we get |f(x)-f(y)| <= K |x-y|^a * d so the constant is C = k*d, correct?
Last edited: May 30, 2009
10. May 31, 2009
### snipez90
Well since a and b are endpoints, and x,y are any points in [a,b], the inequality |x-y| =< |b -a| holds, so I don't think you need any extra variables beyond what's given in the problem statement. | 1,157 | 3,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-43 | longest | en | 0.81339 |
https://classroom.thenational.academy/lessons/identifying-fractions-of-quantity-and-shape-part-1-ccu32t?activity=exit_quiz&step=4 | 1,632,159,558,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00234.warc.gz | 236,881,525 | 35,253 | # Identifying fractions of quantity and shape (Part 1)
In this lesson, we will find one half, one quarter and one third of a rectangle by counting the squares within the shape.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Half of 8 is less than half of 16.
1/3
Q2.One quarter of 8 is the same as half of 16.
2/3
Q3.One quarter of 12 is the same as one third of nine.
3/3
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Half of 8 is less than half of 16.
1/3
Q2.One quarter of 8 is the same as half of 16.
2/3
Q3.One quarter of 12 is the same as one third of nine.
3/3
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Identifying unit fractions of a quantity and shape
Q1.What fraction of the shape is shaded?
1/3
Q2.What fraction of the shape is shaded?
2/3
Q3.What fraction of the shape is shaded?
3/3
Quiz:
# Identifying unit fractions of a quantity and shape
Q1.What fraction of the shape is shaded?
1/3
Q2.What fraction of the shape is shaded?
2/3
Q3.What fraction of the shape is shaded?
3/3
# Lesson summary: Identifying fractions of quantity and shape (Part 1)
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Dance | 520 | 1,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-39 | longest | en | 0.910142 |
http://www.herongyang.com/crypto/des_encryption_operation_modes_3.html | 1,726,123,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00842.warc.gz | 42,050,314 | 3,441 | Cryptography Tutorials - Herong's Tutorial Notes
Dr. Herong Yang, Version 4.00
This site Web DES Algorithm - Operation Modes and JCE SUN Implementation Part: 1 2 3 (Continued from previous part...) ``` KeySpec ks = new DESKeySpec(theKey); SecretKeyFactory kf = SecretKeyFactory.getInstance("DES"); SecretKey ky = kf.generateSecret(ks); Cipher cf = Cipher.getInstance(algorithm); if (theIVp == null) { cf.init(Cipher.ENCRYPT_MODE, ky); } else { AlgorithmParameterSpec aps = new IvParameterSpec(theIVp); cf.init(Cipher.ENCRYPT_MODE, ky, aps); } byte[] theCph = cf.doFinal(theMsg); System.out.println("Key : "+bytesToHex(theKey)); if (theIVp != null) { System.out.println("IV : "+bytesToHex(theIVp)); } System.out.println("Message : "+bytesToHex(theMsg)); System.out.println("Cipher : "+bytesToHex(theCph)); System.out.println("Expected: "+bytesToHex(theExp)); } catch (Exception e) { e.printStackTrace(); return; } } public static byte[] hexToBytes(String str) { if (str==null) { return null; } else if (str.length() < 2) { return null; } else { int len = str.length() / 2; byte[] buffer = new byte[len]; for (int i=0; i
Dr. Herong Yang, updated in 2007
Cryptography Tutorials - Herong's Tutorial Notes - DES Algorithm - Operation Modes and JCE SUN Implementation | 349 | 1,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.403395 |
https://theracingforum.co.uk/forums/topic/speedratings/page/2/ | 1,610,751,530,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00295.warc.gz | 604,096,656 | 26,382 | The home of intelligent horse racing discussion
Speedratings
Home Forums Archive Topics Speedratings
Viewing 13 posts - 16 through 28 (of 28 total)
• Author
Posts
• #99165
Daylight
Member
• Total Posts 369
My speed ratings are a little bit more basic and the backbone of the program although there are a hundred other things that have been added and disregarded over the years the main platform has never really changed much. It is my own design and one that has served me well over the last 3 seasons over distances 5 furlongs to a mile in competative handicaps:<br>
• <br>
• Expected Time in Seconds = (Standard Time + Going Allowance)
• Lbs in One second = ((30 / race distance) x (15 / race distance))
• Expected time in Lbs = (Expected Time in Seconds x Lbs in One second )
• Actual Winners Time in Lbs = (Winners time x Lbs in One second)
• Beaten by winner in Lbs = (distance beaten x (15 / race distance))
• Actual time in Lbs = (Actual Winners Time in Lbs + Beaten by winner in Lbs)
• Raw rating = ((Expected time in Lbs – Actual time in Lbs) + weight carried in Lbs)<br>
<br>That’s the calculation from what I can remember – it may look daunting and complicated but it isn’t really. This was all done from memory so don’t quote me on this as I may have put a wrong figure in somewhere along the lines, sorry can’t be more exact without searching my program for a few hours (which I haven’t got right now).<br>The going allowances I can’t remember but I think if it’s heavy going I add 11% of the standard time on.
#99167
guskennedy
Member
• Total Posts 759
To paul101…the following is an extract from something I wrote a few years ago explaining how I calculate a going allowance:
"How is the going allowance calculated for any particular day’s racing at any particular track? The first step is to sub-divide the returns into individual ‘courses’ at the track e.g. the straight course and the round course at Doncaster, the sprint course and the round course at Sandown etc. Don’t assume that the going allowance on each of these ‘courses’ will be the same on any given day. Next, carry out the calculations previously set out. This leaves you with a ‘raw’ timerating based on a theoretical ‘nil’ going allowance.
As an example on 16th July 1994 there were five races on the straight course at Newmarket ( July course). A ‘nil’ allowance would produce the following timeratings: Jawaal (official rating 73) 129, Lion’s Mane (no official rating) 100, Bintalshaati (official rating 88) 129, Princess Oberon (official rating 70) 96 and Loyalize (no official rating) 120. Remember that there is a 14lb difference between the two scales so normally Jawaal could be expected to produce in a truly-run race a timerating in the region of 87. A timerating of 129 would represent improvement of 42 lbs! Although Jawaal was certainly a lightly-raced and improving colt at that time that degree of improvement is just not feasible and so we can clearly see that conditions on that day were fast and a ‘minus’ going allowance is indicated but to what degree?
Esentially what is involved at this point is exercising a degree of judgement based on experience, looking at the timeratings which would be produced by various going allowances and reaching a ‘best fit’ decision i.e. which going allowance best fits what we know about the horses which raced on this particular day? Anyone who remembers slide-rules will understand what I mean when I say that I am in effect sliding a cursor along the range of possible options to come up with as accurate an allowance as possible."
[to be continued]
#99168
guskennedy
Member
• Total Posts 759
Paul101…it’s going to take me too long to retype everything I typed all those years ago so I’ll change tack by analysing Saturday’s Doncaster card from a time viewpoint.
It’s actually quite a good example as there was no rain during the meeting and six of the seven races were over the straight course and we’ll concentrate on those.
In race order, the winners, their ages, weights carried, official ratings and times relative to standard were as follows:-
1.50 (5f) Intellibet One 2 7-13 (no OR) 6.49 secs slow<br>2.20 (8f) Ellen Mooney 3 8-6 (OR 78) 10.59 secs slow<br>3.35 (8f) Zucchero 6 8-13 (OR 91) 8.94 secs slow<br>4.10 (6f) Falcon Hill 3 8-6 (OR 108) 5.26 secs slow<br>4.40 (6f) Lady of Gdansk 3 8-9 (OR 55) 7.24 secs slow<br>5.15 (7f) Lingo 3 7-11 (OR 60) 8.02 secs slow
Do the calculations set out previously (i.e. with no allowance for the going) and the basic timeratings are IO 21, EM 14, Z 20, FH 47, LoG 23, L 18.
These are obviously far too slow and we know the ground was soft anyway and so some adjustment by way of a going allowance is clearly needed.
Assume that the horses were slowed down by the conditions to the tune of 0.70 seconds per furlong and re-do the calculations. In other words, treat Intellibet One as having run 2.99 seconds slower than standard (i.e. 6.49 minus [5 x 0.70] seconds), Ellen Mooney as having run 4.99 seconds slower (i.e. 10.59 minus [8 x 0.70] seconds) etc. The adjusted timeratings are IO 80, EM 72, Z 78, FH 106, LoG 81 and L 76. Is 0.70 the correct going allowance?
My view is that it isn’t and that the correct going allowance is a bit higher than that. Again, I’m a bit short of time and I’ll have to go into my reasons a bit later on in the week. I’ll also explain the use that can be made of the ratings once they’re finalised.
#99170
paul101
Member
• Total Posts 27
Hi Guskennedy, thanks for all the input, its really interesting, how I would love to be able to calculate time ratings myself, my hobby is reading form, I can spend 20hours or more a week just looking at form,time and pace, and digging deep, allthough I only show a small profit on my punting, I enjoy the time I put in reading, but Iam afraid to make my own time figures would be way way above me, I will be looking forward to your next post.
cheers Paul
#99171
thebairn
Member
• Total Posts 72
Gus / Paul
Very interesting thread – I’ve just checked my going allowance for Doncaster on Saturday, and it came out at 0.69 seconds/furlong.
#99172
buggy
Member
• Total Posts 2
A good bit of reading GUS wanted to do my own speed ratings for a long time and will now have a go<br>hope everybody is fine been away for a while good to be back
#99173
guskennedy
Member
• Total Posts 759
What you end up with is a sort of grid which looks like this:-
<br> 0.60 0.65 0.70 0.75 0.80 0.85
Intellibet One (no OR) 72 76 80 84 88 92<br>Ellen Mooney (OR 78) 64 68 72 76 80 84<br>Zucchero (OR 91) 70 74 78 82 86 90<br>Falcon Hill (OR 108) 98 102 106 110 114 118<br>Lady of Gdansk (OR 55) 73 77 81 85 89 93<br>Lingo (OR 60) 68 72 76 80 84 88
<br>So, if the going allowance is 0.60 seconds per furlong it would produce the timeratings in the first column. If it was 0.65 it would give those in the second column etc.
Bear in mind that there’s a difference of 14lbs between the two scales (see earlier on this thread) so if, for example, the Lincoln was truly-run Zucchero could be expected to record a timerating of about 105. The going allowance would need to be about 1.04 seconds per furlong to produce that timerating for Zucchero. The problem is that that sort of going allowance would produce timeratings for Falcon Hill of about 133 – champion sprinter class – and for Lingo of about 103 and that’s not feasible. It would mean that Lingo had recorded a timerating 29lbs in excess of his handicap mark. Lingo is very probably improving but if you accept as I do that at seven furlongs there are two pounds to a length this would mean that all horses that finished within fourteen and a half lengths of Lingo had shown improved form relative to their handicap marks. Exactly half the field finished within that distance of Lingo – eleven out of twenty two. It’s too many. Conclusion: the Lincoln wasn’t truly-run nor was Ellen Mooney’s race.
Lingo’s race might have been truly-run and so might Lady of Gdansk’s. The latter was a 33/1 shot and was probably showing improved form. Lingo won by two lengths with the third horse a further three and a half lengths away in a big field, distances which often indicate a truly-run race.
It’s impossible to seize on any one race or any one horse through which to "timerate" this Doncaster card. The "best fit" I can come up with is a going allowance of 0.75 seconds per furlong. This allows for both Lingo and Northern Nymph (the first two in the last) having put up performances on time in advance of their handicap marks and it credits Lady of Gdansk with 16lbs improvement on her handicap mark. The timeratings for these two races could in theory have been a little higher but that would have resulted in a corresponding increase in the timeratings of the juveniles in the first race which was, after all, an early season maiden auction event in which many horses finished quite close up.
The problem is that the handicap marks of Lady of Gdansk, Lingo and Northern Nymph are likely to shoot up when the new marks are published tomorrow. I’ll report back then.
#99174
guskennedy
Member
• Total Posts 759
God knows what happened to the grid. It looked all right when I posted it.
I must give up drinking…
#99175
buggy
Member
• Total Posts 2
Gus then how do you work out the rating for the second / third horses etc
#99176
guskennedy
Member
• Total Posts 759
buggy…in answer to your question, it’s simply a case of adjusting the rating according to how far the other horses were beaten and what weight they carried relative to the winner. I allow 3lbs per length at sprint distances, 2lbs between 7f and 10f, 1.5 lbs from 11f-14f and 1lb a length 15f and above. In Lingo’s race, for example, Northern Nymph carried 19lbs more than the winner and was beaten two lengths which equals 4lbs at 7f. Eagles High carried 10lbs more than Lingo and was beaten a total of five and a half lengths (equals 11lbs). Lingo’s timerating was 80, so NN’s was 95 (80+19-4) and EH’s 79 (80+10-11). You have to remember to allow for weight-for-age where the horses are different ages.
The new handicap ratings were published today and Lingo goes up to 69, Northern Nymph to 83 and Lady of Gdansk to 70. From a pure time perspective, Lady of Gdansk is the most interesting as this new rating means that she will be attractively weighted in future handicaps and, as she comes from an unfashionable yard, she is likely to be a decent price. She would also be of interest in any 0-70 rated stakes as she would receive the fillies’ allowance.
#99177
guskennedy
Member
• Total Posts 759
In case anybody is interested, the fastest timerating of the Flat season so far is the 122 recorded by Kyllachy at Newbury on 19th April. I see he’s entered for the Palace House at Newmarket on Saturday.
The official handicapper has underestimated the six-furlong maiden for three-year-olds run at the Craven Meeting. The speedratings for the first four were Indian Country 108, Greenslades 103, Green Line 100 and Faiza 92. Indian Country has been given an official rating of 83. He’s entered in a 7f handicap at HQ on Friday although I’m not sure he’ll be suited by that trip. The second and third haven’t been given a handicap mark yet but Faiza has been left on 69 and she’ll be interesting in ratings-related races as well as handicaps.
The handicapper often gets this race wrong and in the past couple of years San Salvador and Polar Kingdom have got into handicaps on winning marks after running in this race.
#99178
guskennedy
Member
• Total Posts 759
Blowing your own trumpet isn’t the done thing but nobody else is going to do it so here goes…
Of the five horses mentioned above, four have now run. Green Line won a maiden at odds-on last Friday. Kyllachy was backed from 4/1 and won the Palace House and today Faiza won a Brighton handicap at 5/1 (from 7/1), having been beaten in a similar race at Bath last week. She did so from a bad draw and will not go up too much in the handicap. Indian Country is the only non-winner of the four. He was very heavily-backed in a Haydock handicap twelve days ago but flopped. The ground was soft and it had been soft also when he was beaten first time out at odds-on. By contrast, his Newmarket win was on fast ground and I think that’s the key to this horse.
#99179
robgomm
Member
• Total Posts 224
<br>Well done Gus!! You should be TRF’s speed ratings expert, they are obviously working well…i hope you’ve made some money this season :) Rob.
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• You must be logged in to reply to this topic. | 3,410 | 12,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-04 | latest | en | 0.928656 |
https://www.iflscience.com/brain/japanese-mathematician-creates-optical-illusions-nothing-ever-seen-before/ | 1,568,554,549,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00080.warc.gz | 794,008,130 | 16,429 | # This Japanese Professor Creates Optical Illusions Like Nothing You’ve Ever Seen Before
Are this cylinders round or square? Image: The Illusion Contest/Youtube
Internet, it’s time you met Kokichi Sugihara: master mind-muddler, black belt in brain boggling, and optical offender. He’s also a professor of mathematical engineering at Meiji University in Tokyo, and two-time winner of the Illusion of the Year award.
For safety reasons, it’s best to ease yourself into his collection of geometric misdemeanors, so we’ll start with his so-called Ambiguous Cylinders – for which he was awarded second place in the 2016 Illusion of the Year contest.
The secret to this visual riot lies in the fact that the cylinders are in fact a perfect blend between a circle and a square, with one side sloping upwards and the other downwards.
In Sugihara’s own words, “Because the images do not contain information about depth, the brain must guess by entering additional information such as squareness or symmetry,” the Daily Mail reports.
With that in mind, get a load of some of his other illusions, and have your mind added to the long list of those to have been blown by the Japanese genius.
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In practical terms, this arrow that only ever points one way is pretty useless, but as an optical illusion it’s a total banger. Sugihara even wrote a paper explaining how he created it, using what he calls “anomalous mirror symmetry”.
This absolute scandal, entitled Magnet-Like Slopes, won Sugihara his first Illusion of the Year award in 2010.
His next triumph in the annual contest came in 2018, thanks to this work called Triply Ambiguous Objects.
In a document explaining how this illusion was created, Sugihara reveals that “when we see the artwork in a slanted direction, we feel that we are seeing something in the 3D space instead of just seeing a picture facing toward ourselves. In this situation, we are conscious about the direction of the gravity, which is emphasized by the pin standing vertically.”
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Synchronous Machine-I
Synchronous Machine-I
Contents
Recall the Synchronous Machines and Types of Synchronous Machines 1
Describe the working of Synchronous Generator 3
Describe the working of Synchronous Motor 5
Describe the Construction of Synchronous Machines 6
Describe the Rotor Construction of Synchronous Machines 8
Recall the concept of the excitation system and prime mover of synchronous machines 9
Recall the windings and types of Synchronous Machine windings 9
Recall the concept of armature reaction in Synchronous Machines 9
Describe the equivalent circuit of a Cylindrical Rotor Generator 9
Draw the phasor diagram of a cylindrical rotor generator 9
Describe the various Tests performed on Synchronous Generator i. DC Resistance Test ii. Open-Circuit Test iii. Short-Circuit Test iv. Concept of Short-Circuit Ratio 9
List the methods to determine Voltage Regulation and explain the concept of Voltage Regulation 9
Explain the Voltage Regulation by EMF Method 9
Explain the Voltage Regulation by MMF Method 9
Recall the Concept of Zero Power Factor Characteristics or Potier Triangle Characteristics 9
Determine the Voltage Regulation by Zero Power Factor or Potier Triangle 9
Determine the Voltage Regulation by ASA (American Standards Association) Method 9
Recall the conditions to be satisfied for Parallel operation of Synchronous Generators 9
Recall Advantages of Parallel operation of Synchronous Generators 9
Describe the methods of Synchronising the Generators for Parallel Operation 9
Recall the Synchronizing Power of: i. Two identical machines on No-Load, floating with respect to each other ii. Machines connected to Infinite Bus 9
Recall the concept of two machines in Parallel Operation 9
Recall the Synchronous Machines and Types of Synchronous Machines
Learning Outcome:
At the end of this learning outcome, learners should be able to recall the basic concepts of synchronous machines and understand the different types of synchronous machines.
Synchronous Machines:
Synchronous machines are electric machines that operate at synchronous speed, which is the speed at which the rotating magnetic field in the stator of the machine rotates. The speed of the rotating magnetic field is determined by the frequency of the supply voltage and the number of poles in the stator winding.
Synchronous machines can operate either as generators or as motors. When a synchronous machine operates as a generator, it converts mechanical energy into electrical energy, while when it operates as a motor, it converts electrical energy into mechanical energy.
Types of Synchronous Machines:
There are two main types of synchronous machines: synchronous generators and synchronous motors.
1. Synchronous Generators:
Synchronous generators are electric machines that convert mechanical energy into electrical energy. They operate at synchronous speed and are used to generate electrical power in power plants. Synchronous generators are also used in renewable energy systems such as wind turbines and hydroelectric power plants.
Synchronous generators are characterized by their ability to produce a constant voltage at a fixed frequency, which makes them suitable for use in power systems. They are also highly efficient and have a high power factor.
1. Synchronous Motors:
Synchronous motors are electric machines that convert electrical energy into mechanical energy. They operate at synchronous speed and are used in applications that require constant speed such as large industrial fans, pumps, and compressors.
Synchronous motors are characterized by their ability to operate at a constant speed, which makes them suitable for applications that require precise speed control. They are also highly efficient and have a high power factor.
Synchronous motors can be further classified into two types: non-excited synchronous motors and excited synchronous motors. Non-excited synchronous motors have a permanent magnet rotor and do not require a separate excitation source, while excited synchronous motors have a wound rotor and require a separate excitation source.
Example:
An example of a synchronous generator is a hydroelectric power plant. In a hydroelectric power plant, water is used to turn a turbine connected to a synchronous generator. The synchronous generator converts the mechanical energy from the turbine into electrical energy.
An example of a synchronous motor is a large industrial fan used in a manufacturing plant. The synchronous motor operates at a constant speed and provides precise control of the airflow generated by the fan.
In conclusion, synchronous machines are electric machines that operate at synchronous speed and can operate as generators or motors. The two main types of synchronous machines are synchronous generators and synchronous motors, which are characterized by their ability to produce a constant voltage or operate at a constant speed, respectively. Synchronous machines are widely used in power systems and industrial applications due to their high efficiency and precise control capabilities.
Describe the working of Synchronous Generator
Learning Outcome:
At the end of this learning outcome, learners should be able to describe the working of a synchronous generator.
Synchronous Generator:
A synchronous generator is an electrical machine that converts mechanical energy into electrical energy. It operates at synchronous speed and is widely used in power systems to generate electrical power. The basic components of a synchronous generator are the stator, rotor, and excitation system.
Working Principle:
The working principle of a synchronous generator is based on Faraday’s law of electromagnetic induction. According to this law, when a conductor is moved through a magnetic field, a voltage is induced in the conductor. In a synchronous generator, the rotor rotates in a magnetic field produced by the stator windings, which induces a voltage in the stator windings.
The voltage induced in the stator windings is an alternating voltage that varies sinusoidally with time. The frequency of the voltage is determined by the speed of the rotor and the number of poles in the stator winding. The voltage magnitude and phase angle of the induced voltage depend on the magnetic field strength and the position of the rotor relative to the stator.
Excitation System:
The excitation system of a synchronous generator is responsible for producing the magnetic field in the rotor. The magnetic field in the rotor can be produced by either a DC excitation system or an AC excitation system.
In a DC excitation system, a DC current is passed through the rotor winding, which produces a magnetic field. The DC current is supplied by a separate DC source such as a battery or a rectifier.
In an AC excitation system, the rotor winding is connected to a separate AC source, which produces a magnetic field. The AC source can be either a separate generator or a transformer connected to the stator winding of the synchronous generator.
Synchronisation:
Before a synchronous generator can be connected to a power system, it must be synchronised with the system. Synchronisation is the process of matching the frequency, phase angle, and voltage magnitude of the generator with the power system.
During synchronisation, the speed of the generator is adjusted until it matches the frequency of the power system. The phase angle and voltage magnitude of the generator are adjusted until they match the phase angle and voltage magnitude of the power system.
Example:
An example of a synchronous generator is a thermal power plant. In a thermal power plant, steam is used to turn a turbine connected to a synchronous generator. The synchronous generator converts the mechanical energy from the turbine into electrical energy.
In conclusion, a synchronous generator is an electrical machine that converts mechanical energy into electrical energy. Its working principle is based on Faraday’s law of electromagnetic induction. The excitation system is responsible for producing the magnetic field in the rotor, and synchronisation is required before the generator can be connected to a power system. Synchronous generators are widely used in power systems to generate electrical power.
Describe the working of Synchronous Motor
Learning Outcome:
At the end of this learning outcome, learners should be able to describe the working of a synchronous motor.
Synchronous Motor:
A synchronous motor is an AC motor that runs at a constant speed and is synchronised with the frequency of the power system. It is a type of electric motor that operates on the same principles as a synchronous generator. The basic components of a synchronous motor are the stator, rotor, and excitation system.
Working Principle:
The working principle of a synchronous motor is based on the interaction of a magnetic field and a magnetic flux. The stator windings of the motor produce a magnetic field, while the rotor winding is excited by DC current to produce a magnetic flux. The magnetic field produced by the stator and the magnetic flux produced by the rotor interact to produce a torque that causes the rotor to rotate.
The speed of a synchronous motor is determined by the frequency of the power system and the number of poles in the stator winding. The rotor of a synchronous motor must rotate at the same speed as the rotating magnetic field produced by the stator winding. If the rotor speed is different from the speed of the magnetic field, the motor will not produce any torque and will not start.
Excitation System:
The excitation system of a synchronous motor is responsible for producing the magnetic flux in the rotor. The magnetic flux in the rotor can be produced by either a DC excitation system or an AC excitation system.
In a DC excitation system, a DC current is passed through the rotor winding, which produces a magnetic flux. The DC current is supplied by a separate DC source such as a battery or a rectifier.
In an AC excitation system, the rotor winding is connected to a separate AC source, which produces a magnetic flux. The AC source can be either a separate generator or a transformer connected to the stator winding of the synchronous motor.
Example:
An example of a synchronous motor is a hydroelectric power plant. In a hydroelectric power plant, water is used to turn a turbine connected to a synchronous motor. The synchronous motor runs at a constant speed and is synchronised with the frequency of the power system. The synchronous motor drives the generator, which produces electrical power.
In conclusion, a synchronous motor is an AC motor that runs at a constant speed and is synchronised with the frequency of the power system. Its working principle is based on the interaction of a magnetic field and a magnetic flux. The excitation system is responsible for producing the magnetic flux in the rotor. Synchronous motors are widely used in power systems to drive various types of equipment such as pumps, fans, and compressors.
Describe the Construction of Synchronous Machines
Learning Outcome:
At the end of this learning outcome, learners should be able to describe the construction of synchronous machines.
