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https://www.coursehero.com/file/p54fhct/What-is-the-probability-that-a-registered-voter-did-not-vote-in-the-election/ | 1,642,942,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00545.warc.gz | 766,002,570 | 56,538 | What is the probability that a registered voter did
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What is the probability that a registered voter did not vote in the election?
The probability that a registered voter did not vote in the election is approximately .
(Round to three decimal places as needed.)
36. Use the bar graph below, which shows the highest level The probability that the highest level of education for of education received by employees of a company, to find employee chosen at random is B i s 0.263 the probability that the highest level of education for an employee chosen at random is B. (Round to the nearest thousandth as needed.) an . 40 35 30 25 20 15 10 5 5 Level of education 32 21 16 5 1 0 A B C D E F 37. Use the bar graph below, which shows the highest level The probability that the highest level of education for of education received by employees of a company, to find employee chosen at random is E i s 0.078 the probability that the highest level of education for an employee chosen at random is E. (Round to the nearest thousandth as needed.) an . 40 35 30 25 20 15 10 7 5 Level of education 34 25 16 7 1 0 A B C D E F 38. The frequency distribution to the right shows the number of voters (in millions) according to age. Consider the event below. Can it be considered unusual? A voter chosen at random is between 21 and 24 years old Ages of voters Frequency 18 to 20 11.7 21 to 24 10.2 25 to 34 22.9 35 to 44 23.4 45 to 64 19.8 65 and over 55.9 Choose the correct answer below. Number of employees Number of employees
39. When two light yellow flowers (YW) are crossed, there are four equally likely possible outcomes for the geneti makeup of the offspring: yellow (YY), light yellow(YW), light yellow(WY), and white (WW). If two light yellow snapdragons are crossed, what is the probability that the offspring will be (a) light yellow, (b) yellow, and (c) white? (a) If two light yellow snapdragons are crossed, the probability that the offspring will be light yellow i s 0.5 (Do not round.) (b) If two light yellow snapdragons are crossed, the probability that the offspring will be yellow i s 0.25 (Do not round.) (c) If two light yellow snapdragons are crossed, the probability that the offspring will be white i s 0.25 (Do not round.) 40. Use the pie chart at the right, which shows the number of tulip purchased from a nursery. Find the probability that a tulip bulb chosen at random is purple. Tulips purchased from a nursery Red Tulip Bulbs 35 Yellow Tulip Bulbs 20 Purple Tulip Bulbs 45 The probability that a tulip bulb chosen at random is purple i (Type an integer or a decimal.) .45 41. Use the pie chart at the right, which shows the number of workers (in thousands) b y industry for a certain country. Find the probability that a worker chosen at random was not employed in the agriculture, forestry, fishing, and hunting industry. c . . . s s . | 760 | 2,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-05 | latest | en | 0.942013 |
https://www.epcc.ed.ac.uk/discover-and-learn/resources-and-activities/what-is-a-supercomputer/wee-archie | 1,632,190,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00374.warc.gz | 810,713,606 | 5,729 | Wee Archie
Wee Archie is a suitcase-sized supercomputer designed and built to explain what a supercomputer is.
Wee Archie’s big brother, ARCHER, takes up an entire building, with special floors to take its weight because it is so heavy. Find out more about Wee Archie from this video.
How Wee Archie works
Wee Archie is constructed from 18 Raspberry Pi 2’s, a network switch, a power supply unit (PSU) and Ethernet cables. Each Pi can calculate 93 million (that’s 93,000,000) instructions per second. Theoretically Wee Archie can therefore do 1,674,000,000 instructions per second if all of the Pi’s are used together at the same time!!
Wee Archie's big brother ARCHER has 4920 nodes, many network switches and its own special power supply from the power companies! ARCHER can calculate 1.6 x 1015 (1,600,000,000,000,000 or 16 followed by 15 zeroes!) instructions per second.
Each Pi has four cores, each capable of calculating, which can all see the same information, as they all reside on the same device. This is similar to your multicore laptop or phone: multiple processes run at the same time, allowing more than one thing to happen at once. Each core can see all the same information.
The nodes on ARCHER each have 24 cores. That means that ARCHER has a total of 118,080 cores.
16 of the Pi’s are ‘worker’ nodes, resulting in 16x4 (or 64) cores for any job to run on. To use all 64 cores we run a problem in ‘parallel’ so that we use parallel computing.
The final two Raspberry Pi’s are management nodes: they help us run the other 16 nodes, schedule jobs and allow us to interact with the ‘worker’ nodes. This is the same as on a supercomputer where a person cannot interact directly with worker nodes, but instead submits jobs to be run on the supercomputer to the login or management nodes.
Addendum: you can find instructions on how to configure your very own Raspberry Pi cluster here.
Running a parallel simulation
Coming soon: find out how you program a supercomputer using parallel computation. | 481 | 2,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.87891 |
https://reasonandscience.catsboard.com/t2772-the-big-baffling-number-at-the-heart-of-a-cosmic-coincidence | 1,618,974,102,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00030.warc.gz | 559,995,493 | 19,878 | Defending the Christian Worlview, Creationism, and Intelligent Design
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Defending the Christian Worlview, Creationism, and Intelligent Design
This is my personal virtual library, where i collect information, which leads in my view to the Christian faith, creationism, and Intelligent Design as the best explanation of the origin of the physical Universe, life, and biodiversity
The big baffling number at the heart of a cosmic coincidence
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Otangelo
The big baffling number at the heart of a cosmic coincidence
In the 1920s, British astronomer Sir Arthur Eddington became fixated on a curious coincidence regarding a huge ratio number, 10^40 2
The ratio of the electric force to gravitational force (presumably a constant), is a large number (about 10^40), while the ratio of the observable size of the universe (when reaching an equilibrium radius, see later in this article) to the size of an elementary particle is also a large number, surprisingly close to the first number (also about 10^40) 3 It is hard to imagine that two very large and unrelated numbers would turn out to be so close to each other. Why are they?
1937, Dirac wrote a 650-word letter to the journal Nature. The letter considered the number 10^40, the ratio of the strength of the electromagnetic force to the gravitational force. But Dirac compared this number to the ratio of the radius of the universe to the radius of a proton. The results were very similar, certainly similar enough to convince Dirac that there was a connection. It is certainly unusual to find such a huge number in science, and even more surprising to find approximately the same number arising from two different calculations. As John Barrow said: "There must exist some undiscovered mathematical formula linking the quantities involved. They must be consequences rather than coincidences." This is called the Dirac large numbers hypothesis.
The Dirac large numbers hypothesis (LNH) is an observation made by Paul Dirac relating ratios of size scales in the Universe to that of force scales. The ratios constitute very large, dimensionless numbers: some 40 orders of magnitude in the present cosmological epoch. According to Dirac's hypothesis, the apparent similarity of these ratios might not be a mere coincidence but instead could imply a cosmology with these unusual features. 1
The ratio of the strength of the electromagnetic force to the gravitational force is 10^40. The modified gravity hypothesis (MGH) suggests that the universe has a certain equilibrium radius, and the universe will expand until it reaches that radius.
So the strength of the electromagnetic force controls the size of the atom, and the strength of gravity determines the size of the universe. We are dealing with objects which are in an equilibrium state, meaning the forces
holding them together are precisely equal to the forces pulling them apart. This results in an equilibrium radius for the object, and it is this equilibrium distance which interests us. The MGH suggests the universe has an equilibrium radius in much the same way that an atom has an equilibrium radius.
Gravity dominates the universe and determines its size in much the same way as the electromagnetic force determines the overall size of atoms. It is as though the universe is a scaled-up atom!
Neil Turok considered this simplicity in his 2015 talk at the Perimeter Institute called The Amazing Simplicity of Everything
http://tinyurl.com/turoklecture
"The astonishing thing about recent discoveries in physics is that they tell us the universe is surprisingly simple and regular, on the tiniest scale and on the hugest scale. It's only complicated in the middle. To first approximation, the universe is absolutely uniform in all directions. The whole universe is as simple as the simplest atom. If you think about a hydrogen atom, how many numbers do you need to describe an atom? An atom is a pretty simple thing: you have a nucleus, you have an electron going around it, you have the force of electrical attraction between the nucleus and the electron. Well, it turns out to describe the universe you need just one number. That number describes the universe — fewer numbers than you need to describe a single atom. So the universe turns out to be the simplest thing we know."
At these two extremes of scale — the atom and the universe — we find similar situations. We find objects in stable, equilibrium situations dominated by a single particular force. This results in simplicity at the two extremes of scale.
The equilibrium radius which reveals deep truths about the universe. The ratio of this radius of the universe, AND the radius of an atom ARE BOTH 10^40. This is a prediction of a well-founded hypothesis, constructed from
first principles, a hypothesis which makes predictions and agrees with known measurements.
So far, we have calculated the value for the ratio of the size of the universe to the size of an atom, and found the value equal to 10^40. Crucially, this value does not necessarily have to alter with time. For the next step, Dirac considered this 10^40 value and found it was the same as the ratio of the strength of the electromagnetic force to the gravitational force. Is this a just a crazy coincidence? Or could the radiuses of the universe and an atom be related to the strengths of the electromagnetic and gravitational forces in some way?
Firstly, considering the atom. What forces control the size of the atom? Well, the strong force is short range, confined to the nucleus, and so does not play a role. And gravity is too weak at these scales. So it is the electromagnetic force which controls the overall shape of the atom, holding electrons in orbit around the nucleus.
Secondly, considering the universe, the electromagnetic force tends to cancel in atoms, positive charge equalling negative charge, so large objects become electrically neutral. Hence, the electromagnetic force does not play a role in shaping the universe. It is gravity which is the dominant force in the overall size of the universe, even though it is by far the weakest of the four forces. Gravity dominates because all mass has the same gravitational charge (there is no such thing as negative mass). Hence, for very large objects, the force of gravity steadily accumulates until it becomes the dominant force.
The problem of fine tuning is one of the biggest embarrassments facing modern physical and biological science. These “coincidences” may be indicating the existence of some deep, underlying unity involving the fundamental constants, linking the microcosm to the macrocosm just as the ancients saw without mathematics. 3 The anthropic principle just states the obvious: “We are here because we are here.” It has little explanatory power.
1. http://vixra.org/pdf/1811.0478v1.pdf
2. https://cosmosmagazine.com/mathematics/the-big-baffling-number-at-the-heart-of-a-cosmic-coincidence
3. https://www.huffingtonpost.com/deepak-chopra/why-the-universe-is-our-h_1_b_2950189.html
Otangelo
Jesus said: I am the truth, the way, and life.
Jesus is described in Revelation as the alpha and omega.
Alpha is one of the fundamental constants in physics. If it had not the precise value that it has, there would be no life in the universe.
One of these fundamental constants is the fine-structure constant, or alpha, which is the coupling constant for the electromagnetic force and equal to about 1/137.0359. If alpha were just 4% bigger or smaller than it is, stars wouldn't be able to make carbon and oxygen, which would have made it impossible for life as we know it to exist.
The reason 137 has obsessed so many thinkers to try and find the math behind it is clear. Einstein wrote, “In a reasonable theory there are no numbers whose values are only empirically determinable.”
“It is impossible for human minds to construct something as infinitely rich as mathematics, that so successfully describes our universe down to the finest detail. Given that it is impossible for anything else to describe our world so accurately, it is also impossible to conclude that the universe is not mathematical.
It’s often said that the whole of biology is explained by chemistry, the whole of chemistry by physics and the whole of physics by mathematics, the queen of the sciences. If all physics is contained within mathematics, it has the most radical consequences.
If, however, there’s only one physics, defined entirely by ontological mathematics (which is a single, interlocked edifice) then the fact that there is life in our universe is no accident at all.
But it points straightforward to the one, which has the name:
Alpha & Omega.
Revelation 22:13
I am the Alpha and the Omega, the First and the Last, the Beginning and the End.
https://www.secretsinplainsight.com/why-137/
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Easy Accuracy: 50.08% Submissions: 16451 Points: 2
Given two linked lists, your task is to complete the function makeUnion(), that returns the union of two linked lists. This union should include all the distinct elements only.
Example 1:
Input:
L1 = 9->6->4->2->3->8
L2 = 1->2->8->6->2
Output: 1 2 3 4 6 8 9
The task is to complete the function makeUnion() which makes the union of the given two lists and returns the head of the new list.
Note: The new list formed should be in non-decreasing order.
Expected Time Complexity: O(N * Log(N))
Expected Auxiliary Space: O(N)
Constraints: | 269 | 795 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | longest | en | 0.680358 |
https://www.brighthub.com/office/career-planning/articles/77741.aspx | 1,579,744,672,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00264.warc.gz | 786,680,425 | 25,290 | # How to Calculate Hourly Rate from Annual Salary: Comparing Job Wages
## Calculate Hourly Rate from Annual Salary
Perhaps you’re trying to compare your current pay rate to the compensation you would receive in a different position or even at a different company. Or maybe you’re viewing a salary survey and trying to figure out where you stand, but the data is provided in terms of hourly rates and you’re on a salary. Whatever your reason for wanting to calculate hourly rate from annual salary information, know that you can do it in a few simple steps.
Note that if you’re earning \$40,000 a year and have three weeks of combined vacation and sick time, you receive that amount for 49 weeks of work if you use all your vacation and sick time. So the calculation is simplest if you’re comparing jobs with comparable vacation and sick time benefits. You don’t even need to factor in days off, as the company is essentially paying you for both the days you work and the days you take as vacation or sick days.
Performing the Calculation
With 52 weeks in a year and 40 work hours per week, there are 52×40=2080 work hours in a year. So you simply divide your \$40,000 by 2080 hours and get 40,000/2080=\$19.23. So the hourly rate that is equivalent to your salary of \$40,000, all other things being equal, is \$19.23 per hour.
Replacing these specifics with a more general equation, you can calculate an hourly rate from an annual salary using this equation:
Salary (in dollars) / Hours worked per year = Hourly rate
or, more specifically:
Salary (in dollars) / ( Hours worked per week x 52 ) = Hourly rate
Perhaps you want to compare your salaried position to a contract position that pays hourly, or you are determining how to set hourly rates as a contractor to be equivalent to a specific salary. In those cases, you may want to consider that you actually receive a salary for working only the days you don’t take off as sick or vacation days.
For example, if you have two weeks of vacation time and two weeks of sick time, you receive that \$40,000 for 48 weeks of work a year. Thus you are actually paid \$40,000/(40 x 49) = \$40,000/1920 = \$20.83. This is a difference of over \$1.50 an hour from the original calculation. If you were to do contract work at a rate of \$19.23 per hour, you would either have to work 40 hours a week all 52 weeks of the year to earn \$40,000, or work 43.3 hours a week to earn \$40,000 and still get four weeks off each year. If you were to work only 48 weeks a year at 40 hours a week, you would earn only \$36,923.
So while it is quite simple to calculate hourly rate from annual salary, be sure you understand the assumptions you make when you make the comparison so you’ll actually be answering the right question. | 647 | 2,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-05 | latest | en | 0.948879 |
https://www.daniweb.com/programming/software-development/threads/458487/calculate-the-x-and-y-coordinate-of-a-point-with-the-polar-coordinates | 1,726,501,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00121.warc.gz | 668,762,292 | 15,055 | ``````public class PolarCoordinates {
public static void main(String[] args){
int r=10;
double x=0,y=0,theta=30;
x=r*Math.cos(theta);
y=r*Math.sin(theta);
System.out.println("The value of x is :" +x);
System.out.println("The value of y is :" +y);
}
}
``````
help please?
it gives me wrong answer.
