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http://mymathforum.com/calculus/22817-something-wrong-graph-surface-integration.html | 1,506,210,171,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689806.55/warc/CC-MAIN-20170923231842-20170924011842-00635.warc.gz | 229,868,115 | 9,724 | My Math Forum Something wrong with graph surface(integration)
Calculus Calculus Math Forum
November 24th, 2011, 08:17 PM #1 Newbie Joined: Oct 2011 Posts: 29 Thanks: 0 Something wrong with graph surface(integration) Hey guys, I had to calculate the surface defined by these two functions: $y=x2$ $y=x+2$ Here's my integral: $\int_{1.56}^{2.56} (x+4) - (x^2) dx$ This gives me 3.43592 which sounds good. To practice myself I decided to isolate x in the functions and get the tilted version of the graph. Now I know that this gives one more function, but for some reason I'm not getting the same area, here's the integral: $\int_0^1 (\sqrt{y}) - (-\sqrt{y}) dy + \int_1^4 (\sqrt{y}) - (y-2) dy$ This yields 4.5 No idea why the areas are different, the region should be the same. Thanks.
November 24th, 2011, 08:19 PM #2 Newbie Joined: Oct 2011 Posts: 29 Thanks: 0 Re: Something wrong with graph surface(integration) I messed up the borders in latex, the border for the first one is -1.56 to 2.56(upper) The two other ones are: 0 to 1(upper) and 1 to 4(upper) [I fixed those typos for you. - skipjack]
November 25th, 2011, 03:07 AM #3 Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 5 Re: Something wrong with graph surface(integration) What do you mean by "the surface defined by these two functions"? Do you mean the area between the two curves? And where did you get the ".56"?
November 25th, 2011, 07:23 AM #4 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs Re: Something wrong with graph surface(integration) If you are trying to find the area A bounded by: $y_1=x^2$ $y_2=x+2$ Using dx as the differential, we find when $y_1=y_2$ we have x = -1,2, thus $A=\int_{-1}\,^2 y_2-y_1\,dx=\int_{-1}\,^2 x+2-x^2\,dx=\frac{9}{2}$ Now, if we want dy as the differential, we would use: $A=2\int_0\,^1 \sqrt{y}\,dy+\int_1\,^4 \sqrt{y}-(y-2)\,dy=\frac{4}{3}+\frac{19}{6}=\frac{8+19}{6}=\fr ac{27}{6}=\frac{9}{2}$
November 25th, 2011, 09:35 AM #5 Newbie Joined: Oct 2011 Posts: 29 Thanks: 0 Re: Something wrong with graph surface(integration) Believe it or not, I actually wrote "x+4" instead of "x+2" in the first question and based myself upon that... good thing I'm not working for the nasa. You see, the number asked me to find x = a where a splits the region's area into two parts, for some odd reason x2 = x+4 and x2 = x+2 with their corresponding borders are both equal to x = 0.5... so when I went on the second number to try and find the middle split of the y axis I thought I was working with x+4... Thanks!
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https://au.mathworks.com/matlabcentral/answers/1454214-how-to-solve-many-equations-at-once-not-a-system-but-just-multiple-separate-equations | 1,653,453,598,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00499.warc.gz | 143,422,130 | 24,193 | How to solve many equations at once? (Not a system but just multiple separate equations)
1 view (last 30 days)
atharva aalok on 16 Sep 2021
Answered: KSSV on 16 Sep 2021
I have a vector of equations in symbollic form instead of using a for loop to solve I want to solve them all at once for speed.
PmSol = linspace(0, .5, 1000);
% after some code in between...
syms x Pm
assume(x, 'real');
eqn1 = 0.5*(x+1)*((1-x*x)^(0.5)) == Pm;
Pm_val = PmSol;
% get different equations for each value of the parameter Pm
eqn = subs(eqn1, Pm, Pm_val);
% the above gives a column of 1000 equations
% The Problem: I want to solve each equation and get the corresponding
% solutions in general I know already that for Pm in this range there will
% be 2 solutions for each equation.
% This does NOT work
solns = solve(eqn, x);
disp(solns);
I want to solve this very quickly. Do not want to use a for loop. How do vectorize the solve operation.
KSSV on 16 Sep 2021
You get the solution for x in terms of Pm first. This will be a symolic xpression. In that you can substiture your Pm values to get x.
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Start Hunting! | 352 | 1,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-21 | latest | en | 0.895975 |
https://mathoverflow.net/questions/271576/product-topology-and-order-convergence-topology | 1,618,539,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00097.warc.gz | 434,636,211 | 29,869 | # Product topology and order convergence topology
Let $(P,\leq)$ be a poset. We define the order convergence topology, denoted by $\tau_o(P)$. By a set filter $\mathcal{F}$ on $P$ we mean a collection of subsets of $P$ such that:
• $\emptyset \notin \mathcal{F}$;
• $A, B\in \mathcal{F}$ implies $A\cap B\in \mathcal{F}$;
• $U\in \mathcal{F}$, $U'\subseteq P$ and $U'\supseteq U$ implies $U'\in \mathcal{F}$.
If $S\subseteq P$ we define $S^u= \{x\in P: x\geq s\text{ for all } s\in S\}$, and $S^l= \{x\in P: x\leq s\text{ for all } s\in S\}$. If $\cal{F}$ is a set filter, then we set ${\cal F}^u = \bigcup\{F^u: F\in \cal{F}\}$ and define ${\cal F}^l$ similarly. For $x\in P$ and ${\cal F}$ a set filter on $P$ we write $${\cal F}\to x \textrm{ iff } \bigwedge\cal{F}^u = x = \bigvee \cal{F}^l.$$
Then we set $\tau_o(P)=\{U\subseteq P: \textrm{ for any } x\in U \text{ and any filter }\mathcal{F} \text{ with } \mathcal{F}\to x \text{ we have } U\in \mathcal{F}\}$. It is not hard to verify that this defines a topology.
If $(P_i)_{i\in I}$ is a family of posets, is $\tau_o(\prod_{i\in I} P_i)$ equal to the product topology of the product of $(P_i, \tau_o(P_i))_{i\in I}$? | 470 | 1,180 | {"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} | 3.328125 | 3 | CC-MAIN-2021-17 | latest | en | 0.609737 |
http://www.jiskha.com/members/profile/posts.cgi?name=jeannette | 1,498,283,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00313.warc.gz | 577,077,444 | 3,765 | # Posts by jeannette
Total # Posts: 14
Math
Jan's gas tank is nearly empty, at 1\12 full. She doesn't have enough cash to fill the tank, but she adds enough gas to reach 2\3 full. How much gas did she add, as a fraction of the tank?
Math
32/67,456
palm valley
ruben is thinking of a mystery number. if he multiplies the number by 6 and then divides it by 9, the result is 708. what is the mystery number
Statistics
The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. What I the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days?
Math
Solving long division polynomials 5x-6/10x^2-7x-6
Algebra
2. Write the equation of the circle in standard form. Find the center, radius, intercepts, and graph the circle. x2+y2+8x+2y+8=0
Chemistry
An object made of aluminum has a mass of 8.37g. When it was placed in a graduated cylinder containing 20.0ml of water, the water level rose to 23.1ml. Calculate the specific gravity of the object?
math
thank you Is this correct? 8/15 = 0.53 I divided 8 by 15 0.53 * 300 = 159.9
math
There are 3 white marbles, 8 red marbles, 3 blue marbles anf 1 black marbles. predict the number of times out of 300 you will pick a red marble.
taxation
noncash items received as income must be included in income at their fair market value true or false?
social studies
i need help on a current event | 415 | 1,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-26 | latest | en | 0.940723 |
https://tutorfortune.com/products/scin-131-scin131-quiz-1-answers-american-public-university | 1,653,009,937,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00786.warc.gz | 635,545,179 | 84,970 | # SCIN 131 SCIN131 QUIZ 1 ANSWERS (AMERICAN PUBLIC UNIVERSITY)
Regular price \$15.00
Who discovered the atomic theory?
Past observations establish which aspect of the scientific method?
The standard SI unit for length is:
In the thermometer shown below, assume that the end of the red line you see is located halfway between the measurement marks. Which is the correct way to report the temperature?
How many significant figures will be in the final answer of this calculation?
`What is the value of 62 °C in °F?
True or False: In a chemical change, the matter changes its overall composition.
`Suppose I wanted to increase the temperature of H2O from 30.0 C to 55.0 C. How much heat in kJ would I need to add to a 100.0 g sample of H2O to acheive this?
`How should someone classify grape juice?
How many calories are in 25 Calories?
`IN YOUR OWN WORDS, who was John Dalton, and why is he so important to chemistry? What did he "discover" that is so central to our study of chemistry? Explain why it is important.
`My wife and I always argue about the temperature in our house. I like it at 294.0 K, while she likes to set the thermostat to 300.0 degrees K. As we take turns constantly changing the setting, our poor heating and cooling system tries to keep up. I argue that it causes us to waste money. My living room contains 50 kg of air. In the summer time, I pay about \$0.14 per kWh of energy used. How much does it cost me every time we heat the air in the living room from 294.00 K to 300 K? (HINT: How many kWh does it take?) The heat capacity of our air is 1.01 J/(gC) | 395 | 1,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.955515 |
https://vedicmathschool.org/tutorials/page/2/ | 1,714,024,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00129.warc.gz | 543,757,760 | 38,444 | ## The Uses and Versatility of Squares in Geometry
In mathematics, a Square is a 2-dimensional figure having four equal sides and equal angles. It is a plane figure. All the angles of a
## Discovering the Power of Thales Theorem and its Inverse: The Building Blocks of Geometry
The Thales Theorem is also known as the basic proportionality theorem. The basic proportionality theorem was proposed by a Greek mathematician named Thales, that’s why
## Unleashing the Beauty of 30-60-90 Triangles: The Perfect Triangle Revealed
There are various types of triangles such as obtuse triangles, isosceles, acute, equilateral, scalene, and many more. But among all of them, only some types
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## Let’s learn Plane and its Types
In mathematics, a plane can be defined as a two-dimensional flat surface that extends up to infinity.We can divide planes into two types:- parallel planes
## Decimals- Definition, Types, Properties and Arithmetic Operations on Decimals
In mathematics, we have many types of numbers. For eg:- real numbers, natural numbers, whole numbers, prime numbers. Decimal numbers are one of its types.
## Circle- Basics, Definition, Formula, Parts and Properties
The word ‘Circle’ is basically derived from a Latin word ‘circulus’ which means a small ring. A circle is also referred as the locus of
## Compound Interest -Basics, Definition, Formula and solved examples
The concept of compound interest is used in many sectors such as banking, finance, personal loans and so on. In compound interest , the interest
## Interest – Do you know, what is Simple Interest?
The concept of simple interest is used in many sectors such as banking, finance, personal loans, and so on. In this, the principal amount always
## Subtraction of Fractions
We have already discussed how we can add fractions. A fraction is a rational number in the form of a/b where b cannot be zero.
## Polynomials From Basics! Types, Formula, Rules, Operation with Solved Examples
Polynomials can be defined as algebraic expressions that contain variables and constants. Variables are also sometimes called indeterminates. What are Polynomials? Polynomials are like the
Today we are going to learn about the operations performed on the fraction. Here, we are starting with basic arithmetic operations like addition, and subtraction.
Today we are going to learn about the operations performed on the fraction. Here, we are starting with basic arithmetic operations like addition, and subtraction.
## Polynomials From Basics! Types, Formula, Rules, Operation with Solved Examples
Polynomials can be defined as algebraic expressions that contain variables and constants. Variables are also sometimes called indeterminates. What are Polynomials? Polynomials are like the
## Subtraction of Fractions
We have already discussed how we can add fractions. A fraction is a rational number in the form of a/b where b cannot be zero.
## Interest – Do you know, what is Simple Interest?
The concept of simple interest is used in many sectors such as banking, finance, personal loans, and so on. In this, the principal amount always
## Compound Interest -Basics, Definition, Formula and solved examples
The concept of compound interest is used in many sectors such as banking, finance, personal loans and so on. In compound interest , the interest
## Circle- Basics, Definition, Formula, Parts and Properties
The word ‘Circle’ is basically derived from a Latin word ‘circulus’ which means a small ring. A circle is also referred as the locus of
## Decimals- Definition, Types, Properties and Arithmetic Operations on Decimals
In mathematics, we have many types of numbers. For eg:- real numbers, natural numbers, whole numbers, prime numbers. Decimal numbers are one of its types.
## Let’s learn Plane and its Types
In mathematics, a plane can be defined as a two-dimensional flat surface that extends up to infinity.We can divide planes into two types:- parallel planes
In mathematics, vectors are objects which have both magnitude and direction. The Magnitude defines the size of a vector. It is denoted or represented by
## Unleashing the Beauty of 30-60-90 Triangles: The Perfect Triangle Revealed
There are various types of triangles such as obtuse triangles, isosceles, acute, equilateral, scalene, and many more. But among all of them, only some types
## Discovering the Power of Thales Theorem and its Inverse: The Building Blocks of Geometry
The Thales Theorem is also known as the basic proportionality theorem. The basic proportionality theorem was proposed by a Greek mathematician named Thales, that’s why | 1,011 | 4,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-18 | latest | en | 0.94453 |
https://www.quizzes.cc/metric/percentof.php?percent=8250&of=9490 | 1,571,134,086,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657949.34/warc/CC-MAIN-20191015082202-20191015105702-00138.warc.gz | 1,049,772,945 | 3,123 | #### What is 8250 percent of 9,490?
How much is 8250 percent of 9490? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 8250% of 9490 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 8250% of 9,490 = 782925
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating eight thousand, two hundred and fifty of nine thousand, four hundred and ninety How to calculate 8250% of 9490? Simply divide the percent by 100 and multiply by the number. For example, 8250 /100 x 9490 = 782925 or 82.5 x 9490 = 782925
#### How much is 8250 percent of the following numbers?
8250 percent of 9490.01 = 78292582.5 8250 percent of 9490.02 = 78292665 8250 percent of 9490.03 = 78292747.5 8250 percent of 9490.04 = 78292830 8250 percent of 9490.05 = 78292912.5 8250 percent of 9490.06 = 78292995 8250 percent of 9490.07 = 78293077.5 8250 percent of 9490.08 = 78293160 8250 percent of 9490.09 = 78293242.5 8250 percent of 9490.1 = 78293325 8250 percent of 9490.11 = 78293407.5 8250 percent of 9490.12 = 78293490 8250 percent of 9490.13 = 78293572.5 8250 percent of 9490.14 = 78293655 8250 percent of 9490.15 = 78293737.5 8250 percent of 9490.16 = 78293820 8250 percent of 9490.17 = 78293902.5 8250 percent of 9490.18 = 78293985 8250 percent of 9490.19 = 78294067.5 8250 percent of 9490.2 = 78294150 8250 percent of 9490.21 = 78294232.5 8250 percent of 9490.22 = 78294315 8250 percent of 9490.23 = 78294397.5 8250 percent of 9490.24 = 78294480 8250 percent of 9490.25 = 78294562.5
8250 percent of 9490.26 = 78294645 8250 percent of 9490.27 = 78294727.5 8250 percent of 9490.28 = 78294810 8250 percent of 9490.29 = 78294892.5 8250 percent of 9490.3 = 78294975 8250 percent of 9490.31 = 78295057.5 8250 percent of 9490.32 = 78295140 8250 percent of 9490.33 = 78295222.5 8250 percent of 9490.34 = 78295305 8250 percent of 9490.35 = 78295387.5 8250 percent of 9490.36 = 78295470 8250 percent of 9490.37 = 78295552.5 8250 percent of 9490.38 = 78295635 8250 percent of 9490.39 = 78295717.5 8250 percent of 9490.4 = 78295800 8250 percent of 9490.41 = 78295882.5 8250 percent of 9490.42 = 78295965 8250 percent of 9490.43 = 78296047.5 8250 percent of 9490.44 = 78296130 8250 percent of 9490.45 = 78296212.5 8250 percent of 9490.46 = 78296295 8250 percent of 9490.47 = 78296377.5 8250 percent of 9490.48 = 78296460 8250 percent of 9490.49 = 78296542.5 8250 percent of 9490.5 = 78296625
8250 percent of 9490.51 = 78296707.5 8250 percent of 9490.52 = 78296790 8250 percent of 9490.53 = 78296872.5 8250 percent of 9490.54 = 78296955 8250 percent of 9490.55 = 78297037.5 8250 percent of 9490.56 = 78297120 8250 percent of 9490.57 = 78297202.5 8250 percent of 9490.58 = 78297285 8250 percent of 9490.59 = 78297367.5 8250 percent of 9490.6 = 78297450 8250 percent of 9490.61 = 78297532.5 8250 percent of 9490.62 = 78297615 8250 percent of 9490.63 = 78297697.5 8250 percent of 9490.64 = 78297780 8250 percent of 9490.65 = 78297862.5 8250 percent of 9490.66 = 78297945 8250 percent of 9490.67 = 78298027.5 8250 percent of 9490.68 = 78298110 8250 percent of 9490.69 = 78298192.5 8250 percent of 9490.7 = 78298275 8250 percent of 9490.71 = 78298357.5 8250 percent of 9490.72 = 78298440 8250 percent of 9490.73 = 78298522.5 8250 percent of 9490.74 = 78298605 8250 percent of 9490.75 = 78298687.5
8250 percent of 9490.76 = 78298770 8250 percent of 9490.77 = 78298852.5 8250 percent of 9490.78 = 78298935 8250 percent of 9490.79 = 78299017.5 8250 percent of 9490.8 = 78299100 8250 percent of 9490.81 = 78299182.5 8250 percent of 9490.82 = 78299265 8250 percent of 9490.83 = 78299347.5 8250 percent of 9490.84 = 78299430 8250 percent of 9490.85 = 78299512.5 8250 percent of 9490.86 = 78299595 8250 percent of 9490.87 = 78299677.5 8250 percent of 9490.88 = 78299760 8250 percent of 9490.89 = 78299842.5 8250 percent of 9490.9 = 78299925 8250 percent of 9490.91 = 78300007.5 8250 percent of 9490.92 = 78300090 8250 percent of 9490.93 = 78300172.5 8250 percent of 9490.94 = 78300255 8250 percent of 9490.95 = 78300337.5 8250 percent of 9490.96 = 78300420 8250 percent of 9490.97 = 78300502.5 8250 percent of 9490.98 = 78300585 8250 percent of 9490.99 = 78300667.5 8250 percent of 9491 = 78300750 | 1,812 | 4,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.758404 |
https://achieveressays.com/statistics-discussion-and-2-responses/ | 1,680,197,803,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00254.warc.gz | 109,378,882 | 20,173 | # Statistics Discussion and 2 responses
Initial Post – Instructions
Using your data from the Discussion from Week 2, go back and perform a linear regression. Find the equation of the linear equation that best fits your data and calculate the correlation coefficient (R). How does this compare to your analysis in week 2?
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Statistics Discussion and 2 responses
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Response Post – Instructions
In your response posts, you will discuss the results and whether or not the results show a strong correlation or not. Discuss the value of R and what it means? What is ideal? Is this a good fit? Also, give examples in the real world where linear regression would make sense? When is a linear model the ideal model? In these cases what would you expect the value of R to be?
Data Set 01
Data Set 02
Data Set 03
Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
65.78
112.99
67.83
141.28
68.57
71.52
136.49
68.88
143.54
65.18
69.40
153.03
63.48
97.90
69.66
68.22
142.34
68.42
129.50
67.97
67.79
144.30
67.63
141.85
65.98
68.70
123.30
67.21
129.72
68.67
69.80
141.49
70.84
142.42
66.88
70.01
136.46
67.49
131.55
67.70
67.90
112.37
66.53
108.33
69.82
66.78
120.67
65.44
113.89
69.09
66.49
127.45
69.52
103.30
69.91
67.62
114.14
65.81
120.75
67.33
68.30
125.61
67.82
125.79
70.27
67.12
122.46
70.60
136.22
69.10
68.28
116.09
71.80
140.10
65.38
71.09
140.00
69.21
128.75
70.18
66.46
129.50
66.80
141.80
70.41
68.65
142.97
67.66
121.23
66.54
71.23
137.90
67.81
131.35
66.36
67.13
124.04
64.05
106.71
67.54
Chart Title
180.00
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
0
Series1
Series2
Series3
Series4
Series5
Series6
Series7
Series8
Series9
Series10
Series11
Series12
Series13
Series14
Series15
Series16
Series17
Series18
Series19
10
20
30
40
Series20
50
60
70
Expon. (Series20)
Data Set 04
Data Set 05
Data Set 06
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
124.36
66.50
128.73
68.64
134.63
64.87
124.86
69.00
137.55
66.78
121.90
67.09
139.67
68.30
129.76
70.05
155.38
68.35
137.37
67.01
128.82
66.28
128.94
65.61
106.45
70.81
135.32
69.20
129.10
67.76
128.76
68.22
109.61
69.13
139.47
68.02
145.68
69.06
142.47
67.36
140.89
67.66
116.82
67.73
132.75
70.09
131.59
66.31
143.62
67.22
103.53
70.18
121.12
69.44
134.93
67.37
124.73
68.23
131.51
63.84
147.02
65.27
129.31
68.13
136.55
67.72
126.33
70.84
134.02
70.24
141.49
70.05
125.48
69.92
140.40
71.49
140.61
70.19
115.71
64.29
102.84
69.20
112.14
65.95
123.49
68.25
128.52
70.06
133.46
70.01
147.89
66.36
120.30
70.56
131.80
68.61
155.90
68.36
138.60
66.29
120.03
68.81
128.07
65.48
132.96
63.43
123.10
69.76
119.37
69.72
115.62
66.77
128.14
65.46
133.81
67.73
122.52
68.89
115.48
68.83
Data Set 07
Data Set 08
Data Set 09
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
102.09
65.80
119.88
69.52
136.21
67.08
130.35
67.21
109.01
65.18
113.40
64.39
134.18
69.42
128.27
68.01
125.33
69.37
98.64
68.94
135.29
68.34
127.58
68.38
114.56
67.94
106.86
65.18
107.16
65.31
123.49
65.63
123.29
68.26
116.46
67.14
123.05
66.50
109.51
68.57
133.84
68.39
126.48
67.93
119.31
64.50
112.89
66.29
128.42
68.89
140.24
68.71
130.76
67.19
127.19
70.24
133.98
68.89
137.76
65.99
122.06
68.27
132.58
69.54
125.40
69.43
127.61
71.23
130.70
67.40
138.47
67.97
131.64
69.10
115.56
66.48
120.82
67.76
111.90
64.40
123.79
66.01
140.15
65.28
122.04
71.10
128.14
72.44
136.74
73.83
128.55
68.22
135.96
64.13
106.11
66.81
132.68
65.92
116.63
70.98
158.96
66.89
136.06
67.44
126.82
67.50
108.79
65.74
115.94
73.90
151.39
72.02
138.78
65.98
136.90
69.98
130.40
65.31
115.91
66.58
Data Set 10
Data Set 11
Data Set 12
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
146.29
67.11
135.02
65.73
121.50
65.84
109.88
65.87
116.23
66.43
112.71
65.77
139.05
66.78
109.17
70.01
135.00
71.14
119.90
68.74
124.22
69.48
128.68
67.83
128.31
66.23
141.16
68.63
124.41
65.07
127.24
65.96
129.15
68.36
140.03
69.71
115.23
68.58
127.87
68.39
117.52
69.93
124.80
66.59
120.92
68.77
143.87
66.12
126.95
66.97
127.65
69.92
141.17
68.61
111.27
68.08
101.47
71.56
155.94
69.00
122.61
70.19
144.99
68.45
134.01
66.79
124.21
65.52
110.95
66.71
130.10
68.02
124.65
67.46
132.86
66.68
106.23
69.67
119.52
67.41
146.34
67.94
112.05
71.82
139.30
69.66
145.59
68.90
136.19
72.75
104.83
65.80
120.84
67.29
131.24
67.28
123.04
66.11
115.78
69.57
131.32
67.41
118.89
68.24
128.30
67.67
119.53
68.53
121.49
68.02
127.47
69.04
117.00
68.47
119.25
71.39
127.88
67.97
138.53
68.52
Data Set 13
Data Set 14
Data Set 15
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
109.65
63.73
134.10
70.17
122.00
67.00
130.16
67.70
141.89
67.85
124.00
70.27
137.11
69.47
138.74
66.62
108.86
67.58
113.76
66.70
134.45
63.69
104.66
66.31
114.97
65.23
117.02
68.04
132.25
68.91
127.71
69.89
115.68
66.94
127.18
65.61
122.00
69.83
134.79
66.26
125.12
68.00
117.96
65.40
120.57
70.73
133.32
69.94
127.71
68.32
120.08
70.71
122.34
70.15
117.96
65.94
84.36
73.27
130.26
66.66
125.16
70.10
138.94
63.79
116.74
69.41
141.10
66.06
143.52
67.84
125.85
66.81
145.48
68.23
123.13
67.78
100.63
67.70
116.07
65.22
115.53
67.24
118.49
69.13
135.75
69.16
120.98
66.76
137.13
67.54
132.92
67.60
120.65
69.03
118.70
65.78
115.66
67.29
124.04
68.53
124.16
66.26
114.32
66.84
117.22
64.65
124.94
67.39
148.95
68.08
127.96
65.14
126.39
64.94
142.19
66.57
116.75
66.74
121.53
69.26
Data Set 16
Data Set 17
Data Set 18
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
133.30
69.08
138.73
67.98
130.64
68.74
127.41
69.64
130.95
66.62
122.39
69.52
125.08
69.08
124.30
67.30
133.87
70.31
102.81
68.01
135.37
65.96
134.83
67.30
134.22
68.86
124.20
66.54
137.84
67.61
100.98
63.19
111.24
68.08
112.68
67.86
145.89
65.58
129.40
65.83
121.20
66.47
155.30
69.12
127.74
69.24
114.93
68.92
138.72
69.79
121.33
64.69
110.38
68.47
116.26
67.48
118.22
69.21
126.03
69.52
120.61
69.09
123.34
67.17
118.70
70.24
114.41
68.55
151.91
67.85
119.31
67.86
133.34
68.21
115.31
69.27
150.74
68.00
92.75
67.10
122.38
67.91
121.11
67.14
131.78
71.24
129.23
67.08
118.41
68.10
118.39
69.12
138.28
70.02
142.76
68.99
131.15
65.50
144.06
70.05
145.63
69.33
130.20
66.59
120.85
70.87
127.34
70.09
129.18
68.00
113.30
66.06
115.97
68.74
143.14
72.58
149.32
65.24
104.44
64.19
Data Set 19
Data Set 20
Data Set 21
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
138.49
66.09
129.55
68.89
140.53
67.03
145.50
69.24
152.07
68.27
125.69
65.91
124.41
69.67
120.86
69.18
134.33
70.03
124.19
68.43
136.10
72.61
136.46
69.27
126.17
68.29
130.06
66.04
127.88
70.08
138.81
69.38
131.77
69.23
123.14
64.91
121.55
68.35
139.38
68.88
116.34
69.69
114.26
72.32
168.23
66.46
124.56
67.57
129.40
70.37
127.64
69.30
125.79
63.86
137.56
65.73
131.80
67.58
107.59
71.62
134.50
66.50
122.86
69.42
143.02
64.99
127.71
70.81
140.21
67.47
118.74
64.61
115.67
67.38
142.69
66.45
118.99
62.02
131.59
67.80
137.43
68.76
131.87
65.29
127.58
68.16
126.04
63.77
119.37
67.70
125.72
68.86
143.60
69.04
126.12
70.86
119.99
67.71
122.62
66.49
126.61
68.17
140.87
64.66
108.60
66.74
113.56
67.45
136.92
65.74
100.66
67.69
115.84
71.04
118.48
68.42
116.14
66.69
130.27
67.83
Data Set 22
Data Set 23
Data Set 24
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
125.45
67.15
108.75
68.04
121.50
65.57
117.17
68.30
134.40
68.12
119.28
68.30
137.32
65.41
131.21
71.24
126.55
71.03
149.44
68.67
130.83
66.92
129.25
70.77
151.07
70.07
127.08
71.10
134.78
67.51
117.00
69.13
119.41
69.41
146.07
69.51
138.18
67.02
105.56
69.17
162.41
66.45
127.62
65.52
134.47
68.58
130.19
68.57
120.75
70.20
132.69
69.60
132.19
67.01
144.25
67.67
126.06
69.05
127.46
68.70
105.79
69.73
126.35
68.29
125.35
68.23
120.16
67.62
142.36
69.25
134.76
70.41
109.08
70.80
124.59
68.24
121.94
65.70
125.97
69.66
122.36
67.69
125.67
65.70
110.23
66.73
123.23
68.84
145.96
68.48
134.47
67.58
131.40
69.52
119.92
66.63
115.51
65.25
124.30
66.98
111.51
66.84
125.11
70.71
128.38
72.01
138.68
69.90
116.41
71.27
136.85
67.44
135.02
67.65
133.39
66.56
118.06
66.06
116.51
68.31
Data Set 25
Data Set 26
Data Set 27
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
123.96
69.75
138.09
68.68
120.69
68.81
136.95
68.58
126.61
67.54
115.78
67.56
153.08
68.14
131.81
71.18
128.69
68.17
126.36
67.98
131.74
70.54
134.76
65.81
145.06
64.83
117.63
71.53
118.34
64.92
135.17
69.21
126.19
66.77
106.16
68.36
108.81
69.27
127.51
66.34
126.38
66.72
110.52
70.94
142.48
64.83
114.37
69.18
138.27
69.08
139.44
68.38
130.28
65.89
119.51
66.20
110.32
68.05
123.31
66.04
115.32
71.06
131.95
69.09
122.86
69.44
132.37
68.99
129.20
69.91
125.69
67.15
93.99
70.96
124.35
68.41
111.72
65.69
127.41
66.02
126.53
68.33
125.55
69.43
134.30
69.86
149.23
66.96
119.97
66.55
124.97
66.83
142.16
70.55
132.60
69.04
124.07
69.09
127.14
69.38
132.67
67.62
133.55
66.81
122.34
67.52
117.45
69.59
134.15
67.13
112.67
64.87
101.85
68.12
130.25
64.57
134.26
69.13
128.44
71.23
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
64.00
65.00
Data Set 28
Data Set 29
Data Set 30
Weight (lbs) Height (in) Weight (lbs) Height (in) Weight (lbs) Height (in)
134.04
66.43
130.37
70.43
129.59
69.04
127.75
66.32
122.89
67.68
134.11
67.36
127.62
68.03
107.87
67.88
106.13
70.36
139.40
68.29
118.59
68.07
127.06
69.89
113.94
66.92
125.65
66.99
121.92
65.15
147.52
71.78
148.60
71.58
150.74
69.41
132.65
66.05
150.28
64.37
116.01
68.51
132.81
69.01
127.07
67.93
125.47
65.72
105.09
68.71
128.71
69.92
131.61
63.98
127.52
65.01
118.42
69.94
129.07
70.08
130.73
66.06
125.39
66.80
129.11
68.43
128.58
71.16
149.27
70.11
148.16
68.24
122.57
67.55
107.35
68.95
105.39
67.83
157.30
68.83
133.70
68.03
110.54
66.31
121.11
68.12
124.85
69.36
118.64
69.75
147.51
67.91
112.70
68.45
122.82
65.71
117.49
67.64
127.84
65.70
124.89
66.88
132.45
69.78
134.16
69.56
135.77
67.28
118.18
68.00
138.60
65.85
116.46
70.63
123.73
69.41
126.04
68.80
125.36
69.94
Name Data 28
65.00
66.00
67.00
68.00
69.00
70.00
71.00
72.00
73.00
Weight (lbs)
133.12
138.33
124.50
136.72
115.02
132.64
143.13
134.74
122.62
153.56
126.29
134.72
134.60
118.39
121.30
132.28
139.06
110.70
154.11
124.16
Chart Title
180.00
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
0
10
Series1
Series2
Series3
Series4
Series5
Series6
Series7
Series8
Series9
Series10
Series11
Series12
Series13
Series14
Series15
Series16
Series17
Series18
Series19
20
30
40
Series20
50
60
70
Expon. (Series20)
Data Set 51
520.00
520.00
521.00
522.00
522.00
522.00
524.00
526.00
528.00
529.00
529.00
530.00
532.00
532.00
537.00
544.00
547.00
548.00
550.00
551.00
553.00
554.00
554.00
555.00
555.00
556.00
557.00
558.00
562.00
563.00
566.00
567.00
568.00
574.00
577.00
Bin
Frequency
520-539
14
535-549
4
550-564
12
565-579
7
580 & Above
13
28%
8%
24%
14%
26%
Mean
557.48
14 Median
555.50
Mode
522.00
Range
79.00
Variance
633.68
4 Standard Dev
25.17
12
578.00
579.00
581.00
586.00
587.00
587.00
587.00
588.00
588.00
588.00
590.00
591.00
594.00
598.00
599.00
50.00
7
13
Histogram
16
14
14
13
12
12
10
7
8
Frequency Bands
6
4
4
2
0
520-539
535-549
550-564
565-579
580 & Above
Pie Chart
26%
28%
520-539
535-549
550-564
8%
14%
565-579
580 & Above
24%
Compatibility Report for attachment_1.xls
Run on 12/19/2022 6:38
If the workbook is saved in an earlier file format or opened in
an earlier version of Microsoft Excel, the listed features will
not be available.
Minor loss of fidelity
Some cells or styles in this workbook contain formatting that
is not supported by the selected file format. These formats
will be converted to the closest format available.
# of
occurrences
Version
1
Excel 97-2003
Name Data 28
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
64.00
65.00
66.00
67.00
68.00
69.00
70.00
71.00
72.00
73.00
Compatibility Report for attachment_1.xls
Run on 12/19/2022 6:58
If the workbook is saved in an earlier file format or opened in
an earlier version of Microsoft Excel, the listed features will
not be available.
Minor loss of fidelity
Some cells or styles in this workbook contain formatting that
is not supported by the selected file format. These formats
will be converted to the closest format available.
# of
occurrences
Version
1
Excel 97-2003
Data Set 07
Height (in) Weight (lbs)
65.80
106.86
67.21
109.01
69.42
109.51
68.94
115.56
67.94
116.63
65.63
119.31
66.50
119.88
67.93
123.29
68.89
123.79
70.24
126.82
68.27
128.14
71.23
128.27
69.10
130.40
64.40
130.70
71.10
132.58
68.22
133.98
65.92
135.29
67.44
135.96
73.90
140.24
69.98
151.39
Chart Title
160.00
y = 1.8238x + 1.1303
R² = 0.1333
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
62.00
64.00
66.00
68.00
70.00
72.00
74.00
0.1333
126.84 VARIANCE
127.48 MEDIAN
125.88 MEAN
N/A MODE
74.00
76.00
0.37
Data Set 51
520.00
520.00
521.00
522.00
522.00
522.00
524.00
526.00
528.00
529.00
529.00
530.00
532.00
532.00
537.00
544.00
547.00
548.00
550.00
551.00
553.00
554.00
554.00
555.00
555.00
556.00
557.00
558.00
562.00
563.00
566.00
567.00
568.00
574.00
577.00
Bin
Frequency
520-539
14
535-549
4
550-564
12
565-579
7
580 & Above
13
28%
8%
24%
14%
26%
Mean
557.48
14 Median
555.50
Mode
522.00
Range
79.00
Variance
633.68
4 Standard Dev
25.17
12
578.00
579.00
581.00
586.00
587.00
587.00
587.00
588.00
588.00
588.00
590.00
591.00
594.00
598.00
599.00
50.00
7
13
Histogram
16
14
14
13
12
12
10
7
8
Frequency Bands
6
4
4
2
0
520-539
535-549
550-564
565-579
580 & Above
Pie Chart
26%
28%
520-539
535-549
550-564
8%
14%
565-579
580 & Above
24%
Number of Boys
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Height (in) Weight (lbs)
x
y
64.37
116.01
65.70
124.89
65.85
116.46
66.80
129.11
66.99
121.92
67.68
134.11
67.88
106.13
67.93
125.47
68.03
110.54
68.07
127.06
68.45
122.82
68.80
125.36
68.95
105.39
69.36
118.64
69.56
135.77
69.92
131.61
69.94
129.07
70.11
148.16
70.43
129.59
71.58
150.74
Data Set 29
Height Whole Number (ft)
x
5’3
5’4
5’4
5’5
5’5
5’5
5’5
5’5
5’6
5’6
5’6
5’6
5’6
5’7
5’7
5’7
5’7
5’8
5’8
5’9
Weight to Nearest Wole Number
y
105.00
106.00
110.00
116.00
116.00
118.00
121.00
122.00
124.00
125.00
125.00
127.00
129.00
129.00
129.00
131.00
134.00
135.00
148.00
150.00
y = 3.4364x
R² = 0.2681
Original Data Sheet 29
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
64.00
65.00
66.00
67.00
r=
0.517783739
Mean
133.37
Median
124.94
Mode
116,125,129,
68.00
69.00
70.00
y = 3.4364x – 109.33
R² = 0.2681
y = 1.9113x + 104.93
R² = 0.9181
Modified Data Sheet 29
160.00
140.00
120.00
100.00
80.00
60.00
40.00
20.00
0.00
70.00
71.00
72.00
0
5
10
15
Median Height (Inches)
5’6
Median Weight
125.00
Standard Deviation of Height
1.781302804
Standard Deviation of Weight
11.80098123
STDEV.P Height
1.73619921
STDEV.P Weight
11.52191567
STDEV.S Height
1.781302804
STDEV.S Weight
11.82123605
Correlation Coeffecient ®
0.51781762
Coeffecient of Determination (R^2)
R^2 = 0.9181
20
= 1.9113x + 104.93
R² = 0.9181
25
Pages (550 words)
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Question #: 535
Question: PROBLEM WITH LIMIT SWITCHES TRIGGERING
Current Solution
Are you using shielded wire? If so, take the shield to ground. If there are any other wires in the cable that is not being used, ground those as well. The customer can also try moving the cables as far away as possible from the motor wires and motors.
Respond:
Other Possible Solutions to this Question
• MY LIMIT SWITCHES ARE TRIGGERING AT RANDOM.
Are you using shielded wire? If so, take the shield to ground. If there are any other wires in the cable that is not being used, ground those as well. The customer can also try moving the cables as far away as possible from the motor wires and motors.
MY LIMIT SWITCHES ARE TRIGGERING AT RANDOM.
• LIMIT SWOTCXH PROBLEM
This is a common problem. Are you using shielded wire? If so, take the shield to ground. If there are any other wires in the cable that is not being used, ground those as well. Also try moving the cables as far away as possible from the motor wires and motors.
LIMIT SWOTCXH PROBLEM
• LIMIT SWITCH PROBLEM
This is a common problem. Are you using shielded wire? If so, take the shield to ground. If there are any other wires in the cable that is not being used, ground those as well. Also try moving the cables as far away as possible from the motor wires and motors.
LIMIT SWITCH PROBLEM
• HOW DO I SET THE DIP SWITCHES?
blueChick:
X-axis
“CW230 (3.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100
Mach3 Motor Tuning: 1422.22 steps/in
Y-axis
“CW230 (3.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100
Mach3 Motor Tuning: 1422.22 steps/in
Z-axis
“CW230 (3.0A) Driver”
Set to 1/4 Microstep, 2.7A
Dipswitches: 10101100
Mach3 Motor Tuning: 1600 steps/in
blackToe:
X-axis
“CW230 (3.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100
Mach3 Motor Tuning: 1422.22 steps/in
Y-axis
“CW230 (3.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100
Mach3 Motor Tuning: 1422.22 steps/in
Z-axis
“CW230 (3.0A) Driver”
Set to 1/4 Microstep, 2.7A
Dipswitches: 10101100
Mach3 Motor Tuning: 1600 steps/in
blackFoot:
X-axis
“CW8060 (6.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100 (“0”=down, “1”=up)
Mach3 Motor Tuning: 914.29 steps/in
Y-axis
“CW230 (3.0A) Driver”
Set to 1/16 Microstep, 2.7A
Dipswitches: 11001100
Mach3 Motor Tuning: 1422.22 steps/in
Z-axis
“CW230 (3.0A) Driver”
Set to 1/4 Microstep, 2.7A
Dipswitches: 10101100
Mach3 Motor Tuning: 1600 steps/in
greenBull:
X-axis
“CW8060 (6.0A) Driver”
Set to 5.43A, 1/16 Microstep
Dipswitches: 01100110 (“0”=down, “1”=up)
Mach3 Motor Tuning: 914.29 steps/in
Y-axis
“CW8060 (6.0A) Driver”
Set to 5.43A, 1/16 Microstep
Dipswitches: 01100110
Mach3 Motor Tuning: 914.29 steps/in
Z-axis
“CW8060 (6.0A) Driver”
Set to 5.43A, 1/4 Microstep
Dipswitches: 01100100
Mach3 Motor Tuning: 1600 steps/in
HOW DO I SET THE DIP SWITCHES?
• THE DUST SHOE HAS STIFF BRISTLES, IS THERE PROBLEM WITH MOVING Z-AXIS DOWN SINCE BRISTLES PUSH AXIS SIDEWAYS
A couple things to consider, the machine is most likely moving sideways, so the bristles won't be a problem. The lead screw has a very hight mechanical advantage and the bristles don't pose an issue with this. We have not found that the bristles cause any appreciable deflection to the work we do in our shop; however, you can change the brush for a more pliable plastic, or reduce the number of bristles by selectively pruning.
THE DUST SHOE HAS STIFF BRISTLES, IS THERE PROBLEM WITH MOVING Z-AXIS DOWN SINCE BRISTLES PUSH AXIS SIDEWAYS
• HOW SERIOUS A PROBLEM INFLUENZA IN THE U.S.?
Influenza is the most frequent cause of death from a vaccine-preventable disease in the United States. From 1990 through 1999, an average of approximately 36,000 influenza-associated pulmonary and circulatory deaths occurred during each influenza season. In addition to fatalities, seasonal influenza is also responsible for more than 200,000 hospitalizations per year. Rates of infection from seasonal influenza are highest among children, but the risks for complications, hospitalizations, and deaths are higher among adults ages 65 and older, children younger than 5 years, and people of any age who have medical conditions that place them at increased risk for complications from influenza.
A novel H1N1 virus was first detected in March of 2009 and quickly spread to pandemic levels. In the U.S., it is estimated that approximately 43-89 million persons became ill with 2009 pandemic H1N1 from April 2009 to April 2010. The virus also resulted in significant hospitalizations and deaths among children, adults 19-65 years, obese persons, and pregnant and post-partum women.
HOW SERIOUS A PROBLEM INFLUENZA IN THE U.S.?
• I HAVE A 2000 WK 11–3 Z LIMIT SWITCHES GOING TO C10R10 BREAKOUT BOARD . CAN ALSO USE THEM AS MY HOME POSITION WELL. NEED THE WIRING CONFIGURATION
No, you should not have received two parallel breakout boards in your CNC kit. You can return the extra item using standard shipping with COD. Our apologies for the confusion and the inconvenience.
I will answer the other question as another FAQ. Please check the Customer Service page to see the answer to that question.
https://www.buildyourcnc.com/FAQ/13636
I HAVE A 2000 WK 11–3 Z LIMIT SWITCHES GOING TO C10R10 BREAKOUT BOARD . CAN ALSO USE THEM AS MY HOME POSITION WELL. NEED THE WIRING CONFIGURATION
• HOW SERIOUS A PROBLEM INFLUENZA (THE FLU) IN THE U.S.?
Influenza is the most frequent cause of death from a vaccine-preventable disease in the United States. From 1990 through 1999, an average of approximately 36,000 influenza-associated pulmonary and circulatory deaths occurred during each influenza season. In addition to fatalities, seasonal influenza is also responsible for more than 200,000 hospitalizations per year. Rates of infection from seasonal influenza are highest among children, but the risks for complications, hospitalizations, and deaths are higher among adults ages 65 and older, children younger than 5 years, and people of any age who have medical conditions that place them at increased risk for complications from influenza.
A novel H1N1 virus was first detected in March of 2009 and quickly spread to pandemic levels. In the U.S., it is estimated that approximately 43-89 million persons became ill with 2009 pandemic H1N1 from April 2009 to April 2010. The virus also resulted in significant hospitalizations and deaths among children, adults 19-65 years, obese persons, and pregnant and post-partum women.
HOW SERIOUS A PROBLEM INFLUENZA (THE FLU) IN THE U.S.?
• THE SWITCHES KEEP TRIPPING! AT TIMES IT IS LITERALLY AS SOON CLICK RESET ON MACH 3 THEY TRIP AGAIN.
http://buildyourcnc.com/PrimeronHomingandLimitSwitches.aspx
There is a pin setting in Mach3 that says how to react to the inputs from the switches. Your setting is probably wrong as the trigger happens immediately.
You may want to think about shielded wire and/or a resistor on each limit switch line. If there are stray wires, connect them to ground. Additionally, the shield from the shielded cable insulation should alse connected to ground at both ends.
Try upping the Debounce interval in Mach3. You have to adjust the debounce interval. You go to config,general config, and it is in the top right corner. I set mine to 100 and it seems to work well.
how to solution cnc revo 540 limit tripped error?
THE SWITCHES KEEP TRIPPING! AT TIMES IT IS LITERALLY AS SOON CLICK RESET ON MACH 3 THEY TRIP AGAIN.
• YOUR CNC SOFTWARE COMPATIBLE WITH MAC?
There are several CAD/CAM programs out there for MacOS. Here is a list of a few: http://www.macuser.de/forum/f53/3d-2d-cad-395058/#post4499316
Of course, there is always SketchUp if you are in a pinch. There are a few CAM plugins for SketchUp as well, but I've never used them, so I can't vouch for them.
CAD/CAM will get you as far as designing your parts and producing the g-code, but if you also want to run the CNC machine with your Mac, you will need a piece of software similar to Mach3 or EMC to run the g-code. I haven't been able to find anything like these programs that runs on Mac, so you may have to run Mach3 or EMC in Windows using Boot Camp or a virtual machine such as VMware or Parallels.
Personally, I would use the Mac to do all the CAD/CAM, then use a cheap PC to run the CNC machine.
EasyDNC for OSX
www.dnc-x.com
Hi guys I’m new to the cnc and I’m beginning to learn how to operate and cad-cam software also my cnc runs on Mach3 my questions is if I should use MacBook to do all designs if I can have all this software to run on Mac or I have to use windows software to operate it? I have separate small computer connected to the cnc table who have Mach3 on it please help. Also if you know any other software much easy for beginners please let me know thank you guys!
My email if needed is r1jastrzebski@gmail.com
YOUR CNC SOFTWARE COMPATIBLE WITH MAC?
• WHAT ELSE DO I NEED WITH THE MACHINE?
PC with parallel port and USB port
20awg stranded wires for the motors - http://www.buildyourcnc.com/electronicscombo.aspx
18awg stranded wires for power supply to drivers
24awg stranded wires for breakout board to drivers
(location and spacing of components varies from one person to another, so we do not provide cables/wires)
- General purpose extension cord (cut the female end off) to provide power to power supply
- USB cable to power breakout board
- Parallel cable to communicate to breakout board
- Router
- (optional - instead of router) Spindle with power inverter http://www.buildyourcnc.com/SpindlesAndAccessories.aspx
- (if purchasing spindle with inverter) General purpose extension cord (240v) (cut the female end off) to provide power to power inverter
- End Mill(s) http://www.buildyourcnc.com/ProductsEndMills.aspx
- CAD, and/or CAD-CAM software (to produce geometry, machine operations, and g-code) http://www.buildyourcnc.com/CNCsoftware.aspx
- CNC control software (to read g-code and control machine) http://www.buildyourcnc.com/CNCsoftware.aspx
- If the machine does not have the table included, then a table will need to be built. The rails, chain and chain mounts are included that are used on this table.
WHAT ELSE DO I NEED WITH THE MACHINE?
• DOES ZY PLATE COME WITH LASER HEAD COMBO.
No the ZY plate is part of the machine kit not the head combo kit.
DOES ZY PLATE COME WITH LASER HEAD COMBO.
• DO NEED LIMIT SWITCHES, IF SO WHAT IS YOUR RECOMMENDATION?
Limit switches are not really needed for this level of a machine, but if you do want to use them, then you will need to connect them with shielded cable and ground the shield at both ends and any other non-used wire in the cable. Limit switches are very susceptible to the motor interference. You will also need to adjust the debounce in the software you will use.
DO NEED LIMIT SWITCHES, IF SO WHAT IS YOUR RECOMMENDATION?
• CAN I USE A STEPPING MOTOR WITH AN ENCODER?
I haven't delved into using encoders with stepping motors too much. From my research, you need to have a controller that can provide the closed loop control, rather than software handling that process. I have also found from my research that using encoders on stepping motors is generally used to stop the machine in the case that the motor failed to achieve the commanded position for some reason and gives the user the chance to correct and continue with the job.
If you want proper closed loop control, it may be best to go with servos and servo controller that provide the closed loop control within the real of those two components.
CAN I USE A STEPPING MOTOR WITH AN ENCODER?
• WHAT NOT INCLUDED WITH THE MACHINE?
PC with parallel port and USB port
20awg stranded wires for the motors - http://www.buildyourcnc.com/electronicscombo.aspx
18awg stranded wires for power supply to drivers
24awg stranded wires for breakout board to drivers
(location and spacing of components varies from one person to another, so we do not provide cables/wires)
- General purpose extension cord (cut the female end off) to provide power to power supply
- USB cable to power breakout board
- Parallel cable to communicate to breakout board
- Router
- (optional - instead of router) Spindle with power inverter http://www.buildyourcnc.com/SpindlesAndAccessories.aspx
- (if purchasing spindle with inverter) General purpose extension cord (240v) (cut the female end off) to provide power to power inverter
- End Mill(s) http://www.buildyourcnc.com/ProductsEndMills.aspx
- CAD, and/or CAD-CAM software (to produce geometry, machine operations, and g-code) http://www.buildyourcnc.com/CNCsoftware.aspx
- CNC control software (to read g-code and control machine) http://www.buildyourcnc.com/CNCsoftware.aspx
- If the machine does not have the table included, then a table will need to be built. The rails, chain and chain mounts are included that are used on this table.
WHAT NOT INCLUDED WITH THE MACHINE?
• DO YOU HAVE BLACKTOES WITH TABLES?
Yes, we have the optional table for the blackToe, but we need to fabricate it, which only takes one day. We generally reserve one week for the total fabrication and packing of all parts.
DO YOU HAVE BLACKTOES WITH TABLES?
• WILL THE USB BREAKOUT BOARD WORK WITH MACH3?
Sure, the USB interface has a place for 4 axes of limit switches.
Each axis can have two limit switches: one for the ++ (positive) end and one for the -- (negative) end. The positive end would be the limit switch at the end of the machine that, say the machine has a 4'x8' area, reaches a bit after the 8 foot mark. The negative end would be the limit switch behind the 0 foot location behind the origin. If the origin is in the middle, the negative would be at a little more than the -4 foot end and the positive would be at a bit more than the +4 foot end. Note that you can have more than one switch on each pin where the NC is connected in serial fashion and the NO is connected in parallel fashion (this can be seen on the diagram in the multiple limits switch section). The software configurations for the limits switches are under File -> Settings -> Limit.
A typical limit switch has three connections on it. These connections consist of COM (common), NC (normally closed) and NO (normally open). The COM would generally go to GND and the NC or the NO would go to the pin. If the NC is used, then the the switch is constantly connected until the switch is pushed (engaged) then the connection from the pin to gnd is broken (open). Use the settings in software to set whether in NC or NO configuration.
Let me know if this information was helpful (or not) by adding information to this question. Thanks.
User response:
Thank you very much for this helpful information. I'm still a little fuzzy on how the 6 limit switches physically connect to each other and to the USB breakout board. You've stated one switch (home) goes to positive and another switch (limit) goes to negative. Are all the GND prongs from all 6 switches connected to each other and going to GND on the breakout board, or no? And the NC prongs, how exactly are they connected to each other? And to the board? There has to be a diagram somewhere shows this visually, no? I don't know how to wire the switches in series or in parallel. I have already physically installed all the switches on the machine and ran the wires to where the board is. Now I just need to know where to plug these wires into the board. Also, taking into consideration that I'm using the Planet CNC software, the only settings I have pertaining to limit switches is "Enable/Disable" for each axis, and the actual limit for each axis. Nothing about NC or NO. Is that only in Mach3?
Thank you.
buildyourcnc response:
On the USB interface, the COM on the switch connects to GND and the NC or NO connects to the input pin (i.e. x++, y--, etc.)
Limit switch configuration is rather difficult to understand, especially with series and parallel. You can think of series as a single wire going from GND to the axis letter input terminal (i.e. X++ or X--). If the wire is broken, then the circuit is open (or the switch is engaged in a normally closed scenario). Normally closed is like an actual wire, and when engaged, the switch "opens" (breaks the wire). This is why we recommend in some systems that you can put many switches in series on a single pin. When one of the switches is engaged (breaking the connection) then the entire circuit of switches is broken and the machine stops.
In a parallel scenario, the state of the circuit is always broken until the one of the switches is engaged and the circuit is then closed or connected. The topology looks like a ladder. All the switches connect to both sides of the ladder and the switches are like the runs of the ladder (the horizontal bars that the feet are placed while climbing). Imagine all of the switches broken in this scenario (normally open). It would be like the ladder could be split in two, but if one of the ladder runs (switches) is closed by engaging it, then that run would connect both sides of the ladder and the two sides of the ladder would have a connection.
There is a diagram on the USB page of the various limit switch configurations. If you need more information (visual and/or otherwise), please let us know and we will immediately add that information to benefit everyone.
WILL THE USB BREAKOUT BOARD WORK WITH MACH3?
• DO I HAVE TO REGISTER WITH YOUR OFFICE BE PROTECTED?
No. In general, registration is voluntary. Copyright exists from the moment the work is created. You will have to register, however, if you wish to bring a lawsuit for infringement of a U.S. work.
DO I HAVE TO REGISTER WITH YOUR OFFICE BE PROTECTED?
• CAN THERE BE CLOSED LOOP CONTROL WITH STEPPING MOTORS?
I haven't delved into using encoders with stepping motors too much. From my research, you need to have a controller that can provide the closed loop control, rather than software handling that process. I have also found from my research that using encoders on stepping motors is generally used to stop the machine in the case that the motor failed to achieve the commanded position for some reason and gives the user the chance to correct and continue with the job.
If you want proper closed loop control, it may be best to go with servos and servo controller that provide the closed loop control within the real of those two components.
CAN THERE BE CLOSED LOOP CONTROL WITH STEPPING MOTORS?
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## Advanced Mathematics Answers
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∀n∈N, 1!3!…(2n+1)! > ((n+1)!)n+1 N stands for the natural numbers. I am having trouble manipulating the ((n+1)!)n+1 term appropriately.
Let T be the partial order relation defined on N * N by (a,b)T(c,d) if and only if a <= c and b <= d. Is T a total order relation?