Synchronous Machines:
Synchronous machines are electrical machines that convert mechanical energy into electrical energy or vice versa. They are widely used in power systems for power generation and transmission. The two main types of synchronous machines are synchronous generators and synchronous motors.
Construction:
The construction of a synchronous machine consists of two main parts: the stator and the rotor.
Stator:
The stator is the stationary part of the synchronous machine and consists of a cylindrical frame made of steel laminations. The frame is designed to provide support for the stator core and to provide a path for the magnetic flux. The stator core is made of laminations of high-grade silicon steel and is designed to minimize iron losses due to hysteresis and eddy currents.
The stator winding is made of copper or aluminium conductors and is wound around the stator core. The winding can be either a distributed winding or a concentrated winding. In a distributed winding, the winding is spread over the entire surface of the stator core, while in a concentrated winding, the winding is concentrated in a few slots.
Rotor:
The rotor is the rotating part of the synchronous machine and consists of a cylindrical shaft made of steel laminations. The rotor core is also made of laminations of high-grade silicon steel and is designed to minimize iron losses. The rotor winding is made of copper or aluminium conductors and is wound around the rotor core.
The rotor winding can be either a salient pole winding or a non-salient pole winding. In a salient pole winding, the rotor poles are made of steel laminations and are bolted to the rotor shaft. In a non-salient pole winding, the rotor poles are made of a solid piece of steel and are integral to the rotor shaft.
Excitation System:
The excitation system is responsible for producing the magnetic field in the rotor. The excitation system can be either a DC excitation system or an AC excitation system. In a DC excitation system, a DC current is passed through the rotor winding, which produces a magnetic field. In an AC excitation system, the rotor winding is connected to a separate AC source, which produces a magnetic field.
Examples:
Synchronous generators are used in power plants to generate electrical power. They are also used in wind turbines and hydroelectric power plants to convert mechanical energy into electrical energy.
Synchronous motors are used in industries for driving large machinery such as pumps, compressors, and fans. They are also used in power systems for power factor correction and load balancing.
In conclusion, synchronous machines are electrical machines that convert mechanical energy into electrical energy or vice versa. The construction of a synchronous machine consists of two main parts: the stator and the rotor. The stator is the stationary part of the machine, while the rotor is the rotating part. The excitation system is responsible for producing the magnetic field in the rotor. Synchronous machines are widely used in power systems for power generation and transmission.
Describe the Rotor Construction of Synchronous Machines
Learning Outcome:
At the end of this learning outcome, learners should be able to describe the rotor construction of synchronous machines.
The rotor is the rotating part of the synchronous machine and is responsible for producing the magnetic field that interacts with the stator winding to produce torque. The rotor construction of synchronous machines can be classified into two types: salient pole and non-salient pole rotors.
Salient Pole Rotor:
A salient pole rotor is also known as a projected pole rotor or a spider rotor. It is a type of rotor construction that consists of a number of poles that are bolted to the rotor shaft. The poles are made of laminated steel, and their shape is similar to a salient pole, hence the name.
The rotor poles are designed to produce a strong magnetic field, which is necessary for the efficient operation of the machine. The poles are separated from each other by slots, which are used to hold the rotor winding. The winding is made of insulated copper or aluminum conductors and is wound around the pole faces.
The winding can be either a concentrated winding or a distributed winding. In a concentrated winding, the winding is concentrated in a few slots, while in a distributed winding, the winding is spread over the entire surface of the pole faces.
Non-salient Pole Rotor:
A non-salient pole rotor is also known as a cylindrical rotor or a smooth rotor. It is a type of rotor construction that consists of a solid cylindrical rotor that rotates inside the stator winding. The rotor is made of laminated steel, and its surface is smooth and cylindrical.
The rotor winding is made of insulated copper or aluminum conductors and is embedded in slots that are cut into the surface of the rotor. The winding can be either a distributed winding or a concentrated winding.
The non-salient pole rotor is used in high-speed applications because it has a low inertia, which makes it easy to accelerate and decelerate. It is also used in applications where a smooth and uniform torque is required, such as in electric vehicles and aircraft.
Examples:
Salient pole rotors are commonly used in low-speed applications, such as hydroelectric generators, where the rotor speed is low and the torque is high. They are also used in large turbo generators used in power plants, where the power output is high.
Non-salient pole rotors are commonly used in high-speed applications, such as in aircraft generators and electric vehicles, where the rotor speed is high and the torque is low. They are also used in small- to medium-sized generators and motors where a smooth and uniform torque is required.
In conclusion, the rotor construction of synchronous machines can be classified into two types: salient pole and non-salient pole rotors. Salient pole rotors consist of a number of poles that are bolted to the rotor shaft, while non-salient pole rotors consist of a solid cylindrical rotor with a smooth surface. The rotor winding is made of insulated copper or aluminum conductors and is either a distributed or concentrated winding. Salient pole rotors are commonly used in low-speed applications, while non-salient pole rotors are used in high-speed applications.
Recall the concept of the excitation system and prime mover of synchronous machines
Excitation System:
The excitation system of a synchronous machine is used to supply the direct current to the rotor winding of the generator or motor. The magnetic field produced by the rotor winding interacts with the stator winding and induces voltage or current in the stator winding. The excitation system can be either static or rotating.
Static excitation systems use solid-state components such as diodes, thyristors, and transistors to control the excitation voltage. These systems provide more precise control of the excitation voltage and are more efficient.
Rotating excitation systems use a DC generator mounted on the shaft of the synchronous machine to provide the excitation voltage. The output of the DC generator is connected to the rotor winding through slip rings and brushes. These systems are simpler and less expensive than static systems, but they are less efficient and require regular maintenance.
Prime Mover:
The prime mover of a synchronous machine is the source of mechanical energy that drives the rotor of the machine. The prime mover can be any device that converts mechanical energy into rotational energy, such as a steam turbine, gas turbine, water turbine, or diesel engine.
The choice of prime mover depends on the application and the availability of fuel or energy sources. For example, a hydroelectric power plant may use a water turbine as the prime mover, while a gas-fired power plant may use a gas turbine. In some cases, multiple prime movers may be connected to a single synchronous machine to provide redundancy or backup power.
Overall, the excitation system and prime mover are essential components of synchronous machines that work together to produce electrical power. The excitation system provides the necessary current to the rotor winding to create the magnetic field, while the prime mover supplies the mechanical energy to drive the rotor.
Recall the windings and types of Synchronous Machine windings
Windings:
A synchronous machine has two types of windings, namely stator winding and rotor winding. The stator winding is stationary, while the rotor winding rotates with the rotor. The stator winding is used to produce a magnetic field that interacts with the rotor winding to produce the output voltage or torque.
Types of Synchronous Machine Windings:
There are two types of windings in synchronous machines: salient pole winding and non-salient pole winding.
1. Salient Pole Windings:
A synchronous machine with a salient pole winding has a rotor with poles that are projected outward from the rotor surface. The rotor winding is wound around these poles. Salient pole machines are typically used in applications where low-speed operation is required, such as hydroelectric generators.
1. Non-Salient Pole Windings:
A synchronous machine with a non-salient pole winding has a rotor with smooth cylindrical surface, and the rotor winding is embedded in the rotor slots. Non-salient pole machines are typically used in high-speed applications, such as gas turbine generators.
Both types of windings can have two types of connections: star (Y) and delta (Δ). In a star connection, the winding ends are connected to a common point, while in a delta connection, the winding ends are connected in a closed loop. The choice of connection depends on the application and the desired output voltage.
In summary, synchronous machines have two types of windings, and the type of winding and its connection type are chosen based on the application and the operating conditions of the machine.
Recall the concept of armature reaction in Synchronous Machines
Armature reaction is the magnetic field produced by the stator current that interacts with the main magnetic field produced by the rotor. In a synchronous machine, the stator winding produces a magnetic field that rotates at the synchronous speed. The rotor winding, which is excited by a DC current, produces a magnetic field that interacts with the stator magnetic field to generate the output voltage or torque.
When the machine is loaded, the armature current flows through the stator winding, creating a magnetic field that interacts with the rotor magnetic field. The interaction between these two fields causes a shift in the position of the resultant magnetic field. The direction and magnitude of the shift depend on the power factor of the load and the excitation level of the rotor winding.
If the load power factor is lagging, the armature reaction magnetic field will weaken the rotor field and shift it in the opposite direction of rotation. This results in a decrease in the output voltage and an increase in the armature current. Conversely, if the load power factor is leading, the armature reaction magnetic field will strengthen the rotor field and shift it in the direction of rotation. This results in an increase in the output voltage and a decrease in the armature current.
To compensate for the armature reaction effect, synchronous machines are equipped with an excitation system that regulates the DC current supplied to the rotor winding. This ensures that the rotor magnetic field remains constant, and the output voltage and current are stable under varying load conditions.
In summary, armature reaction is the interaction between the stator magnetic field and the rotor magnetic field in a synchronous machine. It can cause a shift in the position of the resultant magnetic field, affecting the output voltage and current. The excitation system is used to compensate for this effect and maintain stable machine performance under varying load conditions.
Describe the equivalent circuit of a Cylindrical Rotor Generator
The equivalent circuit of a cylindrical rotor generator is a simplified representation of the generator that enables the calculation of the machine’s electrical performance. The equivalent circuit consists of three parts: the armature circuit, the field circuit, and the core losses.
The armature circuit includes the stator winding, the armature resistance, and the armature reactance. The stator winding is modelled as a voltage source, which produces the output voltage, and the armature resistance represents the losses due to the flow of current in the stator winding. The armature reactance represents the inductive reactance of the stator winding.
The field circuit includes the rotor winding, the field resistance, and the field reactance. The rotor winding is excited by a DC current, which produces the magnetic field in the rotor. The field resistance represents the losses due to the flow of current in the rotor winding, and the field reactance represents the inductive reactance of the rotor winding.
The core losses include the hysteresis losses and the eddy current losses. Hysteresis losses occur due to the magnetization and demagnetization of the core material, while eddy current losses occur due to the flow of current in the core material.
The equivalent circuit is used to calculate the machine’s performance under different operating conditions. For example, the output voltage can be calculated by subtracting the voltage drop due to the armature resistance and reactance from the voltage produced by the stator winding. The output power can be calculated by multiplying the output voltage with the output current.
In summary, the equivalent circuit of a cylindrical rotor generator consists of the armature circuit, the field circuit, and the core losses. It is a simplified representation of the generator that enables the calculation of the machine’s electrical performance under different operating conditions.
Draw the phasor diagram of a cylindrical rotor generator
The phasor diagram of a cylindrical rotor generator shows the relationship between the electrical quantities of the stator and rotor windings. The phasor diagram is a graphical representation of the complex numbers used to represent the voltage, current, and impedance of the generator.
Let us assume that,
• Ef=Excitation voltage
• V= Terminal voltage per phase applied to the armature
• Ia=Armature current per phase drawn by the motor from the supply
• Ra= Effective armature resistance per phase
• XS=Synchronous reactance per phase of armature winding
• Cosφ= Power factor
• δ=Torque angle
The voltage equation of a cylindrical rotor synchronous motor is
The phasor diagrams of a 3-phase cylindrical rotor synchronous motor operating at different power factors can be drawn with the help of equation 2.
Phasor Diagram at Lagging Power Factor
The phasor diagram of the synchronous motor operating at a lagging power factor Cosφ is shown in Figure-1.
Consider the synchronous motor is taking a lagging current from the supply. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V. For lagging power factor cosφ, the armature current (Ia) lags behind the supply voltage (V) by an angle φ along OB where OB = Ia.
The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD.
The voltage drop per phase in the synchronous reactance is IaXS .The phasor is in a direction perpendicular to the phasor IaRa and is represented by DA.
Therefore, the phasor V is equal to the phasor sum of Ef, and IaRa. The angle δ between V and Ef is called the torque angle. It plays an important role in the power transfer and the stability of the synchronous motor operation.
Phasor Diagram at Unity Power Factor
The phasor diagram of the synchronous motor operating at unity power factor is shown in Figure-2.
Suppose that the synchronous motor is drawing the current (Ia) from the supply at unity power factor. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V.
For unity power factor, the armature current (Ia) drawn by the motor is in phase with the supply voltage (V) and is represented by OB where OB = Ia.
The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD.
The voltage drop per phase in the synchronous reactance is IaXS. The phasor is IaXS in a direction perpendicular to the phasor IaRa and is represented by DA.
Therefore, the phasor V is equal to the phasor sum of Ef,IaRa and jIaXS.The angle δ between V and Ef is called the torque angle. It plays an important role in the power transfer and the stability of the synchronous motor operation
Phasor Diagram at Leading Power Factor
The phasor diagram of the synchronous motor operating at leading power factor cosφ is shown in Figure-3.
Suppose that the synchronous motor is drawing the current (Ia) from the supply at leading power factor cosφ. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V.
For leading power factor, the armature current(Ia) drawn by the motor leads the supply voltage (V) by the phase angle φ and is represented by OB where OB = Ia.
The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD.
The voltage drop per phase in the synchronous reactance isIaXS. The phasor is IaXS in a direction perpendicular to the phasor IaRa and is represented by DA.
Therefore, the phasor V is equal to the phasor sum of Ef,IaRa and jIaXS.
The phasor diagram of a cylindrical rotor generator provides a visual representation of the electrical quantities and their relationship with each other. It is useful in understanding the behavior of the generator under different operating conditions, such as varying loads or changes in excitation current.
Describe the various Tests performed on Synchronous Generator i. DC Resistance Test ii. Open-Circuit Test iii. Short-Circuit Test iv. Concept of Short-Circuit Ratio
Synchronous generators are essential electrical machines used to produce electrical energy. To ensure the proper functioning and performance of these generators, various tests are conducted before their installation and during their operation. Here are the different tests conducted on synchronous generators:
i. DC Resistance Test:
DC resistance test is performed to measure the resistance of the field winding and armature winding of the generator. The purpose of this test is to ensure that the winding resistance is within the specified range to avoid excessive losses and overheating of the machine. The test is conducted by passing a DC current through the winding and measuring the voltage drop across the winding using a voltmeter and ammeter.
ii. Open-Circuit Test:
The open-circuit test is performed to determine the no-load magnetization characteristics of the synchronous generator. In this test, the generator is run at synchronous speed with the field winding excited at a rated voltage and the armature winding left open-circuited. The field current and voltage are varied to obtain the open-circuit characteristic curve of the generator. This curve helps in determining the voltage regulation and efficiency of the generator.
iii. Short-Circuit Test:
The short-circuit test is performed to determine the copper losses and leakage reactance of the synchronous generator. In this test, the field winding is excited at a rated voltage and the armature winding is short-circuited with the help of a switch. The current is then gradually increased until the generator is operating at its rated current. The voltage drop across the armature winding is measured, and the short-circuit characteristic curve is plotted. The curve helps in determining the impedance and efficiency of the generator.
iv. Concept of Short-Circuit Ratio:
The short-circuit ratio (SCR) of a synchronous generator is defined as the ratio of its open-circuit voltage to its short-circuit current at a given power factor. It is a measure of the ability of the generator to deliver current under short-circuit conditions. The SCR is an important parameter in determining the suitability of a generator for a particular application.
In conclusion, these tests are crucial for determining the performance and efficiency of synchronous generators. The results obtained from these tests help in identifying any faults or defects in the machine, ensuring that the generator operates safely and efficiently.
List the methods to determine Voltage Regulation and explain the concept of Voltage Regulation
Voltage regulation is an important parameter of a synchronous generator that determines the ability of the generator to maintain a stable voltage output under different load conditions. It is defined as the change in voltage from no load to full load expressed as a percentage of the rated voltage. A low voltage regulation indicates that the generator can maintain its output voltage within a narrow range, even under varying load conditions.
There are several methods to determine the voltage regulation of a synchronous generator. Some of the commonly used methods are:
1. Direct Load Test Method: This method involves applying a load to the generator and measuring the terminal voltage at different load conditions. The voltage regulation can be calculated by comparing the no-load voltage and the voltage at full load.
2. EMF (Electromotive Force) Method: This method involves measuring the generated EMF of the generator at no-load and full-load conditions. The voltage regulation can be calculated by dividing the difference between the generated EMFs at no-load and full-load by the rated voltage of the generator.
3. ZPF (Zero Power Factor) Method: This method involves connecting a load to the generator that draws a current at unity power factor. The voltage regulation can be calculated by measuring the voltage drop between no-load and full-load conditions and dividing it by the rated voltage of the generator.
4. Potier Reactance Method: This method involves inserting a reactance in series with the generator and adjusting its value until the generator output voltage at full load is equal to the rated voltage. The voltage regulation can be calculated by dividing the reactance value by the synchronous reactance of the generator.
The voltage regulation of a synchronous generator is an important parameter that affects the stability of the power system. A low voltage regulation indicates that the generator can maintain a stable voltage output under varying load conditions, while a high voltage regulation indicates that the generator may not be able to maintain a stable voltage output and may cause instability in the power system.
Explain the Voltage Regulation by EMF Method
Learning Outcome: ALO:
1. Perform an open-circuit test on the synchronous generator to determine the open-circuit voltage and the field current required to produce it.
2. Calculate the synchronous reactance of the generator using the open-circuit test results.
3. Perform a short-circuit test on the synchronous generator to determine the short-circuit current and the field current required to produce it.
4. Calculate the synchronous impedance of the generator using the short-circuit test results.
5. From the synchronous impedance, synchronous reactance and the resistance of the generator can be calculated.
6. Calculate the voltage drop in the armature resistance at full-load current.
7. Calculate the full-load voltage of the generator.
8. Calculate the voltage regulation using the following formula:
Voltage regulation = (E0 – Vf) / Vf x 100%
where E0 is the open-circuit voltage of the generator, Vf is the full-load voltage of the generator.
The EMF method is widely used as it is relatively easy to perform and gives accurate results. However, it is important to note that the EMF method assumes a constant power factor and neglects the armature reaction effects, which may lead to inaccurate results in certain situations.
In conclusion, the voltage regulation of a synchronous generator is an important parameter, and the EMF method is one of the methods used to determine it. The method is based on the open-circuit and short-circuit tests of the generator, and it assumes a constant power factor and neglects the armature reaction effects.
Explain the Voltage Regulation by MMF Method
The voltage regulation of a synchronous generator is defined as the change in voltage magnitude from no-load to full-load conditions, expressed as a percentage of the rated voltage. The voltage regulation is a critical parameter that determines the ability of a generator to maintain a steady voltage output.
The MMF (magneto-motive force) method is one of the methods used to determine the voltage regulation of a synchronous generator. The method is based on the principle that the voltage regulation of a generator is proportional to the ratio of the armature reaction MMF to the field excitation MMF.
The armature reaction MMF is the demagnetizing effect produced by the armature current on the magnetic field of the rotor. The armature reaction MMF is proportional to the armature current and the synchronous reactance of the machine. The field excitation MMF is the magnetising effect produced by the field current on the rotor magnetic field. The field excitation MMF is proportional to the field current and the field winding turns.
The MMF method involves performing two tests on the generator: the open-circuit test and the short-circuit test. In the open-circuit test, the generator is run at synchronous speed with no load connected to the output terminals. The field current is adjusted until the terminal voltage is equal to the rated voltage. The field excitation MMF is calculated from the field current and the number of field winding turns.
In the short-circuit test, the generator is run at synchronous speed with the output terminals short-circuited. The field current is adjusted until the short-circuit current is equal to the rated armature current. The armature reaction MMF is calculated from the armature current and the synchronous reactance of the machine.
The voltage regulation is then calculated using the following formula:
Voltage regulation = (Eo – Ef) / Ef x 100%
where Eo is the open-circuit voltage and Ef is the voltage across the armature terminals during the short-circuit test.
The MMF method is widely used in industry to determine the voltage regulation of synchronous generators. However, the method assumes a sinusoidal waveform for the armature current, which may not be valid in some cases. Other methods, such as the Z- or impedance method, may be more appropriate in such cases.
Recall the Concept of Zero Power Factor Characteristics or Potier Triangle Characteristics
In electrical power systems, it is essential to maintain the power factor close to unity to avoid energy losses and improve the efficiency of the system. The Zero Power Factor (ZPF) characteristics or Potier triangle characteristics is a graphical representation of the voltage-current phase relationship in synchronous generators. It is used to determine the synchronous generator’s excitation system’s operating point, which ensures that the generator operates at a power factor close to unity.
Potier Triangle:
The Potier triangle is a right-angled triangle, with the hypotenuse representing the apparent power (S) of the generator. The horizontal leg represents the active power (P) generated by the generator, while the vertical leg represents the reactive power (Q) generated by the generator. The angle between the horizontal leg and the hypotenuse is the load angle (δ), which represents the phase difference between the voltage and current in the generator.
Zero Power Factor Characteristics:
The Zero Power Factor (ZPF) characteristics or Potier triangle characteristics is a graphical representation of the generator’s performance concerning the excitation system’s operating point. The ZPF characteristic is obtained by plotting the generator’s current and voltage at zero power factor on the Potier triangle. The zero power factor condition is achieved by overexciting the generator, resulting in a leading power factor.
Importance:
The ZPF characteristics are used to determine the excitation system’s operating point that results in the generator operating at a power factor close to unity. This improves the generator’s efficiency and reduces energy losses in the system.
Conclusion:
The ZPF characteristics or Potier triangle characteristics is a useful tool for determining the excitation system’s operating point in synchronous generators. It ensures that the generator operates at a power factor close to unity, resulting in improved efficiency and reduced energy losses in the system.
Determine the Voltage Regulation by Zero Power Factor or Potier Triangle
Zero power factor characteristics, also known as Potier triangle characteristics, is a graphical method used to determine the voltage regulation of synchronous generators. It is based on the concept that when the generator is loaded with a non-inductive or unity power factor load, the power angle is 0 degrees, and the armature current is in phase with the terminal voltage. The excitation voltage is then adjusted to maintain the terminal voltage at the rated value.
The Potier triangle is constructed by plotting three points on the V-I diagram: the open-circuit voltage at the rated excitation, the short-circuit current at rated excitation, and the load current at zero power factor. These three points are connected by lines to form a triangle. The sides of the triangle represent the open-circuit voltage, the short-circuit impedance, and the load impedance. The angle between the load impedance and the short-circuit impedance is the power angle.
To determine the voltage regulation of the synchronous generator using the Potier triangle, the following steps are followed:
1. Draw the Potier triangle by plotting the open-circuit voltage, short-circuit current, and load current at zero power factor.
2. Measure the load impedance on the Potier triangle by drawing a line from the origin to the load current point and measuring the length of the line.
3. Measure the short-circuit impedance on the Potier triangle by drawing a line from the short-circuit current point to the origin and measuring the length of the line.
4. The ratio of the load impedance to the short-circuit impedance is the voltage regulation.
For example, let’s say we have a synchronous generator rated at 100 MVA, 13.8 kV, and 0.85 power factor. The open-circuit voltage is 13.8 kV, and the short-circuit current at rated excitation is 4 kA. The load current at zero power factor is 65 A. The load impedance on the Potier triangle is measured to be 190 ohms, and the short-circuit impedance is measured to be 0.1 ohms. Therefore, the voltage regulation is:
Load Impedance/Short Circuit Impedance = 190/0.1 = 1900
Voltage Regulation = (Open Circuit Voltage – Load Voltage)/Load Voltage = (13.8 kV – 12.9 kV)/12.9 kV = 6.98%
Therefore, the voltage regulation of the synchronous generator is 6.98%.
Determine the Voltage Regulation by ASA (American Standards Association) Method
The American Standards Association (ASA) method is another method used to determine the voltage regulation of synchronous generators. This method involves plotting the open-circuit characteristic (OCC) and the short-circuit characteristic (SCC) of the generator on the same graph. The ASA method is also known as the Z method.
The OCC is obtained by conducting an open-circuit test on the synchronous generator. In this test, the synchronous generator is driven at its rated speed and the field current is gradually increased until the terminal voltage reaches its rated value. The armature circuit is kept open during the test. The voltage and field current readings are noted and used to plot the OCC.
The SCC is obtained by conducting a short-circuit test on the synchronous generator. In this test, the synchronous generator is driven at its rated speed and the armature circuit is short-circuited through an ammeter. The field current is gradually increased until the rated armature current is reached. The voltage and field current readings are noted and used to plot the SCC.
To determine the voltage regulation of the synchronous generator using the ASA method, the following steps are followed:
1. Plot the OCC and SCC on the same graph, with field current on the x-axis and terminal voltage on the y-axis.
2. Draw a line connecting the origin of the graph and the point where the SCC intersects the x-axis. This line is known as the Z line.
3. Draw a line parallel to the Z line, passing through the point where the OCC intersects the x-axis. This line is known as the A line.
4. Draw a line parallel to the x-axis, passing through the point where the OCC intersects the y-axis. This line is known as the B line.
5. The intersection point of the A and B lines is the operating point of the generator under load conditions.
6. The voltage regulation can be determined by calculating the ratio of the difference between the no-load voltage and the load voltage to the load voltage, as a percentage.
The ASA method is useful in determining the voltage regulation of the generator at different power factors and load conditions. However, this method is not commonly used nowadays due to the availability of more advanced methods, such as the EMF and MMF methods.
Recall the conditions to be satisfied for Parallel operation of Synchronous Generators
Synchronous generators are commonly used to provide electrical power to a power system. In order to meet the increasing demand for electrical power, it is often necessary to connect two or more generators in parallel. Parallel operation of synchronous generators is a complex process that requires careful consideration of various factors, including the operating characteristics of the generators, the load demand, and the control systems.