10cos(30) should equal to 8.7, but it gives me 11.7 ans.
10sin(30) should equal to 5, and it gives me 2.63 ans.
whats wrong with my code?
sin. cos etc use angles in radians, your data (30) looks like degrees.
30° equals pi/6 rad
oh yes, its a degrees.
what's wrong with my formula?
To convert yout theta value from degrees to radians use: theta = (theta * 3.141592)/360;
what's wrong with my formula?
sin. cos etc use angles in radians (Google it), you have supplied an angle in degrees
Pi radians is 180 degrees, so ddanbe' formula should be theta = theta * Math.PI / 180;
(sorry ddanbe)
Oops!!! Thanks for correcting me JamesCherrill, 2*pi rad = 360° and not pi rad = 360° of course!
ohh, ok! thanks! i got it now :)
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https://practicaldev-herokuapp-com.global.ssl.fastly.net/reduncan/my-journey-through-javascript-flowcharting-pt-2-4k2 | 1,695,422,271,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00451.warc.gz | 536,757,041 | 20,097 | ## DEV Community
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# My Journey through Javascript: Flowcharting Pt. 2
Last post we looked at the basics of flowcharting and how to structure them. This post we will be expanding the basics and getting into more in-depth processes. So let's get started...
FLOWCHARTING Part Deux:
Remember from Part 1, the only real standard in flowcharting is that Loops/Conditionals are denoted by a diamond. With that being said, I do not follow this standard to a "T". In my own little OCD world, I prefer to have a different symbol for each part of my flowchart. So I will be using a diamond for Loops and a rectangle for conditionals. You can do it whichever way your prefer, but your company/employer may have different requirements. As long as you and other developers on your team understand what is going on, you are doing it correctly.
We will be looking at 2 separate flowcharts today, one using a Loop and another using a Loop and a Conditional. Let's start by looking at one with just a Loop...
We are going to flowchart a program that will take in an array of numbers from the user. The program will sum the numbers in the array and render the sum.
Step 2: We create an empty array and call it numbers
Step 3: We create a variable called sum and set it equal to zero (we must do this so we have something to begin adding to, otherwise we will add our first number onto itself and our sum will not be accurate)
step 4: We use a listener to obtain our first number (we are also creating the variable userInput1 in this step)
Step 5: We use a second listener to obtain another number (we are also creating the variable userInput2 in this step)
Step 6: We use push to add all of our user input numbers into our array
Step 7: Now we create our Loop, and inside of the Loop we specify the parameters in which the loop will run until it stops
Step 8: We have to add in an operator that will reassign the variable sum as we run our loop
Step 9: We render the sum of the array once the Loop has finished running
Here's what this will look like in an actual flowchart...
Now let's look at an example where we have a pre-made array of student names and their GPAs. We want to run a function where we look through the Object Array and find all students that have a GPA of 3.0 or higher and then render the student names.
Step 2: We have a pre-made Object List named studentList and inside of it we have object pairs for name and GPA (note we notate the object pairs in the object out to the right of the array box)
Step 3: We create our Loop that will run through our student list
Step 4: We use a Conditional to check that the student's GPA is 3.0 or higher (note this is inside of the loop)
Step 5: We create another variable, we will call it studentName inside the loop and set it equal to studentList[i].name
Step 6: We render the var studentName, note this is outside of the loop (if the render is inside of the loop, we would run into the issue of it rendering a value each time it ran through the loop)
Here's what this will look like in an actual flowchart...
You can see that this is a much more technical program than the previous one but the flowchart is somewhat easier to chart.
Next post we will take the 3 flowcharts we have created so far and get into the really fun stuff...CODING!
Until next time :) | 780 | 3,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.914827 |
https://studylib.net/doc/9829910/physics-12-and-ap | 1,618,575,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00483.warc.gz | 641,693,109 | 13,131 | Physics 12 & AP
```Physics 12
Mr. Jean
February 26th, 2014
The plan:
•
•
•
•
Video of the day
Quiz #1 Outline
Final Project details
Final Project examples
Quiz outline:
1. Scientific Notation
•
Will be built into questions
2. Projectiles
•
Level & not level projectiles
3. Uniform Circular Motion
•
•
•
•
Horizontal circles
Vertical circles
Conical Pendulums
Banked Turns without Friction
Quiz Format:
• 1) True & False
– Typically off videos shown in class
• 2) Multiple Choice
– Combination of videos shown in class and
theory
– Situational problems and applying theory
Quiz Date:
• March 5th, 2014
• Other details:
– Formula sheet
– Calculator
– Pencil & Eraser
The Final Project:
• A Rube Goldberg machine is your final
project for physics 12.
• It’s due date for this class is April 28th,
2014.
• This project will have a value of 10%
Rube Goldberg Examples:
• Here are a few professionally designed
RGB projects.
– This too shall pass.
Real life RGB:
• 45 seconds of constant motion
• No cuts to video.
• No “time wasters”…
– You’ll know what I mean.
Rube Goldberg Options:
• Examples of Real Life RGB projects
– Paul, Kevin, Tom & Brandon
Digital World RGB:
• You can use any program to create a digital
world ranging from “Phun”, “Algodoo”, Halo,
Little Big planet, Linerider, Portal, etc….
• Constant motion must continue for 2 minutes if
you choose the digital format.
– No cuts to video.
– No “time wasters”…
– Must be creative
Digital RGB Projects:
Rube Goldberg:
RGB - Project
• More details: About the groups
– Soft cap of groups with 4 students.
• For every group member above 4, you need to add
15 seconds to the project if real life or 30 seconds
if digital.
– No more than 6 students will be in one group.
This is the absolute max
RGB - Project
• More details: Planning time
– Three classes of planning will be given.
• Today is one of them.
Suggested schedule:
• By mid-March have all of the machines
and layout planned
• By late-March test these different
machines individually to make sure they
function like you expect.
• By early April have a weekend in mind to
put everything together.
• By mid April have the video and project
complete
``` | 581 | 2,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | latest | en | 0.761718 |
http://etd.ncsi.iisc.ernet.in/handle/2005/411 | 1,418,903,501,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802766267.61/warc/CC-MAIN-20141217075246-00152-ip-10-231-17-201.ec2.internal.warc.gz | 104,822,813 | 5,271 | Please use this identifier to cite or link to this item: `http://hdl.handle.net/2005/411`
Title: Development And Validation Of Two-Dimensional Mathematical Model Of Boron Carbide Manufacturing Process Authors: Kumar, Rakesh Advisors: Gupta, Govind S Keywords: Boron Carbide - ManufacturingBoron Carbide Manufacturing - ModelingCarbothermal ReductionThermodynamicsCarbothermal Reduction - ModelingBoron-Carbon System - ThermochemistryB4C Submitted Date: 2006 Series/Report no.: G21043 Abstract: Boron carbide is produced in a heat resistance furnace using boric oxide and petroleum coke as the raw materials. In this process, a large current is passed through the graphite rod located at the center of the cylindrical furnace, which is surrounded by the coke and boron oxide mixture. Heat generated due to resistance heating is responsible for the reaction of boron oxide with coke which results in the formation of boron carbide. The whole process is highly energy intensive and inefficient in terms of the production of boron carbide. Only 15% charge gets converted into boron carbide. The aim of the present work is to develop a mathematical model for this batch process and validate the model with experiments and to optimize the operating parameters to increase the productivity. To mathematically model the process and understand the influence of various operating parameters on the productivity, existing simple one-dimensional (1-D) mathematical model in radial direction is modified first. Two-dimensional (2-D) model can represent the process better; therefore in second stage of the project a 2-D mathematical model is also developed. For both, 1-D and 2-D models, coupled heat and mass balance equations are solved using finite volume technique. Both the models have been tested for time step and grid size independency. The kinetics of the reaction is represented using nucleation growth mechanism. Conduction, convection and radiation terms are considered in the formulation of heat transfer equation. Fraction of boron carbide formed and temperature profiles in the radial direction are obtained computationally. Experiments were conducted on a previously developed experimental setup consisting of heat resistance furnace, a power supply unit and electrode cooling device. The heating furnace is made of stainless steel body with high temperature ceramic wool insulation. In order to validate the mathematical model, experiments are performed in various conditions. Temperatures are measured at various locations in the furnace and samples are collected from the various locations (both in radial and angular directions) in the furnace for chemical analysis. Also, many experimental data are used from the previous work to validate the computed results. For temperatures measurement, pyrometer, C, B and K type thermocouple were used. It is observed that results obtained from both the models (1-D and 2-D) are in reasonable agreement with the experimental results. Once the models are validated with the experiments, sensitivity analysis of various parameters such as power supply, initial percentage of B4C in the charge, composition of the charge, and various modes of power supply, on the process is performed. It is found that linear power supply mode, presence of B4C in the initial mixture and increase in power supply give better productivity (fraction reacted). In order to have more confidence in the developed models, the parameters of one the computed results in the sensitivity analysis parameters are chosen (in present case, linear power supply is chosen) to perform the experiment. Results obtained from the experiment performed under the same simulated conditions as computed results are found in excellent match with each other. URI: http://hdl.handle.net/2005/411 Appears in Collections: Materials Engineering (formely known as Metallurgy) (materials)
Files in This Item:
File Description SizeFormat | 752 | 3,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.886887 |
http://en.wikipedia.org/wiki/Second_category | 1,410,943,283,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657121798.11/warc/CC-MAIN-20140914011201-00106-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 92,850,256 | 12,196 | # Meagre set
(Redirected from Second category)
In the mathematical fields of general topology and descriptive set theory, a meagre set (also called a meager set or a set of first category) is a set that, considered as a subset of a (usually larger) topological space, is in a precise sense small or negligible. The meagre subsets of a fixed space form a sigma-ideal of subsets; that is, any subset of a meagre set is meagre, and the union of countably many meagre sets is meagre.
General topologists use the term Baire space to refer to a broad class of topological spaces on which the notion of meagre set is not trivial (in particular, the entire space is not meagre). Descriptive set theorists mostly study meagre sets as subsets of the real numbers, or more generally any Polish space, and reserve the term Baire space for one particular Polish space.
The complement of a meagre set is a comeagre set or residual set.
## Definition
Given a topological space X, a subset A of X is meagre if it can be expressed as the union of countably many nowhere dense subsets of X. Dually, a comeagre set is one whose complement is meagre, or equivalently, the intersection of countably many sets with dense interiors.
A subset B of X is nowhere dense if there is no neighbourhood on which B is dense: for any nonempty open set U in X, there is a nonempty open set V contained in U such that V and B are disjoint.
The complement of a nowhere dense set is a dense set. More precisely, the complement of a nowhere dense set is a set with dense interior. Not every dense set has a nowhere dense complement. The complement of a dense set can have nowhere dense, and dense regions.
### Relation to Borel hierarchy
Just as a nowhere dense subset need not be closed, but is always contained in a closed nowhere dense subset (viz, its closure), a meagre set need not be an Fσ set (countable union of closed sets), but is always contained in an Fσ set made from nowhere dense sets (by taking the closure of each set).
Dually, just as the complement of a nowhere dense set need not be open, but has a dense interior (contains a dense open set), a comeagre set need not be a Gδ set (countable intersection of open sets), but contains a dense Gδ set formed from dense open sets.
## Terminology
A meagre set is also called a set of first category; a nonmeagre set (that is, a set that is not meagre) is also called a set of second category. Second category does not mean comeagre – a set may be neither meagre nor comeagre (in this case it will be of second category).
## Properties
• Any subset of a meagre set is meagre; any superset of a comeagre set is comeagre.
• The union of countable many meagre sets is also meagre; the intersection of countably many comeagre sets is comeagre.
This follows from the fact that a countable union of countable sets is countable.
## Banach–Mazur game
Meagre sets have a useful alternative characterization in terms of the Banach–Mazur game. If $Y$ is a topological space, $W$ is a family of subsets of $Y$ which have nonempty interior such that every nonempty open set has a subset in $W$, and $X$ is any subset of $Y$, then there is a Banach-Mazur game corresponding to $X, Y, W$. In the Banach-Mazur game, two players, $P_1$ and $P_2$, alternate choosing successively smaller (in terms of the subset relation) elements of $W$ to produce a descending sequence $W_1 \supset W_2 \supset W_3 \supset \dotsb$. If the intersection of this sequence contains a point in $X$, $P_1$ wins; otherwise, $P_2$ wins. If $W$ is any family of sets meeting the above criteria, then $P_2$ has a winning strategy if and only if $X$ is meagre.
## Examples
### Function spaces
• The set of functions which have a derivative at some point is a meagre set in the space of all continuous functions.[1]
## Notes
1. ^ Banach, S. (1931). "Über die Baire'sche Kategorie gewisser Funktionenmengen". Studia. Math. (3): 174–179. | 1,007 | 3,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2014-41 | latest | en | 0.939118 |
http://www.weegy.com/home.aspx?ConversationId=839AE4FA | 1,529,301,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00379.warc.gz | 552,089,988 | 8,965 | You have new items in your feed. Click to view.
Q: If each portrait is offered in one of 5 different sizes, one of 3 different colors and one of 8 different borders, how many different portrait combinations are there?
A: 5 (choices of sizes) x 3 (choices of colors) x 8 (choices of borders) = 120 choices
Original conversation
User: If each portrait is offered in one of 5 different sizes, one of 3 different colors and one of 8 different borders, how many different portrait combinations are there?
Weegy: 5 (choices of sizes) x 3 (choices of colors) x 8 (choices of borders) = 120 choices
procyon|Points 65|
User: A \$200.00 sweater has been marked down to \$120.00. What percentage discount from the original price does the markdown represent?
Weegy: 40%
amreen|Points 394|
User: If a customer buys 3 wallet-sized prints for \$1.99 each (total price including tax), how much change will she get from a \$20 bill.
Weegy: \$14.03. Any other question? Please click "Good" if satisfied with the answer
KeenResearcher|Points 1447|
User: thank you
Weegy: You are welcome. If you are satisfied with my answer, please select good response. Thank you and all the best!
MasterOfTheGame|Points 553|
User: Amanda received a discount of 30 % off of the price of a portrait product that costs \$75. How much money did Amanda save?
Weegy: 22.50
zch24|Points 438|
User: After picking apples, 4 friends compared the amount of apples in their buckets. Dan’s bucket was 2/3 full, Mary’s bucket was 4/5 full, Julie’s bucket was 3/4 full, and Jane’s bucket was 1/8 full. How would these amounts be arranged in order from greatest to least?
Weegy: 4/5, 2/3, 1/8
zch24|Points 438|
User: A customer has an offer for 5 portrait sheets for 9.99 with additional portrait sheets priced at 19.99 each. If the customer desires to purchase 10 portrait sheets, how much would she save if she instead purchases a 10 portrait sheet package for \$99.99?
Weegy: The cost of the first offer is offer1 = \$9.99 + (10-5)(\$19.99) = \$9.99 + \$99.95 = \$109.94 The cost of the second offer is \$99. [ The savings using the second offer is offer1 - \$99 = \$109.94 - \$99.00 = \$10.94 ]
IfNotSo|Points 7291|
User: 7 3/4 - 3/8 =
Weegy: 0.03125
debnjerry|Points 10831|
User: what is 0.03125 as a fraction
User: :) thanks
Weegy: Your welcome! any more question weegy is here to help!
loveydovey311|Points 208|
User: A photographer conducted three sessions yesterday. She completed the first session in 56 minutes, the second in 42 minutes and the third in 1 hour and 16 minutes. How many hours did it take her to complete all three sessions?