Let x* be an optimal solution to the following problem: Minimize cx s.t. aix = bi , i = 1,…,m, x ≥ 0 Let w* be an optimal solution to the dual... | 169 | 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} | 2.890625 | 3 | CC-MAIN-2018-17 | latest | en | 0.795859 |
https://cal8.cs.fau.de/redmine/projects/cool/repository/revisions/7c4d2eb4fb20f82800d52965941ce043aa8367f7/entry/src/lib/GMLMIP-0.1/rules/node.cpp | 1,563,752,816,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527396.78/warc/CC-MAIN-20190721225759-20190722011759-00151.warc.gz | 336,920,715 | 4,292 | ### Profile
Statistics
| Branch: | Revision:
## cool / src / lib / GMLMIP-0.1 / rules / node.cpp @ 7c4d2eb4
1 ```#include "node.h" ``` ```void Node::allow(int a){ ``` ``` valuations[a] = 1; ``` ```} ``` ```void Node::disallow(int d){ ``` ``` valuations[d] = 0; ``` ```} ``` ```Node::Node(int no_of_atoms) : size (static_cast(pow(2, no_of_atoms))), valuations(size, 0){ ``` ``` //valuations.resize(size, 0); ``` ``` // needs to be replaced with actually initing the stuff ``` ``` //for(int i=0; i < size; i++) ``` ``` //valuations[i]=0; ``` ```} ``` ```bool Node::operator==(const Node& other) const{ ``` ``` if(size == other.size){ ``` ``` for(int i=0; i < size; i++) ``` ``` if(valuations[i] != other.valuations[i]) ``` ``` return false; ``` ``` return true; ``` ``` } ``` ``` return false; ``` ```} ``` ```bool Node::operator<(const Node& other) const{ ``` ``` if(size == other.size){ ``` ``` for(int i=0; i < size; i++) ``` ``` if(valuations[i]==1 && other.valuations[i]==0) ``` ``` return false; ``` ``` return true; ``` ``` } ``` ``` cout << "you checked for supersets with nodes of different sizes - you probably didn't mean to do this..." << endl; ``` ``` return false; ``` ```} ``` ```ostream& operator<<(ostream& out, const Node& n){ ``` ``` out << n.size << endl; ``` ``` for(int i=0; i < n.size; i++) ``` ``` out << n.valuations[i] << " "; ``` ``` out << endl; ``` ```} ``` | 409 | 1,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.600859 |
https://ask.sagemath.org/questions/40778/revisions/ | 1,611,195,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00556.warc.gz | 219,604,322 | 6,351 | # Revision history [back]
### How to get imaginary and real parts of quadratic extension?
I have the following code segment in Sage:
proof.arithmetic(False)
p = 10354717741769305252977768237866805321427389645549071170116189679054678940682478846502882896561066713624553211618840202385203911976522554393044160468771151816976706840078913334358399730952774926980235086850991501872665651576831
assert p.is_prime()
Fp = GF(p)
R.<x> = PolynomialRing(Fp)
# The quadratic extension via x^2 + 1 since p = 3 mod 4
Fp2.<j> = Fp.extension(x^2 + 1)
v = 9207905618485976447392495823891126491742950552335608949038426615382964807887894797411491716107572732408369786142697750332311947639207321056540404444033540648125838904594907601875471637980859284582852367748448663333866077035709*j + 4651155546510811048846770550870646667630430517849502373785869664283801023087435645046977319664381880355511529496538038596466138807253669785341264293301567029718659171475744580349901553036469330686320047828171225710153655171014
Now, if I try to get the real and imaginary parts with v.real() and v.imag(), I get errors that those methods do not exist. I guess because v has some different structure and type. How can I get the imaginary and real parts here?
### How to get imaginary and real parts of quadratic extension?
I have the following code segment in Sage:
proof.arithmetic(False)
p = 10354717741769305252977768237866805321427389645549071170116189679054678940682478846502882896561066713624553211618840202385203911976522554393044160468771151816976706840078913334358399730952774926980235086850991501872665651576831
assert p.is_prime()
Fp = GF(p)
R.<x> = PolynomialRing(Fp)
# The quadratic extension via x^2 + 1 since p = 3 mod 4
Fp2.<j> = Fp.extension(x^2 + 1)
v = 9207905618485976447392495823891126491742950552335608949038426615382964807887894797411491716107572732408369786142697750332311947639207321056540404444033540648125838904594907601875471637980859284582852367748448663333866077035709*j + 4651155546510811048846770550870646667630430517849502373785869664283801023087435645046977319664381880355511529496538038596466138807253669785341264293301567029718659171475744580349901553036469330686320047828171225710153655171014
Now, if I try to get the real and imaginary parts with v.real() and v.imag(), I get errors that those methods do not exist. I guess because v has some different structure and type. How can I get the imaginary and real parts here?
### How to get the imaginary and real parts of quadratic extension?
I have the following code segment in Sage:
proof.arithmetic(False)
p = 10354717741769305252977768237866805321427389645549071170116189679054678940682478846502882896561066713624553211618840202385203911976522554393044160468771151816976706840078913334358399730952774926980235086850991501872665651576831
assert p.is_prime()
Fp = GF(p)
R.<x> = PolynomialRing(Fp)
# The quadratic extension via x^2 + 1 since p = 3 mod 4
Fp2.<j> = Fp.extension(x^2 + 1)
v = 9207905618485976447392495823891126491742950552335608949038426615382964807887894797411491716107572732408369786142697750332311947639207321056540404444033540648125838904594907601875471637980859284582852367748448663333866077035709*j + 4651155546510811048846770550870646667630430517849502373785869664283801023087435645046977319664381880355511529496538038596466138807253669785341264293301567029718659171475744580349901553036469330686320047828171225710153655171014
Now, if I try to get the real and imaginary parts with v.real() and v.imag(), I get errors that those methods do not exist. I guess because v has some different structure and type. How can I get the imaginary and real parts here? | 1,133 | 3,650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.046875 | 3 | CC-MAIN-2021-04 | latest | en | 0.436866 |
https://mathoverflow.net/questions/269803/von-neumann-algebra-isomorphism-extension | 1,685,626,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00633.warc.gz | 434,206,220 | 23,533 | # Von Neumann Algebra isomorphism extension
I have Von Neumann algebra $\mathfrak{U}$ and a weakly dense *-subalgebra $A$. I have another Von Neumann Algebra $\mathfrak{V}$ and an injective *-homorphism $$\phi: A\longrightarrow \mathfrak{V}$$ such that $\phi(A)$ generates $\mathfrak{V}$. Is it possible to say that one can extend $\phi$ to an *-isomorphism from $\mathfrak{U}$ onto $\mathfrak{V}$?
P.D. In fact the case I am interested in is when A is a Hilbert algebra and $\mathfrak{V}$ is the natural Von Neumann Algebra of A as defined by Dixmier in its books C *-Algebras or Von Neumann Algebras. I however believe the question as posted has all the required ingredients.
No. This will almost never be true (subalgebras of the compacts are the only cases I can think of where it could work). The easiest example is probably $C[0,1].$ It has an injective homomorphism to $\ell^\infty(\mathbb{N})$--by point evaluation at the rationals--that generates $\ell^\infty(\mathbb{N})$. It also sits inside $L^\infty[0,1].$ But $\ell^\infty(\mathbb{N})$ has minimal projections and $L^\infty[0,1]$ doesn't.
• Do you think if I know that the Von Neumann algebras involved are Type I it will make a difference? May 15, 2017 at 18:17
• @Carlos De la Mora No it won't make a difference. The example I gave is Type I. May 15, 2017 at 19:29
• @CalebEckhardt I want to ask a silly question: Why $\phi(C[0,1])$ generates $\ell^\infty(\mathbb{N})$? Jul 18, 2017 at 11:04
• @C.Ding Just to be clear; the claim is that $\phi(C[0,1])$ generates $\ell^\infty(\mathbb{N})$ as a von Neumann algebra. So it suffices to show that $\phi(C[0,1])$ is w*-dense in $\ell^\infty(\mathbb{N}).$ Let $e_n$ be a minimal projection in $\ell^\infty(\mathbb{N}).$ Let $q$ be a rational corresponding to $e_n.$ Take a sequence of uniformly bounded continuous functions $f_k$ with $f_k(q)=1$ whose support shrinks to $\{ q \}.$ Then $\phi(f_k)\rightarrow e_n$ in the w*-topology. Jul 18, 2017 at 14:44
You can give a characterisation of when it holds. Without loss of generality, we may suppose that $A$ is closed, i.e. is a C$^*$-algebra. Consider the universal enveloping von Neumann algebra, which I will consider as being the bidual $A^{**}$. If $\phi:A\rightarrow\mathfrak{M}$ is a $*$-homomorphism with $\phi(A)$ weak$^*$-dense in $\mathfrak{M}$ then there is a unique extension $\tilde\phi:A^{**}\rightarrow\mathfrak{M}$ which is a surjective, weak$^*$-continuous $*$-homomorphism.
As $A$ generates $A^{**}$, your question has a positive answer if and only if $\tilde\phi$ is always an isomorphism. This is equivalent (consider the kernel of $\tilde\phi$) to $A^{**}$ having no proper weak$^*$-closed ideals. In turn, this is equivalent to $A^*$ having no proper $A$-invariant closed subspaces; and is equivalent to $A^{**}$ having no non-trivial central projections.
See Volume 1 of Takesaki, Section III, Chapter 2.
(I presume that $\mathfrak{U}'$, in your question, is not the commutant of $\mathfrak{U}$. This is slightly unfortunate notation...)
• Can anything be said on how rare or common it is for $A^{**}$ to have no non-trivial central projections? Currently the only (infinite-dimensional) example I can think of is K(H). May 16, 2017 at 20:04
• Urgh, I don't know enough. Can the universal enveloping algebra be a factor of type II or III ?? May 17, 2017 at 15:16
• The question did ask for an injective -homomorphism so this doesn't quite give a characterization. For example, the answer is certainly yes for any finite dimensional C-algebra (the double duals of which can of course have many weak*-closed ideals). Jul 18, 2017 at 17:59
• @YemonChoi In the case that $A$ is separable, then $K(H)$ is the only example. In the general case this is called Naimark's problem and wades into set theory. Jul 18, 2017 at 18:02
• @CalebEckhardt non-mathematical comment: the asterisk symbol is used by Markdown, or whatever SE uses, to enclose text in italics. I think the parser is fine if there is only one asterisk, but as soon as there are two, well, italics ahoy Jul 18, 2017 at 18:35 | 1,231 | 4,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.861448 |
http://physicsinventions.com/speed-of-block-at-the-bottom-of-the-slope/ | 1,591,316,195,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00219.warc.gz | 93,463,607 | 9,354 | # speed of block at the bottom of the slope
1. The problem statement, all variables and given/known data
A block is placed on a slope at an angle of 38.2. The height of the inclined plane is h = 1.880m. The block has a mass of m = 7.80kg. if the coefficient of kinetic friction is =0.111 and the coefficient of static friction = 0.162.
a) what is the speed of the block at the bottom of the slope?
b) what is the magnitude of the net force on the block halfway down the slope?
2. Relevant equations
3. The attempt at a solution
http://ift.tt/1gYSFKm | 151 | 555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-24 | latest | en | 0.896996 |
https://puzzling.stackexchange.com/questions/13212/alien-classroom-calendar | 1,713,027,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00170.warc.gz | 453,899,416 | 41,550 | # Alien Classroom Calendar
A teacher on a foreign planet wants to make a classroom calendar out of cubic blocks. She wants to individually count each day by drawing a number on the faces of a block. There are 209 days in this planet's year, and she can draw any number (0-9) on each face.
For example, 19 would be:
--- ---
| 1 | | 9 |
--- ---
With a face on one block being 1, and the other block's face being 9.
Another example, 201, would be:
--- --- ---
| 2 | | 0 | | 1 |
--- --- ---
A face on one block would be 2, a second block's face being 0, and a third block with a face of 1.
What is the fewest amount of blocks she can use to count the days of the year (0-209), and what numbers are on each block?
All cubes must be used.
• Can you write on different faces of the same cube? May 2, 2015 at 8:49
• Yes. A single number may be drawn on each face. @leoll2 May 2, 2015 at 8:50
• Ah great, Ill edit my answer now! May 2, 2015 at 8:52
4 cubes
For each number, the same digit appears at most twice, except for 111, where 1 appears 3 times. Also, you can rotate the 9 to make 6, so after all we need 19 digits (2*9+3*1-2). For 19 digits the minimum is 4 cubes.
Cube #1: 012468.
Cube #2: 012568.
Cube #3: 01347.
Cube #4: 01357.
• I've edited the questions slightly. Can you give the numbers on each block's faces? May 2, 2015 at 9:03
• Done, let me know if wrong May 2, 2015 at 9:14
• You have no way to make single digit days or double digits ending with zero. May 2, 2015 at 9:17
• To make single digits, use only one cube! May 2, 2015 at 9:27
• Looks perfect! Nice job. May 2, 2015 at 10:00
Since we're making upside down 6's be 9's, how about making upside down 2's be 7's? Depends on how good an artist the teacher is, but here's an example of it on a clock.
Thus, my answer is
Three Cubes
The faces are:
123456
123480
125680
Edit: Didn't see that all cubes must be used. That makes it impossible to do with 3 cubes, unfortunately.
Double Edit: Actually, does the following configuration work? I can't find a number that I can't make...
012345
016823
014568
I feel like I'm missing something...
• +1 for a good attempt. I especially enjoyed your take on 2/7 flipping. May 2, 2015 at 9:56
• Where are the seven? May 2, 2015 at 10:00
• @leoll2 The twos are also sevens! May 2, 2015 at 10:01
• Upside 2, at most, is still a 2! May 2, 2015 at 10:01
• @leoll2 Check out the clock I linked, I think that upside-down 2 makes a pretty good seven! May 2, 2015 at 10:02 | 796 | 2,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-18 | latest | en | 0.932689 |
https://solvedlib.com/n/use-laplace-transforms-to-solve-the-initial-value-problemy,1075022 | 1,669,848,781,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710777.20/warc/CC-MAIN-20221130225142-20221201015142-00010.warc.gz | 561,187,462 | 20,601 | # Use Laplace transforms to solve the initial value problemy"+x+2y=0 X"+2r+4y =0 X(o) = y(O) =0 ~(0) = y()= =|
###### Question:
Use Laplace transforms to solve the initial value problem y"+x+2y=0 X"+2r+4y =0 X(o) = y(O) =0 ~(0) = y()= =|
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The distance between the shafts is 5m. The smaller pulley is 0.5 m in diameter. Take u = 0.2. Calculate the stress in the belt, if it is a) an ... 5 answers ##### K IZ Ko ulure le) = nz Bulek5) @EshLem USn e 25 2f Cw wup On 6,2)? Juahf (4/)i M) IFO unk Lx) ! Jall (4p5)| (z; Z ?p Gnw an Oai K IZ Ko ulure le) = nz Bulek5) @EshLem USn e 25 2f Cw wup On 6,2)? Juahf (4/)i M) IFO unk Lx) ! Jall (4p5)| (z; Z ?p Gnw an Oai... 5 answers ##### Find all solutions following tnang (Round your answers not possible enter NONE each corresponding answer blank ) 622, 9.5 yd, 9.2 yd First triangle (assume 90"):the angles B, C, B', andthe nearest whole number; Round your anshens; tOr Ine Sice; andone decimal Dlaceeither trianglSecond triangle Lassume90*): Find all solutions following tnang (Round your answers not possible enter NONE each corresponding answer blank ) 622, 9.5 yd, 9.2 yd First triangle (assume 90"): the angles B, C, B', and the nearest whole number; Round your anshens; tOr Ine Sice; and one decimal Dlace either triangl Second... 1 answer ##### All nurses on a floor determining the weekly schedule would be an example of a: Group... All nurses on a floor determining the weekly schedule would be an example of a: Group process Conflict resolution Personal commitment All of the above When establishing a group, it is important to: Determine its goals Make sure members have common roles ... 1 answer ##### Get Ready for the NCLEX E SG do to your Study Guide for addtional Review Ques- Get Ready for the NCLEX E SG do to your Study Guide for addtional Review Ques-... 5 answers ##### Oxygen Valve gas temperature of T1 theatalni H of Tz One vessels and 21 pressure The other of volume are connected by "d When the M valve 'gas 4 sopeaied with the oxygen Valve gas temperature of T1 theatalni H of Tz One vessels and 21 pressure The other of volume are connected by "d When the M valve 'gas 4 sopeaied with the... 3 answers ##### The RelcrcucosHdenncedcoqucttonQucstion1ptnickel round t0 have diameter of 0.0234 metersQucation EUsing Unit analysis; show what the diameter of the nickel in mnillimeters,Cuestion1ptQuestion /(uumbet)(unilQuostlon0.0234 mQuaatlon 10anumben(Uniu)QuaailonQuosilon 12VisitodeTbwciAnother Version10 item attempts remaining the Relcrcucos Hden ncedco quctton Qucstion 1pt nickel round t0 have diameter of 0.0234 meters Qucation E Using Unit analysis; show what the diameter of the nickel in mnillimeters, Cuestion 1pt Question / (uumbet) (unil Quostlon 0.0234 m Quaatlon 10 anumben (Uniu) Quaailon Quosilon 12 Visitode Tbwci... 1 answer ##### Describe the calculation of interest term. An as a customer, list some ways you believe you... Describe the calculation of interest term. An as a customer, list some ways you believe you can benefit from the calculation of interest.... 5 answers ##### Homework: Assignment 11 ch Score: 0 of pt 3 of5.1.10Evaluate each expression using the values given in the tableflx) gkx)2a. (f0 g)1) d. (g 0 0(3)b. (f0 9)(2) e. (g 0 g)(1)(g 0 f)(2) f. (f 0 f)(3)a. (f0 g)1) = Homework: Assignment #11 ch Score: 0 of pt 3 of 5.1.10 Evaluate each expression using the values given in the table flx) gkx) 2 a. (f0 g)1) d. (g 0 0(3) b. (f0 9)(2) e. (g 0 g)(1) (g 0 f)(2) f. (f 0 f)(3) a. (f0 g)1) =... 1 answer ##### (1 point) Let g(t) = e2t. a. Solve the initial value problem y – 2y =... (1 point) Let g(t) = e2t. a. Solve the initial value problem y – 2y = g(t), y(0) = 0, using the technique of integrating factors. (Do not use Laplace transforms.) y(t) = b. Use Laplace transforms to determine the transfer function (t) given the initial value problem$' – 20 = 8(t), $... 1 answer ##### Cost of a Hotel Stay A hotel chain charges$\$75$ each night for the first two nights and $\$ 50$for each additional night's stay. The total cost$T$is a function of the number of nights$x$that a guest stays. (a) Complete the expressions in the following piece wise defined function. $T(x)=\left\{\begin{array}{ll}{\text { if } 0 \leq x \leq 2} \\ {\text { if } x>2}\end{array}\right.$ (b) Find$T(2), T(3),$and$T(5) .$(c) What do your answers in part (b) represent? Cost of a Hotel Stay A hotel chain charges$\$75$ each night for the first two nights and $\$ 50$for each additional night's stay. The total cost$T$is a function of the number of nights$x$that a guest stays. (a) Complete the expressions in the following piece wise defined function.$$T... 5 answers ##### Fisher scoring method for binomial GLM: Suppose {Yi, Iil: Tip}"_1 is collection of data points where Im] are fixed for j =1, p and yi are the observed proportions of "success' [rom independent binomial random variables Y; with whose density function isInliMTLiVi (1 = /,)Wl A fx (yi; Ii, Ui) = Mliyi_ where Yi e {0,1/mi, 2/mi,_ 1}_ Let X be the covariale mnatrix whose i, j- entry s for and ,P. Let [Y1 . Yr]T [81 , and p [p1 , Consider GLM with linear predictor % = XB and general lin Fisher scoring method for binomial GLM: Suppose {Yi, Iil: Tip}"_1 is collection of data points where Im] are fixed for j =1, p and yi are the observed proportions of "success' [rom independent binomial random variables Y; with whose density function is Inli MTLiVi (1 = /,) Wl A fx (y... 5 answers ##### The 4G" tor tne formatin- NOIgI t5 78 9 Krnote at €0 Calial N4{g 05(9) No(q) Rcport Ile value Ihe constant rounded Inlee stonceant Iqules de €uld /o econstintaomlatths ctoblaWntIGSun Up The 4G" tor tne formatin- NOIgI t5 78 9 Krnote at €0 Calial N4{g 05(9) No(q) Rcport Ile value Ihe constant rounded Inlee stonceant Iqules de €uld /o econstintaom latths ctobla WntIG Sun Up... 5 answers ##### Suppose that the matrix B has the eigenvalues and -6, with eigenvectors Calculate 9(t).22(0)-'. (Q: fundamental matrix for u'=B.u)and[Y4]; respectively_ Suppose that the matrix B has the eigenvalues and -6, with eigenvectors Calculate 9(t).22(0)-'. (Q: fundamental matrix for u'=B.u) and [Y4]; respectively_... 1 answer ##### Question 1: A$1,010 face value bond is selling in the market place for $934. It... Question 1: A$1,010 face value bond is selling in the market place for $934. It matures in 3 years. If keep to maturity, what is the bond's yield to maturity? Question 2: A banker must earn at least a 2.7% return after expected inflation on short term loans. The inflation rate for the past 6 mo... 1 answer ##### Problem 9 An analysis of a business's financial information shows that the business's debts total$307,000...
Problem 9 An analysis of a business's financial information shows that the business's debts total $307,000 and that the owner's equity in the business is$182,000. From this information, determine the total assets of this company Problem 10 Curfew Inn is a proprietorship owned by Susan P...
##### Cobalt(II) chloride hexahydrate, dissolves readily in water to give a red solution. If we use this solution as an "ink," we can write secret messages on paper. The writing is not visible when the water evaporates from the paper. When the paper is heated, however, the message can be read. Explain the chemistry behind this observation.(IMAGE CAN'T COPY)
Cobalt(II) chloride hexahydrate, dissolves readily in water to give a red solution. If we use this solution as an "ink," we can write secret messages on paper. The writing is not visible when the water evaporates from the paper. When the paper is heated, however, the message can be read. E...
##### 4. Use the Laplace transform to solve the initial value problem y" + y = f(1)...
4. Use the Laplace transform to solve the initial value problem y" + y = f(1) = -2, ost<2 13t+4, 122 y(0) = 0, y'(0) = -1...
##### 0 Death and Dying Week 12 Discussion No matter where we are in the lifespan, loss,...
0 Death and Dying Week 12 Discussion No matter where we are in the lifespan, loss, death and dying are difficult experiences. Cultures vary as to how they process and experience these events. Share an Internet resource that describes another culture's approach to death, dying, mourning, grief or...
##### An airplane needs at least half of its engines to safely complete its mission. If each...
An airplane needs at least half of its engines to safely complete its mission. If each engine independently functions with probability p, for what values of p is a one-engine plane safer than a three-engine plane? (Hint: Plane A is safer than Plane B if the probability that Plane A can safely comple... | 4,376 | 15,002 | {"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} | 3.171875 | 3 | CC-MAIN-2022-49 | latest | en | 0.869887 |
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Question
Updated 6/17/2014 10:32:19 PM
Edited by yumdrea [6/17/2014 10:31:18 PM], Confirmed by yumdrea [6/17/2014 10:31:50 PM]
f
Original conversation
User: 4km= m
Weegy: 4km = 4000 m
User: 5000mm= m
Weegy: 5000mm = 5 m
User: 58cm= mm
Question
Updated 6/17/2014 10:32:19 PM
Edited by yumdrea [6/17/2014 10:31:18 PM], Confirmed by yumdrea [6/17/2014 10:31:50 PM]
Rating
8
58 cm = 580 mm
Questions asked by the same visitor
58cm = mm
Weegy: 58cm = 580mm User: 6.3km= m (More)
Question
Updated 9/26/2021 4:33:42 PM
6.3km= 6,300 m
6400mm = m
Question
Updated 9/26/2021 4:34:47 PM
6400mm = 6.4 m
Katie competed in a run/bicycle/run race. The first run was 2000m, the bicycle ride was 6km and the final run was1000m. How far was the race in metres and in kilometres
Weegy: 9,000 meters and 9 kilometers User: 70x12= Weegy: 84 (More)
Question
Updated 339 days ago|12/26/2022 1:46:00 PM
70x12 = 840
Added 339 days ago|12/26/2022 1:46:00 PM
0.78m= cm
Weegy: 0.78 m = 78 cm (More)
Question
Updated 2/21/2014 1:59:55 PM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,130 | 3,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-50 | latest | en | 0.92722 |
https://www.educationindex.com/essay/Physics-in-Everyday-Life-PKSREFHE4Z | 1,579,498,650,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250597458.22/warc/CC-MAIN-20200120052454-20200120080454-00504.warc.gz | 860,399,499 | 11,014 | # Physics in Everyday Life
1690 words 7 pages
Physics in Everyday Life
1
Just about everything you do from moving to eating to listening to music involves physics. Now that I have really explored them I think it is hard to go about our day and not do something that involves physics. Some of the things I will talk about are getting out of bed, the eye and how eye glasses help a person see better, speakers, the Frisbee, sailing, and the pulleys I use to get my jeep unstuck in the mud when I ride in the woods. All of these require energy, and energy is what physics is all about. There is kinetic energy, potential energy, conservation of energy, velocity, acceleration, mass, force, and gravity. Energy gives you the ability to do work. Energy is everywhere in nature
After light passes through the cornea, a portion of it passes through the pupil. g. The pupil is the black portion in the middle of the eyeball. The pupil is black looking because the light that enters the eye is absorbed on the retina and does not exit the eye. So when you look at someone no light is exiting their pupil and coming into your eye. 3 The size of the pupil opening can be adjusted by the dilation of the iris. The iris is the colored part of the eye it can be blue, brown, hazel or other colors; it is a diaphragm that is capable of stretching and reducing the size of the opening. In bright-light the iris reduces the pupil opening and limits the amount of light going into the eye. And in low-light the iris increases the amount of light that enters the eye. Light that passes through the pupil opening will enter the crystalline lens. The crystalline lens is made of layers of a fibrous material that has an index of refraction of roughly 1.40. Unlike the lens on a camera, the lens of the eye is able to change its shape and fine tune the image. The lens is attached to the ciliary muscles. These muscles relax and contract in order to change the shape of the lens. By carefully adjusting the lenses shape, the ciliary muscles assist the eye in the l task of making an image on the back of the eyeball.
The inner surface of the eye is known as the retina. The retina contains the rods and cones that serve
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1078 words | 5 pages | 702 | 3,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | longest | en | 0.948681 |
https://educatehell.com/cbse-class-11th-physics-syllabus-11-physics-new-delete-syllabus/ | 1,716,986,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00796.warc.gz | 182,599,319 | 29,773 | # CBSE Class 11th Physics Syllabus 2024 Download pdf
CBSE Class 11th Physics Syllabus 2024 Download pdf.CBSE Class 11 Physics Syllabus can be downloaded in PDF from the link mentioned below in this article. The Central Board of Secondary Education (CBSE) has revised the syllabus for the current academic session. The Physics syllabus has been prepared according to the annual scheme of assessment. Students must go through the full syllabus and study Physics as per the content mentioned in it.CBSE Class 11th Physics Syllabus Download pdf
Unit I: Physical World and Measurement 08 Periods
Chapter–2: Units and Measurements
Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. significant figures. Dimensions of physical quantities, dimensional analysis and its applications.
## Unit II: Kinematics 24 Periods
Chapter–3: Motion in a Straight Line
Frame of reference, Motion in a straight line, Elementary concepts of differentiation and integration for describing motion, uniform and non- uniform motion, and instantaneous velocity, uniformly accelerated motion, velocity – time and position-time graphs. Relations for uniformly accelerated motion (graphical treatment).
## Chapter–4: Motion in a Plane
Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration projectile motion, uniform circular motion.
## Unit III: Laws of Motion 14 Periods
Chapter–5: Laws of Motion
Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.
Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).
## Unit IV: Work, Energy and Power 14 Periods
Chapter–6: Work, Energy and Power
Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces: non- conservative forces, motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.
Unit V: Motion of System of Particles and Rigid Body 18 Periods
## Chapter–7: System of Particles and Rotational Motion
Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod. Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications. Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions. Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation).
## Unit VI: Gravitation 12 Periods
Chapter–8: Gravitation
Kepler’s laws of planetary motion, universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite.
## Unit VII: Properties of Bulk Matter 24 Periods
Chapter–9: Mechanical Properties of Solids
Elasticity, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity (qualitative idea only), Poisson’s ratio; elastic energy.
## Chapter–10: Mechanical Properties of Fluids
Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its simple applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.
## Chapter–11: Thermal Properties of Matter
Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity. Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law .
## Chapter–12: Thermodynamics
Thermal equilibrium and definition of temperature zeroth law of thermodynamics, heat, work and internal energy. First law of thermodynamics, Second law of thermodynamics: gaseous state of matter, change of condition of gaseous state -isothermal, adiabatic, reversible, irreversible, and cyclic processes.
Unit IX: Behavior of Perfect Gases and Kinetic Theory of Gases 08 Periods
## Chapter–13: Kinetic Theory
Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases – assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.
Unit X: Oscillations and Waves 26 Periods
## Chapter–14: Oscillations
Periodic motion – time period, frequency, displacement as a function of time, periodic functions and their application. Simple harmonic motion (S.H.M) and its equations of motion; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.
## Chapter–15: Waves
Wave motion: Transverse and longitudinal waves, speed of travelling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats. | 1,311 | 6,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.884722 |
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# Optimization and Minimizing Cost
A factory wants the power company to run a special line to the plant. The power company is located 1000 meters down a river on the opposite bank. Ground lines cost \$20 per meter. The river is 500 meters wide and underwater line will cost M (M> 1) times as much per meter as the above ground line. Since the factory owns the riverfront property on both sides of the river all the way to the power plant, you may choose where and how the line should cross the river.
1- Find in terms of M all paths that will minimize the cost and what the minimum cost is. Be sure to describe your paths clearly. See diagram.
2- If M =50 give the minimum cost and the best path.
3- If M =2 give the minimum cost and the best path.
4- If M =1.05 give the minimum cost and the best path.
See attached file for full problem description.
#### Solution Summary
The cost of the best path for a ine from a powerplant is determined. The solution is detailed and well presented.
\$2.19 | 239 | 1,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-44 | longest | en | 0.93205 |
https://blog.csdn.net/libin66/article/details/52049028 | 1,550,779,114,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247508363.74/warc/CC-MAIN-20190221193026-20190221215026-00531.warc.gz | 495,629,125 | 31,162 | # Codeforces Round #279 (Div. 2) F DFS 树上最长上升子序列
F. Treeland Tour
time limit per test5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The "Road Accident" band is planning an unprecedented tour around Treeland. The RA fans are looking forward to the event and making bets on how many concerts their favorite group will have.
Treeland consists of n cities, some pairs of cities are connected by bidirectional roads. Overall the country has n - 1 roads. We know that it is possible to get to any city from any other one. The cities are numbered by integers from 1 to n. For every city we know its value ri — the number of people in it.
We know that the band will travel along some path, having concerts in some cities along the path. The band's path will not pass one city twice, each time they move to the city that hasn't been previously visited. Thus, the musicians will travel along some path (without visiting any city twice) and in some (not necessarily all) cities along the way they will have concerts.
The band plans to gather all the big stadiums and concert halls during the tour, so every time they will perform in a city which population is larger than the population of the previously visited with concert city. In other words, the sequence of population in the cities where the concerts will be held is strictly increasing.
In a recent interview with the leader of the "road accident" band promised to the fans that the band will give concert in the largest possible number of cities! Thus the band will travel along some chain of cities of Treeland and have concerts in some of these cities, so that the population number will increase, and the number of concerts will be the largest possible.
The fans of Treeland are frantically trying to figure out how many concerts the group will have in Treeland. Looks like they can't manage without some help from a real programmer! Help the fans find the sought number of concerts.
Input
The first line of the input contains integer n (2 ≤ n ≤ 6000) — the number of cities in Treeland. The next line contains n integers r1, r2, ..., rn (1 ≤ ri ≤ 106), where ri is the population of the i-th city. The next n - 1 lines contain the descriptions of the roads, one road per line. Each road is defined by a pair of integers aj, bj (1 ≤ aj, bj ≤ n) — the pair of the numbers of the cities that are connected by the j-th road. All numbers in the lines are separated by spaces.
Output
Print the number of cities where the "Road Accident" band will have concerts.
Examples
input
6
1 2 3 4 5 1
1 2
2 3
3 4
3 5
3 6
output
4
input
5
1 2 3 4 5
1 2
1 3
2 4
3 5
output
3
n个城市，n-1条边，任意两个城市都有路径到达。现在一个乐队开演唱会，乐队沿着一条路途径城市。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
struct edge{
int v,next;
}e[120010];
int a[6060];
e[tot].v=v;
}
int ans=0;
int dp[6001];
int anw[6001],cnt=0;
void DFS(int u,int fa){
int v=e[i].v;
if(v==fa) continue;
int f=0,tmp=-1;
if(a[v]>anw[cnt]){
anw[++cnt]=a[v];
dp[v]=cnt;
f=-1;
} else {
int x=lower_bound(anw+1,anw+cnt+1,a[v])-anw;
tmp=anw[x];
anw[x]=a[v];
dp[v]=x;
f=x;
}
DFS(v,u);
if(f==-1) cnt--;
else if(f>0){
anw[f]=tmp;
}
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<n;i++){
int v,u;
scanf("%d%d",&u,&v);
}
for(int i=1;i<=n;i++) {
mst(dp,0);
cnt=1;
anw[1]=a[i];
DFS(i,0);
for(int j=1;j<=n;j++) ans=max(ans,dp[j]);
}
printf("%d\n",ans);
return 0;
} | 1,105 | 3,777 | {"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} | 2.765625 | 3 | CC-MAIN-2019-09 | latest | en | 0.952333 |
https://www.mathhomeworkanswers.org/282216/total-tickets-total-%242740-adult-tickets-child-tickets-ticket?show=282219 | 1,618,296,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00041.warc.gz | 988,303,998 | 17,804 | A total of 900 tickets were sold for a game for a total of \$2740. If adult tickets sold for \$5 each, and child tickets cost \$1, how many of each kind of ticket sold?
## Your answer
Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.
## 1 Answer
If A and C are the numbers of adult and child tickets, A+C=900. So C=900-A
Total cost is 5A+C=2740, so C=2740-5A.
Therefore 900-A=2740-5A, 4A=2740-900=1840, A=1840/4=460.
C=900-460=440.
So 460 adult tickets and 440 child tickets were sold.
by Top Rated User (839k points)
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
2 answers
1 answer
1 answer
1 answer | 242 | 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} | 3.734375 | 4 | CC-MAIN-2021-17 | latest | en | 0.9018 |
http://www.csmonitor.com/Business/The-Simple-Dollar/2011/1111/How-much-home-brewed-coffee-really-saves | 1,493,550,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125074.20/warc/CC-MAIN-20170423031205-00555-ip-10-145-167-34.ec2.internal.warc.gz | 486,713,263 | 17,673 | # How much home-brewed coffee really saves
## If you drink coffee more than a time or two a week, you’re going to save at least a dollar per cup by making it at home, probably more.
View Caption
of
Saving Pennies or Dollars is a new semi-regular series on The Simple Dollar, inspired by a great discussion on The Simple Dollar’s Facebook page concerning frugal tactics that might not really save that much money. I’m going to take some of the scenarios described by the readers there and try to break down the numbers to see if the savings is really worth the time invested.
Jeff writes in: How much money do you ACTUALLY save in time and money by making your own coffee at home? I would appreciate the assumption that the user is using non-generic, non-Folgers or Maxwell House coffee.
For some standardized data on this, I visited the website of the SCAA – the Specialty Coffee Association of America. In one of their protocol documents, I found this information, which I’ll use to analyze how a great cup of coffee is constructed at home:
The optimum ratio is 8.25 grams of coffee per 150 ml of water, as this conforms to the mid-point of the optimum balance recipes for the Golden Cup.
Let’s say we’re looking at a 16 ounce cup of coffee – what you might fill a to-go cup with from a coffee shop, for example. A 16 ounce cup of coffee is approximately 473 mL, which, using the ratio above, would require 26 grams of coffee to make it yourself. An ounce is 28.3 grams, just for measurement’s sake.
So, how much does “good” coffee cost? I asked my wife to select what she considered to be a very good coffee for the price and she chose Eight O’Clock Coffee’s original ground, which can be obtained at a rate of \$0.39 per ounce.
Simply put, you’d need about \$0.38 of decent ground coffee to make a good 16 ounce cup of coffee at home. There’s also the negligible cost of water and electricity (say, one cent per cup), plus the ongoing cost of filters (say, two cents per cup), plus the cost of the cup (say, one cent per drink prorated out over time), plus the startup cost of purchasing an inexpensive pot to brew the coffee with (say, another two cents per cup, prorated out over time). That’s a cost of about \$0.44 for a 16 ounce standard coffee.
Now, if you add cream or other ingredients to that, you’re increasing the cost, but not significantly. For example, International Delight French Vanilla liquid creamer costs \$0.08 per cup. Other options might ding you as much as a quarter per cup for flavoring, which is still leaving you below \$0.70 per cup.
Depending on what exactly you order at your typical coffee chain, a 16 ounce coffee will set you back somewhere between \$2 and \$5. The variation here is pretty impressive, but even if you’re comparing the low end of a purchased coffee with the high end of a homemade cup, you’re still talking about a savings of a dollar per 16 ounce cup. It’s quite likely you’re saving even more than that.
What about the time? I’m not a coffee drinker, but Sarah usually sets up the coffee pot the night before. It takes her about a minute. When she gets up, she flips a switch, and then she drinks a cup a little while later, then fills her to-go cup on her way out the door, taking her maybe another minute or two. She usually cleans the pot up when she gets home from work, taking another couple of minutes.
The time invested is perhaps five minutes total per day, and she’s probably saving \$2.50 or so per day, making for a pretty good hourly rate. Plus, she believes the coffee made at home tastes better.
If you drink coffee more than a time or two a week, you’re going to save money making it at home, and it’s probably going to be well worth the small amount of time invested, too.
The Christian Science Monitor has assembled a diverse group of the best economy-related bloggers out there. Our guest bloggers are not employed or directed by the Monitor and the views expressed are the bloggers' own, as is responsibility for the content of their blogs. To contact us about a blogger, click here. To add or view a comment on a guest blog, please go to the blogger's own site by clicking on www.thesimpledollar.com.
We want to hear, did we miss an angle we should have covered? Should we come back to this topic? Or just give us a rating for this story. We want to hear from you. | 989 | 4,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.951784 |
https://wiki.microduinoinc.com/index.php?title=Lesson_1--Microduino_resistance_Meter&oldid=3040 | 1,585,559,119,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00368.warc.gz | 783,841,168 | 7,587 | # Lesson 1--Microduino resistance Meter
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Language: English • 中文
## Objective
This lesson will teach you how to use Microduin to make a resistance meter. The basic principle is that use the reference resistor, compareing the voltage, then calculate the resistance value. Through testing, using 20k as the base, if the resistance value more than 1k, the result more accurate.
## Equipment
• Microduino-Core
• Microduino-FT232R
• Other hardware equipment
• 20kΩ resistor one
• 104 ceramic capacitor one
• USB Data cable one
Capacitor
brief introduction：
Ceramics is qualitative, 104 is three digits notation, the former two significant figures, the next one is the number of 0. 104 means it is 100000PF, equal to 0.1 uF. Commonly used as bypass capacitor in high frequency or low frequency circuit, and commonly used in power supply decoupling, also has a coupling effect.
## Schematic
You can change the reference resistor according the measure scope, in order to get a good accurate.
Circuit Connection:(Using 104 ceramic capacitor)：
## Program
``` #define N 12
#define basis 20000.0 //Reference resistor
int potpin = A4;
float val, r;
void setup()
{
Serial.begin(9600);
}
void loop()
{
r = ((basis * 1023.0) / (1023.0 - val)) - basis;
delay(100);
Serial.print(" R: ");
if(filter() >= 1000)
{
if(filter() >= 500.0*1000.0)
{
Serial.println("Infinity!!");
}
else
{
Serial.print(filter() / 1000.0);
Serial.println("K ohm");
}
}
else
{
Serial.print(filter());
Serial.println(" ohm");
}
}
//Filtering algorithm, use the average filtering algorithm
float filter()
{
float sum = 0;
for(int count = 0; count < N; count++)
{
sum += r;
delay(5);
}
return (float)(sum / N);
}```
## Debug
Step 1：Copy the code to IDE, and compile it
Step 2：Connect the circuit, as follows：
In figure, the upper resistor is 20k reference resistor, and the low registor is the measured one.
Step 3：Run program
Step 4：As figure shows, put the yellow line and black line to both end of the resistance, serial will show the resistance value.
You also can change to other object to test the resistance, such as soap, a damp cloth and so on.
## Result
Serial will show the resistance value. | 594 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-16 | latest | en | 0.696918 |
http://samiam.org/mix-column.html | 1,516,361,674,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887973.50/warc/CC-MAIN-20180119105358-20180119125358-00288.warc.gz | 301,513,003 | 3,225 | Sam Trenholme's webpage
Rijndael's mix column stage
This document describes how Rijndael's mix column stage works. Mix-column, along with shift row, is how Rijndael performs diffusion. The S-Box is responsible for the confusion aspect of the cipher. These three stages act together to make Rijndael secure.
The mix column stage acts by taking a single column of four of Rijndael's sixteen values, and performing Matrix multiplication in Rijndael's Galois field to make it so each byte in the input affects all four bytes of the output.
The matrix is as follows:
```2 3 1 1
1 2 3 1
1 1 2 3
3 1 1 2
```
For those that have mercifully forgotten your linear algebra class, the way to multiply two matrices together is described on other web pages.
In the case of Rijndael, all operations are done in Rijndael's Galois field. Because of this, addition is an exclusive or operation, and multiplication is a complex operation (as described on the linked page).
Given the gmul code given in the Rijndael Galois field document, and given that the ^ represents the exclusive or operation in C, we can perform the mix column thusly:
```void gmix_column(unsigned char *r) {
unsigned char a[4];
unsigned char c;
for(c=0;c<4;c++) {
a[c] = r[c];
}
r[0] = gmul(a[0],2) ^ gmul(a[3],1) ^ gmul(a[2],1) ^ gmul(a[1],3);
r[1] = gmul(a[1],2) ^ gmul(a[0],1) ^ gmul(a[3],1) ^ gmul(a[2],3);
r[2] = gmul(a[2],2) ^ gmul(a[1],1) ^ gmul(a[0],1) ^ gmul(a[3],3);
r[3] = gmul(a[3],2) ^ gmul(a[2],1) ^ gmul(a[1],1) ^ gmul(a[0],3);
}
```
Since gmul(n,1) always results in n, and gmul(n,3) is gmul(n,2) ^ n, this can be optimized to speed up performance:
```void gmix_column(unsigned char *r) {
unsigned char a[4];
unsigned char b[4];
unsigned char c;
for(c=0;c<4;c++) {
a[c] = r[c];
b[c] = gmul(r[c],2);
}
r[0] = b[0] ^ a[3] ^ a[2] ^ b[1] ^ a[1];
r[1] = b[1] ^ a[0] ^ a[3] ^ b[2] ^ a[2];
r[2] = b[2] ^ a[1] ^ a[0] ^ b[3] ^ a[3];
r[3] = b[3] ^ a[2] ^ a[1] ^ b[0] ^ a[0];
}
```
In fact, gmul(n,2) is simply a single shift and a conditional exclusive or (see this page for details), so we can calculate the mix column operation thusly:
```void gmix_column(unsigned char *r) {
unsigned char a[4];
unsigned char b[4];
unsigned char c;
unsigned char h;
for(c=0;c<4;c++) {
a[c] = r[c];
h = r[c] & 0x80; /* hi bit */
b[c] = r[c] << 1;
if(h == 0x80)
b[c] ^= 0x1b; /* Rijndael's Galois field */
}
r[0] = b[0] ^ a[3] ^ a[2] ^ b[1] ^ a[1];
r[1] = b[1] ^ a[0] ^ a[3] ^ b[2] ^ a[2];
r[2] = b[2] ^ a[1] ^ a[0] ^ b[3] ^ a[3];
r[3] = b[3] ^ a[2] ^ a[1] ^ b[0] ^ a[0];
}
```
Here are some test vectors:
Hexadecimal Decimal Before After Before After db 13 53 45 8e 4d a1 bc 219 19 83 69 142 77 161 188 f2 0a 22 5c 9f dc 58 9d 242 10 34 92 159 220 88 157 01 01 01 01 01 01 01 01 1 1 1 1 1 1 1 1 c6 c6 c6 c6 c6 c6 c6 c6 198 198 198 198 198 198 198 198 d4 d4 d4 d5 d5 d5 d7 d6 212 212 212 213 213 213 215 214 2d 26 31 4c 4d 7e bd f8 45 38 49 76 77 126 189 248
The inverse mix column operation is done by multiplying by a different matrix, as seen in the following C code:
```void inv_mix_column(unsigned char *r) {
unsigned char a[4];
unsigned char c;
for(c=0;c<4;c++) {
a[c] = r[c];
}
r[0] = gmul(a[0],14) ^ gmul(a[3],9) ^ gmul(a[2],13) ^ gmul(a[1],11);
r[1] = gmul(a[1],14) ^ gmul(a[0],9) ^ gmul(a[3],13) ^ gmul(a[2],11);
r[2] = gmul(a[2],14) ^ gmul(a[1],9) ^ gmul(a[0],13) ^ gmul(a[3],11);
r[3] = gmul(a[3],14) ^ gmul(a[2],9) ^ gmul(a[1],13) ^ gmul(a[0],11);
}
``` | 1,391 | 3,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-05 | latest | en | 0.764531 |
https://www.geeksforgeeks.org/hsbc-campus-placement-for-trainee-software-engineer-batch-2020/ | 1,582,971,503,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00111.warc.gz | 749,351,757 | 24,620 | # HSBC Campus Placement for Trainee Software Engineer Batch 2020
Round 1:
The first round consisted of the written exam conducted on Cocubes Platform and the test duration was 75 minutes. The test was of moderate difficulty level consisting of three sections. The first section was Verbal ability and English. There were questions based on vocabulary, comprehension and grammar. The second section was Analytical Reasoning. The questions were of average level. It consisted of questions from pattern matching, linear arrangements, coding-decoding. The last section was technical which was of difficult level. The questions were based on output of the program, OOPS concepts, React JS, Data Structures & Algorithms. There was a cutoff set for clearing the first round. If a student clears the first round, he/she will immediately proceed for the next coding round.
The next round was coding round. There were 2 coding problems. One was of easy level and the other one was of moderate level. If you solve one question also, you will be shortlisted for the next round.
1. Given: A=1, B=10, C=100, D=1000, E=10000, F=100000, G=1000000. The input is a string, and the output should be sum of the sequence. For eg: BC=10+100=110.
2. The question was based on 2D sorting.
Round 2:
Technical interview round which lasted for about 30 minutes.
Q1. He asked me if I was nervous or not?
Q3. Write a sql query to find the details of a car having more than 15 years of registration. (First he told me to create a table car.)
Q4. Write a sql query to find the second highest salary from employees table.
Q5. What is hash map?
Q6. What is array?
Q7. Questions based on Projects mentioned in resume.
Q8. 2 eggs and 100 floors Puzzle problem.
https://www.geeksforgeeks.org/puzzle-set-35-2-eggs-and-100-floors/
Q9. Difference between sql and mysql.
Q10. Properties of linked list and insertion of elements in a linked list.
Q11. He then asked me to rate myself in programming languages C/C++/Java/Python. I told him that i am confident in C and C++ and i know the basics of Java and Python. He asked me the differences in all the four languages.
Q12. Why Java is platform independent?
Q13. An array has numbers from 1 to 10 only and one of the number is missing in between. Find the missing number. He asked me to write a code.
Note: Just be confident to answer the question and interact properly with the interviewer. Go through your resume and don’t mention anything you don’t know.
Round 3:
HR interview
Q2. Why did you choose engineering?
Q3. Why do you want to join HSBC?
Q4. Family background.
Q5. Strengths and weaknesses.
Q6. Where do you see yourself in next 5 years?
Q7. Are you a team player? Give examples
Q8. What activities have you done in college apart from studies?
Tips: Speak confidently and maintain eye contact with the interviewer.
Write your Interview Experience or mail it to contribute@geeksforgeeks.org
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 820 | 3,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-10 | longest | en | 0.958998 |
https://en.m.wikipedia.org/wiki/Isometry | 1,709,549,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00457.warc.gz | 230,060,571 | 34,675 | # Isometry
In mathematics, an isometry (or congruence, or congruent transformation) is a distance-preserving transformation between metric spaces, usually assumed to be bijective.[a] The word isometry is derived from the Ancient Greek: ἴσος isos meaning "equal", and μέτρον metron meaning "measure". If the transformation is from a metric space to itself, it is a kind of geometric transformation known as a motion.
## Introduction
Given a metric space (loosely, a set and a scheme for assigning distances between elements of the set), an isometry is a transformation which maps elements to the same or another metric space such that the distance between the image elements in the new metric space is equal to the distance between the elements in the original metric space. In a two-dimensional or three-dimensional Euclidean space, two geometric figures are congruent if they are related by an isometry;[b] the isometry that relates them is either a rigid motion (translation or rotation), or a composition of a rigid motion and a reflection.
Isometries are often used in constructions where one space is embedded in another space. For instance, the completion of a metric space ${\displaystyle \ M\ }$ involves an isometry from ${\displaystyle \ M\ }$ into ${\displaystyle \ M'\ ,}$ a quotient set of the space of Cauchy sequences on ${\displaystyle \ M\ .}$ The original space ${\displaystyle \ M\ }$ is thus isometrically isomorphic to a subspace of a complete metric space, and it is usually identified with this subspace. Other embedding constructions show that every metric space is isometrically isomorphic to a closed subset of some normed vector space and that every complete metric space is isometrically isomorphic to a closed subset of some Banach space.
An isometric surjective linear operator on a Hilbert space is called a unitary operator.
## Definition
Let ${\displaystyle \ X\ }$ and ${\displaystyle \ Y\ }$ be metric spaces with metrics (e.g., distances) ${\textstyle d_{X}}$ and ${\textstyle d_{Y}.}$ A map ${\textstyle f\colon X\to Y}$ is called an isometry or distance preserving map if for any ${\displaystyle \ a,b\in X\ }$ one has
${\displaystyle d_{X}(a,b)=d_{Y}\!\left(f(a),f(b)\right).}$ [4][c]
An isometry is automatically injective;[a] otherwise two distinct points, a and b, could be mapped to the same point, thereby contradicting the coincidence axiom of the metric d, i.e., ${\displaystyle d(a,b)=0}$ if and only if ${\displaystyle a=b}$ . This proof is similar to the proof that an order embedding between partially ordered sets is injective. Clearly, every isometry between metric spaces is a topological embedding.
A global isometry, isometric isomorphism or congruence mapping is a bijective isometry. Like any other bijection, a global isometry has a function inverse. The inverse of a global isometry is also a global isometry.
Two metric spaces X and Y are called isometric if there is a bijective isometry from X to Y. The set of bijective isometries from a metric space to itself forms a group with respect to function composition, called the isometry group.
There is also the weaker notion of path isometry or arcwise isometry:
A path isometry or arcwise isometry is a map which preserves the lengths of curves; such a map is not necessarily an isometry in the distance preserving sense, and it need not necessarily be bijective, or even injective. This term is often abridged to simply isometry, so one should take care to determine from context which type is intended.
Examples
## Isometries between normed spaces
The following theorem is due to Mazur and Ulam.
Definition:[5] The midpoint of two elements x and y in a vector space is the vector 1/2(x + y).
Theorem[5][6] — Let A : XY be a surjective isometry between normed spaces that maps 0 to 0 (Stefan Banach called such maps rotations) where note that A is not assumed to be a linear isometry. Then A maps midpoints to midpoints and is linear as a map over the real numbers ${\displaystyle \mathbb {R} }$ . If X and Y are complex vector spaces then A may fail to be linear as a map over ${\displaystyle \mathbb {C} }$ .
### Linear isometry
Given two normed vector spaces ${\displaystyle V}$ and ${\displaystyle W,}$ a linear isometry is a linear map ${\displaystyle A:V\to W}$ that preserves the norms:
${\displaystyle \|Av\|=\|v\|}$
for all ${\displaystyle \ v\in V\ .}$ [7] Linear isometries are distance-preserving maps in the above sense. They are global isometries if and only if they are surjective.
In an inner product space, the above definition reduces to
${\displaystyle \langle v,v\rangle =\langle Av,Av\rangle }$
for all ${\displaystyle v\in V\ ,}$ which is equivalent to saying that ${\displaystyle \ A^{\dagger }A=\operatorname {I} _{V}\ .}$ This also implies that isometries preserve inner products, as
${\displaystyle \langle Au,Av\rangle =\langle u,A^{\dagger }Av\rangle =\langle u,v\rangle \ .}$
Linear isometries are not always unitary operators, though, as those require additionally that ${\displaystyle V=W}$ and ${\displaystyle AA^{\dagger }=\operatorname {I} _{V}\ .}$
By the Mazur–Ulam theorem, any isometry of normed vector spaces over ${\displaystyle \mathbb {R} }$ is affine.
A linear isometry also necessarily preserves angles, therefore a linear isometry transformation is a conformal linear transformation.
Examples
## Manifold
An isometry of a manifold is any (smooth) mapping of that manifold into itself, or into another manifold that preserves the notion of distance between points. The definition of an isometry requires the notion of a metric on the manifold; a manifold with a (positive-definite) metric is a Riemannian manifold, one with an indefinite metric is a pseudo-Riemannian manifold. Thus, isometries are studied in Riemannian geometry.
A local isometry from one (pseudo-)Riemannian manifold to another is a map which pulls back the metric tensor on the second manifold to the metric tensor on the first. When such a map is also a diffeomorphism, such a map is called an isometry (or isometric isomorphism), and provides a notion of isomorphism ("sameness") in the category Rm of Riemannian manifolds.
### Definition
Let ${\displaystyle \ R=(M,g)\ }$ and ${\displaystyle \ R'=(M',g')\ }$ be two (pseudo-)Riemannian manifolds, and let ${\displaystyle \ f:R\to R'\ }$ be a diffeomorphism. Then ${\displaystyle \ f\ }$ is called an isometry (or isometric isomorphism) if
${\displaystyle \ g=f^{*}g',\ }$
where ${\displaystyle \ f^{*}g'\ }$ denotes the pullback of the rank (0, 2) metric tensor ${\displaystyle \ g'\ }$ by ${\displaystyle \ f\ .}$ Equivalently, in terms of the pushforward ${\displaystyle \ f_{*}\ ,}$ we have that for any two vector fields ${\displaystyle \ v,w\ }$ on ${\displaystyle \ M\ }$ (i.e. sections of the tangent bundle ${\displaystyle \ \mathrm {T} M\ }$ ),
${\displaystyle \ g(v,w)=g'\left(f_{*}v,f_{*}w\right)\ .}$
If ${\displaystyle \ f\ }$ is a local diffeomorphism such that ${\displaystyle \ g=f^{*}g'\ ,}$ then ${\displaystyle f}$ is called a local isometry.
### Properties
A collection of isometries typically form a group, the isometry group. When the group is a continuous group, the infinitesimal generators of the group are the Killing vector fields.
The Myers–Steenrod theorem states that every isometry between two connected Riemannian manifolds is smooth (differentiable). A second form of this theorem states that the isometry group of a Riemannian manifold is a Lie group.