The conditions to be satisfied for parallel operation of synchronous generators are as follows:
1. Voltage Level: The voltage levels of the generators must be equal. This is necessary to ensure that the loads are supplied with a constant voltage level, and to avoid overloading of any individual generator.
2. Frequency: The frequency of the generators must be the same. This is necessary to ensure that the loads are supplied with a constant frequency, and to avoid phase differences between the generators.
3. Phase Sequence: The phase sequence of the generators must be the same. This is necessary to ensure that the loads are supplied with the correct phase sequence, and to avoid problems such as reverse power flow and excessive voltage drops.
4. Phase Angle: The phase angles of the generators must be close to each other. This is necessary to ensure that the generators do not oppose each other’s voltages and cause a sudden change in the power flow direction.
5. Synchronous Impedance: The synchronous impedance of the generators must be the same. This is necessary to ensure that the generators share the load equally and prevent unequal distribution of reactive power.
6. Power Factor: The power factor of the generators must be the same. This is necessary to ensure that the generators share the load equally and prevent unequal distribution of reactive power.
7. Governor Response: The governors of the generators must respond in the same way to changes in load demand. This is necessary to ensure that the generators share the load equally and prevent overloading of any individual generator.
8. Excitation System Response: The excitation systems of the generators must respond in the same way to changes in load demand. This is necessary to ensure that the generators share the load equally and prevent overloading of any individual generator.
By ensuring that these conditions are met, synchronous generators can be connected in parallel to provide the required amount of electrical power to the system.
Recall Advantages of Parallel operation of Synchronous Generators
When two or more synchronous generators are connected in parallel, they are said to be operating in parallel. This is done in order to provide increased power output, improved system reliability, and easier maintenance. Some of the key advantages of parallel operation of synchronous generators are:
1. Increased power output: When two or more synchronous generators are operated in parallel, their combined output is greater than the sum of their individual outputs. This allows for increased power output to meet the demands of the electrical grid.
2. Improved system reliability: Parallel operation of synchronous generators improves system reliability by providing backup power in case one generator fails. This helps to ensure continuous power supply to customers and prevents downtime.
3. Efficient use of resources: Parallel operation of synchronous generators allows for the efficient use of resources, such as fuel and maintenance costs. By operating multiple generators in parallel, each generator can be run at its most efficient load point, reducing fuel consumption and maintenance costs.
4. Load sharing: When synchronous generators are connected in parallel, they automatically share the load based on their individual capacities. This ensures that each generator operates at its rated capacity and prevents overloading of any one generator.
5. Redundancy: Parallel operation of synchronous generators provides redundancy, which is essential for maintaining system reliability. If one generator fails, the remaining generators can continue to operate and meet the power demand.
6. Flexibility: Parallel operation of synchronous generators provides flexibility in system design, allowing for the addition or removal of generators as needed to meet changing demand.
In summary, parallel operation of synchronous generators provides increased power output, improved system reliability, efficient use of resources, load sharing, redundancy, and flexibility.
Describe the methods of Synchronising the Generators for Parallel Operation
Synchronisation is the process of connecting two or more synchronous generators in parallel so that they operate together and share the load. Before synchronising the generators, they must be at the same voltage level, frequency, and phase sequence. The synchronisation process ensures that the voltage, frequency, and phase sequence of the incoming generator are in line with the system parameters.
The following are the methods used for synchronising the generators for parallel operation:
1. Dark Lamp Method: In this method, a dark lamp is used as a visual indicator to detect the phase difference between the incoming generator and the bus bar. The dark lamp is connected between the incoming generator and the bus bar. If the phases are in synchronisation, the lamp will not glow, indicating that the phases are in line. If the lamp glows, it means there is a phase difference, and the phase sequence of the incoming generator needs to be adjusted.
2. Synchroscope Method: The synchroscope is a device that indicates the phase difference between the incoming generator and the bus bar. The synchroscope is connected between the incoming generator and the bus bar, and it displays the relative phase angle between the two systems. The operator adjusts the phase sequence of the incoming generator by observing the movement of the synchroscope’s pointer.
3. Three-Lamp Method: In this method, three lamps are used to indicate whether the voltage, frequency, and phase sequence of the incoming generator are in line with the system parameters. The lamps are connected between the incoming generator and the bus bar, and each lamp represents a different parameter. If all three lamps glow, it means that the parameters are not in line, and adjustments need to be made.
4. Digital Synchronizer: The digital synchronizer is a device that automatically synchronises the incoming generator with the bus bar. The device measures the voltage, frequency, and phase angle of the incoming generator and compares them to the bus bar’s parameters. If the parameters are within the acceptable range, the digital synchronizer closes the circuit breaker to connect the generator to the bus bar.
Overall, synchronisation is an essential step in parallel operation of synchronous generators to ensure safe and efficient operation of the power system.
Recall the Synchronizing Power of: i. Two identical machines on No-Load, floating with respect to each other ii. Machines connected to Infinite Bus
i. Two identical machines on No-Load, floating with respect to each other
When two identical machines are on no-load and are floating with respect to each other, the rotor of each machine will rotate at its synchronous speed. In this situation, there is no mechanical coupling between the two rotors, and they are free to rotate independently. However, if there is a small disturbance in one of the machines, such as a sudden change in mechanical torque or a voltage dip, it can cause the rotor to deviate slightly from its synchronous speed.
When this happens, the rotor of the other machine will also experience a similar disturbance, causing it to deviate from its synchronous speed. The result is that the two rotors will tend to move closer together, as the forces between them increase, until they are back in synchronism. This phenomenon is known as the synchronising power of two identical machines on no-load floating with respect to each other.
For example, consider two identical synchronous generators that are connected to a common bus, but are not electrically connected to each other. If one of the generators experiences a sudden change in mechanical torque, it will cause a momentary change in its rotational speed. This change in speed will cause the generator to produce a transient voltage that will appear across its terminals. As the voltage across the terminals of the second generator is zero, the transient voltage produced by the first generator will appear across the gap between the two machines. This will cause a current to flow between the two machines, creating a force that will tend to bring the rotors back into synchronism.
ii. Machines connected to Infinite Bus
When synchronous generators are connected to an infinite bus, they are electrically and mechanically coupled to each other. An infinite bus is an idealised power system with an infinite amount of power available, which means that the voltage and frequency of the bus remain constant regardless of the amount of power generated or consumed by the connected generators.
When a synchronous generator is connected to an infinite bus, it experiences a constant electrical load, which keeps the rotor rotating at the synchronous speed. If there is a small deviation in the rotor speed, the electrical power output of the generator will also deviate, causing a temporary power imbalance in the system. This power imbalance will result in a torque that tends to bring the rotor back into synchronism with the infinite bus.
For example, consider a synchronous generator that is connected to an infinite bus through a transmission line. If there is a sudden increase in the mechanical torque applied to the generator, it will cause the rotor to speed up slightly. This will result in a temporary increase in the power output of the generator, which will create a power imbalance in the system. The excess power will flow through the transmission line and cause a momentary voltage drop at the generator’s terminals. This voltage drop will cause a current to flow through the line, creating a torque that tends to slow down the rotor and bring it back into synchronism with the infinite bus. Similarly, if there is a sudden decrease in the mechanical torque applied to the generator, it will cause the rotor to slow down slightly, resulting in a temporary decrease in the power output and a momentary voltage rise at the generator’s terminals. This will cause a current to flow in the opposite direction through the transmission line, creating a torque that tends to speed up the rotor and bring it back into synchronism with the infinite bus.
Recall the concept of two machines in Parallel Operation
When two or more synchronous generators are connected to a common electrical bus and are operating in parallel, they share the total electrical load and produce power together. Parallel operation is commonly used in power systems to increase the overall system capacity and provide redundancy. However, proper synchronisation and load sharing between the generators are essential for efficient and safe operation.
In parallel operation, the frequency and voltage of each generator must be identical, and the phase angle between them must be zero. Any deviation in frequency or voltage can result in an unbalanced load sharing between the generators and can cause voltage instability and other undesirable effects. Therefore, a synchronising panel is used to monitor the frequency and voltage of each generator and adjust them as necessary to achieve proper synchronisation.
The load sharing between the generators is determined by their respective droop characteristics. Droop is a characteristic that describes how the generator output voltage changes with respect to the load. In a droop system, as the load on the generator increases, the output voltage decreases, and as the load decreases, the output voltage increases. This means that the generator with a higher droop characteristic will provide more power for a given load than the generator with a lower droop characteristic.
For example, consider two synchronous generators that are operating in parallel and connected to a common bus. If the total load on the system is 100 MW, and each generator has a capacity of 60 MW, the generators will share the load in proportion to their droop characteristics. If generator 1 has a droop characteristic of 3% and generator 2 has a droop characteristic of 4%, generator 1 will provide 40 MW of power, and generator 2 will provide 60 MW of power to the load. If the load increases to 120 MW, the generators will adjust their output voltage to share the load accordingly, with generator 1 providing 48 MW and generator 2 providing 72 MW.
In summary, parallel operation of two or more generators requires proper synchronization and load sharing to ensure efficient and safe operation of the power system. The synchronising panel monitors the frequency and voltage of each generator and adjusts them as necessary, while the droop characteristic determines the load sharing between the generators. | 11,261 | 58,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.794038 |
http://www.piclist.com/techref/postbot.asp?by=time&id=piclist/2004/04/21/024050a&tgt=post | 1,537,372,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156252.31/warc/CC-MAIN-20180919141825-20180919161825-00044.warc.gz | 382,060,533 | 5,018 | piclist 2004\04\21\024050a >
Thread: Looking for an AVR FFT tutorial
www.piclist.com/techref/logic/dsps.htm?key=fft
BY : Richard.ProsserTakeThisOuT@STOPspamPOWERWARE.COM
Thanks - will have to have a play with it!
RP
On 21 Apr 2004 at 8:52, Richard.ProsserPOWERWARE.COM wrote:
> The web site I used for my initial code was (I think)
> http://www.8052.com/users/steve/FFTC.C, which is an FFT example for the
> 8052 micro.
Thanks, that seems to be a very good example!
> As far as comparing the Goertzel algorithm with the FFT I think it is a
> question of what you are actually trying to do.
> My impression is that the Goertzel algorithm aims at identifying a single
> frequency component in a signal, basically using a IIR filter. It is
> capable of running more or less "on the fly" once a certain number of
> samples have been obtained.
I do the eight filters at every sample. When I reach N samples,
I compute the eight magnitudes and get the two DTMF frequencies. The
whole process used less than 20mS at 4MHz clock.
>
> Incidentally, I'd be interested in looking at you Goertzel implementation
> if possible!
Here is a MACRO I used at every sample, based on that article at:
http://www.embedded.com/story/OEG20020819S0057
;
; At every sample, calculate for each frequency:
;
; Q0 = (coeff*Q1)/32768-Q2+sample
; Move Q1 to Q2
; Move Q0 to Q1
;
macro SAMPLE ; Q1l,Q1h,Q2l,Q2h,Coeff
ldi auxl, LOW(@4)
ldi auxh, HIGH(@4)
lds matl, @0
lds math, @1
muls math, auxh ; (signed)Q1h *
(signed)Coeffh
movw genl, prdl
mul matl, auxl ; Q1l * Coeffl
mov intemp, prdh
mulsu auxh, matl ; (signed)Coeffh * Q1l
sbc genh, zero
mulsu math, auxl ; (signed)Q1h * Coeffl
sbc genh, zero
lsl genl
rol genh ; (Q1*Coeff)<<1
lds auxl, @2
lds auxh, @3
sub genl, auxl
sbc genh, auxh ; Sub Q2
sample
sts @0, genl
sts @1, genh ; Q1 = Q0
sts @2, matl ; Q2 = Q1
sts @3, math
endmacro ; 45 clock cycles
;
;
; After N samples, calculate for each frequency:
;
; MAG = SQRT(Q1*Q1+Q2*Q2-(Q1*Q2*coeff)/32768)
;
Mark Jordan
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https://blog.dreamshire.com/solutions/project_euler/project-euler-problem-010-solution/ | 1,701,576,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00721.warc.gz | 168,679,273 | 3,100 | ## Project Euler Problem 10 & HackerRank Solution
Summation of primes
by {BetaProjects} | Project Euler & HackerRank
### Project Euler Problem 10 Statement
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
### Solution
#### A simple approach
Using the function `prime_sieve(N)`, introduced in Project Euler Problem 7, to find prime numbers less than N, we can solve this problem in less than 100ms using a couple lines of Python:
``````from Euler import prime_sieve
print (sum(prime_sieve(2000000)))
``````
#### A more extensible solution
HackerRank poses a more demanding problem by having us find the sum of primes not greater than N (same as less than or equal to), where 1 ≤ N ≤ 106 for up to 10,000 queries.
A reasonable way to approach solving this problem is to set up a list, `primeSum[]`, with the intention of building a one-to-one relationship with the query's value as the index and the sum of primes as the element. For example, `primeSum[10]` = 17, the sum of all prime numbers not greater than 10.
The list would have to be big enough to handle all possible queries within the specified range. In this case we will increase it from one million to two million to accommodate the original problem's limit.
Basically, this is done using the principles of the sieve of Eratosthenes: find the next zero value in `primeSum[]`, set the prime sum, `s`, at that index and the next, and mark all the primes's multiples as composites (i.e., set to -1).
Marking multiples is achieved by using list slicing in the form `primeSum[start:stop:step]`. Specifically, the Python statement `primeSum[i::i] = [−1]*(N//i)` fills every `step` index of the list, starting from `start` to the end of the list1 with −1; our value to indicate composite numbers. The choice for −1 was arbitrary and was selected because it stands out.
1When `stop` is omitted, as it is in this case, it fills to the end of the list.
#### A step–by–step breakdown of this algorithm
There’s a lot going on with the few lines of code in this algorithm. The table below illustrates the method for N=10 and s=2.
``` primeSum[]
i primeSum[i]==0? s [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
3 Y 5 [0, 0, 2, −1, 0, 0, −1, 0, 0, −1, 0]
primeSum[3],[4] = s [0, 0, 2, 5, 5, 0, −1, 0, 0, −1, 0]
5 Y 10 [0, 0, 2, 5, 5, −1, −1, 0, 0, −1, −1]
primeSum[5],[6] = s [0, 0, 2, 5, 5, 10, 10, 0, 0, −1, −1]
7 Y 17 [0, 0, 2, 5, 5, 10, 10, −1, 0, −1, −1]
primeSum[7],[8] = s [0, 0, 2, 5, 5, 10, 10, 17, 17, −1, −1]
9 N 17 [ all multiples of 9 already set by multiples of 3 ]
primeSum[9],[10]= s [0, 0, 2, 5, 5, 10, 10, 17, 17, 17, 17]
```
At the end of the main loop the list is ready to accept as many queries as required.
#### HackerRank version
HackerRank asks us to run up to 10,000 test cases and sum prime numbers not greater than N, where 1 ≤ N ≤ 106.
### Python Source Code
``````
s = 2; N = 2000000; primeSum = [0]*(N+1)
primeSum[2] = s
for i in range(3, N, 2):
if primeSum[i]==0:
s+= i
primeSum[i::i] = [-1]*(N//i)
primeSum[i] = s; primeSum[i+1] = s #each odd index and the following even index
L = int(input("Sum of primes not greater than? "))
print ("is", primeSum[L])
``````
### Last Word
Have a look at the `prime_sieve()` function listed in Common Functions and Routines for Project Euler. | 1,193 | 3,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-50 | latest | en | 0.863572 |
https://cs.brown.edu/courses/cs1951x/docs/algebra/lie/of_associative.html | 1,718,820,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00407.warc.gz | 153,979,614 | 39,481 | # mathlibdocumentation
algebra.lie.of_associative
# Lie algebras of associative algebras #
This file defines the Lie algebra structure that arises on an associative algebra via the ring commutator.
Since the linear endomorphisms of a Lie algebra form an associative algebra, one can define the adjoint action as a morphism of Lie algebras from a Lie algebra to its linear endomorphisms. We make such a definition in this file.
## Main definitions #
• lie_algebra.of_associative_algebra
• lie_algebra.of_associative_algebra_hom
• lie_module.to_endomorphism
• lie_algebra.ad
• linear_equiv.lie_conj
• alg_equiv.to_lie_equiv
## Tags #
lie algebra, ring commutator, adjoint action
@[protected, instance]
def ring.has_bracket {A : Type v} [ring A] :
A
The bracket operation for rings is the ring commutator, which captures the extent to which a ring is commutative. It is identically zero exactly when the ring is commutative.
Equations
theorem ring.lie_def {A : Type v} [ring A] (x y : A) :
x,y = x * y - y * x
theorem commute_iff_lie_eq {A : Type v} [ring A] {x y : A} :
y x,y = 0
theorem commute.lie_eq {A : Type v} [ring A] {x y : A} (h : y) :
@[protected, instance]
def lie_ring.of_associative_ring {A : Type v} [ring A] :
An associative ring gives rise to a Lie ring by taking the bracket to be the ring commutator.
Equations
theorem lie_ring.of_associative_ring_bracket {A : Type v} [ring A] (x y : A) :
x,y = x * y - y * x
@[simp]
theorem lie_ring.lie_apply {A : Type v} [ring A] {α : Type u_1} (f g : α → A) (a : α) :
f,g a = f a,g a
@[reducible]
def lie_ring_module.of_associative_module {A : Type v} [ring A] {M : Type w} [ M] :
We can regard a module over an associative ring A as a Lie ring module over A with Lie bracket equal to its ring commutator.
Note that this cannot be a global instance because it would create a diamond when M = A, specifically we can build two mathematically-different has_bracket A As:
1. @ring.has_bracket A _ which says ⁅a, b⁆ = a * b - b * a
2. (@lie_ring_module.of_associative_module A _ A _ _).to_has_bracket which says ⁅a, b⁆ = a • b (and thus ⁅a, b⁆ = a * b)
Equations
theorem lie_eq_smul {A : Type v} [ring A] {M : Type w} [ M] (a : A) (m : M) :
a,m = a m
@[protected, instance]
def lie_algebra.of_associative_algebra {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] :
A
An associative algebra gives rise to a Lie algebra by taking the bracket to be the ring commutator.
Equations
def lie_module.of_associative_module {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {M : Type w} [ M] [ M] [ M] :
A M
A representation of an associative algebra A is also a representation of A, regarded as a Lie algebra via the ring commutator.
See the comment at lie_ring_module.of_associative_module for why the possibility M = A means this cannot be a global instance.
Equations
@[protected, instance]
def module.End.lie_ring_module {R : Type u} [comm_ring R] {M : Type w} [ M] :
M
Equations
@[protected, instance]
def module.End.lie_module {R : Type u} [comm_ring R] {M : Type w} [ M] :
M) M
Equations
def alg_hom.to_lie_hom {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] (f : A →ₐ[R] B) :
The map of_associative_algebra associating a Lie algebra to an associative algebra is functorial.
Equations
@[protected, instance]
def alg_hom.lie_hom.has_coe {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] :
Equations
@[simp]
theorem alg_hom.to_lie_hom_coe {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] (f : A →ₐ[R] B) :
@[simp]
theorem alg_hom.coe_to_lie_hom {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] (f : A →ₐ[R] B) :
theorem alg_hom.to_lie_hom_apply {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] (f : A →ₐ[R] B) (x : A) :
@[simp]
theorem alg_hom.to_lie_hom_id {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] :
@[simp]
theorem alg_hom.to_lie_hom_comp {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} {C : Type w₁} [ring B] [ring C] [ B] [ C] (f : A →ₐ[R] B) (g : B →ₐ[R] C) :
(g.comp f) = g.comp f
theorem alg_hom.to_lie_hom_injective {A : Type v} [ring A] {R : Type u} [comm_ring R] [ A] {B : Type w} [ring B] [ B] {f g : A →ₐ[R] B} (h : f = g) :
f = g
def lie_module.to_endomorphism (R : Type u) (L : Type v) (M : Type w) [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] :
A Lie module yields a Lie algebra morphism into the linear endomorphisms of the module.
See also lie_module.to_module_hom.
Equations
@[simp]
theorem lie_module.to_endomorphism_to_linear_map_apply_apply (R : Type u) (L : Type v) (M : Type w) [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (x : L) (m : M) :
@[simp]
theorem lie_module.to_endomorphism_apply_apply (R : Type u) (L : Type v) (M : Type w) [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (x : L) (m : M) :
( M) x) m = x,m
def lie_algebra.ad (R : Type u) (L : Type v) [comm_ring R] [lie_ring L] [ L] :
The adjoint action of a Lie algebra on itself.
Equations
@[simp]
theorem lie_algebra.ad_apply (R : Type u) (L : Type v) [comm_ring R] [lie_ring L] [ L] (x y : L) :
( L) x) y = x,y
@[simp]
theorem lie_module.to_endomorphism_module_End (R : Type u) (M : Type w) [comm_ring R] [ M] :
theorem lie_subalgebra.to_endomorphism_eq (R : Type u) (L : Type v) (M : Type w) [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (K : L) {x : K} :
M) x = M) x
@[simp]
theorem lie_subalgebra.to_endomorphism_mk (R : Type u) (L : Type v) (M : Type w) [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (K : L) {x : L} (hx : x K) :
M) x, hx⟩ = M) x
theorem lie_submodule.coe_map_to_endomorphism_le {R : Type u} {L : Type v} {M : Type w} [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] {N : M} {x : L} :
theorem lie_submodule.to_endomorphism_comp_subtype_mem {R : Type u} {L : Type v} {M : Type w} [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (N : M) (x : L) (m : M) (hm : m N) :
(linear_map.comp ( M) x) N.subtype) m, hm⟩ N
@[simp]
theorem lie_submodule.to_endomorphism_restrict_eq_to_endomorphism {R : Type u} {L : Type v} {M : Type w} [comm_ring R] [lie_ring L] [ L] [ M] [ M] [ L M] (N : M) (x : L) (h : (∀ (m : M) (hm : m N), (linear_map.comp ( M) x) N.subtype) m, hm⟩ N) := _) :
theorem lie_algebra.ad_eq_lmul_left_sub_lmul_right {R : Type u} [comm_ring R] (A : Type v) [ring A] [ A] :
theorem lie_subalgebra.ad_comp_incl_eq {R : Type u} {L : Type v} [comm_ring R] [lie_ring L] [ L] (K : L) (x : K) :
def lie_subalgebra_of_subalgebra (R : Type u) [comm_ring R] (A : Type v) [ring A] [ A] (A' : A) :
A subalgebra of an associative algebra is a Lie subalgebra of the associated Lie algebra.
Equations
def linear_equiv.lie_conj {R : Type u} {M₁ : Type v} {M₂ : Type w} [comm_ring R] [add_comm_group M₁] [ M₁] [add_comm_group M₂] [ M₂] (e : M₁ ≃ₗ[R] M₂) :
M₁ ≃ₗ⁅R M₂
A linear equivalence of two modules induces a Lie algebra equivalence of their endomorphisms.
Equations
@[simp]
theorem linear_equiv.lie_conj_apply {R : Type u} {M₁ : Type v} {M₂ : Type w} [comm_ring R] [add_comm_group M₁] [ M₁] [add_comm_group M₂] [ M₂] (e : M₁ ≃ₗ[R] M₂) (f : M₁) :
(e.lie_conj) f = (e.conj) f
@[simp]
theorem linear_equiv.lie_conj_symm {R : Type u} {M₁ : Type v} {M₂ : Type w} [comm_ring R] [add_comm_group M₁] [ M₁] [add_comm_group M₂] [ M₂] (e : M₁ ≃ₗ[R] M₂) :
def alg_equiv.to_lie_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [comm_ring R] [ring A₁] [ring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₁ ≃ₗ⁅R A₂
An equivalence of associative algebras is an equivalence of associated Lie algebras.
Equations
@[simp]
theorem alg_equiv.to_lie_equiv_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [comm_ring R] [ring A₁] [ring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₁) :
@[simp]
theorem alg_equiv.to_lie_equiv_symm_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [comm_ring R] [ring A₁] [ring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₂) : | 2,908 | 7,904 | {"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} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.634585 |
https://www.free-dc.org/showthread.php?2884-A-New-Estimate-for-When-Will-Find-Primes-above-n-3-million/page2&s=5a5558ae92359037449a127e3554d877 | 1,600,881,012,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400211096.40/warc/CC-MAIN-20200923144247-20200923174247-00590.warc.gz | 840,658,087 | 16,408 | # Thread: A New Estimate for When Will Find Primes above n = 3 million
1. There is a 50% probability that we will get all 12 primes by n=2.9 x 10^13
Does this mean that there is a 50% probability that we will
NOT see the sierpinski problem solved in our lifetimes,
assuming that no quantum computers are invented?
You think we'll reach n=2.9 x 10^13 in our lifetimes?
2. You think we'll reach n=2.9 x 10^13 in our lifetimes?
Depends on the lifetime we got left.
I think it is widely distributed among the participants...
3. Depends on the lifetime we got left.
Lets assume the folowing:
1) We tested 12 K's from 1 to 3M in 1 year
2) Computer speed doubles every 1 years
3) Number of searchers doubles every year
4) a test twice as large takes 4 times longer
All of the above are very optimistic to make up with the removal of K's
This means we'll have a progress of 3M a year.