Weegy: 56 minutes + 42 minutes + 76 minutes (1 hour=60 minutes, so 1hr, 16 minutes is 76 minutes), which totals 174 minutes. 174 minutes divided by 60 minutes per hour equals 2 hours, 54 minutes.
ramkitten|Points 7200|
User: weegy rocks thanks
Weegy: Weegy.com is a community-based expert system that belongs to all of us. Got a question? Go to weegy.com. Have a problem? Go to weegy.com. Weegy will usually have the answer.
eyezofarebel|Points 15|
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,216 | 4,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.931219 |
https://statproofbook.github.io/P/duni-qf | 1,716,531,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00304.warc.gz | 484,982,024 | 4,050 | Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Discrete uniform distribution ▷ Quantile function
Theorem: Let $X$ be a random variable following a discrete uniform distribution:
$\label{eq:duni} X \sim \mathcal{U}(a, b) \; .$
Then, the quantile function of $X$ is
$\label{eq:duni-qf} Q_X(p) = \left\{ \begin{array}{rl} -\infty \; , & \text{if} \; p = 0 \\ a (1-p) + (b+1) p - 1 \; , & \text{when} \; p \in \left\lbrace \frac{1}{n}, \frac{2}{n}, \ldots, \frac{b-a}{n}, 1 \right\rbrace \; . \end{array} \right.$
with $n = b - a + 1$.
$\label{eq:duni-cdf} F_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < a \\ \frac{\left\lfloor{x}\right\rfloor - a + 1}{b - a + 1} \; , & \text{if} \; a \leq x \leq b \\ 1 \; , & \text{if} \; x > b \; . \end{array} \right.$
The quantile function $Q_X(p)$ is defined as the smallest $x$, such that $F_X(x) = p$:
$\label{eq:qf} Q_X(p) = \min \left\lbrace x \in \mathbb{R} \, \vert \, F_X(x) = p \right\rbrace \; .$
Because the CDF only returns multiples of $1/n$ with $n = b - a + 1$, the quantile function is only defined for such values. First, we have $Q_X(p) = -\infty$, if $p = 0$. Second, since the cumulative probability increases step-wise by $1/n$ at each integer between and including $a$ and $b$, the minimum $x$ at which
$\label{eq:duni-cdf-p} F_X(x) = \frac{c}{n} \quad \text{where} \quad c \in \left\lbrace 1, \ldots, n \right\rbrace$
is given by
$\label{eq:duni-qf-p} Q_X\left( \frac{c}{n} \right) = a + \frac{c}{n} \cdot n - 1 \; .$
Substituting $p = c/n$ and $n = b - a + 1$, we can finally show:
$\label{eq:duni-qf-qed} \begin{split} Q_X(p) &= a + p \cdot (b-a+1) - 1 \\ &= a + pb - pa + p - 1 \\ &= a (1-p) + (b+1) p - 1 \; . \end{split}$
Sources:
Metadata: ID: P142 | shortcut: duni-qf | author: JoramSoch | date: 2020-07-28, 06:17. | 737 | 1,862 | {"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} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.548754 |
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May 23, 2013
ARTICLE: Chasing Constant Change [back to article]
Reason McLucus wrote on Jun. 3, 2007 @ 07:13 GMT
Physicists can be so naive. The constants in physics equations are not necessarily "constant" They may represent unknown factors which do not appear to change, at least not in the short run. Constants are placed in equations because the known factors by themselves cannot produce an answer when specific values are placed in the equation. Placing the "constant" in the equation allows physicists to get an answer.
In the case of electromagnetic radiation over long distances, constants might not be the only factor which could be associated with differences in data. Conditions in space may change radiation between the time it leaves a distant galaxy and the time it reaches earth.
The values of many physical constants were determined with equipment that was not as sensitive as the equipment in use today. New equipment may produce different values or detect differences in potential factors that could not be determined even a few decades ago. Today's equipment may allow identification of factors that could not be identified before.
report post as inappropriate
paul valletta wrote on Sep. 3, 2007 @ 01:23 GMT
The Constants of Nature are "fixed" into physical Laws within the Universe. The fixing paramiter is of course Time, we place a value on testable quantities "now", in the present time.
In GR and within QM, both allow the Constants of Nature to be fixed, but both QM and GR allow the "effects" of the constants to be varied!
In QM, there is the Law of observation, in GR there is also the Law of observation, both theories are WRT position and momenum, encased within a reference frame of Time, present-time.
The constants of Nature are Observationally varied by the Observers who are detecting them.
report post as inappropriate
paul valletta wrote on Sep. 3, 2007 @ 01:26 GMT
Sorry, this is missing from the above post:The constants of Nature are Observationally varied by the Observers who are detecting them, at a certain instant in Time.
report post as inappropriate
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insert equation into post at cursor | 753 | 3,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.948416 |
https://www.jiskha.com/questions/14997/which-measurement-is-the-largest-14mm-or-1cm-1cm-is-10-mm-it-is-time-you-learned-this | 1,621,178,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991224.58/warc/CC-MAIN-20210516140441-20210516170441-00484.warc.gz | 869,957,992 | 4,406 | # pre-algebra
which measurement is the largest.
14mm or 1cm
1cm is 10 mm.
It is time you learned this.
how do you do this??? you know??
1. 👍
2. 👎
3. 👁
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4. ### physics
A trucker sees the image of a car passing her truck in her diverging rear-view mirror, whose focal length is 60.0 cm. If the car is 1.85 high and 6.75 away, what is the location and height of the image. 1/p + 1/q = 1/f 1/675cm + | 610 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-21 | latest | en | 0.849795 |
http://mathhelpforum.com/algebra/37126-inequality.html | 1,524,154,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936981.24/warc/CC-MAIN-20180419150012-20180419170012-00354.warc.gz | 204,255,741 | 10,534 | 1. Inequality
Solve for x in the given inequality: $\displaystyle 6x - x^2 \ge\;9$
My Work:
$\displaystyle -x^2 + 6x - 9 \ge\; 0$
$\displaystyle (-x + 3)(x - 3) \ge\; 0$
$\displaystyle -x + 3\ge\; 0\Longrightarrow -x \ge\; -3\Longrightarrow x\le\; 3$
$\displaystyle x - 3 \ge\; 0\Longrightarrow x \ge\; 3$
What did I do wrong?
2. Hello,
Originally Posted by R3ap3r
Solve for x in the given inequality: $\displaystyle 6x - x^2 \ge\;9$
My Work:
$\displaystyle -x^2 + 6x - 9 \ge\; 0$
$\displaystyle {\color{red}(-x + 3)(x - 3) \ge\; 0}$
$\displaystyle -x + 3\ge\; 0\Longrightarrow -x \ge\; -3\Longrightarrow x\le\; 3$
$\displaystyle x - 3 \ge\; 0\Longrightarrow x \ge\; 3$
What did I do wrong?
Until the red thing, it's OK
But you could notice that $\displaystyle -x^2+6x-9=-(x^2-6x+9)=-(x-3)^2$
Is it possible that $\displaystyle -(x-3)^2 \ge 0$ ?
3. how exactly is that wrong though? It's how it factors isn't it?
4. Originally Posted by R3ap3r
how exactly is that wrong though? It's how it factors isn't it?
No, your working was correct, but there are two problems :
- there is no solution (but x=3), so that's why you thought you were wrong
- you forgot a situation : -x+3<0 and x-3<0 (their product is positive ). It'll also yield to an impossibility, but you mustn't forget this case.
5. So the final answer is just x=3? That makes sense.
6. Yep.
7. hmmm amazing how a simple problem can trick me even when my working is correct | 513 | 1,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.872748 |
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From Jesus Christ@21:1/5 to All on Thu Sep 13 11:02:33 2018
Quantum of Action = 9.670554000 x 10^-36 J-s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-29 erg-s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-27 erg-s/rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 1 * (4.803204712 x 10^-10 esu)^2 / 2.997924580 x 10^10 cm/s
hbar = 9.670554000 x 10^-29 erg-s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * 1
e^2 = (4.803204712 x 10^-10 esu)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * (4.803204712 x 10^-10 esu)^2
k = 1
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
MKS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-36 kg-m^2/s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-34 kg-m^2/s-rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi) * (1.602176634 x 10^-19 A-s)^2 / 2.997924580 x 10^8 m/s
hbar = 9.670554000 x 10^-36 kg-m^2/s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
e^2 = (1.602176634 x 10^-19 A-s)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * (1.602176634 x 10^-19 A-s)^2
k = 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = (4pi)*e^2/(2*h*c)
alpha = (4pi) * (4.803204712 x 10^-10 esu)^2 / (2.000000000 x 10^0 rad/sr * 6.626070150 x 10−27 erg-s/rad * 2.997924580 x 10^10 cm/s)
alpha = 7.297352566 x 10^-3 sr
MKS
--
alpha = e^2/(2*e0*h*c)
alpha = (1.602176634 x 10^-19 A-s)^2 / (2.000000000 x 10^0 rad/sr * 8.854187817 x 10^-12 A^2-s^4/kg-m^3 * 6.626070150 x 10^-34 kg-m^2/s-rad * 2.997924580 x 10^8 m/s)
alpha = 7.297352566 x 10^-3 sr
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 J-s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
"Fix the planet Earth's physics Errors"
Jesus Christ
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Holy Ghost@21:1/5 to Jesus Christ on Thu Sep 13 13:37:27 2018
On Thursday, September 13, 2018 at 1:02:34 PM UTC-5, Jesus Christ wrote:
Quantum of Action = 9.670554000 x 10^-36 J-s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-29 erg-s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-27 erg-s/rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 1 * (4.803204712 x 10^-10 esu)^2 / 2.997924580 x 10^10 cm/s hbar = 9.670554000 x 10^-29 erg-s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * 1
e^2 = (4.803204712 x 10^-10 esu)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * (4.803204712 x 10^-10 esu)^2
k = 1
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
MKS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-36 kg-m^2/s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-34 kg-m^2/s-rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi) * (1.602176634 x 10^-19 A-s)^2 / 2.997924580 x 10^8 m/s
hbar = 9.670554000 x 10^-36 kg-m^2/s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
e^2 = (1.602176634 x 10^-19 A-s)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * (1.602176634 x 10^-19 A-s)^2
k = 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = (4pi)*e^2/(2*h*c)
alpha = (4pi) * (4.803204712 x 10^-10 esu)^2 / (2.000000000 x 10^0 rad/sr * 6.626070150 x 10−27 erg-s/rad * 2.997924580 x 10^10 cm/s)
alpha = 7.297352566 x 10^-3 sr
MKS
--
alpha = e^2/(2*e0*h*c)
alpha = (1.602176634 x 10^-19 A-s)^2 / (2.000000000 x 10^0 rad/sr * 8.854187817 x 10^-12 A^2-s^4/kg-m^3 * 6.626070150 x 10^-34 kg-m^2/s-rad * 2.997924580 x 10^8 m/s)
alpha = 7.297352566 x 10^-3 sr
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 J-s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
"Fix the planet Earth's physics Errors"
Jesus Christ
Planck Constant is Angular Momentum & Angular Frequency
Quantum of Action = 9.670554000 x 10^-36 J-s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-29 erg-s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-27 erg-s/rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 1 * (4.803204712 x 10^-10 esu)^2 / 2.997924580 x 10^10 cm/s
hbar = 9.670554000 x 10^-29 erg-s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * 1
e^2 = (4.803204712 x 10^-10 esu)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-29 erg-s * 2.997924580 x 10^10 cm/s / (4pi) * (4.803204712 x 10^-10 esu)^2
k = 1
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
MKS
--
alpha = hbar/2*h
alpha = 9.670554000 x 10^-36 kg-m^2/s / 2.000000000 x 10^0 rad/sr * 6.626070150 x 10^-34 kg-m^2/s-rad
alpha = 7.297352566 x 10^-3 sr
hbar = (4pi)*k*e^2/c
hbar = (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi) * (1.602176634 x 10^-19 A-s)^2 / 2.997924580 x 10^8 m/s
hbar = 9.670554000 x 10^-36 kg-m^2/s
e^2 = hbar*c/(4pi)*k
e^2 = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
e^2 = (1.602176634 x 10^-19 A-s)^2
k = hbar*c/(4pi)*e^2
k = 9.670554000 x 10^-36 kg-m^2/s * 2.997924580 x 10^8 m/s / (4pi) * (1.602176634 x 10^-19 A-s)^2
k = 8.987551787 x 10^9 kg-m^3/A^2-s^4-(4pi)
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
CGS
--
alpha = (4pi)*e^2/(2*h*c)
alpha = (4pi) * (4.803204712 x 10^-10 esu)^2 / (2.000000000 x 10^0 rad/sr * 6.626070150 x 10−27 erg-s/rad * 2.997924580 x 10^10 cm/s)
alpha = 7.297352566 x 10^-3 sr
MKS
--
alpha = e^2/(2*e0*h*c)
alpha = (1.602176634 x 10^-19 A-s)^2 / (2.000000000 x 10^0 rad/sr * 8.854187817 x 10^-12 A^2-s^4/kg-m^3 * 6.626070150 x 10^-34 kg-m^2/s-rad * 2.997924580 x 10^8 m/s)
alpha = 7.297352566 x 10^-3 sr
Quantum of Action = 9.670554000 x 10^-36 kg-m^2/s
Quantum of Action = 9.670554000 x 10^-29 erg-s
Quantum of Action = 9.670554000 x 10^-36 J-s
https://www.bipm.org/en/worldwide-metrology/cgpm/ https://www.bipm.org/en/cgpm-2018/
"Fix the planet Earth's physics Errors"
Holy Ghost
--- SoupGate-Win32 v1.05
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https://uk.answers.yahoo.com/question/index?qid=20200924160340AAnb1f2 | 1,604,169,979,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00228.warc.gz | 592,583,576 | 25,481 | Anonymous
Anonymous asked in Science & MathematicsEngineering · 1 month ago
circuit transient?
having no idea how to deal with 2 voltage sources, how should it be analysis?
Relevance
• 6 days ago
Unionwell micro switch from https://unionwellgermany.com/
• 1 month ago
After a long time the cap charges to 15V.