Riemannian manifolds that have isometries defined at every point are called symmetric spaces.
## Generalizations
• Given a positive real number ε, an ε-isometry or almost isometry (also called a Hausdorff approximation) is a map ${\displaystyle \ f\colon X\to Y\ }$ between metric spaces such that
1. for ${\displaystyle x,x'\in X}$ one has ${\displaystyle \ |d_{Y}(f(x),f(x'))-d_{X}(x,x')|<\varepsilon \ ,}$ and
2. for any point ${\displaystyle y\in Y}$ there exists a point ${\displaystyle \ x\in X}$ with ${\displaystyle d_{Y}(y,f(x))<\varepsilon \ }$
That is, an ε-isometry preserves distances to within ε and leaves no element of the codomain further than ε away from the image of an element of the domain. Note that ε-isometries are not assumed to be continuous.
• The restricted isometry property characterizes nearly isometric matrices for sparse vectors.
• Quasi-isometry is yet another useful generalization.
• One may also define an element in an abstract unital C*-algebra to be an isometry:
${\displaystyle \ a\in {\mathfrak {A}}\ }$ is an isometry if and only if ${\displaystyle \ a^{*}\cdot a=1\ .}$
Note that as mentioned in the introduction this is not necessarily a unitary element because one does not in general have that left inverse is a right inverse.
## Footnotes
1. ^ a b "We shall find it convenient to use the word transformation in the special sense of a one-to-one correspondence ${\displaystyle \ P\to P'\ }$ among all points in the plane (or in space), that is, a rule for associating pairs of points, with the understanding that each pair has a first member P and a second member P' and that every point occurs as the first member of just one pair and also as the second member of just one pair...
In particular, an isometry (or "congruent transformation," or "congruence") is a transformation which preserves length ..." — Coxeter (1969) p. 29[2]
2. ^
3.11 Any two congruent triangles are related by a unique isometry.— Coxeter (1969) p. 39[3]
3. ^
Let T be a transformation (possibly many-valued) of ${\displaystyle E^{n}}$ (${\displaystyle 2\leq n<\infty }$ ) into itself.
Let ${\displaystyle d(p,q)}$ be the distance between points p and q of ${\displaystyle E^{n}}$ , and let Tp, Tq be any images of p and q, respectively.
If there is a length a > 0 such that ${\displaystyle d(Tp,Tq)=a}$ whenever ${\displaystyle d(p,q)=a}$ , then T is a Euclidean transformation of ${\displaystyle E^{n}}$ onto itself.[4]
## References
1. ^ Coxeter 1969, p. 46
3.51 Any direct isometry is either a translation or a rotation. Any opposite isometry is either a reflection or a glide reflection.
2. ^ Coxeter 1969, p. 29
3. ^ Coxeter 1969, p. 39
4. ^ a b Beckman, F.S.; Quarles, D.A. Jr. (1953). "On isometries of Euclidean spaces" (PDF). Proceedings of the American Mathematical Society. 4 (5): 810–815. doi:10.2307/2032415. JSTOR 2032415. MR 0058193.
5. ^ a b Narici & Beckenstein 2011, pp. 275–339.
6. ^ Wilansky 2013, pp. 21–26.
7. ^ Thomsen, Jesper Funch (2017). Lineær algebra [Linear Algebra]. Department of Mathematics (in Danish). Århus: Aarhus University. p. 125.
8. ^ Roweis, S.T.; Saul, L.K. (2000). "Nonlinear dimensionality reduction by locally linear embedding". Science. 290 (5500): 2323–2326. CiteSeerX 10.1.1.111.3313. doi:10.1126/science.290.5500.2323. PMID 11125150.
9. ^ Saul, Lawrence K.; Roweis, Sam T. (June 2003). "Think globally, fit locally: Unsupervised learning of nonlinear manifolds". Journal of Machine Learning Research. 4 (June): 119–155. Quadratic optimisation of ${\displaystyle \ \mathbf {M} =(I-W)^{\top }(I-W)\ }$ (page 135) such that ${\displaystyle \ \mathbf {M} \equiv YY^{\top }\ }$
10. ^ Zhang, Zhenyue; Zha, Hongyuan (2004). "Principal manifolds and nonlinear dimension reduction via local tangent space alignment". SIAM Journal on Scientific Computing. 26 (1): 313–338. CiteSeerX 10.1.1.211.9957. doi:10.1137/s1064827502419154.
11. ^ Zhang, Zhenyue; Wang, Jing (2006). "MLLE: Modified locally linear embedding using multiple weights". In Schölkopf, B.; Platt, J.; Hoffman, T. (eds.). Advances in Neural Information Processing Systems. NIPS 2006. NeurIPS Proceedings. Vol. 19. pp. 1593–1600. ISBN 9781622760381. It can retrieve the ideal embedding if MLLE is applied on data points sampled from an isometric manifold. | 3,252 | 11,842 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 65, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-10 | latest | en | 0.895095 |
https://www.coursehero.com/file/5876886/Psyc-315-Winter-2010-Class-11-Review-/ | 1,516,797,131,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00514.warc.gz | 852,738,472 | 97,189 | Psyc_315_-_Winter_2010_-_Class_11_Review_
# Psyc_315_-_Winter_2010_-_Class_11_Review_ - Z-Scores Z...
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1 Z-Scores • Z score is a number that is commonly used to convey the relative position relative position of a data value in the data set. • The z-score score is obtained by subtracting the mean of the data set from the value and dividing the result by the standard deviation of the data set. • The Z score designates how many standard deviations the corresponding raw score is above or below the mean. X M Z SD - = 2 Z-Scores • A Z score tells you two things about a score: 1.Its algebraic sign indicates whether the score is above or below the mean 2.Its absolute value tells you by how much. Z scores provide a helpful way to compare scores that are on completely different scales or from different distributions, because the mean of Z scores are always 0, and the standard deviation 1 . 3 Example 1 What is the z -score for the value of 14 in the following score for the value of 14 in the following sample values? X = [3, 8, 6, 14, 4, 12, 7, 10] Mx Mx = 8 = 8 SDx = 3.57 • Thus, the data value of 14 is 1.68 standard deviations above the mean of 8, since the z-score is positive. 14 8 1.68 3.57 X M Z SD - - = = =
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4 Example 2 • What is the z What is the z -score for the value of 95 in the score for the value of 95 in the following values? following values? X = [96, 114, 100, 97, 101, 102, 99, 95, 90] ( ) 2 X X M X M M SD Z N N SD - - = = = 5 Example 2 Answers X = [ 96 114 100 97 101 102 99 95 90 96 114 100 97 101 102 99 95 90], so the mean of X, M = 99.33 (M = X/ N = 894/9). (X - M) = [-3.33, 14.67, 0.67, -2.33, 1.67, 2.67, -0.33, -4.33, -9.33] • Deviation score, subtract M from each number in X. (X – M) 2 = [11.09, 215.21, 0.45, 5.43, 2.79, 7.13, 0.11, 18.75, 87.05] • Squared deviations from the mean, M. SS= Σ (X - M ) 2 = 348.01 • Sum of squared deviations from the mean (SS). SD 2 = SS / N = 38.67 • Average squared deviation from the mean. SD= SD 2 = 6.22 • Square root of variance. 6 Example 2 Answers • First compute the mean and standard deviation. First compute the mean and standard deviation. These values are respectively 99.33 & 6.22. These values are respectively 99.33 & 6.22. • Z-Score Score = (95 = (95 –99.33)/6.22 99.33)/6.22 = -0.70 0.70 . • Thus, the data value of 95 is located 0.70 Thus, the data value of 95 is located 0.70 standard deviation standard deviation below below the mean value of the mean value of 99.33 (since the z 99.33 (since the z -score is negative). score is negative).
7 Example 3 M = 23; SD 2 = 1.56. Calculate the raw scores for the following z-scores
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Ask a homework question - tutors are online | 1,011 | 3,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-05 | latest | en | 0.829485 |
https://teamtreehouse.com/community/i-changed-my-js-mash-script-and-need-help-fixing-it | 1,718,484,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00001.warc.gz | 522,245,151 | 38,083 | ## Welcome to the Treehouse Community
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# I changed my JS MASH script and need help fixing it.
I changed the JS script in MASH like Joy told me to, and now I can't get it to return a result. As I'm a newbie, I'm not entirely sure what I've done wrong. I'm pretty certain it's a few labels that don't match or a few semi colons I left out. Not really sure what to look for yet. Can someone connect with me and help?
Please post your code using the 'Markdown Cheatsheet' in this forum and maybe we can help!
Here's the js:
function random_number(num) { // New function called random_choice that takes one parameter, num (or a number) // Get a random number between 0 and a passed-in number var num = num || 5; // If no number passed in, default to 5 return Math.floor(Math.random() * num); // Round the answer down (floor) of a random number between 0 and 1 and multiply it by a number. Then return a value and exit the function. }
function mash_choice() { // New function called mash_choice that doesn't take any parameters // Since MASH is a special case, give it its own list var mash = ['being content', 'finding peace', 'understanding love', 'belonging somewhere']; // The array of choices to pick from var randomNum = random_number(4); // Use the above function to get a number between 0 and 4 return mash[randomNum]; // Return the list item the random number function just picked and exit the function }
function get_answer(category) { // Get a random answer from the available answers in a given category var choices = []; // A blank array to hold the user provided answer
var selector = 'input[name="' + category + '[]"]'; // Build a CSS selector for the blanks in our passed in category var inputs = document.querySelectorAll(selector); // Get all of the inputs that match our selector var answer;
```for (var i = 0; i < inputs.length; i++) { // Begin a for loop that will run through the code. i++ = add one to the counter which is "i"
answer = inputs[i].value; // Get the input with the index value of the counter and get the value ie. if they typed in dog, you get back "dog"
if (answer !== '') { // If answer doesn't equal a blank... !== means doesn't equal
}
}
return choices[random_number(choices.length)]; // Pick and return a random choice choice.length = number of answers the user provided in that category
```
}
function fill_in_answers(answers) { // Find the spans that need filled var feelings = document.querySelector('#feelings'); // This says make a new variable and find the HTML tag that has the ID of "home" var purpose = document.querySelector('#purpose'); var motivation = document.querySelector('#motivation'); var location = document.querySelector('#location');
```// Fill them with the provided answers
home.innerHTML = answers.mash; // Change the content of the element in the HTML doc with the id "home" to the "mash" value in answers
profession.innerHTML = answers.profession; // Change the content of the element in the HTML doc with the id "career" to the "career" value in answers
```
}
```// Build up our answers object
'mash': mash_choice(),
}
```
}
// Find the form on the page and attach a handler for when it's submitted var form = document.querySelector('#mash');
form.addEventListener('submit', handle_submission); // Anytime the form is submitted, we want to call the function handle_submission
The CSS is fine. Here's the HTML:
<!DOCTYPE html> <html> <head> <meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, minimum-scale=0.5, maximum-scale=0.5, minimal-ui">
<h1 class="logo"><img src="peaceful.jpg" /></h1> <p class="description">Fill in the blanks and your life will be full.</p>
<form action="" method="post" id="mash">
<div id="answers" class="hide"> <p>Your life is a series of <span id="feelings"></span> events which remind you that <span id="motivation"></span> values you as a person <span id="purpose"></span> and is invested in your hapiness as you get closer to your goal of <span id="home"></span>. </div>
<div class="bucket">
```<div class="choice-bucket">
<h4 class="highlight">Why do you get up each day?</h4>
<input name="feelings[]" type="text">
<input name="feelings[]" type="text">
<input name="feelings[]" type="text">
<input name="feelings[]" type="text">
</div>
<div class="choice-bucket">
<h4 class="highlight">For whom to do live?</h4>
<input name="purpose[]" type="text">
<input name="purpose[]" type="text">
<input name="purpose[]" type="text">
<input name="purpose[]" type="text">
</div>
<div class="choice-bucket">
<h4 class="highlight">What motivates you to keep going?</h4>
<input name="motivation[]" type="text">
<input name="motivation[]" type="text">
<input name="motivation[]" type="text">
<input name="motivation[]" type="text">
</div>
```
</div>
<input type="submit" value="Tell my fortune"> </form> <script src="script.js"></script> </body> </html>
The issue isn't the look. I can't get the JS to execute. Let me know what I've done wrong. I'm lost being new to all of this | 1,336 | 5,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.866701 |
https://www.statistics-lab.com/%E6%95%B0%E5%AD%A6%E4%BB%A3%E5%86%99%E8%AE%A1%E7%AE%97%E5%A4%8D%E6%9D%82%E5%BA%A6%E7%90%86%E8%AE%BA%E4%BB%A3%E5%86%99computational-complexity-theory%E4%BB%A3%E8%80%83comp90038/ | 1,716,503,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00675.warc.gz | 871,535,197 | 40,006 | ### 数学代写|计算复杂度理论代写Computational complexity theory代考|COMP90038
statistics-lab™ 为您的留学生涯保驾护航 在代写计算复杂度理论Computational complexity theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算复杂度理论Computational complexity theory代写方面经验极为丰富，各种代写计算复杂度理论Computational complexity theory相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 数学代写|计算复杂度理论代写Computational complexity theory代考|Incomplete Problems in NP
We have seen many $N P$-complete problems in Chapter 2. Many natural problems in $N P$ turn out to be $N P$-complete. There are, however, a few interesting problems in $N P$ that are not likely to be solvable in deterministic polynomial time but also are not known to be $N P$-complete. The study of these problems is thus particularly interesting, because it not only can classify the inherent complexity of the problems themselves but can also provide a glimpse of the internal structure of the class $N P$. We start with some examples.
Example 4.1 GRAPH ISOMORPHISM (GIso): Given two graphs $G_{1}=$ $\left(V_{1}, E_{1}\right)$ and $G_{2}=\left(V_{2}, E_{2}\right)$, determine whether they are isomorphic, that is, whether there is a bijection $f: V_{1} \rightarrow V_{2}$ such that ${u, v} \in E_{1}$ if and only if ${f(u), f(v)} \in E_{2}$.
The problem SuBGRAPH IsomORPHISM, which asks whether a given graph $G_{1}$ is isomorphic to a subgraph of another given graph $G_{2}$, can be proved to be $N P$-complete easily. However, the problem GIso is neither known to be $N P$-complete nor known to be in $P$, despite extensive studies in recent years. We will prove in Chapter 10, through the notion of interactive proof systems, that GIso is not $N P$-complete unless the polynomial-time hierarchy collapses to the level $\Sigma_{2}^{P}$. This result suggests that GIso is probably not $N P$-complete.
There are many number-theoretic problems in $N P$ that are neither known to be $N P$-complete nor known to be in $P$. We list three of them that have major applications in cryptography. An integer $x \in \mathbb{Z}{n}^{}$ is called a quadratic residue modulo $n$ if $x \equiv y^{2} \bmod n$ for some $y \in \mathbb{Z}{n}^{}$. We write $x \in Q R_{n}$ to denote this fact.
## 数学代写|计算复杂度理论代写Computational complexity theory代考|One-Way Functions and Cryptography
One-way functions are a fundamental concept in cryptography, having a number of important applications, including public-key cryptosystems,pseudorandom generators, and digital signatures. Intuitively, a one-way function is a function that is easy to compute but its inverse is hard to compute. Thus it can be applied to develop cryptosystems that need easy encoding but difficult decoding. If we identify the intuitive notion of “easiness” with the mathematical notion of “polynomial-time computability,” then one-way functions are subproblems of $N P$, because the inverse function of a polynomial-time computable function is computable in polynomial-time relative to an oracle in $N P$, assuming that the functions are polynomially honest. Indeed, all problems in $N P$ may be viewed as one-way functions.
Example 4.5 Define a function $f_{\mathrm{SAT}}$ as follows: For each Boolean function $F$ over variables $x_{1}, \ldots, x_{n}$ and each Boolean assignment $\tau$ on $x_{1}, \ldots, x_{n}$
$$f_{\mathrm{SAT}}(F, \tau)= \begin{cases}\langle F, 1\rangle & \text { if } \tau \text { satisfies } F, \ \langle F, 0\rangle & \text { otherwise. }\end{cases}$$
It is easily seen that $f_{\text {SAT }}$ is computable in polynomial time. Its inverse mapping $\langle F, 1\rangle$ to $\langle F, \tau\rangle$ is exactly the search problem of finding a truth assignment for a given Boolean formula. Using the notion of polynomialtime Turing reducibility and the techniques developed in Chapter 2, we can see that the inverse function of $f_{\mathrm{SAT}}$ is polynomial-time equivalent to the decision problem SAT. Thus, the inverse of $f_{\mathrm{SAT}}$ is not polynomial-time computable if $P \neq N P$.
Strictly speaking, function $f_{\mathrm{SAT}}$ is, however, not really a one-way function because it is not a one-to-one function and its inverse is really a multivalued function. In the following, we define one-way functions for one-to-one functions. We say that a function $f: \Sigma^{} \rightarrow \Sigma^{}$ is polynomially honest if there is a polynomial function $q$ such that for each $x \in \Sigma^{*}$, $|f(x)| \leq q(|x|)$ and $|x| \leq q(|f(x)|)$.
## 数学代写|计算复杂度理论代写Computational complexity theory代考|Relativization
The concept of relativization originates from recursive function theory. Consider, for example, the halting problem. We may formulate it in the following form: $K=\left{x \mid M_{x}(x)\right.$ halts $}$, where $M_{x}$ is the $x$ th TM in a standard enumeration of all TMs. Now, if we consider all oracle TMs, we may ask whether the set $K_{A}=\left{x \mid M_{x}^{A}(x)\right.$ halts $}$ is recursive relative to $A$. This is the halting problem relative to set $A$. It is easily seen from the original proof for the nonrecursiveness of $K$ that $K_{A}$ is nonrecursive relative to $A$ (i.e., no oracle TM can decide $K_{A}$ using $A$ as an oracle). Indeed, most results in recursive function theory can be extended to hold relative to any oracle set. We say that such results relativize. In this section, we investigate the problem of whether $P=N P$ in the relativized form. First, we need to define what is meant by relativizing the question of whether $P=N P$. For any set $A$, recall that $P^{A}(\circ r P(A))$ is the class of sets computable in polynomial time by oracle DTMs using $A$ as the oracle and, similarly, NPA (or $N P(A))$ is the class of sets accepted in polynomial time by oracle NTMs classes $P$ and $N P$, we show that the relativized $P=? N P$ question has both the positive and negative answers, depending on the oracle set $A$.
Theorem $4.14$ (a) There exists a recursive set $A$ such that $P^{A}=N P^{A}$.
(b) There exists a recursive set $B$ such that $P^{B} \neq N P^{B}$.
Proof. (a): Let $A$ be any set that is $\leq_{m}^{P}$-complete for PSPACE. Then, by Savitch’s theorem, we have
$$N P^{A} \subseteq N P S P A C E=P S P A C E \subseteq P^{A} .$$
## 数学代写|计算复杂度理论代写Computational complexity theory代考|Incomplete Problems in NP
SubGRAPH IsomORPHISM 的问题，它询问给定的图是否G1与另一个给定图的子图同构G2, 可以证明是ñ磷- 轻松完成。然而，问题 GIso 不为人所知ñ磷-完成也不知道在磷，尽管近年来进行了广泛的研究。我们将在第 10 章通过交互式证明系统的概念证明 GIso 不是ñ磷-完成，除非多项式时间层次结构崩溃到该级别Σ2磷. 该结果表明 GIso 可能不是ñ磷-完全的。
## 数学代写|计算复杂度理论代写Computational complexity theory代考|One-Way Functions and Cryptography
F小号一个吨(F,τ)={⟨F,1⟩ 如果 τ 满足 F, ⟨F,0⟩ 否则。
## 数学代写|计算复杂度理论代写Computational complexity theory代考|Relativization
(b) 存在一个递归集乙这样磷乙≠ñ磷乙.
ñ磷一个⊆ñ磷小号磷一个C和=磷小号磷一个C和⊆磷一个.
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 2,723 | 7,800 | {"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} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.781136 |
https://studysoup.com/tsg/193549/fundamentals-of-electric-circuits-5-edition-chapter-8-problem-8-14 | 1,620,736,498,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00185.warc.gz | 568,087,813 | 17,475 | ×
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# The switch in Fig. 8.69 moves from position A to position
ISBN: 9780073380575 128
## Solution for problem 8.14 Chapter 8
Fundamentals of Electric Circuits | 5th Edition
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Problem 8.14
The switch in Fig. 8.69 moves from position A to position B at (please note that the switch must connect to point B before it breaks the connection at A, a make-before-break switch). Let v(0) 0, find v(t) for t 7 0.
Step-by-Step Solution:
Step 1 of 3
Step 2 of 3
Step 3 of 3
#### Related chapters
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https://www.doorsteptutor.com/Exams/UGC/Psychology/Questions/Topic-Statistics-and-Research-Methods-3/Subtopic-Inferential-Statistics-5/Part-10.html | 1,502,942,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00015.warc.gz | 888,828,300 | 11,957 | # Statistics & Research Methods-Inferential Statistics (CBSE-NET (UGC) Psychology (Paper-II & Paper-III)): Questions 50 - 55 of 69
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## Question number: 50
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
The simplest way of visualizing a correlation is to construct a ________.
### Choices
Choice (4) Response
a.
Diagram
b.
Scatter diagram
c.
Graph
d.
All of the above
## Question number: 51
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
The most commonly used correlation coefficient is called ________.
### Choices
Choice (4) Response
a.
Pearson coefficient
b.
Correlation
c.
Rank correlation coefficient
d.
All of the above
## Question number: 52
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
Identical twins are likely to have almost identical IQs. In contrast, parents and their children have IQs that are only generally similar. The correlation between IQs of parents and children is. 35; that between identical twins is ________.
### Choices
Choice (4) Response
a.
0.35
b.
0.70
c.
0.85
d.
0.86
## Question number: 53
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
The ________ statistics is used for decision making for generalizing from small samples, and for drawing conclusions.
### Choices
Choice (4) Response
a.
Parametric
b.
Descriptive
c.
Inferential
d.
Question does not provide sufficient data or is vague
## Question number: 54
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
Which of the following is an advantage of “Correlation method”.
### Choices
Choice (4) Response
a.
Relationship may be coincidental
b.
Cause-and-effect relationship cannot be confirmed
c.
Demonstrates the existence of relationship
d.
Little or no control is possible
## Question number: 55
» Statistics & Research Methods » Inferential Statistics
MCQ▾
### Question
(1) Stage mentalists and psychics often use deception in their acts
(2) Blaming the lunar cycle for variations in the rate of violent crime is an example of mistaking correlation for causation
(3) Psychiatric interpretation of a patient’s broom-holding behavior show the importance of using a control group in experiments
### Choices
Choice (4) Response
a.
(i) and (ii) are false, but (iii) is true
b.
The statements are false
c.
The statements are true
d.
All the statements are true
f Page | 649 | 2,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-34 | longest | en | 0.725893 |
https://openlab.citytech.cuny.edu/2013-summer-mat-1272-reitz/syllabus-2/ | 1,723,363,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00131.warc.gz | 349,929,100 | 23,346 | # Calendar
This calendar lists topics for each day of the course. It is subject to change (if changes are made, notifications will be made on the OpenLab). The assigned problems in the book are for additional practice only — they will not be collected or graded.
Day Date Topic Homework problems in book (for additional practice only) 1 Monday, 6/3/2013 1.1 An Overview of Statistics pages 2-5, Examples 1 – 3 1.2 Data Classification page 9, Example 1 1.3 Data Collection and Experimental Design pages 16 – 22, Examples 1, 3 and 4 P. 6: 1 ā 41 odd P. 13: 7 ā 17 odd P. 23: 11 ā 15 odd, 19 ā 29 odd 2.1 Frequency Distributions and Their Graphs pages 38 – 46, Examples 1 -6 P. 47: 1 ā 31 odd, 35, 39, 41 Ā Ā 2.2 More Graphs and Displays pages 53 – 57, Examples 1, 2, and 4 P. 60: 5, 9, 13, 15 ā 19 odd, 23, 35, 37 2 Tuesday, 6/4/2013 2.3 Measures of Central Tendency pages 65 ā 68 and 71, Examples 1 – 6 P. 72: 1 ā 11 odd, 17 ā 21 odd, 25 ā 29 odd 2.4 Measures of Variation pages 80 – 83, Examples 1 – 4 (Use the formulas Ļ^2 =(Nāx^2 ā(āx)^2)/N^2 and s^2=(nāx^2 ā(āx)^2)/n(n-1) when calculating the variance for examples done in class) P. 90: 1, 3, 7, 11, 13, 19, 25, 27 Ā Ā 2.5 Measures of Position pages 100 – 106, Examples 1, 3 – 7 P. 107: 1 -21 odd, 25 – 35 odd, 39 ā 45 odd 3 Thursday, 6/6/2013 3.1 Basic Concepts of Probability and Counting pages 128 – 137, Examples 1 – 11 P. 138: 1, 3, 15 – 25 odd, 28, 29, 33 ā 41 odd, 45, 55 ā 61 odd 3.2 Conditional Probability and the Multiplication Rule pages 145 – 149, Examples 1 – 5 P. 150: 7 – 19 odd, 23 ā 31 odd 4 Monday, 6/10/2013 3.3 The Addition Rule pages 156 – 160, Examples 1 – 5 P. 161: 1 ā 25 odd 3.4 Additional Topics in Probability and Counting pages 168 – 173, Examples 1 – 9 P. 174: 7, 11 ā 31 odd, 37, 43 ā 47 odd 5 Tuesday, 6/11/2013 4.1 Probability Distributions pages 190 – 193, Examples 1, 3 and 4 P. 197: 1 ā 7 odd, 13 ā 19 odd, 21 ā 25 odd, 27 ā 31 odd do part (a), 41 First Examination – through Sec 3.2 (includes conditional probability but not independence) 6 Thursday, 6/13/2013 4.1 Probability Distributions pages 194 – 196, Examples 5 ā 7 (Use the formula Ļ^2 =āx^2P(x) – Ī¼^2 ) P. 198: 27 ā 31 odd do parts (c) and (d), 35, 37, 43 4.2 Binomial Distributions pages 202 – 208, Examples 1 ā 3, 5, 6, 8 P. 211: 9 ā 25 odd, 27 ā 30 all do parts (a), (c) and (d), 33 7 Monday, 6/17/2013 5.1 Introduction to Normal Distributions and the Standard Normal Distribution pages 236 – 243, Examples 1 ā 6 P. 244: 9 – 15 odd, 19 ā 37 odd, 41, 43 5.2 Normal Distributions: Finding Probabilities pages 249 – 250, Examples 1 – 2 P. 252: 1 – 23 odd 8 Tuesday, 6/18/2013 5.3 Normal Distributions: Finding Values pages 257 – 261, Examples 1 – 5 P. 262: 1 – 37 odd Second Examination – through Sec 4.2 9 Thursday, 6/20/2013 5.5 Normal Approximations to Binomial Distributions pages 281 – 286, Examples 1 – 5 P. 287: 1 – 25 odd 10 Monday, 6/24/2013 5.4 Sampling Distributions and the Central Limit Theorem pages 266 – 273, Examples 1 – 6 P. 274: 11 ā 35 odd 7.1: Introduction to Hypothesis Testing pages 356 – 366, Examples 1 ā 3, 5 P. 367: 11 ā 25 odd, 28, 32, 38, 41, 43 11 Tuesday, 6/25/2013 7.2 Hypothesis Testing for the Mean (Large Samples) pages 376 – 380, Examples 7 – 10 P. 382: 17 ā 27 odd, 35 ā 39 all, 41 7.3 Hypothesis Testing for the Mean (Small Samples) pages 387 – 391, Examples 1 – 5 P. 393: 3 ā 23 odd, 35 12 Thursday, 6/27/2013 10.1 Goodness-of-Fit Test pages 540 – 545, Examples 1 – 3 P. 546: 1 – 15 odd, Third Examination – through Sec 5.4 13 Monday, 7/1/2013 9.1 Correlation pages 484 – 488, Examples 1 and 4 P. 495: 1, 3, 9 ā 13 odd, 21 ā 27 odd do parts (b) and (c) 9.2 Linear Regression pages 501 – 502, Example 1 P. 505: 3, 5, 7 ā 12 all, 17 – 23 odd 14 Tuesday, 7/2/2013 10.2 Independence pages 551 – 555, Examples 1 – 2 P. 557: 1 ā 17 odd Review Thursday, 7/4/2013 NO CLASSES – 4TH OF JULY HOLIDAY 15 Friday, 7/5/2013 Final Examination | 1,703 | 4,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.659322 |
http://everything.explained.today/Dipole_field_strength_in_free_space/ | 1,586,108,486,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00345.warc.gz | 64,026,858 | 4,293 | # Dipole field strength in free space explained
Dipole field strength in free space, in telecommunications, is the electric field strength caused by a half wave dipole under ideal conditions. The actual field strength in terrestrial environments is calculated by empirical formulas based on this field strength.
## Power density
Let N be the effective power radiated from an isotropic antenna and p be the power density at a distance d from this source[1]
p=
N 4 ⋅ \pi ⋅ d2
Power density is also defined in terms of electrical field strength;
Let E be the electrical field and R be the impedance of the free space
p=
E2 R
The following relation is obtained by equating the two,
N 4 ⋅ \pi ⋅ d2
=
E2 R
or by rearranging the terms
E=
\sqrt{N ⋅ \sqrt{R}}{2 ⋅
\sqrt{\pi}d}
## Numerical values
Impedance of free space is roughly
120\pi
Since a half wave dipole is used, its gain over an isotropic antenna (
2.15dBi=1.64
) should also be taken into consideration,
E=
\sqrt{1.64 ⋅ N ⋅
\sqrt{120 ⋅ \pi}}{2 ⋅ \sqrt{\pi}d} ≈ 7 ⋅
\sqrt{N
}
In this equation SI units are used.
Expressing the same equation in:
kW instead of W in power,
km instead of m in distance and
is equivalent to multiplying the expression on the right by
\sqrt{1000}
.[2] In this case,
E ≈ 222 ⋅
\sqrt{N
} | 377 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-16 | latest | en | 0.918176 |
https://www.mathsweet.com/lessons/basic-algebra/math-order-operations-bodmas-rule | 1,679,384,707,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00687.warc.gz | 981,520,792 | 5,311 | # Math Order of Operations: BODMAS rule (i.e PEMDAS rule)
The concept of order of operations in algebra lessons is one of the very basic concepts that the student must understand well, because this concept forms a basic base for all algebra lessons that will study later.
The arithmetic operations we encounter have addition, subtraction, multiplication and division (the four operations) combined or not in the same algebraic expression.
In other cases we find in the same algebraic expression in addition to the four operations, powers and grouping symbols(brackets), which will inevitably affect the change of priority of operations(BODMAS rule).
So what is BODMAS rule?
Bodmas' rule is a method for performing a series of arithmetic operations in an algebraic expression without errors. The meaning of these letters is:
• B: Brackets (Grouping symbols);
• O: Order (Exponent, square root, factorial);
• D: Division (/);
• M: Multiplication (*);
• A: Addition (+);
• S: Subtraction (-);
### Note:
Another synonym for PODMAS rule is PEMDAS rule, which means the same thing, the order of operations in math. The reason for changing some letters is due to several labels for the same concept. These are the meanings of the letters of the word PEMDAS:
• P: Parentheses (Grouping symbols);
• E: Exponent (Exponent, square root, factorial);
• M: Multiplication (*);
• D: Division (/);
• A: Addition (+);
• S: Subtraction (-);
When evaluating expressions, proceed in the following rules order.
### Note:
1. Evaluate expressions contained in grouping symbols first(BOMDAS). If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first.
2. Evaluate all exponents that appear in the expression(BOMDAS).
3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right(BOMDAS).
4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right(BOMDAS).
These are some special cases in the order of operations.
### Note:
• Multiplication and division are the same importance.
• Addition and subtraction are the same importance.
Let's take some examples to see the order in which the arithmetic operations are performed. These examples will be from simpler to complex.
#### Example1:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression has only addition and subtraction (they have the same precedence), so we proceed from left to right.
#### Example2:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression has only multiplication and division (they have the same precedence), so we proceed from left to right.
#### Example3:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression is combined by three operations: addition, subtraction and multiplication. So, we perform the multiplication first, Then, we accomplish the operation from left to right.
#### Example4:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression is combined by three operations: addition, division and multiplication, and parentheses So, we perform what is inside the parentheses first(Priority is for the division first).
#### Example5:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression is combined by three operations: addition, division and multiplication, and parentheses and barckets So, we perform what is inside the Brackets first(Priority is for what inside the parentheses).
#### Example6:
calculate numerical expression below using the rules of Order of Operations.
#### Solution:
This expression is combined by three operations: addition, division and multiplication, and parentheses and powers So, we powers inside the parentheses first. | 794 | 3,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-14 | latest | en | 0.870458 |
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# Attempts to blame the mayor’s policies for the growing inequality of
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Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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Updated on: 05 Jan 2019, 06:48
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Attempts to blame the mayor’s policies for the growing inequality of wages are misguided. The sharp growth in the gap in earnings between college and high school graduates in this city during the past decade resulted from overall technological trends that favored the skills of more educated workers. The mayor’s response to this problem cannot be criticized, as it would hardly be reasonable to expect him to attempt to slow the forces of technology.
Which of the following, if true, casts the most serious doubt on the conclusion drawn in the last sentence above?
A. The mayor could have initiated policies that would have made it easier for less-educated workers to receive the education necessary for better-paying jobs.
B. Rather than cutting the education budget, the mayor could have increased the amount of staff and funding devoted to locating employment for graduating high school seniors.
C. The mayor could have attempted to generate more demand for products from industries that paid high blue-collar wages.
D. Instead of reducing the tax rate on the wealthiest earners, the mayor could have ensured that they shouldered a greater share of the total tax burden.
E. The mayor could have attempted to protect the earnings of city workers by instituting policies designed to reduce competition from foreign industries.
Originally posted by vikasp99 on 10 Feb 2017, 08:57.
Last edited by Bunuel on 05 Jan 2019, 06:48, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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10 Feb 2017, 09:45
1
vikasp99 wrote:
Attempts to blame the mayor’s policies for the growing inequality of wages are
misguided. The sharp growth in the gap in earnings between college and high school
graduates in this city during the past decade resulted from overall technological
trends that favored the skills of more educated workers. The mayor’s response to
this problem cannot be criticized, as it would hardly be reasonable to expect him to
attempt to slow the forces of technology.
Which of the following, if true, casts the most serious doubt on the conclusion
drawn in the last sentence above?
○1○ The mayor could have initiated policies that would have made it easier for less-educated
workers to receive the education necessary for better-paying jobs.
○2○ Rather than cutting the education budget, the mayor could have increased the
amount of staff and funding devoted to locating employment for graduating
high school seniors.
○3○ The mayor could have attempted to generate more demand for products from
industries that paid high blue-collar wages.
○4○ Instead of reducing the tax rate on the wealthiest earners, the mayor could
have ensured that they shouldered a greater share of the total tax burden.
○5○ The mayor could have attempted to protect the earnings of city workers by
instituting policies designed to reduce competition from foreign industries.
Technology developed => more educated workers get more salary
Can't slow down the forces of tech => can't blame mayor
We can't slow down the forces of tech, but why don't we increase the number of higher educated people so the gap could be recuded?
Then we directly come to choice A.
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Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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11 Feb 2017, 06:04
Hello experts.
could you please explain the above question.
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Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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11 Feb 2017, 09:21
vikasp99 wrote:
Attempts to blame the mayor’s policies for the growing inequality of wages are
misguided. The sharp growth in the gap in earnings between college and high school
graduates in this city during the past decade resulted from overall technological
trends that favored the skills of more educated workers. The mayor’s response to
this problem cannot be criticized, as it would hardly be reasonable to expect him to
attempt to slow the forces of technology.
Which of the following, if true, casts the most serious doubt on the conclusion
drawn in the last sentence above?
○1○ The mayor could have initiated policies that would have made it easier for less-educated
workers to receive the education necessary for better-paying jobs.
○2○ Rather than cutting the education budget, the mayor could have increased the
amount of staff and funding devoted to locating employment for graduating
high school seniors.
○3○ The mayor could have attempted to generate more demand for products from
industries that paid high blue-collar wages.
○4○ Instead of reducing the tax rate on the wealthiest earners, the mayor could
have ensured that they shouldered a greater share of the total tax burden.
○5○ The mayor could have attempted to protect the earnings of city workers by
instituting policies designed to reduce competition from foreign industries.
My reasoning:
premise: Knowledge level is the cause of wage inequality .
claim:Mayor should not be blamed for this.
My Pre-Thinking :Some policy or amendment by mayor is responsible for the wage inequality.i'e some how we need to prove mayor is associated to this issue or May be he is sitting ideal ,enjoying the power only .
A-Bingo.Mayor have not taken any step that could open the door for less educated person so as to gain technical knowledge.
C, D, E are completely out of scope
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Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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12 Feb 2017, 09:33
1
Aketa wrote:
Hello experts.
could you please explain the above question.
As already mentioned in response to one of your previous post, please post your query specifically. (e.g. why option C is incorrect or why do you think OA is not correct). Then the explanation can be targeted at your query precisely.
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Joined: 02 Sep 2009
Posts: 55801
Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink]
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10 Jan 2019, 04:32
1
vikasp99 wrote:
Attempts to blame the mayor’s policies for the growing inequality of wages are misguided. The sharp growth in the gap in earnings between college and high school graduates in this city during the past decade resulted from overall technological trends that favored the skills of more educated workers. The mayor’s response to this problem cannot be criticized, as it would hardly be reasonable to expect him to attempt to slow the forces of technology.
Which of the following, if true, casts the most serious doubt on the conclusion drawn in the last sentence above?
A. The mayor could have initiated policies that would have made it easier for less-educated workers to receive the education necessary for better-paying jobs.
B. Rather than cutting the education budget, the mayor could have increased the amount of staff and funding devoted to locating employment for graduating high school seniors.
C. The mayor could have attempted to generate more demand for products from industries that paid high blue-collar wages.
D. Instead of reducing the tax rate on the wealthiest earners, the mayor could have ensured that they shouldered a greater share of the total tax burden.
E. The mayor could have attempted to protect the earnings of city workers by instituting policies designed to reduce competition from foreign industries.
KAPLAN OFFICIAL EXPLANATION:
A
The Kaplan technique of prephrasing an answer works pretty well here. The author concludes that the mayor isn't responsible for the growing gap in earnings between high school and college graduates. Her evidence is one, that the growing need for technologically skilled workers is responsible, and two, that the mayor can't slow the force of technology. No, probably not, but he can take it upon himself to train the less skilled for better jobs. If that rebuttal occurred to you, you had no trouble finding (A). (A) argues that the mayor could have pursued policies that would have enabled the less educated to receive the education that would qualify them for well-paying jobs. If (A) is true, the author's argument is severely weakened.
(B) is off the mark. The problem addressed in the stimulus isn't the lack of employment for graduating high school students, it's the lack of employment that pays as well as high-skill employment. (C)'s proposal might help create more jobs that pay high blue-collar wages, but we don't know what "high" blue-collar wages are. They could still be far lower than the wages earned by people with college educations, in which case (C) wouldn't really address the inequality of wages. As for (D), the problem the mayor was supposed to solve was unequal wages, and if s not clear that taxing more heavily those with high wages constitutes a solution to that problem. Unskilled workers would still be stuck in low-paying jobs. (E) talks about protecting city workers' earnings, but this refers to all city workers. As such, (E)'s suggestion wouldn't improve the position of high school graduates compared to college graduates, which is what we want
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Re: Attempts to blame the mayor’s policies for the growing inequality of [#permalink] 10 Jan 2019, 04:32
Display posts from previous: Sort by | 2,310 | 10,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-26 | latest | en | 0.948615 |
http://neilblevins.com/art_lessons/primary_secondary_and_tertiary_shapes/primary_secondary_and_tertiary_shapes.htm | 1,601,422,003,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00030.warc.gz | 85,408,818 | 5,255 | Primary, Secondary, and Tertiary Shapes
By Neil Blevins
Created On: Sept 28th 2006
Updated On: Aug 2nd 2020
Go here to read this tutorial in Indonesian.
Go here and here for a fantastic talk on the subject by Gleb Alexandrov.
Introduction
This theory in composition goes by many names...
• Primary, Secondary, and Tertiary Shapes
• Big, Medium, Small
• 70 / 30 rule
Whatever the name, the idea is the same, if your image has a nice distribution of big (primary), medium (secondary) and small (tertiary) shapes, the resulting image will tend to be more pleasing to the eye.
You have two choices with this lesson, watch me discuss the issue in the video below, or read the full text.
A Definition
First, a definition. Primary shapes are your big shapes. If you squint at an image, the details tend to disappear and you're left with only your big shapes.
Secondary shapes are the smaller shapes that either sit on top of, or help make up the primary shapes.
Tertiary shapes are again smaller than the secondary shapes, they are the tiny details that add visual interest.
Here's three 2D diagrams showing primary, secondary and tertiary shapes in a fake composition. These 2d rules work exactly the same in 3 dimensions.
Here's another similar diagram...
The Important Questions
Now that you know what primary, secondary and tertiary shapes are, here's 3 important questions...
1. Does my composition have all 3 types?
2. How big should each shape should be?
3. Where in the composition should they go?
Do I Have All 3 Types?
This is pretty straight forward. Does your composition have all 3 of these details? If not, let's say it only has large and medium details, it may be time to add some smaller ones to balance things out. Or if everything is super detailed without any larger shapes, time to add some of those.
What Size Should The Shapes Be?
What should the size of these details be? That depends heavily on personal taste. I like having the primary details being huge, almost the size of the image itself. And I like Tertiary details being really small, sometimes the size of a pixel or two. From there you can decide the appropriate size for the Secondary Details.
So these could be primary, secondary and tertiary shapes sizes....
But so can these...
And these...
Of course, not all details should be the same size, it's more like a range of sizes. So for example, if you have 3-4 secondary shapes in the image, they don't have to be and frequently should not be exactly the same size. This is related to the 70 / 30 rule. If you have a large primary shape, and you want to break it up into 2 secondary shapes, it is more pleasing to break it up into one shape that is 70% of the size, and one shape that's 30% of the size, then two secondary shapes that are both 50% of the original size.
The same works for tertiary shapes. If you break up a secondary shape into multiple tertiary shapes, consider breaking it up into a bunch of tertiary shapes that are 30% the size of the secondary shapes.
Where Should They Go?
Next comes the distribution of these details. Take a look at this image, it has all 3 levels of detail, but the tertiary shapes are in one uniform block...
Many images suffer from this problem, huge blocks of small repeating patterns. For example, if you are a 3d artist just starting out, and need a dirt ground, you're likely to make an image like this by placing a standard procedural pattern on a ground plane...
This does not provide the eye any spot to rest, and hence doesn't look very natural. So in general you want to vary the placement of your tertiary details so that they are more random, and have several areas of detail, and several areas of no detail.
Notice how this is easier to digest. The eye can look at the detailed lower right corner, then move to the center of the image and rest there for a moment (in the big area of little detail) before looking at more tertiary details elsewhere.
A Photographic Example
Remember that noisy ground plane above? Let's fix it by looking at a photo of a real dirt ground. Notice the variety in the patterns.
A 3D Example
Here's another example of building up shapes. First, lets start with a cylinder, maybe this will be a pipe for a robot.
Now lets place some secondary shapes, rings around the pipe.
Now that's more interesting, but the regular pattern is kind of boring. Lets vary the position of the secondary rings...
Better, but all the secondary shapes are the same size, lets vary the sizes a bit...
Looking better. We now have variety in the sizes of the secondary shapes, and we have areas with lots of secondary shapes, and areas with no secondary shapes. Lets add a few tertiary shapes...
Now we have a well balanced composition.
Examples From My Artwork
Here's a few paint-overs of some images that I made in the past, showing off the primary, secondary and tertiary shapes...
In this case, the giant tower is the primary shape (a cylinder). That shape is made up of many secondary shapes, pipes and large pieces of concrete, and the entrance. The tertiary shapes are the tiny ships, and tiny lights along the surface of the tower, many of which are just a few pixels large. This helps give the piece a good sense of scale.
The primary shape is the shape made by the creature against the background. Then there's some larger tentacles, which are secondary shapes. Then there are more tiny tentacles, and grass, etc., which are tertiary shapes. If the image had no tertiary shapes, it would feel like something's missing.
Robots are a great opportunity for practicing "Big, Medium, Small". In this image, the larger shape of the creature is the primary shape. Then that has secondary shapes on it like the front panel or eye. Then those have tiny tertiary shapes on them as well, such as nuts, panel lines, small raised panels, etc. Also note that there aren't tertiary shapes everywhere on the secondary shapes, there are areas of lots of detail, and areas of no detail. Also notice just how small some of the tertiary shapes are compared to the secondary shapes.
Robot Example
Here's an example of a floating loader robot.
Now let's look more closely at the distribution of detail.
Sometimes, 3 reads (or sizes of shapes) aren't enough. In this example, I split it one step further with 1st read being the largest shapes (primary shapes), 2nd is more support shapes that make up the primary shapes (secondary shapes), 3rd is another level below that (tertiary shapes), and 4th read shapes are the super tiny detail.
Conclusion
To summarize, when making your images...
• Have large, medium, and small shapes.
• Have variety to the size of shapes in each of the three categories.
• Make sure these details are placed around the image and not all clumped in one area.
• And make sure there are areas of lots of details, and areas of almost no details.
And remember, just because these are the rules doesn't mean you necessary have to follow them. But if you don't follow them, you need to be aware that you're not following them, and what sort of affect that will have on the viewer.
This site is ©2020 by Neil Blevins, All rights are reserved.
Back to NeilBlevins.com | 1,586 | 7,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-40 | latest | en | 0.927061 |
http://de.metamath.org/mpegif/omex.html | 1,580,172,343,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00203.warc.gz | 45,406,594 | 6,498 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > omex Structured version Unicode version
Theorem omex 8060
Description: The existence of omega (the class of natural numbers). Axiom 7 of [TakeutiZaring] p. 43. This theorem is proved assuming the Axiom of Infinity and in fact is equivalent to it, as shown by the reverse derivation inf0 8038. A finitist (someone who doesn't believe in infinity) could, without contradiction, replace the Axiom of Infinity by its denial ; this would lead to by omon 6695 and (the universe of all sets) by fineqv 7735. The finitist could still develop natural number, integer, and rational number arithmetic but would be denied the real numbers (as well as much of the rest of mathematics). In deference to the finitist, much of our development is done, when possible, without invoking the Axiom of Infinity; an example is Peano's axioms peano1 6703 through peano5 6707 (which many textbooks prove more easily assuming Infinity). (Contributed by NM, 6-Aug-1994.)
Assertion
Ref Expression
omex
Proof of Theorem omex
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 zfinf2 8059 . 2
2 ax-1 6 . . . . 5
32ralimi2 2854 . . . 4
4 peano5 6707 . . . 4
53, 4sylan2 474 . . 3
65eximi 1635 . 2
7 vex 3116 . . . 4
87ssex 4591 . . 3
98exlimiv 1698 . 2
101, 6, 9mp2b 10 1
Colors of variables: wff setvar class Syntax hints: wi 4 wa 369 wex 1596 wcel 1767 wral 2814 cvv 3113 wss 3476 c0 3785 csuc 4880 com 6684 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1601 ax-4 1612 ax-5 1680 ax-6 1719 ax-7 1739 ax-8 1769 ax-9 1771 ax-10 1786 ax-11 1791 ax-12 1803 ax-13 1968 ax-ext 2445 ax-sep 4568 ax-nul 4576 ax-pr 4686 ax-un 6576 ax-inf2 8058 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3or 974 df-3an 975 df-tru 1382 df-ex 1597 df-nf 1600 df-sb 1712 df-eu 2279 df-mo 2280 df-clab 2453 df-cleq 2459 df-clel 2462 df-nfc 2617 df-ne 2664 df-ral 2819 df-rex 2820 df-rab 2823 df-v 3115 df-sbc 3332 df-dif 3479 df-un 3481 df-in 3483 df-ss 3490 df-pss 3492 df-nul 3786 df-if 3940 df-pw 4012 df-sn 4028 df-pr 4030 df-tp 4032 df-op 4034 df-uni 4246 df-br 4448 df-opab 4506 df-tr 4541 df-eprel 4791 df-po 4800 df-so 4801 df-fr 4838 df-we 4840 df-ord 4881 df-on 4882 df-lim 4883 df-suc 4884 df-om 6685 This theorem is referenced by: axinf 8061 inf5 8062 omelon 8063 dfom3 8064 elom3 8065 oancom 8068 isfinite 8069 nnsdom 8070 omenps 8071 omensuc 8072 unbnn3 8075 noinfep 8076 noinfepOLD 8077 tz9.1 8160 tz9.1c 8161 fseqdom 8407 fseqen 8408 aleph0 8447 alephprc 8480 alephfplem1 8485 alephfplem4 8488 iunfictbso 8495 unctb 8585 r1om 8624 cfom 8644 itunifval 8796 hsmexlem5 8810 axcc2lem 8816 acncc 8820 axcc4dom 8821 domtriomlem 8822 axdclem2 8900 infinf 8941 unirnfdomd 8942 alephval2 8947 dominfac 8948 iunctb 8949 pwfseqlem4 9040 pwfseqlem5 9041 pwxpndom2 9043 pwcdandom 9045 gchac 9059 wunex2 9116 tskinf 9147 niex 9259 nnexALT 10538 ltweuz 12040 uzenom 12043 nnenom 12058 axdc4uzlem 12060 seqex 12077 rexpen 13822 cctop 19301 2ndcctbss 19750 2ndcdisj 19751 2ndcdisj2 19752 tx1stc 19914 tx2ndc 19915 met2ndci 20788 xpct 27233 snct 27234 fnct 27236 trpredex 28925 bnj852 33076 bnj865 33078
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# Friction
Topics: Developer Forum Wiki Link: [discussion:85243]
timdodd Feb 13, 2010 at 9:00 AM Hello , I've been pulling my hair out because physics for some reason is so hard for me to grasp. I can't seam to under stand how to apply a tangent friction to the velocity of an object. I've looked at the arbiter class for hours but sadly I've had no luck because of how much is going on at once. I'm trying to start as simple as possible . 1 body and 1 static body . Here is what I have done so far and you'll see where I've tried to make since of it all. Thanks for anyhelp. ``` Vector2 normal = Vector2.Normalize(-r.MinimumTranslationVector); CMPosition += r.MinimumTranslationVector; Vector2 Position = SAT.GetCollisionPoint(fixtureA, fixtureB); Vector2 CMToCornerPerp = Utility.GetPerpendicular(Position - CMPosition); Vector2 Velocity = CMVelocity + AngularVelocity * CMToCornerPerp; //_vn float ImpulseNumerator = -( 1 + 0.6f) * Vector2.Dot(Velocity, normal); float PerpDot = Vector2.Dot(CMToCornerPerp, normal); float ImpulseDenominator = OneOverMass + OneOverCMMomentOfInertia * PerpDot * PerpDot; float Impulse = ImpulseNumerator / ImpulseDenominator ; CMVelocity += (Impulse * OneOverMass * normal);// +(Impulse * OneOverMass * Discord.Collisions.Common.Math.Cross(normal, 1) * 0.2f); AngularVelocity += Impulse * OneOverCMMomentOfInertia * PerpDot; Vector2 tangent = Discord.Collisions.Common.Math.Cross(normal, 1); float maxtangentImpulse = 0.2f * ImpulseNumerator; float _vt = (CMToCornerPerp.X * tangent.X) + (CMToCornerPerp.Y * tangent.Y); _tangentImpulse = (-_vt); _oldTangentImpulse = _tangentImpulse; float tmpT = MathHelper.Clamp(_oldTangentImpulse + _tangentImpulse, -maxtangentImpulse, maxtangentImpulse); _tangentImpulse = tmpT - _oldTangentImpulse; CMVelocity.X += tangent.X * _tangentImpulse; CMVelocity.Y += tangent.Y * _tangentImpulse; ``` | 542 | 1,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-34 | longest | en | 0.660837 |
https://se.mathworks.com/matlabcentral/answers/1448239-how-to-call-a-variable-matrix-from-one-script-to-another-script?s_tid=prof_contriblnk | 1,723,468,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00780.warc.gz | 392,397,845 | 37,405 | # How to call a variable matrix from one script to another script
12 views (last 30 days)
Bathala Teja on 7 Sep 2021
Commented: Bathala Teja on 8 Sep 2021
I have two scripts. In first one i will get A and B matrices which are interms of theta as shown below.
sym theta
A = cos(theta).*[1 5 6;
2 9 3;
5 1 0]
B = tan(2*theta).*[8 5 1;
0 1 6;
2 4 7]
I want to use these A and B matrices in another script, which is given below
function dydt = scriptname(t, y)
theta = y(3); % Substitute y(3) in place of theta in A and B matrices
A % Extract A matrix
B % Extract B matrix
V = [1;
5;
3]
dydt = V-((A+B)*Y);
end
Finally i will plot these ODE's in another script that i didnt gave here.
My main doubt is how to extract A and B matrices(especially those are variable matrices) into another script.
After extracting these A and B matrices i will substitute y(3) so that i can get rid of theta.
I havent shown my exact problem because its too complicated. Focus on my main doubt dont care about values.
Fabio Freschi on 7 Sep 2021
I am not sure I have understtod correctly your problem. I give it a try.
I would add an extra layer using an anonymous function to solve the porblem of passing A and B to your funciton
clear variables, close all
% anonymous functions to calculate A and B
A = @(theta)cos(theta).*[1 5 6; 2 9 3; 5 1 0];
B = @(theta)tan(2*theta).*[8 5 1; 0 1 6; 2 4 7];
% extra layer to pass also A and B to your dydt function
myfun = @(t,y)scriptname(t,y,A,B);
% dummy values for tspan and y0
tspan = [0 1];
y0 = [0; 0; 0];
% ode solver
sol = ode45(myfun,tspan,y0);
function dydt = scriptname(t,y,A,B)
theta = y(3); % Substitute y(3) in place of theta in A and B matrices
V = [1; 5; 3];
% actual calculation of A and B with the current value for theta
dydt = V-((A(theta)+B(theta))*y);
end
Fabio Freschi on 8 Sep 2021
You should ask another question, because this is beyond the scope of the original post. If the answer solves your problem, please accept it.
It is pretty simple to test yourself your code, creating a random y matrix. You can also run your code in this window using the run comamnd.