(2.9 x 10^13) / (3 x 10^6) is about 10 million years
Even i'f i'm a couple of magnitudes off.................
4. (2.9 x 10^13) / (3 x 10^6) is about 10 million years
If you want to die that early...
5. 2) Computer speed doubles every 1 years
That would be true if we assume that every new user has a faster computer than the current searchers have and/or if we assume that the current searchers will bring newer machines to the project.
3) Number of searchers doubles every year
This is a very very very optimistic assumpition. Yes, the number of new users grows a bit every day (and will grow a lot when a new prime is discovered), but the number of active users hasn't increased significantly in the last months (even, I think it has decreased).
It would be nice to have in the stats an "active users in the last months" graph.
6. Originally posted by Moo_the_cow
Does this mean that there is a 50% probability that we will
NOT see the sierpinski problem solved in our lifetimes, assuming that no quantum computers are invented?
First let’s update that estimate. The bad news is that several of Yves Gallot’s Proth Weights were smaller than my estimates, and this substantially increases the time until we are likely to have found all of the primes. Using his Proth Weights and assuming no primes found by n=3.5M, the 50% probability estimate for finding all the primes is n=10^14.
Let’s assume this Proth Weight model is good enough for such ridiculous estimates, and assume no new algorithmic discoveries over the course of Seventeen or Bust, and assume that Moore’s Law continues to double processing power every eighteen months, and then do a back-of-the-envelope estimation.
On the resource allocation thread we are discussing models of the time needed to complete a primality test. One of the models says it grows with n as n^2 * (log(n))^2. If that’s right, then the 50 percentile primality test will take 4*10^15 times as long as the present tests around n=3.5M. If we want to finish that test in the same amount of time as today, we need about 52 doublings of computer power, which will take about 78 years. But there is another problem: clearing each successive power of 10 takes about 10 times as many tests. We cleared about a half-power of 10 in a year. To clear powers of 10 at the same pace at the end, we would need to complete each primality test about 3*10^7 faster. Maintaining that pace through powers of 10 would finish in only 15 years, though, so we can slow down by a factor of 5 to 6*10^6. In addition, we will only be testing one k value instead of 17, and it will be a low proth weight value, so a speed increase of only 2*10^5 is probably sufficient. This is another 18 doublings, requiring an additional 26 years.
So, assuming no new algorithmic improvements, and assuming all these modeling assumptions, it looks like there is a 50% probability that Seventeen or Bust will require more than 104 years to finish. Under these assumptions, I would say that yes, there is a greater than 50% chance the Sierpinski problem will not be solved in our lifetimes.
Note to smh – you missed that computer speed overtakes you because if you are only completing 3M tests per year, you don’t double the n’s fast enough, so computing power goes to faster tests rather than harder tests in the same time.
I don’t despair, though. It’s silly to assume the probability model is still useful at such extremes, and it’s even sillier to assume no new algorithmic breakthroughs will occur.
7. ## almost year passed
well, how about CURRENT estimations...
8. In the near and longterm future advances in optical processes, nanotechnology, quantumcomputers, ........... will certainly help in solving this and many more problems much faster!!!!!
Dont you agreeee???????
9. Ah things I forgot, the development of grid computing or utility computing or whatever you call it, these developments will certainly reduce the amount of cpu cycles wasted in the future by the millions of computerusers in the world.
Another thing, possible alliances in the near future with platforms such as BOINC, so you can reach much more people.....
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• | 1,255 | 5,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.942488 |
https://www.jiskha.com/display.cgi?id=1326668530 | 1,503,067,719,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104681.22/warc/CC-MAIN-20170818140908-20170818160908-00263.warc.gz | 931,828,429 | 5,449 | # Calculus
posted by .
A piece of elastic is attached to two nails on a flat board, with a button attached to the midpoint of the elastic. The nails are 5 cm apart. You stretch the elastic by pulling the button along the board in a direction that is perpendicular to the line between the nails.
A. Fnd an equation that relates the total length of the elastic x to the distance y that the button has moved.
B. You pull the button at a constant 3 cm/sec. Find the rate at which the length of the elastic is increasing when it is 12 cm long.
• Calculus -
you have two right triangles of height y
2.5^2 + y^2 = (x/2)^2
or
25 + 4y^2 = x^2
8y dy/dt = 2x dx/dt
when x = 12, y = 5.454
8(5.454) = 2(12) dx/dt
dx/dt = 1.818
• Calculus -
First, let's sketch what we can derive geometrically.
|dw:1355290130364:dw|
(A) Given we know those two triangles are right, we can relate the hypotenuse $$\frac12x$$ to the altitude $$y$$ and base $$\frac52$$ using the Pythagorean theorem, which we can rearrange to yield an adequate relation:$$\left(\frac12x\right)^2=y^2+\left(\frac52\right)^2\\\frac14x^2=y^2+\frac{25}4\\x^2=4y^2+25$$
(B) We're given that the button is moving at a rate of 3 cm/s, which can be expressed using a time derivative as $$\frac{dy}{dt}=3$$. We're told that the elastic (at the instant we're interested in) is 12 cm long, i.e. $$x=12$$; given this, we can determine the distance of the button from its initial position with relative ease:$$(12)^2=4y^2+25\\4y^2=144-25=119\\y^2=\frac{119}{4}\\y=\frac{\sqrt{119}}2$$Let's use implicit differentiation on our formula above to relate the rates of elongation:$$2x\frac{dx}{dt}=8y\frac{dy}{dt}\\24\frac{dx}{dt}=12\sqrt{119}\\2\frac{dx}{dt}=\sqrt{119}\\\frac{dx}{dt}=\frac{\sqrt{119}}2\approx5.4544$$
• Calculus -
each side of the elastic is the hypotenuse of a triangle with legs 5/2 and y, so
x = 2√(2.5^2 + y^2)
dx/dt = 2y/√(2.5^2+y^2) dy/dt
when x=12, y=5.45, so
dx/dt = 2(5.45)/6 (-3) = -1/5.45 = -0.18 cm/s
• Calculus -
A.
x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)
B.
x^2 = 4y^2 + 25
=> 2x dx/dt = 8y dy/dt
=> dx/dt = 4 (y/x) dy/dt
When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
=> dx/dt = 4 (2.727/12) x 3 cm/s
= 2.727 cm/s.
• Calculus -
A.
x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)
B.
x^2 = 4y^2 + 25
=> 2x dx/dt = 8y dy/dt
=> dx/dt = 4 (y/x) dy/dt
When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
=> dx/dt = 4 (2.727/12) x 3 cm/s
= 2.727 cm/s.
• Calculus -
Pythagorous Theorem,
z^2 = y^2 + x^2, where y = 2.5 cm and x is the legth of the pull.
substituting z^2 = 2.5^2 +x^2, z =sqrt(x^2 +6.25)
dz/dx (2z) = 2x
dz/dx = (x/z) ......... (a)
but dz/dt = dz/dx * dx/dt
Thus, dz/dt = (x/z) * dx/dt = x/z * 3
when z = 12 , y = 5, x= sqrt(144-25) = sqrt(119)
Thus dz/dt = sqrt(119)/12 * 3
= sqrt(119)/4 =10.91 /4 =2.73 cm/sec
Rate of increase in length of elastic = 2.73 cm/sec
## Similar Questions
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A force of 50.0N is used to stretch an elastic so that it had 40.0 J of elastic potential energy. Through what distance was the elastic stretched?
2. ### Physics
A 0.25-kg ball is attached to a 26-cm piece of string. The ball is first raised so that the string is taut and horizontal, then the ball is released so that, at the bottom of its swing, it undergoes an elastic head-on collision with …
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Wheat farmers in Kansas would benefit from a devastating crop failure in N.Dakota if the U.S.demand for wheat?
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5. ### Calculus (demand)
18) The demand equation is x + 1/6p - 10+0. Compute the elasticity of demand and determine whether the demand is elastic, unitary, or inelastic at p=50. a) 5; elastic b) 1/9; inelastic c) 7/6; elastic d) 6; elastic I choose answer …
6. ### calculus
18) The demand equation is x + 1/6p - 10+0. Compute the elasticity of demand and determine whether the demand is elastic, unitary, or inelastic at p=50. a) 5; elastic b) 1/9; inelastic c) 7/6; elastic d) 6; elastic I choose answer …
7. ### Chemistry
Typically, there is a scale provided for weighing the nails. For example, a notice placed above the nail bin might read, “For the nails in the bin below, there are 500 nails per kg.” Using this conversion factor, perform the following …
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More Similar Questions | 1,750 | 5,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2017-34 | latest | en | 0.828002 |
https://www.reference.com/science/geologists-locate-epicenter-earthquake-e05e19246110dab8 | 1,623,900,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00555.warc.gz | 863,090,369 | 25,100 | # How Do Geologists Locate the Epicenter of an Earthquake?
By Staff WriterLast Updated Apr 6, 2020 7:06:39 PM ET
Geologists locate the epicenter of an earthquake by taking measurements from three seismograms. These are measured at seismic stations and give the distance that the earthquake's waves traveled in order to reach the station. These three distances are then used to triangulate the epicenter.
The following shows the steps used:
1. Get three seismograms
2. The first step to locate the epicenter of an earthquake is to get three seismograms. They must come from seismic stations located in different places.
3. Measure the distance
4. For each of the seismograms it is necessary to measure the S - P time interval. "S" stands for secondary waves, and "P" stands for primary waves. Then, the S - P interval is used to measure the distance that the waves traveled to reach the seismic station. This will not indicate the direction it came from though, just the distance the waves traveled.
5. Triangulate
6. To get the epicenter, it is necessary to triangulate the results worked out in the previous step. This is done by drawing a circle around the location of the seismic station that has an equal radius between the station and the distance the waves traveled. This should be done with all three and it should produce an overlap. The epicenter of the earthquake is in the area represented by the overlap.
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https://www.quizover.com/course/section/practice-makes-perfect-use-properties-of-rectangles-by-openstax | 1,540,229,497,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00153.warc.gz | 1,059,445,025 | 24,089 | 9.4 Use properties of rectangles, triangles, and trapezoids (Page 8/17)
Page 8 / 17
The height of a trapezoid is $7$ centimeters and the bases are $4.6$ and $7.4$ centimeters. What is the area?
42 sq. cm
The height of a trapezoid is $9$ meters and the bases are $6.2$ and $7.8$ meters. What is the area?
63 sq. m
Vinny has a garden that is shaped like a trapezoid. The trapezoid has a height of $3.4$ yards and the bases are $8.2$ and $5.6$ yards. How many square yards will be available to plant?
Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the area of a trapezoid Step 3. Name. Choose a variable to represent it. Let A = the area Step 4. Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Is this answer reasonable? Yes. The area of the trapezoid is less than the area of a rectangle with a base of 8.2 yd and height 3.4 yd, but more than the area of a rectangle with base 5.6 yd and height 3.4 yd. Step 7. Answer the question. Vinny has 23.46 square yards in which he can plant.
Lin wants to sod his lawn, which is shaped like a trapezoid. The bases are $10.8$ yards and $6.7$ yards, and the height is $4.6$ yards. How many square yards of sod does he need?
40.25 sq. yd
Kira wants cover his patio with concrete pavers. If the patio is shaped like a trapezoid whose bases are $18$ feet and $14$ feet and whose height is $15$ feet, how many square feet of pavers will he need?
240 sq. ft
Key concepts
• Properties of Rectangles
• Rectangles have four sides and four right (90°) angles.
• The lengths of opposite sides are equal.
• The perimeter, $P$ , of a rectangle is the sum of twice the length and twice the width.
• $P=2L+2W$
• The area, $A$ , of a rectangle is the length times the width.
• $A=L\cdot W$
• Triangle Properties
• For any triangle $\Delta ABC$ , the sum of the measures of the angles is 180°.
• $m\angle A+m\angle B+m\angle C=180°$
• The perimeter of a triangle is the sum of the lengths of the sides.
• $P=a+b+c$
• The area of a triangle is one-half the base, b, times the height, h.
• $A=\frac{1}{2}bh$
Practice makes perfect
Understand Linear, Square, and Cubic Measure
In the following exercises, determine whether you would measure each item using linear, square, or cubic units.
amount of water in a fish tank
cubic
length of dental floss
living area of an apartment
square
floor space of a bathroom tile
height of a doorway
linear
capacity of a truck trailer
In the following exercises, find the perimeter and area of each figure. Assume each side of the square is $1$ cm.
1. 10 cm
2. 4 sq. cm
1. 8 cm
2. 3 sq. cm
1. 10 cm
2. 5 sq. cm
Use the Properties of Rectangles
In the following exercises, find the perimeter and area of each rectangle.
The length of a rectangle is $85$ feet and the width is $45$ feet.
1. 260 ft
2. 3825 sq. ft
The length of a rectangle is $26$ inches and the width is $58$ inches.
A rectangular room is $15$ feet wide by $14$ feet long.
1. 58 ft
2. 210 sq. ft
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? | 1,962 | 7,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-43 | longest | en | 0.925406 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-and-chemical-reactivity-9th-edition/chapter-6-the-structure-of-atoms-study-questions-page-247f/84 | 1,685,576,536,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00021.warc.gz | 900,401,338 | 13,291 | ## Chemistry and Chemical Reactivity (9th Edition)
a) $\psi=\dfrac 1{\sqrt{\pi.(52.9\times10^{-12}\ m)^3}}.e^{-\frac{r=a_0}{a_0}}$ $\psi=5.39\times10^{14}\ m^{-3/2}$ $P=4\pi.(52.9\times10^{-12}\ m)^2\times(5.39\times10^{14}\ m^{-3/2})^2\times1.0\times10^{-12}\ m$ $P=0.01\approx 1\%$ b) Similarly as in a: $0.50 a_0: P=6.95\times10^{-3}$ $4a_0: P=4.06\times10^{-4}$ The decline in probability away from the most probable distance is in accord with figure 6.12b. | 202 | 462 | {"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} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.520751 |
https://optionsavym.web.app/casamayor32508ler/market-value-futures-contract-qaxe.html | 1,638,309,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00436.warc.gz | 515,186,182 | 6,150 | ## Market value futures contract
HG00 | Copper Continuous Contract Overview | MarketWatch HG00 | A complete Copper Continuous Contract futures overview by MarketWatch. View the futures and commodity market news, futures pricing and futures trading.
How to Calculate Futures Value. In order to show how to calculate Futures value, we must start with an example. Say you own \$240,000 of stock in the S&P 500 Index market at the price of 1400.00, and you would like to “hedge”, or protect your long position because you’re wary of the economy going into a tailspin. It's January and you enter into a futures contract to purchase 100 shares of IBM stock at \$50 a share on April 1. The contract has a price of \$5,000. But if the market value of the stock goes up before April 1, you can sell the contract early for a profit. Let's say the price of IBM stock rises to \$52 a share on March 1. The price at which the contract is traded is not pre-set, but is determined by market forces. It is possible to calculate a theoretical fair value for a futures contract. The fair value of a futures contract should approximately equal the current value of the underlying shares or index, plus an amount referred to as the 'cost of carry'. Futures markets trade futures contracts. A futures contract is an agreement between a buyer and seller of the contract that some asset--such as a commodity, currency or index--will bought/sold for a specific price, on a specific day, in the future (expiration date). Current Value If the current price of WTI futures is \$54, the current value of the contract is determined by multiplying the current price of a barrel of oil by the size of the contract. In this example, the current value would be \$54 x 1000 = \$54,000. Value of a One-Tick Move MTM is used to price futures contracts, which is very important for investors who trade futures in margin accounts. MTM pricing accurately reflects the true value of an asset. In Mark-to-Market accounting the asset values are determined according to market prices at the end of each day in order to arrive at the profit or loss status of the parties in a futures transaction.
## The Notional Value Calculation for a Futures Contract
15 May 2017 The pricing for futures contracts starts at a baseline figure of 100, and declines based on the implied interest rate in a contract. For example, if a liquid market in the world, with daily value around. US\$10 billion and, as such, is the most liquid of all assets. What is a futures contract? Definition. A currency 13 Feb 2020 This agency ensures the integrity of the futures market pricing. Any brokerage firms that engage in futures trading are regulated by the Commodity Futures contracts are used to hedge risk and to speculate in the market. used to minimize risks for producers from fluctuating market prices for their goods. This strategy involves buying the underlying asset of a futures contract in the spot market and holding [carrying] it for the duration of the arbitrage. Basic Steps: (1)
### An index future is essentially a contract to buy/sell a certain value of the Apart from stock market index futures, options on a stock market index are an
27 Feb 2019 trading activity in Arabica and Robusta futures markets over time, and. (ii) strong relationship between futures contract and spot prices for all 15 May 2017 The pricing for futures contracts starts at a baseline figure of 100, and declines based on the implied interest rate in a contract. For example, if a
### 12 Sep 2009 As the market value of the futures contract changes, the change is reflected in the enterprise's account with the broker on a regular basis.
Mark to Market (MTM) Definition - Investopedia Mark To Market - MTM: Mark to market (MTM) is a measure of the fair value of accounts that can change over time, such as assets and liabilities. Mark to market aims to provide a realistic Calculate the Size of a Futures Market Trade
## forward contracts, futures contracts are marked to market daily. As futures prices change daily cash flows are made, and the contract rewritten in such a way that
Mark to Market (MTM) Definition - Investopedia Mark To Market - MTM: Mark to market (MTM) is a measure of the fair value of accounts that can change over time, such as assets and liabilities. Mark to market aims to provide a realistic Calculate the Size of a Futures Market Trade For futures markets, the trade size is the number of contracts that are traded (with the minimum being one contract). The trade size is calculated using the tick value, the maximum account risk and the trade risk (size of the stop loss in ticks). Assume you have a \$10,000 future account, and are risking 1% per trade. Margins are determined on the basis of market risk and contract value. Also referred to as performance bond margin. Initial margin is the 6 Jan 2020 Futures Contracts and the Market. There are two main participants in the futures markets. Hedgers seek to manage their price risk for commodities
Calculating Futures Contract Profit or Loss | 1,084 | 5,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.936207 |
https://it.tradingview.com/script/TDx6il5e-Strategy-PnL-Library/ | 1,723,185,330,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00440.warc.gz | 253,296,443 | 61,103 | PINE LIBRARY
Strategy PnL Library
Library "Strategy_PnL_Library"
TODO: This is a library that helps you learn current pnl of open position and use it to create your own dynamic take profit or stop loss rules based on current level of your profit. It should only be used with strategies.
inTrade: Checks if a position is currently open.
Returns: bool: true for yes, false for no.
inTrade: Checks if a position is currently open. Interchangeable with inTrade but just here for simple semantics.
Returns: bool: true for yes, false for no.
pnl()
pnl: Calculates current profit or loss of position after the commission. If the strategy is not in trade it will always return na.
Returns: float: Current Profit or Loss of position, positive values for profit, negative values for loss.
entryBars()
entryBars: Checks how many bars it's been since the entry of the position.
Returns: int: Returns a int of strategy entry bars back. Minimum value is always corrected to 1 to avoid lookback errors.
pnlvelocity()
pnlvelocity: Calculates the velocity of pnl by following the change in open profit compared to previous bar. If the strategy is not in trade it will always return na.
Returns: float: Returns a float value of pnl velocity.
pnlacc()
pnlacc: Calculates the acceleration of pnl by following the change in profit velocity compared to previous bar. If the strategy is not in trade it will always return na.
Returns: float: Returns a float value of pnl acceleration.
pnljerk()
pnljerk: Calculates the jerk of pnl by following the change in profit acceleration compared to previous bar. If the strategy is not in trade it will always return na.
Returns: float: Returns a float value of pnl jerk.
pnlhigh()
pnlhigh: Calculates the highest value the pnl has reached since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float highest value the pnl has reached.
pnllow()
pnllow: Calculates the lowest value the pnl has reached since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float lowest value the pnl has reached.
pnldev()
pnldev: Calculates the deviance of the pnl since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float deviance value of the pnl.
pnlvar()
pnlvar: Calculates the variance value of the pnl since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float variance value of the pnl.
pnlstdev()
pnlstdev: Calculates the stdev value of the pnl since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float stdev value of the pnl.
pnlmedian()
pnlmedian: Calculates the median value of the pnl since the start of the current position. If the strategy is not in trade it will always return na.
Returns: float: Returns a float median value of the pnl.
dynamicPNLstatisticsstoplossstrategiesstrategytakeprofit
One does not simply win every trade.
Libreria Pine
In pieno spirito TradingView, l'autore ha pubblicato questo codice Pine come libreria open-source in modo che altri programmatori Pine della nostra comunità possano riutilizzarlo. Un saluto all'autore! È possibile utilizzare questa libreria privatamente o in altre pubblicazioni open-source, ma il riutilizzo di questo codice in una pubblicazione è regolato dal nostro Regolamento. | 798 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.833589 |
https://www.boredofstudies.org/threads/predict-your-raw-mark.330082/ | 1,685,654,893,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00403.warc.gz | 732,625,718 | 17,577 | # Predict Your Raw Mark (1 Viewer)
• ### 14 - 24
• Total voters
87
Poll Above.
comment below...
#### rumbleroar
##### Survivor of the HSC
where is the poll lel
#### Swaan
##### Stupid Fat Hobbit
i think i got low to mid 50's. pretty content
bump bump
#### Kurosaki
##### True Fail Kid
Where is the $\bg_white 0 \leq x \leq 70$ option?? D:
#### rumbleroar
##### Survivor of the HSC
Where is the $\bg_white 0 \leq x \leq 70$ option?? D:
maths is over, inequalities are unnecessary
#### mreditor16
##### Well-Known Member
Where is the $\bg_white 0 \leq x \leq 70$ option?? D:
do you think you're funny?
#### Kurosaki
##### True Fail Kid
maths is over, inequalities are unnecessary
did you square both sides
no you didn't
1 mark off!!!!!!!!
Haters </3
do you think you're funny?
I'm sure you do .
To answer thread Q: I think I got 70, checked Carrot's solutions and couldn't find anything wrong.
#### mreditor16
##### Well-Known Member
Haters </3
I'm sure you do .
To answer thread Q: I think I got 70, checked Carrot's solutions and couldn't find anything wrong.
great job buddy
#### rumbleroar
##### Survivor of the HSC
Haters </3
I'm sure you do .
To answer thread Q: I think I got 70, checked Carrot's solutions and couldn't find anything wrong.
fuark beast
see you at the state rank party then
#### Kurosaki
##### True Fail Kid
fuark beast
see you at the state rank party then
Nah my internal rank is too bad :/.
#### SquareZ
##### New Member
So close.... 69/70 after reading carrot's solutions.
##### i'm the cook
Haters </3
I'm sure you do .
To answer thread Q: I think I got 70, checked Carrot's solutions and couldn't find anything wrong.
are you first internally?
#### mreditor16
##### Well-Known Member
fuark beast
see you at the state rank party then
they don't have one. there is only one for first in the state.
#### rumbleroar
##### Survivor of the HSC
they don't have one. there is only one for first in the state.
i was talking about the one where bos throws one for its own!!
70 ez
#### mreditor16
##### Well-Known Member
i was talking about the one where bos throws one for its own!!
legit? could you elaborate?
#### Kurosaki
##### True Fail Kid
fuark beast
see you at the state rank party then
i was talking about the one where bos throws one for its own!!
Given the first quote, can I safely assume you think you state ranked something?
are you first internally?
Nope, too prone to sillies and the people beating me aren't.
#### rumbleroar
##### Survivor of the HSC
legit? could you elaborate?
idk i was just making stuff up but hey we should have a 2014 hsc celebration but thread derail here, so good luck to everyone who thought they got 70! (hope to see your names when the lists are released) | 748 | 2,759 | {"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.909208 |
https://www.studypool.com/discuss/1244682/Who-can-determine-the-Ka-for-CH3NH3-at-250C-?free | 1,480,942,333,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541696.67/warc/CC-MAIN-20161202170901-00504-ip-10-31-129-80.ec2.internal.warc.gz | 1,016,103,574 | 14,289 | ##### Who can determine the Ka for CH3NH3+ at 250C?
Chemistry Tutor: None Selected Time limit: 1 Day
Determine the Ka for CH3NH3+ at 250C. The Kb for CH3NH2 is 4.4 X 10 -4.
Oct 31st, 2015
First of all, Ka*Kb = Kw (the dissociation constant of water at 25 degrees Celsius) = 10^-14
Ka = Kw/Kb = 10^-14/ (4/4 * 10-4)
Ka = 2.3 * 10^-11
Please let me know if you need any clarification. I'm always happy to answer your questions. Feel free to recommend Studypool to your friends!
Oct 31st, 2015
...
Oct 31st, 2015
...
Oct 31st, 2015
Dec 5th, 2016
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ICSE chat | 3,578 | 13,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-51 | latest | en | 0.914416 |
http://mathhelpforum.com/trigonometry/14628-factorials-print.html | 1,529,476,119,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00300.warc.gz | 203,573,438 | 2,634 | factorials
• May 6th 2007, 11:57 AM
jeph
factorials
im not sure if this is trig or precal. for factorial a_n = (2n)! what does a_n+1 = ?
i thought you just add 1 to the end and get (2n+1)! but its not...
• May 6th 2007, 12:01 PM
CaptainBlack
Quote:
Originally Posted by jeph
im not sure if this is trig or precal. for factorial a_n = (2n)! what does a_n+1 = ?
i thought you just add 1 to the end and get (2n+1)! but its not...