After a long time, with the switch closed the cap is @ 9.375V.
from t=0 closed the cap discharges Vi-Vf*1-e^-t/RC
15-9.375=5.625v
find RC
Circuit Simulator Applet says:
@ t=.5s V=11.441v | 149 | 508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.821741 |
https://www.sqlservercentral.com/Forums/Topic1124040-2900-1.aspx | 1,506,276,113,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690112.3/warc/CC-MAIN-20170924171658-20170924191658-00119.warc.gz | 866,573,242 | 33,254 | Returning 0's in a SUM (SQL Spackle)
Author Message Patrick Cahill SSCarpal Tunnel Group: General Forum Members Points: 4164 Visits: 1016 Comments posted to this topic are about the item Returning 0's in a SUM (SQL Spackle) berzat.museski SSC Rookie Group: General Forum Members Points: 37 Visits: 102 Looks too much complicated for the goal, isn't it ? DavidBridgeTechnology.com SSC Veteran Group: General Forum Members Points: 265 Visits: 238 There may be other ways but apart from using a dedicated number table rather than spt_values, this is exactly the way I do it.I am intrigued by the comment that this looks too complicated. I would like to see a simpler example.In short, create a calendar table that holds the date / time / base value range and then embellish it with the actual data. How can this be achieved any simpler? Please share.A good article in my opinion.Dave David BridgeDavid Bridge Technology Limitedwww.davidbridgetechnology.com harry9katz Valued Member Group: General Forum Members Points: 52 Visits: 234 HiCan use cross-join with hour tableThis is easier solution by more costly on the execution plan Matjaz Justin SSC Veteran Group: General Forum Members Points: 256 Visits: 543 This should be a simpler solution.`select N2.n as dan, N.n - 1 as Ura, isnull(sum(SaleQTY), 0) as Dfrom dbo.GetNums(24) as N cross join dbo.GetNums(31) as N2 left join (select datepart(day, SaleTime) as Dan, datepart(hh, SaleTime) as Ura, sum(SaleQTY) as SaleQTY from dbo.#SalesTestData group by datepart(day, SaleTime), datepart(hour, SaleTime))as D on D.Dan = N2.n and D.Ura = (N.n - 1)group by N2.n, N.norder by 1, 2`Uncle Goole will tell you more about function GetNums ( = Virtual Auxiliary Table of Numbers). Jeff Moden SSC Guru Group: General Forum Members Points: 340519 Visits: 42644 Matjaz Justin (6/13/2011)This should be a simpler solution.`select N2.n as dan, N.n - 1 as Ura, isnull(sum(SaleQTY), 0) as Dfrom dbo.GetNums(24) as N cross join dbo.GetNums(31) as N2 left join (select datepart(day, SaleTime) as Dan, datepart(hh, SaleTime) as Ura, sum(SaleQTY) as SaleQTY from dbo.#SalesTestData group by datepart(day, SaleTime), datepart(hour, SaleTime))as D on D.Dan = N2.n and D.Ura = (N.n - 1)group by N2.n, N.norder by 1, 2`Uncle Goole will tell you more about function GetNums ( = Virtual Auxiliary Table of Numbers).Although I'm sure that Uncle Google will tell us about the GetNums, it would be real handy if you'd simply provide the URL in the future since you're the one that brought up the function. ;-) --Jeff ModenRBAR is pronounced ree-bar and is a Modenism for Row-By-Agonizing-Row.First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column.If you think its expensive to hire a professional to do the job, wait until you hire an amateur. -- Red Adair Helpful Links:How to post code problemsHow to post performance problemsForum FAQs Jeff Moden SSC Guru Group: General Forum Members Points: 340519 Visits: 42644 harry9katz (6/13/2011)HiCan use cross-join with hour tableThis is easier solution by more costly on the execution planCool! Got code? ;-) --Jeff ModenRBAR is pronounced ree-bar and is a Modenism for Row-By-Agonizing-Row.First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column.If you think its expensive to hire a professional to do the job, wait until you hire an amateur. -- Red Adair Helpful Links:How to post code problemsHow to post performance problemsForum FAQs Jeff Moden SSC Guru Group: General Forum Members Points: 340519 Visits: 42644 Folks, the GetNums function that Matjaz reference is Itzek Ben-Gan's cascading Cross Join code wrapped in a function. If you decide to use that instead of some other method, make sure you use the code that has "TOP" in it for the very reasons that Itzek mentions and the fact that I've also confirmed it's a bit faster than his previous rendition. Itsek's article can be found at the following URL:http://www.sqlmag.com/article/sql-server/virtual-auxiliary-table-of-numbers --Jeff ModenRBAR is pronounced ree-bar and is a Modenism for Row-By-Agonizing-Row.First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column.If you think its expensive to hire a professional to do the job, wait until you hire an amateur. -- Red Adair Helpful Links:How to post code problemsHow to post performance problemsForum FAQs tlewis-993411 SSC-Enthusiastic Group: General Forum Members Points: 145 Visits: 80 Although the "GetNums" function would indeed make this simpler in-line, it simply moves the complexity to another routine. Combined, the two parts have nearly the same code. That having been said, standardizing common functions like getnums() is a good practice. The article is probably best viewed as a general approach to reporting on "sparse data". Instead of hours or dates, consider regions, possible questionnaire responses, etc.I've found that using in-memory tables and "building" results incrementally to be much clearer and less error-prone than highly complex joins, and with very acceptable performance (consider the use case...) I've seen many cases where the added developer time in debugging and maintaining far exceeds the additional performance time over the life of the application. berzat.museski SSC Rookie Group: General Forum Members Points: 37 Visits: 102 No need for uncle Google in this case, spt_values is quite enough, unless you are querying some large data warehouse.However, I agree that GetNums is a cool trick to generate large order numbers.I was thinking of the following solution:I first add one computed column in the fact table (I assume that it is ok to do that) in order to keep the query SARGable. The new column will have the sales datetime rounded up to hour percision.ALTER TABLE #SalesTestDataADD SaleTimeHour AS CONVERT(DATETIME, CONVERT(VARCHAR,SaleTime,112)+' '+CAST(DATEPART(HOUR,SaleTime) AS VARCHAR(2))+':00:00.000',112) PERSISTEDThe report will then be formed by this single query:DECLARE @StartDate DATETIME, @EndDate DATETIMESET @StartDate='2011-01-01 00:00:00.000'SET @EndDate='2011-12-01 00:00:00.000'SELECT CAST(AllDates.ReportDayHour AS DATE) AS ReportDay, CAST(AllDates.ReportDayHour AS TIME) AS ReportHour, ISNULL(SUM(s.SaleQty),0) AS TotalSaleQtyFROM ( SELECT DATEADD(hh,h.number,DATEADD(dd,d.number,DATEADD(mm,m.number,@StartDate))) AS ReportDayHour FROM master..spt_values m CROSS JOIN master..spt_values d CROSS JOIN master..spt_values h WHERE m.type='p' AND m.number BETWEEN 0 AND DATEDIFF(mm,@StartDate,@EndDate) AND d.type='p' AND d.number BETWEEN 0 AND 30 AND h.type='p' AND h.number BETWEEN 0 AND 23) AS AllDatesLEFT JOIN #SalesTestData s ON s.SaleTimeHour=AllDates.ReportDayHourGROUP BY AllDates.ReportDayHourORDER BY AllDates.ReportDayHourThe parameters are the start and end moments of the reports, as any manager would want | 1,814 | 7,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-39 | latest | en | 0.862304 |
http://www.gametheorystrategies.com/2011/05/13/introduction-to-game-theory-prisoners-dilemma-part-1/ | 1,579,277,461,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589861.0/warc/CC-MAIN-20200117152059-20200117180059-00536.warc.gz | 218,281,714 | 7,096 | # Introduction to Game Theory – Prisoners’ Dilemma (Part 1)
The prisoners’ dilemma is usually the first thing that people come across when they start to get interested in game theory. This post will give a (maths free) explanation of the idea.
It is called the prisoners’ dilemma because the standard example is one where two suspects have been arrested by the police and are being held in separate cells, there is no way for them to communicate. Both prisoners were involved in the same crime. There are three possible scenarios:
1 – If both prisoners keep quiet then the police will have limited evidence to convict them and they will both go to prison for two years.
2 – If one gives the police more evidence against the other then they will go free and the other will go to prison for ten years.
3 – If they both decide to give extra evidence about the other suspect to the police then they both get convicted for longer, but not as long as if there is evidence against just one of them, say five years each.
One of the basic elements to game theory is to take into account what you think the other person will do when you are working out your own strategy. In this case the first prisoner needs to think about what the second prisoner will do. There are two possibilities
a – if the second prisoner keeps quiet then the first should give evidence to the police as this will mean that they go free while the second prisoner gets convicted for 10 years.
b – if the second prisoner gives evidence then the first prisoner should also give evidence as this will mean a five year sentence rather than the 10 if they keep quiet.
We can see here that in both situations the prisoner should give evidence to the police as the outcome is better for them if they do that. This is the same for both players so the logical outcome of the game is that they both give evidence to the police and end up getting five years in jail. Although this is the outcome from the prisoners acting logically it is not the best outcome they could have achieved which would have been two years each in prison if they had both kept quiet.
The key thing to understand from the prisoners’ dilemma is that even though each person involved in a situation might act logically and in their own best interests the overall outcome might not be the best one for everyone.
There are lots of real situations where this can occur and I will look at a few of these in the second installment of this introduction to the prisoners’ dilemma (click here for part 2 of this post, and here for part 3)
This entry was posted in Game theory, Introduction and tagged , , . Bookmark the permalink. | 543 | 2,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-05 | longest | en | 0.972205 |
https://www.experts-exchange.com/questions/21919994/Convert-Decimal-Number-to-HH-MM-SS.html | 1,529,879,988,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00072.warc.gz | 830,879,444 | 16,870 | # Convert Decimal Number to HH:MM:SS
Hello Experts:
We have a FireBird DB from which we're using CR XI to pull a decimal number that represents hours spent on a project.
I'd like to convert this decimal number to appear in my reports in the format of: "HH:MM:SS" (without any reference to AM/PM, as that is irrelevant)
But I will also need to summarize the column that holds these times into the same format, thus performing math on the converted numbers. (without conversion to days at HH > 24.)
Examples:
Decimal Hours= Desired result
0.03 = 00:01:48
2.25 = 02:15:00
Summarized as:
02:16:48
I'm sure I can work out the math, multiplying the initial decimal by 3600 for total SS, and working my way from there to MM and HH, but am wondering if there is a shorter method within CR XI.
TIA!
LVL 1
###### Who is Participating?
Commented:
Conversion of the detail items is easy: TimeValue ({YourHours}/24).
Assuming you accuimulate a total of {YourHours}, conversion of the total is a bit more tricky because of the HH > 24 situation; the following formula converts the accumulated hours into a string in the format hh:mm:ss:
Totext({YourHours},0,'') + ':' + totext(timevalue({YourHours} / 24),'mm:ss')
0
Software Solutions ConsultantCommented:
CR has a function called TIME(number)
the number can be decimal
Use a forumla to convert the field(type number) to type time
Then use the fomula where ever required.
ex: 2.83 converts to 7:55:12PM
18.87 converts to 8:52:48PM
You might use CTime because you can only use Time in Crystal syntax as it is a type name in Basic syntax.
CTime (number) converts the given number to a Time value; the given number is in units of 24 hours, can be fractional or negative as well.
Hope it helps
MikeV
0
Author Commented:
While neither comment hit exactly what I was looking for KenTurner's gave me a new perspective. Thank you both for responding.
0
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Have a better answer? Share it in a comment. | 527 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-26 | latest | en | 0.88023 |
https://bsci-ch.org/how-many-pounds-is-9-kg/ | 1,652,770,603,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00415.warc.gz | 210,341,295 | 4,838 | ## Conversion formula exactly how to convert 9 kilograms to pounds?
We know (by definition) that:1kg≈2.2046226lb
We can collection up a relationship to fix for the variety of pounds.
You are watching: How many pounds is 9 kg
1kg9kg≈2.2046226lbxlb
Now, us cross multiply to fix for our unknown x:
xlb≈9kg1kg*2.2046226lb→xlb≈19.8416034lb
Conclusion:9kg≈19.8416034lb
## Approximation
An almost right numerical result would be: nine kilograms is around nineteen point eight 3 pounds, or alternatively, a lb is around zero suggest zero 5 times nine kilograms.
## Units involved
This is how the devices in this conversion space defined:
### Kilograms
The kilogram is the basic unit of fixed in the global System of devices (the Metric system) and also is defined as being equal to the massive of the international Prototype of the Kilogram. The avoirdupois (or international) pound, used in both the imperial and also US customary systems, is identified as precisely 0.45359237 kg, making one kilogram about equal come 2.2046 avoirdupois pounds. Other timeless units that weight and mass roughly the human being are also defined in regards to the kilogram, make the IPK the major standard for virtually all devices of mass on Earth. | 342 | 1,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-21 | latest | en | 0.894823 |
https://www.mathworks.com/matlabcentral/cody/problems/167-pizza/solutions/274906 | 1,508,681,900,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825264.94/warc/CC-MAIN-20171022132026-20171022152026-00092.warc.gz | 949,124,299 | 11,412 | Cody
# Problem 167. Pizza!
Solution 274906
Submitted on 7 Jul 2013 by Kumar Sandeep
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
%% z = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))
2 Pass
%% z = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))
3 Pass
%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
4 Pass
%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct)) | 198 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-43 | latest | en | 0.586766 |
https://nl.mathworks.com/matlabcentral/cody/problems/2063-a-matrix-of-extroverts/solutions/1453398 | 1,585,835,671,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00467.warc.gz | 597,854,406 | 16,467 | Cody
# Problem 2063. A matrix of extroverts
Solution 1453398
Submitted on 1 Mar 2018 by Noriko HOUNOKI
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = magic(3); y = extroverts(x); y_c = [4.2500 4.7500 ; 5.2500 5.7500]; assert(max(max(abs(y-y_c)))<1e-9);
y = [] rt = [] temp = 4.2500 rt = 4.2500 temp = 4.7500 rt = 4.2500 4.7500 y = 4.2500 4.7500 temp = 5.2500 rt = 5.2500 temp = 5.7500 rt = 5.2500 5.7500 y = 4.2500 4.7500 5.2500 5.7500
2 Pass
x = [1 2 3 ; 4 5 6]; y = extroverts(x); y_c = [3 4]; assert(max(max(abs(y-y_c)))<1e-9);
y = [] rt = [] temp = 3 rt = 3 temp = 4 rt = 3 4 y = 3 4
3 Pass
x=[magic(4) -magic(4)]; y = extroverts(x); y_c=[8.5 6.5 8.5 0 -8.5 -6.5 -8.5 8 8.5 9 1.5 -8 -8.5 -9 8.5 10.5 8.5 0 -8.5 -10.5 -8.5]; assert(max(max(abs(y-y_c)))<1e-9);
y = [] rt = [] temp = 8.5000 rt = 8.5000 temp = 6.5000 rt = 8.5000 6.5000 temp = 8.5000 rt = 8.5000 6.5000 8.5000 temp = 0 rt = 8.5000 6.5000 8.5000 0 temp = -8.5000 rt = 8.5000 6.5000 8.5000 0 -8.5000 temp = -6.5000 rt = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 temp = -8.5000 rt = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 -8.5000 y = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 -8.5000 temp = 8 rt = 8 temp = 8.5000 rt = 8.0000 8.5000 temp = 9 rt = 8.0000 8.5000 9.0000 temp = 1.5000 rt = 8.0000 8.5000 9.0000 1.5000 temp = -8 rt = 8.0000 8.5000 9.0000 1.5000 -8.0000 temp = -8.5000 rt = 8.0000 8.5000 9.0000 1.5000 -8.0000 -8.5000 temp = -9 rt = 8.0000 8.5000 9.0000 1.5000 -8.0000 -8.5000 -9.0000 y = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 -8.5000 8.0000 8.5000 9.0000 1.5000 -8.0000 -8.5000 -9.0000 temp = 8.5000 rt = 8.5000 temp = 10.5000 rt = 8.5000 10.5000 temp = 8.5000 rt = 8.5000 10.5000 8.5000 temp = 0 rt = 8.5000 10.5000 8.5000 0 temp = -8.5000 rt = 8.5000 10.5000 8.5000 0 -8.5000 temp = -10.5000 rt = 8.5000 10.5000 8.5000 0 -8.5000 -10.5000 temp = -8.5000 rt = 8.5000 10.5000 8.5000 0 -8.5000 -10.5000 -8.5000 y = 8.5000 6.5000 8.5000 0 -8.5000 -6.5000 -8.5000 8.0000 8.5000 9.0000 1.5000 -8.0000 -8.5000 -9.0000 8.5000 10.5000 8.5000 0 -8.5000 -10.5000 -8.5000
4 Pass
x = ones(20); y = extroverts(x); y_c = ones(19); assert(max(max(abs(y-y_c)))<1e-9);
y = [] rt = [] temp = 1 rt = 1 temp = 1 rt = 1 1 temp = 1 rt = 1 1 1 temp = 1 rt = 1 1 1 1 temp = 1 rt = 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 temp = 1 rt = 1 1 temp = 1 rt = 1 1 1 temp = 1 rt = 1 1 1 1 temp = 1 rt = 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 temp = 1 rt = 1 1 temp = 1 rt = 1 1 1 temp = 1 rt = 1 1 1 1 temp = 1 rt = 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 temp = 1 rt = 1 1 temp = 1 rt = 1 1 1 temp = 1 rt = 1 1 1 1 temp = 1 rt = 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 temp = 1 rt = 1 1 temp = 1 rt = 1 1 1 temp = 1 rt = 1 1 1 1 temp = 1 rt = 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 temp = 1 rt = 1 temp = ... | 4,307 | 6,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.162742 |
https://www.slideserve.com/search/biostatistics-office-ppt-presentation | 1,642,619,221,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00082.warc.gz | 994,770,906 | 6,407 | # 'Biostatistics office' presentation slideshows
## Interim Analysis in Clinical Trials: A Bayesian Approach in the Regulatory Setting
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## Biostatistics
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## Biostatistics
Biostatistics. Unit 6 – Confidence Intervals. Statistical inference.