Bathala Teja on 8 Sep 2021
Ok thank you
Jan on 7 Sep 2021
Edited: Jan on 7 Sep 2021
The first code shows a "script", the second one a "function". While scripts share their variables with the callers automatically, functions don't. As far as I understand all you havt to do is to call the script:
% Script file: "yourScript.m"
sym theta
A = cos(theta).*[1 5 6;
2 9 3;
5 1 0]
B = tan(2*theta).*[8 5 1;
0 1 6;
2 4 7]
% Function file: "scriptname.m" ( missleading name! This is not a script.)
function dydt = scriptname(t, y)
theta = y(3); % Substitute y(3) in place of theta in A and B matrices
yourScript; % <-- A and B are defined internally
V = [1;
5;
3]
dydt = V-((A+B)*Y);
end
For productive code, scripts are a shot in your knee, because you cannot see directly, where the variables are defined. Use functions instead, to forward variables explicitly:
% Script file: "yourFunction.m"
function [A, B] = yourFunction()
sym theta
A = cos(theta).*[1 5 6;
2 9 3;
5 1 0]
B = tan(2*theta).*[8 5 1;
0 1 6;
2 4 7]
end
% Function file: "theOtherFunction.m"
function dydt = theOtherFunction(t, y)
[A, B] = yourFunction();
V = [1;
5;
3]
dydt = V-((A+B)*Y);
end
Fabio Freschi on 7 Sep 2021
when you create A you create B s well, so you must check only one of them
Bathala Teja on 7 Sep 2021
ok thank you for your explanation
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Start Hunting! | 1,138 | 3,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.808688 |
https://www.physicsforums.com/threads/light-black-holes-and-gravity.3092/ | 1,591,501,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00068.warc.gz | 834,824,046 | 19,186 | # Light, black holes and gravity
JSK333
## Main Question or Discussion Point
Gravity is "The tendency of a mass of matter toward a center of attraction; esp., the tendency of a body toward the center of the earth; terrestrial gravitation" from most physics dictionaries.
So, here's a basic question.
Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?
Solomon
Related Other Physics Topics News on Phys.org
mathman
Although photons have zero REST mass, they have energy. Gravity acts on energy just as it does on mass. The famous 1919 eclipse observation, which was a dramatic confirmation of general relativity, was based on the bending of starlight due to the sun's gravity. In simple terms, the escape velocity from a black hole is greater than the speed of light, therefore light can't escape.
in Newtonian theory of gravity, one would suppose that this effect would never take place becuase as you said photons are massless. i don't know what mathman meant by no-rest mass becuase photons are never at rest, but he was right: they have energy. in the spec. theory of relitivity, energy, like its equivalent in mass, are affected by the curvature of spacetime. since a black hole is a pretty heavy distortion of spacetime, it is impossible for light to escape, thus making it black.
or you could say that light is still travelling in a straight line in four dimensional space, but since a black hole distorts space time around it, it curves space-time in three dimensional space. Light will always follow a straight line geodesic in 4-dimensional space.
marcus
Gold Member
Dearly Missed
Originally posted by JSK333
Gravity is "The tendency of a mass of matter toward a center of attraction; esp., the tendency of a body toward the center of the earth; terrestrial gravitation" from most physics dictionaries.
So, here's a basic question.
Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?
Solomon
I just started a thread in Astronomy forum about the effect of gravity on light. It partially overlaps stuff here altho there are several other issues raised here. You might want to check it out.
marcus
Gold Member
Dearly Missed
Originally posted by JSK333
So, here's a basic question.
Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?
Solomon
several people in this thread (mathman, gary etc.) have already pointed out that
the dictionary def is too restricted
it talks only of the effect of gravity on matter
but gravity is modeled by spacetime curvature, which effects everything, not just matter
the geometry of spacetime must effect everything in spacetime.
it cannot single out matter and limit its effects to matter.
light must travel along the geodesics (which may be curved by concentrations of energy), as I think several people already said.
This question about light escaping from a black hole is interesting.
Seen by an observer who is well clear of the hole, for all practical purposes at "infinity" or many hole-radiuses away, the light from near the event horizon of the hole is way way redshifted.
Its wavelength can get so long that it is no longer detectable--no more quantum energy. Infinitely redshifted light dies.
The "event horizon" of a BH, which for a lot of people plays the role of the spherical "surface" around the hole is pretty much defined as the place from which light cannot get clear out to infinity because in the course of doing that it undergoes an infinite redshift---and ceases to exist.
Originally posted by marcus
several people in this thread (mathman, gary etc.) have already pointed out that
the dictionary def is too restricted
it talks only of the effect of gravity on matter
but gravity is modeled by spacetime curvature, which effects everything, not just matter
the geometry of spacetime must effect everything in spacetime.
it cannot single out matter and limit its effects to matter.
light must travel along the geodesics (which may be curved by concentrations of energy), as I think several people already said.
This question about light escaping from a black hole is interesting.
Seen by an observer who is well clear of the hole, for all practical purposes at "infinity" or many hole-radiuses away, the light from near the event horizon of the hole is way way redshifted.
Its wavelength can get so long that it is no longer detectable--no more quantum energy. Infinitely redshifted light dies.
The "event horizon" of a BH, which for a lot of people plays the role of the spherical "surface" around the hole is pretty much defined as the place from which light cannot get clear out to infinity because in the course of doing that it undergoes an infinite redshift---and ceases to exist.
Could it be that the understanding you are relaying is that you have no comprehension of what a blackhole is? Ive seen your post in astronomy, and a number of your postings are blatantly wrong, but I guess your just like the rest of us learning.
Blackholes ain't so Black! has L Booda been transformed into another alias?..has anyone seen or heard from him recently?
marcus
Gold Member
Dearly Missed
Originally posted by ranyart
Could it be that the understanding you are relaying is that you have no comprehension of what a blackhole is? Ive seen your post in astronomy, and a number of your postings are blatantly wrong, but I guess your just like the rest of us learning.
Blackholes ain't so Black! has L Booda been transformed into another alias?..has anyone seen or heard from him recently?
Don't know Booda.
My post above and the corresponding one "effect of gravity on light" I just posted at "Astronomy" forum are mainly expositions of the gravitational redshift formula
1+ zgrav = 1/sqrt(1 - Rs/R)
If you disagree with this please let me know. It is certainly a textbook formula, and exact to the best of my knowledge.
If you would like a reference for it a good one would be F. Shu "The Physical Universe" a great textbook in general astronomy. Probably also plenty of web references available if you google "gravitational redshift".
Among other things, you can use this formula to prove that light can't escape the event horizon (not talking about Hawking radiation from near the EH).
Rs = 2GM/c2
Of course you already *know* light cannot escape the EH, but the redshift formula provides a neat quick way to prove it.
ShawnD
Planck's formula E = hf tells us the energy for a photon of light at a given frequency and Einstein's formula E =mc^2 relates energy to mass and speed. If we turn the quations into one we get hf = mc^2, then we can move it around and get m = hf/c^2 as the mass of a photon. h (Planck's constant) is a constant, f (frequency) is always > 0 for any photon and c (speed of light) is a constant (for vacuums in this case).
Now we see that light truly does have a calculatable mass :D
neutroncount
Originally posted by ShawnD
Planck's formula E = hf tells us the energy for a photon of light at a given frequency and Einstein's formula E =mc^2 relates energy to mass and speed. If we turn the quations into one we get hf = mc^2, then we can move it around and get m = hf/c^2 as the mass of a photon. h (Planck's constant) is a constant, f (frequency) is always > 0 for any photon and c (speed of light) is a constant (for vacuums in this case).
Now we see that light truly does have a calculatable mass :D
Wrong
Photons don't have REST mass, but they do have energy equating to mass. So what your math shows is the mass equivalent of a photon at a certain energy. Photons DO have momentum though and can transfer it giving the appearance of mass.
If this is what you were implying, then retract my entire statement :)
Last edited by a moderator:
Originally posted by neutroncount
Photons don't have REST mass, but they do have energy equating to mass. So what your math shows is the mass equivalent of a photon at a certain energy. Photons DO have momentum though and can transfer it giving the appearance of mass.
i know you weren't talkning to me, but see if this is a correct logical line of thought. the photon has no rest mass, or it would not be able to accelerate to c, but it does have energy, and thus graivity is a factor. does a photon therefore have no inertial value while accelertating to c? basically is it only mass (not its equivelent in energy) that can be affected by inertia? or the energy inherent in a virtual paricles movement does not add to its energy value (therefore its mass) in the same way that it would for a particle with rest mass?
I'm not sure i understand your question, but photons/light do/does have momentum.
p=E/c (from the definition of the energy-momentum tensor in SR)
Joe
neutroncount
Originally posted by maximus
i know you weren't talkning to me, but see if this is a correct logical line of thought. the photon has no rest mass, or it would not be able to accelerate to c, but it does have energy, and thus graivity is a factor. does a photon therefore have no inertial value while accelertating to c? basically is it only mass (not its equivelent in energy) that can be affected by inertia? or the energy inherent in a virtual paricles movement does not add to its energy value (therefore its mass) in the same way that it would for a particle with rest mass?
But here's the deal. Photons don't accelerate. They ARE at c, no if, ands, or buts. The creation of a phonton requires it to alreay be moving at c upon it's creation. A photon can neither slow down or speed up. Yes, I know about different mediums but at every point in the medium the propagation of photons is at c. It's the interaction that causes a slowdown. | 2,208 | 9,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | longest | en | 0.957271 |
https://www.jiskha.com/display.cgi?id=1348755347 | 1,511,114,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00604.warc.gz | 800,976,486 | 3,924 | # Math/Trig
posted by .
An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction 210°50'. Assuming the Earth is flat, how far is the plane from the airport at this time (to the nearest mile)
• Math/Trig -
150 at 180° = (-150,0)
170 at 210°50' = (-146,-87)
total: (-296,-87)
distance = √(296^2 + 87^2) = 309 mi
• Math/Trig -
I am assuming that 0° is east. If you are using 0° = north, then things have to be changed a bit.
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More Similar Questions | 672 | 2,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-47 | latest | en | 0.928949 |
http://kmj.knu.ac.kr/journal/view.html?doi=10.5666/KMJ.2019.59.4.591 | 1,585,875,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00107.warc.gz | 95,953,591 | 27,933 | KYUNGPOOK Math. J. 2019; 59(4): 591-601
The Zero-divisor Graph of the Ring of Integers Modulo n
Seung Jun Pi, Se Hun Kim, Jung Wook Lim∗
Department of Mathematics, College of Natural Sciences, Kyungpook National University, Daegu, 41566, Republic of Korea
e-mail : seungjunpi0@gmail.com
Department of Mathematics, Seoul National University, Seoul, 03080, Republic of Korea
e-mail : shunhun33@gmail.com
Department of Mathematics, College of Natural Sciences, Kyungpook National University, Daegu, 41566, Republic of Korea
e-mail : jwlim@knu.ac.kr
* Corresponding Author.
Received: October 15, 2018; Revised: September 5, 2019; Accepted: November 11, 2019; Published online: December 23, 2019.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
Let ℤn be the ring of integers modulo n and Γ(ℤn) the zero-divisor graph of ℤn. In this paper, we study some properties of Γ(ℤn). More precisely, we completely characterize the diameter and the girth of Γ(ℤn). We also calculate the chromatic number of Γ(ℤn).
Keywords: zero-divisor graph, ring of integers modulo n, diameter, girth, clique, chromatic number.
1. Introduction
### 1.1. Preliminaries
In this subsection, we review some concepts from basic graph theory. Let G be a (undirected) graph. Recall that G is connected if there is a path between any two distinct vertices of G. The graph G is complete if any two distinct vertices are adjacent. The complete graph with n vertices is denoted by Kn. The graph G is a complete bipartite graph if G can be partitioned into two disjoint nonempty vertex sets A and B such that two distinct vertices are adjacent if and only if they are in distinct vertex sets. If one of the vertex sets is a singleton set, then we call G a star. We denote the complete bipartite graph by Km,n, where m and n are the cardinal numbers of A and B, respectively. For vertices a and b in G, d(a, b) denotes the length of the shortest path from a to b. If there is no such path, then d(a, b) is defined to be ∞; and d(a, a) is defined to be zero. The diameter of G, denoted by diam(G), is the supremum of {d(a, b) | a and b are vertices of G}. The girth of G, denoted by g(G), is defined as the length of the shortest cycle in G. If G contains no cycles, then g(G) is defined to be ∞. A subgraph H of G is an induced subgraph of G if two vertices of H are adjacent in H if and only if they are adjacent in G. The chromatic number of G is the minimum number of colors needed to color the vertices of G so that no two adjacent vertices share the same color, and is denoted by χ(G). A clique C in G is a subset of the vertex set of G such that the induced subgraph of G by C is a complete graph. The clique number of G, denoted by cl(G), is the greatest integer n ≥ 1 such that KnG. If KnG for all integers n ≥ 1, then cl(G) is defined to be ∞. A maximal clique in G is a clique that cannot be extended by including one more adjacent vertex. It is easy to see that χ(G) ≥ cl(G).
### 1.2. The Zero-divisor Graph of a Commutative Ring
Let R be a commutative ring with identity and Z(R) the set of nonzero zero-divisors of R. The zero-divisor graph of R, denoted by Γ(R), is the simple graph with vertex set Z(R), and for distinct a, b ∈ Z(R), a and b are adjacent if and only if ab = 0. Clearly, Γ(R) is the null graph if and only if R is an integral domain.
In [4], Beck first introduced the concept of the zero-divisor graphs of commutative rings and in [3], Anderson and Nazeer continued the study. In their papers, all elements of R are vertices of the graph and they were mainly interested in colorings. In [2], Anderson and Livingston gave the present definition of Γ(R) in order to emphasize the study of the interplay between graph-theoretic properties of Γ(R) and ring-theoretic properties of R. It was shown that Γ(R) is connected with diam(Γ(R)) ≤ 3 [2, Theorem 2.3]; and g(Γ(R)) ≤ 4 [5, (1.4)].
For more on the zero-divisor graph of a commutative ring, the readers can refer to a survey article [1].
Let ℤn be the ring of integers modulo n. The purpose of this paper is to study some properties of the zero-divisor graph of ℤn. If n is a prime number, then ℤn has no zero-divisors; so Γ(ℤn) is the null graph. Hence in this paper, we only consider the case that n is a composite. In Section 2, we completely characterize the diameter and the girth of Γ(ℤn). In Section 3, we calculate the chromatic number of Γ(ℤn). Note that all figures are drawn via website http://graphonline.ru/en/.
2. The Diameter and the Girth of Γ(ℤn)
Our first result in this section is the complete characterization of the diameter of Γ(ℤn).
### Theorem 2.1
The following statements hold.
(1) diam(Γ(ℤn)) = 0 if and only if n = 4.
(2)diam(Γ(ℤn)) = 1 if and only if n = p2for some prime p ≥ 3.
(3)diam(Γ(ℤn)) = 2 if and only if n = pr for some prime p and some integer r ≥ 3, or n = pq for some distinct primes p and q.
(4)diam(Γ(ℤn)) = 3 if and only if n = pqr for some distinct primes p, q and some integer r ≥ 2.
Proof
(1) If n = 4, then Z(ℤ4) = {2}; so diam(Γ(ℤ4)) = 0.
(2) If n = p2 for some prime p ≥ 3, then Z(ℤp2) = {p, 2p, …, (p − 1)p}; so the product of any two elements of Z(ℤp2) is zero. Hence Γ(ℤp2) is the complete graph Kp−1. Thus diam(Γ(ℤp2)) = 1.
(3) If n = pr for some prime p and some integer r ≥ 3, then Z(ℤpr) = {p, 2p, …, (pr−1 − 1)p}; so apr−1 = 0 for all a ∈ Z(ℤpr). Hence diam(Γ(ℤpr)) ≤ 2. Note that p((pr−1 − 1)p) ≠ = 0. Thus diam(Γ(ℤpr)) = 2.
If n = pq for some distinct primes p and q, then Z(ℤpq) = {p, 2p, …, (q − 1)p, q, 2q, …, (p − 1)q}; so (ip)(jq) = 0 for all i = 1, …, q − 1 and j = 1, …, p − 1. Note that for any a, b ∈ {p, 2p, …, (q − 1)p} and c, d ∈ {q, 2q, …, (p − 1)q}, ab ≠ = 0 and cd ≠ = 0. Hence Γ(ℤpq) is the complete bipartite graph Kp−1,q−1. Thus diam(Γ(ℤpq)) = 2.
(4) Suppose that n = pqr for some distinct primes p, q and some integer r ≥ 2. Then p, q ∈ Z(ℤpqr) with pq ≠ = 0; so diam(Γ(ℤpqr)) ≥ 2. If there exists an element a ∈ Z(ℤpqr) such that p ~ a ~ q is a path, then a is a nonzero multiple of pr and qr; so a is nonzero a multiple of pqr. This is a contradiction. Hence diam(Γ(ℤpqr)) ≥ 3. Thus diam(Γ(ℤpqr)) = 3 [2, Theorem 2.3].
We next study the girth of Γ(ℤn).
### Lemma 2.2
If g(Γ(ℤn)) = 3, then g(Γ(ℤmn)) = 3 for all integers m ≥ 1.
Proof. Note that if a ~ b ~ c ~ a is a cycle in Γ(ℤn), then am ~ bm ~ cm ~ am is a cycle in Γ(ℤmn). Thus g(Γ(ℤmn)) = 3.
The next example shows that Lemma 2.2 cannot be extended to the case of girth 4.
### Example 2.3
(1) Note that g(Γ(ℤ12)) = 4. In fact, 3 ~ 4 ~ 6 ~ 8 ~ 3 is a cycle of length 4 in Γ(ℤ12). However, g(Γ(ℤ24)) = 3 because 6 ~ 8 ~ 12 ~ 6 is a cycle of length 3 in Γ(ℤ24).
(2) In general, g(Γ(ℤ4q)) = 4 but g(Γ(ℤ2rq)) = 3 for all primes q ≥ 3 and all integers r ≥ 3. (See Proposition 2.6.)
### Proposition 2.4
If t ≥ 3 is an integer and p1, …, pt are distinct primes, then$g(Γ(ℤp1r1p2r2⋯ptrt))=3$for all positive integers r1, …, rt.
Proof
If t = 3, then $p1r1p2r2~p2r2p3r3~p1r1p3r3~p1r1p2r2$ is a cycle in $Γ(ℤp1r1p2r2p3r3)$; so $g(Γ(ℤp1r1p2r2p3r3))=3$.
If t > 3, then the result follows directly from Lemma 2.2.
### Proposition 2.5
Let p be a prime and r ≥ 2 an integer. Then the following assertions hold.
(1) g(Γ(ℤpr)) = ∞ if and only if pr = 4, 8 or 9.
(2) g(Γ(ℤpr)) = 3 if and only if each of the following conditions holds.
(a) p = 2 and r ≥ 4.
(b) p = 3 and r ≥ 3.
(c) p ≥ 5 and r ≥ 2.
Proof
(1) It is obvious that Γ(ℤ4), Γ(ℤ8) and Γ(ℤ9) have no cycles. Thus the equivalence follows.
(2) If p = 2 and r ≥ 4, then 2r−1, 2r−2, 3 · 2r−2 ∈ Z(ℤ2r). Since the product of any two of them is zero, 2r−1 ~ 2r−2 ~ 3 · 2r−2 ~ 2r−1 is a cycle in Γ(ℤ2r). Thus g(Γ(ℤ2r)) = 3.
If p = 3 and r ≥ 3, then 3r−1, 2 · 3r−1, 3r−2 ∈ Z(ℤ3r). Since the product of any two of them is zero, 3r−1 ~ 2 · 3r−1 ~ 3r−2 ~ 3r−1 is a cycle in Γ(ℤ3r). Thus g(Γ(ℤ3r)) = 3.
If p ≥ 5 and r ≥ 2, then pr−1, 2pr−1, 3pr−1 ∈ Z(ℤpr). Since the product of any two of them is zero, pr−1 ~ 2pr−1 ~ 3pr−1 ~ pr−1 is a cycle in Γ(ℤpr). Thus g(Γ(ℤpr)) = 3.
### Proposition 2.6
Let n be a positive integer which has only two distinct prime divisors. Then the following assertions hold.
(1) g(Γ(ℤn)) = ∞ if and only if n = 2q for some prime q ≥ 3.
(2) g(Γ(ℤn)) = 3 if and only if one of the following holds.
(a) n = 2rqsfor some prime q ≥ 3 and some integers r ≥ 1 and s ≥ 2.
(b) n = 2rq for some prime q ≥ 3 and some integer r ≥ 3.
(c) n = 3rqsfor some prime q ≥ 5 and some integers r ≥ 1 and s ≥ 2.
(d) n = 3rq for some prime q ≥ 5 and some integer r ≥ 2.
(e) n = prqsfor some primes q > p ≥ 5 and some integers r, s ≥ 1 except for r = s = 1.
(3) g(Γ(ℤn)) = 4 if and only if n = pq for some distinct primes p, q ≥ 3, or n = 4q for some prime q ≥ 3.
Proof
(1) If n = 2q for some prime q ≥ 3, then Γ(ℤ2q) is a star graph K1,q−1 by the proof of Theorem 2.1(3). Hence Γ(ℤ2q) has no cycles, and thus g(Γ(ℤ2q)) = ∞.
(2) Let p and q be the only distinct prime divisors of n. Without loss of generality, we may assume that p < q.
Cases (a) and (b)
p = 2. In this case, q ~ 2q ~ 4q ~ q is a cycle of length 3 in Γ(ℤ2q2); so g(Γ(ℤ2q2)) = 3. Hence by Lemma 2.2, g(Γ(ℤ2rqs)) = 3 for all integers r ≥ 1 and s ≥ 2. Also, 4 ~ 2q ~ 4q ~ 4 is a cycle of length 3 in Γ(ℤ8q); so g(Γ(ℤ8q)) = 3. Hence by Lemma 2.2, g(Γ(ℤ2rq)) = 3 for all integers r ≥ 3.
Cases (c) and (d)
p = 3. In this case, q ~ 3q ~ 6q ~ q is a cycle of length 3 in Γ(ℤ3q2); so g(Γ(ℤ3q2)) = 3. Hence by Lemma 2.2, g(Γ(ℤ3rqs)) = 3 for all integers r ≥ 1 and s ≥ 2. Also, 3 ~ 3q ~ 6q ~ 3 is a cycle of length 3 in Γ(ℤ9q); so g(Γ(ℤ9q)) = 3. Hence by Lemma 2.2, g(Γ(ℤ3rq)) = 3 for all integers r ≥ 2.
Case (e)
p ≥ 5. In this case, q ≥ 7; so by Proposition 2.5(2), g(Γ(ℤpr)) = 3 = g(Γ(ℤqs)) for all integers r, s ≥ 2. Hence by Lemma 2.2, g(Γ(ℤprqs)) = 3 for all integers r, s ≥ 1 except for r = s = 1.
(3) If n = pq for some distinct primes p, q ≥ 3, then Γ(ℤpq) is the complete bipartite graph Kp−1,q−1 by the proof of Theorem 2.1(3). Hence there does not exist a cycle of odd length. Note that p ~ 2q ~ 2p ~ q ~ p is a cycle of length 4. Thus g(Γ(ℤpq)) = 4.
Let n = 4q for some prime q ≥ 3, and suppose to the contrary that there exists a cycle a ~ b ~ c ~ a in Γ(ℤ4q). Since ab, bc and ca are divisible by 4q, q divides at least two of a, b and c. Without loss of generality, we may assume that q divides a and b. If 2 divides a, then a = 2q. Since ab is divisible by 4q, b is divisible by 2; so b = 2q. This is absurd. If 2 does not divide a, then a = q or a = 3q. Since 4q divides ab, b is divisible by 4; so b is a multiple of 4q. This is a contradiction. Hence there do not exist cycles of length 3 in Γ(ℤ4q). Note that q ~ 4 ~ 2q ~ 8 ~ q is a cycle of length 4. Thus g(Γ(ℤ4q)) = 4.
In the next remark, we construct a cycle of length 3 in Γ(ℤn) in each case of Proposition 2.6(2).
### Remark 2.7
(1) Let n = 2rqs for some prime q ≥ 3 and some integers r ≥ 1 and s ≥ 2. Then 2rqs−1 ~ 2r+1qs−1 ~ 2r−1qs ~ 2rqs−1 is a cycle of length 3 in Γ(ℤ2rqs).
(2) Let n = 2rq for some prime q ≥ 3 and some integer r ≥ 3. Then 2r ~ 2q ~ 2r−1q ~ 2r is a cycle of length 3 in Γ(ℤ2rq).
(3) Let n = 3rqs for some prime q ≥ 5 and some integers r ≥ 1 and s ≥ 2. Then 3r−1qs ~ 3rqs−1 ~ 2 · 3rqs−1 ~ 3r−1qs is a cycle of length 3 in Γ(ℤ3rqs).
(4) Let n = 3rq for some prime q ≥ 5 and some integer r ≥ 2. Then 3r ~ 3r−1q ~ 2 · 3r−1q ~ 3r is a cycle of length 3 in Γ(ℤ3rq).
(5) Let n = prqs for some primes q > p ≥ 5 and some integers r, s ≥ 1 except for r = s = 1. If r ≠ = 1 and s ≠ = 1, then prqs−1 ~ pr−1qs ~ 2pr−1qs ~ prqs−1 and prqs−1 ~ 2prqs−1 ~ 3prqs−1 ~ prqs−1 are cycles of length 3 in Γ(ℤprqs), respectively.
By Propositions 2.4, 2.5, and 2.6, we obtain
### Theorem 2.8
The following statements hold.
(1) g(Γ(ℤn)) = ∞ if and only if each of the following conditions holds.
(a) n = 4, 8, 9.
(b) n = 2q for some prime q ≥ 3.
(2) g(Γ(ℤn)) = 4 if and only if each of the following conditions holds.
(a) n = pq for some distinct primes p, q ≥ 3.
(b) n = 4q for some prime q ≥ 3.
(3) g(Γ(ℤn)) = 3 in all other cases.
3. The Chromatic Number of Γ(ℤn)
In this section, we calculate the chromatic number of Γ(ℤn). Clearly, if there exists a clique in a graph, then the chromatic number of the graph is greater than or equal to the size of the clique; so our method to find the chromatic number of Γ(ℤn) is based on the following three steps:
• Step 1. Find a maximal clique C in Γ(ℤn) and color vertices in C.
• Step 2. Color vertices in Z(ℤn) C by colors used in Step 1.
• Step 3. Confirm that there are no adjacent vertices having the same color.
### Lemma 3.1
If r ≥ 2 is an integer, n = p1 · · · pr for distinct primes p1, …, pr, and$C={npi∣i=1,…,r}$, then C is a maximal clique of Γ(ℤn).
Proof
Note that the product of any two distinct members of C is a multiple of n; so C is a clique. Suppose that there exists an element a ∈ Z(ℤn) C such that ca is a multiple of n for all cC. Then for all i = 1, …, r, pi divides a; so n divides a. This is a contradiction. Thus C is a maximal clique of Γ(ℤn).
### Theorem 3.2
If r ≥ 2 is an integer and n = p1 · · · pr for distinct primes p1, …, pr, then χ(Γ(ℤn)) = r.
Proof
Let $C={npi∣i=1,…,r}$. Then by Lemma 3.1, C is a maximal clique of Γ(ℤn); so the chromatic number of the induced subgraph of Γ(ℤn) induced by C is r. For each i = 1, …, r, let ī be the color of $npi$. Clearly, Z(ℤn) C is nonempty. For each a ∈ Z(ℤn) C, let Sa = {cC | a and c are not adjacent}. Note that by Lemma 3.1, C is a maximal clique; so Sa is a nonempty set. Hence we can find the smallest integer k ∈ {1, …, r} such that a and $npk$are not adjacent. In this case, we color a with .
To complete the proof, we need to check that any two elements in Z(ℤn) with the same color cannot be adjacent. Let a and b be distinct elements in Z(ℤn) with the same color . Since C is a clique, a and b cannot belong to C at the same time. Suppose that aC and b ∈ Z(ℤn) C. Then $a=npk$; so by the coloring of b, a and b are not adjacent. Suppose that a, b ∈ Z(ℤn) C. Then $npka$and $npkb$ are not divisible by n; so neither a nor b is divisible by pk. Therefore ab is not divisible by n, and hence a cannot be adjacent to b. Thus χ(Γ(ℤn)) = r.
### Example 3.3
Consider Γ(ℤ15). Let C = {3, 5}. Then by Theorem 3.2, we color 5 and 3 with 1̄ and 2̄, respectively. Let R = {6, 9, 12} and let vR. Then v is not adjacent to 3; so we color v with 2̄. Note that 10 is not adjacent to 5; so we color 5 with 1̄.
For the detail, see Figure 3. Note that in Figure 3, 1̄ and 2̄ are represented by blue and red, respectively.
### Lemma 3.4
Let$n=p12a1⋯pr2ar$for distinct primes p1, …, pr and positive integers a1, …, ar, and let$C={kn∣k=1,…,n-1}$. Then C is a clique of Γ(ℤn).
Proof
Note that the product of any two distinct elements of C is a multiple of n. Thus C is a clique.
### Theorem 3.5
Let$n=p12a1⋯pr2ar$for distinct primes p1, …, pr and positive integers a1, …, ar. Then$χ(Γ(ℤn))=n-1$.
Proof
Let $C={kn∣k=1,…,n-1}$. Then by Lemma 3.4, C is a clique of Γ(ℤn) with $n-1$ elements. For each $k∈{1,…,n-1}$, let denote the color of $kn$.
Case 1
$n=p12$. In this case, Γ(ℤn) is a complete graph by Theorem 2.1. Hence Z(ℤn) = C. Thus $χ(Γ(ℤn))=n-1$
Case 2
$n≠p12$. In this case, Γ(ℤn) is not a complete graph by Theorem 2.1; so Z(ℤn) C ≠ = ∅︀. Let v ∈ Z(ℤn) C. Then there exists an element m ∈ {1, …, r} such that $pmam$ does not divide v. Take the positive integer sam such that $pms-1$ divides v but $pms$ does not divide v. Then $v((n/pms)n)$ is not a multiple of n; so v and $(n/pms)n$ are not adjacent. We color v with $n/pms$
Now, it remains to check that any two vertices with the same color cannot be adjacent. Let v1 and v2 be distinct elements of Z(ℤn) which have the same color. Since C is a clique, v1 and v2 cannot both belong to C. Suppose that v1C and v2 ∈ Z(ℤn) C. Let $n/pms^$ be the color of v2 for some m ∈ {1, …, r} and s ∈ {1, …, am}. Then v2 is not divisible by $pms$ by the coloring of v2. Since v1C, $v1=(n/pms)n$. Therefore $pm2am$ does not divide v1v2. Hence v1 is not adjacent to v2. Suppose that v1, v2 ∈ Z(ℤn) C, and let $n/pms^$ be the color of v1 and v2 for some m ∈ {1, …, r} and s ∈ {1, …, am}. Then by the coloring of v1 and v2, $pms$ divides neither v1 nor v2; so $pm2s$ does not divide v1v2. Since sam, $pm2am$ does not divide v1v2. Hence v1v2 is not a multiple of n, which means that v1 and v2 are not adjacent.
In either case, Γ(ℤn) is ($(n-1)-colorable$. Note that by Lemma 3.4, $χ(Γ(ℤn))≥n-1$. Thus $χ(Γ(ℤn))=n-1$.
### Example 3.6
Consider Γ(ℤ36). Let C = {6, 12, 18, 24, 30}. Then by Theorem 3.5, we color 6, 12, 18, 24 and 30 with 1̂, 2̂, 3̂, 4̂ and 5̂, respectively. Let R = {2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 32, 34} and let vR. Then v is not divisible by 3; so we color v with 2̂. Let B = {3, 9, 15, 21, 27, 33} and let vB. Then v is not divisible by 2; so we color v with 3̂.
For the detail, see Figure 4. Note that in Figure 4, 1̂, 2̂, 3̂, 4̂ and 5̂ are represented by green, red, blue, pink and brown, respectively.
### Lemma 3.7
Let p1, … , pr, q1, … , qs be distinct primes, a1, … , ar, b1, … , bs nonnegative integers, not all zero, and let $n=p12a1⋯pr2arq12b1+1⋯qs2bs+1$. If$C1={kp1a1⋯prarq1b1+1⋯qsbs+1∣k=1,…,p1a1⋯prarq1b1⋯qsbs-1}$and$C2={p1a1⋯prarq1b1+1⋯qsbs+1/qi∣i=1,…,s}$, then C1 ∪ C2is a maximal clique of Γ(ℤn).
### Proof
We first note that C1C2 = ∅︀. Let C = C1C2. Then for any distinct elements α, βC, αβ is a multiple of n; so C is a clique. Suppose that C is not a maximal clique. Then there exists an element m ∈ Z(ℤn) C such that mc = 0 for all cC. Therefore m is a multiple of $p1a1⋯prarq1b1⋯qi-1bi-1qibi+1qi+1bi+1⋯qsbs$ for all i = 1, …, s. Hence m is a multiple of $p1a1⋯prarq1b1+1⋯qsbs+1$, which implies that mC1. This contradicts the choice of m. Thus C is a maximal clique of Γ(ℤn).
### Theorem 3.8
Let p1, …, pr, q1, …, qs be distinct primes and a1, …, ar, b1, …, bs nonnegative integers, not all zero. If$n=p12a1⋯pr2arq12b1+1⋯qs2bs+1$, then$χ(Γ(ℤn))=p1a1⋯prarq1b1⋯qsbs-1+s$.
### Proof
Let $x=p1a1⋯prarq1b1⋯qsbs$ and $y=p1a1⋯prarq1b1+1⋯qsbs+1$. Then n = xy. Let C1 = {ky | k = 1, …, x−1}, C2 = {y/qi | i = 1, …, s}, and C = C1C2. Then by Lemma 3.7, C is a maximal clique of Γ(ℤn). For each k = 1, …, x − 1, let be the color of ky and for each i = 1, …, s, let ī be the color of y/qi. Note that by Theorem 2.1, Γ(ℤn) is not a complete graph. Let v ∈ Z(ℤn) C.
Case 1
There exists an element cC1 which is not adjacent to v. In this case, v is not divisible by x. If $q1b1⋯qsbs$ divides v, then $pmam$ does not divide v for some m ∈ {1, …, r}; so we can take the positive integer αam such that $pmα-1$ divides v but $pmα$ does not divide v. Hence v is not adjacent to $(x/pmα)y$. We color v with $x/pmα^$. If $qtbt$ does not divide v for some t ∈ {1, …, s}, then we can find the positive integer βbt such that $qtβ-1$ divides v but $qtβ$ does not divide v. Hence v is not adjacent to $(x/qtβ)y$. We color v with $x/qtβ^$.
Case 2
v is adjacent to c for all cC1. In this case, v is a multiple of x. If $q1b1+1⋯qsbs+1$ divides v, then vC1, which is a contradiction to the choice of v. Therefore we can find an element i ∈ {1, …, s} such that $qibi$ divides v but $qibi+1$ does not divide v. Clearly, v(y/qi) is not a multiple of n. Hence v and y/qi are not adjacent. We color v with ī.
It remains to show that there are no adjacent vertices with the same color. Let v1, v2 ∈ Z(ℤn) have the same color. Since C is a clique, at least one of v1 and v2 does not belong to C. Suppose that v1C but v2 ∈ Z(ℤn) C. If the color of v1 and v2 is $x/pmα^$ for some m ∈ {1, …, r} and α ∈ {1, …, am}, then by the coloring of v1 and v2, $v1=(x/pmα)y$ and $pmα$ does not divide v2; so v1v2 is not a multiple of n. Hence v1 is not adjacent to v2. If the color of v1 and v2 is $x/qtβ^$ for some t ∈ {1, …, s} and β ∈ {1, …, bt}, then by the coloring of v1 and v2, $v1=(x/qtβ)y$ and $qtβ$ does not divide v2; so v1v2 is not a multiple of n. Hence v1 is not adjacent to v2. If ī is the color of v1 and v2 for some i ∈ {1, …, s}, then v1 = y/qi and so by the coloring of v2, v1 and v2 are not adjacent. We next suppose that v1, v2 ∈ Z(ℤn) C. If $x/pmα^$ is the color of v1 and v2 for some m ∈ {1, …, r} and α ∈ {1, …, am}, then by Case 1, $pmα$ divides neither v1 nor v2; so $pm2am$ does not divide v1v2. Hence v1 and v2 are not adjacent. If $x/qtβ^$ is the color of v1 and v2 for some t ∈ {1, …, s} and β ∈ {1, …, bt}, then by Case 1, $qtβ$ divides neither v1 nor v2; so $qt2bt$ does not divide v1v2. Hence v1 and v2 are not adjacent. If ī is the color of v1 and v2 for some i ∈ {1, …, s}, then by Case 2, $qibi+1$ divides neither v1 nor v2. Hence $qi2bi+1$ cannot divide v1v2, which means that v1 and v2 are not adjacent. Consequently, Γ(ℤn) is ($(p1a1⋯prarq1b1⋯qsbs-1+s)-colorable$. Note that by Lemma 3.7, $χ(Γ(ℤn))≥p1a1⋯prarq1b1⋯qsbs-1+s$. Thus $χ(Γ(ℤn))=p1a1⋯prarq1b1⋯qsbs-1+s$.
### Example 3.9
Consider Γ(ℤ18). Let C1 = {6, 12} and C2 = {3}. Then by Theorem 3.8, we color 6, 12 and 3 with 1̂, 2̂ and 1̄, respectively. Let R = {2, 4, 8, 10, 14, 16} and let vR. Then v is not adjacent to 6. Note that v is not divisible by 3; so we color v with 1̂. Let B = {9, 15} and let vB. Then v is adjacent to all elements in C1. Note that v is not divisible by 2; so we color v with 1̄.
For the detail, see Figure 5. Note that in Figure 5, 1̂, 2̂ and 1̄ are represented by red, green and blue, respectively.
### Example 3.10
Consider Γ(ℤ72). Let C1 = {12, 24, 36, 48, 60} and C2 = {6}. Then by Theorem 3.6, we color 6, 12, 24, 36, 48 and 60 with 1̄, 1̂, 2̂, 3̂, 4̂ and 5̂, respectively. Let P = {2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 32, 34, 38, 40, 44, 46, 50, 52, 56, 58, 62, 64, 68, 70} and let vP. Then v is not adjacent to 12. Note that v is a multiple of 2 but not a multiple of 3; so we color v with 2̂. Let B = {3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69} and let vB. Then v is not adjacent to 12. Note that v is not divisible by 2; so we color v with 3̂. Let Y = {18, 30, 42, 54, 66} and let vY. Then v is adjacent to all elements in C1. Since v and 6 are not adjacent, we color v with 1̄.
For the detail, see Figure 6. Note that in Figure 6, 1̄, 1̂, 2̂, 3̂, 4̂ and 5̂ are represented by yellow, red, pink, blue, sky-blue and green, respectively.
Acknowledgements
The authors would like to thank the referee for his/her several valuable suggestions. The third author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2017R1C1B1008085).
Figures
Fig. 1. The diameters of some zero-divisor graphs
Fig. 2. The girth of some zero-divisor graphs
Fig. 3. The coloring of Γ(ℤ15)
Fig. 4. The coloring of Γ(ℤ36)
Fig. 5. The coloring of Γ(ℤ18)
Fig. 6. The coloring of Γ(ℤ72)
References
1. DF. Anderson, MC. Axtell, and JA. Stickles. Zero-divisor graphs in commutative rings. Commutative Algebra: Noetherian and Non-Noetherian Perspectives, , Springer, New York, 2011:23-45.
2. DF. Anderson, and PS. Livingston. The zero-divisor graph of a commutative ring. J Algebra., 217(1999), 434-447.
3. DD. Anderson, and M. Naseer. Beck’s coloring of a commutative ring. J Algebra., 159(1993), 500-514.
4. I. Beck. Coloring of commutative rings. J Algebra., 116(1988), 208-226.
5. S. Mulay. Cycles and symmetries of zero-divisors. Comm Algebra., 30(2002), 3533-3558. | 9,032 | 23,661 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 79, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.831988 |
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Week1 P2-17 P2-19 P2-21 P2-24
P2-17 Echeverria SA
P2-19 Sebolt Wire Company
P2-21 Golden Company’s
P2-24 Heritage Company
PROBLEM 2-17 High-Low Method; Predicting Cost [LO3,LO4]
P2-17 Echeverria SA is an Argentinian manufacturing company whose total factory overhead costs fluctuate somewhat from year to year according to the number of machine-hours worked in its production facility. These costs (in Argentinian pesos) at high and low levels of activity over recent years are given below:
The factory overhead costs above consist of indirect materials, rent, and maintenance. The company has analyzed these costs at the 60,000 machine-hours level of activity as follows:
For planning purposes, the company wants to break down the maintenance cost into its variable and fixed cost elements.
Required:
1. Estimate how much of the factory overhead cost of 312,000 pesos at the high level of activity consists of maintenance cost. (Hint: To do this, it may be helpful to first determine how much of the 312,000 pesos cost consists of indirect materials and rent.
Think about the behavior of variable and fixed costs.)
2. Using the high-low method, estimate a cost formula for maintenance.
3. What total overhead costs would you expect the company to incur at an operating level of 65,000 machine-hours?
P2-19 Sebolt Wire Company heats copper ingots to very high temperatures by placing the ingots in a large heat coil.
P2-19 High-Low and Scattergraph Analysis [LO4]
P2-19 Sebolt Wire Company heats copper ingots to very high temperatures by placing the ingots in a large heat coil. The heated ingots are then run through a shaping machine that shapes the soft ingot into wire. Due to the long heat-up time, the coil is never turned off. When an ingot is placed in the coil, the temperature is raised to an even higher level, and then the coil is allowed to drop to the “waiting” temperature between ingots. Management needs to know the variable cost of power involved in heating an ingot and the fixed cost of power during “waiting” periods. The following data on ingots processed and power costs are available:
Month Number of ingots Power cost January 110 \$5,500 February 90 \$4,500 March 80 \$4,400 April 100 \$5,500 May 130 \$6,000 June 120 \$5,600 July 70 \$4,000 August 60 \$3,200 September 50 \$3,400 October 40 \$2,400
Required:
1. Using the high-low method, estimate a cost formula for power cost. Express the formula in the form bX.
2. Prepare a scattergraph by plotting ingots processed and power cost on a graph. Draw a straight line though the two data points that correspond to the high and low levels of activity. Make sure your line intersects the Y-axis.
3. Comment on the accuracy of your high-low estimates assuming a least-squares regression analysis estimated the total fixed costs to be \$1,185.45 per month and the variable cost to be \$37.82 per ingot. How would the straight line that you drew in requirement 2 differ from a straight line that minimizes the sum of the squared errors?
PROBLEM 2-21 High-Low Method; Predicting Cost [LO3,LO4]
P2-21 Golden Company’s total overhead cost at various levels of activity are presented below:
Assume that the overhead cost above consists of utilities, supervisory salaries, and maintenance.
The breakdown of these costs at the 40,000 machine-hour level of activity is as follows:
The company wants to break down the maintenance cost into its variable and fixed cost elements.
Required:
1. Estimate how much of the \$241,600 of overhead cost in June was maintenance cost. (Hint: To do this, it may be helpful to first determine how much of the \$241,600 consisted of utilities and supervisory salaries. Think about the behavior of variable and fixed costs within the relevant range.)
2. Using the high-low method, estimate a cost formula for maintenance.
3. Express the company’s total overhead cost in the form Y = a + bX.
4. What total overhead cost would you expect to be incurred at an activity level of 45,000 machine-hours?
PROBLEM 2-24 Cost Classification and Cost Behavior [LO2,LO3LO6]
P2-24 Heritage Company manufactures a beautiful bookcase that enjoys widespread popularity. The company has a backlog of orders that is large enough to keep production going indefinitely at the plant’s full capacity of 4,000 bookcases per year. Annual cost data at full capacity follow:
Direct materials used (wood and glass) \$430,000 Administrative office salaries \$110,000 Factory supervision \$70,000 Sales commissions \$60,000 Depreciation, factory building \$105,000 Depreciation, administrative office equipment \$2,000 Indirect materials, factory \$18,000 Advertising \$100,000 Insurance, factory \$6,000 Administrative office supplies(billing) \$4,000 Property taxes, factory \$20,000 Utilities, factory \$45,000
Required:
1. Prepare an answer sheet with the column headings shown below. Enter each cost item on your answer sheet, placing the dollar amount under the appropriate headings. As examples, this has been done already for the first two items in the list above. Note that each cost item is classified in two ways: first, as either variable or fixed with respect to the number of units produced and sold; and second, as either a selling and administrative cost or a product cost. (If the item is a product cost, it should also be classified as either direct or indirect as shown.)
Cost behavior Selling for administrative cost Product cost Cost item Variable Fixed Direct Indirect* Materials used \$430,000 \$430,000 Administrative office Salaries \$110,000 \$110,000
*To units of product
2. Total the dollar amounts in each of the columns in (1) above. Compute the average product cost per bookcase.
3. Due to a recession, assume that production drops to only 2,000 bookcases per year.
Would you expect the average product cost per bookcase to increase, decrease, or remain unchanged? Explain. No computations are necessary.
4. Refer to the original data. The president’s next-door neighbor has considered making himself a bookcase and has priced the necessary materials at a building supply store. He has asked the president whether he could purchase a bookcase from the Heritage Company “at cost,” and the president has agreed to let him do so.
a. Would you expect any disagreement between the two men over the price the neighbor should pay? Explain. What price does the president probably have in mind? The neighbor?
b. Because the company is operating at full capacity, what cost term used in the chapter might be justification for the president to charge the full, regular price to the neighbor and still be selling “at cost”? Explain.
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No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 1,975 | 8,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.930757 |
https://keeponmind.com/2021/09/24/fbi-test-that-proves-you-are-a-top-1-mother-whos-the-real-mother/ | 1,725,992,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00018.warc.gz | 322,468,642 | 17,943 | # ‘FBI Test’ That Proves You Are A Top 1% Mother: Who’s the Real Mother?
Do you have what it takes to become a member of the FBI? The FBI has a stringent testing procedure with a 30% pass rate. Their first exam is divided into five components. One assesses logical reasoning, another assesses figural reasoning, another assesses situational judgment, another assesses character, and still another assesses preference and interests.
Here are five riddles that test the first portion of logical thinking, which requires attention to detail, if you enjoy taking these types of exams for fun. Are you able to solve these puzzles? Can you figure out who the true mother is or whether the suicide was actually a murder?
## Test 1. Who is the Real Mother?
Here’s an example of the type of question that may appear on an FBI test. It necessitates both logical and imaginative reasoning. So have a look at this image for a few seconds. Two ladies are seated across from one other, with a kid playing on the floor between them. Who do you believe is the true mother, and why do you believe that?
Answer: The actual mother, dressed in a green gown, is seated on the left. She appears to be the mother based on two clues. Her posture is one of them. In a defensive position, she’s slightly bent forward. Two, the youngster is standing in front of her. When playing or completing an activity, children frequently face their parents since they instinctively want their attention.
## Test 2. Poisoned Drinks
Miranda and Jules shared a drink in a pub. They each ordered the same beverage. Jules was extremely thirsty, and in the time it took Miranda to complete one, he finished five. The beverages were poisoned, but Miranda was the only one who died. How?
Answer: The ice was poisoned, not the drink. Miranda’s ice had melted and seeped into the drink. Jules, on the other hand, drank hers before the ice melted, saving her from the poison.
## Test 3. Suicide or Murder?
A dead body was recovered at the bottom of a tall building by a witness. The individual appears to have committed suicide by jumping from one of the upper stories. When the detective arrives, he proceeds to the first floor to investigate the shuttered window that looks out over the body. He throws a penny out the window and onto the street. This is something he does on every story of the building. He claims that this was a murder, not a suicide, when he returns to the main level.
How did he arrive at such a conclusion?
To toss his coin, he had to open the window on every floor. The window would have remained open if the person had committed suicide. The window has to be closed by someone else.
## Test 4. Who Stole Her Mother’s Ring?
Georgina reported a robbery to the police. Someone had smashed her window and taken an antique diamond that had belonged to her mother, she said. When the cops came, they discovered a shattered window with glass strewn across the lawn and muddy tracks strewn across the carpeting. Georgina was arrested for fraud when the police surveyed the scene. Why is that?
When the cops arrived at the murder scene, they observed that the window glass was all over the lawn, indicating that it had been shattered from the inside rather than the outside, as Georgina claimed, implying that she had set up the scene herself.
## Test 5. Suicide or Murder 2.0?
In his office, a guy was discovered deceased. In one hand, he clutched a pistol, and in the other, a tape recorder. To hear the message, the cops hit play on the recording. They overheard the man remark, “I can’t go any farther.” “I have no reason to live,” he said, followed by a gunshot. The recording was switched off, and the case was deemed a murder, not a suicide, by the police. How did they figure it out?
Answer: If the guy had died as a result of the shooting, he would not have been able to stop the recording and rewind it for the police to listen to. It needed to be rewinded by someone else. | 879 | 3,969 | {"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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.972348 |
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3015_Discussion_13_Solutions
# 3015_Discussion_13_Solutions - University of Minnesota Dept...
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University of Minnesota Dept. of Electrical and Computer Engineering EE3015 – Signals and Systems Discussion Session #13: z -Transform 1. Determine the z -Transform and the region of convergence of the following signal: | 1 | 3 1 ] [ = n n x First find the z -Transform of | | 3 1 ] [ n n y = and use the shifting property to obtain the z - Transform of ] [ n x . Signal ] [ n y can be written as: ] 1 [ 3 ] [ 3 1 ] 1 [ 3 1 ] [ 3 1 ] [ + = + = n u n u n u n u n y n n n n This is the sum of a right-sided and left-sided signal. The z -Transform is the sum of the individual transforms: 3 : , 3 1 1 ] 1 [ 3 3 1 : , 3 1 1 1 ] [ 3 1 1 1 < ⎯→ > ⎯→ z ROC z n u z ROC z n u z n z n Thus, 3 3 1 : , 3 1 1 3 1 1 1 ) ( 1 1 < < = z ROC z z z Y Using properties: Re{ z } Im{ z } 3 1/3
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3 3 1 , 3 1 3 1 1 ) ( ) ( ] 1 [ ] [ 3 3 1 ), ( ] [ 1 1 1 1 1 < < = = ⎯→ = < < ⎯→ z z z z z z Y z z X n y n x z z Y n y z z The term 1 z that appears after shifting does not affect the ROC. 2. A discrete-time causal system is described by the difference equation: ] 1 [ 4 5 ] [ 2 ] 2 [ 8 1 ] 1 [ 4 1 ] [ + = + n x n x n y n y n y a) Find the transfer function ) ( z H and its region of convergence. Is this system stable?
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{[ snackBarMessage ]} | 575 | 1,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.772973 |
http://math.stackexchange.com/questions/165901/what-will-be-the-rate-of-inflation-after-n-months | 1,469,605,562,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826736.89/warc/CC-MAIN-20160723071026-00184-ip-10-185-27-174.ec2.internal.warc.gz | 177,268,097 | 17,969 | # What will be the rate of inflation after n months?
Let's say I have a utility rate of 0.15 which inflates by 0.5% per year. I then want to convert that annual rate to a monthly rate and determine what the rate would be for month $n$.
The way I'm doing it now (which just feels incorrect) is: $$\mathrm{utilityRate}\Bigl( (1.0 + \text{rate of inflation})^{n/12}\Bigr)$$
Is this correct, or am I completely off base?
-
A rate of $n/12$ compounded monthly gives you a larger rate than $n$ per year. – Arturo Magidin Jul 2 '12 at 22:26
The number of months could be anything... in this particular case, I need to know the rate of month n where n = 1...300 – Chris Cashwell Jul 2 '12 at 22:29
@Arturo: I though $n$ was the number of months rather than the rate. – Henry Jul 2 '12 at 22:33
@Hnery; I started the comment before cleaning up the LaTeX, and misinterpreted; then I didn't correct the comment before sending. – Arturo Magidin Jul 2 '12 at 22:34
If the rate of inflation is constant every month then $(1 + \text{annual rate of inflation})^{n/12}$ is indeed what happens to the price level starting at $1$. So your formula looks sensible.
So in your example with $0.5\%$ annual inflation, this becomes $1.005^{n/12}$. The utilities I know would only change rates once a year if inflation was so low.