I'll assume that we are after a_{n+1}, so we have:
a_n = (2n)!
so:
a_{n+1} = (2(n+1))! = (2n+2)!
RonL
• May 6th 2007, 12:02 PM
Jhevon
Quote:
Originally Posted by jeph
im not sure if this is trig or precal. for factorial a_n = (2n)! what does a_n+1 = ?
i thought you just add 1 to the end and get (2n+1)! but its not...
a_n = (2n)!
a_n+1 = (2(n+1))! = (2n + 2)!
EDIT: You're too quick for me Captain! | 319 | 837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-26 | latest | en | 0.937041 |
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Bank Clerk :: CR Quiz 148
Home > Bank Clerk > CR Quiz 148 > General Questions
1 .
Direction (Q. 1 - 5): Study the following information carefully and answer the questions given below:
A, B, C, D, E, F, G and H are sitting around a circle facing at the centre. F is third to the right of B who is third to the right of H. A is third to the left of H. C is fourth to the left of A. E is third to the right of D who is not a neighbor of A.
In which of the following pairs the second person is to the immediately right of the first person ?
HC BE GB FA
2 .
Who is second to the right of D ?
F G A None of these
3 .
Who is third to the left of G ?
H D C F
4 .
Who is fourth to the left of C ?
5 .
What is B’s position with respect to D ?
1. Fourth to the right
2. Fourth to the left
3. Fifth to the left
4. Fifth to the right
1 only 2 only 1 and 2 only 3 and 4 only
6 .
Direction (Q. 6 - 10): In each question below is given a group of letters followed by four combinations of digits / symbols lettered (1), (2), (3) and (4). You have to find out which of the combinations correctly represents the group of letters based on the following coding system and mark the number of that combination as the answer. If none of the four combinations correctly represents the group of letters, mark (4) i.e. ‘None of these’ as the answer.
Letter P M A K T I J E R V D F U W B Code / Symbol 7 # 8 % 1 9 2 @ 3 © \$ 4 « 5 6
Conditions:
(i) If both the first and the last letters of the group are consonants, both are to be coded as the code for the last letter.
(ii) If the first letter is a consonant and the last letter is a vowel, the codes are to be interchanged.
Q. BDATFE
6\$8146 6\$814@ @\$814@ @\$8146
7 .
Q. AWBRND
8 .
Q. EMNTKU | 525 | 1,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-13 | latest | en | 0.89837 |
https://howkgtolbs.com/convert/95.3-kg-to-lbs | 1,656,486,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00098.warc.gz | 343,532,527 | 12,059 | # 95.3 kg to lbs - 95.3 kilograms to pounds
Do you need to learn how much is 95.3 kg equal to lbs and how to convert 95.3 kg to lbs? Here it is. You will find in this article everything you need to make kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to emphasize that whole this article is devoted to only one amount of kilograms - exactly one kilogram. So if you need to know more about 95.3 kg to pound conversion - read on.
Before we get to the practice - that is 95.3 kg how much lbs calculation - we want to tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 95.3 kg to lbs? 95.3 kilograms it is equal 210.100535686 pounds, so 95.3 kg is equal 210.100535686 lbs.
## 95.3 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).
At times the kilogram is written as kilogramme. The symbol of this unit is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but totally impractical to use.
Later, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by another definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 95.3 kilogram to pounds
You learned some facts about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to point out that there are not only one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Naturally, this unit is in use also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 95.3 kg?
95.3 kilogram is equal to 210.100535686 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 95.3 kg in lbs
Theoretical section is already behind us. In this section we want to tell you how much is 95.3 kg to lbs. Now you learned that 95.3 kg = x lbs. So it is high time to get the answer. Just look:
95.3 kilogram = 210.100535686 pounds.
It is an accurate result of how much 95.3 kg to pound. You can also round off the result. After it your outcome will be as following: 95.3 kg = 209.66 lbs.
You learned 95.3 kg is how many lbs, so look how many kg 95.3 lbs: 95.3 pound = 0.45359237 kilograms.
Naturally, in this case you can also round off this result. After rounding off your outcome will be as following: 95.3 lb = 0.45 kgs.
We are also going to show you 95.3 kg to how many pounds and 95.3 pound how many kg outcomes in tables. Have a look:
We want to start with a chart for how much is 95.3 kg equal to pound.
### 95.3 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
95.3 210.100535686 209.660
Now look at a table for how many kilograms 95.3 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
95.3 0.45359237 0.45
Now you learned how many 95.3 kg to lbs and how many kilograms 95.3 pound, so it is time to go to the 95.3 kg to lbs formula.
### 95.3 kg to pounds
To convert 95.3 kg to us lbs a formula is needed. We are going to show you two formulas. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 210.100535686 result in pounds
The first version of a formula will give you the most accurate outcome. In some situations even the smallest difference can be significant. So if you want to get a correct outcome - this version of a formula will be the best solution to convert how many pounds are equivalent to 95.3 kilogram.
So move on to the another formula, which also enables calculations to learn how much 95.3 kilogram in pounds.
The shorter formula is as following, have a look:
Amount of kilograms * 2.2 = the outcome in pounds
As you see, the second version is simpler. It could be better option if you need to make a conversion of 95.3 kilogram to pounds in quick way, for instance, during shopping. Just remember that your outcome will be not so accurate.
Now we want to show you these two formulas in practice. But before we will make a conversion of 95.3 kg to lbs we are going to show you another way to know 95.3 kg to how many lbs without any effort.
### 95.3 kg to lbs converter
An easier way to know what is 95.3 kilogram equal to in pounds is to use 95.3 kg lbs calculator. What is a kg to lb converter?
Converter is an application. It is based on first formula which we showed you above. Due to 95.3 kg pound calculator you can easily convert 95.3 kg to lbs. You only have to enter amount of kilograms which you want to calculate and click ‘convert’ button. You will get the result in a flash.
So try to calculate 95.3 kg into lbs using 95.3 kg vs pound converter. We entered 95.3 as a number of kilograms. This is the result: 95.3 kilogram = 210.100535686 pounds.
As you see, our 95.3 kg vs lbs converter is intuitive.
Now we can move on to our chief topic - how to convert 95.3 kilograms to pounds on your own.
#### 95.3 kg to lbs conversion
We are going to begin 95.3 kilogram equals to how many pounds calculation with the first formula to get the most correct outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 210.100535686 the result in pounds
So what have you do to learn how many pounds equal to 95.3 kilogram? Just multiply amount of kilograms, in this case 95.3, by 2.20462262. It gives 210.100535686. So 95.3 kilogram is equal 210.100535686.
You can also round off this result, for instance, to two decimal places. It is exactly 2.20. So 95.3 kilogram = 209.660 pounds.
It is high time for an example from everyday life. Let’s convert 95.3 kg gold in pounds. So 95.3 kg equal to how many lbs? And again - multiply 95.3 by 2.20462262. It is exactly 210.100535686. So equivalent of 95.3 kilograms to pounds, when it comes to gold, is equal 210.100535686.
In this example it is also possible to round off the result. This is the outcome after rounding off, this time to one decimal place - 95.3 kilogram 209.66 pounds.
Now we can move on to examples calculated using a short version of a formula.
#### How many 95.3 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 209.66 the outcome in pounds
So 95.3 kg equal to how much lbs? And again, you need to multiply number of kilogram, in this case 95.3, by 2.2. Let’s see: 95.3 * 2.2 = 209.66. So 95.3 kilogram is equal 2.2 pounds.
Let’s make another conversion with use of shorer version of a formula. Now calculate something from everyday life, for example, 95.3 kg to lbs weight of strawberries.
So convert - 95.3 kilogram of strawberries * 2.2 = 209.66 pounds of strawberries. So 95.3 kg to pound mass is 209.66.
If you learned how much is 95.3 kilogram weight in pounds and are able to calculate it with use of two different versions of a formula, let’s move on. Now we want to show you these results in tables.
#### Convert 95.3 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in charts for your convenience. Due to this you can quickly compare 95.3 kg equivalent to lbs outcomes.
Let’s start with a 95.3 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
95.3 210.100535686 209.660
And now look 95.3 kg equal pound table for the second version of a formula:
Kilograms Pounds
95.3 209.66
As you see, after rounding off, if it comes to how much 95.3 kilogram equals pounds, the outcomes are the same. The bigger number the more significant difference. Keep it in mind when you need to make bigger amount than 95.3 kilograms pounds conversion.
#### How many kilograms 95.3 pound
Now you learned how to calculate 95.3 kilograms how much pounds but we want to show you something more. Are you curious what it is? What about 95.3 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it little by little. Start. How much is 95.3 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 95.3, by 2.20462262. So 95.3 * 2.20462262 = 210.100535686. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 95.3 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So your outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is 2 pounds and 33 ounces.
As you see, conversion 95.3 kilogram in pounds and ounces quite simply.
The last calculation which we want to show you is calculation of 95.3 foot pounds to kilograms meters. Both of them are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we show you it, see:
• 95.3 kilograms meters = 7.23301385 foot pounds,
• 95.3 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to calculate 95.3 foot pounds to kilograms meters you have to multiply 95.3 by 0.13825495. It gives 0.13825495. So 95.3 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 95.3 foot pounds is equal 0.14 kilogram meters.
We hope that this conversion was as easy as 95.3 kilogram into pounds calculations.
We showed you not only how to do a conversion 95.3 kilogram to metric pounds but also two another conversions - to know how many 95.3 kg in pounds and ounces and how many 95.3 foot pounds to kilograms meters.
We showed you also other way to do 95.3 kilogram how many pounds calculations, it is using 95.3 kg en pound converter. This is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you are able to make 95.3 kilogram equal to how many pounds conversion - on your own or with use of our 95.3 kgs to pounds converter.
It is time to make your move! Let’s calculate 95.3 kilogram mass to pounds in the way you like.
Do you want to do other than 95.3 kilogram as pounds conversion? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 95.3 kilogram equal many pounds.
### How much is 95.3 kg in pounds
To quickly sum up this topic, that is how much is 95.3 kg in pounds , we prepared one more section. Here you can see the most important information about how much is 95.3 kg equal to lbs and how to convert 95.3 kg to lbs . It is down below.
How does the kilogram to pound conversion look? It is a mathematical operation based on multiplying 2 numbers. How does 95.3 kg to pound conversion formula look? . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 95.3 kilogram to pounds? The accurate answer is 210.100535686 lbs.
There is also another way to calculate how much 95.3 kilogram is equal to pounds with another, easier type of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 95.3 kg equal to how much lbs ? The answer is 210.100535686 pounds.
How to convert 95.3 kg to lbs in just a moment? You can also use the 95.3 kg to lbs converter , which will do all calculations for you and give you an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,565 | 13,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.945345 |
https://simplywall.st/news/brook-crompton-holdings-ltd-sgxawc-earns-a-nice-return-on-capital-employed/ | 1,568,749,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573105.1/warc/CC-MAIN-20190917181046-20190917203046-00479.warc.gz | 510,471,986 | 16,217 | latest
# Brook Crompton Holdings Ltd. (SGX:AWC) Earns A Nice Return On Capital Employed
Today we are going to look at Brook Crompton Holdings Ltd. (SGX:AWC) to see whether it might be an attractive investment prospect. Specifically, we’ll consider its Return On Capital Employed (ROCE), since that will give us an insight into how efficiently the business can generate profits from the capital it requires.
First of all, we’ll work out how to calculate ROCE. Second, we’ll look at its ROCE compared to similar companies. Last but not least, we’ll look at what impact its current liabilities have on its ROCE.
### Understanding Return On Capital Employed (ROCE)
ROCE measures the ‘return’ (pre-tax profit) a company generates from capital employed in its business. Generally speaking a higher ROCE is better. Ultimately, it is a useful but imperfect metric. Author Edwin Whiting says to be careful when comparing the ROCE of different businesses, since ‘No two businesses are exactly alike.’
### So, How Do We Calculate ROCE?
The formula for calculating the return on capital employed is:
Return on Capital Employed = Earnings Before Interest and Tax (EBIT) ÷ (Total Assets – Current Liabilities)
Or for Brook Crompton Holdings:
0.14 = S\$5.1m ÷ (S\$50m – S\$14m) (Based on the trailing twelve months to June 2019.)
Therefore, Brook Crompton Holdings has an ROCE of 14%.
### Does Brook Crompton Holdings Have A Good ROCE?
ROCE is commonly used for comparing the performance of similar businesses. In our analysis, Brook Crompton Holdings’s ROCE is meaningfully higher than the 3.5% average in the Trade Distributors industry. I think that’s good to see, since it implies the company is better than other companies at making the most of its capital. Independently of how Brook Crompton Holdings compares to its industry, its ROCE in absolute terms appears decent, and the company may be worthy of closer investigation.
It is important to remember that ROCE shows past performance, and is not necessarily predictive. Companies in cyclical industries can be difficult to understand using ROCE, as returns typically look high during boom times, and low during busts. ROCE is only a point-in-time measure. If Brook Crompton Holdings is cyclical, it could make sense to check out this free graph of past earnings, revenue and cash flow.
### Brook Crompton Holdings’s Current Liabilities And Their Impact On Its ROCE
Liabilities, such as supplier bills and bank overdrafts, are referred to as current liabilities if they need to be paid within 12 months. The ROCE equation subtracts current liabilities from capital employed, so a company with a lot of current liabilities appears to have less capital employed, and a higher ROCE than otherwise. To counter this, investors can check if a company has high current liabilities relative to total assets.
Brook Crompton Holdings has total liabilities of S\$14m and total assets of S\$50m. Therefore its current liabilities are equivalent to approximately 28% of its total assets. A fairly low level of current liabilities is not influencing the ROCE too much.
### The Bottom Line On Brook Crompton Holdings’s ROCE
With that in mind, Brook Crompton Holdings’s ROCE appears pretty good. Brook Crompton Holdings shapes up well under this analysis, but it is far from the only business delivering excellent numbers . You might also want to check this free collection of companies delivering excellent earnings growth.
If you like to buy stocks alongside management, then you might just love this free list of companies. (Hint: insiders have been buying them).
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 898 | 4,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-39 | longest | en | 0.92864 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-5-section-5-3-polynomials-and-polynomial-functions-exercise-set-page-281/99 | 1,524,181,566,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937074.8/warc/CC-MAIN-20180419223925-20180420003925-00048.warc.gz | 421,968,187 | 13,668 | ## Intermediate Algebra (6th Edition)
$-3x-7.9$
We are given $(12x-1.7)-(15x+6.2)=12x-1.7-15x+6.2$. However, this is an incorrect use of the distributive property to distributive the negative sign to the second term. $(12x-1.7)-1\times(15x+6.2)=12x-1.7-15x-6.2=-3x-7.9$ | 113 | 270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-17 | latest | en | 0.749448 |
https://www.brainscape.com/flashcards/airplane-systems-10993240/packs/19573179 | 1,716,143,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00639.warc.gz | 621,489,243 | 52,301 | # Airplane Systems Flashcards Preview
## Private Pilot > Airplane Systems > Flashcards
Flashcards in Airplane Systems Deck (133)
1
Q
The basic purpose of adjusting the fuel/air mixture at altitude is to
A
decrease the fuel flow in order to compensate for decreased air density.
The higher an aircraft climbs, the less air there is to mix with the available fuel. To maintain the correct ratio of fuel to air as an aircraft increases in altitude, the pilot adjusts the amount of fuel with the manual mixture control.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
2
Q
What should be the indication on the magnetic compass as you roll into a standard rate turn to the right from a south heading in the Northern Hemisphere?
A
The compass will indicate a turn to the right, but at a faster rate than is actually occurring.
When making a turn from a southerly heading, the compass gives an indication of a turn in the correct direction, but leads the actual heading in the Northern hemisphere.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
3
Q
Carburetor ice is most likely to occur when temperatures are:
A
Below 70 degrees Fahrenheit (°F) and the relative humidity is above 80 percent.
Carburetor ice is most likely to occur when temperatures are below 70 degrees Fahrenheit (°F) or 21 degrees Celsius (°C) and the relative humidity is above 80 percent. Carb icing can occur even in outside air temperatures as high as 100 °F (38 °C).
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
4
Q
(Refer to figure 7.) The proper adjustment to make on the attitude indicator during level flight is to align the
View Figure 7
A
miniature airplane to the horizon bar.
Normally, the miniature airplane is adjusted so that the wings overlap the horizon bar when the aircraft is in straight-and-level flight.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
5
Q
Under what condition is pressure altitude and density altitude the same value?
A
At standard temperature.
Pressure altitude and density altitude are the same value in standard conditions.
Aviation Weather (AC 00-6) Ch. 3
6
Q
How do variations in temperature affect the altimeter?
A
Pressure levels are raised on warm days and the indicated altitude is lower than true altitude.
On a warm day the atmosphere expands. An aircraft will actually be higher than the altimeter is indicating (higher pressure level). On a cold day the atmosphere contracts and an aircraft will be lower than the altimeter is indicating.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
7
Q
Which would most likely cause the cylinder head temperature and engine oil temperature gauges to exceed their normal operating ranges?
A
Using fuel that has a lower-than-specified fuel rating.
Fuel with a lower-than-specified rating can lead to detonation. Detonation produces extreme heat and leads to high operating temperatures.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
8
Q
(Refer to figure 9, area C.) How should the flight controls be held while taxiing a tricycle-gear equipped airplane with a left quartering tailwind?
A
Left aileron down, elevator down.
When taxiing a tricycle-gear equipped airplane with a left quartering tailwind, use down aileron on the left hand wing and down elevator.
Airplane Flying Handbook Ch. 2
9
Q
(Refer to figure 4.) Which marking identifies the never-exceed speed?
A
Never-exceed speed is indicated on the airspeed indicator with a red line. This is the maximum speed that an aircraft can be operated in smooth air.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
10
Q
(Refer to figure 4.) What is the full flap operating range for the airplane?
A
55 to 100 KTS.
The white arc on the airspeed indicator represents the flap operating range for an aircraft. It extends from 55 to 100 knots.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
11
Q
For internal cooling, reciprocating aircraft engines are especially dependent on
A
the circulation of lubricating oil.
Airflow over the cylinders and other parts of the engine provides some measure of cooling. Additional heat is absorbed by the oil and carried to the oil cooling system for dissipation.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
12
Q
If the engine oil temperature and cylinder head temperature gauges have exceeded their normal operating range, the pilot may have been operating with
A
too much power and with the mixture set too lean.
High power settings with a lean mixture may lead to overheating. The additional fuel of a richer mixture has a cooling effect.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
13
Q
(Refer to figure 4.) Which color identifies the power-off stalling speed with wing flaps and landing gear in the landing configuration?
A
Lower limit of the white arc.
The lower limit of the white arc represents power-off stalling speed with the wing flaps and landing gear in the landing position.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
14
Q
A disconnected ground wire from a magneto to the ignition switch:
A
Could allow the engine to continue to run after the ignition switch is turned off.
Even with the ignition switch in the OFF position, if the ground wire between a magneto and the ignition switch becomes disconnected or broken, the engine could accidentally start if the propeller is moved with residual fuel in the cylinder. If this occurs, the only way to stop the engine is to move the mixture lever to the idle cutoff position.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
15
Q
In the Northern Hemisphere, the magnetic compass will normally indicate a turn toward the south when
A
the aircraft is decelerated while on a west heading.
Acceleration / deceleration error is most pronounced on a heading of east or west. ANDS, Accelerate, North, Decelerate, South.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
16
Q
Under what condition is indicated altitude the same as true altitude?
A
When at sea level under standard conditions.
When the altimeter is at sea level under standard conditions, indicated altitude and true altitude are equal.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
17
Q
Detonation may occur at high-power settings when
A
the fuel mixture ignites instantaneously instead of burning progressively and evenly.
Detonation is a sudden explosion or shock to a small area of the piston top. Detonation may occur at high-power settings when the fuel mixture ignites instantaneously instead of burning progressively and evenly.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
18
Q
Most of the heat caused by internal combustion is eliminated via:
A
The exhaust system.
The burning fuel within the cylinders produces intense heat, most of which is expelled through the exhaust system.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
19
Q
The possibility of carburetor icing exists even when the ambient air temperature is as
A
high as 70 °F and the relative humidity is high.
Be alert for carburetor icing if the temperature is between 20° F and 70° F with visible moisture or high humidity.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
20
Q
What is the purpose of the rudder on an airplane?
A
To control yaw.
The rudder controls the aircraft around the vertical or yaw axis.
Pilot’s Handbook of Aeronautical Knowledge Ch. 6
21
Q
What is one of the advantages of an alternator over a generator in an airplane engine?
A
Alternators have several advantages over generators. Alternators produce sufficient current to operate the entire electrical system, even at slower engine speeds, by producing alternating current (AC), which is converted to DC. The electrical output of an alternator is more constant throughout a wide range of engine speeds.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
22
Q
The presence of carburetor ice in an aircraft equipped with a fixed-pitch propeller can be verified by applying carburetor heat and noting
A
a decrease in RPM and then a gradual increase in RPM.
The RPM will decrease with the application of carburetor heat due to the decreased air density of the warmer air. As the ice melts, the RPM will increase due to better fuel/air mixture flow. Some engine roughness may occur as the ice melts.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
23
Q
If the grade of fuel used in an aircraft engine is lower than specified for the engine, it will most likely cause
A
detonation.
Detonation is the explosive burning of the fuel/air mixture. This phenomenon is often caused by the use of a lower than recommended grade of fuel.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
24
Q
The pitot system provides impact pressure for which instrument?
A
Airspeed indicator.
The pitot tube provides ram air to operate the airspeed indicator. It does not affect the other two pitot-static instruments.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
25
Q
Which aileron positions should a pilot generally use when taxiing in strong quartering headwinds?
A
Aileron up on the side from which the wind is blowing.
When taxiing in a strong quartering headwind, use an up aileron on the side from which the wind is blowing.
Airplane Flying Handbook Ch. 2
26
Q
During flight, when are the indications of a magnetic compass accurate?
A
Only in straight-and-level unaccelerated flight.
To reduce errors, the magnetic compass should be read only when the aircraft is flying straight and level and at a constant speed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
27
Q
(Refer to figure 9, area C.) How should the flight controls be held while taxiing a tailwheel airplane with a left quartering tailwind?
A
Left aileron down, elevator down.
When taxiing a tailwheel airplane with a left quartering tailwind, use down aileron on the left hand wing and down elevator.
Airplane Flying Handbook Ch. 2
Airplane Flying Handbook Ch. 14
28
Q
Which instrument(s) will become inoperative if the static vents become clogged?
A
Airspeed, altimeter, and vertical speed.
All three pitot-static instruments are affected by a clogged static vent.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
29
Q
(Refer to figure 9, area A.) How should the flight controls be held while taxiing a tricycle-gear equipped airplane into a left quartering headwind?
A
Left aileron up, elevator neutral.
When taxiing a tricycle-gear equipped airplane into a left quartering headwind, use up aileron on the left hand wing and neutral elevator.
Airplane Flying Handbook Ch. 2
30
Q
Why would an aviation piston engine continue running after the ignition switched is placed in the OFF position?
A
The magneto’s grounding wire is broken.
Even with the ignition switch in the OFF position, if the ground wire between the magneto and the ignition switch becomes disconnected or broken, the only way to stop the engine is to move the mixture lever to the idle cutoff position, then have the system checked by a qualified AMT.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
31
Q
What is absolute altitude?
A
The vertical distance of the aircraft above the surface.
Absolute altitude is the vertical distance of an aircraft above the terrain.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
32
Q
What is an important airspeed limitation that is not color coded on airspeed indicators?
A
Maneuvering speed.
Maneuvering speed is the maximum speed for abrupt maneuvers. It is also sometimes called the “rough air” speed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
33
Q
At the beginning of a turn from a northerly heading:
A
The compass will show a turn in the opposite direction.
When starting a turn from a northerly heading, the compass lags behind the turn by a slight turn in the opposite direction.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
34
Q
When taxiing with strong quartering tailwinds, which aileron positions should be used?
A
Aileron down on the side from which the wind is blowing.
When taxiing with a strong quartering tailwind, use down aileron on the side from which the wind is blowing to keep the wind from pushing under the wing, and flipping the airplane.
Airplane Flying Handbook Ch. 2
35
Q
The angular difference between true north and magnetic north is
A
magnetic variation.
Magnetic variation is the angular difference between the true, or geographic, north pole and the magnetic north pole at a given point.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
36
Q
Detonation occurs in a reciprocating aircraft engine when
A
the unburned charge in the cylinders explodes instead of burning normally.
Detonation or knock is the explosive burning of the fuel/air mixture inside the cylinder.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
37
Q
An electrical system failure (battery and alternator) occurs during flight. In this situation, you would
A
experience avionics equipment failure.
A complete electrical system failure would result in the loss of the avionics equipment as well as most other electrical equipment.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
38
Q
Which wind condition would be most critical when taxiing a nosewheel equipped high-wing airplane?
A
Quartering tailwind.
When taxiing a nosewheel equipped high-wing airplane the most critical wind position is a quartering tailwind because it could lift the tail and push the airplane over on its prop and far wingtip.
Airplane Flying Handbook Ch. 2
39
Q
Which condition would cause the altimeter to indicate a lower altitude than true altitude?
A
Air temperature warmer than standard.
When air is warmer than average, the aircraft is higher than the altimeter indicates. Therefore, the altimeter will read lower than the aircraft’s actual altitude.