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## Biostatistics
Biostatistics. A foundation for analysis in the health science. Yongli YANG Ph.D, Associate Professor Department of Biostatistics & Epidemiology, college of public health TEL: 67781249 E-mail: ylyang377@126.com. STATISTICS IN LIFE.
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## Biostatistics
Biostatistics. Frank H. Osborne, Ph. D. Professor. Biostatistics. Unit 1 Introduction. Biostatistics. Biostatistics can be defined as the application of the mathematical tools used in statistics to the fields of biological sciences and medicine.
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## BIOSTATISTICS
Qualitative variable (Categorical). Quantitative variable (Continuous). BIOSTATISTICS. DESCRIPTIVE. ANALYTIC. INFERENTIAL. Proportion, Ratio. Qualitative v. Prevalence, incidence. Quantitative v. BIOSTATISTICS. DESCRIPTIVE. ANALYTIC. INFERENTIAL. Min. Max. Range. Quantitative v.
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## Biostatistics
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## Biostatistics
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## Biostatistics
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## Biostatistics
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## Biostatistics
Biostatistics. Carsten Dahl Mørch Fredrik Bajersvej 7 A2-212 Tel: 9635 8757 Mail: cdahl@hst.aau.dk Web: http://www.hst.aau.dk/~cdahl/biostat_9BME/. Biostatistics. Biostatistics is statistics applied to biology Design of experiments The limitations when working with human subjects
By fayer (0 views) | 800 | 3,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-05 | latest | en | 0.734071 |
https://how-what-instructions.com/13220473-how-to-choose-a-violin-size-for-a-child-13-steps | 1,679,670,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00526.warc.gz | 373,079,693 | 11,015 | # How to choose a violin size for a child: 13 steps
When learning to play the violin, choosing the correct size is crucial. In total, there are 7 different sizes of violins, ranging from 1/16, 1/10, 1/8, 1/4, 1/2, 3/4, and 4/4. You can measure a violin for a child by measuring the length of his arm and selecting the corresponding violin size. Have the child hold the violin and hold the headstock to check the size. If you can comfortably wrap your fingers around the headstock, the violin is the right size.
## Steps
### Part 1 of 3: Measure a Child's Arm
#### Step 1. Have the child extend his arm to the side, parallel to the floor
It is best to measure the size of the violin when the player's arm is straight. To do this, ask the child to extend his dominant arm outward.
### Make sure they hold their arm as straight as possible. If your arm is bent at the elbow, the measurement may not be accurate
#### Step 2. Begin by measuring from the center of the child's hand
Ask the child to spread his fingers and flatten his hand. Next, hold the edge of a tape measure in the center of your hand.
• The middle part of the hand is approximately where the base of the thumb meets the hand.
• If you don't have a tape measure, you can use a ruler instead.
#### Step 3. Stretch the tape measure up to the child's neck to find out how long his arm is
Hold the tape measure in the child's hand, and stretch it up to his neck. Take the measurement where the neck meets the shoulder.
• This area covers the region with which the violin is held, providing a correct measurement.
• Be careful with the metric tape, do not accidentally hurt the child. The edges can be sharp!
### Part 2 of 3: Choosing the Right Violin
#### Step 1. Try a 1/10 or 1/16 violin if your arm is less than 16 inches (41 cm)
Use a 1/10 size violin if the child's arm is between 15 and 16 inches (38 to 41 cm). Go for a 1/16 violin if the child's arm is only 14 to 15 inches (36 to 38 cm). If your child is 4 years old or younger, these are good size options.
### While these sizes are not very common, they are available for young aspiring violinists
#### Step 2. Go for a 1/8 violin if the child's arm length is 16 inches
Young children most often begin to learn the violin at this size. This is a great option for children ages 4-6 or in kindergarten.
#### Step 3. Try a ¼-size violin if the child's arm length is 16.5 to 18 inches (42 to 46 cm)
On average, children between the ages of 5 and 7 use this size violin. This is the typical size for first grade violinists.
### The body of a ¼ violin is between 42 and 46 cm (16.5 and 18 inches). This measurement does not include the fingerboard or headstock
#### Step 4. Go for a half violin if its arm is 18.5 to 20 inches (47 to 51 cm)
The ½ violin is the middle point on the spectrum of violin sizes. This size is recommended for most children ages 7-9 and second and third grade.
### The average body measurement of a medium violin is between 19 to 20 inches (48 to 51 cm)
#### Step 5. Select a size ¾ violin for an arm length between 52 and 56 cm (20.5 and 22 inches)
If your child's arm length is too long for a medium-sized violin, try a ¾-size violin. On average, this size works well for children between the ages of 9 and 12. It is also common among fifth and sixth grade orchestra students.
### Some adults prefer to use a ¾-size violin
#### Step 6. Choose a full-size violin if its arm length is 23 inches (58 cm) or more
If your measurement is over 23 inches (58 cm), your child should use a full-size violin. Full-size violins typically fit children 12 and older. They are generally used from the seventh grade onwards.
### Part 3 of 3: Testing the Violin
#### Step 1. Have the child hold the violin in position to play
Ask the student to reach out and hold the end of the violin. Then ask him to hold the violin between his chin and shoulder.
### This way, you can choose a violin size based on how it fits in your hand
#### Step 2. Ask the child to close his hand around the head of the violin
The child must place his fingers around the head, which is decorative, carved on the end of the violin. If the child can comfortably play the pegbox (where the tuning pegs are), the violin is a good size.
### The violin headstock normally has a spiral wound shape
#### Step 3. Choose a smaller size if your child has trouble reaching the head
If the child cannot comfortably cover the headstock with his fingers, the violin is the wrong size. Your fingers should easily reach the tuner without any effort or discomfort.
### If the child plays a violin that is too large, it may be difficult for him to balance the weight. This could lead to incorrect playing posture and technique
#### Step 4. Try a larger violin size if the child's fingers go over the headstock
If your child's fingers extend to the pegbox, try holding a larger violin. Your fingers should reach the end of the head. | 1,228 | 4,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-14 | latest | en | 0.938281 |
http://perminc.com/resources/fundamentals-of-fluid-flow-in-porous-media/chapter-5-miscible-displacement/fluid-phase-behavior/pressure-temperature-diagram-p-t-diagram/ | 1,568,875,977,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573444.87/warc/CC-MAIN-20190919060532-20190919082532-00120.warc.gz | 146,275,413 | 82,100 | # Pressure-Temperature Diagram (P-T Diagram)
Pressure-Temperature Diagram (P-T Diagram) 2016-10-25T11:54:38+00:00
# Fluid Phase Behavior: Pressure-Temperature Diagram (P-T Diagram)
Figure 5‑2 shows a P-T diagram for a pure component. The line connecting the triple point and critical points is the vapor pressure curve; the extension below the triple point is sublimation point. As this figure shows in pure materials, by decreasing the pressure at a fixed temperature, phase change happens just at a point (vapor pressure curve is a line). According to Figure 5‑2, the phase boundary between liquid and gas does not continue indefinitely. Instead, it terminates at a point on the phase diagram called the critical point. This reflects the fact that, at extremely high temperatures and pressures, the liquid and gaseous phases become indistinguishable.
Figure 5-2: A Typical Phase Diagram for a Pure Component
The phase behavior of a multi-component system is more elaborate than that of a pure compound. the complexity generally compounds as components with widely different structure and molecular sizes comprise the system. Reservoir fluids are mainly composed of hydrocarbons with similar structure so their phase behavior is not generally highly complex. Figure 5‑3 shows an idealized P-T diagram for a multi component with a fixed overall composition. As it shows, there is a transition zone between the complete liquid phase and complete vapor phase. In other words in contrast to the pure substance, phase change from liquid to vapor happens, by decreasing the pressure at a fixed temperature, on a line. So there is a region that two phases are at equilibrium. Two phases region that is bounded by the bubble point and dew point curves is called “phase envelope”. The bubble point and dew point curves meet at the critical point. Two phases can exist at a pressure greater than critical pressure and at a temperature greater than critical temperature, unlike a pure component system. Cricondonbar for a multicomponent system is defined as the maximum pressure that two phases (vapor-liquid) can exist in equilibrium and cricondontherm is the maximum temperature that two phases can exist in equilibrium.
Figure 5‑3 shows the general behavior for a fixed composition of a system consisting of two or more components. If the composition were changed then the position of the phase envelope on the P-T plot would shift.
Figure 5-3: Typical P-T Diagram for a Multi-component System
Figure 5‑4.a shows the phase diagram of Ethane-normal Heptane system. As it shows, the critical temperature of different mixtures lies between the critical temperatures of the two pure components. The critical pressure, however, exceeds the values of both components as pure, in most cases. The locus of critical or “plait” points is shown by the dashed line (Figure 5‑4.a). The difference between the critical pressure of two components system and each pure component critical pressure increases by increasing the difference between the critical points of the two pure components (Figure 5‑4.b). No binary mixture can exist as a two-phase system outside the region bounded by the locus of critical points.
Figure 5-4: (a) Phase Diagram of Ethane-normal Heptane
[1], (b) Critical Loci for Binary Mixture
Extracted data from the graphs such as Figure 5‑4 indicate conditions at which mixture of binary pairs are miscible. To have a miscibility of two compounds of ethane and normal pentane with any composition, at each temperature the pressure should be higher than the pressure indicated by the locus of critical pressure line for that specific temperature. At lower pressure there are possible concentrations at which the system will separate into two phases. So a minimum miscible pressure for each temperature could be determined according to graph’s data such as Figure 5‑4. For systems containing a large number of components no maps of critical points are available.
Example 5‑1[50]
A FCM miscible displacement is to be designed in which a slug of butane is the primary displacing solvent. The butane slug is to be displaced by dry gas consisting essentially of methane. Assume that the crude oil could be represented by n-decane. The reservoir conditions are 160°F and 2500 psi. Use Figure 5‑4.b to determine whether miscible conditions would exist at the front and back of the solvent (butane) slug. Is the miscible condition would be exist at p=1500 psi and T=160°F?
Solution
The primary slug is butane displacing oil. Refer to Figure 5‑4.a. the locus of critical points for C4-C10 is not given, however at 160°F and 2500 psi C4 and C10 are in liquid state. C4 and C10 will be miscible. The secondary slug is methane displacing butane. In Figure 5‑4.a at 160°F, a pressure of 2600 psi is well above the locus of critical points for C1-C4 mixtures. Therefore C1 and C4 will be miscible. So at 160°F and P=2500 psi proposed displacement will be miscible at both the leading and trailing edges.
If the reservoir pressure was 1500 psi then the displacement of butane by methane would be at conditions below the critical locus. At this condition, at some points in the process two phases would form and the dry gas would displace butane immiscibility at reduced microscopic displacement efficiency. So some amount of the butane may be bypassed during the displacement by dry gas and trapped in the reservoir. This would result in a relatively rapid degradation of the integrity of the primary butane slug and drastically influence oil production.
## References
[1] “PVT and Phase Behavior of Petroleum Reservoir Fluids”, A. Danesh, 2003
## Questions?
If you have any questions at all, please feel free to ask PERM! We are here to help the community. | 1,244 | 5,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-39 | latest | en | 0.895094 |
https://aras-p.info/blog/2023/02/01/Float-Compression-3-Filters/ | 1,719,040,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00083.warc.gz | 82,621,700 | 7,122 | # Float Compression 3: Filters
Introduction and index of this series is here.
In the previous parts we saw that using generic data compression libraries, we can get our 94.5MB data down to 33.8MB (zstd level 7) or 29.6MB (oodle kraken level 2) size, if we’re not willing to spend more than one second compressing it.
That’s not bad, but is there something else we can do? Turns out, there is, and in fact it’s quite simple. Enter data filtering.
#### Prediction / filtering
We saw filtering in the past (EXR and SPIR-V), and the idea is simple: losslessly transform the data so that it is more compressible. Filtering alone does nothing to reduce the data size, but (hopefully!) it decreases data randomness. So the process is: filter the data, then compress that. Decompression is reverse: decompress, then un-filter it.
Here’s some simple filters that I’ve tried (there are many, many other filters possible, I did not try them all!).
#### Reorder floats array-of-structures style
Recall that in our data, we know that water simulation has four floats per “element” (height, velocity x, velocity y, pollution); snow simulation similarly has four floats per element; and other data is either four or three floats per element. Instead of having the data like that (“array of structures” style), we can try to reorder it into “structure of arrays” style. For water simulation, that would be all heights first, then all x velocities, then all y velocities, etc.
So this:
becomes this:
Completely unoptimized code to do that could look like this (and our data is floats, i.e. 4 bytes, so you’d call these templates with a 4-byte type e.g. `uint32_t`):
``````// channels: how many items per data element
// dataElems: how many data elements
template<typename T>
static void Split(const T* src, T* dst, size_t channels, size_t dataElems)
{
for (size_t ich = 0; ich < channels; ++ich)
{
const T* ptr = src + ich;
for (size_t ip = 0; ip < dataElems; ++ip)
{
*dst = *ptr;
ptr += channels;
dst += 1;
}
}
}
template<typename T>
static void UnSplit(const T* src, T* dst, size_t channels, size_t dataElems)
{
for (size_t ich = 0; ich < channels; ++ich)
{
T* ptr = dst + ich;
for (size_t ip = 0; ip < dataElems; ++ip)
{
*ptr = *src;
src += 1;
ptr += channels;
}
}
}
``````
Does that help? The results are interesting (click for an interactive chart):
• It does help LZ4 to achieve a bit higher compression ratios.
• Makes zstd compress faster, and helps the ratio at lower levels, but hurts the ratio at higher levels.
• Hurts oodle kraken compression.
• Hurts the decompression performance quite a bit (for lz4 and kraken, slashes it in half). In all cases the data still decompresses under 0.1 seconds, so acceptable for my case, but the extra pass over memory is not free.
Ok, so this one’s a bit “meh”, but hey, now that the data is grouped together (all heights, then all velocities, …), we could try to exploit the fact that maybe neighboring elements are similar to each other?
#### Reorder floats + XOR
In the simulation data example, it’s probably expected that usually the height, or velocity, or snow coverage, does not vary “randomly” over the terrain surface. Or in an image, you might have a color gradient that varies smoothly.