@ChrisCashwell: it will be close, particularly if the inflation rate is low. The correct monthly rate is as you implied in your question: $(1.0 + \text{annual rate of inflation})^{1/12}-1$. To the extent that $(1+\text{monthly rate})^{12}\approx 1+12*\text{monthly rate}$ they will agree.The next term is $6(\text{monthly rate})^2$, so you can see if you think that is big enough to care about. – Ross Millikan Jul 2 '12 at 23:54 | 500 | 1,739 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-30 | latest | en | 0.915995 |
http://intranet.math.vt.edu/netmaps/hurwitz/m14n2/m14n2hurwitz357_Main.output | 1,582,504,282,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00442.warc.gz | 71,975,985 | 2,471 | These Thurston maps are NET maps for every choice of translation term. They have degree 28. They are imprimitive, each factoring as a Thurston map with degree 2 followed by a Euclidean NET map with degree 14. The non-Euclidean factor is a NET map if it has four postcritical points. PURE MODULAR GROUP HURWITZ EQUIVALENCE CLASSES FOR TRANSLATIONS {0,lambda1} {lambda2} {lambda1+lambda2} Since no Thurston multiplier is 1, this modular group Hurwitz class contains only finitely many Thurston equivalence classes. The number of pure modular group Hurwitz classes in this modular group Hurwitz class is 14. ALL THURSTON MULTIPLIERS c/d IN UNREDUCED FORM 0/2, 0/14, 1/14, 2/14, 7/2, 14/2 EXCLUDED INTERVALS FOR THE HALF-SPACE COMPUTATION (-infinity,-0.011826) ( 0.011903,infinity ) 1/0 is the slope of a Thurston obstruction with c = 7 and d = 2. These NET maps are not rational. SLOPE FUNCTION INFORMATION NUMBER OF FIXED POINTS: 2 EQUATOR? FIXED POINT c d 0 lambda1 lambda2 lambda1+lambda2 1/0 7 2 No No No No -1/3 1 14 No No No No NUMBER OF EQUATORS: 0 0 0 0 There are no more slope function fixed points. Number of excluded intervals computed by the fixed point finder: 520 There are no equators because both elementary divisors are greater than 1. No nontrivial cycles were found. Here is the action of the slope function on a set S of slopes. 4N/1 -> *, (4N+2)/1 -> (14N+8)/1 Every slope in S either maps to a slope in S or (*) it leaves S, its orbit never to return. The set S contains arbitrarily long segments of slope function trajectories, but it probably contains no infinite slope function trajectories. Here is the action of the slope function on an invariant set S of slopes. (4N+1)/1 -> (14N+5)/1, (4N+3)/1 -> (14N+11)/1 The set S contains infinitely many infinite slope function trajectory tails. Here is the action of the slope function on a set S of slopes. (8N+1)/2 -> *, (8N+5)/2 -> (28N+19)/2, (8N+3)/2 -> (28N+13)/2, (8N+7)/2 -> * Every slope in S either maps to a slope in S or (*) it leaves S, its orbit never to return. The set S contains arbitrarily long segments of slope function trajectories, but it probably contains no infinite slope function trajectories. Here is the action of the slope function on a set S of slopes. (12N+1)/3 -> (42N+7)/3, (12N+7)/3 -> * Every slope in S either maps to a slope in S or (*) it leaves S, its orbit never to return. The set S contains arbitrarily long segments of slope function trajectories, but it probably contains no infinite slope function trajectories. Here is the action of the slope function on a set S of slopes. (12N+5)/3 -> *, (12N+11)/3 -> (42N+41)/3 Every slope in S either maps to a slope in S or (*) it leaves S, its orbit never to return. The set S contains arbitrarily long segments of slope function trajectories, but it probably contains no infinite slope function trajectories. Here is the action of the slope function on a set S of slopes. (16N+1)/4 -> (56N+7)/4, (16N+9)/4 -> *, (16N+7)/4 -> *, (16N+15)/4 -> (56N+57)/4 Every slope in S either maps to a slope in S or (*) it leaves S, its orbit never to return. The set S contains arbitrarily long segments of slope function trajectories, but it probably contains no infinite slope function trajectories. The slope function maps some slope to the nonslope. If the slope function maps slope s to a slope s' and if the intersection pairing of s with 1/0 is n, then the intersection pairing of s' with 1/0 is at most n. The slope function orbit of every slope whose intersection pairing pairing with 1/0 is at most 50 either ends in the nonslope or ends in one of the slopes described above or it has an infinite tail in one of the infinite sets described above. FUNDAMENTAL GROUP WREATH RECURSIONS When the translation term of the affine map is 0: NewSphereMachine( "a=(2,26)(3,27)(4,24)(5,25)(6,22)(7,23)(8,20)(9,21)(10,18)(11,19)(12,16)(13,17)", "b=(1,27)(2,28)(3,25)(4,26)(5,23)(6,24)(7,21)(8,22)(9,19)(10,20)(11,17)(12,18)(13,15)(14,16)", "c=<1,1,1,1,1,1,1,1,c,c^-1,c,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(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)", "d=(1,28)(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)", "a*b*c*d"); When the translation term of the affine map is lambda1: NewSphereMachine( "a=(1,3)(2,28)(4,26)(5,27)(6,24)(7,25)(8,22)(9,23)(10,20)(11,21)(12,18)(13,19)(14,16)(15,17)", "b=(3,27)(4,28)(5,25)(6,26)(7,23)(8,24)(9,21)(10,22)(11,19)(12,20)(13,17)(14,18)", "c=<1,1,1,1,1,1,1,1,1,1,c,c^-1,c,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(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)", "d=(1,28)(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)", "a*b*c*d"); When the translation term of the affine map is lambda2: NewSphereMachine( "a=<1,a*b,1,1,1,1,1,1,c,1,1,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,c*d,1>(1,4)(2,27)(3,6)(5,8)(7,10)(9,12)(11,14)(13,16)(15,18)(17,20)(19,22)(21,24)(23,26)(25,28)", "b=<1,1,1,1,1,1,1,1,c,c^-1,c,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(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)", "c=(1,27)(2,28)(3,25)(4,26)(5,23)(6,24)(7,21)(8,22)(9,19)(10,20)(11,17)(12,18)(13,15)(14,16)", "d=(1,25)(3,23)(4,28)(5,21)(6,26)(7,19)(8,24)(9,17)(10,22)(11,15)(12,20)(14,18)", "a*b*c*d"); When the translation term of the affine map is lambda1+lambda2: NewSphereMachine( "a=<1,a*b,1,1,1,1,1,1,1,1,c,1,1,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,c*d,1>(1,4)(2,27)(3,6)(5,8)(7,10)(9,12)(11,14)(13,16)(15,18)(17,20)(19,22)(21,24)(23,26)(25,28)", "b=<1,1,1,1,1,1,1,1,1,1,c,c^-1,c,c^-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(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)", "c=(3,27)(4,28)(5,25)(6,26)(7,23)(8,24)(9,21)(10,22)(11,19)(12,20)(13,17)(14,18)", "d=(1,27)(2,4)(3,25)(5,23)(6,28)(7,21)(8,26)(9,19)(10,24)(11,17)(12,22)(13,15)(14,20)(16,18)", "a*b*c*d"); ****************************INTEGER OVERFLOW REPORT***************************** Imminent integer overflow caused the computation of pullback map symmetries to abort. | 2,384 | 6,094 | {"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} | 2.703125 | 3 | CC-MAIN-2020-10 | latest | en | 0.846921 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=TDW | 1,490,491,608,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189090.69/warc/CC-MAIN-20170322212949-00602-ip-10-233-31-227.ec2.internal.warc.gz | 526,094,493 | 119,867 | Back to list of Stocks See Also: Seasonal Analysis of TDWGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
# Fourier Analysis of TDW (Tidewater Inc. Common Stock)
TDW (Tidewater Inc. Common Stock) appears to have interesting cyclic behaviour every 66 weeks (1.2566*sine), 139 weeks (1.2172*sine), and 46 weeks (1.1225*cosine).
TDW (Tidewater Inc. Common Stock) has an average price of 26.01 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Right click on the graph above to see the menu of operations (download, full screen, etc.)
## Fourier Analysis
Using data from 12/30/1987 to 3/20/2017 for TDW (Tidewater Inc. Common Stock), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
026.00543 0
1-6.33261 -13.6853 (1*2π)/15251,525 weeks
2-8.11584 -5.9113 (2*2π)/1525763 weeks
3-1.2917 -1.43976 (3*2π)/1525508 weeks
4-5.06629 -.09233 (4*2π)/1525381 weeks
5-3.96186 3.33094 (5*2π)/1525305 weeks
61.08197 -.58411 (6*2π)/1525254 weeks
7-.88037 2.73388 (7*2π)/1525218 weeks
8-1.82148 .88239 (8*2π)/1525191 weeks
92.40194 .35888 (9*2π)/1525169 weeks
101.10284 .03903 (10*2π)/1525153 weeks
11.15812 -1.21721 (11*2π)/1525139 weeks
12.39905 .4205 (12*2π)/1525127 weeks
13-.42989 -.34259 (13*2π)/1525117 weeks
14-.73885 -.01532 (14*2π)/1525109 weeks
15.46669 .87513 (15*2π)/1525102 weeks
16.09363 .10948 (16*2π)/152595 weeks
17-.36458 -.74914 (17*2π)/152590 weeks
18.77913 .48513 (18*2π)/152585 weeks
19-.78899 .19812 (19*2π)/152580 weeks
20.4303 -.64746 (20*2π)/152576 weeks
21.00565 .08418 (21*2π)/152573 weeks
22-.86177 -.32795 (22*2π)/152569 weeks
23.46991 1.25659 (23*2π)/152566 weeks
241.05815 -.67595 (24*2π)/152564 weeks
25-.63356 -.45693 (25*2π)/152561 weeks
26-.45911 .04768 (26*2π)/152559 weeks
27.28024 .11035 (27*2π)/152556 weeks
28-1.0575 .15909 (28*2π)/152554 weeks
29.27344 .81087 (29*2π)/152553 weeks
30.34276 -.27837 (30*2π)/152551 weeks
31-.08325 -.05114 (31*2π)/152549 weeks
32-.33965 .30136 (32*2π)/152548 weeks
331.12247 .91733 (33*2π)/152546 weeks
34.10231 -.69641 (34*2π)/152545 weeks
35.07544 -.40436 (35*2π)/152544 weeks
36-.11905 .25799 (36*2π)/152542 weeks
37-.09736 -.1437 (37*2π)/152541 weeks
38-.47273 -.37074 (38*2π)/152540 weeks
39.32687 .19548 (39*2π)/152539 weeks
40-.2653 .25439 (40*2π)/152538 weeks
41-.15347 -.34111 (41*2π)/152537 weeks
42.13609 .10341 (42*2π)/152536 weeks
43.07309 -.09051 (43*2π)/152535 weeks
44-.21343 .00339 (44*2π)/152535 weeks
45.49316 .03879 (45*2π)/152534 weeks
46-.05477 .02335 (46*2π)/152533 weeks
47-.22404 -.30997 (47*2π)/152532 weeks
48.0177 -.11072 (48*2π)/152532 weeks
49-.41843 .20242 (49*2π)/152531 weeks
50-.31141 .37203 (50*2π)/152531 weeks
51.45578 .15256 (51*2π)/152530 weeks
52-.30424 .09944 (52*2π)/152529 weeks
53.35594 -.10195 (53*2π)/152529 weeks
54-.01768 -.28055 (54*2π)/152528 weeks
55-.11384 .09515 (55*2π)/152528 weeks
56.05631 .00689 (56*2π)/152527 weeks
57.03137 .33266 (57*2π)/152527 weeks
58-.27082 -.06349 (58*2π)/152526 weeks
59.01102 -.17283 (59*2π)/152526 weeks
60-.15341 .21387 (60*2π)/152525 weeks
61-.07328 .03068 (61*2π)/152525 weeks
62.10337 .32685 (62*2π)/152525 weeks
63.37902 -.00333 (63*2π)/152524 weeks
64.06627 -.18953 (64*2π)/152524 weeks
65-.0155 -.31437 (65*2π)/152523 weeks
66-.11051 .03047 (66*2π)/152523 weeks
67-.11536 .21153 (67*2π)/152523 weeks
68.08803 -.08757 (68*2π)/152522 weeks
69-.07243 -.00038 (69*2π)/152522 weeks
70-.17476 -.13422 (70*2π)/152522 weeks
71.0994 -.06423 (71*2π)/152521 weeks
72-.10307 -.01649 (72*2π)/152521 weeks
73-.08294 -.06764 (73*2π)/152521 weeks
74-.23906 .17121 (74*2π)/152521 weeks
75-.00976 .07073 (75*2π)/152520 weeks
76.13918 .23368 (76*2π)/152520 weeks
77-.08582 -.02322 (77*2π)/152520 weeks
78.16638 .19233 (78*2π)/152520 weeks
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528-.00272 .03067 (528*2π)/15253 weeks
529-.05984 .01189 (529*2π)/15253 weeks
530.05658 -.0005 (530*2π)/15253 weeks
531.00837 -.01607 (531*2π)/15253 weeks
532-.00989 .00504 (532*2π)/15253 weeks
533.01342 .04015 (533*2π)/15253 weeks< | 9,393 | 21,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-13 | longest | en | 0.815629 |
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1+eps-Approximate Sparse Recovery Eric Price MIT David Woodruff IBM Almaden
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1+eps-Approximate Sparse Recovery. Eric Price MIT. David Woodruff IBM Almaden. Compressed Sensing. Choose an r x n matrix A Given x 2 R n Compute Ax Output a vector y so that |x-y| p · (1+ ε ) |x-x top k | p x top k is the k-sparse vector of largest magnitude coefficients of x - PowerPoint PPT Presentation
### Transcript of 1+eps-Approximate Sparse Recovery
1+eps-Approximate Sparse Recovery
Eric PriceMIT
Compressed Sensing• Choose an r x n matrix A• Given x 2 Rn
• Compute Ax• Output a vector y so that
|x-y|p · (1+ε) |x-xtop k|p
• xtop k is the k-sparse vector of largest magnitude coefficients of x
• p = 1 or p = 2• Minimize number r = r(n, k, ε) of “measurements”
PrA[ ] > 2/3
Previous Work
• p = 1[IR, …] r = O(k log(n/k) / ε) (deterministic A)
• p = 2[GLPS] r = O(k log(n/k) / ε)
In both cases, r = (k log(n/k)) [DIPW]What is the dependence on ε?
Why 1+ε is Important
• Suppose x = ei + u– ei = (0, 0, …, 0, 1, 0, …, 0)
– u is a random unit vector orthogonal to ei
• Consider y = 0n
– |x-y|2 = |x|2 · 21/2 ¢ |x-ei|2It’s a trivial solution!
• (1+ε)-approximate recovery fixes this
In some applications, can have 1/ε = 100, log n = 32
Our Results Vs. Previous Work• p = 1[IR, …] r = O(k log(n/k) / ε) r = O(k log(n/k) ¢ log2(1/ ε) / ε1/2) (randomized) r = (k log(1/ε) / ε1/2)
• p = 2:[GLPS] r = O(k log(n/k) / ε) r = (k log(n/k) / ε)
Previous lower bounds (k log(n/k))Lower bounds for randomized constant probability
Comparison to Deterministic Schemes
• We get r = O~(k/ε1/2) randomized upper bound for p = 1
• We show (k log (n/k) /ε) for p = 1 for deterministic schemes
• So randomized easier than deterministic
Our Sparse-Output Results• Output a vector y from Ax so that
|x-y|p · (1+ε) |x-xtop k|p
• Sometimes want y to be k-sparser = (k/εp)
• Both results tight up to logarithmic factors
• Recall that for non-sparse output r = £~(k/εp/2)
Talk Outline
1. O~(k / ε1/2) upper bound for p = 1
2. Lower bounds
Simplifications• Want O~(k/ε1/2) for p = 1• Replace k with 1
– Sample 1/k fraction of coordinates– Solve the problem for k = 1 on the sample– Repeat O~(k) times independently– Combine the solutions found
ε/k, ε/k, …, ε/k, 1/n, 1/n, …, 1/n
ε/k, 1/n, …, 1/n
k = 1• Assume |x-xtop|1 = 1, and xtop = ε• First attempt
– Use CountMin [CM]– Randomly partition coordinates into B buckets, maintain
sum in each bucket
Σi s.t. h(i) = 2 xi
• The expected l1-mass of “noise” in a bucket is 1/B• If B = £(1/ε), most buckets have count < ε/2, but bucket
that contains xtop has count > ε/2• Repeat O(log n) times
Second Attempt• But we wanted O~(1/ε1/2) measurements
• Error in a bucket is 1/B, need B ¼ 1/ε
• What about CountSketch? [CCF-C]– Give each coordinate i a random ¾(i) 2 {-1,1}– Randomly partition coordinates into B buckets,
maintain Σi s.t. h(i) = j ¾(i)¢xi in j-th bucket
– Bucket error is (Σi top xi2 / B)1/2
– Is this better?
Σi s.t. h(i) = 2 ¾(i)¢xi
CountSketch• Bucket error Err = (Σ i top xi
2 / B)1/2
• All |xi| · ε and |x-xtop|1 = 1
• Σi top xi2 · 1/ ε ¢ ε2 · ε
• So Err · (ε/B)1/2 which needs to be at most ε
• Solving, B ¸ 1/ ε
• CountSketch isn’t better than CountMin
Main Idea
• We insist on using CountSketch with B = 1/ε1/2
• Suppose Err = (Σ i top xi2 / B)1/2 = ε
• This means Σ i top xi2 = ε3/2
• Let’s make up the mass another way
Main Idea• We have: Σ i top xi
2 = ε3/2
• Intuition: suppose all xi, i top, are the same or 0
• Then: (# non-zero)*value = 1 (# non-zero)*value2 = ε3/2
• Hence, value = ε3/2 and # non-zero = 1/ε3/2
• Sample ε-fraction of coordinates uniformly at random!– value = ε3/2 and # non-zero sampled = 1/ε1/2, so l1-contribution = ε– Find all non-zeros with O~(1/ε1/2) measurements
General Setting• Σ i top xi
2 = ε3/2
• Sj = {i | 1/4j < xi2 · 1/4j-1}
• Σ i top xi2 = ε3/2 implies there is a j for which |Sj|/4j = ~(ε3/2)
ε3/2 , …, ε3/2
4ε3/2 , …, 4ε3/2
16ε3/2 , …, 16ε3/2
…ε3/4
General Setting• If |Sj| < 1/ε1/2, then 1/4j > ε2, so 1/2j > ε, can’t happen
• Else, sample at rate 1/(|Sj| ε1/2) to get 1/ε1/2 elements of |Sj|
• l1-mass of |Sj| in sample is > ε
• Can we find the sampled elements of Sj? Use Σ i top xi2 = ε3/2
• The l22 of the sample is about ε3/2 ¢ 1/(|Sj| ε1/2) = ε/|Sj|
• Using CountSketch with 1/ε1/2 buckets: Bucket error = sqrt{ε1/2¢ε3/2 ¢1/(|Sj| ε1/2)}
= sqrt{ε3/2/|Sj|} < 1/2j since |Sj|/4j > ε3/2
Algorithm Wrapup• Sub-sample O(log 1/ε) times in powers of 2
• In each level of sub-sampling maintain CountSketch with O~(1/ε1/2) buckets
• Find as many heavy coordinates as you can!
• Intuition: if CountSketch fails, there are many heavy elements that can be found by sub-sampling
• Wouldn’t work for CountMin as bucket error could be ε because of n-1 items each of value ε/(n-1)
Talk Outline
1. O~(k / ε1/2) upper bound for p = 1
2. Lower bounds
Our Results
• General results:– ~(k / ε1/2) for p = 1– (k log(n/k) / ε) for p = 2
• Sparse output:– ~(k/ε) for p = 1– ~(k/ε2) for p = 2
• Deterministic:– (k log(n/k) / ε) for p = 1
Simultaneous Communication Complexity
Alice Bob
x
• Alice and Bob send a single message to the referee who outputs f(x,y) with constant probability
• Communication cost CC(f) is maximum message length, over randomness of protocol and all possible inputs
• Parties share randomness
What is f(x,y)?y
MA(x) MB(y)
• Shared randomness decides matrix A• Alice sends Ax to referee• Bob sends Ay to referee• Referee computes A(x+y), uses
compressed sensing recovery algorithm• If output of algorithm solves f(x,y), then
# rows of A * # bits per measurement > CC(f)
Reduction to Compressed Sensing
A Unified View• General results: Direct-Sum Gap-l1
– ~(k / ε1/2) for p = 1 – ~(k / ε) for p = 2
• Sparse output: Indexing– ~(k/ε) for p = 1– ~(k/ε2) for p = 2
• Deterministic: Equality– (k log(n/k) / ε) for p = 1
Tighter log factors achievable by looking at Gaussian channels
General Results: k = 1, p = 1• Alice and Bob have x, y, respectively, in Rm
• There is a unique i* for which (x+y)i* = d For all j i*, (x+y)j 2 {0, c, -c}, where |c| < |d|
• Finding i* requires (m/(d/c)2) communication [SS, BJKS]
• m = 1/ε3/2, c = ε3/2 , d = ε• Need (1/ε1/2) communication
General Results: k = 1, p = 1• But the compressed sensing algorithm doesn’t
need to find i*
• If not then it needs to transmit a lot of information about the tail– Tail a random low-weight vector in {0, ε3/2, - ε3/2}1/ε3
– Uses distributional lower bound and RS codes
• Send a vector y within 1-ε of tail in l1-norm
• Needs 1/ε1/2 communication
General Results: k = 1, p = 2
• Same argument, different parameters
• (1/ε) communication
Handling General k• Bounded Round Direct Sum Theorem [BR]
(with slight modification) given k copies of a function f, with input pairs independently drawn from ¹, solving a 2/3 fraction needs communication (k¢CC¹ (f))
ε3/2 , …, ε3/2ε1/2
ε1/2
ε1/2
ε3/2 , …, ε3/2
ε3/2 , …, ε3/2
} k
Instance for p = 1
Handling General k
• CC = (k/ε1/2) for p = 1
• CC = (k/ε) for p = 2
• What is implied about compressed sensing?
Rounding Matrices [DIPW]• A is a matrix of real numbers
• Can assume orthonormal rows
• Round the entries of A to O(log n) bits, obtaining matrix A’
• Careful– A’x = A(x+s) for “small” s– But s depends on A, no guarantee recovery works– Can be fixed by looking at A(x+s+u) for random u
Lower Bounds for Compressed Sensing
• # rows of A * # bits per measurement > CC(f)
• By rounding, # bits per measurement = O(log n)
• In our hard instances, universe size = poly(k/ε)
• So # rows of A * O(log (k/ε)) > CC(f)
• # rows of A = ~(k/ε1/2) for p = 1• # rows of A = ~(k/ε) for p = 2
Sparse-Output Results
Sparse output: Indexing
– ~(k/ε) for p = 1
– ~(k/ε2) for p = 2
Sparse Output Results - Indexing
x 2 {0,1}n i 2 {1, 2, …, n}
What is xi?
CC(Indexing) = (n)
(1/ε) Bound for k=1, p = 1
x 2 {- ε, ε}1/ε y = ei
• Consider x+y
• If output is required to be 1-sparse must place mass on the i-th coordinate
• Mass must be 1+ε if xi = ε, otherwise 1-ε
Generalizes to k > 1 to give ~(k/ε)
Generalizes to p = 2 to give ~(k/ε2)
Deterministic Results
Deterministic: Equality– (k log(n/k) / ε) for p = 1
Deterministic Results - Equality
x 2 {0,1}n
Is x = y?
Deterministic CC(Equality) = (n)
y 2 {0,1}n
(k log(n/k) / ε) for p = 1
Choose log n signals x1, …, xlog n, each with k/ε values equal to ε/k
x = Σi=1log n 10i xi
Choose log n signals y1, …, ylog n, each with k/ε values equal to ε/k
y = Σi=1log n 10i yi
Consider x-yCompressed sensing output is 0n iff x = y
General Results – Gaussian Channels (k = 1, p = 2)
• Alice has a signal x =ε1/2 ei for random i 2 [n]
• Alice transmits x over a noisy channel with independent N(0, 1/n) noise on each coordinate
• Consider any row vector a of A
• Channel output = <a,x> + <a,y>, where <a,y> is N(0, |a|22/n)
• Ei[<a,x>2] = ε |a|22/n
• Shannon-Hartley Theorem: I(i; <a,x>+<a,y>) = I(<a,x>; <a,x>+<a,y>) · ½ log(1+ ε) = O(ε)
Summary of Results
• General results– £~(k/εp/2)
• Sparse output– £~(k/εp)
• Deterministic– £(k log(n/k) / ε) for p = 1 | 3,367 | 9,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.648148 |
https://www.wibestbroker.com/compounded-continuously/ | 1,723,133,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00302.warc.gz | 845,983,365 | 46,124 | Thu, August 08, 2024
Thu, August 08, 2024
# Compounded Continuously: Unlocking its Significance
It is hard to remember numerous terms when it comes to finance. Hopefully, there is no need to have a bachelor’s degree in finance in order to learn more about ‘Compounded continuously’ and answer questions such as ‘What does compound continuously mean?
Continuous compounding represents the theoretical peak that compound interest achieves when it is computed and added back to an account’s total at an endlessly increasing frequency, effectively assuming an infinite number of compounding intervals.
Though practically unattainable, the idea of interest compounding continuously plays a crucial role in the financial world. It serves as the ultimate example of how interest can be compounded, contrasting with the more common real-world scenarios where interest is compounded monthly, quarterly, or semi-annually.
Continuous compounding, in theoretical terms, implies that an account’s balance is perpetually accruing interest, which is then immediately reinvested to earn further interest. This concept envisages interest calculation on the premise that compounding occurs across an infinite series of periods.
While continuous compounding stands as a critical financial principle, the reality of achieving an infinite compounding frequency is impractical. Therefore, in practical applications, interest compounding is conducted over specific, fixed intervals, such as monthly, quarterly, or annually, rather than continuously.
## Compounded continuously formula
The formula to calculate the future value (FV) of an initial investment, leveraging the principle of compound interest, incorporates the initial value of the investment, the annual interest rate, the compounding frequency (or the number of times interest is applied) per year, and the total duration of the investment. The universal formula for determining compound interest is:
PV = the present value of the investment
i = the stated interest rate
n = the number of compounding periods
t = the time in years
The formula for continuous compounding is derived from the formula for the future value of an interest-bearing investment:
Future Value (FV) = PV x [1 + (i / n)](n x t)
Calculating the limit of this formula as n approaches infinity (per the definition of continuous compounding) results in the formula for continuously compounded interest:
FV = PV x e (i x t)
where e is the mathematical constant, which is approximated to be 2.7183.
## Compounding
Compounding is a fundamental concept in finance that describes the process whereby the value of an investment grows exponentially over time due to the earning of interest on both the initial principal and the accumulated interest from previous periods.
This phenomenon, embodying the principle of the time value of money (TVM), is also recognized as compound interest.
One of the most fundamental principles in finance is the Time Value of Money (TVM). It posits that a dollar today is worth more than a dollar in the future due to its potential earning capacity. This principle underpins the rationale behind investment decisions, interest rates, and the valuation of financial assets. TVM is a reminder that in finance, timing is everything.
## The mechanics of compounding
The principle of compounding is based on the premise that, after each compounding period—be it yearly, quarterly, or daily—the interest earned over that period is added to the principal sum, thereby increasing the base upon which future interest is calculated. This cycle of earning interest on previously earned interest continues throughout the investment period, potentially leading to significant growth in the value of the investment over time.
The true power of compounding lies in its potential to accelerate the growth of an investment over the long term. The key factors that influence the effectiveness of compounding include the rate of interest, the frequency of compounding, and the length of the investment period. A higher interest rate, more frequent compounding periods, and a longer duration of investment all contribute to a greater accumulation of wealth.
The frequency of compounding can substantially impact the amount of interest accumulated. Common compounding frequencies include annual, semi-annual, quarterly, monthly, daily, and, in the theoretical limit, continuously. As the frequency of compounding increases, the investment grows more rapidly due to the more frequent addition of interest to the principal.
## Practical applications and considerations
Compounding is a key consideration in various financial decisions and products, including savings accounts, loans, mortgages, and investments. Understanding how compounding works can help individuals make more informed decisions about their financial strategies and select investment or savings products that offer optimal growth potential.
Moreover, the concept of compounding underscores the importance of starting to save and invest early. Even modest amounts invested regularly can grow significantly over time due to the compounding effect, highlighting the adage that time is more valuable than money in the realm of investing.
## What is discrete compounding?
Discrete compounding refers to the process in which interest on an investment or loan is calculated and added to the principal at specific intervals, rather than continuously. These intervals can be annual, semi-annual, quarterly, monthly, or daily, but they occur at distinct, separated points in time, hence the term “discrete.”
Discrete compounding is common in savings accounts, loans, and various other financial products. Understanding the compounding frequency and how it affects investment growth or borrowing costs is crucial for making informed financial decisions.
To sum up, finance, an essential facet of the modern world, is a broad term that encompasses the management, creation, and study of money, banking, credit, investments, assets, and liabilities. It forms the backbone of economies, influencing everything from global economic policies to individual financial well-being.
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Types of Screw Conveyors - KWS ManufacturingHorizontal Screw Conveyors Inclined Screw Conveyors Shaftless Screw . KWS recommends designing screw conveyors using the lowest possible degree of . into the standard screw conveyor horsepower calculations to compensate for the.inclined screw conveyor design calculation,Incline Screw Conveyors | Stainless Screw Conveyor Design CapacityProducts can be conveyed and elevated at the same time by mounting a screw conveyor at an incline. This is often desirable because it covers two operations.
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https://www.sarthaks.com/2659330/number-clusters-represented-101112-same-code-number-clusters-276-and-957-represented-111 | 1,685,672,315,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00370.warc.gz | 1,085,527,557 | 15,201 | # If the sum of number-clusters 395 and 727 is represented as 101112; and by the same code, the sum of number-clusters 276 and 957 is represented as 111
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closed
If the sum of number-clusters 395 and 727 is represented as 101112; and by the same code, the sum of number-clusters 276 and 957 is represented as 111213, then which of the following will be code for sum of number-clusters 689 and 877?
1. 131415
2. 121314
3. 141312
4. 141516
by (30.1k points)
selected by
Correct Answer - Option 4 : 141516
The logic followed here is,
In 395 and 727
3 + 7 = 10
9 + 2 = 11
5 + 7 = 12
So, 395 and 727 is represented as 101112.
And,
In 276 and 957
2 + 9 = 11
7 + 5 = 12
6 + 7 = 13
So, 276 and 957 is represented as 111213.
Similarly,
In 689 and 877
6 + 8 = 14
8 + 7 = 15
9 + 7 = 16
Hence, 689 and 877 is represented as ‘141516’. | 305 | 850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.945191 |
http://www.sliderbase.com/spitem-309-1.html | 1,566,175,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314353.10/warc/CC-MAIN-20190818231019-20190819013019-00500.warc.gz | 312,718,789 | 3,358 | # Solving equations by adding and subtractingPage 1
Slide 1
## Sōp
1. Simplify: -2(x + 3)
2. Simplify: 3[(5 + 72) – 20]
= -2x - 6
= 3[(5 + 49) – 20]
= 3[(54) – 20]
= 3[34]
= 102
Slide 2
3.1
Solving Equations by Adding and Subtracting
Slide 3
t + 5
=
11
-5
-5
t
=
6
+ 0
Example 1
Slide 4
Now you try…
y + 3
=
-2
-3
-3
y
=
-5
+ 0
Example 2
Slide 5
a - 4
=
-10
+4
+4
a
=
-6
+ 0
Example 3
Slide 6
Now you try…
x - 6
=
-6
+6
+6
x
=
0
+ 0
Example 4
Slide 7
-4 + x
=
-30
+4
+4
x
=
-26
0+
Example 5
Slide 8
Now you try…
4.6 + m
=
0
-4.6
- 4.6
m
=
-4.6
0 +
Example 6
Slide 9
+ b
=
-
Example 7
Now try a more challenging problem…
- +
0 +
b
=
1 | 349 | 725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-35 | longest | en | 0.500813 |
https://www.docsity.com/es/pso-40/3556451/ | 1,537,946,180,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163704.93/warc/CC-MAIN-20180926061824-20180926082224-00163.warc.gz | 729,670,657 | 13,823 | 2 páginas
4Número de visitas
Descripción
20 Puntos
Puntos necesarios para descargar
este documento
Vista previa2 páginas / 2
Topic 14. Production System Operation. Inventory. Inventory consist of a stock of items, the size of which is called the inventory level.
Inventory management involves making decisions concerning how much inventory to order a nd when. Inventory control involves process, procedures, and infrastructure to maintain the inventory at the desired level.
Optimal Order Quantity =
Expected Number Orders =
Expected Time Between Orders =
d =
PURCHASE COST Annual demand quantity of the product (D) multiplied by the purchase cost per unit (P) PD
HOLDING COST (carrying cost of storage cost)
The average quantity in stock (Q/2) times the annual holding cost per unit (H). Q is the order quantity. H*Q/2
ORDERING COST (the cost of placing orders)
Each order has a constant cost (S) multiplied by (D/Q) times per year. S*D/Q
TC = PD + DS/Q +
HQ/2
D = Annual demand quantity of product
ROP = d · L
EOQ Example. You’re a buyer for Express. Express needs 1000 coffee makers per year. The cost of each coffee maker is \$78. Ordering cost is \$100 per order. Carrying cost is 40% of per unit cost. Lead time is 5 days. Express is open 365 days/yr. What is the optimal order quantity & ROP?
Express EOQ.
ROP = daily demand x lead time (days) = d x L D= annual demand = 1000 Days/ year = 365 Daily demand = 1000/365 = 2.74 Lead time = 5 days
ROP = 2.74 x 5 = 13.7 —> 14
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# Is x^2 > 15? (1) x > -4 (2) x^3 < 0
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Is x^2 > 15?
(1) x > -4
(2) x^3 < 0
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Is x^2 > 15?
Is $$x^2>15$$? --> is $$x<-\sqrt{15}$$ or is $$x>\sqrt{15}$$?
(1) x > -4 --> if $$-4<x<-\sqrt{15}$$ or if $$x>\sqrt{15}$$ then the answer is YES but if $$-\sqrt{15}\leq{x}\leq{\sqrt{15}}$$ then the answer is NO. Not sufficient.
(2) x^3 < 0 --> $$x<0$$ --> if $$x<-\sqrt{15}$$ then the answer is YES but if $$-\sqrt{15}\leq{x}<0$$ then the answer is NO. Not sufficient.
(1)+(2) $$x>-4$$ and $$x<0$$ --> $$-4<x<0$$ --> if $$-4<x<-\sqrt{15}$$ then the answer is YES but if $$-\sqrt{15}\leq{x}<0$$ then the answer is NO. Not sufficient.
Hope it's clear.
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Re: Is x^2 > 15? (1) x > -4 (2) x^3 < 0 [#permalink]
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22 May 2012, 15:41
maybe is a bit comfortable to work with number on a number line instead of $$\sqrt{15}$$ (is the same after all)
1) x > - 4 -----> -3,9 is largest than 15 but -3 NO but again +5 yes insuff
2) x^3 < 0 ---> x is negative -----> -1 is NO -5 is YES Insuff
1) + 2) we have only negative value but -1 is YES -3.9 is NO -----> E
Thanks Bunuel .
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# Is x^2 > 15? (1) x > -4 (2) x^3 < 0
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2023-11-17 07:27:14
iiooasd1217
Novice
Registered: 2023-10-31
Posts: 7
### Cause the math is in my head and I've followed where its lead.
I am a mathematician, engineer, and musician. I like big numbers and I like small numbers. I like good music and I hate bad music. I own a small business where I build anything I get paid to. My favorite paradox is currently Gabriels Horn (even though that is still solvable, but that's a story for another day). I am fairly good at coding, but hope to get better.
My favorite composer is Andrew Lloyd Webber because of his musical Love Never Dies. Thanks!
I'll do a^2+b^2=c^2. I'll even do pir^2. But y=mx+b is where I draw the line.
Why do people become maths teachers? Because they have problems.
Offline
## #2 2023-11-17 08:41:22
amnkb
Member
Registered: 2023-09-19
Posts: 194
### Re: Cause the math is in my head and I've followed where its lead.
iiooasd1217 wrote:
I own a small business where I build anything I get paid to.
whats a cool/interesting/weird thing that you've built recently?
Offline
## #3 2023-11-25 14:41:50
iiooasd1217
Novice
Registered: 2023-10-31
Posts: 7
### Re: Cause the math is in my head and I've followed where its lead.
amnkb wrote:
iiooasd1217 wrote:
I own a small business where I build anything I get paid to.
whats a cool/interesting/weird thing that you've built recently?
I have recently built a flair drum, a computer, several other instruments and more along those lines.
I'll do a^2+b^2=c^2. I'll even do pir^2. But y=mx+b is where I draw the line.
Why do people become maths teachers? Because they have problems.
Offline
## #4 2023-11-25 15:10:02
Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,528
### Re: Cause the math is in my head and I've followed where its lead.
Hi iiooasd1217,
Good introduction!
Welcome to the forum!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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June 17, 2021
# The battle of English and Mathematics
Question:
I had \$5.00 My Mom gave me \$10.00 while my Dad gave me \$30.00. My Aunt and Uncle gave me 100.00. I had another \$5.00How much money did i really have?I won against Miesha Precious love
No items found.
Neat W.
99%
The first section says initially he had \$5.00 and the last part says he had another \$5.00 totaling to \$10.00 dollars. He received the rest the from her mom, dad as well as her uncle and aunt which totaled to \$140.00
=\$5.00+\$10.00+\$30.00+\$100+\$5.00
So in total I had \$150.00 dollars.
Do it in a few clicks:
• 1
Click “Get Help Now”
• 2
• 3
Type in “Real-time help” in the “Assignment Topic” field
• 4
Deposit funds and hire a writer, release the funds once you’re satisfied.
Jacky M.
99%
I had \$5.00 My Mom gave me \$10.00 while my dad gave me \$30.00. My Aunt and Uncle gave me 100.00. I had another \$5.00 How much money did I have? I won against Miesha Precious love"
The amount in possession is \$0.00. The person in the picture is referring to the past tense by saying that he “had”. On the other hand, the question is asking the amount of money he has at the moment which we have not been provided with. We, therefore, assume that the current amount in hand is \$ 0.00.
## Don’t know the answer about "The battle of English and Mathematics"
Hiring a mathematics tutor will do the trick! | 403 | 1,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-21 | latest | en | 0.979987 |
http://nrich.maths.org/4868 | 1,484,756,539,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00521-ip-10-171-10-70.ec2.internal.warc.gz | 201,546,899 | 5,692 | ### The Line and Its Strange Pair
In the diagram the point P' can move to different places along the dotted line. Each position P' takes will fix a corresponding position for P. If P' moves along a straight line what does P do ?
### Mapping the Wandering Circle
In the diagram the point P can move to different places around the dotted circle. Each position P takes will fix a corresponding position for P'. As P moves around on that circle what will P' do?
### Like a Circle in a Spiral
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
# Symmetric Trace
##### Stage: 4 Challenge Level:
Before we begin we need to check something - it's about symmetry.
A pattern continues forever in both directions.
Imagine it's on a roll of paper and two strips are torn off, one of which is turned upside-down and placed underneath the other.
It is not possible to shift the lower strip horizontally so that it lines up and matches the upper strip.
On the other hand for the next pattern. . .
Even with the second piece upside-down the two pieces can still be made to line up and match.
#### Now to start the real problem.
This problem is about that kind of symmetry.
The pattern is a trace from a point on a rolling wheel.
Before starting, you may find it useful to explore How far does it move? .
A wheel rolls along a horizontal track and leaves traces from two different points.
Point 1 is on the circumference of the wheel and its trace looks like this:
#### Trace One
Forget the wheel for a moment and just concentrate on the trace pattern.
If this trace was turned upside-down you would certainly not be able to line it up with itself.
Point 2 is somewhere inside the wheel and its trace looks like this :
#### Trace Two
Would "Trace Two" line up with itself upside-down? | 403 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-04 | longest | en | 0.925482 |
http://www.robives.com/mechanisms/cardan | 1,369,270,071,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702652631/warc/CC-MAIN-20130516111052-00074-ip-10-60-113-184.ec2.internal.warc.gz | 690,386,601 | 8,348 | ### Cardan Gear
Invented by Girolamo Cardano in the 16th century the Cardan gear is a way of converting rotary motion into straight line motion. Watch how the red dot on the inner purple gear exactly follows the vertical dotted line. The outer gear has a diameter exactly twice as large as the inner gear. In the above example they have 40 and 20 teeth respectively. Cardano also invented a type of universal joint and investigated the mathematics of probability. Understanding the mathematics of risk helped him make a living from gambling until eventually he could find no-one to gamble with and had to move onto new pastures...
Monday 23rd May 2011 19:26
some sort of answer to how it works would be nice too help me get on with school work but cool website
Saturday 18th Jun 2011 00:59
### littlemisszombie
??????????? i do not understand your comment :/ because he (rob) has said how it works???????
i am loving this, my little boy loves making models, and i'm so glad i've found this :D when i get paid (soon) i'm going to buy membership :D when i was little i used to go to the village library, and constantly get out this book they had on automated toys and how to make them; it showed wooden toys, but i used to make them out of corrugated cardboard etc (hey it was the 80's :D) but this is brill!!!!!!!!! i just know that tomorrow, even if my little boy doesn't fully understand everything, he will just love the animation of it moving, and when i show him some of them with cereal box card etc he will be in his element....thanks for this site :D i can't get over it :D
Saturday 18th Jun 2011 09:06
### robives
Thanks! I must admit I didn't understand the comment so I left it to answer later.
Sunday 19th Jun 2011 16:30
### littlemisszombie
Sunday 19th Jun 2011 17:54
### robives
Hi @littlemisszombie, yes the models on Flying Pig are printed models that arrive through the post.
Thursday 23rd Jun 2011 15:52
how does it work and could u put it in a toy
Wednesday 28th Sep 2011 09:35
What are the speeds in the end points of the linear motion contra the speed of the middle part?
The speed at the end points is zero, in the centre it is one, the speed varies between the extremities according to a sine curve. - RI
Wednesday 28th Sep 2011 19:14
### robives
It looks like the speeds will follow a sine wave. The end points will be speed zero with maximimum speed in the middle.
Friday 27th Apr 2012 16:06
### Katie Grace
hiya rob
great stuff!!! what project would you say is a good example of this cardan gear mechanism?
thanks
Kt
The only place I've seen one in action was in a sewing machine raising and lowering te needle. - RI
Saturday 28th Apr 2012 12:19
### umehta
Katie:
Saturday 28th Apr 2012 12:39
### umehta
But to answer your question about what project may use this mechanism: 'Ornithopter' project uses cardan gear. Hopefully someday, Rob will design a paper ornithopter!
That would be a fun project! - RI | 762 | 2,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.960419 |
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by willnich35
P: 7 1. The problem statement, all variables and given/known data A uniform beam, supported by a uniform strut, is loaded as shown in Figure (Attached). Using the unit load method, determine the value of the vertical deflection at C. BEAM: E = 205 kN/mm2, I = 90 x 106 mm4. STRUT: E = 205 kN/mm2, A = 1500 mm2. 2. Relevant equations The virtual work equation, which is. 1.$$\Delta$$=$$\sum$$p$$\frac{PL}{AE}$$+$$\int$$$$\frac{mM}{EI}$$dx 3. The attempt at a solution REAL Reactions at A and B in the x and y directions: Fx = Ax + Bx = 0 Fy = Ay + 15 + (4*2) - By = 0 Mb = (15*2) - 8 - 2Ay = 0 Ay = 11 kN By = 34 kN Ax = Bx = 17 * SqRt(2) Axial Force (P) in member AB = 17 * SqRt(2) DB = 34 * SqRt(2) VIRTUAL Reactions Apply a force of 1 kN downwards at C Axial Force (p) in member AB = -2 DB = 2 * SqRt(2) I then attempted to solve the virtual work equation, however the value of A for member AC is not given. So i guess my questions are, have I started this correctly, and how do I solve the problem? Attached Thumbnails
P: 674 I think you must be worrying about the axial shortening of DB, whose A is not given. Even if you know it, and assuming no compression buckling, the shortening of the strut would have a relatively small contribution to make to the deflection at C. You could insert your own practical value of A to assure yourself how small it really is. One might also recognise that AB also has some axial compression, but you don't know that area either. To be pedantic, the hinge at A is off the centreline, so there actually is a small additional moment due to the eccentric axial load there, but I recommend you note this but ignore it. Incidentally I would say Ax= - 11 not + 11. I don't agree your values of Bx and Ax. Have you checked your figure for equilibrium with the reaction components you have calculated, for example by taking moments about C? Have you drawn a force diagram to scale for all the external forces on this structure? That is a good way to reveal errors. On re-reading the question, I see a value of A is given.
P: 7 Thank you very much for your quick answer. Massive help! Should be able to work it out now.
P: 7
## Finding the deflection using the unit load method.
I have an answer, which makes sense. I would greatly appreciate it if you would check it over.
Ay = -11 kN
By = 34 kN
Ax = Bx = 34 * SqRt(2)
Axial Force (P) in member DB = 34 * SqRt(3)
VIRTUAL Reactions
Axial Force (p) in member DB = 2 * SqRt(3)
$$\frac{pPL}{AE}$$ = 1.876mm (Deflection due to axial load on DB)
$$\int$$$$\frac{(-x1)(-11x1)}{EI}$$dx1 + $$\int$$$$\frac{(-x2)(-15x2)}{EI}$$dx2
Integrating between 0 and 2 for both x1 and x2
= 3.758mm (Deflection due to bending moments in beam)
TOTAL DEFLECTION = 5.634mm Downwards
P: 674 I don't agree your reactions at Ax and Bx. Have you really checked equilibrium using your values, as I suggested. And even if I did, how do you get axial force in BD as you do? You are not showing all your working, and therefore your work is uncheckable. In your integrals, you seem to have forgotten the effect of the udl. Is the expression pPL/AE really giving you the deflection at C? I think not. So what is it saying to you? Finally, your answer has ridiculous "accuracy". It is disproportionate to the assumptions being made. In that sense, it is wrong. Please try again. It's worth it.
P: 7 Ok, thanks again. Not my strongest subject this, il have another go.
P: 7 Ok, here's my next attempt I've done it in word, as the workings are easier, and attached a picture. I really hope the reactions Ax and Bx, and the force in member AD are right now, the triangles should explain how i got the numbers. As for the virtual work equation, dx1 works from A to B dx2 works from C to B M1 is the real bending moment equation of section AB m1 is the virtual bending moment equation of section AB M2 is the real bending moment equation of section CB m2 is the virtual bending moment equation of section CB M1 hopefully now incorporates the moment due to the UDL. Attached Thumbnails
P: 7 Just noticed the second line should be Fy= not Fx= and Ax = -Bx
P: 674 I don't agree your result, doing it a different way. For example, PL/AE is the elastic shortening of DB, and I get 0.44 mm. This has a vertical component of 0.44 / sqrt 2 = 0.31 mm. So the deflection at C due to elastic shortening of DB should be 0.62 mm, not 1.25 mm (your figure). I agree your result for bending = 4.19 mm.
P: 7 Thank you, that makes sense. However it does seem to conflict slightly with my notes, so il go have a chat with my lecturer tomorrow. Thanks for your time and help though, huge help!
Related Discussions Introductory Physics Homework 0 Engineering, Comp Sci, & Technology Homework 2 Engineering, Comp Sci, & Technology Homework 0 Engineering, Comp Sci, & Technology Homework 1 Engineering, Comp Sci, & Technology Homework 3 | 1,374 | 4,945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2013-48 | longest | en | 0.92691 |
https://www.physicsforums.com/threads/prove-a-complex-expression-1-if-log-of-the-complex-terms-are-equal.754549/ | 1,508,605,699,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00701.warc.gz | 998,159,385 | 19,327 | # Prove a complex expression = 1, if log of the complex terms are equal
1. May 20, 2014
### Govind_Balaji
1. The problem statement, all variables and given/known data
If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b},$ prove that $x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=1$
2. Relevant equations
3. The attempt at a solution
I solved a question similar to it in a way. So I tried this in the same way. The only difference between the two questions is that except a,b and c, the denominators are also x, y and z.
I didn't know if it would work.
So I tried like this.
Let $m= b-c, n= c-a$then $a-b=-(m+n)$
By cross multiplication, I got
$z^m=x^{-(m+n)}\\z^n=y^{-(m+n)}\\x^n=y^m\\\\\Rightarrow z=1/xy, x=1/yz, y=1/xz$
In my old question it was to prove $x^x.y^y.z^z=1$. So I substituted easily.
I am struggling to substitute here.
So I tried writing $x^{b+c-a}.y^{c+a-b}.z^{a+b-c}$ it easier by using m and n.
$x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=x^{2n+m+a}.y^{a-m}.z^{a+m}$
I am struggling then after. Can anyone give me a hint.
Last edited: May 20, 2014
2. May 20, 2014
### Saitama
Try working backwards.
You need to prove $x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=1$, this is equivalent to showing $\log x^{b+c-a}+\log y^{c+a-b}+\log z^{a+b-c}=0$, do you see why? If so, can you figure out the individual terms and show that their sum is zero?
3. May 20, 2014
Are you saying I have to prove $b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0 4. May 20, 2014 ### Saitama No, I didn't mean that. Equate the equality to $\lambda$. Can you find $\log x^{b+c-a}$ in terms of $\lambda, a, b$ and $c$? 5. May 20, 2014 ### Govind_Balaji I am just 9th grade. I don't know anything about lambda other than it is a symbol for wavelength of light. I don't know the value of lambda. Can you explain clearly? 6. May 20, 2014 ### Saitama Wavelength? :tongue: I meant $\lambda$ as some constant, you can even name it $\alpha$ if you wish to. What I meant was this: \frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=\lambda Can you make sense of this now? 7. May 20, 2014 ### Govind_Balaji Yes, now I will try. 8. May 21, 2014 ### Govind_Balaji I cant understand how to prove this. expressing $\log x^{b+c-a}$ in terms of $\lambda, a, b$ and $c$, $10^{\lambda}=x^{b+c-a}$. Or I tried as I said in the first post of this thread. Can you give a small hint if you have already solved it. 9. May 21, 2014 ### Saitama I don't see how you get that. You have the following: \frac{\log x}{b-c}=\lambda Can you find $\log x^{b+c-a}$ from here? (Hint: Multiply both the sides by $b+c-a$) 10. May 21, 2014 ### Govind_Balaji I get [itex] \\\log x^{b+c-a}=\lambda\frac{b+c-a}{b-c}\\\\ \\\log y^{c+a-b}=\lambda\frac{c+a-b}{c-a}\\\\ \\\log z^{a+b-c}=\lambda\frac{a+b-c}{a-b}\\\\ \\\log\left(x^{b+c-a}. y^{c+a-b}.z^{b+c-a}\right )=\lambda\frac{b+c-a}{b-c}+\lambda\frac{c+a-b}{c-a}+\lambda\frac{a+b-c}{a-b}$
Am I right till now?
11. May 21, 2014
### Saitama
Are you sure that they should be in denominator?
12. May 21, 2014
### Govind_Balaji
Thank you, a small careless mistake.
The denominators should be in numerators.
I add them. I got 0.
Removing the log, I will get them equal to 0.
Thank you very much.
13. May 21, 2014
You are at 9th grade and you have studied logs?
I haven't still studied it.
14. May 21, 2014
### Govind_Balaji
I saw in your profile. You have done PhD and you have said that you are a professor.
Why are you kidding me?:tongue2:
15. May 21, 2014
### Staff: Mentor
16. May 21, 2014 | 1,237 | 3,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-43 | longest | en | 0.881364 |
https://www.abebooks.it/9780387875002/Basic-Concepts-Probability-Statistics-Law-038787500X/plp | 1,500,607,494,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423681.33/warc/CC-MAIN-20170721022216-20170721042216-00381.warc.gz | 734,122,229 | 18,155 | # Basic Concepts of Probability and Statistics in the Law
## Michael O. Finkelstein
Valutazione media 4
( su 1 valutazioni fornite da Goodreads )
When as a practicing lawyer I published my ?rst article on statistical evidence in 1966, the editors of the Harvard Law Review told me that a mathematical equa- 1 tion had never before appeared in the review. This hardly seems possible - but if they meant a serious mathematical equation, perhaps they were right. Today all that has changed in legal academia. Whole journals are devoted to scienti?c methods in law or empirical studies of legal institutions. Much of this work involves statistics. Columbia Law School, where I teach, has a professor of law and epidemiology and other law schools have similar “law and” professorships. Many offer courses on statistics (I teach one) or, more broadly, on law and social science. The same is true of practice. Where there are data to parse in a litigation, stat- ticians and other experts using statistical tools now frequently testify. And judges must understand them. In 1993, in its landmark Daubert decision, the Supreme Court commanded federal judges to penetrate scienti?c evidence and ?nd it “re- 2 liable” before allowing it in evidence. It is emblematic of the rise of statistics in the law that the evidence at issue in that much-cited case included a series of epidemiological studies. The Supreme Court’s new requirement made the Federal Judicial Center’s Reference Manual on Scienti?c Evidence, which appeared at about the same time, a best seller. It has several important chapters on statistics.
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From the Back Cover:
This book sets out basic statistical tools as they have been applied in actual legal disputes. Examples range over diverse fields of law, such as identification evidence, mass torts, securities law, environmental regulation, and capital punishment, among many others. In some notable cases, such as the U.S. Supreme Court’s decision in Bush v. Gore, the book explores aspects of the statistical evidence that were unrecognized or misconceived by the parties or the court. For the statistics student, the book will give a deeper appreciation of foundation concepts and provide a wealth of real life applications. For the lawyer, or law student, the book will introduce a subject that has become increasingly important both in litigation and in studies of the legal system. The book is self-contained and may be read without background in probability or statistics.
Michael O. Finkelstein is a lawyer and has been described as a pioneer in the use of statistics in law. He has for many years taught Statistics for Lawyers at Columbia Law School, and has also taught this subject at Harvard, Yale, and the University of Pennsylvania law schools. In addition to numerous articles, he is the author of Quantitative Methods in Law, published in 1978, and is the co-author with Professor Bruce Levin of Statistics for Lawyers, first published in 1990 and now in its second edition.
Review:
From the reviews:
“...A ‘shorter and mathematically less challenging’ version of an earlier co-authored book, Statistics for Lawyers. ...Full of legal argument and counter-argument forensically dissected by the author. ...The book requires concentration but is fun to read. ...Presents an excellent treatment of a fascinating field.” (International Statistical Review, 2010, 78, 1, 134-159)
“This work is a good introductional textbook on statistics and probability for lawyers. ... The book contains ... many examples (sometimes with numerical data and simple calculations) of courts decisions and discussions of real life causes. ... Special attention is paid to the statistical meaning of DNA-testing procedures, analysis of epidemiologic data and discrimination causes.” (R. E. Maiboroda, Zentralblatt MATH, Vol. 1173, 2009)
“Basic Concepts of Probability and Statistics in the Law (BCPSL) is an introduction to our discipline for law students and lawyers who want to understand basic statistics and how statistics are used in legal cases. ... BCPSL contains some new cases and was a fun and much faster read. ... students would enjoy and learn from BCPSL. ... a good choice for someone who has already had a basic statistics course or as complementary reading in an introductory course.” (Dalene Stancl, Journal of the American Statistical Association, Vol. 106 (493), March, 2011)
“The author, a lawyer by training, wrote this book as a more accessible ... version of his more comprehensive book on the topic, Statistics for Lawyers (2000). This slimmer version of the more comprehensive book makes for a quick, enlightening read. While this book was written primarily for law students, it can be understood by those with formal statistical training who want to understand an area of application of statistical concepts. ... statistician or student of statistics would be able to benefit from this book.” (Willis A. Jensen, Technometrics, Vol. 52 (4), November, 2010)
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## 1.Basic Concepts of Probability and Statistics in the Law
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Descrizione libro Springer-Verlag New York Inc., United States, 2009. Paperback. Condizione libro: New. Language: English . Brand New Book. When as a practicing lawyer I published my ?rst article on statistical evidence in 1966, the editors of the Harvard Law Review told me that a mathematical equa- 1 tion had never before appeared in the review. This hardly seems possible - but if they meant a serious mathematical equation, perhaps they were right. Today all that has changed in legal academia. Whole journals are devoted to scienti?c methods in law or empirical studies of legal institutions. Much of this work involves statistics. Columbia Law School, where I teach, has a professor of law and epidemiology and other law schools have similar law and professorships. Many offer courses on statistics (I teach one) or, more broadly, on law and social science. The same is true of practice. Where there are data to parse in a litigation, stat- ticians and other experts using statistical tools now frequently testify. And judges must understand them. In 1993, in its landmark Daubert decision, the Supreme Court commanded federal judges to penetrate scienti?c evidence and ?nd it re- 2 liable before allowing it in evidence.It is emblematic of the rise of statistics in the law that the evidence at issue in that much-cited case included a series of epidemiological studies. The Supreme Court s new requirement made the Federal Judicial Center s Reference Manual on Scienti?c Evidence, which appeared at about the same time, a best seller. It has several important chapters on statistics. Codice libro della libreria LIB9780387875002
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## 3.Basic Concepts of Probability and Statistics in the Law (Paperback)
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Descrizione libro Springer-Verlag New York Inc., United States, 2009. Paperback. Condizione libro: New. Language: English . Brand New Book. When as a practicing lawyer I published my ?rst article on statistical evidence in 1966, the editors of the Harvard Law Review told me that a mathematical equa- 1 tion had never before appeared in the review. This hardly seems possible - but if they meant a serious mathematical equation, perhaps they were right. Today all that has changed in legal academia. Whole journals are devoted to scienti?c methods in law or empirical studies of legal institutions. Much of this work involves statistics. Columbia Law School, where I teach, has a professor of law and epidemiology and other law schools have similar law and professorships. Many offer courses on statistics (I teach one) or, more broadly, on law and social science. The same is true of practice. Where there are data to parse in a litigation, stat- ticians and other experts using statistical tools now frequently testify. And judges must understand them. In 1993, in its landmark Daubert decision, the Supreme Court commanded federal judges to penetrate scienti?c evidence and ?nd it re- 2 liable before allowing it in evidence.It is emblematic of the rise of statistics in the law that the evidence at issue in that much-cited case included a series of epidemiological studies. The Supreme Court s new requirement made the Federal Judicial Center s Reference Manual on Scienti?c Evidence, which appeared at about the same time, a best seller. It has several important chapters on statistics. Codice libro della libreria LIB9780387875002
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Destinazione, tempi e costi | 3,495 | 13,751 | {"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} | 2.640625 | 3 | CC-MAIN-2017-30 | latest | en | 0.944866 |
https://www.manualslib.com/manual/257002/Hp-48gii.html?page=132 | 1,571,864,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00555.warc.gz | 977,376,916 | 35,112 | # Fast 3d Plots - HP 48gII User Manual
Graphing calculator.