Aviation Weather (AC 00-6) Ch. 3
40
Q
The common heading indicator requires periodic adjustment. It is important to check its indications frequently against the:
A
Magnetic compass.
The magnetic compass provides an indication of the magnetic heading. When setting the gyroscopic heading indicator to agree with the magnetic compass, use the average indication between the swings.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
41
Q
Should it become necessary to handprop an airplane engine, it is extremely important that a competent pilot
A
be at the controls in the cockpit.
It is extremely important that a competent pilot be at the controls in the cockpit because of the hazards involved in hand propping.
Airplane Flying Handbook Ch. 2
42
Q
With regard to carburetor ice, float-type carburetor systems in comparison to fuel injection systems are generally considered to be
A
more susceptible to icing.
Float-type carburetors are more prone to carburetor ice than fuel injection systems. In a fuel injected system, mixing of the fuel & air occurs just prior to the intake valve or in the cylinder. The cooling effect of fuel vaporization has less influence.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
43
Q
What is pressure altitude?
A
The altitude indicated when the barometric pressure scale is set to 29.92.
Pressure altitude is indicated when the altimeter setting window (barometric scale) is adjusted to 29.92” Hg.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
44
Q
(Refer to figure 3.) Which altimeter(s) indicate(s) more than 10,000 feet?
A
1 and 2 only.
The shortest hand indicates tens of thousands of feet. The middle length hand represents thousands of feet and the longest hand indicates hundreds of feet. The shortest hand is past 10,000’ on altimeters 1 & 2.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
45
Q
How is engine operation controlled on an engine equipped with a constant-speed propeller?
A
The throttle controls power output as registered on the manifold pressure gauge and the propeller control regulates engine RPM.
On an engine equipped with a constant-speed propeller, throttle controls the power output registered on a manifold pressure gauge and the propeller control regulates the engine RPM.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
46
Q
What is one procedure to aid in cooling an engine that is overheating?
A
Enrichen the fuel mixture.
A rich mixture will usually tend to cool the engine. Leftover fuel, caused by a rich mixture, absorbs some heat from the cylinders, resulting in a cooler engine.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
47
Q
As altitude increases, the indicated airspeed at which a given airplane stalls in a particular configuration will
A
remain the same regardless of altitude.
Indicated airspeed is dependent on the amount of ram air pressure entering the pitot tube. Regardless of an aircraft’s altitude, the indicated airspeed for a particular maneuver remains the same.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
48
Q
One of the main functions of flaps during approach and landing is to
A
increase the angle of descent without increasing the airspeed.
Flaps increase lift (with an increase in induced drag) enabling the pilot to make steeper approaches without an increase in airspeed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 6
49
Q
On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?
A
In the event engine-driven fuel pump fails.
The auxiliary or “boost” pump is used as a backup for the engine-driven pump.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
50
Q
Why are reciprocating engines preferred over other types for small aircraft.
A
They are less expensive to operate and they are simple in design.
Most small aircraft are designed with reciprocating engines. reciprocating engine technology has improved dramatically over the past two decades.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
51
Q
(Refer to figure 4.) Which color identifies the power-off stalling speed in a specified configuration?
A
Lower limit of the green arc.
The lower limit of the green arc represents the aircraft’s power-off stalling speed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
52
Q
What is true altitude?
A
The vertical distance of the aircraft above sea level.
True altitude is the true vertical distance of the aircraft above sea level. It is the actual altitude of the aircraft.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
53
Q
Which factor would tend to increase the density altitude at a given airport?
A
An increase in ambient temperature.
The primary factor that would tend to increase density altitude at a given airport is an increase in ambient temperature. The hotter the air temperature, the less dense the air. This is equivalent to increasing density altitude.
Pilot’s Handbook of Aeronautical Knowledge Ch. 10
Aviation Weather (AC 00-6) Ch. 3
54
Q
If a pilot changes the altimeter setting from 30.11 to 29.96, what is the approximate change in indication?
A
Altimeter will indicate 150 feet lower.
The difference between 30.11 and 29.96 is .15” Hg. The standard lapse rate is 1” Hg per 1,000’ in altitude. Therefore, .15 x 1,000’ = 150’ and the indication would be lower.
Aviation Weather (AC 00-6) Ch. 3
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
55
Q
Excessively high engine temperatures will
A
cause loss of power, excessive oil consumption, and possible permanent internal engine damage.
Operating at an excessively high engine temperature can lead to loss of power, excessive oil consumption, and detonation. Damage to the cylinders, pistons, piston rings, and valves may occur.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
56
Q
How is power controlled on an airplane equipped with a constant-speed propeller?
A
Power output is controlled by the throttle and indicated by a manifold pressure gauge.
On aircraft equipped with a constant-speed propeller, power output is controlled by the throttle and indicated by a manifold pressure gauge.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
57
Q
The air/fuel ratio (AFR) is the measurement of:
A
The ratio of weight of air to the weight of fuel in the mixture.
The ratio of air/fuel is expressed in terms of weight (mass).
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
58
Q
(Refer to figure 4.) The maximum speed at which the airplane can be operated in smooth air is
A
208 knots.
Never-exceed speed is indicated on the airspeed indicator with a red line. This is the maximum speed that an aircraft can be operated in smooth air.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
59
Q
Which instrument is affected if the pitot tube is blocked?
A
Airspeed indicator.
When dynamic pressure cannot enter the pitot tube opening, the airspeed indicator no longer operates.
Instrument Flying Handbook Ch. 5
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
60
Q
The only north seeking instrument in a typical training airplane is:
A
The magnetic compass.
The magnetic compass inherently seeks north.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
61
Q
(Refer to figure 3.) Altimeter 3 indicates
A
9,500 feet.
The shortest hand indicates tens of thousands of feet. The middle length hand represents thousands of feet and the longest hand indicates hundreds of feet. The shortest hand is just before 10,000’. The hands read 9,500’.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
62
Q
The pitot-static system drives the:
A
Airspeed indicator, altimeter, and vertical speed indicator.
The pitot-static system is a combined system that utilizes the static air pressure and the dynamic pressure due to the motion of the aircraft through the air. These combined pressures are utilized for the operation of the airspeed indicator (ASI), altimeter, and vertical speed indicator (VSI).
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
63
Q
Which gyroscopic instrument is the foundation for all instrument flight?
A
Attitude indicator.
The attitude indicator is the foundation for all instrument flight, this instrument reflects the airplane’s attitude in relation to the horizon.
Pilot’s Handbook of Aeronautical Knowledge
64
Q
The order of operation regarding the strokes of a typical four-stroke per cycle airplane engine is:
A
Intake, compression, power, and exhaust.
The intake, compression, power, and exhaust processes occur in four separate strokes of the piston.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
65
Q
If you land without the use of flaps, the approach will be:
A
Faster and shallower.
Landing without flaps, lift is reduced resulting in a faster and shallower approach. Flaps increase the airfoil camber, resulting in a significant increase in the coefficient of lift at a given AOA.
Pilot’s Handbook of Aeronautical Knowledge Ch. 6
66
Q
What action can a pilot take to aid in cooling an engine that is overheating during a climb?
A
Reduce rate of climb and increase airspeed.
Engine temperature can be reduced by lowering the nose of the aircraft to increase the flow of air over the cylinders and reduce engine strain.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
67
Q
Under what condition will true altitude be lower than indicated altitude?
A
In colder than standard air temperature.
True altitude will be lower than indicated altitude when the aircraft is in an air temperature colder than standard.
Aviation Weather (AC 00-6) Ch. 3
68
Q
If the outside air temperature (OAT) at a given altitude is warmer than standard, the density altitude is
A
higher than pressure altitude.
If the temperature is above standard, the density altitude will be higher than the pressure altitude.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
69
Q
For a given power setting with a constant-speed, variable pitch propeller:
A
Low pitch results in high RPM.
When maximum power and thrust are required, the constant-speed propeller is at a low propeller blade angle or pitch allowing it to pass through the air faster.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
70
Q
(Refer to figure 4.) Which color identifies the normal flap operating range?
A
The white arc.
The white arc on the airspeed indicator represents the flap operating range for an aircraft. It extends from 55 to 100 knots.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
71
Q
If it is necessary to set the altimeter from 29.15 to 29.85, what change occurs?
A
700-foot increase in indicated altitude.
One inch of Hg is equal to approximately 1,000’ of altitude. The difference between 29.15 and 29.85 is an increase of .70” Hg, or 700’ of increased altitude indication.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
72
Q
Why do most standard certificated aircraft incorporate a dual ignition system with two individual magnetos, separate sets of wires, and spark plugs?
A
To increase reliability of the ignition system.
Dual systems are intended to increase reliability of the ignition system. If one of the magnetos fails, the other is unaffected. The engine continues to operate normally, although a slight decrease in engine power can be expected.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
73
Q
You have been running an excessively rich mixture for some time now. As a result:
A
The spark plugs may become fouled.
Excessively rich mixtures will cause additional carbon buildup on spark plugs, which may cause engine roughness.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
74
Q
(Refer to figure 4.) What is the maximum flaps-extended speed?
A
100 knots.
The upper limit of the white arc represents the aircraft’s maximum flap extension speed, or 100 knots in this example.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
75
Q
To minimize the side loads placed on the landing gear during touchdown, the pilot should keep the
A
longitudinal axis of the aircraft parallel to the direction of its motion.
It is important that the touchdown of an aircraft occur with the aircraft’s longitudinal axis parallel to the direction in which it is moving along the runway to reduce side loads on the landing gear.
Airplane Flying Handbook Ch. 9
76
Q
(Refer to figure 3.) Altimeter 1 indicates
A
10,500 feet.
The shortest hand indicates tens of thousands of feet. The middle length hand represents thousands of feet and the longest hand indicates hundreds of feet. The shortest hand is just past 10,000’. The hands read 10,500’.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
77
Q
While cruising at 9,500 feet MSL, the fuel/air mixture is properly adjusted. What will occur if a descent to 4,500 feet MSL is made without readjusting the mixture?
A
The fuel/air mixture may become excessively lean.
Air becomes more dense with a decrease in altitude. If the mixture is not adjusted, the fuel to air ratio will probably become too lean (insufficient fuel for the amount of air).
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
78
Q
(Refer to figure 4.) What is the caution range of the airplane?
A
165 to 208 knots.
The caution range of an aircraft is resented by the yellow arc on the airspeed indicator. It extends from 165 to 208 knots.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
79
Q
As you are climbing to your cruise altitude, you realize you forgot to lean the mixture control. What happens to fuel/air mixture entering the engine?
A
The fuel-air mixture becomes richer because the density of air decreases while the amount of fuel remains constant.
As altitude increases, the density of air entering the carburetor/intake manifold decreases, while the density of the fuel remains the same. This creates a progressively richer mixture that can result in engine roughness and an appreciable loss of power.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
80
Q
Generally speaking, the use of carburetor heat tends to
A
decrease engine performance.
Warm air is less dense than cold air. Applying carb heat will decreases air density causing a richer mixture (higher fuel to air ratio). This condition causes a decrease in engine performance.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
81
Q
A precaution for the operation of an engine equipped with a constant-speed propeller is to
A
avoid high manifold pressure settings with low RPM.
For any given RPM there is a manifold pressure that should not be exceeded. If manifold pressure is excessive for a given RPM, the pressure within the cylinders could be exceeded, placing undue stress on them. Follow the manufacturer’s recommendations.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
82
Q
Using a turn and slip indicator, how do you know that you are in a coordinated turn?
A
Centering the ball results in a coordinated turn.
If inadequate right rudder is applied in a right turn, a slip results. Too much right rudder causes the aircraft to skid through the turn. Centering the ball results in a coordinated turn.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
83
Q
(Refer to figure 3.) Altimeter 2 indicates
A
14,500 feet.
The shortest hand indicates tens of thousands of feet. The middle length hand represents thousands of feet and the longest hand indicates hundreds of feet. The shortest hand is past 10,000’. The hands read 14,500’.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
84
Q
Filling the fuel tanks after the last flight of the day is considered a good operating procedure because this will
A
prevent moisture condensation by eliminating airspace in the tanks.
Moisture can condense from the air and contaminate the fuel. If there is no space for air inside the tank, this cannot occur.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
85
Q
What change occurs in the fuel/air mixture when carburetor heat is applied?
A
The fuel/air mixture becomes richer.
Warm air is less dense than cold air. Applying carb heat will decreases air density causing a richer mixture (higher fuel to air ratio).
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
86
Q
If the pitot tube and outside static vents become clogged, which instruments would be affected?
A
The altimeter, airspeed indicator, and vertical speed indicator.
The pitot-static instruments are the altimeter, vertical speed indicator, and airspeed indicator.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
87
Q
What is the purpose of the airplane engine’s mixture control?
A
To regulate the ratio of gasoline to air entering the fuel distribution system.
To maintain the correct fuel-air mixture, the mixture must be leaned using the mixture control. Leaning the mixture decreases fuel flow, which compensates for the decreased air density at high altitude.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
88
Q
Excessively high engine temperatures, either in the air or on the ground, will
A
cause loss of power, excessive oil consumption, and possible permanent internal engine damage.
Whenever an engine overheats, there can be permanent damage caused as well as loss of power and excessive oil consumption.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
89
Q
Applying carburetor heat will
A
enrich the fuel/air mixture.
Warm air is less dense than cold air. Applying carb heat will decreases air density causing a richer mixture (higher fuel to air ratio).
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
90
Q
During the run-up at a high-elevation airport, a pilot notes a slight engine roughness that is not affected by the magneto check but grows worse during the carburetor heat check. Under these circumstances, what would be the most logical initial action?
A
Check the results obtained with a leaner setting of the mixture.
The aircraft is at a higher altitude airport, causing the proper ratio of fuel to air to be too rich (too much fuel for available air). By leaning out the fuel with the mixture control, the proper ratio will be attained and the roughness will smooth out.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
91
Q
What is one purpose of wing flaps?
A
To enable the pilot to make steeper approaches to a landing without increasing the airspeed.
Wing flaps increase lift and drag allowing the pilot to make steeper approaches without increasing airspeed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 6
92
Q
The uncontrolled firing of the fuel/air charge in advance of normal spark ignition is known as
A
pre-ignition.
The uncontrolled firing of the fuel/air charge in advance of normal spark ignition is the definition of preignition.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
93
Q
If a flight is made from an area of low pressure into an area of high pressure without the altimeter setting being adjusted, the altimeter will indicate
A
lower than the actual altitude above sea level.
When flying from an area of low pressure to an area of high pressure, the aircraft will be higher than the indication on the altimeter. Therefore, the altimeter will read lower than the aircraft’s actual altitude.
Aviation Weather (AC 00-6) Ch. 3
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
94
Q
What would happen if your airplane experienced a complete electrical failure during flight?
A
The airplane will lose all electrical equipment.
When the alternator fails, the entire electrical load is placed on the battery. If the battery also fails, the airplane will lose all electrical equipment, including position lights, instrument lights, and radio equipment.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
95
Q
On a turn from south:
A
The compass exaggerates the rate of turn.
When starting a turn from a southerly heading, the compass leads the turn or exaggerates the rate of turn.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
96
Q
What is an advantage of a constant-speed propeller?
A
Permits the pilot to select the blade angle for the most efficient performance.
A controllable-pitch propeller permits the pilot to select the blade angle that will result in the most efficient performance for a particular flight condition.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
97
Q
A too rich mixture:
A
Will create spark plug fouling.
If the mixture is too rich it can result in engine roughness and an appreciable loss of power. The roughness normally is due to spark plug fouling from excessive carbon buildup on the plugs.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
98
Q
Position (navigation) lights on an aircraft are:
A
Red for the left wingtip, green for the right wing tip, and white for the tail.
Lights on an aircraft consist of a red light on the left wing, a green light on the right wing, and a white light on the tail.
Airplane Flying Handbook Ch. 11
99
Q
In the Northern Hemisphere, a magnetic compass will normally indicate a turn toward the north if
A
an aircraft is accelerated while on an east or west heading.
Acceleration / deceleration error is most pronounced on a heading of east or west. ANDS, Accelerate, North, Decelerate, South.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
100
Q
(Refer to figure 9, area B.) How should the flight controls be held while taxiing a tailwheel airplane into a right quartering headwind?
A
Right aileron up, elevator up.
When taxiing a tailwheel airplane into a right quartering headwind, use up aileron on the right hand wing and up elevator.
Airplane Flying Handbook Ch. 2
Airplane Flying Handbook Ch. 14
101
Q
An abnormally high engine oil temperature indication may be caused by
A
the oil level being too low.
Oil absorbs and dissipates engine heat caused by the internal combustion process.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
102
Q
Why is it a good idea to visually inspect to make sure that the crankcase breather lines are free of ice?
A
Ice may have formed as a result of the crankcase vapors freezing after the engine has been turned off.
You should always visually inspect to make sure that the crankcase breather lines are free of ice. The ice may have formed as a result of the crankcase vapors freezing in the lines after the engine has been turned off.
Cold Weather Operation of Aircraft (AC 91-13)
103
Q
What is density altitude?
A
The pressure altitude corrected for nonstandard temperature.
Density altitude is pressure altitude corrected for nonstandard temperature variations.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
104
Q
On a turn from a northerly heading the compass will:
A
Lag behind the airplane.
When starting a turn from a northerly heading, the compass lags behind the turn.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
105
Q
The magnetic compass:
A
Is self powered.
The magnetic compass is the simplest instrument in the panel and needs no outside power to operate.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
106
Q
The only instrument that provides an instantaneous, direct indication of the airplane’s pitch and bank attitude is the:
A
Attitude indicator.
The relationship of the miniature aircraft to the horizon bar is the same as the relationship of the real aircraft to the actual horizon. The instrument gives an instantaneous indication of even the smallest changes in pitch and bank attitude.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
107
Q
To properly purge water from the fuel system of an aircraft equipped with fuel tank sumps and a fuel strainer quick drain, it is necessary to drain fuel from the
A
fuel strainer drain and the fuel tank sumps.
If water shows up in any of the samples from the fuel tank drains, the fuel strainer drain, or other sump drains, they all must be drained until no more water is present in any of the samples to properly purge the water from the fuel system.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
108
Q
The airspeed indicator has various color markings. The green arc is:
A
The normal operating range of the airplane.
The green arc identifies the normal operating range of the aircraft. Most flying occurs within this range.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
109
Q
(Refer to figure 3) Altimeter 2 indicates
A
14,500 feet.
The smallest hand represents 10s of thousands of feet. Here the smallest hand is beyond the 10,000 foot mark. The next smallest hand, the 1,000 foot pointer, is past the 4,000 foot mark and the largest hand, the 100 foot pointer, is on 500 feet—14,500 feet.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
110
Q
If an aircraft is equipped with a fixed-pitch propeller and a float-type carburetor, the first indication of carburetor ice would most likely be
A
loss of RPM.
The formation of ice in the carburetor reduces the amount of fuel reaching the engine. This causes a reduction of RPM.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
111
Q
What should be the first action after starting an aircraft engine?
A
Adjust for proper RPM and check for desired indications on the engine gauges.
As soon as the engine starts, ensure that the aircraft is not moving, set the power to the recommended RPM, and check the engine gauges for proper indications.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
112
Q
The rotating propeller of an airplane makes a very good gyroscope and thus has similar properties. In a prop driven airplane, a decrease in pitch attitude results in:
A
A yawing moment to the left around the vertical axis.
A decrease in pitch attitude has the same effect as applying a force to the top of the propeller’s plane of rotation. The resultant force acting 90° ahead in the direction of the rotation causes a yawing moment to the left around the vertical axis.
Pilot’s Handbook of Aeronautical Knowledge Ch. 5
113
Q
Static pressure, also known as ambient pressure, is:
A
Always present whether an aircraft is moving or at rest.
Static pressure, also known as ambient pressure, is always present whether an aircraft is moving or at rest. It is the barometric pressure in the local area.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
114
Q
In the Northern Hemisphere, if an aircraft is accelerated or decelerated, the magnetic compass will normally indicate
A
correctly when on a north or south heading.
Acceleration / deceleration error is most pronounced on a heading of east or west. This error does not occur on a north or south heading.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
115
Q
Which condition is most favorable to the development of carburetor icing?
A
Temperature between 20 and 70 °F and high humidity.
Be alert for carburetor icing if the temperature is between 20° F and 70° F with visible moisture or high humidity.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
116
Q
(Refer to figure 7.) How should a pilot determine the direction of bank from an attitude indicator such as the one illustrated?
A
By the relationship of the miniature airplane (C) to the deflected horizon bar (B).
The relationship of the miniature airplane to the horizon bar is the same as the relationship of the real aircraft to the actual horizon.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
117
Q
A constant-speed propeller is more efficient than other propellers because:
A
It allows selection of the most efficient engine rpm for the given conditions.
A constant-speed propeller converts a high percentage of brake horsepower into thrust horsepower over a wide range of rpm and airspeed combinations. Therefore, it is more efficient than other propellers.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
118
Q
In the Northern Hemisphere, a magnetic compass will normally indicate initially a turn toward the west if
A
a right turn is entered from a north heading.
On a northerly heading, if a right turn is made toward the east, the compass will initially indicate a turn in the opposite direction or toward the west.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
119
Q
In the Northern Hemisphere, a magnetic compass will normally indicate initially a turn toward the east if
A
a left turn is entered from a north heading.
On a northerly heading, if a left turn is made toward the west, the compass will initially indicate a turn in the opposite direction or toward the east.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
120
Q
One purpose of the dual ignition system on an aircraft engine is to provide for
A
improved engine performance.
The dual ignition system provides improved safety through redundancy and improved engine performance through more efficient burning of the fuel/air mixture. This efficiency is derived through dual flame fronts within the cylinder.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
121
Q
If a pilot suspects that the engine (with a fixed-pitch propeller) is detonating during climb-out after takeoff, the initial corrective action to take would be to
A
lower the nose slightly to increase airspeed.
Engine detonation results in an overheated engine. Engine temperature can be reduced by lowering the nose of the aircraft to increase the flow of air over the cylinders and reduce engine strain.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
122
Q
Under which condition will pressure altitude be equal to true altitude?
A
When standard atmospheric conditions exist.
True altitude is the actual altitude above mean sea level. Pressure altitude is the altitude in the standard atmosphere where pressure is the same as where you are. Pressure altitude is the same as true altitude in standard atmosphere.
Aviation Weather (AC 00-6) Ch. 3
123
Q
Altimeter setting is the value to which the barometric pressure scale of the altimeter is set so the altimeter indicates
A
true altitude at field elevation.
True altitude is the true vertical distance of the aircraft above sea level. It is the actual altitude of the aircraft.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
124
Q
There is no acceleration/deceleration error on a heading of:
A
North or south.
The magnetic dip and the forces of inertia cause magnetic compass errors when accelerating and decelerating on easterly and westerly headings.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
125
Q
What does the red line on an airspeed indicator represent?
A
Never-exceed speed.
Never-exceed speed is indicated on the airspeed indicator with a red line. This is the maximum speed that an aircraft can be operated in smooth air. It is abbreviated as VNE.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
FAR 1
126
Q
Deviation in a magnetic compass is caused by the
A
magnetic fields within the aircraft distorting the lines of magnetic force.
Magnetic interference, from magnetic fields produced by metals and electrical accessories in the aircraft, disturb the compass needles and produce additional error.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
127
Q
Deviation error of the magnetic compass is caused by
A
certain metals and electrical systems within the aircraft.
Magnetic fields caused by aircraft electronics and wiring can affect the accuracy of the magnetic compass - called compass deviation.
Instrument Flying Handbook Ch. 5
128
Q
If a flight is made from an area of high pressure into an area of lower pressure without the altimeter setting being adjusted, the altimeter will indicate
A
higher than the actual altitude above sea level.
When flying from an area of high pressure to an area of low pressure, the aircraft will be lower than the indication on the altimeter. Therefore, the altimeter will read higher than the aircraft’s actual altitude.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
Aviation Weather (AC 00-6) Ch. 3
129
Q
Extending the flaps:
A
Lowers the stalling speed.
Flaps allow a compromise between high cruising speed and low landing speed because they may be extended when needed and retracted into the wing’s structure when not needed.
Pilot’s Handbook of Aeronautical Knowledge Ch. 6
130
Q
The operating principle of float-type carburetors is based on the
A
difference in air pressure at the venturi throat and the air inlet.
When air flows through a venturi, a low-pressure area is created. This lower pressure causes fuel to flow from the discharge nozzle and mix with the air in the throat of the venturi.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
131
Q
(Refer to figure 4.) What is the maximum structural cruising speed?
A
165 knots.
The upper limit of the green arc represents the maximum structural cruising speed of the aircraft, or 165 knots.
Pilot’s Handbook of Aeronautical Knowledge Ch. 8
132
Q
(Refer to figure 5.) A turn coordinator provides an indication of the
A
movement of the aircraft about the yaw and roll axis.
The turn coordinator shows the rate of turn about the yaw or vertical axis and the rate of roll about the roll or longitudinal axis.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7
133
Q
(Refer to figure 6.) To receive accurate indications during flight from a heading indicator, the instrument must be
A
periodically realigned with the magnetic compass as the gyro precesses.
The heading indicator is not direction-seeking. It is important to check its indication frequently and reset it with reference to the magnetic compass.