“But can’t data compressors already compress that really well?!”
Yes and no. Usually generic data compressors can’t. Most of them are very much oriented at finding repeated sequences of bytes. So if you have a very smoothly varying surface height or image pixel color, e.g. a sequence of bytes `10, 11, 12, 13, 14, 15`, well that is not compressible at all! There are no repeating byte sequences.
But, if you transform the sequence using some sort of “difference” between neighboring elements, then repeated byte sequences might start appearing. At first I tried XOR’ing the neighboring elements together (interpreting each float as an `uint32_t`), since at some point I saw that trick being mentioned in some “time series database” writeups (e.g. Gorilla).
A completely unoptimized code to do that:
``````template<typename T>
static void EncodeDeltaXor(T* data, size_t dataElems)
{
T prev = 0;
for (size_t i = 0; i < dataElems; ++i)
{
T v = *data;
*data = v ^ prev;
prev = v;
++data;
}
}
template<typename T>
static void DecodeDeltaXor(T* data, size_t dataElems)
{
T prev = 0;
for (size_t i = 0; i < dataElems; ++i)
{
T v = *data;
v = prev ^ v;
*data = v;
prev = v;
++data;
}
}
``````
And that gives (faint dashed line: raw compression, thin line: previous attempt (split floats), thick line: split floats + XOR):
• Compression ratio is way better for zstd and lz4 (for kraken, only at lower levels).
• zstd pretty much reaches kraken compression levels! The lines almost overlap in the graph.
• Decompression speed takes a bit of a hit, as expected. I might need to do something about it later.
So far we got from 33.8MB (zstd) / 29.6MB (kraken) at beginning of the post down to 28MB (zstd, kraken), while still compressing in under 1 second. Nice, we’re getting somewhere.
#### Reorder floats + Delta
The “xor neighboring floats” trick from Gorilla database was in the context of then extracting the non-zero sequences of bits from the result and storing that in less space than four bytes. I’m not doing any of that, so how about this: instead of XOR, do a difference (“delta”) between the neighboring elements? Note that delta is done by reinterpreting data as if it were unsigned integers, i.e. these templates are called with `uint32_t` type (you can’t easily do completely lossless floating point delta math).
``````template<typename T>
static void EncodeDeltaDif(T* data, size_t dataElems)
{
T prev = 0;
for (size_t i = 0; i < dataElems; ++i)
{
T v = *data;
*data = v - prev;
prev = v;
++data;
}
}
template<typename T>
static void DecodeDeltaDif(T* data, size_t dataElems)
{
T prev = 0;
for (size_t i = 0; i < dataElems; ++i)
{
T v = *data;
v = prev + v;
*data = v;
prev = v;
++data;
}
}
``````
And that gives (faint dashed line: raw compression, thin line: previous attempt (split floats + XOR), thick line: split floats + delta):
Now that’s quite an improvement! All three compressors tested get their compression ratio lifted up. Good! Let’s keep on going.
#### Reorder bytes
Hey, how about instead of splitting each data point into 4-byte-wide “streams”, we split into 1-byte-wide ones? After all, general compression libraries are oriented at finding byte sequences that would be repeating. This is also known as a “shuffle” filter elsewhere (e.g. HDF5). Exactly the same `Split` and `UnSplit` functions as above, just with `uint8_t` type.
Faint dashed line: raw compression, thin line: previous attempt (split floats + Delta), thick line: split bytes:
• kraken results are almost the same as with “split floats and delta”. Curious!
• zstd ratio (and compression speed) is improved a bit.
• lz4 ratio is improved a lot (it’s beating original unfiltered kraken at this point!).
I’ll declare this a small win, and let’s continue.
#### Reorder bytes + Delta
Split by bytes as previous, and delta-encode that. Faint dashed line: raw compression, thin line: previous attempt (split bytes), thick line: split bytes + delta:
Holy macaroni grated potato dumplings!
• Another compression ratio increase. Both zstd and kraken get our data to 23MB in about one second (whereas it was 33.8MB and 29.6MB at the start of the post).
• zstd actually slightly surpasses kraken at compression ratios in the area (“under 1 sec”) that I care about. 😮
• lz4 is not too shabby either, being well ahead of unfiltered kraken.
• Downside: decompression is slightly longer than 0.1 seconds now. Not “terrible”, but I’d want to look into whether all this reordering and delta could be sped up.
### Conclusion and what’s next
There’s lots of other data filtering approaches and ideas I could have tried, but for now I’m gonna call “reorder bytes and delta” a pretty good win; it’s extremely simple to implement and gives a massive compression ratio improvement on my data set.
I did actually try a couple other filtering approaches. Split data by bits (using bitshuffle library) was producing worse ratios than splitting by bytes. Rotating each float left by one bit, to make the mantissa & exponent aligned on byte boundaties, was also not an impressive result. Oh well!
Maybe at some point I should also test filters specifically designed for 2D data (like the water and snow simulation data files in my test), e.g. something like PNG Paeth filter or JPEG-LS LOCO-I (aka “ClampedGrad”).
Next up, I’ll look at an open source library that does not advertise itself as a general data compressor, but I’m gonna try it anyway :) Until then! | 2,201 | 8,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.815874 |
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New Contributor
## Dynamically position the reference line on a scatter plot
Dear ,
I'm putting together a dashboard with a scatter plot that will represent a graph quadrants . To mount the quadrant graph have to put two reference lines dividing the graph in half on the x axis y . In the creation of the reference line box has the option to set the percentage of the same position but does not work correctly because I put in 50 % and the line appears near the zero point and not in the middle as expected . Can also be defined by expression but can not calculate what is the value that must be present at the time of presentation for not knowing how to get what maximum values of the axes ( x and y) being presented at the time on the chart.
My question is: how to get these values dynamically ?
Tags (1)
1 Solution
Accepted Solutions
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Not applicable
## Re: Dynamically position the reference line on a scatter plot
Try this
Max(Aggr(Sum(Valor_X),Chart_Dimension))/2
Regards,
KKR
3 Replies
Contributor III
## Re: Dynamically position the reference line on a scatter plot
=Max(the_expression_in the chart)/2 u used in the chart in the reference line
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New Contributor
## Re: Dynamically position the reference line on a scatter plot
Dear Yusuf
The expression used was Sum ( Valor_X ) . When I use MAX ( Sum ( Valor_X ) ) / 2 gives syntax error. Not to give error have to use MAX ( TOTAL Sum ( Valor_X ) ) / 2. But this does not work because it does not position the bar on site. It does not appear.
Is getting the most value plotted in the graph to the axis of x and y axis ? some function that returns these values.
Emmanoel Gomes da Silva
Highlighted
Not applicable
## Re: Dynamically position the reference line on a scatter plot
Try this
Max(Aggr(Sum(Valor_X),Chart_Dimension))/2
Regards,
KKR | 483 | 2,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-10 | latest | en | 0.851114 |
https://www.coursehero.com/file/4495166/Bakery/ | 1,516,752,859,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892802.73/warc/CC-MAIN-20180123231023-20180124011023-00066.warc.gz | 869,059,561 | 77,358 | # Bakery - Pierre's Bakery bakes and sells french bread Each...
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Demand distribution by demand categories Distribution of demand type Demand probability distribution Type of dem probability Demand High Average Low Average 0.45 36 0.05 0.10 0.15 High 0.30 48 0.10 0.20 0.25 Low 0.25 60 0.25 0.30 0.35 72 0.30 0.25 0.15 84 0.20 0.10 0.05 96 0.10 0.05 0.05 Pierre's Bakery bakes and sells french bread. Each morning, the bakery satisfies the demand for the day using freshly baked bread.Pierre's can bake the bread only in batches of a dozen loaves each. From past observations, the firm obtained the probability distribution with respect to loaves and three demand levels; high, average and low. The probabilities are given below. Each loaf costs \$8 to make and it sells \$12 for each. It wants to determine the daily production size.
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Demand distribution by demand categories Cummulative distribution Demand probability distribution Demand probability dis Demand High Average Low Demand High Average 36 0.05 0.10 0.15 36 0.05 0.10 48 0.10 0.20 0.25 48 0.15 0.30 60 0.25 0.30 0.35 60 0.40 0.60 72 0.30 0.25 0.15 72 0.70 0.85 84 0.20 0.10 0.05 84 0.90 0.95 96 0.10 0.05 0.05 96 1.00 1.00 Distribution of demand type Type of demand probability cumm. dist. Intervals Price
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• Spring '09
• EyupCetin
• Harshad number, Highly composite number, \$8, Highly totient number, 60 0.25 0.30 0.35 72 0.30 0.25 0.15 84, High Average
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Ask a homework question - tutors are online | 592 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.683489 |
http://math.stackexchange.com/questions/52384/while-calculating-why-do-i-get-1-pac-neq-pa/52388 | 1,448,455,304,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445142.9/warc/CC-MAIN-20151124205405-00180-ip-10-71-132-137.ec2.internal.warc.gz | 146,136,059 | 20,601 | While calculating: Why do I get $1-P(A^c)\neq P(A)$?
I have one single dice. And I throw this dice 4 times. What is the probability to get at least once a "6"? Let $A$ denote the desired event.
Sure this is easy, but I got my problems trying to understand why this
$$P(A)=1-P(A^c)=1-\left(\frac{5}{6}\right)^4$$
is correct. What I do understand is that $\frac{5}{6}$ is the probability that the dice does not show a "6" after being thrown. But why the power of 4? Because the events "to throw a six (or not)" are independent?
Assuming my guess is correct. I tryed to calculate the probability without considering the complement $A^c$:
$$P(A)=\left(\frac{1}{6}\right)^4$$
Clearly this is not the same result. I would be glad if someone could enlighten me. I clearly do miss something. :-)
-
You are correct, the power of 4 is because each dice roll is independent.
However, the number $\left(\frac{1}{6}\right)^4$ represents the probability of getting four 6's (or, in fact, any specific outcome - in other words, $\left(\frac{1}{6}\right)^4$ is also the probability of the outcomes being 1, then 2, then 5, then 3). You want to calculate the probability of getting at least one 6, which is much higher. Naively calculating, we might think $$P(\text{at least one }6)=P(6\text{ on 1st roll})+P(6\text{ on 2nd roll})+P(6\text{ on 3rd roll})+P(6\text{ on 4th roll})$$ $$P(\text{at least one }6)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{4}{6}$$ But! We have double-counted some occurrences. For example, (6,6,2,1) is one of the possible combinations with at least one 6, but we've counted it in both $P(6\text{ on 1st roll})$ and in $P(6\text{ on 2nd roll})$. So we need to subtract off the amount we double-counted. $$P(\text{at least one }6)=\frac{4}{6}-\left(P(\text{6 on 1st and 2nd rolls})+\cdots+P(6\text{ on 3rd and 4th rolls})\vphantom{g^{G}}\right)$$ $$P(\text{at least one }6)=\frac{4}{6}-\left(\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^2\right)=\frac{1}{2}$$ But! We now have double-subtracted some occurrences. For example, (6,6,6,2) is one of the possible combinations with at least one 6. We counted it 3 times initially (in $P(6\text{ on 1st roll})$, $P(6\text{ on 2nd roll})$, and $P(6\text{ on 3rd roll})$), but now have subtracted it three times (in $P(6\text{ on 1st and 2nd roll})$, $P(6\text{ on 1st and 3rd roll})$, and $P(6\text{ on 2nd and 3rd roll})$). We need to add the occurrences with three 6's back in now. $$P(\text{at least one }6)=\frac{1}{2}+\left(P(\text{6 on 1st, 2nd, 3rd rolls})+\cdots+P(6\text{ on 2nd, 3rd, 4th rolls})\vphantom{g^{G}}\right)$$ $$\frac{1}{2}+\left(\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3+\left(\frac{1}{6}\right)^3\right)=\frac{14}{27}$$ Finally, due to the same problem, we must subtract off the probability of the sole combination with four sixes, (6,6,6,6). $$P(\text{at least one }6)=\frac{14}{27}-P(6\text{ on 1st, 2nd, 3rd, 4th roll})=\frac{14}{27}-\frac{1}{1296}=\frac{671}{2196}$$ which is the correct answer. But consider what we've just calculated: $$4\left(\frac{1}{6}\right)-6\left(\frac{1}{6}\right)^2+4\left(\frac{1}{6}\right)^3-\left(\frac{1}{6}\right)^4=1-\left(1-\frac{1}{6}\right)^4=1-\left(\frac{5}{6}\right)^4$$ by the binomial theorem. So computing this probability via the complement is really a very clever, time-saving trick!
In general, the reasoning I gave above is referred to as the "inclusion-exclusion principle", and it is very useful in keeping track of combinatorics or probability arguments. For us, the sets $A$, $B$, $C$, and $D$ were the sets of possible dice rolls where a 6 was rolled in the first, second, third, or fourth roll, respectively. Then $|A\cup B\cup C\cup D|$ is the number of rolls with at least one 6, and we calculated it using inclusion-exclusion.
-
$1-\left(\frac 5 6\right)^4$ is the probability that you throw at least one six in four throws (i.e. one minus the probability that you don't throw a six in any of them). On the other hand, $\left(\frac 1 6\right)^4$ is the probability that you throw a six every time in four throws. Clearly the latter probability must be considerably smaller than the former.
And yes, the $\,^4$ is basically there because the throws are independent and identically distributed — if the probability of getting a six was different for different throws, you'd have to multiply all the different probabilities together; and if they weren't even independent, you'd end up with a complicated expression involving conditional probabilities.
Specifically, $1-\left(\frac 5 6\right)^4$ is just shorthand for $1 - \left( \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \right)$ $= 1 - P(A_1^c) \cdot P(A_2^c) \cdot P(A_3^c) \cdot P(A_4^c)$, where the event $A_n$ denotes getting a six on the $n$-th throw.
-
So if I don't want to consider $A^c$: $(1/6)^4$ denotes only the probability of throwing the sequence $(6,6,6,6)$. That means I have to consider all the probabilitys of the other sequences like $(6,6,x,x)$ or $(6,x,x,x)$ or $(x,6,x,6)$ (and all the other combinations) and sum them up, right? – Aufwind Jul 19 '11 at 12:14
Yes, that will also work; you just have to be careful to count all the possible combinations once and none of them twice. There are two basic approaches you can try: either a) enumerate all the possible basic outcomes like (6,6,6,6), (6,6,6,1), (6,6,1,6), etc. and sum their probabilities (you can save some work by only considering the throws "six" and "not six"), or b) use the inclusion–exclusion principle like Zev Chonoles does in his answer. Both, however, are a lot more work than just using $P(A)=1-P(A^c)$. – Ilmari Karonen Jul 19 '11 at 12:23
The following is another approach to the "at least one $6$" problem, grossly less efficient than the approach through first finding the probability of the complement. But it introduces ideas that will be very important later, so it is worth thinking about.
We will be happy if we get at least one $6$. What is the probability that we will be happy? Well, we will be happy if we get exactly $1$ $6$. We will also be happy if we get exactly $2$ $6$'s. Also if we get exactly $3$ $6$'s. And also with exactly $4$ $6$'s.
These events are (pairwise) disjoint. For example, it is impossible to get exactly $2$ $6$'s and (simultaneously) also exactly $4$ $6$'s.
Now I will introduce a bit of notation that looks initially mysterious, and that is not really needed here, but will become important to you later. Let $X$ be the number of $6$'s that we get. Then, for example, $P(X=2)$ means the probability that we get exactly $2$ $6$'s.