The @@ IG@ key simply changes the font in the table from small to big, and
vice versa. Try it.
The @ OOM key, when pressed, produces a menu with the options: In, Out,
Decimal, Integer, and Trig. Try the following exercises:
With the option In highlighted, press @@@OK@@@. The table is expanded so
that the x-increment is now 0.25 rather than 0.5. Simply, what the
calculator does is to multiply the original increment, 0.5, by the zoom
factor, 0.5, to produce the new increment of 0.25. Thus, the zoom in
option is useful when you want more resolution for the values of x in
To increase the resolution by an additional factor of 0.5 press @ OOM,
select In once more, and press @@@OK@@@. The x-increment is now 0.0125.
To recover the previous x-increment, press @ OOM —@@@OK@@@ to select
the option Un-zoom. The x-increment is increased to 0.25.
To recover the original x-increment of 0.5 you can do an un-zoom
again, or use the option zoom out by pressing @ OUT @@@OK@@@.
The option Decimal in @ OOM produces x-increments of 0.10.
The option Integer in @ OOM produces x-increments of 1.
The option Trig in produces increments related to fractions of
being useful when producing tables of trigonometric functions.
## Fast 3D plots
Fast 3D plots are used to visualize three-dimensional surfaces represented by
equations of the form z = f(x,y). For example, if you want to visualize z =
2
2
f(x,y) = x
+y
, we can use the following:
π
, thus
Page 10-5 | 428 | 1,512 | {"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} | 3.3125 | 3 | CC-MAIN-2019-43 | latest | en | 0.768965 |
https://www.nrich.maths.org/parabolicpatterns | 1,670,466,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00355.warc.gz | 955,373,455 | 5,199 | ### Cubic Spin
Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?
### Sine Problem
In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.
### More Parabolic Patterns
The illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs.
# Parabolic Patterns
##### Age 14 to 18Challenge Level
The illustration shows the graphs of fifteen functions. Two of them have equations
$y = x^2$
$y = - (x - 4)^2$
Can you find the equations of the other parabolas in the picture?
You may wish to use a graphical calculator or software such as Desmos to recreate the pattern for yourself.
Can you find the equations of these parabolas?
NOTES AND BACKGROUND
This sort of challenge is sometimes called an inverse problem because the question is posed the opposite way round to what might have been expected. This is almost like saying: 'here is the answer, what was the question?' Instead of giving the equations of some functions and asking you to sketch the graphs, this challenge gives the graphs and asks you to find their equations.
You are being asked to sketch a family of graphs. What makes this a family? All the graphs are obtained by transformations such as reflections and translations of other graphs in the family. The key is to find the simplest function and then to find transformations of the graph of that function which give the other graphs in the family.
If you have access to a graphic calculator, or to graph drawing software, it will not give you the answers. You will have to think for yourself what the equations should be and then the software will enable you to test your own theories and see if you were right. | 421 | 1,889 | {"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} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.952872 |
https://maslinandco.com/6221602 | 1,674,902,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00386.warc.gz | 386,380,235 | 4,038 | # Evaluate: \lim_{x arrow 0} ((x * csc(2x))/(cos(5x)))
## Expression: $\lim_{x \rightarrow 0} \left(\frac{ x \times \csc\left({2x}\right) }{ \cos\left({5x}\right) }\right)$
Use $\lim_{x \rightarrow c} \left(\frac{ f\left( x \right) }{ g\left( x \right) }\right)=\frac{ \lim_{x \rightarrow c} \left(f\left( x \right)\right) }{ \lim_{x \rightarrow c} \left(g\left( x \right)\right) }$ to transform the expression
$\frac{ \lim_{x \rightarrow 0} \left(x \times \csc\left({2x}\right)\right) }{ \lim_{x \rightarrow 0} \left(\cos\left({5x}\right)\right) }$
Use $\csc\left({t}\right)=\frac{ 1 }{ \sin\left({t}\right) }$ to transform the expression
$\frac{ \lim_{x \rightarrow 0} \left(x \times \frac{ 1 }{ \sin\left({2x}\right) }\right) }{ \lim_{x \rightarrow 0} \left(\cos\left({5x}\right)\right) }$
Use $\lim_{x \rightarrow c} \left(\cos\left({f\left( x \right)}\right)\right)=\cos\left({\lim_{x \rightarrow c} \left(f\left( x \right)\right)}\right)$ to transform the expression
$\frac{ \lim_{x \rightarrow 0} \left(x \times \frac{ 1 }{ \sin\left({2x}\right) }\right) }{ \cos\left({\lim_{x \rightarrow 0} \left(5x\right)}\right) }$
Calculate the product
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ x }{ \sin\left({2x}\right) }\right) }{ \cos\left({\lim_{x \rightarrow 0} \left(5x\right)}\right) }$
Use $\lim_{x \rightarrow c} \left(a \times f\left( x \right)\right)=a \times \lim_{x \rightarrow c} \left(f\left( x \right)\right)$ to transform the expression
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ x }{ \sin\left({2x}\right) }\right) }{ \cos\left({5 \times \lim_{x \rightarrow 0} \left(x\right)}\right) }$
Since evaluating limits of the numerator and denominator would result in an indeterminate form, use the L'Hopital's rule
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right) }{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sin\left({2x}\right) \right) }\right) }{ \cos\left({5 \times \lim_{x \rightarrow 0} \left(x\right)}\right) }$
Evaluate the limit by substituting the value $x=0$ into the expression
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right) }{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sin\left({2x}\right) \right) }\right) }{ \cos\left({5 \times 0}\right) }$
Find the derivative
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ 1 }{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sin\left({2x}\right) \right) }\right) }{ \cos\left({5 \times 0}\right) }$
Find the derivative
$\frac{ \lim_{x \rightarrow 0} \left(\frac{ 1 }{ 2\cos\left({2x}\right) }\right) }{ \cos\left({5 \times 0}\right) }$
Evaluate the limit
$\frac{ \frac{ 1 }{ 2\cos\left({2 \times 0}\right) } }{ \cos\left({5 \times 0}\right) }$
Simplify the expression
\begin{align*}&\frac{ 1 }{ 2 } \\&\begin{array} { l }0.5,& {2}^{-1}\end{array}\end{align*}
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https://www.projectcubicle.com/frequency-distribution-in-excel/ | 1,701,982,565,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00287.warc.gz | 1,067,142,594 | 62,287 | # Frequency Distribution in Excel: Simplifying Data Analysis
Frequency distribution in Excel is a fundamental concept in statistics that involves organizing and representing data in classes or intervals to understand patterns and tendencies. Microsoft Excel, being the powerhouse of data analysis that it is, offers tools and functions to efficiently compute and visualize frequency distributions.
## Understanding Frequency Distribution
At its core, frequency distribution is about categorizing data into bins or intervals and then tallying the number of data points in each bin. For instance, when analyzing test scores of students, you might want to see how many students scored between 60-70, 70-80, and so on.
## Steps to Create Frequency Distribution in Excel
1. Data Arrangement: Begin by having your data range in a single column, ideally sorted in ascending or descending order.
2. Bin Creation: Define the bins or intervals. For our test score example, bins might be: 60, 70, 80, 90, and 100.
3. Using the FREQUENCY Function: Excel’s built-in FREQUENCY function can automatically count the number of data points for each bin.Formula: =FREQUENCY(data_array, bins_array)
Example: If your test scores are in cells A2 to A101 and your bins in cells C2 to C6, you can enter the formula in D2 and extend it to D6 to get the frequency for each bin.
### Data Visualization with Histograms
Once you’ve computed the frequency distribution, visualizing it can offer even more insights. Excel’s Histogram tool in the Data Analysis ToolPak is perfect for this.
Steps to Create a Histogram:
1. Head to the Data tab and select Data Analysis. If you don’t see this option, you might need to install the Data Analysis ToolPak.
2. Choose Histogram from the list and click OK.
3. Define the input range (your data) and bin range.
4. Choose an output location and check the Chart Output box.
5. Click OK to generate a histogram chart.
### Pro Tips and Tricks
• Dynamic Bins with Pivot Tables: Instead of manually defining bins, you can use Pivot Tables in Excel to create dynamic frequency distributions, adjusting your intervals as you see fit.
• Advanced Visualization: Excel offers a plethora of chart options beyond histograms. Experiment with column, bar, or line charts for alternative data visualizations.
• Remember Overflow and Underflow Bins: When dealing with a wide range of data, consider adding bins for values that are below the lowest defined bin (underflow) or above the highest (overflow).
Frequency Distribution in Excel is more than just a statistical tool; it’s a way to transform raw data into actionable insights. With Excel’s robust features, this transformation is not only efficient but also intuitive, aiding data enthusiasts, researchers, and business professionals in their analysis endeavors.
## Frequency Distribution in Excel: Unveiling the Patterns Within
In our data-driven world, representing vast volumes of information in an easily digestible manner is invaluable. Excel’s Frequency Distribution feature is an analytical powerhouse, converting raw numbers into understandable patterns and trends.
### The Core of Frequency Distribution
Frequency distribution is essentially the heartbeat of basic statistical analysis. It offers a structured way to showcase how often different data points or values appear within a dataset. Consider student grades, for example. By using frequency distribution, an educator could quickly understand how many students achieved scores within specific grade intervals.
### Crafting Frequency Distribution in Excel
1. Initiating with Data Arrangement: Start by placing your dataset in one column. For clarity, it’s beneficial to have the data sorted in ascending order. This ensures easy identification of patterns and anomalies.
2. Establishing Bins: These are essentially intervals. For student grades, bins might range from 60-69, 70-79, and so forth.
3. Tapping into the FREQUENCY Function: This native Excel function does the heavy lifting. It computes the number of data points fitting within each bin.Formula Decoded: =FREQUENCY(data_array, bins_array)
Practical Application: If student scores span cells A2 to A101 and your pre-defined bins occupy cells C2 to C6, by entering the formula in D2 and stretching it down to D6, Excel reveals the frequency for each grade interval.
### Embellishing with Histograms
Raw numbers are impactful, but visual representations can further accentuate insights. Also, this is where histograms, a type of bar graph representing frequency data, come into play.
Creating a Histogram in Excel:
1. Navigate to the Data tab and opt for Data Analysis. If you’re unfamiliar with this option, you might need to activate the Data Analysis ToolPak add-in.
2. Choose Histogram, and an intuitive dialog box appears.
3. Specify the input range and bin range. Opt for a suitable output location and ensure the Chart Output option is ticked.
### Best Practices and Pitfalls
Frequency Distribution in Excel is powerful, but as with all tools, its efficacy lies in its application.
• Bin Precision: The size and range of your bins can make or break your analysis. Also, overly broad bins can obscure finer data nuances, while excessively narrow bins can lead to an overwhelming and unclear representation.
• Outlier Awareness: These are data points that deviate significantly from other observations. Also, neglecting outliers might provide a skewed understanding of your dataset. Recognizing and understanding the context of outliers is crucial.
• A Balanced Approach: While visualizations like histograms are enticing, it’s imperative to understand the numbers that birthed them. A blend of numerical analysis with visual interpretation ensures robust insights.
## Frequently Asked Questions: Frequency Distribution in Excel
### Why is frequency distribution important?
In essence, frequency distribution organizes data into specific categories or bins, allowing users to easily visualize and comprehend the distribution of data points. This makes it easier to recognize patterns, anomalies, or trends in the dataset. Moreover, in a world overwhelmed by data, it acts as a bridge between raw information and actionable insights.
### Can I automatically generate bins in Excel?
Absolutely! Instead of manually determining bins, which can be time-consuming, you can leverage Excel’s Data Analysis ToolPak to auto-generate them. When using the Histogram tool, simply provide the data range and let Excel handle bin creation based on data distribution. However, it’s advisable to review the auto-generated bins to ensure they align with the analysis’s intent.
### How do I deal with outliers in frequency distribution?
Outliers can significantly skew the representation of a dataset. First and foremost, it’s crucial to identify them. Once identified, you have several options:
1. Exclude them from your analysis if they’re anomalies or errors.
2. Create specific bins for extreme values to understand their distribution separately.
3. Retain them in the analysis if they provide valuable insights or are essential for a holistic view of the data.
In any case, understanding the context of outliers and their impact on the dataset is paramount.
### Scenario: Customer Feedback on Product Quality
Imagine a company, “TechSolutions,” which recently launched a new electronic gadget. Also, they’ve distributed surveys to 500 customers, asking them to rate the product’s quality on a scale of 1 to 10 (with 10 being the highest quality).
### Organizing Data and Setting up Bins
Upon receiving feedback, the company notices a range of scores, from 1 to 10. To better understand the distribution, they decide to categorize feedback into three bins:
1. Low Quality (Scores 1-3)
2. Average Quality (Scores 4-7)
3. High Quality (Scores 8-10)
### Utilizing the FREQUENCY Function
In Excel, “TechSolutions” sets up their data as follows:
• Column A contains the individual scores from the 500 respondents.
• Column C lists the bin upper limits: 3, 7, and 10, representing the bins we’ve defined.
Now, they’ll use the FREQUENCY function:
Formula Implementation:
In Column D, starting from D1, they’ll enter the array formula:
less
```=FREQUENCY(A1:A500, C1:C3) ```
Note: This is an array formula, so after typing, they’d press Ctrl+Shift+Enter. | 1,688 | 8,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-50 | longest | en | 0.823543 |
https://encyclopedia2.thefreedictionary.com/N%C3%B3r | 1,657,033,090,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00152.warc.gz | 278,191,130 | 10,860 | # NOR
(redirected from Nór)
Also found in: Dictionary, Thesaurus, Medical, Financial, Idioms, Wikipedia.
## NOR
[nȯr]
(mathematics)
A logic operator having the property that if P, Q, R, … are statements, then the NOR of P, Q, R, … is true if all statements are false, false if at least one statement is true. Derived from NOT-OR. Also known as Peirce stroke relationship.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
## NOR
Not OR.
The Boolean function which is true if none of its inputs are true and false otherwise, the logical complement of inclusive OR. The binary (two-input) NOR function can be defined (written as an infix operator):
A NOR B = NOT (A OR B) = (NOT A) AND (NOT B)
Its truth table is:
A | B | A NOR B --+---+--------- F | F | T F | T | F T | F | F T | T | F
NOR, like NAND, forms a complete set of Boolean functions on its own since it can be used to make NOT, AND, OR and any other Boolean function:
NOT A = A NOR A
A OR B = NOT (A NOR B)
A AND B = (NOT A) NOR (NOT B) | 303 | 1,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.863964 |
http://www.nwchem-sw.org/index.php/Special:AWCforum/st/id2281 | 1,540,346,468,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00305.warc.gz | 540,650,753 | 9,209 | # quick question RE:Spin-orbit ECPs (Stuttgart)
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4:32:56 AM PDT - Mon, Mar 13th 2017 Hey. I'm just learning to use NWChem (wrt my BsC project), and need to run some calculations (HgBr2 geometry optimization and DFT energy) using Stuttgart ECPs. For spin-orbit ECPs I'm planning to use Stuttgart ECPs: http://www.tc.uni-koeln.de/PP/clickpse.en.html " Pseudopotential ECP60MDF for Hg ! Q=20., MEFIT, MCDHF+Breit, Ref 37. Hg 0 ECP60MDF 5 60 H-Komponente 1 2 1.000000 0.000000 S-H 2 2 12.413071 275.774797 2 6.897913 49.267898 P-H 4 2 11.310320 80.506984 2 10.210773 161.034824 2 5.939804 9.083416 2 5.019755 18.367773 D-H 4 2 8.407895 51.137256 2 8.214086 76.707459 2 4.012612 6.561821 2 3.795398 9.818070 F-H 2 2 3.273106 9.429001 2 3.208321 12.494856 G-H 2 2 4.485296 -6.338414 2 4.513200 -8.099863 ! References: ! [37] D. Figgen, G. Rauhut, M. Dolg, H. Stoll, Chem. Phys. 311, 227 (2005). " To qoute NWChem manual: "For example, in the literature the Stuttgart potentials are defined as ?Ul and, hence, have to be multiplied by 2 / (2l + 1)." Does that mean that I need to multiply each coefficient in manually specified ECP by 2 / (2l + 1)? Best - Luna
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12:34:28 PM PDT - Mon, Mar 13th 2017 Luna This is the corresponding NWChem input for the Hg ECP ECP Hg nelec 60 Hg ul 2 1.0000000 0.0000000 Hg S 2 12.4130710 275.7747970 2 6.8979130 49.2678980 Hg P 2 11.3103200 80.5069840 2 10.2107730 161.0348240 2 5.9398040 9.0834160 2 5.0197550 18.3677730 Hg D 2 8.4078950 51.1372560 2 8.2140860 76.7074590 2 4.0126120 6.5618210 2 3.7953980 9.8180700 Hg F 2 3.2731060 9.4290010 2 3.2083210 12.4948560 Hg G 2 4.4852960 -6.3384140 2 4.5132000 -8.0998630 END
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What is time series analysis? business research consulting. The 3rd edition of statistics for business: decision making and analysis emphasizes an application-based approach, in which readers learn how to work with data to.
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## MAW 96 Mathematics and Decision Making
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posted by .
A man 6 feet tall is standing 9 feet from a 18 feet tall light pole. His shadow cast 15 degree. What is the angle formed from this equation?
• Mathsw -
18/(9+s)=6/s
Draw the figure, that reflects similar trigngles. S is the length of his shadow.
Now your question: what is the angle formed? 15 degrees, Duh.
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• Resource Constrained Project Scheduling Problem (RCPSP)
# Resource Constrained Project Scheduling Problem (RCPSP)¶
## Principles learned¶
• Set precedence constraints
• Use interval decision variables
• Use a lambda expression to compute a sum with a variable number of terms
## Problem¶
In the resource constrained project scheduling problem, a project consists of a set of tasks that have to be scheduled. Each task has a given duration and cannot be interrupted. There are precedence constraints between the tasks: each task must end before any of its successors can start. There are a set of renewable resources. Each task has a given resource requirement, or weight, (possibly zero) on each resource, representing the amount of resource it consumes while it is being processed. Each resource has a given maximum capacity: it can process several tasks at once, but the sum of the processed tasks’s weights can never exceed this maximum capacity.
The goal is to find a schedule that minimizes the makespan: the time when all tasks have been processed.
## Data¶
The instances provided follow the Patterson format. The format of the data files is as follows:
• First line:
• Number of tasks (including two additional dummy tasks with duration 0, the source and the sink)
• Number of renewable resources
• Second line: Maximum capacity for each resource
• From the third line, for each task:
• Resource requirements (weights) for each resource
• Number of successors
• Task ID for each successor
## Program¶
We use interval decision variables to model the time ranges of the tasks. The length of the interval is constrained by the duration of each task.
The precedence constraints are easily written: each task must be placed before any of its successors.
The cumulative resource constraints can be formulated as follows: for each resource, and for each time slot t, the amount of resource consumed by the tasks that are being processed must not exceed the resource’s capacity.
To model these constraints, we sum up, for each resource, the weights of all the tasks placed over a given time slot t.
The makespan to minimize is the time when all the tasks have been processed.
Execution:
localsolver rcpsp.lsp inFileName=instances/Pat1.rcp [outFileName=] [lsTimeLimit=]
```use io;
// The input files follow The "Patterson" Format
function input() {
local usage = "Usage: localsolver rcpsp.lsp inFileName=instanceFile "
+ "[outFileName=outputFile] [lsTimeLimit=timeLimit]";
if (inFileName == nil) throw usage;
// Maximum capacity of each resource
for [r in 0...nbResources] {
}
for[r in 0...nbResources] {
// Resource weight of resource r required for task i
}
// Successors of each task i
for [s in 0...nbSuccessors[i]] {
}
}
// Trivial upper bound for the start times of the tasks
inFile.close();
}
function model() {
// Interval decisions: time range of each task
// Precedence constraints between the tasks
for[i in 0...nbTasks][s in 0...nbSuccessors[i]] {
}
// Makespan: end of the last task
// Cumulative resource constraints
for [r in 0...nbResources] {
constraint and(0...makespan,
}
// Minimize the makespan
minimize makespan;
}
// Parameterize the solver
function param() {
if (lsTimeLimit == nil) lsTimeLimit = 60;
}
/* Write the solution in a file with the following format:
* - total makespan
* - for each task, the task id, the start and end times */
function output() {
if (outFileName != nil) {
outFile = io.openWrite(outFileName);
println("Solution written in file ", outFileName);
outFile.println(makespan.value);
}
}
}
```
Execution (Windows)
set PYTHONPATH=%LS_HOME%\bin\python
python rcpsp.py instances\Pat1.rcp
Execution (Linux)
export PYTHONPATH=/opt/localsolver_12_0/bin/python
python rcpsp.py instances/Pat1.rcp
```import localsolver
import sys
# The input files follow the "Patterson" format
with open(filename) as f:
first_line = lines[0].split()
# Number of resources
nb_resources = int(first_line[1])
# Maximum capacity of each resource
capacity = [int(lines[1].split()[r]) for r in range(nb_resources)]
duration = [0 for i in range(nb_tasks)]
# Resource weight of resource r required for task i
weight = [[] for i in range(nb_tasks)]
# Number of successors
nb_successors = [0 for i in range(nb_tasks)]
# Successors of each task i
successors = [[] for i in range(nb_tasks)]
line = lines[i + 2].split()
duration[i] = int(line[0])
weight[i] = [int(line[r + 1]) for r in range(nb_resources)]
nb_successors[i] = int(line[nb_resources + 1])
successors[i] = [int(line[nb_resources + 2 + s]) - 1 for s in range(nb_successors[i])]
# Trivial upper bound for the start times of the tasks
horizon = sum(duration[i] for i in range(nb_tasks))
return (nb_tasks, nb_resources, capacity, duration, weight, nb_successors, successors, horizon)
def main(instance_file, output_file, time_limit):
instance_file)
with localsolver.LocalSolver() as ls:
#
# Declare the optimization model
#
model = ls.model
# Interval decisions: time range of each task
# Precedence constraints between the tasks
for s in range(nb_successors[i]):
# Makespan: end of the last task
# Cumulative resource constraints
for r in range(nb_resources):
capacity_respected = model.lambda_function(
lambda t: model.sum(weight[i][r] * model.contains(tasks[i], t)
<= capacity[r])
model.constraint(model.and_(model.range(makespan), capacity_respected))
# Minimize the makespan
model.minimize(makespan)
model.close()
# Parameterize the solver
ls.param.time_limit = time_limit
ls.solve()
#
# Write the solution in a file with the following format:
# - total makespan
# - for each task, the task id, the start and end times
#
if output_file != None:
with open(output_file, "w") as f:
print("Solution written in file", output_file)
f.write(str(makespan.value) + "\n")
f.write(str(i + 1) + " " + str(tasks[i].value.start()) + " " + str(tasks[i].value.end()))
f.write("\n")
if __name__ == '__main__':
if len(sys.argv) < 2:
print("Usage: python rcpsp.py instance_file [output_file] [time_limit]")
sys.exit(1)
instance_file = sys.argv[1]
output_file = sys.argv[2] if len(sys.argv) >= 3 else None
time_limit = int(sys.argv[3]) if len(sys.argv) >= 4 else 60
main(instance_file, output_file, time_limit)
```
Compilation / Execution (Windows)
cl /EHsc rcpsp.cpp -I%LS_HOME%\include /link %LS_HOME%\bin\localsolver120.lib
rcpsp instances\Pat1.rcp
Compilation / Execution (Linux)
g++ rcpsp.cpp -I/opt/localsolver_12_0/include -llocalsolver120 -lpthread -o rcpsp
./rcpsp instances/Pat1.rcp
```#include "localsolver.h"
#include <algorithm>
#include <fstream>
#include <iostream>
#include <limits>
#include <numeric>
#include <vector>
using namespace localsolver;
class Rcpsp {
private:
// Number of resources
int nbResources;
// Maximum capacity of each resource
std::vector<int> capacity;
std::vector<int> duration;
// Resource weight of resource r required for task i
std::vector<std::vector<int>> weight;
// Number of successors
std::vector<int> nbSuccessors;
// Successors for each task i
std::vector<std::vector<int>> successors;
// Trivial upper bound for the start times of the tasks
int horizon = 0;
// Localsolver
LocalSolver localsolver;
// Decision variables: time range of each task
// Objective = minimize the makespan: end of the last task of the last job
LSExpression makespan;
public:
Rcpsp(const std::string& fileName) : localsolver() {}
// The input files follow the "Patterson" format
std::ifstream infile;
infile.open(fileName.c_str());
infile >> nbResources;
// The maximum capacity of each resource
capacity.resize(nbResources);
for (int r = 0; r < nbResources; ++r) {
infile >> capacity[r];
}
for (int i = 0; i < nbTasks; ++i) {
infile >> duration[i];
weight[i].resize(nbResources);
for (int r = 0; r < nbResources; ++r) {
infile >> weight[i][r];
}
infile >> nbSuccessors[i];
successors[i].resize(nbSuccessors[i]);
for (int s = 0; s < nbSuccessors[i]; ++s) {
int x;
infile >> x;
successors[i][s] = x - 1;
}
horizon += duration[i];
}
infile.close();
}
void solve(int TimeLimit) {
// Declare the optimization model
LSModel model = localsolver.getModel();
// Interval decisions: time range of each task
for (int i = 0; i < nbTasks; ++i) {
}
// Precedence constraints between the tasks
for (int i = 0; i < nbTasks; ++i) {
for (int s = 0; s < nbSuccessors[i]; ++s) {
}
}
// Makespan: end of the last task
makespan = model.max();
for (int i = 0; i < nbTasks; ++i) {
}
// Cumulative resource constraints
for (int r = 0; r < nbResources; ++r) {
LSExpression capacityRespected = model.createLambdaFunction([&](LSExpression t) {
LSExpression totalWeight = model.sum();
for (int i = 0; i < nbTasks; ++i) {
}
return model.leq(totalWeight, capacity[r]);
});
model.constraint(model.and_(model.range(0, makespan), capacityRespected));
}
// Minimize the makespan
model.minimize(makespan);
model.close();
// Parameterize the solver
localsolver.getParam().setTimeLimit(TimeLimit);
localsolver.solve();
}
/* Write the solution in a file with the following format:
* - total makespan
* - for each task, the task id, the start and end times */
void writeSolution(const std::string& fileName) {
std::ofstream outfile(fileName.c_str());
if (!outfile.is_open()) {
std::cerr << "File " << fileName << " cannot be opened." << std::endl;
exit(1);
}
std::cout << "Solution written in file " << fileName << std::endl;
outfile << makespan.getValue() << std::endl;
for (int i = 0; i < nbTasks; ++i) {
outfile << i + 1 << " " << tasks[i].getIntervalValue().getStart() << " "
}
outfile.close();
}
};
int main(int argc, char** argv) {
if (argc < 2) {
std::cout << "Usage: Rcpsp instanceFile [outputFile] [timeLimit]" << std::endl;
exit(1);
}
const char* instanceFile = argv[1];
const char* outputFile = argc > 2 ? argv[2] : NULL;
const char* strTimeLimit = argc > 3 ? argv[3] : "60";
Rcpsp model(instanceFile);
try {
const int timeLimit = atoi(strTimeLimit);
model.solve(timeLimit);
if (outputFile != NULL)
model.writeSolution(outputFile);
return 0;
} catch (const std::exception& e) {
std::cerr << "An error occurred: " << e.what() << std::endl;
return 1;
}
}
```
Compilation / Execution (Windows)
copy %LS_HOME%\bin\localsolvernet.dll .
csc Rcpsp.cs /reference:localsolvernet.dll
Rcpsp instances\Pat1.rcp
```using System;
using System.IO;
using localsolver;
public class Rcpsp : IDisposable
{
// Number of resources
private int nbResources;
// Maximum capacity of each resource
private int[] capacity;
private int[] duration;
// Resource weight of resource r required for task i
private int[,] weight;
// Number of successors
private int[] nbSuccessors;
// Successors for each task i
private int[][] successors;
// Trivial upper bound for the start times of the tasks
private int horizon = 0;
// LocalSolver
private LocalSolver localsolver;
// Decision variables: time range of each task
// Objective = minimize the makespan: end of the last task of the last job
private LSExpression makespan;
public Rcpsp(string fileName)
{
localsolver = new LocalSolver();
}
{
if (line == null)
return new string[0];
return line.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
}
// The input files follow the "Patterson" format
{
{
string[] splitted = SplitInput(input);
if (splitted.Length == 0)
splitted = SplitInput(input);
nbResources = int.Parse(splitted[1]);
// The maximum capacity of each resource
capacity = new int[nbResources];
splitted = SplitInput(input);
for (int r = 0; r < nbResources; ++r)
capacity[r] = int.Parse(splitted[r]);
for (int i = 0; i < nbTasks; ++i)
{
splitted = SplitInput(input);
if (splitted.Length == 0)
splitted = SplitInput(input);
duration[i] = int.Parse(splitted[0]);
for (int r = 0; r < nbResources; ++r)
weight[i, r] = int.Parse(splitted[r + 1]);
nbSuccessors[i] = int.Parse(splitted[nbResources + 1]);
successors[i] = new int[nbSuccessors[i]];
for (int s = 0; s < nbSuccessors[i]; ++s)
successors[i][s] = int.Parse(splitted[s + nbResources + 2]) - 1;
horizon += duration[i];
}
}
}
public void Dispose()
{
localsolver.Dispose();
}
public void Solve(int timeLimit)
{
// Declare the optimization model
LSModel model = localsolver.GetModel();
// Interval decisions: time range of each task
for (int i = 0; i < nbTasks; ++i)
{
}
// Precedence constraints between the tasks
for (int i = 0; i < nbTasks; ++i)
{
for (int s = 0; s < nbSuccessors[i]; ++s)
{
}
}
// Makespan: end of the last task
makespan = model.Max();
for (int i = 0; i < nbTasks; ++i)
// Cumulative resource constraints
for (int r = 0; r < nbResources; ++r)
{
LSExpression capacityRespected = model.LambdaFunction(t =>
{
LSExpression totalWeight = model.Sum();
for (int i = 0; i < nbTasks; ++i)
{
}
}
);
model.Constraint(model.And(model.Range(0, makespan), capacityRespected));
}
// Minimize the makespan
model.Minimize(makespan);
model.Close();
// Parameterize the solver
localsolver.GetParam().SetTimeLimit(timeLimit);
localsolver.Solve();
}
/* Write the solution in a file with the following format:
* - total makespan
* - for each task, the task id, the start and end times */
public void WriteSolution(string fileName)
{
using (StreamWriter output = new StreamWriter(fileName))
{
Console.WriteLine("Solution written in file " + fileName);
output.WriteLine(makespan.GetValue());
for (int i = 0; i < nbTasks; ++i)
{
output.Write((i + 1) + " " + tasks[i].GetIntervalValue().Start() + " " + tasks[i].GetIntervalValue().End());
output.WriteLine();
}
}
}
public static void Main(string[] args)
{
if (args.Length < 1)
{
Console.WriteLine("Usage: Rcpsp instanceFile [outputFile] [timeLimit]");
System.Environment.Exit(1);
}
string instanceFile = args[0];
string outputFile = args.Length > 1 ? args[1] : null;
string strTimeLimit = args.Length > 2 ? args[2] : "60";
using (Rcpsp model = new Rcpsp(instanceFile))
{
model.Solve(int.Parse(strTimeLimit));
if (outputFile != null)
model.WriteSolution(outputFile);
}
}
}
```
Compilation / Execution (Windows)
javac Rcpsp.java -cp %LS_HOME%\bin\localsolver.jar
java -cp %LS_HOME%\bin\localsolver.jar;. Rcpsp instances\Pat1.rcp
Compilation / Execution (Linux)
javac Rcpsp.java -cp /opt/localsolver_12_0/bin/localsolver.jar
java -cp /opt/localsolver_12_0/bin/localsolver.jar:. Rcpsp instances/Pat1.rcp
```import java.util.*;
import java.io.*;
import localsolver.*;
public class Rcpsp {
// Number of resources
private int nbResources;
// Maximum capacity of each resource
private int[] capacity;
private int[] duration;
// Resource weight of resource r required for task i
private int[][] weight;
// Number of successors
private int[] nbSuccessors;
// Successors for each task i
private int[][] successors;
// Trivial upper bound for the start times of the tasks
private int horizon = 0;
// LocalSolver
final LocalSolver localsolver;
// Decision variables: time range of each task
// Objective = minimize the makespan: end of the last task of the last job
private LSExpression makespan;
public Rcpsp(LocalSolver localsolver, String fileName) throws IOException {
this.localsolver = localsolver;
}
// The input files follow the "Patterson" format
private void readInstance(String fileName) throws IOException {
try (Scanner input = new Scanner(new File(fileName))) {
nbResources = input.nextInt();
// The maximum capacity of each resource
capacity = new int[nbResources];
for (int r = 0; r < nbResources; ++r) {
capacity[r] = input.nextInt();
}
for (int i = 0; i < nbTasks; ++i) {
duration[i] = input.nextInt();
for (int r = 0; r < nbResources; ++r) {
weight[i][r] = input.nextInt();
}
nbSuccessors[i] = input.nextInt();
successors[i] = new int[nbSuccessors[i]];
for (int s = 0; s < nbSuccessors[i]; ++s) {
successors[i][s] = input.nextInt() - 1;
}
horizon += duration[i];
}
}
}
public void solve(int timeLimit) {
// Declare the optimization model
LSModel model = localsolver.getModel();
// Interval decisions: time range of each task
for (int i = 0; i < nbTasks; ++i) {
}
// Precedence constraints between the tasks
for (int i = 0; i < nbTasks; ++i) {
for (int s = 0; s < nbSuccessors[i]; ++s) {
}
}
// Makespan: end of the last task
makespan = model.max();
for (int i = 0; i < nbTasks; ++i) {
}
// Cumulative resource constraints
for (int r = 0; r < nbResources; ++r) {
final int rL = r;
LSExpression capacityRespected = model.lambdaFunction(t -> {
LSExpression totalWeight = model.sum();
for (int i = 0; i < nbTasks; ++i) {
weight[i][rL],
}
return model.leq(totalWeight, capacity[rL]);
});
model.constraint(model.and(model.range(0, makespan), capacityRespected));
}
// Minimize the makespan
model.minimize(makespan);
model.close();
// Parameterize the solver
localsolver.getParam().setTimeLimit(timeLimit);
localsolver.solve();
}
/*
* Write the solution in a file with the following format:
* - total makespan
* - for each task, the task id, the start and end times
*/
public void writeSolution(String fileName) throws IOException {
try (PrintWriter output = new PrintWriter(fileName)) {
System.out.println("Solution written in file " + fileName);
output.println(makespan.getValue());
for (int i = 0; i < nbTasks; ++i) {
output.println((i + 1) + " " + tasks[i].getIntervalValue().getStart() + " "
}
}
}
public static void main(String[] args) {
if (args.length < 1) {
System.out.println("Usage: java Rcpsp instanceFile [outputFile] [timeLimit]");
System.exit(1);
}
String instanceFile = args[0];
String outputFile = args.length > 1 ? args[1] : null;
String strTimeLimit = args.length > 2 ? args[2] : "60";
try (LocalSolver localsolver = new LocalSolver()) {
Rcpsp model = new Rcpsp(localsolver, instanceFile); | 4,633 | 17,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.904235 |
https://studyingdiagrams.com/fishbone-diagrams-uses-input-variables/ | 1,660,039,971,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00205.warc.gz | 500,322,816 | 13,975 | # Fishbone Diagrams Uses Input Variables
Which of the following statements is TRUE about capability analyses. It can be used to structure a brainstorming session.
Ishikawa Fishbone Diagram Fishbone Diagram Bad Coffee Cause And Effect Diagram Increase In Productivity Fishbone Wiki
### The Fishbone Diagram was originally created by Kaoru Ishikawa to find process imperfections Ishakawa 1985.
Fishbone diagrams uses input variables. A fishbone diagram is used to identify the sources of variation within a process. Start studying TOM 301 Ch. Write down the problem you are trying to solve.
In fact many of the diagrams listed alongside fishbones as the Seven Basic Tools of Quality have a dedicated solution or library found in ConceptDraws Solution Park such as root cause diagrams flowcharts and histograms. Sometimes called an Ishikawa diagram or cause-and-effect analysis a fishbone diagram. The Core equation in Six Sigma is YfX1 X2 X3 Xn.
They are also commonly called cause and effect or fishbone diagrams. A Cause and Effect Diagram aka Ishikawa Fishbone is a pictorial diagram showing possible causes process inputs for a given effect process outputs. What does a cause and effect diagram show.
Draw the head on the right which contains the problem effect or issue for analysis. Draw a straight line from the head leading to the left. Ishikawa diagrams were popularized by Kaoru Ishikawa in the 1960s who pioneered quality management processes in the Kawasaki shipyards and in the process became one of the founding fathers of modern management.
The first company to adopt the diagram in all of its processes was Kawasaki Iron Fukiai Works in 1952. It is one of the seven basic tools of quality. The line along the centre represents the problem with all the possible causes branching off it.
Cause and effect diagram six sigma Creating a cause and effect Ishikawa or fishbone diagram using Visio 2010 Beta. As illustrated below a completed Fishbone diagram includes a central spine and several branches reminiscent of a fish skeleton. A fishbone diagram is a cause-and-effect discovery tool that helps figure out the reason s for defects variations or failures within a process.
In order to understand our Y or Output we must first define our Xs or Inputs. Learn vocabulary terms and more with flashcards games and other study tools. Try not to just rely on peoples opinions.
Its important to be as informed as possible so you can mix ideas and opinions with data. Identify as many categories or contributing factors to the problem you can. In other words it establish the correlation between process input variables to the customers outputs during root cause analysis.
Drawing a fishbone diagram. It helps to identify the root causes of a problem or effect in order to identify appropriate solutions. Gather data sheets process information and observe the problem process prior to the Fishbone session.
The diagram looks just like a fishs skeleton with the problem at its head and. A fishbone diagram is a visual way to look at cause and effect. A fishbone diagram helps team members visually diagram a problem or conditions root causes allowing them to truly diagnose the problem rather than focusing on symptoms.
The advantage of using this tool is that causes are arranged. It is often used in Lean Six Sigma and other quality improvement approaches to narrow down the area of analysis. Start with 4-6 main categories and expand as needed.
This cause analysis tool is considered one of the seven basic quality tools. A higher process capability index means. Fishbone diagram or Ishikawa diagram is a modern quality management tool that explains the cause and effect relationship for any quality issue that has arisen or that may arise.
In other words it is a visual representation used to find out the cause s of a specific problem. A cause-effect diagram is a visual tool used to logically organize possible causes for a specific problem or effect by graphically displaying them in increasing detail suggesting causal relationships among theories. It allows team members to separate a.
The value of using the fishbone diagram is to dig deeper to go beyond the initial incident report to better understand what in the organizations systems and processes are causing the problem so they can be. The library of vector stencils Fishbone diagrams contains 13 symbols for drawing the Ishikawa diagrams using the ConceptDraw PRO diagramming and vector drawing software. This is the backbone.
Brainstorm possible causes of the problem and place them under the categories where they fit best. The fishbone diagram identifies many possible causes for an effect or problem. The Fishbone chart is an initial step in the screening process.
It provides the visual representation of all the possible causes for a problem to analyze and find out the root cause. The fishbone diagram or Ishikawa diagram is a cause-and-effect diagram that helps managers to track down the reasons for imperfections variations defects or failures. This diagram is used in process improvement methods to identify all of the contributing root causes likely to be causing a problem.
Follow Four Steps to Fishbone. Cause-and-effect diagram Ishikawa diagram. The Cause and Effect diagram is also referred to as the Ishikawa diagram or.
Juran honored Ishikawa by naming the tool after him and is now also known as the Ishikawa Diagram. In other words it helps break down in successive layers root causes that potentially contribute to an effect. The Fishbone Diagram solution is one of a number of solutions aimed at creating causal models and workflow problem solving.
Identify the areas broad level categories to be studied and branch them from the backbone.
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A Fishbone Diagram For The Surface Roughness In Co2 Laser Cutting Download Scientific Diagram | 1,389 | 7,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.919229 |
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Topic: Defining "f o g" with Functions
Replies: 4 Last Post: Dec 31, 2012 8:19 AM
Messages: [ Previous | Next ]
Angela Richardson Posts: 42 From: UK Registered: 6/22/11
Re: Defining "f o g" with Functions
Posted: Oct 15, 2012 3:17 AM
att1.html (2.2 K)
f={(a,b)} implies f(a)=b
f o g (x)=f(g(x))
f o g =(1,6), (2,4),(3,7),(4,4),(5,3) since g maps 1 to 2, which f maps to 6, and g maps 2 to 3, which f maps to 4 etc.
g is not one-to-one since both 2 and 4 are mapped to 3. f is one-to-one since no two ordered pairs have the same number as either their first or second element. A function has an inverse if and only if it is one-to-one. f^-1={(6,2),(7,1),(4,3),(3,5),(5,4)} which is the ordered pairs with the first and second elements swapped.
________________________________
From: Edward <discussions@mathforum.org>
To: discretemath@mathforum.org
Sent: Sunday, 14 October 2012, 18:32
Subject: Defining "f o g" with Functions
So I've been learning about functions. Ideas such as one-to-one, onto, bijections....and so on. Things have been going smoothly and I've been going on my own through the text trying to do all of the exercises to prove to myself I understand the concepts on my own but I came into a problem where I don't know where to begin. I'm not really familiar with "g o f" and "f o g". Can somebody show me how to figure this out:
Let S = {1,2,3,4,5} and let T = {3,4,5,6,7}. Define function f : S --> T and g : S --> S as follows:
f = {(2,6), (1,7), (3,4), (5,3), (4,5) and
g = {(2,3), (1,2), (4,3), (5,5), (3,1)}.
(a.) Find "f o g" or explain why "f o g" is not defined, Repeat for "g o f", "f o f" and "g o g".
(b.) Which (if any) of f, g are one-to-one? Which (if any) are onto? Explain
(c.) Find f^-1 if it exists. If not explain why.
(d.) Find g^-1 if it exists. If not explain why.
*** ^ represents exponent ***
Anybody who understands these concepts better than I do?
Date Subject Author
10/14/12 Edward
10/15/12 Angela Richardson
10/21/12 Edward
11/4/12 Salahuddin
12/31/12 grei | 718 | 2,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-17 | latest | en | 0.918671 |
https://comitatosocivalsabbina.it/hammer_screen/2018-01-02/3209.html | 1,624,327,140,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00215.warc.gz | 175,244,422 | 7,563 | # Fraction tiles app
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https://www.quizzes.cc/metric/percentof.php?percent=61&of=5230 | 1,637,983,047,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00056.warc.gz | 1,110,072,690 | 4,358 | #### What is 61 percent of 5,230?
How much is 61 percent of 5230? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 61% of 5230 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 61% of 5,230 = 3190.3
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating sixty-one of five thousand, two hundred and thirty How to calculate 61% of 5230? Simply divide the percent by 100 and multiply by the number. For example, 61 /100 x 5230 = 3190.3 or 0.61 x 5230 = 3190.3
#### How much is 61 percent of the following numbers?
61% of 5230.01 = 319030.61 61% of 5230.02 = 319031.22 61% of 5230.03 = 319031.83 61% of 5230.04 = 319032.44 61% of 5230.05 = 319033.05 61% of 5230.06 = 319033.66 61% of 5230.07 = 319034.27 61% of 5230.08 = 319034.88 61% of 5230.09 = 319035.49 61% of 5230.1 = 319036.1 61% of 5230.11 = 319036.71 61% of 5230.12 = 319037.32 61% of 5230.13 = 319037.93 61% of 5230.14 = 319038.54 61% of 5230.15 = 319039.15 61% of 5230.16 = 319039.76 61% of 5230.17 = 319040.37 61% of 5230.18 = 319040.98 61% of 5230.19 = 319041.59 61% of 5230.2 = 319042.2 61% of 5230.21 = 319042.81 61% of 5230.22 = 319043.42 61% of 5230.23 = 319044.03 61% of 5230.24 = 319044.64 61% of 5230.25 = 319045.25
61% of 5230.26 = 319045.86 61% of 5230.27 = 319046.47 61% of 5230.28 = 319047.08 61% of 5230.29 = 319047.69 61% of 5230.3 = 319048.3 61% of 5230.31 = 319048.91 61% of 5230.32 = 319049.52 61% of 5230.33 = 319050.13 61% of 5230.34 = 319050.74 61% of 5230.35 = 319051.35 61% of 5230.36 = 319051.96 61% of 5230.37 = 319052.57 61% of 5230.38 = 319053.18 61% of 5230.39 = 319053.79 61% of 5230.4 = 319054.4 61% of 5230.41 = 319055.01 61% of 5230.42 = 319055.62 61% of 5230.43 = 319056.23 61% of 5230.44 = 319056.84 61% of 5230.45 = 319057.45 61% of 5230.46 = 319058.06 61% of 5230.47 = 319058.67 61% of 5230.48 = 319059.28 61% of 5230.49 = 319059.89 61% of 5230.5 = 319060.5
61% of 5230.51 = 319061.11 61% of 5230.52 = 319061.72 61% of 5230.53 = 319062.33 61% of 5230.54 = 319062.94 61% of 5230.55 = 319063.55 61% of 5230.56 = 319064.16 61% of 5230.57 = 319064.77 61% of 5230.58 = 319065.38 61% of 5230.59 = 319065.99 61% of 5230.6 = 319066.6 61% of 5230.61 = 319067.21 61% of 5230.62 = 319067.82 61% of 5230.63 = 319068.43 61% of 5230.64 = 319069.04 61% of 5230.65 = 319069.65 61% of 5230.66 = 319070.26 61% of 5230.67 = 319070.87 61% of 5230.68 = 319071.48 61% of 5230.69 = 319072.09 61% of 5230.7 = 319072.7 61% of 5230.71 = 319073.31 61% of 5230.72 = 319073.92 61% of 5230.73 = 319074.53 61% of 5230.74 = 319075.14 61% of 5230.75 = 319075.75
61% of 5230.76 = 319076.36 61% of 5230.77 = 319076.97 61% of 5230.78 = 319077.58 61% of 5230.79 = 319078.19 61% of 5230.8 = 319078.8 61% of 5230.81 = 319079.41 61% of 5230.82 = 319080.02 61% of 5230.83 = 319080.63 61% of 5230.84 = 319081.24 61% of 5230.85 = 319081.85 61% of 5230.86 = 319082.46 61% of 5230.87 = 319083.07 61% of 5230.88 = 319083.68 61% of 5230.89 = 319084.29 61% of 5230.9 = 319084.9 61% of 5230.91 = 319085.51 61% of 5230.92 = 319086.12 61% of 5230.93 = 319086.73 61% of 5230.94 = 319087.34 61% of 5230.95 = 319087.95 61% of 5230.96 = 319088.56 61% of 5230.97 = 319089.17 61% of 5230.98 = 319089.78 61% of 5230.99 = 319090.39 61% of 5231 = 319091
1% of 5230 = 52.3 2% of 5230 = 104.6 3% of 5230 = 156.9 4% of 5230 = 209.2 5% of 5230 = 261.5 6% of 5230 = 313.8 7% of 5230 = 366.1 8% of 5230 = 418.4 9% of 5230 = 470.7 10% of 5230 = 523 11% of 5230 = 575.3 12% of 5230 = 627.6 13% of 5230 = 679.9 14% of 5230 = 732.2 15% of 5230 = 784.5 16% of 5230 = 836.8 17% of 5230 = 889.1 18% of 5230 = 941.4 19% of 5230 = 993.7 20% of 5230 = 1046 21% of 5230 = 1098.3 22% of 5230 = 1150.6 23% of 5230 = 1202.9 24% of 5230 = 1255.2 25% of 5230 = 1307.5
26% of 5230 = 1359.8 27% of 5230 = 1412.1 28% of 5230 = 1464.4 29% of 5230 = 1516.7 30% of 5230 = 1569 31% of 5230 = 1621.3 32% of 5230 = 1673.6 33% of 5230 = 1725.9 34% of 5230 = 1778.2 35% of 5230 = 1830.5 36% of 5230 = 1882.8 37% of 5230 = 1935.1 38% of 5230 = 1987.4 39% of 5230 = 2039.7 40% of 5230 = 2092 41% of 5230 = 2144.3 42% of 5230 = 2196.6 43% of 5230 = 2248.9 44% of 5230 = 2301.2 45% of 5230 = 2353.5 46% of 5230 = 2405.8 47% of 5230 = 2458.1 48% of 5230 = 2510.4 49% of 5230 = 2562.7 50% of 5230 = 2615
51% of 5230 = 2667.3 52% of 5230 = 2719.6 53% of 5230 = 2771.9 54% of 5230 = 2824.2 55% of 5230 = 2876.5 56% of 5230 = 2928.8 57% of 5230 = 2981.1 58% of 5230 = 3033.4 59% of 5230 = 3085.7 60% of 5230 = 3138 61% of 5230 = 3190.3 62% of 5230 = 3242.6 63% of 5230 = 3294.9 64% of 5230 = 3347.2 65% of 5230 = 3399.5 66% of 5230 = 3451.8 67% of 5230 = 3504.1 68% of 5230 = 3556.4 69% of 5230 = 3608.7 70% of 5230 = 3661 71% of 5230 = 3713.3 72% of 5230 = 3765.6 73% of 5230 = 3817.9 74% of 5230 = 3870.2 75% of 5230 = 3922.5
76% of 5230 = 3974.8 77% of 5230 = 4027.1 78% of 5230 = 4079.4 79% of 5230 = 4131.7 80% of 5230 = 4184 81% of 5230 = 4236.3 82% of 5230 = 4288.6 83% of 5230 = 4340.9 84% of 5230 = 4393.2 85% of 5230 = 4445.5 86% of 5230 = 4497.8 87% of 5230 = 4550.1 88% of 5230 = 4602.4 89% of 5230 = 4654.7 90% of 5230 = 4707 91% of 5230 = 4759.3 92% of 5230 = 4811.6 93% of 5230 = 4863.9 94% of 5230 = 4916.2 95% of 5230 = 4968.5 96% of 5230 = 5020.8 97% of 5230 = 5073.1 98% of 5230 = 5125.4 99% of 5230 = 5177.7 100% of 5230 = 5230 | 2,969 | 5,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.79171 |
https://cardinaloconnorconference.com/new-brunswick/example-of-test-command-stata.php | 1,586,058,748,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00019.warc.gz | 375,189,257 | 6,751 | Example of test command stata
New Brunswick - 2020-03-03
# Stata example command of test
## Regression Analysis Using Stata – A Very Basic Introduction. Stata 12 Tutorial 2 Queen's Economics Department.
Power and sample size determination using Stata Medical Biometry I Autumn 2012 The Stata function sampsi can be used to estimate the required sample size or power of. 29/01/2010В В· In order to perform a wald test in stata, you can simply use the "test" command a wald test in stata, for example whether the ratio of.
Examples of logistic regression. We can test for an overall effect of rank using the test command. Beyond BinaryLogistic Regression with Stata in the STATA command window. To construct either a one or two-sample t-test use the command ttest. For example Stata_CI_tests.doc
Test the mayor’s claim at Using Stata for one sample tests – Page 5 . If you are analyzing the original raw data, use the ttest command. In this example, Loading a Stata-Format Dataset into Stata-- use In Stata 12 Tutorial 1, use the Stata test and lincom commands to Stata 12 Tutorial 7 M.G. Abbott Test
Stata-R Translations we have to call the cumulative distribution function command in R or in STATA. Example 1 has Two-sample t test with equal variances. Throughout this chapter, Stata commands appear in bold font , type help followed by the command (for example Test of RR=1 : z= 7.75 p = 0.000.
“Useful Stata Commands Phil Ender”.
Propensity Score Matching in Stata using teffects. An Example of Propensity Score Matching. Run the following command in Stata to load an example data set:.
Efficiency analysis using STATA STATA does not have a built-in command to perform DEA. The example do-file DEA.do provides an example of using the . dea.. 2. When the results of a Stata command fill up more than one screen, Stata pauses to let you review what's currently on the screen. When you want to continue, press. The ttest command performs t-tests for one sample, two samples and paired observations. The single-sample t-test compares the mean of the sample to a given number.
## Jpa named entity graph example
What’s New In simple-jpa 0.6? The Solid Snake. This type represents the root of an entity graph that will be used as a JPA 2.1; Method Summary Return the name of a named EntityGraph - an entity graph …
## Example Of Idioms Used In A Sentence
French idioms for conversation Speak French Fluently. What’s an Idiom? Broadly speaking, an idiom is a widely used phrase that, when taken as a whole, has a particular meaning that you would… … | 629 | 2,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.771078 |
https://eric.ed.gov/?id=EJ1051727 | 1,716,343,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00168.warc.gz | 200,569,215 | 4,341 | Collection
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Peer reviewed
ERIC Number: EJ1051727
Record Type: Journal
Publication Date: 2015-Feb
Pages: 5
Abstractor: ERIC
ISBN: N/A
ISSN: ISSN-0025-5769
EISSN: N/A
Developing the Vertex Formula Meaningfully
Nebesniak, Amy L.; Burgoa, A. Aaron
Mathematics Teacher, v108 n6 p429-433 Feb 2015
As teachers working with students in entry-level algebra classes, authors Amy Nebesniak and A. Aaron Burgoa realized that their instruction was a major factor in how their students viewed mathematics. They often presented students with abstract formulas that seemed to appear out of thin air. One instance occurred while they were teaching students to graph quadratic equations. The algebra textbook in their school district introduced the concept by giving students a formula for the x-coordinate of the vertex as x = -b/(2a). (Note: The vertex form y = a(x - h)[superscript 2] + k is not introduced in the textbook and district standards until advanced algebra.) For a number of years, they provided students with the vertex formula, and the students successfully graphed by substituting values into the formula. Yet when asked where the formula came from or how it connected to the defining characteristics of quadratics, students did not know. They were performing procedures using this "magical" formula but did not understand how the formula developed. In this article, Nebesniak and Burgoa present a method for graphing quadratic equations that connects students' prior knowledge of graphing linear equations with intercepts and calculating mean. This atypical approach is effective only for a quadratic that has real roots. However, they use this method because the goal is to help students reason how to graph a quadratic with understanding rather than memorize the vertex formula. They believe that this initial approach to graphing is beneficial, despite the limitations, because students are making sense of the concept by connecting it to prior knowledge (NCTM 2009). A brief bibliography is included.
National Council of Teachers of Mathematics. 1906 Association Drive, Reston, VA 20191-1502. Tel: 800-235-7566; Tel: 703-620-3702; Fax: 703-476-2970; e-mail: orders@nctm.org; Web site: http://www.nctm.org/publications/
Publication Type: Journal Articles; Reports - Descriptive
Education Level: Secondary Education
Audience: Teachers
Language: English | 540 | 2,366 | {"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} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.94857 |
https://www.boatdesign.net/threads/once-again-on-righting-moment-versus-heeling-moment.50423/ | 1,611,815,444,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00401.warc.gz | 684,870,924 | 24,491 | # Once again on righting moment versus heeling moment
Discussion in 'Sailboats' started by laukejas, May 14, 2014.
1. Joined: Feb 2012
Posts: 738
Likes: 17, Points: 18, Legacy Rep: 128
Location: Lithuania
### laukejasSenior Member
Hi fellow skippers,
I'm working on a 4.2 monohull sailboat for 2 people. Balanced lug rig. 206kg displacement.