Pilot’s Handbook of Aeronautical Knowledge Ch. 7 | 10,628 | 46,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.858083 |
https://oeis.org/A000691 | 1,481,311,001,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542765.40/warc/CC-MAIN-20161202170902-00342-ip-10-31-129-80.ec2.internal.warc.gz | 863,649,563 | 4,155 | This site is supported by donations to The OEIS Foundation.
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A000691 Ramanujan's approximation to population of x^2 + y^2. (Formerly M0713 N0263) 2
1, 2, 3, 5, 9, 16, 29, 52, 94, 175, 327, 616, 1169, 2231, 4273, 8215, 15842, 30628, 59345, 115208, 224040, 436343, 850981, 1661663, 3248231, 6356076, 12448925, 24402959, 47873156, 93984236, 184632691, 362938014, 713852252, 1404817026 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 REFERENCES D. Shanks, The second-order term in the asymptotic expansion of B(x), Math. Comp., 18 (1964), 75-86. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS MAPLE Digits:=500; K:=.764223653589220662990698731250092328116790541393409514721686673 7496146416587328588384015050131312337219372691207925926341874206467 8084323063315434629380531605171169636177508819961243824994277683469 0516235139218719620569053295644670419176349770659569905712938660289 3858998296105166296089099177929836072973697200640316985128636517347 3921065768550978681981674707359066921; a:=n->round(evalf(K*int(1/sqrt(ln(t)), t=1..2^n))); # Salvador Perez (pies314(AT)hotmail.com), May 08 2005 CROSSREFS K = A064533. Sequence in context: A006788 A054650 A022857 * A192804 A255071 A103285 Adjacent sequences: A000688 A000689 A000690 * A000692 A000693 A000694 KEYWORD nonn AUTHOR EXTENSIONS More terms from Salvador Perez (pies314(AT)hotmail.com), May 08 2005 Corrected by Sean A. Irvine, Feb 24 2011 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 702 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-50 | longest | en | 0.564588 |
https://www.allinterview.com/showanswers/197210/mass-transfer-example-4-1-concentric-counter-current-heat-exchanger-used-cool-lu.html | 1,726,500,894,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00707.warc.gz | 590,517,598 | 8,403 | MASS TRANSFER - EXAMPLE 4.1 : A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO ?TOO) = (FW)(CW)(TOW ?TIW).
MASS TRANSFER - EXAMPLE 4.1 : A concentric, counter-current heat exchanger is used to cool lubricati..
MASS TRANSFER - ANSWER 4.1 : Rearranging the given equation, then FW = (FO)(CO)(TIO – TOO) / [ (CW)(TOW – TIW) ] = (0.1)(2131)(100 – 55) / [ (4178) (42 – 35) ] = 0.327891 kg / s. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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• Engineering AllOther (1379) | 1,161 | 4,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.887555 |
https://books.opencourseware.online/ocw051-one-way-anova/ | 1,627,515,734,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00567.warc.gz | 157,563,087 | 17,652 | # OCW051: One-Way ANOVA
## The F-Test
An F-test is any statistical test in which the test statistic has an F-distribution under the null hypothesis.
### Learning Objectives
Summarize the F-statistic, the F-test and the F-distribution.
### Key Takeaways
#### Key Points
• The F-test is most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled.
• Perhaps the most common F-test is that which tests the hypothesis that the means and standard deviations of several populations are equal. (Note that all populations involved must be assumed to be normally distributed.)
• The F-test is sensitive to non-normality.
• The F-distribution is skewed to the right, but as the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
#### Key Terms
• ANOVA: Analysis of variance—a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups).
• Type I error: Rejecting the null hypothesis when the null hypothesis is true.
• F-Test: A statistical test using the F-distribution, most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled.
An F-test is any statistical test in which the test statistic has an F-distribution under the null hypothesis. It is most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled. Exact F-tests mainly arise when the models have been fitted to the data using least squares. The name was coined by George W. Snedecor, in honour of Sir Ronald A. Fisher. Fisher initially developed the statistic as the variance ratio in the 1920s.
The F-test is sensitive to non-normality. In the analysis of variance (ANOVA), alternative tests include Levene’s test, Bartlett’s test, and the Brown–Forsythe test. However, when any of these tests are conducted to test the underlying assumption of homoscedasticity (i.e., homogeneity of variance), as a preliminary step to testing for mean effects, there is an increase in the experiment-wise type I error rate.
Examples of F-tests include:
1. The hypothesis that the means and standard deviations of several populations are equal. (Note that all populations involved must be assumed to be normally distributed.) This is perhaps the best-known F-test, and plays an important role in the analysis of variance (ANOVA).
2. The hypothesis that a proposed regression model fits the data well (lack-of-fit sum of squares).
3. The hypothesis that a data set in a regression analysis follows the simpler of two proposed linear models that are nested within each other.
4. Scheffé’s method for multiple comparisons adjustment in linear models.
### The F-Distribution
F-distribution: The F-distribution is skewed to the right and begins at the x-axis, meaning that F-values are always positive.
The F-distribution exhibits the following properties, as illustrated in the above graph:
1. The curve is not symmetrical but is skewed to the right.
2. There is a different curve for each set of degrees of freedom.
3. The F-statistic is greater than or equal to zero.
4. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
The F-statistic also has a common table of values, as do zscores and t-scores.
## The One-Way F-Test
The [latex]text{F}[/latex]-test as a one-way analysis of variance assesses whether the expected values of a quantitative variable within groups differ from each other.
### Learning Objectives
Explain the purpose of the one-way ANOVA [latex]text{F}[/latex]-test and perform the necessary calculations.
### Key Takeaways
#### Key Points
• The advantage of the ANOVA [latex]text{F}[/latex]-test is that we do not need to pre-specify which treatments are to be compared, and we do not need to adjust for making multiple comparisons.
• The disadvantage of the ANOVA [latex]text{F}[/latex]-test is that if we reject the null hypothesis, we do not know which treatments can be said to be significantly different from the others.
• If the [latex]text{F}[/latex]-test is performed at level [latex]alpha[/latex] we cannot state that the treatment pair with the greatest mean difference is significantly different at level [latex]alpha[/latex].
• The [latex]text{F}[/latex]– statistic will be large if the between-group variability is large relative to the within-group variability, which is unlikely to happen if the population means of the groups all have the same value.
#### Key Terms
• omnibus: containing multiple items
• F-Test: a statistical test using the [latex]text{F}[/latex] distribution, most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled
• ANOVA: Analysis of variance—a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups).
The [latex]text{F}[/latex] test as a one-way analysis of variance is used to assess whether the expected values of a quantitative variable within several pre-defined groups differ from each other. For example, suppose that a medical trial compares four treatments. The ANOVA [latex]text{F}[/latex]-test can be used to assess whether any of the treatments is on average superior, or inferior, to the others versus the null hypothesis that all four treatments yield the same mean response. This is an example of an “omnibus” test, meaning that a single test is performed to detect any of several possible differences.
Alternatively, we could carry out pairwise tests among the treatments (for instance, in the medical trial example with four treatments we could carry out six tests among pairs of treatments). The advantage of the ANOVA [latex]text{F}[/latex]-test is that we do not need to pre-specify which treatments are to be compared, and we do not need to adjust for making multiple comparisons. The disadvantage of the ANOVA [latex]text{F}[/latex]-test is that if we reject the null hypothesis, we do not know which treatments can be said to be significantly different from the others. If the [latex]text{F}[/latex]-test is performed at level [latex]alpha[/latex] we cannot state that the treatment pair with the greatest mean difference is significantly different at level [latex]alpha[/latex].
The formula for the one-way ANOVA [latex]text{F}[/latex]-test statistic is:
[latex]text{F}=dfrac { text{explained variance} }{ text{unexplained variance} }[/latex]
or
[latex]text{F}=dfrac { text{between-group variability} }{ text{within-group variability} }[/latex]
The “explained variance,” or “between-group variability” is:
[latex]displaystyle sum _{ text{i} }^{ } frac{{ text{n} }_{ text{i} }{ left( { bar { text{Y} } }_{ text{i} }-bar { text{Y} } right) }^{ 2 }}{left( text{K}-1 right)}[/latex]
where [latex]{ bar { text{Y} } }_{ text{i} }[/latex] denotes the sample mean in the [latex]text{i}[/latex]th group, [latex]text{n}_text{i}[/latex] is the number of observations in the [latex]text{i}[/latex]th group, [latex]bar { text{Y} }[/latex] denotes the overall mean of the data, and [latex]text{K}[/latex] denotes the number of groups.
The “unexplained variance”, or “within-group variability” is:
[latex]displaystyle sum _{ text{ij} }^{ }frac{{ left( { bar { text{Y} } }_{ text{ij} }-{ bar { text{Y} } }_{ text{i} } right) }^{ 2 }} {left( text{N}-text{K} right)}[/latex]
where [latex]bar{text{Y}_{text{ij}}}[/latex] is the [latex]text{j}[/latex]th observation in the [latex]text{i}[/latex]th out of [latex]text{K}[/latex] groups and [latex]text{N}[/latex] is the overall sample size. This [latex]text{F}[/latex]-statistic follows the [latex]text{F}[/latex]– distribution with [latex]text{K}-1[/latex], [latex]text{N}-text{K}[/latex] degrees of freedom under the null hypothesis. The statistic will be large if the between-group variability is large relative to the within-group variability, which is unlikely to happen if the population means of the groups all have the same value.
Note that when there are only two groups for the one-way ANOVA [latex]text{F}[/latex]-test, [latex]text{F}=text{t}^2[/latex] where [latex]text{t}[/latex] is the Student’s [latex]text{t}[/latex]-statistic.
### Example
Four sororities took a random sample of sisters regarding their grade means for the past term. The data were distributed as follows:
• Sorority 1: 2.17, 1.85, 2.83, 1.69, 3.33
• Sorority 2: 2.63,1.77, 3.25, 1.86, 2.21
• Sorority 3: 2.63, 3.78, 4.00, 2.55, 2.45
• Sorority 4: 3.79, 3.45, 3.08, 2.26, 3.18
Using a significance level of 1%, is there a difference in mean grades among the sororities?
#### Solution
Let [latex]mu_1[/latex], [latex]mu_2[/latex], [latex]mu_3[/latex], [latex]mu_4[/latex] be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each size 5. Also, note that this is an example of a balanced design, since each factor (i.e., sorority) has the same number of observations.
[latex]text{H}_0: mu_1 = mu_2 = mu_3 = mu_4[/latex]
[latex]text{H}_text{a}: [/latex] Not all of the means [latex]mu_1[/latex], [latex]mu_2[/latex], [latex]mu_3[/latex], [latex]mu_4[/latex] are equal
Distribution for the test: [latex]text{F}_{3, 16}[/latex]
where [latex]text{k}=4[/latex] groups and [latex]text{n}=20[/latex] samples in total
[latex]text{df}_{text{numerator}} = text{k}-1 = 4-1 = 3[/latex]
[latex]text{df}_{text{denominator}} = text{n}-text{k} = 20-4 = 16[/latex]
Calculate the test statistic: [latex]text{F}=2.23[/latex]
Graph:
Graph of [latex]text{p}[/latex]-Value: This chart shows example p-values for two F-statistics: p = 0.05 for F = 3.68, and p = 0.00239 for F = 9.27. These numbers are evidence of the skewness of the F-curve to the right; a much higher F-value corresponds to an only slightly smaller p-value.
Probability statement: [latex]text{p}text{-value} = text{P}(text{F}>2.23) = 0.1241[/latex]
Compare [latex]alpha[/latex] and the [latex]text{p}[/latex]-value: [latex]alpha = 0.01[/latex], [latex]text{p}text{-value} = 0.1241[/latex]
Make a decision: Since [latex]alpha < text{p}text{-value}[/latex], you cannot reject [latex]text{H}_0[/latex].
Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.
## Variance Estimates
The [latex]text{F}[/latex]-test can be used to test the hypothesis that the variances of two populations are equal.
### Learning Objectives
Discuss the [latex]text{F}[/latex]-test for equality of variances, its method, and its properties.
### Key Takeaways
#### Key Points
• This [latex]text{F}[/latex]-test needs to be used with caution, as it can be especially sensitive to the assumption that the variables have a normal distribution.
• This test is of importance in mathematical statistics, since it provides a basic exemplar case in which the [latex]text{F}[/latex]-distribution can be derived.
• The null hypothesis is rejected if [latex]text{F}[/latex] is either too large or too small.
• [latex]text{F}[/latex]-tests are used for other statistical tests of hypotheses, such as testing for differences in means in three or more groups, or in factorial layouts.
#### Key Terms
• F-Test: A statistical test using the [latex]text{F}[/latex] distribution, most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled.
• variance: a measure of how far a set of numbers is spread out
### [latex]text{F}[/latex]-Test of Equality of Variances
An [latex]text{F}[/latex]-test for the null hypothesis that two normal populations have the same variance is sometimes used; although, it needs to be used with caution as it can be sensitive to the assumption that the variables have this distribution.
Notionally, any [latex]text{F}[/latex]-test can be regarded as a comparison of two variances, but the specific case being discussed here is that of two populations, where the test statistic used is the ratio of two sample variances. This particular situation is of importance in mathematical statistics since it provides a basic exemplar case in which the [latex]text{F}[/latex] distribution can be derived.
### The Test
Let [latex]text{X}_1, dots, text{X}_text{n}[/latex] and [latex]text{Y}_1, dots, text{Y}_text{m}[/latex] be independent and identically distributed samples from two populations which each have a normal distribution. The expected values for the two populations can be different, and the hypothesis to be tested is that the variances are equal. The test statistic is:
[latex]displaystyle text{F} = frac{text{S}^{2}_{text{X}}}{text{S}^{2}_{text{Y}}}[/latex]
It has an [latex]text{F}[/latex]-distribution with [latex]text{n}-1[/latex] and [latex]text{m}-1[/latex] degrees of freedom if the null hypothesis of equality of variances is true. The null hypothesis is rejected if [latex]text{F}[/latex] is either too large or too small. The immediate assumption of the problem outlined above is that it is a situation in which there are more than two groups or populations, and the hypothesis is that all of the variances are equal.
### Properties of the [latex]text{F}[/latex] Test
This [latex]text{F}[/latex]-test is known to be extremely sensitive to non-normality. Therefore, they must be used with care, and they must be subject to associated diagnostic checking.
[latex]text{F}[/latex]-tests are used for other statistical tests of hypotheses, such as testing for differences in means in three or more groups, or in factorial layouts. These [latex]text{F}[/latex]-tests are generally not robust when there are violations of the assumption that each population follows the normal distribution, particularly for small alpha levels and unbalanced layouts. However, for large alpha levels (e.g., at least 0.05) and balanced layouts, the [latex]text{F}[/latex]-test is relatively robust. Although, if the normality assumption does not hold, it suffers from a loss in comparative statistical power as compared with non-parametric counterparts.
## Mean Squares and the F-Ratio
Most [latex]text{F}[/latex]-tests arise by considering a decomposition of the variability in a collection of data in terms of sums of squares.
### Learning Objectives
Demonstrate how sums of squares and mean squares produce the [latex]text{F}[/latex]-ratio and the implications that changes in mean squares have on it.
### Key Takeaways
#### Key Points
• The test statistic in an [latex]text{F}[/latex]-test is the ratio of two scaled sums of squares reflecting different sources of variability.
• These sums of squares are constructed so that the statistic tends to be greater when the null hypothesis is not true.
• To calculate the [latex]text{F}[/latex]-ratio, two estimates of the variance are made: variance between samples and variance within samples.
• The one-way ANOVA test depends on the fact that the mean squares between samples can be influenced by population differences among means of the several groups.
#### Key Terms
• pooled variance: A method for estimating variance given several different samples taken in different circumstances where the mean may vary between samples but the true variance is assumed to remain the same.
• null hypothesis: A hypothesis set up to be refuted in order to support an alternative hypothesis; presumed true until statistical evidence in the form of a hypothesis test indicates otherwise.
Most [latex]text{F}[/latex]-tests arise by considering a decomposition of the variability in a collection of data in terms of sums of squares. The test statistic in an [latex]text{F}[/latex]-test is the ratio of two scaled sums of squares reflecting different sources of variability. These sums of squares are constructed so that the statistic tends to be greater when the null hypothesis is not true. In order for the statistic to follow the [latex]text{F}[/latex]– distribution under the null hypothesis, the sums of squares should be statistically independent, and each should follow a scaled chi-squared distribution. The latter condition is guaranteed if the data values are independent and normally distributed with a common variance.
[latex]text{F}[/latex]-Distribution: The [latex]text{F}[/latex] ratio follows the [latex]text{F}[/latex]-distribution, which is right skewed.
There are two sets of degrees of freedom for the [latex]text{F}[/latex]-ratio: one for the numerator and one for the denominator. For example, if [latex]text{F}[/latex] follows an [latex]text{F}[/latex]-distribution and the degrees of freedom for the numerator are 4 and the degrees of freedom for the denominator are 10, then [latex]text{F} sim text{F}_{4, 10}[/latex].
To calculate the [latex]text{F}[/latex]-ratio, two estimates of the variance are made:
1. Variance between samples: An estimate of [latex]sigma^2[/latex] that is the variance of the sample means multiplied by [latex]text{n}[/latex] (when there is equal [latex]text{n}[/latex]). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation.
2. Variance within samples: An estimate of [latex]sigma^2[/latex] that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation.
• [latex]text{SS}_{text{between}}[/latex] is the sum of squares that represents the variation among the different samples.
• [latex]text{SS}_{text{within}}[/latex] is the sum of squares that represents the variation within samples that is due to chance.
To find a “sum of squares” is to add together squared quantities which, in some cases, may be weighted. [latex]text{MS}[/latex] means “mean square. ” [latex]text{MS}_{text{between}}[/latex] is the variance between groups and [latex]text{MS}_{text{within}}[/latex] is the variance within groups.
### Calculation of Sum of Squares and Mean Square
• [latex]text{k}[/latex] is the number of different groups
• [latex]text{n}_text{j}[/latex] is the size of the [latex]text{j}[/latex]th group
• [latex]text{s}_text{j}[/latex] is the sum of the values in the [latex]text{j}[/latex]th group
• [latex]text{n}[/latex] is total number of all the values combined. (Total sample size: [latex]sum_text{j} text{n}_text{j}[/latex])
• [latex]text{x}[/latex] is one value: [latex]sum text{x} = sum_text{j} text{s}_text{j}[/latex]
• Sum of squares of all values from every group combined: [latex]sum text{x}^2[/latex]
• Between group variability: [latex]displaystyle { text{SS} }_{ text{total} }=sum { { text{x} }^{ 2 }- } frac { { left( sum { text{x} } right) }^{ 2 } }{ n }[/latex]
• Total sum of squares: [latex]displaystyle sum { { text{x} }^{ 2 }- } frac { { left( sum { text{x} } right) }^{ 2 } }{ text{n} }[/latex]
• Explained variation: sum of squares representing variation among the different samples [latex]displaystyle { text{SS} }_{ text{between} }=sum { left[ frac { { left( text{s}_text{j} right) }^{ 2 } }{ { text{n} }_{ text{j} } } right] – } frac { { left( sum { { text{s} }_{ text{j} } } right) }^{ 2 } }{ text{n} }[/latex]
• Unexplained variation: sum of squares representing variation within samples due to chance: [latex]text{SS}_{text{within}} = text{SS}_{text{total}} = text{SS}_{text{between}}[/latex]
• [latex]text{df}[/latex]‘s for different groups ([latex]text{df}[/latex]‘s for the numerator): [latex]text{df}_{text{between}} = text{k}-1[/latex]
• Equation for errors within samples ([latex]text{df}[/latex]‘s for the denominator): [latex]text{df}_{text{within}} = text{n}-text{k}[/latex]
• Mean square (variance estimate) explained by the different groups: [latex]displaystyle { text{MS} }_{ text{between} }=frac { { text{SS} }_{ text{between} } }{ { text{df} }_{ text{between} } }[/latex]
• Mean square (variance estimate) that is due to chance (unexplained): [latex]displaystyle{ text{MS} }_{ text{within} }=frac { { text{SS} }_{ text{within} } }{ { text{df} }_{ text{within} } }[/latex]
MSbetween and MSwithin can be written as follows:
• [latex]displaystyle { text{MS} }_{ text{between} }=frac { { text{SS} }_{ text{between} } }{ { text{df} }_{ text{between} } } =frac { { text{SS} }_{ text{between} } }{ text{k}-1 }[/latex]
• [latex]displaystyle { text{MS} }_{ text{within} }=frac { { text{SS} }_{ text{within} } }{ { text{df} }_{ text{within} } } =frac { { text{SS} }_{ text{within} } }{ text{n}-text{k} }[/latex]
The one-way ANOVA test depends on the fact that [latex]text{MS}_{text{between}}[/latex] can be influenced by population differences among means of the several groups. Since [latex]text{MS}_{text{within}}[/latex] compares values of each group to its own group mean, the fact that group means might be different does not affect [latex]text{MS}_{text{within}}[/latex].
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, [latex]text{MS}_{text{between}}[/latex] and [latex]text{MS}_{text{within}}[/latex] should both estimate the same value. Note that the null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution because it is assumed that the populations are normal and that they have equal variances.
### F Ratio
[latex]displaystyle text{F}=frac { { text{MS} }_{ text{between} } }{ { text{MS} }_{ text{within} } }[/latex]
If [latex]text{MS}_{text{between}}[/latex] and [latex]text{MS}_{text{within}}[/latex] estimate the same value (following the belief that Ho is true), then the F-ratio should be approximately equal to one. Mostly just sampling errors would contribute to variations away from one. As it turns out, [latex]text{MS}_{text{between}}[/latex] consists of the population variance plus a variance produced from the differences between the samples. [latex]text{MS}_{text{within}}[/latex] is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, [latex]text{MS}_{text{between}}[/latex] will generally be larger than [latex]text{MS}_{text{within}}[/latex]. Then, the F-ratio will be larger than one. However, if the population effect size is small it is not unlikely that [latex]text{MS}_{text{within}}[/latex] will be larger in a give sample.
## ANOVA
ANOVA is a statistical tool used in several ways to develop and confirm an explanation for the observed data.
### Learning Objectives
Recognize how ANOVA allows us to test variables in three or more groups.
### Key Takeaways
#### Key Points
• ANOVA is a particular form of statistical hypothesis testing heavily used in the analysis of experimental data.
• ANOVA is used in the analysis of comparative experiments —those in which only the difference in outcomes is of interest.
• The statistical significance of the experiment is determined by a ratio of two variances.
• The calculations of ANOVA can be characterized as computing a number of means and variances, dividing two variances and comparing the ratio to a handbook value to determine statistical significance.
• ANOVA statistical significance results are independent of constant bias and scaling errors as well as the units used in expressing observations.
#### Key Terms
• ANOVA: Analysis of variance—a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups).
• null hypothesis: A hypothesis set up to be refuted in order to support an alternative hypothesis; presumed true until statistical evidence in the form of a hypothesis test indicates otherwise.
Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests involving more than two averages, statisticians have developed a method called analysis of variance (abbreviated ANOVA).
ANOVA is a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups). In ANOVA setting, the observed variance in a particular variable is partitioned into components attributable to different sources of variation. In its simplest form, ANOVA provides a statistical test of whether or not the means of several groups are equal, and therefore generalizes t-test to more than two groups. Doing multiple two- sample t-tests would result in an increased chance of committing a type I error. For this reason, ANOVAs are useful in comparing (testing) three or more means (groups or variables) for statistical significance.
ANOVA is a particular form of statistical hypothesis testing heavily used in the analysis of experimental data. In the typical application of ANOVA, the null hypothesis is that all groups are simply random samples of the same population. This implies that all treatments have the same effect (perhaps none). Rejecting the null hypothesis implies that different treatments result in altered effects.
### Characteristics of ANOVA
ANOVA is used in the analysis of comparative experiments—those in which only the difference in outcomes is of interest. The statistical significance of the experiment is determined by a ratio of two variances. This ratio is independent of several possible alterations to the experimental observations, so that adding a constant to all observations, or multiplying all observations by a constant, does not alter significance. Therefore, ANOVA statistical significance results are independent of constant bias and scaling errors as well as the units used in expressing observations.
The calculations of ANOVA can be characterized as computing a number of means and variances, dividing two variances and comparing the ratio to a handbook value to determine statistical significance. Calculating a treatment effect is then trivial; therefore, the effect of any treatment is estimated by taking the difference between the mean of the observations which receive the treatment and the general mean.
### Summary
ANOVA is the synthesis of several ideas and it is used for multiple purposes. As a consequence, it is difficult to define concisely or precisely. In short, ANOVA is a statistical tool used in several ways to develop and confirm an explanation for the observed data. Additionally:
1. It is computationally elegant and relatively robust against violations to its assumptions.
2. ANOVA provides industrial strength (multiple sample comparison) statistical analysis.
3. It has been adapted to the analysis of a variety of experimental designs.
As a result, ANOVA has long enjoyed the status of being the most used (some would say abused) statistical technique in psychological research, and ANOVA is probably the most useful technique in the field of statistical inference. ANOVA with a very good fit and ANOVA with no fit are shown, respectively, in and.
ANOVA With No Fit: This graph shows a representation of a situation with no fit at all in terms of ANOVA statistics.
ANOVA With Very Good Fit: This graph is a representation of a situation with a very good fit in terms of ANOVA statistics
## ANOVA Design
Many statisticians base ANOVA on the design of the experiment, especially on the protocol that specifies the random assignment of treatments to subjects.