We will be happy if $X=1$, also if $X=2$, and so on up to if $X=4$. Since these events are pairwise disjoint, the probability we will be happy is $$P(X=1)+P(X=2)+P(X=3)+P(X=4).$$
So "all" we need to do is to calculate the probabilities of these four events, and add up!
Let us, for instance, find $P(X=2)$. Think about about our $4$ tosses. Write Y (for yes) if on a toss we get a $6$, and N is we don't. So for example NYYN means we got a non-$6$, then a $6$, then a $6$, then a non-$6$. Note that on this sequence of tosses, the total number of $6$ is $2$.
The probability of NYYN is is $(5/6)(1/6)(1/6)(5/6)$. There are several other patterns in which we end up with exactly $2$ $6$'s. For example, there is YYNN, YNYN, some others. The probability of YYNN is $(1/6)(1/6)(5/6)(5/6)$, which is exactly the same as the probability of NYYN. In fact the probability of any single pattern with exactly $2$ $6$'s is $(1/6)^2(5/6)^2$.
So to find $P(X=2)$, we find the number of patterns with exactly $2$ $6$'s, and multiply that by $(1/6)^2(5/6)^2$. The number of patterns is the number of ways of choosing where the two Y will go.
There are $\dbinom{4}{2}$ such ways (I assume you have already met this). So $$P(X=2)=\binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2.$$
We can do a similar analysis for exactly $1$ $6$, exactly $3$, exactly $4$. The probability we will be happy is therefore $$\binom{4}{1}\left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^3+ \binom{4}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2+ \binom{4}{3}\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1+ \binom{4}{4}\left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^0.$$
Now calculate! Really, I almost mean it. Getting the right answer, which you already know from the "quick" method, is a useful test of accuracy and attention to detail.
Of course the above was an inefficient way of calculating the probability. But the same set of ideas shows, for example, that if we toss the die $20$ times, the probability of exactly $4$ $6$'s is $\dbinom{20}{4}(1/6)^4(5/6)^{16}$. The general idea has many applications, and not only to gambling!
- | 2,878 | 8,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2015-48 | longest | en | 0.896011 |
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MatPlus.Net Forum General Double Check Selfmates
### Double Check Selfmates
I had a funky idea today: the most double check selmates in a position. Basically, it’s a selfmate where one side forces the other side to checkmate them with a double check.
I managed to find 9. Three black knights can move to d4 to deliver check. White has only three legal responses to each black knight check-to take with one if their white knights and deliver a doublecheckmate at the same time. 3*3=9.
(= 10+7 )
Is 10 or more possible? Additionally, are there any known problem that features this theme or something similar?
I’m also curious about a problem with a string of double checks as the solution and/or strongest moves.
I did find this nice #s4 on yacpdb.
Seyferth, Paul Karl Edmund
Die Schwalbe (2951) 1934-03
(= 10+5 )
I just threw out my 9 with a damned 28!
(= 13+9 )
Each black queen and rook, all 7 of them, take the white pawn on d4. For each capture, white has only 4 legal moves, and each one results in a doublecheckmate to the black king. 7*4=28. Huzzah!
I severely doubt that this can be improved at all now. Sorry for ruining the fun!
I’ll make a proof game of legality and adjust my position if it is not legal to make it legal.
UPDATE: 1. h4 g5 2. g4 f5 3. f4 e5 4. b4 exf4 5. c4 fxg4 6. d4 gxh4 7. e4 h5 8. a4 b5 9. axb5 c5 10. bxc5 d5 11. exd5 a6
And indeed the position is legal indeed. All needed promoting pawns are now on their way to promotion, the d4 and a6 pawns are in place, and white will be able to attain two light squared bishops with the doubled c-pawns.
I assume that you don't mean BOTH sides keep doublechecking,
although this can fairily :-) easy done with lion, nighthopper
and nightrider.* (Can one do it only with grasshoppers?
You can parry a queen/grasshopper double check with a
non-king move after all!)
In the latter case, OTB game examples exist and I'm willing to bet
you find stuff on Tim Krabbes Chess Curiosities page.
(= 7+8 )
* La7,c7,NHe2,d2,Na8,b8. Draw :-)
Hauke
You’re right that I’ve seen a similar thing on Krabbe’s site before!
From Tim Krabbe #120-Here is a setup with mutual perpetual check that uses Nightriders.
(= 3+3 )
Also, Arisktotle from chess.com has achieved a new record of 32 in the domain of most selfmates via doublecheck, An extra battery, and some knights were added to my matrix for 28. It is a legal position.
https://www.chess.com/forum/view/endgames/triple-checkmate?page=3
BTM, s#1, 32 Variants
(= 16+11 )
(6) Posted by seetharaman kalyan [Friday, Sep 27, 2019 22:47]
when white double checks same it not relevant which piece is captured. so no 32
Yes it does matter, seetharam.The count is for many possible variations there are, which there are 32 of. Each capture is a variation.
After some more back and forth work on the forum, Arisktotle and I have produced this jewel.
BTM, s#1 Via Double Check, 88 Variations
(= 11+14 ) | 888 | 3,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.883251 |
http://www.ccs.neu.edu/home/futrelle/teaching/com1201sp2000/quiz1-info.html | 1,508,805,303,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00321.warc.gz | 410,842,751 | 1,976 | ## Notes for Quiz #1 -- To be given Monday, April 24th
#### (Version of 4/22/00)
The quiz will last the entire class period. Closed book, no calculators allowed. I plan to return the return the graded quiz to you on Thursday the 27th and go over the answers at that time.
Here is a copy of last quarter's quiz. The first four questions are similar to what will appear on the Spring quiz. Ignore the fifth question.
I spent all of Thursday's class, the 20th, reviewing for the quiz. Feel free to email me with questions before the quiz, futrelle@ccs.neu.edu.
Here is a brief description of the some of the topics that may be included. Not all will, of course:
• Quick-Find and Quick-Union algorithms
• The Quick-Find algorithm -- code analysis for data pairs input I will give you and/or drawing diagrams of the algorithm in operation. Any code you might need will be given to you.
• The Quick-Union algorithm -- as above.
• The following two examples were not discussed in the review, but you should be ready to answer such questions.
• Sequential search. How the code works. Complexity of the search -- best, average, and worst-case.
• Binary search -- as above.
• Iteration of a recurrence relation to derive the complexity of an algorithm.
• Graphs
• Understand the relation between adjacency-list representations of graphs and adjacency matrix representations. Similar to question 3 on last quarter's exam.
• Graph traversal, if included, would only be as an extra-credit question.
• Sorting
• You should understand the difference between a stable and unstable sort and be able to produce figures similar to Fig. 6.1 for data I give you.
• For insertion sort, I would give you the code of Prog. 6.1 and/or Prog. 6.3. You need to be able to show a small number of steps in an insertion sort and how they relate to execution of the statements in the code.
Go to Syllabus and Calendar page for COM1201 | 440 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-43 | latest | en | 0.916331 |
https://www.physicsforums.com/threads/rl-circuit-two-loops-problem.10800/ | 1,553,124,651,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202474.26/warc/CC-MAIN-20190320230554-20190321012554-00427.warc.gz | 867,145,116 | 13,615 | # RL Circuit two loops problem
1. Dec 12, 2003
### discoverer02
I have a homework problem with an RL circuit. There are two loops in the circuit. One has the emf and two resistors and a switch and the other has two resistors and an inductor. The resistors, the inductor and the emf are all given values and I'm given that the switch is closed at time = 0 and to find the equation for the current through the inductor and the switch for the time after t = 0.
I tried using Kirchhoff's loop rule to see what I could come up with but I after I related Itotal, I1, I2 and solved for
e^(-t/T) my answers didn't match the ones in the book.
Is my approach correct or am I missing something?
Thanks
2. Dec 12, 2003
### himanshu121
3. Dec 12, 2003
### gnome
Kirchhoff's rules, yes, but the devil is in the details. If you're still stuck, let's see your equations.
4. Dec 12, 2003
### discoverer02
OK, I've posted a diagram.
I = I1 + I2
First Loop: 10V = 4ohmsI + 4ohmsI1
Second Loop: 10V = 4ohmsI + 8ohmsI2 + Ldi2/dt
If I divide both equations by 1 ohm:
First Loop: 10I = 4I + 4I1
Second Loop: 10I = 4I + 8I2 + L/R(di2/dt)
L/R = Time constant = T
4I1 = 6I
I1 = I - I2 so 4I2 = -2I ==> I = -2I2
Tdi2/dt + 8I2 = 6I
Tdi2/dt + 8I2 = 6(-2I2) = -12I2
Tdi2/dt = -20I2
di2/(20I2) = -1/T(dt)
integrate and ==> ln(20I2)/20 = e^(-t/T)
If I'm correct so far then this is where I get confused.
There's no emf in the second loop, so there's an initial current in the second loop?
I'm not sure which form of the equation for I in an RL circuit I should use. (no initial current, no emf, both) I'm also not sure that what I did with L/R above is valid.
#### Attached Files:
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Last edited: Dec 12, 2003
5. Dec 12, 2003
### gnome
First loop should be:
10 = 4I + 4I1 (NOTE: 10, not 10I)
second loop:
10 = 4I + 8I2 + LdI2/dt
and L is 1 so you can leave it off & just say
10 = 4I + 8I2 + dI2/dt
I don't understand what that R is in your second loop equation, but the back emf is just LdI2/dt. No R.
Note also: you can also use as the second equation:
4I1 - 8i2 - dI2/dt
6. Dec 12, 2003
### discoverer02
Thanks gnome.
That was a silly error. I have no idea why I put that 'I' in the equation. For some reason I thought I was dividing through by R instead of 1 ohm. Too tired. I'm glad the semester's almost over. | 818 | 2,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-13 | latest | en | 0.925371 |
https://id.scribd.com/doc/141857754/Tethot | 1,606,893,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00166.warc.gz | 325,940,442 | 67,601 | Anda di halaman 1dari 7
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Rata-rata
## Keanekaragaman hewan mamalia
Ketinggian tempat dari dari permukaan laut
Penutupan tajuk vegetasi
Kemiringan lereng
Y
X1
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65
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Perhitungan
Sigma X1y=
Sigma x1x2=
Sisma x1x3=
sigma x2.x3=
Sigma x2.y=
sigma x3.y=
5,974.7143
11,436.9643
6,428.7857
-1,171.4643
3,458.8214
5,998.9643
1,789.6786
2,632.2143
2,606.1071
3327.92857
## Masukkan nilai yang diperoleh ke dalam persamaan
11,436.9643 b1 -1,171.4643 b2 +
3,458.8214 b3
-1,171.4643 b1 +
5,998.9643 b2 +
1,789.6786 b3
3,458.8214 b1 +
1,789.6786 b2 +
2,606.1071 b3
=
=
=
6,428.7857
2,632.2143
3327.928571
## Copy paste special (value)
1 11436.9643 b1 - -1171.464286 b2 + 3458.821429 b3
-1171.4643 b1 + 5998.964286 b2 + 1789.678571 b3
3458.82143 b1 + 1789.678571 b2 + 2606.107143 b3
=
=
=
6428.785714
2632.214286
3327.928571
1 b1 -0.102427905 b2 + 0.302424782 b3
1 b1 -5.120910948 b2 -1.527727813 b3
1 b1 + 0.517424391 b2 + 0.753466808 b3
=
=
=
0.562105953
-2.246943691
0.962156804
0
0
-5.018483043 b2 -1.830152595 b3
0.619852296 b2 + 0.451042027 b3
=
=
-2.809049644
0.400050851
1 b2 + 0.364682431 b3
1 b2 + 0.727660493 b3
=
=
0.559740786
0.645397062
0 b2 + 0.362978062 b3
0.085656276
b3
0.235981964
b2
=
=
0.559740786
0.47368231
b1
=
=
=
0.562105953
0.562105953
0.539257446
b3
b3
b3
b3
=
=
=
=
=
6428.785714
2632.214286
3327.928571
12388.92857
12388.92857
-11.73196113
b2 + 0.086058476
b1
-0.048518286 +
b1
0.071366794
0.022848508
CEK!!!!!!!!
11436.9643
-1171.4643
3458.82143
13724.3214
CEK!!!!
b1 b1 +
b1 +
b1 +
-1171.464286
5998.964286
1789.678571
6617.178571
Substitusi
b2 +
b2 +
b2 +
b2 +
3458.821429
1789.678571
2606.107143
7854.607143
12388.92857
## berganda: Y = -11.73 +nn
Reduction SS
Total SS
=
=
5498.935028
5,974.7143
Source
Reduction due to X1, X2, and X3
Residual
Total
R2 =
R=
df
F Tabel
0.05
3 5498.935 1832.978 92.46195 3.008787
24 475.7793 19.82414
27 5,974.7143
SS
MS
F hit
F Tabel
0.01
4.718051 | 1,315 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-50 | latest | en | 0.206052 |
https://studysoup.com/tsg/15804/conceptual-physics-12-edition-chapter-20-problem-39e | 1,586,103,116,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00476.warc.gz | 721,706,078 | 11,439 | ×
×
# A pair of loudspeakers on two sides of a stage are
ISBN: 9780321909107 29
## Solution for problem 39E Chapter 20
Conceptual Physics | 12th Edition
• Textbook Solutions
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Conceptual Physics | 12th Edition
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20
5
Problem 39E
A pair of loudspeakers on two sides of a stage are emitting identical pure tones (tones of a fixed frequency and fixed wavelength in air). When you stand in the center aisle, equally distant from the two speakers, you hear the sound loud and clear. Why does the intensity of the sound diminish considerably when you step to one side? (? uggestion:? Use a diagram to make your point.)
Step-by-Step Solution:
Step 1 of 3
Solution 39E STEP 1: Here is an image of two waves coming from two opposite sides and reach the middle point where you are standing. Now you can hear the sounds very clearly because of the constructive interference by the two waves. As they are at the same distance from your position, their phase difference is zero and when they meet, they create constructive interference. By this the amplitude of the final wave gets increased and you hear a louder voice. The diagram of constructive interference of two waves of zero phase difference is shown below. The final amplitude is the sum of two individual amplitudes. So the sound you hear becomes very high and louder. STEP 2: When...
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321909107
This full solution covers the following key subjects: fixed, tones, Sound, loud, considerably. This expansive textbook survival guide covers 45 chapters, and 4650 solutions. This textbook survival guide was created for the textbook: Conceptual Physics, edition: 12. The answer to “A pair of loudspeakers on two sides of a stage are emitting identical pure tones (tones of a fixed frequency and fixed wavelength in air). When you stand in the center aisle, equally distant from the two speakers, you hear the sound loud and clear. Why does the intensity of the sound diminish considerably when you step to one side? (? uggestion:? Use a diagram to make your point.)” is broken down into a number of easy to follow steps, and 69 words. Since the solution to 39E from 20 chapter was answered, more than 407 students have viewed the full step-by-step answer. Conceptual Physics was written by and is associated to the ISBN: 9780321909107. The full step-by-step solution to problem: 39E from chapter: 20 was answered by , our top Physics solution expert on 04/03/17, 08:01AM.
#### Related chapters
Unlock Textbook Solution | 619 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.893821 |
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Every Question Helps You Learn
You can check if 2 x 3 = 6 is correct by dividing 6 by 2. It should give the answer 3.
# Check Multiplication with Division - Using Division Facts
This Math quiz is called 'Check Multiplication with Division - Using Division Facts' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.