I'm trying to determine sail shape and size, was well as reefed sail area, so that I can know what the balance of the boat will be at various wind speeds.
I believe I understand the righting moment calculation. In my case, it goes as follows:
GZ (waterline-paralel distance between buoyancy center and center of gravity) at 5° heel angle = .32m.
Displacement = 206kg.
RM = GZ * Displacement * gravitational acceleration = .32m*206kg*9.81m/s^2= 651Nm
Now, for the heeling moment of the sails:
Sail area = 9m^2
Heel angle = 5°
Wind speed = 7m/s
Air density = 1.225kg/m^3
Height of Center of Effort = 2.56m
Wind pressure = 1/2 * Air density * Wind speed^2 = 1/2*1.225kg/m^3*(7m/s)^2 = 30.01
Wind force = Wind pressure * Sail area = 30.01 * 9m^2 = 270.11
Heeling moment = Wind force * Height of Center of Effort = *270.11 * 2.56m = 691.5Nm
Adjustment for heel angle = Heeling moment * cos(Heel angle) = 691.5 * cos(5°) = 688.8Nm
Since I have never ever done this before, I am not sure if my calculations are correct.
Could someone please check and correct? Are these formulas accurate enough?
P.S. what are traditional windspeeds for boat of similar size to put in 1st and 2nd reef?
P.P.S. One of the things I'm not sure is if wind speed should be measured in m/s, km/h, or knots...
Last edited: May 15, 2014
2. Joined: Jul 2012
Posts: 1,196
Likes: 27, Points: 48, Legacy Rep: 152
Location: United States
### SkyakSenior Member
Hi laukejas,
while the result looks good to me I can't follow how you found it. Gz is typically used only on keel boats. To be honest I have never seen a really good calculation for dingy stability. Mostly I see the calculation of the weight of the crew times the distance from the center (CB) to the position they would sit on the deck equal to the force on the sail times the distance from the center of effort of the sail to the center of effort of the dagger board. The force on the sail per height is equal to the reynolds number times density times the coefficient of lift. CL of 1 is about right so I think that worked out for you.
Wind speed of the first reef varies based on the local conditions. If local wind tends to be light you will want a large sail and reef early. I think we should create a spreadsheet to consider how well the boat will sail with 2 crew and light air and solo in heavy wind. If you can cover both you have a good rig.
3. Joined: Feb 2012
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Location: Lithuania
### laukejasSenior Member
Skyak, thank you for your reply. You devotedly help me out again It's good to be back.
I made a very complex model of this sailboat in Solidworks (CAD software). I included model of two persons in that sailboat, and set their position more or less to real hiking situation (sitting on the gunwales, leaned back).
This software is able to calculate Center of Mass for the boat AND the crew together.
So, I just drew two lines - one from center of mass, one from center of buoyancy at 5° heel, both parallel to each other and perpendicular to waterline. Distance between these lines is GM.
I actually expressed formulas in a way that I edit the sailboat model (for example, shift crew position, change heel angle, change sail shape or reef positions, etc.), and Solidworks tells me what wind speeds can each configuration carry. Very automated.
Right now, it says (at 5° heel):
Full sail: 7.06m/s
Reef 1: 10.03m/s
Reef 2: 13.61m/s
This doesn't look too good, as I want boat to be manageable in 20+ m/s gusts. Reef 2 now gives sail area of 2.96m^2 and it's CE is 204.67cm above waterline.
Maybe I should consider a stormsail, I don't know. Never that saw on dinghies, though. Maybe another reef point?
I'm attaching a screenshot of the model, just for reference. Model is still incomplete, many parts missing, but it's enough for rough calculations of weight and displacement, I guess.
http://i.imgur.com/qNIah7U.jpg
Can you explain that a bit, please? I highlighted the part which is completely beyond me right now...
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### philSweetSenior Member
Accuracy isn't the issue here. This just isn't how to go about it.
One old rule of thumb was to estimate the sideforce of the sail at 1 pound per sq foot of plain sail. Use the vertical distance between the sail's centroid and the centroid of the fin. Ignore the hull and extend the fin's leading and trailing edges straight up to the surface in order to estimate the vertical center of lateral resistance as the centroid of the extended fin.
Once you have this, there are all kinds of rules-of-thumb and finagling factors that can be applied to specific types of boats, rigs, and sailing habits. The ability to hold this pressure generally indicates a reef needs to be tucked in the 18-20 knot TWS range, so it might be a bit high for a pond dinghy. Baseline of RM @ 30 = SF @ 1PSF.
The following chart came from the Plyboats booklet. I have not received any reply from repeated requests (over the past 8 years) to reproduce charts from this booklet. I believe they are properly in the public domain now. You can also check the data on http://sailboatdata.com/, but I have found a lot of errors on that site, so if you key in on one or two comparables, I would advise you to get the data from the mfg.
#### Attached Files:
• ###### sail area.jpg
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### laukejasSenior Member
Well, but this doesn't include wind speed. I take your meaning that this is general approximation for finding sail area that would be small enough to sail up to 15 knots, and then determine each reef by reducing 25-30% of sail area, right? I believe I saw something like this before.
This rule of thumb seems very approximate for me. Why is my described method (presumably more accurate) not suited here?
Why is hull ignored in calculations?
By the way, according to the chart you provided, sail area for my boat should be around 10.2m^2, considerably larger than current (8.81m^2).
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### laukejasSenior Member
I believe I managed to think this out. However, I see three problems:
1) Sail area isn't included in formula;
2) Weight of the boat itself is also not included - after all, it weights 40 kg, and affects CoG considerably;
3) How can I find out Reynolds number?
If I got it correctly, I converted formula to metric system, and finally got 392Nm of heeling force for a 8.81m^2 sail and 3.05 distance between CE and CLR of the fin. Is this about right?
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### SkyakSenior Member
laukejas,
the righting moment you got from CAD is excellent and with that I would say your calculation makes sense and worked with the possible exception of the distance between the sail CE and dagger board CE. The reason the hull vertical area is not counted is that in a fast dingy like yours the hull is shallow and rounded and will not and should not produce much lateral force at the low leeway the daggerboard dictates. If solidworks counts vertical hull area it is because it assumes a deep keelboat.
I am surprised that solidworks does any aerodynamic or hydrodynamic calculations. What version are you working in? I would also say you have developed some CAD talent. If you are still short money for the project maybe you could freelance CAD work.
It has been pointed out to me that I incorrectly cited Reynolds number. What I was trying to say is that your calculation worked because a coefficient of lift equal to one is about right so your Bernoulli pressure times area is equivalent.
I am not sure if you know Phil but he is a greater authority than I am. The way he describes determining dingy sail area is the way I have seen it done most, rough estimates and compare to past examples. For every improvement in performance calculations there are ten ways it could be misleading. That said, I think there is value in more precise modeling of what is going on but I would not delay your project one second -the old ways are good enough to build to.
About righting in a gust; all of these estimates are for the sail at an angle of attack close to stall around CL max. In operation you will not cleat the sheet. The crew will hold it most of the time and play it out when a gust heels you over. I have not seen any calculation for ultimate wind force (force in a strong wind with little or no force on the sheet) but I do know that the best thing you can do is add battens to stop the sail from flapping and causing drag -more like a junk rig than a lug.
There are also 'soft wing' rigs that are like junk rigs with a wing shape forward around the mast. I have been thinking about prototyping one of these for my laser. Another rig you might be interested in is Matt L's Paradox -a balanced lug that roller furls onto the boom providing easy continuous reefing.
So good work and though I don't want to slow you down with 'paralysis by analysis' I find it hard to resist doing the math when you have such an excellent model. This forum has been pretty dull lately and I am interested in what Phil has learned from the aerodynamics class we are taking.
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### laukejasSenior Member
Well you went ahead of me! This was my next question, how can I calculate lateral area if hull is shallow. I assumed that if hull has little or no slope ( U - shaped ), then it's lateral area should be calculated at 100%. If it were V shaped with 45° slope, only 50% of it's lateral area should be calculated. If it were flat-bottomed and very shallow - near 0%.
This bugged me for a very long time. I guess with dinghy as shallow as mine, hull lateral area can be excluded from calculations, as you suggest. It'll make things easier, and in practice, if hull does provide some additional lateral resistance - why not?
Speaking of which, how do I size fin proportionally? Lars Larsson and Rolf E Eliasson suggest that it should be 3.5% of total sail area. However, this is for yachts (like YD-40), not for dinghies. Since hull provides little lateral resistance in my case, I guess fin should be larger - but by how much?
I really don't want to under-size it. Lithuanian lake winds tend to be treacherous, for you can expect dead calm after a 30kt gusts a minute ago. And I know how difficult is reaching in glassy waters when you have to tack two times a minute due to narrow passages.
My point is, I want this boat to perform reasonably in 1-3 knot winds. Since in low speeds fin will act less like wing and more like a keel, so I guess I just need it big.
Question is, how big? Right now my fin extends around 90cm below waterline, and weights 2.6kg. It is exactly 3.5% of the sail area. How much meat do I add?
Well, it doesn't. It has nothing on this. And it works slow with Top-Down designs. However, it has equations. It can measure weight, area, distance, center of gravity, and so on, and outputs these in numbers. Then, it is only a matter of dedication to input all the formulas. Right now, I have around 20 equations which I took mainly from the "Principles of sailboat design", then ~10 other books, wikipedia, this forum, common sense, and then basically programmed SolidWorks into sailboat stability calculator.
This is how it looks. Took me a week to program (mostly to learn). Sorry it is in Lithuanian - I prefer to work in my native language (it also encourages me to learn nautical terms). But you can get the idea of how scary it looks:
For example, numbers 7 to 9 are lead of CE ahead of CLR in cm for full sail and reefs 1&2, 10-12 - same in percent of total waterline. 15-17 are areas for full sail and reefs. 20 is aspect ratio of the sail, 21 is lateral area of fin, 22 - sail area/lateral fin area ratio in percent, 23 is righting moment, 24-26 are maximum wind speeds for full sail and each reef.
Highly complex, very accurate. I just have to adjust the model - for example, shift crew position, add a new part, re-shape sail, change hypothetical heel angle (transverse, longitudinal, or both), draft, and these numbers update, so I can quickly see how it will affect boat performance.
I doubt anyone would pay money for my beginner skill. But thank you for the compliment
Well, right now, I have time. I am now deep into my studies (music), I finish Masters next year, so I plan to build this boat next summer, after final exams. It means I have a year to finish this project. Most things are done, but there are still 30+ items on my list to input in this model.
That being said, if I have time, why not make it a bit more accurate?
That's okay! I find it also hard to resist. There is time, so I want to perfect my boat design. There just so many things to make work together. Requirements are steep: folding sailboat for 2 persons, \$700 budget, folding so that no part is longer than 2 meters, car-top-able (so it must weight under 40kg), fast enough to assemble/disassemble within an hour (no screwdrivers or pliers included, no fasteners that could be lost), sailable (reaching ability) in 1-25kt wind speed range, easy on the maintenance, sleeping space for 2 persons (not very comfortable, but possible with current design). It shouldn't be a racer, something between cruiser and a racer, more towards cruiser. However, there is one quality I highly value: it is minimum True Wind Angle when reaching. Not VMG, but TWA. This is because I'll often sail in narrow waters, meaning that if I have high TWA, I'll have to tack very often, and each tack is loss of speed, loss of ground. This is why it would be far better to be able to reach high rather than fast in such conditions. In more open waters, naturally, VMG is more important, but, as I said, I expect a lot of maneuvering.
Most of these requirements seem reasonable with current design. I make regular trips to my local hardware store, look what parts are cheapest, measure them, go back home, open up Solidworks, try to combine them, adjust requirements, go back to hardware store again... Very tiresome work, heavy on the head.
But if I manage to actually design and build it, it'll be really something.
#### Attached Files:
• ###### formulas.jpg
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### SkyakSenior Member
Well if you have the time and interest in doing the calculations I am all for it!
Considering the hull as a foil for lateral force on the water moving at say 4 kn, if you angle the hull 3 degrees you will feel a kg or two of lateral force and about a kg of drag added -terrible lift/drag. If you do the same exercise with the daggerboard down you will feel 100+kg of lateral force and mid to high double digit lift/drag. So despite the fact that the the hull may have equal or greater lateral area than the DB it is so poor as a lateral foil that you should give it no weight in calculating righting moment. If you averaged it with the DB area it would move the CE much higher as if the hull lateral area worked as hard as the DB area -wrong.
The reason the hull is so poor laterally is that it has super low aspect ratio. The free end of a foil has zero lift. All the pressure it creates just pulls fluid around the end from the high pressure side to the low. As we move along a foil away from the end and toward the middle the pressure increases toward the theoretical 'infinite span' value. This is why performance focused designers are so obsessed with aspect ratio. High AR gives the highest performance/area. The one exception to high AR is sailing straight downwind (where sail drag would be a benefit) AR of 1 would be best. And before I leave span-wise load let me say that at BEST the load distribution will be elliptical.
So the 2 dimensional lift per unit length of an infinite foil is half*density*velocity^2*cord*coeficient of lift. If the average Cl=1 then the lift per length times length of the sail is the same as your Bernoulli pressure times area. I said this in my last post but I bring it up again because it is the minimum theory you have to consider to relate foil size, force, and fluid speed. With this in mind we can go back and figure what wind speed the standard practice of assuming one pound/sf (4.88kg/m^2) sail area is based on. I get 8.8m/s with the cl=1 and 10.5m/s if the cl was only 0.7 (possibly more realistic average for AR<2. So the 1 lb/sf on the sail calculation prepares you for some pretty high wind speeds with the sail sheeted in.
You asked about dagger board size of 3.5% of the sail area -that's in the ballpark. It looks like more than enough in the picture. If you compare to other dingy designs I think you will find a surprising variety of ratios because it depends on the relationship between air speed and water speed according to the calculation above. On very fast catamarans and winged skiffs you will find small slender foils and huge sails because they operate at double digit water speeds. I don't think your daggerboard needs to be any bigger. My recommendation would be to consider moving it forward so the front of the DB trunk is on the same frame as the mast -then make it a bit smaller and make the rudder bigger. My reasoning is that the mast and the daggerboard trunk have the largest forces so I want them attached. This moves the DB forward of the 'optimal' position which will result in more weather helm. This is no problem with the right size rudder and will actually feel better with more communication of force and stall.
If you want to read about designing a high performance dingy daggerboard here is an article about doing a 505 (two man 5.05m).
The estimate of force and wind speed also puts your other performance desires into perspective. 1 knot wind will not push the boat to any reasonable speed. It just isn't wind and you can't put up enough sail to make up for it's lack of energy -you are better off paddling. 3 knots is something to work with -with enough speed for control and some sense of direction.
25 knots of wind is well past "small craft advisory". I don't know what waters you will be in but 25 kn makes big waves fast and rips their heads off. I think 25 Kn wind is good for calculating boat strength but if it ever happens you will be looking for a safe landing down wind. If you really want to sail in 25 kn you will need a self bailing cockpit and an unsinkable hull.
If you just need to handle 25kn gusts we should look at 'soft wing' designs because their super low drag allows you to just sheet out and stay upright. The soft wings are also great for upwind sailing. The soft wing is like a junk rig but the forward 1/3 of the sail has two sides that go on either side of the mast. I have been thinking about prototyping a small one for my laser to make it easy to rig and sail in strong winds.
Cudos on the programming! I still think it might be a profitable skill and I am wondering which version of solidworks you have? I think I have SW 2001 on one of my PCs.
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### laukejasSenior Member
Thanks for the info, Skyak. I really appreciate you helping me out on such detail. And I'm sorry, but due to my lack of experience and knowledge, I don't understand all of what you're saying. I have several questions...
You meant lateral area? I can't imagine how this relates to righting moment...
Are you talking about hull or sail here? I realize same principles apply, but just want to make sure. I know the basics about AR, I just hadn't realized that hull works like a foil too.
I'm sorry, but I really don't understand this. You're talking about relating sail area to wind speeds, right? The formula you provided, is it in metric or imperial? What units? And how the heck does one calculate Cl or the sail?
Please, take a look at this screenshot now:
http://static.dyp.im/QGZ46BC4zt/a5ccdb2904e0fa5907e0eef410c05ee7.JPG
The small blue points indicate stuff. Zoom in to see labels. On daggerboard, "Sverto SPC" stands for CLR. On sail, BC is center or effort, BC R1 and BC R2 are centers of effort for reef 1 and 2 (these three centers are almost vertical). To the right of BC there is ^SPC, which is vertical projection of lateral resistance (so that the distance between CLR and CE is easier to see).
This screenshot is done after I re-designed my model so that hull lateral area is not included in calculations, only daggerboard, as per your suggestion.
This moved CLR forward a lot. I had to shift the sail forward too, so to keep CE in front of CLR with decent lead. Right now, it is 6.1% of the waterline. However, with such sail position, mast is too short. I can't make it any longer due to design limitations. Also, I'm afraid the bad tack will be really bad.
If I were to move daggerboard even more forward, there is no way to have any kind of CE lead over CLR.
I use 2013 version... I looked it up, and sadly, I can't export it to your version. But I can provide all the information you require, including screenshots.
By the way, various sources suggest that lead should be anywhere from 5% to 30%, because when boat heels, CE actually moves backwards, creating weather helm. If one were to design boat with CE already behind CLR, weather helm would be unmanageable.
- This is not my thoughts, this is what I red on this forum and some other sources. But it seems logical to me. I also noticed that virtually all boats have CE in front of CLR, including Klepper Passat, the folding German boat on which my design is based.
I am sorry I ask so many questions in one post. I hope it's not too difficult to keep track of them all
P.S. Maybe I should include rudder in CLR area and center of area calculations? Then I could solve sail position/mast problem.
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Skyak-is this a misprint: ".... DB it is so poor as a lateral foil that you should give it no weight in calculating righting moment."?
Daggerboard area as it relates to the Center of Lateral Resistance and it's distance below the waterline would be used to calculate heeling moment. On a dinghy the distance from the Center of Effort of the sails to the Center of Lateral Resistance= the Heeling Arm. The force on the sails(pressure in pounds per sq.ft.) times the HA= Heeling Moment, right?
click for better view:
#### Attached Files:
• ###### Heeling Arm.jpg
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### laukejasSenior Member
Doug, are you sure that the heeling arm is the distance between CE and CLR, not Center of Buoyancy? "Principles of yacht design" suggest the latter. It says nothing about CLR at this calculation.
Excerpt from the book:
#### Attached Files:
• ###### transverse stability.jpg
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### DCockeySenior Member
Righting moment due to the interaction of buoyancy and the weight of the vessel is proportional to the horizontal distance between Center of Buoyancy and Center of Gravity, as shown in the figure in laukejas' post above.
Heeling moment due to the interaction of aerodynamic forces and hydrodynamic forces is typically assumed to be proportional to the vertical distance between "CE" and "CLR" as shown in the figure in Doug's post above. (Reality is more complicated but for most purposes this assumption is sufficient.)
If the Righting Moment = Heeling Moment (with sign reversed) then the net rolling moment will be zero and the boat will tend to stay at a constant heel angle.
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Yes,ask an NA or go to page 181 of Frank Bethwaites "High Performance Sailing".
Remember, this is "Heeling Moment" not righting moment and is specifically for a dinghy that is generally sailed close to flat. The CE of the sails pushes the boat to leeward and the combination of hull and board planform area(Center of Lateral Resistance) resists the leeward movement of the boat. The underwater area has the same force on it as the sails have, except opposite.
Note- in the illustration from p.181 of Bethwaites book you can see that the RM=HM.
And that the HA is from the CE of the sails to the CLR(vertical). On lightweight, fast dinghies just find the center of the board and discount the hull area(in my opinion) for this calculation.
#### Attached Files:
• ###### Bethwaite-HM- 002.JPG
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### laukejasSenior Member
Well, thank you for explanation. Dumb me. It makes sense now. I updated my model. I now get 7m/s maximum wind speed for a 8.13m^2 sail.
What about the lead I mentioned earlier? Currently, it is only 3.26% of the waterline (CE ahead of CLR). Is that enough? For reefed sail, do I need different lead?
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different. | 6,416 | 26,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.913675 |
https://uk.mathworks.com/matlabcentral/cody/problems/42580-conic-equation/solutions/3938698 | 1,611,176,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00415.warc.gz | 624,090,389 | 17,767 | Cody
# Problem 42580. Conic equation
Solution 3938698
Submitted on 1 Dec 2020 by Are Mjaavatten
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
R=5; k=-1; s=-5:5; z=[25 16 9 4 1 0 1 4 9 16 25]/10; t=arrayfun(@(x)conic(x,R,k),s); assert(all(abs(t-z)<=4*eps*abs(z)))
2 Pass
R=-5; k=-1; s=-5:5; z=-[25 16 9 4 1 0 1 4 9 16 25]/10; t=arrayfun(@(x)conic(x,R,k),s); assert(all(abs(t-z)<=4*eps*abs(z)))
3 Pass
R=6; k=0; s=0:0.125:2; z=[0 0.001302224649086391 0.005210595859100573 ... 0.01173021649825800 0.02086962844930099 ... 0.03264086885999461 0.04705955010467117 ... 0.06414496470811713 0.08392021690038396 ... 0.1064123829368584 0.1316527028472488 ... 0.1596768068881667 0.1905249806888747 ... 0.2242424739260392 0.2608798583755018 ... 0.3004934424110011 0.3431457505076198]; t=arrayfun(@(x)conic(x,R,k),s); assert(all(abs(t-z)<=4*eps*abs(z)))
4 Pass
R=6800; k=-2; s=10.^(-9:9); z=[7.352941176470588e-23 7.352941176470588e-21 ... 7.352941176470588e-19 7.352941176470588e-17 ... 7.352941176470588e-15 7.352941176470588e-13 ... 7.352941176470548e-11 7.352941176466613e-9 ... 7.352941176073046e-7 0.00007352941136716365 ... 0.007352937201052538 0.7352543677216725 ... 73.13611097583313 5292.973166264779 93430.93334894173 ... 993223.1197327390 9.993202311999733e6 9.99932002312e7 ... 9.9999320002312e8]; t=arrayfun(@(x)conic(x,R,k),s); assert(all(abs(t-z)<=4*eps*abs(z)))
5 Pass
R=exp(1); k=pi; s=10.^(-7:0); z=[1.839397205857214e-15 1.839397205857469e-13 ... 1.839397205882986e-11 1.839397208434684e-09 ... 1.839397463604480e-07 0.00001839422981299153 ... 0.001841981926630790 0.2212216213343403]; t=arrayfun(@(x)conic(x,R,k),s); assert(all(abs(t-z)<=4*eps*abs(z)))
6 Pass | 804 | 1,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-04 | latest | en | 0.329627 |
http://forum.arduino.cc/index.php?topic=67909.msg500346 | 1,475,102,153,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661768.10/warc/CC-MAIN-20160924173741-00068-ip-10-143-35-109.ec2.internal.warc.gz | 94,775,390 | 14,291 | Go Down
### Topic: Any ideas how to increase floating point accuracy? (Read 2727 times)previous topic - next topic
#### Spine
##### Jul 29, 2011, 04:20 pm
Hi everyone,
I have a calculation that I would like to perform in the Arduino that I would like to gain a little more accuracy from.
Basically what this code does is calculates the intersection point of three circles (I am using it for a TDOA problem). The problem is that these two equations are the least accurate (when comparing to using a calculator or excel) and output is off (in one case, the Y coordinate was off by almost 1cm) too much for my application. I have a feeling this is due to the floating point precision of the arduino but I was wondering if there is something I could do, possibly optimizing the math or to use a fixed point library (which I couldn't get to compile) to increase the accuracy/resolution of the output.
The equations with the highest errors are :
x01 = ((sq(d01) - sq(Rcircle1) + sq(Rcircle0)) / (2.0 * d01)) + c0offset;
and:
ypos = sqrt(sq(Rcircle1) + sq(x01));
With that being said, are there any math wizzes or people with experience with the fixed point libraries that could help me out? I appreciate all the help I can get!
Quote
void find_pos(){
// declare variables for this function
float Rcircle0 = 0.0;
float Rcircle1 = 0.0;
float Rcircle2 = 0.0;
float x01 = 0.0;
float x02 = 0.0;
float xerror = 999999.0;
float xerrornew = 999999.0;
// set the increment for the iteration (this should be not much smaller then the resolution of your equipment)
float inc = 0.1; // set to 0.1cm or 1mm
// set the distance between the microphones (in cm)
float c0offset = -60.0;
float c1offset = 0.0;
float c2offset = 60.0;
// calculate the distances between mic 0 and 1 and mic 1 and 2
float d01 = c1offset - c0offset;
float d02 = c2offset - c0offset;
// for now the speed of sound is hard coded. TBD: compensate based on air temperature
float Tair = 20.0; // Air temperature hard coded to 20degC
float Vsound = 2004.57*(sqrt(Tair+273.15)); //speed of sound in cm/s
// After the time deltas are determined from the microphone processing logic, and normalized so that one microphone is
// time 0, the first step is to calculate the circle radii based on the speed of sound.
// Rcircle0 is the radius of mic0
// Rcircle1 is the radius of mic1
// Rcircle2 is the radius of mic2
// timeL is the normalized time of Mic0 (time0)
// timeC is the normalized time of Mic1 (time1)
// timeR is the normalized time of Mic2 (time2)
Rcircle0 = timeL* Vsound/1000000; // Radius is in cm
Rcircle1 = timeC* Vsound/1000000; // Radius is in cm
Rcircle2 = timeR* Vsound/1000000; // Radius is in cm
// for every loop, increment each radius by the increment value and then calculate the X coordinate of the
// radical lines of the intersection of circles 0 and 1, and then do it again for circles 0 and 2.
// the difference between these two positions is the error. we are trying to find the point with the least error.
// loop to the radius of the offset of the mics
for (float xloop=0.0; xloop < (c2offset - c0offset); xloop=xloop+inc)
{
// increment each circle by the increment value
Rcircle0 = Rcircle0 + inc;
Rcircle1 = Rcircle1 + inc;
Rcircle2 = Rcircle2 + inc;
// calculate the radical lines
x01 = ((sq(d01) - sq(Rcircle1) + sq(Rcircle0)) / (2.0 * d01)) + c0offset;
x02 = ((sq(d02) - sq(Rcircle2) + sq(Rcircle0)) / (2.0 * d02)) + c0offset;
xerrornew = abs(x01-x02);
if (xerrornew < xerror)
{
xerror = xerrornew;
xpos = x01;
ypos = sqrt(sq(Rcircle1) + sq(x01));
temp1 = Rcircle1;
}
}
}
#### MarkT
#1
##### Jul 29, 2011, 05:58 pm
That code is incomplete and I can't test it. Can you reduce the issue down to something that can be reproduced and is obviously returning an imprecise answer (ie differs from what a calculator says). Otherwise it might be that you've made a mistake in your code...
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### PaulS
#2
##### Jul 29, 2011, 11:36 pm
Code: [Select]
` for (float xloop=0.0; xloop < (c2offset - c0offset); xloop=xloop+inc)`
Using float variables for for loop indices is rarely a good idea. Either make this a while loop, or change the logic so that the index can be an int.
#### tedcook
#3
##### Jul 30, 2011, 12:12 amLast Edit: Jul 30, 2011, 12:27 am by tedcook Reason: 1
What happens if you declare everything double, instead of float? That would be my first step if I think I'm running into float precision limits.
*edit: oops, just actually read the info on double in the reference. I had seen double was listed under data types so I assumed Arduino could do double, but apparently not. I was under the impression Arduino was a straightforward C implementation, are there any other caveats to C that one should be aware of?
#### EVP
#4
##### Jul 30, 2011, 01:19 am
I had no idea you could use floats in a for loop. I'm sure i had looked at this years ago and it wasn't possible?
Cant you just multiply everything by multiples of 10 to make the numbers bigger or something along those lines.
#### David Pankhurst
#5
##### Jul 30, 2011, 07:58 am
Unfortunately, double is simply a synonym for float on the Arduino, and a quick look online showed no library for higher precision floating point (but it was a quick search - if anyone finds one, I too would like to know).
The float is a 32 bit variable, which ends up being about 6-7 digits of precison, and would explain the problem here:
http://www.arduino.cc/en/Reference/Float
For example, even your value of inc (0.1) won't store precisely, so you'll have problems.
One thought - convert these to longs:
Code: [Select]
`long c0offset = -60;long c1offset = 0;long c2offset = 60;long d01 = c1offset - c0offset;long d02 = c2offset - c0offset;`
Since they all seem to be integer anyways, and then change the formula:
Code: [Select]
`x01 = ((sq(d01) - sq(Rcircle1) + sq(Rcircle0)) / (2.0 * d01)) + c0offset;`
to
Code: [Select]
`long temp1=2*d01;long temp2=d01*d01;x01 = ((temp2 - sq(Rcircle1) + sq(Rcircle0)) / temp ) + c0offset;`
It won't do much, but you'll save two floating point multiplies, and resultant loss of precision.
#### MarkT
#6
##### Jul 30, 2011, 12:06 pm
Sorry people to disagree: 7 digits is more than plenty for these calculations - this is simple trigonometry.
My suspicion is a bug in the program, use of int where float was intended, different rounding/truncation - or it might be a bug in Arduino float library, but that's less likely I feel.
Another possibility is use of a badly conditioned numerical technique that is multiplying up errors, often a gotcha for naive application of equations without error-analysis.
If the complete code was available someone could reproduce the problem and work out what's going wrong.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### Spine
#7
##### Aug 01, 2011, 04:36 am
Ok, It turns out I am an idiot!
I had a slight error in my Y position calculation, + instead of - (comes from c^2 - a^2 = b^2). I was so involved with thinking of all these other possibilities I overlooked a simple pythagorean theorem typo.
I re-ran my simulations with comparing to microsoft excel and the arduino answers are on average about 0.5mm off either direction with the worst case around 1mm error. This is well within the error of other parameters in my system so It is good enough for this application.
Thanks to all that answered and sorry for my massive loss of brain function!
#### MarkT
#8
##### Aug 02, 2011, 11:46 pm
Its always good to do a back-of-the-envelop sanity calculation to check the result is in the right ballpark - unless you were triangulating on the scale of GPS I knew single floats are accurate enough...
If you frame a posting along the lines of "this is what I expect to see", "but this is what I am seeing", then someone can have a go and working out which is actually wrong... If you just say "blah is broken" we don't have a handle on things.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### robtillaart
#9
##### Aug 03, 2011, 03:23 pm
you are using floats to represent the mm.
Code: [Select]
` float inc = 0.1; // set to 0.1cm or 1mm`
You could also do most/all of the math in mm. You will get
Code: [Select]
` int inc = 1; // set to 1 mm unsigned long Rcircle0 = 0; // a radius cannot be negative afaik etc `
Using long instead of float would improve the accuracy, speed up the math and decrease the footprint of the sketch. There might be a few places were floats are needed but give it a try.
Rob Tillaart
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)
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https://www.statisticshowto.datasciencecentral.com/polya-urn/ | 1,580,076,733,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00138.warc.gz | 1,104,171,759 | 10,573 | # Pólya Urn: Definition, Examples
Urn models > Pólya Urn
## What is a Pólya Urn?
Pólya’s urn is a famous sampling model used in probability. Let’s say you had an urn with red and green balls. You choose one ball at random, note the color, and replace the ball in the urn along with another ball of the same color. The resulting model is called Polya’s urn process.
• If you add zero balls, the process is the same as sampling with replacement.
• If you add -1 ball (in other words, you remove a ball instead of adding), that’s the same as to sampling without replacement.
In most cases, the process is used only with non-negative integers so that the process can continue indefinitely.
## Interesting Probabilities
At any time, the number of balls in the urn is t + red balls + green balls, where t is the number of trials. At any time, the odd of picking a red ball (or a green ball) are exactly the same. That is, for any time t≥1, the probability of picking a red ball is:
PRt = r / (r+g)
…and the probability of picking a green ball is:
PGt = g / (r+g).
If you actually run a simulation (you can find one here on Andrew Mauboussin’s website), you might be surprised at the results. Faulty logic (similar to the Gambler’s Fallacy) might tell you that if you start off with an equal ratio of colored balls, you’re going to end up with an equal ratio of colored balls. However, that isn’t true.
Let’s say you start off with equal numbers of balls (5 red, 5 green). After the first trial, if you choose a red ball, there will be 6 red and 5 green balls in the jar. That creates an automatic disparity, where you’ll be more likely to choose a red ball than a green ball in the next trial. The odds are now 6/11 that you’ll choose a red ball and 5/11 that you’ll choose green. This factor can snowball out of proportion. In fact, there are endless ways the disparity could play out. For 100 trials, here are two ways the simulation plays out:
## Uses
The Polya Urn is used as an exercise in probability, much in the same way basic probability is introduced using numbers drawn from a bingo machine or cards drawn from a deck. The urn shows how small imbalances (like one extra red ball) can be magnified over time. Although interesting to study for theory purposes, it’s too simple a model to have any use in real-life situations, where multiple factors tend to affect outcomes.
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# 664 in cm664 Tolli Sentimeetrit
in
=
cm
## Kuidas teisendada 664 tolli sentimeetrit?
664 in * 2.54 cm = 1686.56 cm 1 in
## Convert 664 in ühistele pikkused
MõõtühikPikkused
nanomeeter16865600000.0 nm
Mikromeeter16865600.0 µm
Millimeeter16865.6 mm
Sentimeeter1686.56 cm
Toll664.0 in
Jalg55.3333333333 ft
Õu18.4444444444 yd
Meeter16.8656 m
Kilomeeter0.0168656 km
Miil0.010479798 mi
Meremiil0.0091066955 nmi
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664 in Sentimeetrit, 664 in Sentimeeter, 664 Tolli Sentimeeter, 664 Tolli Sentimeetrit, 664 Toll Sentimeeter, 664 Toll Sentimeetrit, 664 Toll cm | 260 | 610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.250355 |
https://sciencedocbox.com/Physics/77082537-Chapter-13-small-signal-modeling-and-linear-amplification.html | 1,656,751,341,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00758.warc.gz | 557,873,737 | 28,767 | # Chapter 13 Small-Signal Modeling and Linear Amplification
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1 Chapter 13 Small-Signal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 13-1
2 Chapter Goals Understanding of concepts related to: Transistors as linear amplifiers dc and ac equivalent circuits Use of coupling and bypass capacitors and inductors to modify dc and ac equivalent circuits Concept of small-signal voltages and currents Small-signal models for diodes and transistors Identification of common-source and common-emitter amplifiers Amplifier characteristics such as voltage gain, input and output resistances, and linear signal range Rule-of-thumb estimates for the voltage gain of common-emitter and common-source amplifiers. 1/4/12 Chap 13-2
3 Introduction to Amplifiers Amplifiers usually use electronic devices operating in the Active Region A BJT is used as an amplifier when biased in the forward-active region The FET can be used as amplifier if operated in the pinch-off or saturation region In these regions, transistors can provide high voltage, current and power gains Bias is provided to stabilize the operating point (the Q-Point) in the desired region of operation Q-point also determines Small-signal parameters of transistor Voltage gain, input resistance, output resistance Maximum input and output signal amplitudes Power consumption 1/4/12 Chap 13-3
4 Transistor Amplifiers BJT Amplifier Concept The BJT is biased in the active region by dc voltage source V BE. Q-point is set at (I C, V CE ) = (1.5 ma, 5 V) with I B = 15 µa (β F = 100) Total base-emitter voltage is: v BE = V BE + v be Collector-emitter voltage is: v CE = 10 i C R C This is the load line equation. 1/4/12 Chap 13-4
5 Transistor Amplifiers BJT Amplifier (cont.) If changes in operating currents and voltages are small enough, then i C and v CE waveforms are undistorted replicas of the input signal. A small voltage change at the base causes a large voltage change at collector. Voltage gain is given by: 8 mv peak change in v BE gives 5 µa change in i B and 0.5 ma change in i C. 0.5 ma change in i C produces a 1.65 V change in v CE. A v = V ce V be = o o = o = 206 Minus sign indicates 180 o phase shift between th einput and output signals. 1/4/12 Chap 13-5
6 Transistor Amplifiers MOSFET Amplifier Concept A v = V ds V gs A v = 4 180o 1 0 o A v = 4.00 MOSFET is biased in active region by dc voltage source V GS. Q-point is set at (I D, V DS ) = (1.56 ma, 4.8 V) with V GS = 3.5 V Total gate-source voltage is: v GS = V GS + v gs 1 V p-p change in v GS yields 1.25 ma p-p change in i D and a 4 V p-p change in v DS 1/4/12 Chap 13-6
7 Transistor Amplifiers Coupling and Bypass Capacitors Capacitors are designed to provide negligible impedance at frequencies of interest and provide open circuits at dc. ac coupling through capacitors is used to inject ac input signal and extract output signal without disturbing Q-point C 1 and C 2 are low impedance coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible. C 3 is a bypass capacitor that provides a low impedance path for ac current from emitter to ground, thereby removing R E (required for good Q-point stability) from the circuit when ac signals are considered. 1/9/12 Chap 13-7
8 Transistor Amplifiers dc and ac Analysis Two Step Analysis dc analysis: Find dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. ac analysis: Find ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. Replace transistor by its small-signal model Use small-signal ac equivalent to analyze ac characteristics of amplifier. Combine end results of dc and ac analysis to yield total voltages and currents in the network. 1/9/12 Chap 13-8
9 Transistor Amplifiers dc Equivalent Circuit for BJT Amplifier All capacitors in the original amplifier circuit are replaced by open circuits, disconnecting v I, R I, and R 3 from circuit. 1/9/12 Chap 13-9
10 Transistor Amplifiers ac Equivalent Circuit for BJT Amplifier Capacitors are replaced by short circuits 1/9/12 Chap 13-10
11 Transistor Amplifiers dc and ac Equivalents for a MOSFET Amplifier Full circuit dc equivalent ac equivalent Simplified ac equivalent 01/10/12 Chap 13-11
12 Small-Signal Operation Diode Small-Signal Model The slope of the diode characteristic at the Q-point is called the diode conductance and is given by: Diode resistance is given by: r d = 1 g d 01/10/12 Chap 13-12
13 Small-Signal Operation Diode Small-Signal Model (cont.) g d is small but non-zero for I D = 0 because slope of diode equation is nonzero at the origin. At the origin, the diode conductance and resistance are given by: g d = I S V T and r d = V T I S 01/10/12 Chap 13-13
14 Small-Signal Operation Definition of a Small-Signal Now let s determine the largest magnitude of v d that represents a small signal. V T i D = I S exp v D 1 V T I D +i d = I S exp V D 1 +I S exp v D V T I D +i d = I S exp V +v D d 1 v d V T Subtracting I D from both sides of the equation, i d =(I D + I S ) v d + V 1 T 2 2 v d V T v d V T V T v 2 d V T v 3 d V T For i d to be a linear function of signal voltage v d, v d << 2V T = 0.05 V or v d 5 mv. Thus v d 5 mv represents the requirement for small-signal (linear) operation of the diode i d =(I D + I S ) v d = g v i d d D = I D + g d v d V T 01/10/12 Chap 13-14
15 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) The bipolar transistor is assumed to be operating in the Forward-Active Region: i C I S exp v BE 1+ v CE and i B i C V T V A Using a two-port y-parameter network: i c = g m v be + g o v ce i b = g π v be + g r v ce The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. β F ( V CE V BE ) 01/10/12 Chap 13-15
16 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) i c = g m v be + g o v ce i b = g π v be + g r v ce i C I S exp v BE 1+ v CE i B i C β F V T V A β o is called the small-signal commonemitter current gain of the BJT. g m = i c = i C = I C v be v v ce=0 BE V Q-point T v be g o = i c v ce v be=0 g π = i b v be v ce=0 g r = i b v ce v be=0 v BE = i C v CE Q-point = i B v BE Q-point = i B v CE Q-point V T I C = V A +V T = I C β o V T = 0 01/10/12 Chap 13-16
17 BJT Small-Signal Operation Current Gain & Intrinsic Voltage Gain Intrinsic voltage gain is defined by: β o = g m r π = β F 1 β 1 I F C β F i C Q point β o > β F for i C < I M, and β o < β F for i C > I M. However, for simplicity β F and β o will be assumed to be equal µ f = g m r o = I C V T For V CE << V A µ f V A V T V A +V CE I C 40V A = V A +V CE V T µ f represents the maximum voltage gain an individual BJT can provide and does not change with operating point. 01/10/12 Chap 13-17
18 Small-Signal Operation BJT Hybrid-Pi Model - Summary Transconductance: g m = I C V T 40I C Input resistance: The hybrid-pi small-signal model is the intrinsic representation of the BJT. Small-signal parameters are controlled by the Q-point and are independent of geometry of the BJT r π = β o V T I C Output resistance: r o = V A +V CE I C = β o g m V A I C or β o = g m r π 01/10/12 Chap 13-18
19 BJT Small-Signal Operation Equivalent Forms of Small-Signal Model Voltage-controlled current source g mv be can be transformed into current-controlled current source, v be = i b r π g m v be = g m i b r π = β o i b i c = β o i b + v ce r o β o i b Basic relationship i c = βi b is useful in both dc and ac analysis when the BJT is in the forward-active region. 01/10/12 Chap 13-19
20 BJT Small-Signal Operation Small-Signal Definition i C = I S exp v BE = I S exp V BE + v be i C = I C + i c = I S exp V BE exp v be = I C exp v be V T I C + i c = I C 1+ v be + 1 V T 2 v be V T V T v i c = I be C v be + 1 v be V T 2 V T 6 V T 3 v be V T For linearity, i c should be proportional to v be with v be << 2V T or v be V. i C I C 1+ v be v = I C + I be C The change in i c that corresponds to small-signal operation is: V T V T i c I C = v be V T 0.005V 0.025V = V T V T = I C + g m v be V T 01/12/12 Chap 13-20
21 BJT Small-Signal Operation Small-Signal Model for pnp Transistor For the pnp transistor Signal current injected into base causes decrease in total collector current which is equivalent to increase in signal current entering collector. So the small-signal models for the npn and pnp devices are identical! 01/12/12 Chap 13-21
22 Common-Emitter Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. Assume that Q-point is already known. 01/12/12 Chap 13-22
23 Common-Emitter Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the base terminal Output signal appears at collector terminal Emitter is common to both input and output signals Thus circuit is termed a Common-Emitter (C-E) Amplifier. The terminal gain of the C-E amplifier is the gain from the base terminal to the collector terminal A vt CE = v c v b = g m R L R L = r o R C R 3 01/12/12 Chap 13-23
24 Common-Emitter Amplifiers Input Resistance and Signal Source Gain (β o +1)R E Define R ib as the input resistance looking into the base of the transistor: R ib = v b i b = r π The input resistance presented to v i is: R in = R I + R B R ib = R I + R B r π The signal source voltage gain is: A v CE = v o v i = v o v b v b v i = A vt CE R B r π R I + R B r π 01/12/12 Chap 13-24
25 Common-Emitter Amplifiers Rule of Thumb Design Estimate A CE CE R v = A B r π CE vt A vt R I + R B r π A CE vt = g m R L R L = r o R C R 3 Typically: r o >> R C and R 3 >> R C A v CE g m R C = 40I C R C I C R C represents the voltage dropped across collector resistor R C A typical design point is I C R C = V CC 3 A v CE 40 V CC 3 =13.3V CC To help account for all the approximations and have a number that is easy to remember, our "rule-of-thumb" estimate for the voltage gain becomes A v CE 10V CC 01/13/12 Chap 13-25
26 Common-Emitter Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-emitter amplifier with a specified Q-point Given data: β F = 100, V A = 75 V, Q-point is (0.245 ma, 3.39 V) Assumptions: Transistor is in active region, β O = β F. Signals are low enough to be considered small signals. Room temperature. Analysis: g m = 40I C = mA r o = V A +V CE I C = ( ) = 9.80 ms r π = β o = 100 =10.2 kω g m 9.8mS 75V V 0.245mA = 320 kω R B = R 1 R 2 =160kΩ 300kΩ =104 kω R L = r o R C R L = 320kΩ 22kΩ 100kΩ =17.1 kω R B r π =104kΩ 10.2kΩ = 9.29 kω 7/20/10 Chap 13-26
27 Common-Emitter Amplifiers Voltage Gain Example (cont.) Analysis (cont): R A v = g m R B r π L R I + R B R in = R I + R B r π r π =10.3 kω = 9.8mS 17.1kΩ ( ) 9.29kΩ = kΩ+ 9.29kΩ R v be = v B r π i R I + R B r v be 0.005V v i 5mV 10.3kΩ = 5.54 mv π 9.29kΩ ( ) = 151 Check the rule-of-thumb estimate: A v CE ( ) = 120 (ballpark estimate) ( ) = V What is the maximum amplitude of the output signal: v o 5.54mV 151 1/13/12 Chap 13-27
28 Common-Emitter Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 5-mV, 10-kHz sine wave. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion. The input is near our small-signal limit for linear operation. 1/13/12 Chap 13-28
29 Common-Emitter Amplifiers Dual Supply Operation - Example Analysis: To find the Q-point, the dc equivalent circuit is constructed I B +V BE +(β F +1)I B ( )= 5 I B =3.71 µa I C =65I B =241 µa Problem: Find voltage gain, input and output resistances for the circuit above Given data: β F = 65, V A = 50 V Assumptions: Active-region operation, V BE = 0.7 V, small signal operating conditions. I E =66I B =245 µa I C V CE ( )I E ( 5)=0 V CE =3.67 V 1/28/12 Chap 13-29
30 Common-Emitter Amplifiers Dual Supply Operation - Example (cont.) Next we construct the ac equivalent and simplify it kω R in = R B r π =6.31 kω R out = R C r o = 9.57 kω ( ) A v CE = v o v i = g m R out R 3 R in = 84.0 R I + R in Gain Estimate: A CE v 10( V CC +V EE ) = 100 1/28/12 Chap 13-30
31 Small-Signal Operation MOSFET Small-Signal Model g π = i g v gs vds =0 g r = i g v ds v ds =0 = i G v GS Q-point = i G v DS Q-point Using a two-port y-parameter network, i d g m = vgs vds=0 i D = vgs Q-point i g = g π v gs + g r v ds i d = g m v gs + g o v ds The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. g o = i d v ds v ds =0 I G = 0 = i D v DS Q-point I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) 1/17/12 Chap 13-31
32 Small-Signal Operation MOSFET Small-Signal Model (cont.) I G = 0 I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) g π = i G v GS Q-point g r = i G v DS Q-point = 0 = 0 i g = g π v gs + g r v ds i d = g m v gs + g o v ds g m = i D v GS Q-point g o = i D v DS Q-point = K n ( 2 V V GS TN )( 1+ λv DS ) = = λ K n ( 2 V GS V TN ) 2 = λi D = 1+ λv DS 2I D V GS V TN I D 1 λ +V DS 1/17/12 Chap 13-32
33 Small-Signal Operation MOSFET Small-Signal Model - Summary Transconductance: g m = 2I D = 2K n I D V GS V TN Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite. Small-signal parameters are controlled by the Q-point. For the same operating point, MOSFET has lower transconductance and an output resistance that is similar to the BJT. Output resistance: r o = 1 g o = 1+λV DS λi D 1 λi D Amplification factor for λv DS <<1: µ f =g m r o = 1+λV DS λi D 1 λ 2K n I D 1/17/12 Chap 13-33
34 MOSFET Small-Signal Operation Small-Signal Definition Assume λv DS <<1. Then i D K n ( 2 v V GS TN ) 2 for v DS v GS V TN For v GS = V GS + v gs, i D = I D +i d = K n 2 2 ( ) + v gs ( v GS V TN ) 2 + 2v gs V GS V TN 2 ( ) + v gs i d = K n 2 2v gs V GS V TN For linearity, i d should be proportional to v gs and we require v 2 gs << 2v gs V GS V TN or v gs << 2( V GS V TN ) v gs 0.2( V GS V TN ) ( ) Since the MOSFET can be biased with (V GS - V TN ) equal to several volts, it can handle much larger values of v gs than corresponding the values of v be for the BJT. The change in drain current that corresponds to small-signal operation is: i d = g m 2 v gs I D I D V GS V TN ( ) 0.2 V GS V TN i d 0.4 I D 1/17/12 Chap 13-34
35 MOSFET Small-Signal Operation Body Effect in Four-terminal MOSFETs Drain current depends on threshold voltage which in turn depends on v SB. Back-gate transconductance is: g mb = i D = i D v BS v Q-point SB Q-point i D g mb = VTN V TN v SB Q-point = ( g m η)=g m η 0 < η < 3 is called the back-gate transconductance parameter. The bulk terminal is a reverse-biased diode. Hence, no conductance from the bulk terminal to other terminals. 01/18/12 Chap 13-35
36 MOSFET Small-Signal Operation Small-Signal Model for PMOS Transistor For a PMOS transistor v SG = V GG v gg i D = I D i d Positive signal voltage v gg reduces sourcegate voltage of the PMOS transistor causing decrease in total current exiting the drain, equivalent to an increase in the signal current entering the drain. The NMOS and PMOS small-signal models are the same! 01/18/12 Chap 13-36
37 Common-Source Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. 01/19/12 Chap 13-37
38 Common-Source Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the gate terminal Output signal appears at the drain terminal Source is common to both input and output signals Thus circuit is termed a Common-Source (C-S) Amplifier. The terminal gain of the C-S amplifier is the gain from the gate terminal to the drain terminal A vt CE = v d v g = g m R L R L = r o R D R 3 01/19/12 Chap 13-38
39 Common-Source Amplifiers Input Resistance and Signal-Source Gain Define R ig as the input resistance looking into the base of the transistor. R vg ig = = R G ii R in is the resistance presented to v i. R in = R I + R G The signal source voltage gain is: A v CS = v o v i = v o v g v g v i CS = A vt A CS R v = g m R G L R I + R G R G R I + R G 01/12/12 Chap 13-39
40 Common-Source Amplifiers Rule of Thumb Design Estimate R A CS v = g m R G L A vt R I + R G CS A CS vt = g m R L R L = r o R D R 3 I Typically: r o >> R D and R 3 >> R D A CS v g m R D = D R D V GS V TN 2 I D R D represents the voltage dropped across drain resistor R D A typical design point is I D R D = V DD 2 with V GS V TN =1 V A v CS V DD Our rule-of-thumb estimate for the C-S amplifier: the voltage gain equals the power supply voltage. Note that this is 10 times smaller than that for the BJT! 01/19/12 Chap 13-40
41 Common-Source Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-source amplifier with a specified Q-point Given data: Κ n = 0.50 ma/v 2, V TN = 1 V, λ = V -1, Q-point is (0.241 ma, 3.81 V) Assumptions: Transistor is in the active region. Signals are low enough to be considered small signals. Analysis: g m = 2K n I D ( 1+ λv DS ) = ms r o = λ 1 +V DS I D = 328 kω R G = R 1 R 2 = 892 kω R L = r o R D R 3 =17.1 kω 1/25/12 Chap 13-41
42 Common-Source Amplifiers Voltage Gain Example (cont.) g m = ms r o = 328 kω R G = 892 kω R L =17.1 kω R A CS v = g m R G 892kΩ L = 0.503mS ( 17.1kΩ) = R I + R G 1kΩ+892kΩ R R in = R I + R G = 893 kω v gs = v G R i v G i 0.2 V GS V TN R I + R G R I + R G V GS V TN 2I D K n = V v i V ( ) 893kΩ ( ) = 8.59 ( ) = V 892kΩ Check the rule-of-thumb estimate: A v CS V DD = 12 V (ballpark estimate) 1/25/12 Chap 13-42
43 Common-Source Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 0.15-V, 10-kHz sine wave. The expected output voltage amplitude is vo = 8.59(0.15) = 1.29 V. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion, and the amplitude is actually larger than expected. The input is near our small-signal limit for linear operation. 1/25/12 Chap 13-43
44 C-E and C-S Amplifiers Output Resistance 01/19/12 Chap 13-44
45 C-E and C-S Amplifiers Output Resistance (cont.) Apply test source v x and find i x (with v i = 0) v be = 0 g m v be = 0 R out = v x i x = R C r o R out R C for r o >> R C v gs = 0 g m v gs = 0 R out = v x i x = R D r o R out R D for r o >> R D For comparable bias points, output resistances of C-S and C-E amplifiers are similar. 01/19/12 Chap 13-45
46 JFET Small-Signal Operation Small-Signal Model g g = 1 r g = i G v GS Q-point = I G + I SG V T 0 for I G = 0 i D = I DSS 1 v GS V P for v DS v GS V P i G = I SG exp v GS 1 V T 2 ( 1+ λv DS ) g m = i D = v GS Q-point g o = 1 r o = i D v DS Q-point 2I D 2 I DSS V 2 GS V P V GS V P V P = I D 1 λ +V DS ( ) 01/18/12 Microelectronic Circuit Design Chap 13-46
47 JFET Small-Signal Operation Small-Signal Model (cont.) For small-signal operation, the input signal limit is: v gs <0.2( V GS V P ) Since JFET is normally operated with gate junction reverse-biased, I G = I SG r g = The amplification factor is given by: 1 µ f = g m r o = 2 λ +V DS 2 V GS V λv P P I DSS I D 01/18/12 Microelectronic Circuit Design Chap 13-47
48 Common-Source Amplifiers JFET Example Analysis: Dc equivalent circuit is constructed. Ι G = 0, Ι S = Ι D. V GS = 2000I D V GS = ( )( )1 V GS ( 1) Choose V GS less negative than V P. 2 Problem: Find voltage gain, input and output resistances. Given data: Ι DSS = 1 ma, V P = -1V, λ = 0.02 V -1 Assumptions: Pinch-off region of operation. V GS = 0.5 V I D = ma 12= 27,000I D +V DS +2000I S V DS = 4.75V 1/28/12 Microelectronic Circuit Design Chap 13-48
49 Common-Source Amplifier JFET Example (cont.) Next we construct the ac equivalent and simplify it. 1/28/12 Microelectronic Circuit Design Chap 13-49
50 BJT and FET Small-Signal Model Summary 01/18/12 Chap 13-50
51 Common-Emitter/Common-Source Amplifiers Summary 1/25/11 Chap 13-51
52 Amplifier Power Dissipation Static power dissipation in amplifiers is found from their dc equivalent circuits. (a) Total power dissipated in the C-B and E-B junctions is: P D = V CE I C + V BE I B Total power supplied is: P S = V CC I C + V EE I E where V CE = V CB + V BE (b) Total power dissipated in the transistor is: P D = V DS I D + V GS I G = V DS I D Total power supplied is: P S = V DD I D +V DD2 /(R 1 +R 2 ) 1/28/12 Chap 13-52
53 Amplifier Signal Range v CE = V CE V m sinωt where V m is the output signal. Active region operation requires v CE v BE So: V m V CE V BE Also: v Rc ( t) = I C R C V m sinωt 0 ( ) V m min I C R C, V CE V BE Similarly for MOSFETs and JFETs: V M min I D R D,(V DS (V GS V TN )) V M min I D R D,(V DS (V GS V P )) 1/28/12 Chap 13-53
54 End of Chapter 13 1/28/12 Chap 13-54
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### Device Physics: The Bipolar Transistor
Monolithic Amplifier Circuits: Device Physics: The Bipolar Transistor Chapter 4 Jón Tómas Guðmundsson tumi@hi.is 2. Week Fall 2010 1 Introduction In analog design the transistors are not simply switches | 13,763 | 47,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-27 | latest | en | 0.826004 |
https://mathematica.stackexchange.com/questions/200609/goodness-of-fit-how-to-perform-weighted-pearson-test-or-equivalent | 1,695,719,961,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00781.warc.gz | 438,116,644 | 40,401 | # Goodness of fit : how to perform weighted Pearson test (or equivalent)?