### Learning Objectives
Differentiate one-way, factorial, repeated measures, and multivariate ANOVA experimental designs; single and multiple factor ANOVA tests; fixed-effect, random-effect and mixed-effect models
### Key Takeaways
#### Key Points
• Some popular experimental designs use one-way ANOVA, factorial ANOVA, repeated measures ANOVA, or MANOVA (multivariate analysis of variance ).
• ANOVA can be performed for a single factor or multiple factors.
• The classes of models use in ANOVA are fixed-effects models, random-effects models, and multi-effects models.
#### Key Terms
• ANOVA: Analysis of variance—a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups).
• blocking: A schedule for conducting treatment combinations in an experimental study such that any effects on the experimental results due to a known change in raw materials, operators, machines, etc., become concentrated in the levels of the blocking variable.
There are several types of ANOVA. Many statisticians base ANOVA on the design of the experiment, especially on the protocol that specifies the random assignment of treatments to subjects. The protocol’s description of the assignment mechanism should include a specification of the structure of the treatments and of any blocking. It is also common to apply ANOVA to observational data using an appropriate statistical model. Some popular designs use the following types of ANOVA.
ANOVA With Fair Fit: This graph shows a representation of a situation with a fair fit in terms of ANOVA statistics.
1. One-way ANOVA is used to test for differences among two or more independent groups. Typically, however, the one-way ANOVA is used to test for differences among at least three groups, since the two-group case can be covered by a [latex]text{t}[/latex]-test. When there are only two means to compare, the [latex]text{t}[/latex]-test and the ANOVA [latex]text{F}[/latex]-test are equivalent.
2. Factorial ANOVA is used when the experimenter wants to study the interaction effects among the treatments.
3. Repeated measures ANOVA is used when the same subjects are used for each treatment (e.g., in a longitudinal study).
4. Multivariate analysis of variance (MANOVA) is used when there is more than one response variable.
### ANOVA for a Single Factor
The simplest experiment suitable for ANOVA analysis is the completely randomized experiment with a single factor. More complex experiments with a single factor involve constraints on randomization and include completely randomized blocks. The more complex experiments share many of the complexities of multiple factors.
### ANOVA for Multiple Factors
ANOVA generalizes to the study of the effects of multiple factors. When the experiment includes observations at all combinations of levels of each factor, it is termed factorial. Factorial experiments are more efficient than a series of single factor experiments, and the efficiency grows as the number of factors increases. Consequently, factorial designs are heavily used.
The use of ANOVA to study the effects of multiple factors has a complication. In a 3-way ANOVA with factors [latex]text{x}[/latex], [latex]text{y}[/latex], and [latex]text{z}[/latex], the ANOVA model includes terms for the main effects ([latex]text{x}[/latex], [latex]text{y}[/latex], [latex]text{z}[/latex]) and terms for interactions ([latex]text{xy}[/latex], [latex]text{xz}[/latex], [latex]text{yz}[/latex], [latex]text{xyz}[/latex]). All terms require hypothesis tests. The proliferation of interaction terms increases the risk that some hypothesis test will produce a false positive by chance. Fortunately, experience says that high order interactions are rare. The ability to detect interactions is a major advantage of multiple factor ANOVA. Testing one factor at a time hides interactions, but produces apparently inconsistent experimental results.
### Classes of Models
There are three classes of models used in the analysis of variance, and these are outlined here.
### Fixed-Effects Models
The fixed-effects model of analysis of variance applies to situations in which the experimenter applies one or more treatments to the subjects of the experiment to see if the response variable values change. This allows the experimenter to estimate the ranges of response variable values that the treatment would generate in the population as a whole.
### Random-Effects Models
Random effects models are used when the treatments are not fixed. This occurs when the various factor levels are sampled from a larger population. Because the levels themselves are random variables, some assumptions and the method of contrasting the treatments (a multi-variable generalization of simple differences) differ from the fixed-effects model.
### Mixed-Effects Models
A mixed-effects model contains experimental factors of both fixed and random-effects types, with appropriately different interpretations and analysis for the two types. For example, teaching experiments could be performed by a university department to find a good introductory textbook, with each text considered a treatment. The fixed-effects model would compare a list of candidate texts. The random-effects model would determine whether important differences exist among a list of randomly selected texts. The mixed-effects model would compare the (fixed) incumbent texts to randomly selected alternatives.
## ANOVA Assumptions
The results of a one-way ANOVA can be considered reliable as long as certain assumptions are met.
### Learning Objectives
List the assumptions made in a one-way ANOVA and understand the implications of unit-treatment additivity
### Key Takeaways
#### Key Points
• Response variables are normally distributed (or approximately normally distributed).
• Samples are independent.
• Variances of populations are equal.
• Responses for a given group are independent and identically distributed normal random variables—not a simple random sample (SRS).
• The randomization-based analysis assumes only the homogeneity of the variances of the residuals (as a consequence of unit-treatment additivity ) and uses the randomization procedure of the experiment.
#### Key Terms
• ANOVA: Analysis of variance—a collection of statistical models used to analyze the differences between group means and their associated procedures (such as “variation” among and between groups).
• unit-treatment additivity: An assumption that states that the observed response from the experimental unit when receiving treatment can be written as the sum of the unit’s response [latex]text{y}_text{i}[/latex] and the treatment-effect [latex]text{t}_text{j}[/latex].
• simple random sample: A sample in which each individual is chosen randomly and entirely by chance, such that each individual has the same probability of being chosen at any stage during the sampling process, and each subset of [latex]text{k}[/latex] individuals has the same probability of being chosen for the sample as any other subset of [latex]text{k}[/latex] individuals.
The results of a one-way ANOVA can be considered reliable as long as the following assumptions are met:
• Response variables are normally distributed (or approximately normally distributed).
• Samples are independent.
• Variances of populations are equal.
• Responses for a given group are independent and identically distributed normal random variables—not a simple random sample (SRS).
Necessary assumptions for randomization-based analysis are as follows.
### Randomization-Based Analysis
In a randomized controlled experiment, the treatments are randomly assigned to experimental units, following the experimental protocol. This randomization is objective and declared before the experiment is carried out. The objective random-assignment is used to test the significance of the null hypothesis, following the ideas of C.S. Peirce and Ronald A. Fisher. This design-based analysis was developed by Francis J. Anscombe at Rothamsted Experimental Station and by Oscar Kempthorne at Iowa State University. Kempthorne and his students make an assumption of unit-treatment additivity.
In its simplest form, the assumption of unit-treatment additivity states that the observed response from the experimental unit when receiving treatment can be written as the sum of the unit’s response [latex]text{y}_text{i}[/latex] and the treatment-effect [latex]text{t}_text{j}[/latex], or
[latex]text{y}_{text{i}, text{j}} = text{y}_text{i}+text{t}_text{j}[/latex]
The assumption of unit-treatment additivity implies that for every treatment [latex]text{j}[/latex], the [latex]text{j}[/latex]th treatment has exactly the same effect [latex]text{t}_text{j}[/latex] on every experiment unit. The assumption of unit-treatment additivity usually cannot be directly falsified; however, many consequences of unit-treatment additivity can be falsified. For a randomized experiment, the assumption of unit-treatment additivity implies that the variance is constant for all treatments. Therefore, by contraposition, a necessary condition for unit-treatment additivity is that the variance is constant. The use of unit-treatment additivity and randomization is similar to the design-based inference that is standard in finite-population survey sampling.
### Derived Linear Model
Kempthorne uses the randomization- distribution and the assumption of unit-treatment additivity to produce a derived linear model, very similar to the one-way ANOVA discussed previously. The test statistics of this derived linear model are closely approximated by the test statistics of an appropriate normal linear model, according to approximation theorems and simulation studies. However, there are differences. For example, the randomization-based analysis results in a small but (strictly) negative correlation between the observations. In the randomization-based analysis, there is no assumption of a normal distribution and certainly no assumption of independence. On the contrary, the observations are dependent.
In summary, the normal model based ANOVA analysis assumes the independence, normality and homogeneity of the variances of the residuals. The randomization-based analysis assumes only the homogeneity of the variances of the residuals (as a consequence of unit-treatment additivity) and uses the randomization procedure of the experiment. Both these analyses require homoscedasticity, as an assumption for the normal model analysis and as a consequence of randomization and additivity for the randomization-based analysis.
Source: Statistics | 9,082 | 39,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.95616 |
https://www.mathworks.com/matlabcentral/answers/491519-hot-to-locate-indices-of-2-d-data-located-out-of-circle?s_tid=prof_contriblnk | 1,579,642,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250605075.24/warc/CC-MAIN-20200121192553-20200121221553-00447.warc.gz | 981,445,530 | 20,666 | MATLAB Answers
# Hot to locate indices of 2-D data located out of circle?
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Abdulaziz Abutunis on 17 Nov 2019
Dear All,
I have x,y data that are extracted from a triangle solution. I wonder if there is a command (no loop) that will find the indices of elements when x^2+y^2 > than a specific value. The reason I want to eliminate them.
Thank you
Aziz
#### 0 Comments
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### Accepted Answer
Walter Roberson on 17 Nov 2019
vector_of_y_values = (1:number_of_rows) - y_center;
vector_of_x_values = (1:number_of_columns) - x_center;
[Yg, Xg] = ndgrid(vector_of_y_values, vector_of_x_values);
idx = find(Xg.^2 + Yg.^2 > specific_value);
#### 4 Comments
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Abdulaziz Abutunis on 17 Nov 2019
I have not used the ndgrid and used the next line. Also, I made a mathematical mistake in the previous script by forgetting the square root
ind_remove = find(sqrt(x.^2 + y.^2) > percentage_added);
Though your answer will do what i want if not using the ndgrid
Thanks
Walter Roberson on 17 Nov 2019
I forgot to account for possible scale factors such as the possibility that the x coordinates are percentage. The x and y vectors should be constructed so that they list the x and y coordinates, such
linspace(firstx, lastx, number of x)
Abdulaziz Abutunis on 18 Nov 2019
Thank you, Walter, for your valuable suggestions
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### More Answers (0)
Sign in to answer this question. | 397 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.836828 |
http://www.netlib.org/lapack/explore-3.2.1-html/dlarrc.f.html | 1,555,591,421,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517639.17/warc/CC-MAIN-20190418121317-20190418143317-00408.warc.gz | 271,890,592 | 3,026 | ```001: SUBROUTINE DLARRC( JOBT, N, VL, VU, D, E, PIVMIN,
002: \$ EIGCNT, LCNT, RCNT, INFO )
003: *
004: * -- LAPACK auxiliary routine (version 3.2) --
005: * -- LAPACK is a software package provided by Univ. of Tennessee, --
006: * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
007: * November 2006
008: *
009: * .. Scalar Arguments ..
010: CHARACTER JOBT
011: INTEGER EIGCNT, INFO, LCNT, N, RCNT
012: DOUBLE PRECISION PIVMIN, VL, VU
013: * ..
014: * .. Array Arguments ..
015: DOUBLE PRECISION D( * ), E( * )
016: * ..
017: *
018: * Purpose
019: * =======
020: *
021: * Find the number of eigenvalues of the symmetric tridiagonal matrix T
022: * that are in the interval (VL,VU] if JOBT = 'T', and of L D L^T
023: * if JOBT = 'L'.
024: *
025: * Arguments
026: * =========
027: *
028: * JOBT (input) CHARACTER*1
029: * = 'T': Compute Sturm count for matrix T.
030: * = 'L': Compute Sturm count for matrix L D L^T.
031: *
032: * N (input) INTEGER
033: * The order of the matrix. N > 0.
034: *
035: * VL (input) DOUBLE PRECISION
036: * VU (input) DOUBLE PRECISION
037: * The lower and upper bounds for the eigenvalues.
038: *
039: * D (input) DOUBLE PRECISION array, dimension (N)
040: * JOBT = 'T': The N diagonal elements of the tridiagonal matrix T.
041: * JOBT = 'L': The N diagonal elements of the diagonal matrix D.
042: *
043: * E (input) DOUBLE PRECISION array, dimension (N)
044: * JOBT = 'T': The N-1 offdiagonal elements of the matrix T.
045: * JOBT = 'L': The N-1 offdiagonal elements of the matrix L.
046: *
047: * PIVMIN (input) DOUBLE PRECISION
048: * The minimum pivot in the Sturm sequence for T.
049: *
050: * EIGCNT (output) INTEGER
051: * The number of eigenvalues of the symmetric tridiagonal matrix T
052: * that are in the interval (VL,VU]
053: *
054: * LCNT (output) INTEGER
055: * RCNT (output) INTEGER
056: * The left and right negcounts of the interval.
057: *
058: * INFO (output) INTEGER
059: *
060: * Further Details
061: * ===============
062: *
063: * Based on contributions by
064: * Beresford Parlett, University of California, Berkeley, USA
065: * Jim Demmel, University of California, Berkeley, USA
066: * Inderjit Dhillon, University of Texas, Austin, USA
067: * Osni Marques, LBNL/NERSC, USA
068: * Christof Voemel, University of California, Berkeley, USA
069: *
070: * =====================================================================
071: *
072: * .. Parameters ..
073: DOUBLE PRECISION ZERO
074: PARAMETER ( ZERO = 0.0D0 )
075: * ..
076: * .. Local Scalars ..
077: INTEGER I
078: LOGICAL MATT
079: DOUBLE PRECISION LPIVOT, RPIVOT, SL, SU, TMP, TMP2
080:
081: * ..
082: * .. External Functions ..
083: LOGICAL LSAME
084: EXTERNAL LSAME
085: * ..
086: * .. Executable Statements ..
087: *
088: INFO = 0
089: LCNT = 0
090: RCNT = 0
091: EIGCNT = 0
092: MATT = LSAME( JOBT, 'T' )
093:
094:
095: IF (MATT) THEN
096: * Sturm sequence count on T
097: LPIVOT = D( 1 ) - VL
098: RPIVOT = D( 1 ) - VU
099: IF( LPIVOT.LE.ZERO ) THEN
100: LCNT = LCNT + 1
101: ENDIF
102: IF( RPIVOT.LE.ZERO ) THEN
103: RCNT = RCNT + 1
104: ENDIF
105: DO 10 I = 1, N-1
106: TMP = E(I)**2
107: LPIVOT = ( D( I+1 )-VL ) - TMP/LPIVOT
108: RPIVOT = ( D( I+1 )-VU ) - TMP/RPIVOT
109: IF( LPIVOT.LE.ZERO ) THEN
110: LCNT = LCNT + 1
111: ENDIF
112: IF( RPIVOT.LE.ZERO ) THEN
113: RCNT = RCNT + 1
114: ENDIF
115: 10 CONTINUE
116: ELSE
117: * Sturm sequence count on L D L^T
118: SL = -VL
119: SU = -VU
120: DO 20 I = 1, N - 1
121: LPIVOT = D( I ) + SL
122: RPIVOT = D( I ) + SU
123: IF( LPIVOT.LE.ZERO ) THEN
124: LCNT = LCNT + 1
125: ENDIF
126: IF( RPIVOT.LE.ZERO ) THEN
127: RCNT = RCNT + 1
128: ENDIF
129: TMP = E(I) * D(I) * E(I)
130: *
131: TMP2 = TMP / LPIVOT
132: IF( TMP2.EQ.ZERO ) THEN
133: SL = TMP - VL
134: ELSE
135: SL = SL*TMP2 - VL
136: END IF
137: *
138: TMP2 = TMP / RPIVOT
139: IF( TMP2.EQ.ZERO ) THEN
140: SU = TMP - VU
141: ELSE
142: SU = SU*TMP2 - VU
143: END IF
144: 20 CONTINUE
145: LPIVOT = D( N ) + SL
146: RPIVOT = D( N ) + SU
147: IF( LPIVOT.LE.ZERO ) THEN
148: LCNT = LCNT + 1
149: ENDIF
150: IF( RPIVOT.LE.ZERO ) THEN
151: RCNT = RCNT + 1
152: ENDIF
153: ENDIF
154: EIGCNT = RCNT - LCNT
155:
156: RETURN
157: *
158: * end of DLARRC
159: *
160: END
161: ``` | 1,722 | 5,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-18 | latest | en | 0.532614 |
http://www.ehow.co.uk/how_5767519_calculate-volt-drop.html | 1,527,302,599,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00523.warc.gz | 361,387,467 | 10,685 | DISCOVER
# How to calculate volt drop
Updated February 21, 2017
A primary concern when installing long runs of wire is voltage drop. Voltage drop is a reduction in the voltage on a circuit between the power source, like a stereo, and the load, like a speaker. Voltage drop over a long run of wire can result in the underperformance of electrical or electronic equipment. You can calculate the voltage drop provided you know the current load in amperes (amps) and the AWG size of the wire. AWG is the wire gauge, expressed in terms of the American Wire Gauge standard, used in the United States since 1857.
Confirm the wire size (AWG, also known as American Wire Gauge) and the current load expressed in amperes (amps). The both of these factors can be determined by direct measurement (by wire gauge and multimeter) or from specifications.
Multiply the current load in amps by 0.2. If, for example, the current load on the wires is 16 amps, then 16 x 0.2 = 3.2.
The wire size component of the formula has a mantissa of 1.26 and an exponent calculated by subtracting 10 from the AWG of the wire. If the wire in the above example is 8AWG, then 8 - 10 = -2. The result, 1.26 E (-2), is 0.0126 in decimal notation.
Multiply these two results to obtain the voltage drop over 100ft.: 3.2 x 0.0126 = 0.04032
Divide the total wire run length by 100; if the wire run in the example is 300 feet, then 300 / 100 = 3. Multiply this result by the voltage drop over 100 feet. If you continue to use numbers given so far in the example, this would result in the following calculation: 3 x 0.04032 = 0.12096, which would equal the total voltage drop over the hypothetical 300-foot wire run.
#### Tip
The AWG standard is based on solid copper wire; however, for the purposes of determining voltage drop, the characteristics of stranded copper wire and solid copper wire are the same.
#### Things You'll Need
• Standard AWG wire gauge
• Multimeter
• Calculator | 490 | 1,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.881432 |
http://www.mathworks.com/matlabcentral/fileexchange/37616-modern-pricing-method-using-transforms/content/Cos_Conv_Methods/CallPricingFFT.m | 1,438,629,879,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990112.92/warc/CC-MAIN-20150728002310-00167-ip-10-236-191-2.ec2.internal.warc.gz | 565,111,089 | 8,427 | Code covered by the BSD License
# Modern Pricing Method using Transforms
### Kienitz Wetterau FinModelling (view profile)
25 Jul 2012 (Updated )
COS, CONV, Lewis Option Pricing Methods including Bermudan and American Options.
CallPricingFFT(model,n,S,K,T,r,d,varargin)
```% This is material illustrating the methods from the book
% Financial Modelling - Theory, Implementation and Practice with Matlab
% source
% Wiley Finance Series
% ISBN 978-0-470-74489-5
%
% Date: 02.05.2012
%
% Authors: Joerg Kienitz
% Daniel Wetterau
%
% kienitzwetterau_FinModelling@gmx.de
%
% (C) Joerg Kienitz, Daniel Wetterau
%
% Since this piece of code is distributed via the mathworks file-exchange
% it is covered by the BSD license
%
% This code is being provided solely for information and general
% illustrative purposes. The authors will not be responsible for the
% consequences of reliance upon using the code or for numbers produced
% from using the code.
function call_price_fft = CallPricingFFT(model,n,S,K,T,r,d,varargin)
lnS = log(S);
lnK = log(K);
%optAlpha = optimalAlpha(model,lnS,lnK,T,r,d,varargin{:});
optAlpha = .75;
DiscountFactor = exp(-r*T);
%-------------------------
%--- FFT Option Pricing --
%-------------------------
% from: Option Valuation Using the Fast Fourier Transform,
% Peter Carr, March 1999, pp 10-11
%-------------------------
% predefined parameters
FFT_N = 2^n; % must be a power of two (2^14)
FFT_eta = 0.05; % spacing of psi integrand
% effective upper limit for integration (18)
% uplim = FFT_N * FFT_eta;
FFT_lambda = (2 * pi) / (FFT_N * FFT_eta); %spacing for log strike output (23)
FFT_b = (FFT_N * FFT_lambda) / 2; % (20)
uvec = 1:FFT_N;
%log strike levels ranging from lnS-b to lnS+b
ku = - FFT_b + FFT_lambda * (uvec - 1); %(19)
jvec = 1:FFT_N;
vj = (jvec-1) * FFT_eta;
% optimal alpha illustration (payoff independent)
% alpharange = -3:0.1:8;
% resultrangef = zeros(1,length(alpharange));
% resultrangef1 = zeros(1,length(alpharange));
% eps = 0.000001;
% for n = 1:length(alpharange)
% resultrangef(n)= (-alpharange(n) * log(K) + log(psialpha(model,alpharange(n),lnS,T,r,d,varargin{:})));
% resultrangef1(n)= (-(alpharange(n) + eps) * log(K) ...
% + log(psialpha(model,alpharange(n)+eps,lnS,T,r,d,varargin{:})));
% resultrangef1(n)= (resultrangef1(n) - resultrangef(n)) / eps;
% end
% plot(alpharange,resultrangef); hold on; plot(alpharange,resultrangef1, 'g'); hold off;
%applying FFT
tmp = DiscountFactor * psi(model,vj,optAlpha,lnS,T,r,d,varargin{:}) .* exp(1i * vj * (FFT_b)) * FFT_eta;
tmp = (tmp / 3) .* (3 + (-1).^jvec - ((jvec - 1) == 0) ); %applying simpson's rule
cpvec = real(exp(-optAlpha .* ku) .* fft(tmp) / pi); %call price vector resulting in equation 24
indexOfStrike = floor((lnK + FFT_b)/FFT_lambda + 1);
iset = max(indexOfStrike)+1:-1:min(indexOfStrike)-1;
xp = ku(iset);
yp = cpvec(iset);
call_price_fft = real(interp1(xp,yp,lnK));
end
%analytical formula for zhi in equation ( 6 ) of Madan's paper
function ret = psi(model,v,alpha,varargin)
ret = exp(feval(@CharacteristicFunctionLib, model, v - (alpha + 1) * 1i,varargin{:})) ./ (alpha.^2 + alpha - v.^2 + 1i * (2 * alpha + 1) .* v);
end
% function ret = psialpha(model,alpha,varargin)
% ret = exp(feval(@CharacteristicFunctionLib, model, - (alpha + 1) * 1i,varargin{:}))./ (alpha.^2 + alpha);
% end``` | 1,077 | 3,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.537892 |
https://www.learnatnoon.com/s/earth-can-be-thought-of-as-a-sphere-of-radius-6400-km/39084/ | 1,680,287,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00421.warc.gz | 967,161,593 | 22,683 | a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2? b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 1011m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2? b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 1011m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
(a) According to the question, we have been given that,
Radius of the earth .
Time period of the motion day
As we know that Centripetal acceleration,
At equator, latitude makes an angle
b) The earth’s orbital radius around the sun is measured in meters.
Time period days
Centripetal acceleration | 337 | 1,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-14 | latest | en | 0.902596 |
https://stats.stackexchange.com/questions/226890/expected-length-of-a-memory-game-of-two-clueless-players | 1,569,172,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575596.77/warc/CC-MAIN-20190922160018-20190922182018-00051.warc.gz | 672,604,822 | 31,154 | Expected length of a memory game of two clueless players
What is the expected length of a memory game when neither player (you may as well play against yourself) remembers any of the cards being uncovered in previous rounds?
Initially and until the first match (because players do not remember), when playing with $N$ pairs, there is a probability of $p=1/(2N-1)$ of a match, as there are $2N-1$ remaining cards which could be a match for the first card you draw.
The game is over when all $N$ pairs are matched, and in principle the game could last forever. Hence, a negative binomial distribution would seem like a good starting point, as we aim to model the probability of $N$ successes in a sequence of Bernoulli experiments.
However, once a match has been found, the success probability increases to $1/(2N-3)$, as the two matches are removed from the game. Hence, the success probability in the trials is not constant. Once there are only two cards left, you are bound to have a match.
I found a paper here that discusses the case in which the players play optimally, i.e., forget nothing.
• A recursive solution is attractive here and easy to work out: the expected length is the expected time to obtain the first match plus the expected length of a game with $N-1$ pairs. The resulting sum simplifies considerably :-). – whuber Aug 2 '16 at 15:19
• @whuber, oh, right, I could have seen that, thanks@ – Christoph Hanck Aug 2 '16 at 16:25
$$E(\text{number of attempts for }N\text{ pairs})=\sum_{j=1}^NE(\text{number of attempts for }j\text{th pair})$$ The first pair has, as mentioned in the question, a success probability of $p=1/(2N-1)$ in any given attempt, and follows a geometric distribution. The expectation therefore is $$E(\text{number of attempts for }1\text{st pair})=2N-1$$ Likewise, $$E(\text{number of attempts for }2\text{nd pair})=2N-2-1$$ and in general $$\sum_{j=1}^NE(\text{number of attempts for }j\text{th pair})=\sum_{j=1}^N2N-2(j-1)-1$$ which indeed simplifies very nicely to $$\sum_{j=1}^N2N-2(j-1)-1=2N^2+N-2N(N+1)/2=N^2$$ | 569 | 2,062 | {"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} | 3.765625 | 4 | CC-MAIN-2019-39 | latest | en | 0.940121 |
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