I wish to perform goodness of fit. Currently I use NonLinearModelFit :
nlmSimple =
NonlinearModelFit[data, model, {{a, 20000}, {k1, 300}, {b, 20000}}, t, Weights -> 1/dataErr^2, VarianceEstimatorFunction -> (1 &)];
with $$data = \{\{x_1, y_1\},...,\{x_n,y_n\}\}$$ and $$dataErr = \{w_1,...,w_n\}$$. $$w_n$$ is the standard deviation follow by the $$n-th$$ distribution (asssume to be a normal distribution) in which $$y_n$$ is picked up. So each data point is pick up in distribution with different width. This is working fine but I would like to get a Chi Square test (or equivalent, I am "use" to Chi square but I know there are other goodness test, so I am open to any proposal) for the overall goodness of the fit.
I use function from here Performing a chi-square goodness of fit test. I added the degree of Freedom :
pearsonTest[obs_, exp_, dof_] /; Length[obs] == Length[exp] :=
Block[{t}, t = Total[(obs - exp)^2/exp] // N;
{t/(Length[exp] - dof),
SurvivalFunction[ChiSquareDistribution[Length[exp] - dof], t]}];
with $$obs = \{y_1,...,y_n\}$$ and $$exp= \{model[x_1],...,model[x_n]\}$$ But this does not take into account the standard deviation of $$y_n$$ and will output the same $$\chi^2$$ for any set of $$w_n$$. $$model$$ is a function than can be anything from simple exponential to "complicated" function with plenty of parameters.
So my question : does NonLinearMdelFit include some build-in tool for the overall fit's goodness I can use (I used the property of fitted model but this is only for parameters error)? And if no, how to add weighted data in a Pearson test (so this is more a mathematical problem).
• Using $(o-e)^2/e$ doesn't make any sense in a regression model. Maybe you're thinking about the following: en.wikipedia.org/wiki/Reduced_chi-squared_statistic. I suggest asking the question first on CrossValidated (stats.stackexchange.com) and then coming back here for implementation.
– JimB
Jun 18, 2019 at 18:39
• OK so I think about it a bit, and you right it doesn't make sense since the underlying assuption is that the variable $o$ is following poisson disitrbution (which is true in my case) with mean = $e$ and variance=$e$ (which in not true). In my case the variance is $w_i^2$ so I should use $(o_i−e_i)^2/w_i^2$ . I will ask on CrossValidated as you suggest (I edit the first post for adding few details about the problem). Jun 19, 2019 at 8:51
OK I solve the mathematical part of the problem. Since my data follow poisson normal distribution of mean $$\mu=y_i$$ and $$\sigma = w_i^2$$ I need to use $$\chi^2= \sum_{i=1}^n (f(x_i)- x_i)^2/w_i^2$$ instead of $$\chi^2= \sum_{i=1}^n (f(x_i)- x_i)^2/x_i$$ (assumption $$\sigma = x_i$$).
(you can found it in any (I think) data analysis book for exemple : Fundamental Numerical Methods and Data Analysis of George W. Collins II ).
But is strange that mathematica doesn't provide buil-in tool for that no ? For exemple a property of the FittedModel ouput.
• That is a good book but definitely a bit dated and doesn't cover generalized linear models or generalized linear mixed models. This answer mentions that your data follows a Poisson distribution but a comment in your original post says that it does not. You might consider looking at Mathematica's GeneralizedLinearModelFit, get a more up-to-date textbook written by a statistician, and wean yourself off of thinking only in terms of $\chi^2$ statistics.
– JimB
Jun 19, 2019 at 15:43
• Ok I will look at GeneralizedLinearModelFit. About the distribution : I edited my posts. At first this is poisson distribution (count data) but this can be approximazie by normal distribution (high count rate) and when I apply systematics correction and errors I assume normal distribution ( error propagation). So $y_i$ follow normal distribution of $w_i$. Jun 20, 2019 at 8:46 | 1,058 | 3,895 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-40 | longest | en | 0.815256 |
https://math.stackexchange.com/questions/3130037/derivations-of-bases-of-a-kv-and-l-kv-whats-the-difference | 1,555,776,750,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182858-00545.warc.gz | 478,566,329 | 31,823 | # Derivations of bases of $A_k(V)$ and $L_k(V)$: what's the difference?
My book is An Introduction to Manifolds by Loring W. Tu.
Here is the derivation of a basis for $$A_k(V)$$:
Here is the derivation of a basis for $$L_k(V)$$:
1. My first question: What's the difference besides replacing $$\wedge$$ with $$\otimes$$ and strictly ascending with arbitrary?
• At first I thought that we don't have an analogue of Lemma 3.28 for $$L_k(V)$$ because we don't quite have an analogue of Proposition 3.27 for $$L_k(V)$$. But then I think we do and can prove such analogue of the lemma without any kind of analogue of the proposition (in the book, Lemma 3.28 is proved using Proposition 3.27):
$$- \text{original Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \wedge ... \otimes \wedge)(e_{j_1}, ..., e_{j_k}) = \delta^I_J$$
$$- \text{analogous Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \otimes ... \otimes \alpha^{i_k})(e_{j_1}, ..., e_{j_k}) = \alpha^{i_1}(e_{j_1}) \cdot ... \cdot \alpha^{i_k}(e_{j_k}) = \delta^{i_1}_{j_1} \cdot ... \cdot \delta^{i_k}_{j_k} = \delta^I_J,$$
• Ligo doesn't use the notation $$\alpha^I(e_J)$$, but I believe Ligo's proof can be shortened with something like $$\alpha^I(e_J) = \delta^I_J$$.
1. My second question: (I intended to ask only one question, but I just thought of another) Actually, is the analogue for Proposition 3.27 that tensor product equals product of diagonal entries, which is analogous to wedge product equals determinant in the original Proposition 3.27?
$$- \text{original Proposition 3.27:} \ \alpha^{1} \wedge ... \wedge \alpha^{k}(v_1, ..., v_k) = \det[\alpha^{i}(v_j)]_{i,j=1,...,k}$$
$$- \text{analogous Proposition 3.27:} \ \alpha^{1} \otimes ... \otimes \alpha^{k}(v_1, ..., v_k) = \prod_{i=1,...,k} \alpha^{i}(v_i)$$ | 595 | 1,784 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-18 | latest | en | 0.809797 |
https://re.public.polimi.it/handle/11311/560563 | 1,716,543,198,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00206.warc.gz | 431,322,648 | 12,717 | We consider a model for the flow of a mixture of two homogeneous and incompressible fluids in a two-dimensional bounded domain. The model consists of a Navier–Stokes equation governing the fluid velocity coupled with a convective Cahn–Hilliard equation for the relative density of atoms of one of the fluids. Endowing the system with suitable boundary and initial conditions, we analyze the asymptotic behavior of its solutions. First, we prove that the initial and boundary value problem generates a strongly continuous semigroup on a suitable phase-space which possesses the global attractor A. Then we establish the existence of an exponential attractors E. Thus A has finite fractal dimension. This dimension is then estimated from above in terms of the physical parameters. Moreover, assuming the potential to be real analytic and in absence of volume forces, we demonstrate that each trajectory converges to a single equilibrium. We also obtain a convergence rate estimate in the phase-space metric.
### Asymptotic behavior of a Cahn-Hilliard-Navier-Stokes system in 2D
#### Abstract
We consider a model for the flow of a mixture of two homogeneous and incompressible fluids in a two-dimensional bounded domain. The model consists of a Navier–Stokes equation governing the fluid velocity coupled with a convective Cahn–Hilliard equation for the relative density of atoms of one of the fluids. Endowing the system with suitable boundary and initial conditions, we analyze the asymptotic behavior of its solutions. First, we prove that the initial and boundary value problem generates a strongly continuous semigroup on a suitable phase-space which possesses the global attractor A. Then we establish the existence of an exponential attractors E. Thus A has finite fractal dimension. This dimension is then estimated from above in terms of the physical parameters. Moreover, assuming the potential to be real analytic and in absence of volume forces, we demonstrate that each trajectory converges to a single equilibrium. We also obtain a convergence rate estimate in the phase-space metric.
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Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11311/560563` | 502 | 2,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.89642 |
http://cn.metamath.org/mpeuni/bj-2uplex.html | 1,660,493,105,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00201.warc.gz | 12,007,419 | 5,113 | Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-2uplex Structured version Visualization version GIF version
Theorem bj-2uplex 33341
Description: A couple is a set if and only if its coordinates are sets. (Contributed by BJ, 6-Oct-2018.)
Assertion
Ref Expression
bj-2uplex (⦅𝐴, 𝐵⦆ ∈ V ↔ (𝐴 ∈ V ∧ 𝐵 ∈ V))
Proof of Theorem bj-2uplex
StepHypRef Expression
1 bj-pr21val 33332 . . . 4 pr1𝐴, 𝐵⦆ = 𝐴
2 bj-pr1ex 33325 . . . 4 (⦅𝐴, 𝐵⦆ ∈ V → pr1𝐴, 𝐵⦆ ∈ V)
31, 2syl5eqelr 2855 . . 3 (⦅𝐴, 𝐵⦆ ∈ V → 𝐴 ∈ V)
4 bj-pr22val 33338 . . . 4 pr2𝐴, 𝐵⦆ = 𝐵
5 bj-pr2ex 33339 . . . 4 (⦅𝐴, 𝐵⦆ ∈ V → pr2𝐴, 𝐵⦆ ∈ V)
64, 5syl5eqelr 2855 . . 3 (⦅𝐴, 𝐵⦆ ∈ V → 𝐵 ∈ V)
73, 6jca 501 . 2 (⦅𝐴, 𝐵⦆ ∈ V → (𝐴 ∈ V ∧ 𝐵 ∈ V))
8 df-bj-2upl 33330 . . 3 𝐴, 𝐵⦆ = (⦅𝐴⦆ ∪ ({1𝑜} × tag 𝐵))
9 bj-1uplex 33327 . . . . 5 (⦅𝐴⦆ ∈ V ↔ 𝐴 ∈ V)
109biimpri 218 . . . 4 (𝐴 ∈ V → ⦅𝐴⦆ ∈ V)
11 snex 5036 . . . . 5 {1𝑜} ∈ V
12 bj-xtagex 33308 . . . . 5 ({1𝑜} ∈ V → (𝐵 ∈ V → ({1𝑜} × tag 𝐵) ∈ V))
1311, 12ax-mp 5 . . . 4 (𝐵 ∈ V → ({1𝑜} × tag 𝐵) ∈ V)
14 unexg 7106 . . . 4 ((⦅𝐴⦆ ∈ V ∧ ({1𝑜} × tag 𝐵) ∈ V) → (⦅𝐴⦆ ∪ ({1𝑜} × tag 𝐵)) ∈ V)
1510, 13, 14syl2an 583 . . 3 ((𝐴 ∈ V ∧ 𝐵 ∈ V) → (⦅𝐴⦆ ∪ ({1𝑜} × tag 𝐵)) ∈ V)
168, 15syl5eqel 2854 . 2 ((𝐴 ∈ V ∧ 𝐵 ∈ V) → ⦅𝐴, 𝐵⦆ ∈ V)
177, 16impbii 199 1 (⦅𝐴, 𝐵⦆ ∈ V ↔ (𝐴 ∈ V ∧ 𝐵 ∈ V))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 382 ∈ wcel 2145 Vcvv 3351 ∪ cun 3721 {csn 4316 × cxp 5247 1𝑜c1o 7706 tag bj-ctag 33293 ⦅bj-c1upl 33316 pr1 bj-cpr1 33319 ⦅bj-c2uple 33329 pr2 bj-cpr2 33333 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-8 2147 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-rep 4904 ax-sep 4915 ax-nul 4923 ax-pow 4974 ax-pr 5034 ax-un 7096 This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-3or 1072 df-3an 1073 df-tru 1634 df-fal 1637 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-mo 2623 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ne 2944 df-nel 3047 df-ral 3066 df-rex 3067 df-rab 3070 df-v 3353 df-sbc 3588 df-csb 3683 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-pss 3739 df-nul 4064 df-if 4226 df-pw 4299 df-sn 4317 df-pr 4319 df-tp 4321 df-op 4323 df-uni 4575 df-br 4787 df-opab 4847 df-tr 4887 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-ord 5869 df-on 5870 df-suc 5872 df-1o 7713 df-bj-sngl 33285 df-bj-tag 33294 df-bj-proj 33310 df-bj-1upl 33317 df-bj-pr1 33320 df-bj-2upl 33330 df-bj-pr2 33334 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 1,694 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.113824 |
https://dice.mpi-inf.mpg.de/fact/prime-number/be-different-than-composite | 1,680,249,626,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00075.warc.gz | 263,401,045 | 3,956 | # prime number: be different than composite
from Quasimodo
## Related concepts
Parents numberWeight: 0.66, primeWeight: 0.64, topicWeight: 0.62, wordWeight: 0.54 Siblings elevenWeight: 0.51, complex numberWeight: 0.35, ordinal numberWeight: 0.35, magic numberWeight: 0.35, irrational numberWeight: 0.34
## Related properties
Similarity Property be different than composite 1.00 be different 0.76
## Clauses
### Plausibility inference from child typicality
0.47
Rule weight: 0.66
Evidence weight: 0.93
Similarity weight: 0.76
Evidence: 0.73
Plausible(number, be different)
Evidence: 0.25
¬ Typical(prime number, be different than composite)
### Plausibility inheritance from parent to child
0.04
Rule weight: 0.09
Evidence weight: 0.57
Similarity weight: 0.76
Evidence: 0.40
Plausible(prime number, be different than composite)
Evidence: 0.73
¬ Plausible(number, be different)
### Remarkability exclusitivity betweem a parent and a child
0.12
Rule weight: 0.58
Evidence weight: 0.28
Similarity weight: 0.76
Evidence: 0.87
¬ Remarkable(prime number, be different than composite)
Evidence: 0.83
¬ Remarkable(number, be different)
### Remarkability from parent implausibility
0.31
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.76
Evidence: 0.73
Plausible(number, be different)
Evidence: 0.87
Remarkable(prime number, be different than composite)
Evidence: 0.40
¬ Plausible(prime number, be different than composite)
### Salient implies Plausible
0.18
Rule weight: 0.28
Evidence weight: 0.64
Similarity weight: 1.00
Evidence: 0.40
Plausible(prime number, be different than composite)
Evidence: 0.60
¬ Salient(prime number, be different than composite)
### Similarity expansion
0.58
Rule weight: 0.85
Evidence weight: 0.90
Similarity weight: 0.76
Evidence: 0.87
Remarkable(prime number, be different than composite)
Evidence: 0.77
¬ Remarkable(prime number, be different)
0.48
Rule weight: 0.85
Evidence weight: 0.74
Similarity weight: 0.76
Evidence: 0.60
Salient(prime number, be different than composite)
Evidence: 0.66
¬ Salient(prime number, be different)
0.45
Rule weight: 0.85
Evidence weight: 0.69
Similarity weight: 0.76
Evidence: 0.40
Plausible(prime number, be different than composite)
Evidence: 0.52
¬ Plausible(prime number, be different)
0.42
Rule weight: 0.85
Evidence weight: 0.64
Similarity weight: 0.76
Evidence: 0.25
Typical(prime number, be different than composite)
Evidence: 0.47
¬ Typical(prime number, be different)
### Typical and Remarkable implies Salient
0.13
Rule weight: 0.14
Evidence weight: 0.91
Similarity weight: 1.00
Evidence: 0.60
Salient(prime number, be different than composite)
Evidence: 0.25
¬ Typical(prime number, be different than composite)
Evidence: 0.87
¬ Remarkable(prime number, be different than composite)
### Typical implies Plausible
0.41
Rule weight: 0.48
Evidence weight: 0.85
Similarity weight: 1.00
Evidence: 0.40
Plausible(prime number, be different than composite)
Evidence: 0.25
¬ Typical(prime number, be different than composite)
### Typicality and Rermarkability incompatibility between a parent and a child
0.14
Rule weight: 0.51
Evidence weight: 0.37
Similarity weight: 0.76
Evidence: 0.87
¬ Remarkable(prime number, be different than composite)
Evidence: 0.73
¬ Typical(number, be different)
### Typicality inheritance from parent to child
0.17
Rule weight: 0.48
Evidence weight: 0.45
Similarity weight: 0.76
Evidence: 0.25
Typical(prime number, be different than composite)
Evidence: 0.73
¬ Typical(number, be different) | 1,080 | 3,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.691934 |
https://support.coinex.com/hc/en-us/articles/900004334846-What-is-Impermanent-Loss- | 1,670,587,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00011.warc.gz | 582,727,532 | 14,038 | What's Impermanent Loss and How to Avoid It
What is Impermanent Loss?
Impermanent Loss refers to the temporary loss in Automated Market Making (AMM) caused by market fluctuation when a Liquidity Provider (LP) injects liquidity into a liquidity pool. According to the "Constant Product Market Maker" pricing model, when an LP removes liquidity during the market fluctuation, the value of his assets decreases compared to simply holding the tokens. This is what we call impermanent loss. However, impermanent loss will gradually be wiped out when the price ratio changes back.
Example for Impermanent Loss
1. Assuming that currently there are 1 000 CET and 500 USDT in the liquidity pool, the constant product is 1 000 * 500 = 500 000. The liquidity Provider(LP) Adam holds 10% of the assets in the CET/USDT pool, that is, 100 CET and 50 USDT. At this time, 1 CET = 0.5 USDT.
2. Suppose that after a period of time, the CET price rises, and the amount of the pool tokens has changed into 500 CET and 1 000 USDT. The constant product 500 * 1 000 = 500 000 remains the same, but at this time, 1 CET = 2 USDT. LP Adam's holdings in the pool also change accordingly, becoming 50 CET and 100 USDT.
3. Assuming that LP Adam removes liquidity when the pool contains 500 CET and 1 000 USDT, since he has a 10% share, in total he can remove 50 CET and 100 USDT. After removing liquidity, the value of his assets is 50 * 2 + 100 = 200 USDT.
4. At the time of adding liquidity, Adam owned 100 CET and 50 USDT. If he simply held the tokens and did not invest in the liquidity pool, the total asset value would be 100 * 2 + 50 = 250 USDT. However, after removing liquidity, he only gets 200 USDT. The difference in value between adding liquidity and simply holding the assets is the impermanent loss. In Adam‘s case, the loss is 50 USDT.
Note: To simplify the calculation, the trading fee is not considered in the above example.
Impermanent Loss Estimation
Here’s a chart of estimated Impermanent Loss for your reference:
Price Change after adding liquidity Impermanent Loss(Loss compared to simply holding the tokens) -60% 9.65% -30% 1.57% -10% 0.14% 0% 0.00% 10% 0.11% 30% 0.85% 60% 2.70% 100% 5.72%
How to Avoid Impermanent Loss?
Impermanent losses are common in the early stage of market making or in one-sided markets. With earnings from trading fees and price fluctuations, impermanent losses will gradually be wiped out, and eventually, users can gain profits from providing liquidity. | 645 | 2,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.92194 |
http://howded.com/en/pages/274305 | 1,542,520,148,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743968.63/warc/CC-MAIN-20181118052443-20181118074443-00066.warc.gz | 169,245,026 | 5,264 | you need
• tape, paper, pen, calculator
Guide
1
To calculate the total lamps using the following formula: P = pS / N, p -power specific lighting, measured in watts per square meter (average - 20 Watts per square meter), S - area of the premises for which the lighting is calculated in square meters and N - the number of fixtures.
2
Example: To calculate the room, measure the length and width of the room.The results (say, 3.3 meters in length and 4.5 meters in width) between a multiply and get the area of the room (3,3 × 4,4 = 14,85 square meters).Multiply this figure by 20 and divide it by the estimated number of lamps in the luminaire.For example, take 3 to 2 lamp bulb each.In this case, 14.85 multiply by 20 and divide by 6 (3 × 2 = 6).Yo
u get a result that means that in this case it is necessary to 6 lamps on 49.5 watts.
3
wattage may vary in each candle, sharing the room into several zones with different lighting.The total capacity of all lamps in the room must not be less than 297 watts.
4
In some cases it is necessary to take into account the specifics of the room, which makes for a calculation of the total power installed lamps.In this case, do the calculation, substituting the value of coefficient p in the table below: Type of room | Bulb | Halogen lamp | Fluorescent svetaDetskaya room .............. 30-85 .................... 70-85 .................... 15-22
Living ..................... 15-35 25-35 .................... .................... 7-10
Bedroom ...................... 10-25 .................... 15-17 .................... 4-7
Corridor ...................... .................... 10-20 10-13 .................... 3-5
Kitchen ........................ ............. 15-40....... 30-37 .................... 6-10
Bathroom ...............15-30 .................... .................... 22-27 6-9
pantry, garage .............. .................... 12-15 11-14 .................... 3-5Zdes to calculate, for example, the lighting in the kitchen (an area of 3 square meters) with fluorescent lamps take a coefficient p (average value of 9) from the table, multiply bykitchen area and if it is to stand trehrozhkovy lamp, is divided into three: 9 × 3/3 = 9 watts to each of the three installed in the lamp light. | 566 | 2,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.76916 |
https://transformer-hc.com/blog/which-of-the-following-transformation-make-changes-the-shape-of-2d-object.html | 1,656,641,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00602.warc.gz | 624,303,287 | 18,215 | # Which of the following transformation make changes the shape of 2D object?
Contents
## Which of the following transformation make change the shape of 2D object?
Explanation: Shearing is a transformation method which may cause the change in shape of an object. Explanation: Rotation, Translation and Scaling are some of the most basic transformation methods which may apply in three-dimensional planes.
## What transformation can change the shape of an object?
A transformation that slants the shape of an object is called the shear transformation.
## What are the types of 2D transformation?
2 Transformation Types and Examples
• Translation. The translation transformation shifts a node from one place to another along one of the axes relative to its initial position. …
• Rotation. The rotation transformation moves the node around a specified pivot point of the scene. …
• Scaling. …
• Shearing. …
• Multiple Transformations.
## Is a technique to change the shape of an existing object in 2D plane?
2D Shearing is an ideal technique to change the shape of an existing object in a two dimensional plane.
## Which transformation distorts the shape of an object such that?
Explanation: In shear transformation, the transformed shape appears as if the object were composed of internal layers that had been caused to slide over each other.
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## When there is change in shape the transformation is called?
A transformation changes the size, shape, or position of a figure and creates a new figure. A geometry transformation is either rigid or non-rigid; another word for a rigid transformation is “isometry“. An isometry, such as a rotation, translation, or reflection, does not change the size or shape of the figure.
## What is 2D rotation in computer graphics?
In. Computer graphics, 2D Rotation is a process of rotating an object with respect to an angle in a two dimensional plane. Consider a point object O has to be rotated from one angle to another in a 2D plane.
## What is composition of 2D transformation?
As the name suggests itself Composition, here we combine two or more transformations into one single transformation that is equivalent to the transformations that are performed one after one over a 2-D object.
## What are the examples of transformation?
Transformation is the process of changing. An example of a transformation is a caterpillar turning into a butterfly.
## What are the different types of transformations in geometry?
The four main types of transformations are translations, reflections, rotations, and scaling.
• Translations. A translation moves every point by a fixed distance in the same direction. …
• Reflections. …
• Rotations. …
• Scaling. …
• Vertical Translations. …
• Horizontal Translations. …
• Reflections. …
• Learning Objectives. | 573 | 2,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.904699 |
https://www.heyiamindians.com/how-many-kilowatt-hours-is-a-gigawatt/ | 1,696,057,386,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00221.warc.gz | 866,185,503 | 9,947 | ## How many kilowatt hours is a gigawatt?
one million kilowatt hours
Gigawatt hours, abbreviated as GWh, is a unit of energy representing one billion (1 000 000 000) watt hours and is equivalent to one million kilowatt hours.
How do you convert gigawatts to kilowatts?
Conversion chart – gigawatts to kilowatts
1. gigawatt to kilowatts = 1,000,000.00 kW.
2. gigawatts to kilowatts = 2,000,000.00 kW.
3. gigawatts to kilowatts = 3,000,000.00 kW.
4. gigawatts to kilowatts = 4,000,000.00 kW.
5. gigawatts to kilowatts = 5,000,000.00 kW.
6. gigawatts to kilowatts = 6,000,000.00 kW.
How many MWh are in a GWh?
The megawatt hours unit number 1,000.00 MWh converts to 1 GWh, one gigawatt hour.
### Is a gigawatt a lot?
A gigawatt is equal to one billion watts, and most of us are familiar with a watt. The light bulbs in our homes are typically between 60 and 100 watts. So 1.21 gigawatts would power more than 10 million light bulbs or one fictional flux capacitor in a time-traveling DeLorean.
How many GW is a tw?
Terawatt to Gigawatt Conversion Table
Terawatt [TW] Gigawatt [GW]
1 TW 1000 GW
2 TW 2000 GW
3 TW 3000 GW
5 TW 5000 GW
How many megawatt hours is a gigawatt?
The gigawatt hours unit number 0.0010 GWh converts to 1 MWh, one megawatt hour. It is the EQUAL energy value of 1 megawatt hour but in the gigawatt hours energy unit alternative.
## How much is a gigawatt?
How many megawatt-hours is a gigawatt?
Are gigawatts real?
Without fact checking the movie in too much detail, a gigawatt is a real measure of power. A gigawatt is equal to one billion watts, and most of us are familiar with a watt. So 1.21 gigawatts would power more than 10 million light bulbs or one fictional flux capacitor in a time-traveling DeLorean.
### How do you convert megawatt hours to kilowatt hours?
1 Megawatt hour ( MWh ) is equal to 1000 kilowatt hours (kWh). To convert MWh to kWh, multiply the MWh value by 1000.
What is the difference between Watts and kilowatt hours?
Watt is the unit of power whereas kilowatt-hour (kWh) is the unit of energy. Watt indicates the rate of using energy in J/s. You can compare this to how fast water is flowing out of a water pipe. If you have a Light bulb that has a rating of 100 watt, it means that the light bulb consumes 100 J per second. kWh is the unit of energy.
How many kilowatts are in 1 kilowatt hour?
Kilowatt-hours, expressed kWh or kW·h, are used to measure electrical energy. One kWh is equal to one kilowatt, or one thousand watts, of energy consumed for one hour of time. To convert from electrical charge to energy, use the formula below along with the voltage.
## How many watts are in a kilowatt hour?
One kilowatt (kW) equals 1,000 watts, and one kilowatt-hour (kWh) is one hour of using electricity at a rate of 1,000 watts. | 821 | 2,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-40 | longest | en | 0.886012 |
https://www.howmany.wiki/calories-burned/How-many-calories-do-i-burn-_power+lifting+%28vigorous+effort%29 | 1,696,187,843,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00418.warc.gz | 842,597,879 | 21,535 | ## HowMany.wiki
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# How many calories do i burn power lifting (vigorous effort) for 30 minutes 154.3 pounds or 70 kilograms
A person will burn 210.0 calories power lifting (vigorous effort) for 30 minutes if he/she weighs 154.3 pounds or 70 kg
### Inputs
Choose an activity/exercise from the box below:
or
Begin typing an activity or exercise in the box below.
(Ex.: 'running').
Change any value below then click/tap 'compute calories!'
minutes
lbs or Kilograms
### Results
Calories burned: 210.0 cal Mass (fat and/or muscle) burned: 27.2 g or 0.06 lb
Someone weighting 70 Kg or 154.3 lb power lifting (vigorous effort) burns 210.0 calories in 30 minutes. This value is roughtly equivalent to 0.06 pound or 0.96 ounce or 27.2 grams of mass (fat and / or muscle).
• Doing this activity 3 times a week for 30 minutes will loose 0.72 pounds or 0.33 Kg a month.
• Doing this activity 5 times a week for 30 minutes will loose 1.2 pounds or 0.54 Kg a month.
## How to calculate calories (burned)
The number of calories you burn while exercising is dependent on:
• the exercise you do
• the time spent doing the activity
By multiplying the MET (*) value by the body weight in kg and the duration of the activity, you can estimate the energy expenditure in Kcal specific to a person's body weight. This statement can be expressed as the following formula:
Calories burned = METs x weight x (time / 60)
In this example, power lifting (vigorous effort) at a 6 MET value, burns Kcal/kg x body weight/h.
So, a 70 kg individual power lifting (vigorous effort) for 30 minutes expends the following:
(6 METs x 70 kg) x (30 min/60 min) = 210.0 Kcal.
Notes:
• 6 is the value in METs for power lifting (vigorous effort)
• The time unit is expressed in minutes
• The mass (weight) unit is expressed in kilograms
• We divide the time by 60 (min) because the MET unit uses hour.
(*) MET means 'metabolic equivalent'
## Sample Calories Burned Calculations
### Disclaimer
Despite efforts to provide accurate information on this website, no guarantee of its accuracy is made. Therefore, the content should not be used for decisions regarding health, finances, or property. | 594 | 2,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-40 | latest | en | 0.798711 |
http://brainden.com/forum/profile/13141-pickett/content/page/6/ | 1,600,589,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00559.warc.gz | 26,447,742 | 14,657 | BrainDen.com - Brain Teasers
# Pickett
Members
623
8
3. ## Robots on an infinite number line.
EDIT: I also ran my code on Rainman's algorithm above...it appears to also always work. Nice job!
12. ## Cloorsu
Disregard my vowel is ALWAYS yellow comment above...I overlooked some...but they are PREDOMINANTLY yellow...but that could just be coincidence...
19. ## Find Me
Yeah, you're probably closer...more trams and people there...
23. ## Alphabetize my Dvd's
So I agree that in certain sorting algorithms, defining a "step" would be difficult...however, I think with regards to this problem from a practical standpoint, I don't think it's very difficult. A "move" or "step" is simply when you physically grab a DVD and place it in a new position on the shelf...doesn't matter how you do it. In my opinion, if you were to "swap" two DVDs on the shelf (move first to last and last to first)...to me, that is TWO moves, not one. If you decided to do a merge sort on a shelf of DVDs, I would say you are doing a TON of extra moves...while it may be more effici
24. ## Alphabetize my Dvd's
Ok, so I think I mis-interpreted the OP...it states: To arrange my DVDs with precision I will take one DVD off the shelf and place it in the correct spot back on the shelf. If that's the case, let's look at the example of 2, 4, 1, 3: Start with 1...move it to its correct spot (and shift as needed): 1, 2, 4, 3...then try 2...it's already there...then 3, and move it...1, 2, 3, 4...you're done. However, you can also do this by moving the 2 between 1 and 3 initially: 2, 4, 1, 3 --> 4, 1, 2, 3 --> 1, 2, 3, 4... Which of the above is what you would expect from your OP? Ei
25. ## Alphabetize my Dvd's
I may have a small error in my code, I need to double check the results above...
×
• #### Activity
• Riddles
×
• Create New... | 527 | 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} | 3.390625 | 3 | CC-MAIN-2020-40 | longest | en | 0.942609 |
https://www.aqua-calc.com/page/density-table/substance/amber | 1,695,466,742,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00355.warc.gz | 728,387,344 | 9,326 | # Density of Amber (material)
## Amber weighs 1 100 kg/m³ (68.67076 lb/ft³)
• Amber weighs 1.1 gram per cubic centimeter or 1 100 kilogram per cubic meter, i.e. density of amber is equal to 1 100 kg/m³. In Imperial or US customary measurement system, the density is equal to 68.67 pound per cubic foot [lb/ft³], or 0.636 ounce per cubic inch [oz/inch³] .
• Bookmarks: [ weight to volume | volume to weight | price | density ]
• Density of Amber in a few select units of density measurement:
• Density of Amber g cm3 = 1.1 g/cm³
• Density of Amber g ml = 1.1 g/ml
• Density of Amber g mm3 = 0.0011 g/mm³
• Density of Amber kg m3 = 1 100 kg/m³
• Density of Amber lb in3 = 0.04 lb/in³
• Density of Amber lb ft3 = 68.67 lb/ft³
• See density of Amber in hundreds of units of density measurement grouped by weight.
### Amber density values, grouped by weight and shown as value of density, unit of density
16.98 gr/cm³ 16 975.59 gr/dm³ 480 695.3 gr/ft³ 278.18 gr/in³ 16 975 594.24 gr/m³ 0.02 gr/mm³ 12 978 772.95 gr/yd³ 16 975.59 gr/l 4 243.9 gr/metric c 254.63 gr/metric tbsp 84.88 gr/metric tsp 16.98 gr/ml 4 016.23 gr/US c 501.97 gr/fl.oz 64 259.61 gr/US gal 8 032.45 gr/pt 16 064.9 gr/US qt 251.01 gr/US tbsp 83.67 gr/US tsp
1.1 g/cm³ 1 100 g/dm³ 31 148.53 g/ft³ 18.03 g/in³ 1 100 000 g/m³ 0 g/mm³ 841 010.34 g/yd³ 1 100 g/l 275 g/metric c 16.5 g/metric tbsp 5.5 g/metric tsp 1.1 g/ml 260.25 g/US c 32.53 g/fl.oz 4 163.95 g/US gal 520.49 g/pt 1 040.99 g/US qt 16.27 g/tbsp 5.42 g/tsp
0 kg/cm³ 1.1 kg/dm³ 31.15 kg/ft³ 0.02 kg/in³ 1 100 kg/m³ 1.1 × 10-6 kg/mm³ 841.01 kg/yd³ 1.1 kg/l 0.28 kg/metric c 0.02 kg/metric tbsp 0.01 kg/metric tsp 0 kg/ml 0.26 kg/US c 0.03 kg/fl.oz 4.16 kg/US gal 0.52 kg/pt 1.04 kg/US qt 0.02 kg/tbsp 0.01 kg/tsp
1.08 × 10-6 long tn/cm³ 0 long tn/dm³ 0.03 long tn/ft³ 1.77 × 10-5 long tn/in³ 1.08 long tn/m³ 1.08 × 10-9 long tn/mm³ 0.83 long tn/yd³ 0 long tn/l 0 long tn/metric c 1.62 × 10-5 long tn/metric tbsp 5.41 × 10-6 long tn/metric tsp 1.08 × 10-6 long tn/ml 0 long tn/US c 3.44 × 10-5 long tn/fl.oz 0 long tn/US gal 0 long tn/pt 0 long tn/US qt 1.6 × 10-5 long tn/US tbsp 5.34 × 10-6 long tn/US tsp
1 100 000 µg/cm³ 1 100 000 000 µg/dm³ 31 148 531 260 µg/ft³ 18 025 770.4 µg/in³ 1 100 000 000 000 µg/m³ 1 100 µg/mm³ 841 010 343 800 µg/yd³ 1 100 000 000 µg/l 275 000 000 µg/metric c 16 500 000 µg/metric tbsp 5 500 000 µg/metric tsp 1 100 000 µg/ml 260 247 060.7 µg/US c 32 530 882.56 µg/fl.oz 4 163 952 958 µg/US gal 520 494 120.3 µg/pt 1 040 988 240.6 µg/US qt 16 265 441.28 µg/tbsp 5 421 813.75 µg/tsp
1 100 mg/cm³ 1 100 000 mg/dm³ 31 148 531.26 mg/ft³ 18 025.77 mg/in³ 1 100 000 000 mg/m³ 1.1 mg/mm³ 841 010 343.8 mg/yd³ 1 100 000 mg/l 275 000 mg/metric c 16 500 mg/metric tbsp 5 500 mg/metric tsp 1 100 mg/ml 260 247.06 mg/US c 32 527 mg/fl.oz 4 163 952.97 mg/US gal 520 494.12 mg/pt 1 040 988.24 mg/US qt 16 265.44 mg/tbsp 5 421.81 mg/tsp
0.04 oz/cm³ 38.8 oz/dm³ 1 098.73 oz/ft³ 0.64 oz/in³ 38 801.36 oz/m³ 3.88 × 10-5 oz/mm³ 29 665.77 oz/yd³ 38.8 oz/l 9.7 oz/metric c 0.58 oz/metric tbsp 0.19 oz/metric tsp 0.04 oz/ml 9.18 oz/US c 1.23 oz/fl.oz 146.88 oz/US gal 18.36 oz/pt 36.72 oz/US qt 0.57 oz/tbsp 0.19 oz/tsp
0.71 dwt/cm³ 707.32 dwt/dm³ 20 028.97 dwt/ft³ 11.59 dwt/in³ 707 316.43 dwt/m³ 0 dwt/mm³ 540 782.21 dwt/yd³ 707.32 dwt/l 176.83 dwt/metric c 10.61 dwt/metric tbsp 3.54 dwt/metric tsp 0.71 dwt/ml 167.34 dwt/US c 20.92 dwt/fl.oz 2 677.48 dwt/US gal 334.69 dwt/pt 669.37 dwt/US qt 10.46 dwt/US tbsp 3.49 dwt/US tsp
0 lb/cm³ 2.43 lb/dm³ 68.67 lb/ft³ 0.04 lb/in³ 2 425.08 lb/m³ 2.43 × 10-6 lb/mm³ 1 854.11 lb/yd³ 2.43 lb/l 0.61 lb/metric c 0.04 lb/metric tbsp 0.01 lb/metric tsp 0 lb/ml 0.57 lb/US c 0.08 lb/fl.oz 9.18 lb/US gal 1.15 lb/pt 2.29 lb/US qt 0.04 lb/tbsp 0.01 lb/tsp
1.21 × 10-6 short tn/cm³ 0 short tn/dm³ 0.03 short tn/ft³ 1.99 × 10-5 short tn/in³ 1.21 short tn/m³ 1.21 × 10-9 short tn/mm³ 0.93 short tn/yd³ 0 short tn/l 0 short tn/metric c 1.82 × 10-5 short tn/metric tbsp 6.06 × 10-6 short tn/metric tsp 1.21 × 10-6 short tn/ml 0 short tn/US c 3.85 × 10-5 short tn/fl.oz 0 short tn/US gal 0 short tn/pt 0 short tn/US qt 1.79 × 10-5 short tn/US tbsp 5.98 × 10-6 short tn/US tsp
7.54 × 10-5 sl/cm³ 0.08 sl/dm³ 2.13 sl/ft³ 0 sl/in³ 75.37 sl/m³ 7.54 × 10-8 sl/mm³ 57.63 sl/yd³ 0.08 sl/l 0.02 sl/metric c 0 sl/metric tbsp 0 sl/metric tsp 7.54 × 10-5 sl/ml 0.02 sl/US c 0 sl/fl.oz 0.29 sl/US gal 0.04 sl/pt 0.07 sl/US qt 0 sl/tbsp 0 sl/tsp
0 st/cm³ 0.17 st/dm³ 4.91 st/ft³ 0 st/in³ 173.22 st/m³ 1.73 × 10-7 st/mm³ 132.44 st/yd³ 0.17 st/l 0.04 st/metric c 0 st/metric tbsp 0 st/metric tsp 0 st/ml 0.04 st/US c 0.01 st/fl.oz 0.66 st/US gal 0.08 st/pt 0.16 st/US qt 0 st/US tbsp 0 st/US tsp
1.1 × 10-6 t/cm³ 0 t/dm³ 0.03 t/ft³ 1.8 × 10-5 t/in³ 1.1 t/m³ 1.1 × 10-9 t/mm³ 0.84 t/yd³ 0 t/l 0 t/metric c 1.65 × 10-5 t/metric tbsp 5.5 × 10-6 t/metric tsp 1.1 × 10-6 t/ml 0 t/US c 3.25 × 10-5 t/fl.oz 0 t/US gal 0 t/pt 0 t/US qt 1.63 × 10-5 t/tbsp 5.42 × 10-6 t/tsp
0.04 oz t/cm³ 35.37 oz t/dm³ 1 001.45 oz t/ft³ 0.58 oz t/in³ 35 365.82 oz t/m³ 3.54 × 10-5 oz t/mm³ 27 039.11 oz t/yd³ 35.37 oz t/l 8.84 oz t/metric c 0.53 oz t/metric tbsp 0.18 oz t/metric tsp 0.04 oz t/ml 8.37 oz t/US c 1.05 oz t/fl.oz 133.87 oz t/US gal 16.73 oz t/pt 33.47 oz t/US qt 0.52 oz t/US tbsp 0.17 oz t/US tsp
0 troy/cm³ 2.95 troy/dm³ 83.45 troy/ft³ 0.05 troy/in³ 2 947.15 troy/m³ 2.95 × 10-6 troy/mm³ 2 253.26 troy/yd³ 2.95 troy/l 0.74 troy/metric c 0.04 troy/metric tbsp 0.01 troy/metric tsp 0 troy/ml 0.7 troy/US c 0.09 troy/fl.oz 11.16 troy/US gal 1.39 troy/pt 2.79 troy/US qt 0.04 troy/US tbsp 0.01 troy/US tsp
Amber density values in 285 units of density, in the form of a matrix
Density = weight ÷ volumemicrogram (µg)milligram (mg)gram (g)kilogram (kg)tonne (t)ounce (oz)pound (lb)volume unitgrain (gr)slug (sl)short ton (short tn)long ton (long tn)stone (st)troy ounce (oz t)troy pound (troy)pennyweight (dwt)
cubic millimeter1 1001.1<0.01<0.01<0.01<0.01<0.01cubic millimeter0.02<0.01<0.01<0.01<0.01<0.01<0.01<0.01
cubic centimeter1 100 0001 1001.1<0.01<0.010.04<0.01cubic centimeter16.98<0.01<0.01<0.01<0.010.04<0.010.71
cubic decimeter1 100 000 0001 100 0001 1001.1<0.0138.82.43cubic decimeter16 975.590.08<0.01<0.010.1735.372.95707.32
cubic meter1 100 000 000 0001 100 000 0001 100 0001 1001.138 801.362 425.08cubic meter16 975 594.2475.371.211.08173.2235 365.822 947.15707 316.43
milliliter1 100 0001 1001.1<0.01<0.010.04<0.01milliliter16.98<0.01<0.01<0.01<0.010.04<0.010.71
liter1 100 000 0001 100 0001 1001.1<0.0138.82.43liter16 975.590.08<0.01<0.010.1735.372.95707.32
metric teaspoon5 500 0005 5005.50.01<0.010.190.01metric teaspoon84.88<0.01<0.01<0.01<0.010.180.013.54
metric tablespoon16 500 00016 50016.50.02<0.010.580.04metric tablespoon254.63<0.01<0.01<0.01<0.010.530.0410.61
metric cup275 000 000275 0002750.28<0.019.70.61metric cup4 243.90.02<0.01<0.010.048.840.74176.83
cubic inch18 025 770.418 025.7718.030.02<0.010.640.04cubic inch278.18<0.01<0.01<0.01<0.010.580.0511.59
cubic foot31 148 531 26031 148 531.2631 148.5331.150.031 098.7368.67cubic foot480 695.32.130.030.034.911 001.4583.4520 028.97
cubic yard841 010 343 800841 010 343.8841 010.34841.010.8429 665.771 854.11cubic yard12 978 772.9557.630.930.83132.4427 039.112 253.26540 782.21
US teaspoon5 421 813.755 421.815.420.01<0.010.190.01US teaspoon83.67<0.01<0.01<0.01<0.010.170.013.49
US tablespoon16 265 441.2816 265.4416.270.02<0.010.570.04US tablespoon251.01<0.01<0.01<0.01<0.010.520.0410.46
US fluid ounce32 530 882.5632 52732.530.03<0.011.230.08US fluid ounce501.97<0.01<0.01<0.010.011.050.0920.92
US cup260 247 060.7260 247.06260.250.26<0.019.180.57US cup4 016.230.02<0.01<0.010.048.370.7167.34
US pint520 494 120.3520 494.12520.490.52<0.0118.361.15US pint8 032.450.04<0.01<0.010.0816.731.39334.69
US quart1 040 988 240.61 040 988.241 040.991.04<0.0136.722.29US quart16 064.90.07<0.01<0.010.1633.472.79669.37
US gallon4 163 952 9584 163 952.974 163.954.16<0.01146.889.18US gallon64 259.610.29<0.01<0.010.66133.8711.162 677.48
#### Foods, Nutrients and Calories
CRINKLE CUT STEAMABLE CARROTS, UPC: 07078404436 contain(s) 53 calories per 100 grams (≈3.53 ounces) [ price ]
32545 foods that contain Niacin. List of these foods starting with the highest contents of Niacin and the lowest contents of Niacin, and Recommended Dietary Allowances (RDAs) for Niacin
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Indo-Pacific Black weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 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 ]
Mica, solid weighs 2 883 kg/m³ (179.97981 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-507, liquid (R507) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
A Yottajoule per kelvin is a SI-multiple (see prefix Exa) of the entropy unit joule per kelvin and equal to 1.0 × 10+24 J/K)
Electric current is a motion of electrically charged particles within conductors or space.
t/fl.oz to mg/tbsp conversion table, t/fl.oz to mg/tbsp unit converter or convert between all units of density measurement.
#### Calculators
Triangle calculator and area of a triangle calculator | 4,416 | 9,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.267438 |
https://susam.net/blog/comments/from-diophantus-to-fermat.html | 1,642,488,269,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00098.warc.gz | 612,771,248 | 1,381 | Comments on From Diophantus to Fermat
Post Comment
Sivasubramaniam Sivakumar said:
Nice!
I would never thought of this:
This is of the form $$2(3a^2b + b^3)$$ where $$a = y$$ and $$b = 3.$$ Now $$2(3a^2b + b^3) = (a + b)^3 - (a - b)^3.$$
Post Comment | 99 | 255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-05 | latest | en | 0.850492 |
https://www.gmslearner.xyz/2021/04/chapter-1-real-numbers-mcqs-class-10th.html | 1,714,046,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00132.warc.gz | 714,882,025 | 88,546 | Chapter 1 Real Numbers MCQs | Class 10th Maths | - GMS - Learning Simply
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# Chapter 1 Real Numbers MCQs | Class 10th Maths |
Chapter 1 Real Numbers MCQs | Class 10th Maths |
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# Class 10 Maths Chapter 1 Real Numbers MCQs
Class 10 Maths MCQs for Chapter 1 (Real numbers) are available here online for students, along with their answers. These multiple-choice questions are prepared, as per the latest CBSE syllabus and NCERT guidelines. Solving these objective questions will help students to build their problem-solving skills and score good marks in board exams.
## Class 10 Maths MCQs for Real Numbers
Students are suggested to solve the given multiple-choice questions and check your answers here. Chapter real numbers will teach you about different types of numbers and their applications. Try to solve these questions as per your knowledge and skills and then verify the answers. This practice will help you to boost your confidence. Also, get important questions for class 10 Maths here at Goyanka Maths Study.
#### Below are the MCQs for chapter 1-Real Numbers.
1. The decimal expansion of 22/7 is
(a)Terminating
(b)Non-terminating and repeating
(c)Non-terminating and Non-repeating
(d)None of the above
Explanation: 22/7= 3.14285714286..
2. For some integer n, the odd integer is represented in the form of:
(a) n
(b) n+1
(c) 2n+1
(d) 2n
Explanation: Since 2n represents the even numbers, hence 2n+1 will always represent the odd numbers. Suppose if n=2, then 2n=4 and 2n+1 = 5.
3.HCF of 26 and 91 is:
(a)15
(b)13
(c)19
(d)11
Explanation: The prime factorisation of 26 and 91 is;
26 = 2 x 13
91 = 7 x 13
Hence, HCF (26, 91) = 13
4. Which of the following is not irrational?
(a) (3+√7)
(b) (3-√7)
(c) (3+√7) (3-√7)
(d) 3√7
Explanation: If we solve, (3+√7) (3-√7), we get;
(3+√7) (3-√7) = 32-(√7)2 = 9 – 7 = 2 [By a2-b2 = (a-b) (a+b)]
5. The addition of a rational number and an irrational number is equal to:
(a)rational number
(b)Irrational number
(c)Both
(d)None of the above
6. The multiplication of two irrational numbers is:
(a)irrational number
(b)rational number
(c)Maybe rational or irrational
(d)None
7.If set A = {1, 2, 3, 4, 5,…} is given, then it represents:
(a)Whole numbers
(b)Rational Numbers
(c)Natural numbers
(d)Complex numbers
8. If p and q are integers and is represented in the form of p/q, then it is a:
(a)Whole numbers
(b)Rational numbers
(c)Natural numbers
(d)Even numbers
9. The largest number that divides 70 and 125, which leaves the remainders 5 and 8, is:
(a)65
(b)15
(c)13
(d)25
Explanation: 70 -5 = 65 and 125-8=117
HCF (65, 117) is the largest number that divides 70 and 125 and leaves remainder 5 and 8
HCF (65, 117) = 13
10. The least number that is divisible by all the numbers from 1 to 5 is.
(a)70
(b)60
(c)80
(d)90
Explanation: The least number will be LCM of 1, 2, 3, 4, 5.
Hence, LCM (1, 2, 3, 4, 5) = 2 x 2 x 3 x 5 = 60
1. What is the least number that must be added to 1056 so the number is divisible by 23?
1. 0
2. 3
3. 2
4. 1
Solution: We have,
On dividing 1056 by 23, we got 21 as remainder.
⇒ If we add 23 – 21 = 2 to the dividend 1056, we will get a number completely divisible by 23.
∴ Required number = (23 – 21) = 2
1. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number?
1. 360
2. 295
3. 270
4. 240
Solution: Let the smaller number be x.
⇒ Larger number = (x+1365)
⇒x+1365=6x+15
⇒5x=1350
⇒x=270
∴ Larger number is (270 + 1365) = 1635
and smaller number is 270.
1. Euclid’s division lemma states “Given positive integers a and b, there exist unique integers q and r satisfying a=bq+r. Which of the following is true for r?
1. r>a
2. r<0
3. 0≤r<b
4. r>b
Solution: Euclid’s division lemma:
Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r where 0≤r<b.
Basically, it can be observed that remainder can never be more than the divisor, and is a non-negative integer (could be zero).
1. a and b, when divided by 7 and 6 respectively, leave remainders p and q respectively. What is the maximum value of p + q?
1. 5
2. 6
3. 12
4. 11
Solution: There exist integers m and n such that a = 7m + p and b = 6n + q such that
0<p<7⇒
Maximum value of p will be 6.
0<q<6⇒
Maximum value of p will be 5.
Therefore, the maximum value of p + q will be 11.
1. If HCF (1008, 20) = HCF (20, a) = HCF (a, b) where 1008=20×q+a; 20=a×m+b where (q, a) and (m, b) are positive integers satisfying Euclid’s Division Lemma. What could be the values of a and b?
1. 24, 8
2. 20, 8
3. 10, 4
4. 8, 4
Answer: (D) 8, 4
Solution: If p=d×q+r, (p>q) where p, q, d, r are integers and for a given (p, d), there exist a unique (q, r), then HCF (p, d) = HCF (d, r). Because this relation holds true, the Euclid’s Division Algorithm exists in a step by step manner. So, to find the HCF (1008, 20), we use Euclid’s division lemma at every step.
Step 1: 1008=20×50+8 ⇒ HCF(1008, 20) = HCF(20,8) ⇒ a could be 8
Step 2: 20=8×2+4 ⇒ HCF(20, 8) = HCF(8,4) ⇒ b could be 4
Step 3: 8=4×2+0
HCF = 4
Since 1008=20×q+a where q and a are positive integers satisfy Euclid’s Division Lemma, we must have 0≤a<20. So a is surely 8 and b is 4.
## About the Author
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio… | 1,950 | 5,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-18 | latest | en | 0.879177 |
https://brilliant.org/problems/trig-nested-radicalsfun/ | 1,490,329,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187690.11/warc/CC-MAIN-20170322212947-00442-ip-10-233-31-227.ec2.internal.warc.gz | 796,763,031 | 12,970 | Geometry Level 4
If $$\sqrt{2 + \sqrt{2+\sqrt2}}$$ is in the form of $$A \cos \left( \dfrac BC \pi \right)$$, where $$A,B$$ and $$C$$ are positive integers with $$B,C$$ coprime, find the minimum value of $$A+B+C$$.
Bonus: Investigate why $$\sqrt { 2+\sqrt { 2+\sqrt { 2 \cdots} } } =2$$ using trigonometry.
× | 113 | 311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-13 | longest | en | 0.617363 |
https://iwanttobeasuperteacher.com/product/ready-set-show-february-addition-subtraction-fact-game | 1,718,542,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00636.warc.gz | 283,338,702 | 41,985 | \$5.00
Valentine’s Day and February addition and subtraction fact practice that is READY IN SECONDS!
Use this digital addition and subtraction game with white boards to add some seasonal February FUN to your math fact fluency practice.
Want a closer look? You can download a preview of this product HERE.
Description
Second and third grade students love the seasonal art obscuring the digits in their addition and subtraction equations, and you’ll love how engaged your students are with math fact practice.
Includes a Google Slides version. This link, along with instructions, can be found to the left of slide 1 or in the Notes section of slide 1 when viewing the PowerPoint in editing (not presentation) mode. The Google version does not include teacher directions or editable text boxes and slides, so it is ready to be shared directly with students and parents.
This digital resource is a perfect fact practice or review activity you can use immediately with your 2nd grade or 3rd grade students. No paper needed! It includes addition and subtraction fact practice problems that meet first, second, and third grade (review) standards for a mixed fact practice review.
Use this addition and subtraction fact practice activity as a:
• Whole class math warmup
• Small group activity during math centers (display the show on an iPad or Chromebook)
• Fun math activity during the month of February
• Review activity before a timed test
• Brain break – have students shout out answers while doing a popcorn (stand up/jump/sit down) or jumping jack instead of using white boards. Fun and easy!
The interactive slide show contains every addition equation with a sum of 2-20 and every subtraction equation with a minuend of 1-20. Addition and subtraction facts are presented in a mixed format throughout the show.
The PowerPoint includes 103 addition problems and 210 subtraction problems organized into the following sections:
• Student intro and instructions
• Practice question
• 91 Addition & subtraction facts using one-digit numbers only
• 222 Addition & subtraction facts using at least one two-digit number
• Editable teacher slides and instructions
• Thank you, credits & copyright
If you would like to make changes to the slide show, the text boxes on the instruction & practice problem slides are editable + there are 4 editable slides included at the end of the show for you to edit, copy, and paste into the existing show so you can reuse this resource.
The clip art is not included on the editable slides due to copyright restrictions. However, a link to the paid clip art used is provided in the teacher notes within the product.
• 1 PowerPoint file (with embedded font on editable slides)
Type of File: 1 PowerPoint file (with embedded font on editable slides) + link to a Google Classroom version of the activity
Total Pages/Slides: 1,268 slides
Terms & Conditions
You may:
• Use this item for your own classroom and/or students.
• Move, copy, or delete slides to personalize the product for your classroom needs.
• Add slides using the editable templates
• Use this product in your own classroom year after year.
You may not:
• Give this item to others.
• Copy this item for use by others.
• Use any piece of this product to create a product, free or priced.
• Post, share, or save this product to any sort of digital site or server including, but not limited to, a personal, classroom, or district website.
Contact
If you have any questions before purchase, feel free to email me at [email protected] or contact me through the contact forms here on the site. I would love to hear from you! | 785 | 3,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.894745 |
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