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Elementary school children should begin to understand the relationship (or facts) between multiplication and division, and learn how they can use that relationship to check their answers. For example, if they calculate that 3 x 4 = 12, than they could check by calculating that 12 shared into 4 equal groups would give 3 in each group. Knowing and using the division facts will help with this.
Try using division to check multiplication calculations.
1.
Which 'family' of facts is correct?
2 x 3 = 5
6 x 2 = 3
6 ÷ 5 = 3
2 ÷ 3 = 6
4 x 5 = 20
5 x 4 = 20
20 ÷ 4 = 5
20 ÷ 5 = 4
2 x 8 = 14
8 x 2 = 14
18 ÷ 2 = 8
18 ÷ 8 = 2
4 x 9 = 50
9 x 4 = 50
55 ÷ 4 = 9
55 ÷ 9 = 4
If the multiplication facts are correct, the division facts should use the same numbers
2.
Which two division facts can be found from knowing that 8 x 5 = 40?
5 ÷ 5 = 8 and 8 ÷ 8 = 5
40 ÷ 5 = 8 and 40 ÷ 8 = 5
40 ÷ 40 = 8 and 5 ÷ 5 = 40
8 ÷ 5 = 40 and 5 ÷ 8 = 40
The largest number comes first, followed by either of the smaller numbers to give the other as the answer
3.
What could you do to check this calculation is correct:
9 x 5 = 45
Multiply 45 by 9
Divide 45 by either 9 or 5
Multiply 45 by 5
Find the total of all 3 numbers
The calculation is correct, but checking it with either division would prove it
4.
Seven groups of 2 makes 14 so 14 divided into groups of 2 would make...
28 groups
14 groups
2 groups
7 groups
If 7 x 2 = 14, then 14 ÷ 2 = 7 or 14 ÷ 7 = 2
5.
What is the opposite, or inverse of multiplication?
Subtraction
Multiplication
Division
Division and multiplication are the inverse of each other, as are addition and subtraction
6.
Which number is missing?
? ÷ 5 = 7
35
57
75
30
5 x 7 = 35 so 35 ÷ 5 = 7
7.
Which number is missing?
90 ÷ ? = 10
19
90
9
109
9 x 10 = 90 so 90 ÷ 9 = 10
8.
Which division calculation could you use to check this is correct:
10 x 5 = 50
50 ÷ 50 = 10
50 ÷ 5 = 10
10 ÷ 5 = 50
10 ÷ 50 = 5
You could also use 50 ÷ 10 = 5
9.
Which division calculation could you use to check this is correct:
5 x 6 = 30
6 ÷ 30 = 5
6 ÷ 5 = 30
30 ÷ 5 = 5
30 ÷ 5 = 6
In the division fact, the largest number always comes first
10.
If 10 groups of 3 is 30...
then 30 multiplied by 30 will give 10
then 30 divided by 30 will give 3
then 30 multiplied by 3 will give 10
then 30 divided into groups of 3 will give 10 groups
A division 'cancels out' the multiplication as it is the opposite
Author: Angela Smith | 1,013 | 3,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.953997 |
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# Que:- A parabola drawn touching the axis of x at the origin and having its vertex at a given distance k from
the x - axis . Prove that the axis of parabola is a tangent to the parabola x2 + 8k(y - 2k) = 0.
## Ans : Let the equation of the parabola be Y2 = 4ax
2 Any tangent to it at the point (at , 2at) is 2 2 Yt = X +at .........(1) The normal at the point( at , 2at) is 3 Y + tX = 2at + at ........(2) Take the equations of transformation -
in xy coordinates p=(0,0) and PT is the axis which is tangent to the parabola at the origin. Now,
## the distance of the vertex v(0,0) in the x,y coordinates from pt
Full Question: A regular polygon with 2n + 1 sides is inscribed in a circle. Three of the polygon's vertices are selected at random and the triangle formed by connecting these vertices is drawn. Prove that the probability that the center of the circle lies inside the triangle is (n+1)/(2(2n-1)). Surely the total number of ways to pick 3 vertices on the (2n+1)-gon without replacement is (2n+1)*(2n)*(2n-1). Now we need to find the number of triples of vertices that have the desired property. Let's start by choosing 1 vertex -- this can surely be done in 2n+1 ways. Now consider the distance between the first point and the second point, where by distance I mean the number of sides in between. As the number of sides is odd, each case has a (unique!) counterpart by reflexive symmetry. Therefore, we now have the number of desired triples as (2n+1)*2*([number of good last points when first two are 1 away] + [number of good last points when first two are 2 away] + ... + [number of good last points when first two are n away]) n is the maximum distance, as a distance of n+1 is simply a distance of n going around the polygon in the opposite direction. Note also that each side between the first two points (in the path of least length) corresponds to a diametrically opposite vertex, and the triangle created by each such point contains the longer portion of the diameter and hence the center! Thus, the number of good last points when the first two are d away is exactly d itself! Hence, the number of triples with the desired property is (2n+1)*2*(1 + 2 + ... + n). (Can that last sum be simplified?) The desired probability, then, is the fraction with the number of desired triples as the numerator, and the number of ALL triples as the denominator. EDIT: How silly of me, I neglected the fact that order does not matter, so the number of good triples and the number of total triples should each be divided by the number of permutations on 3 letters, which is 3! = 6. Fortunately, when forming the probability, this divisor cancels, and so the probability is the same as with the quantities above.
**From the 2n+1 vertices, randomly select a vertex, say A. Then select
vertices B and C clockwise. Denote angle AOB by p*a, angle BOC by q*a, where a=2*pi/(2n+1); p, q are both integers. For the center to lie inside triangle ABC, the triangle must be acute. Therefore, 0 < p*a/2 < pi/2 0 < q*a/2 < pi/2 0 <(2*pi-p*a-q*a) < pi/2 which can be further simplified as 1 <= p <= n 1 <= q <= n n+1 <= p+q <= 2*n The number of (p,q) pairs that satisfy the conditions above can be easily
calculated as n(n+1)/2. Without these requirements, the number of (p,q) pairs can also be conveniently obtained to be n(2n-1). So the probability is [n(n+1)/2]/[n(2n-1)] = (n+1)/[2*(2n-1)]
**The median does not alter because it is only dependent on the middle observation's value. The mean does change,
however, because it is dependent on the average value of all observations. So, in the above example, as the last value of the last observation increases, so too does the mean. In the third data set, the value of 14 is very different from any other values. When an observation is very different from all other observations in a data set, it is called anoutlier. (For more information on outliers, see the Stem and leaf plots section.) The mean is the measure of central tendency most affected by outliers. Outliers can sometimes occur as a result of error or deliberate misinformation. In these cases, the outliers should be excluded from the measure of central tendency. Other times, outliers just show how different one value is, and this can be a very useful piece of data.
## Example 7 Comparing the mean and median
When house prices are referred to in newspapers, generally the median price is quoted. Why is this measure used instead of the mean? There are many moderately priced houses, but there are also some expensive ones and a few very expensive ones. The mean figure could be quite high as it includes the prices of the more expensive houses. But the median gives a more accurate and realistic value of the prices faced by most people. In summary, the median is the central number and is good to use in skewed (or unbalanced) distributions because it is not affected by outliers.
## Example 8 Comparing the mean and median
Suppose you want to know how much money a family could afford to spend on housing. This would depend on the total family income. For a family of five (two parents who work and three children with no income) the mean income of each family member is the total income divided by five (e.g., 60,000 5 = 12,000). However, the median income would be zero, because more than half of the members of the family make nothing. In some situations, the mean can be much more informative than the median.
## Example 9 Comparing the mean and median
If you want to find out whether a country is wealthy or not, you might consider using the median as your measure of central tendency instead of the mean. The mean family income could be quite high if income is highly concentrated in a few very wealthy families (despite the fact that most families might earn essentially nothing). Thus, the median family income would be a more meaningful measureat least half the families would earn the median income or less, and at least half would earn at least as much as the median income or more.
## Example 10 Comparing the mean and median
Suppose you are applying for a job as an accountant at several large firms, and you want to get an idea of how much money you could expect to be making in five years if you join a particular firm. You may want to consider the salaries of accountants in each firm five years after they are hired. One very high salary could make the mean salary higher; that might not reflect a typical salary within these firms. However, half the accountants make the median salary or less, and half make the median salary or more. So, the measure of central tendency that would give you a better idea of a typical salary would be the median.
## Example 11 Comparing the mean and median
By choosing a measure of central tendency favourable to your point of view, you can mislead people with statistics. In fact, this is commonly done. Imagine you are the owner of a bakery that makes and sells individual birthday cakes and huge wedding cakes. It might be in your interest to claim to your customers that the prices have been lowered, and to claim to your shareholders that you have raised the prices. Suppose that last year you sold 100,000 birthday cakes at \$10 each, and 1,000 wedding cakes at \$1,000 each. This year, you sold 100,000 birthday cakes at \$8 each and 1,000 wedding cakes at \$1,200 each.
The median price of the 101,000 cakes sold last year is \$10, because more than half of the items sold were birthday cakes. The median price of the 101,000 cakes sold this year is \$8. The mean price of the 101,000 cakes sold last year is \$19.80. (100,000 x \$10 + 1,000 x \$1,000) 101,000 = \$19.80 The mean price of the 101,000 cakes sold this year is also \$19.80. (100,000 x \$8 + 1,000 x \$1,200) 101,000 = \$19.80
The average price per cake sold is the same in both years. Also, the total revenue and the number of the cakes sold was the same. The idea is that you can make data appear to tell conflicting stories by choosing the appropriate measure of central tendency. It is important to note that you do not have to use only one measure of central tendency. The mean and median can both be used, thus providing more information about the data. | 1,997 | 8,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-39 | latest | en | 0.933334 |
https://resources.quizalize.com/view/quiz/volume-of-solids-bedb7f82-d6d8-4492-a35c-1f8abf694628 | 1,721,859,115,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00739.warc.gz | 423,007,410 | 16,592 | Volume of Solids
Quiz by Paul Dietrich
Mathematics
Common Core
Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
5 questions
• Q1
What is volume?
The measurement around a shape.
The amount of space that a substance or object occupies.
The number of unit squares a shape can contain.
Something you want for your hair.
30s
• Q2
Find the volume of a cylinder with a diameter of 3ft and a height of 10ft
282ft cubed
15ft cubed
70.65ft cubed
22.5ft cubed
30s
8.G.C.9
• Q3
What is the volume of a sphere with a radius of 5in?
62.8in cubed
1,570in cubed
166.67in cubed
523.6in cubed
30s
8.G.C.9
• Q4
What is the volume of cone with a diameter of 5cm and a height of 18cm?
353.25cm cubed
117.81cm cubed
47.1cm cubed
188.4cm cubed
30s
8.G.C.9
• Q5
What is the volume of a rectangular prism with a width of 2m, a length of 5m, and a height of 8m?
56m cubed
26m cubed
80m cubed
15m cubed
30s
8.G.C.9
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16th October 2006, 04:42 AM #21 likeDIY diyAudio Member Join Date: Sep 2006 Hi guys, the output capacitor C103 =10uF can be changed by 22uF? Will the output frequency response changes alot?
16th October 2006, 06:12 AM #22 Mad_K diyAudio Member Join Date: Sep 2002 Location: Norway It will go down, which is a good thing (-3dB from 3 to 1,5Hz nominal) __________________ Mads K
16th October 2006, 06:14 AM #23 Nexus diyAudio Member Join Date: Aug 2003 Location: The Netherlands Hello likeDIY, That will depend on the impedance which loads your BOZ. You can calculate it out with: Fc=1/2*pi*R*C. where Fc= your freq response R=the load impedance for your BOZ C=10 or 22uF. Best Regards: Nexus
18th October 2006, 08:18 AM #24 likeDIY diyAudio Member Join Date: Sep 2006 Ok, I will give it a go with a 22uF. The values of resistors must exactly the same as in the circuit? I'm abit lazy not to make the 33.2K and 22.1k, or 4.75K, in stead I just make it 33k, 22k, 4.7 respectively. Would it be much difference?
18th October 2006, 08:22 AM #25 Mad_K diyAudio Member Join Date: Sep 2002 Location: Norway That's fine __________________ Mads K
19th October 2006, 02:45 AM #26 likeDIY diyAudio Member Join Date: Sep 2006 Help help help. It's blown up !!!!! After connecting to the AC source, it's blown up :-( The two resitors R2 and R3 are the first to get fire... Please help, what is the problem? I'm quite sure that I connect the polar of the components correctly.
19th October 2006, 05:37 AM #27 Mad_K diyAudio Member Join Date: Sep 2002 Location: Norway Your amp is drawing a lot of current! What sets the current is the voltage at the mosfet's Source and the source resistor; are R101, P102, and R108 correct value and placement? Or maybe you have a short circuit (to gnd) somewhere? Did you smoke any other components? __________________ Mads K
19th October 2006, 06:33 AM #28 likeDIY diyAudio Member Join Date: Sep 2006 I think I should check the P102, how can we indicate the CW and CCW for the P102 as shown in the circuit? As soon as it gets smoke, I disconnect the power, so hopefully no other things get burn.
19th October 2006, 06:44 AM #29 likeDIY diyAudio Member Join Date: Sep 2006 One more point, before making the PCB, I have mirrored the image and print it out, It might be a problem with the pin connection?
19th October 2006, 07:13 AM #30 Mad_K diyAudio Member Join Date: Sep 2002 Location: Norway The pcb layout in the original article is bottom viewed from bottom. How did you make the pcb? __________________ Mads K
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All times are GMT. The time now is 12:37 PM. | 1,273 | 5,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-26 | latest | en | 0.914295 |
https://www.clarendonlearning.org/lesson-plans/inches-feet-yards/ | 1,555,985,468,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578584186.40/warc/CC-MAIN-20190423015050-20190423041050-00248.warc.gz | 644,117,250 | 84,795 | # Inches, Feet, Yards
\$0.00
Our Inches, Feet, Yards Lesson Plan further develops measuring skills as students identify, measure and compare the length of objects using inches, feet, and yards, including half and quarter inches. With an interactive activity, students practice through meaningful game play with manipulatives. Generalize the learned concepts by weighing each student and converting their weight into ounces.
## Inches, Feet, Yards Lesson Plan Includes:
• Full Teacher Guidelines with Creative Teaching Ideas
• Instructional Content Pages about Inches, Feet, and Yards.
• Hands on homework activities giving students practice on determining how to measure, identify, and compare the length of different objects.
• Common Core State Standards
*Note: These lessons are PDF downloads. You will be directed to the download once you checkout. Clarendon Learning resources are FREE, we rely 100% on donations to operate our site. Thank you for your support!
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### Description
Our Inches, Feet, Yards Lesson Plan further develops measuring skills as students identify, measure and compare the length of objects using inches, feet, and yards, including half and quarter inches. With an interactive activity, students practice through meaningful game play with manipulatives. Generalize the learned concepts by weighing each student and converting their weight into ounces.
Sample Classroom Procedure / Teacher Instruction
1. Ask students: How do you know the size of shoes you need? How do carpenters build houses correctly? How does a GPS know the distance from your home to another location?
2. Allow for responses and discussion. Introduce measuring in inches, feet, yards, and miles.
3. Distribute Inches, Feet, and Yards content pages. Read and review the information with the students. Use extra examples to help students understand the conversion of inches, feet, etc. Save the final question for lesson closing.
4. Carefully explain how the remainders are turned into inches or feet, and 12 inches change to one foot, and 3 feet to one yard, etc. Use the additional resources to enhance understanding.
5. Distribute Activity pages. Read and review the instructions. Carefully explain the samples and the calculation of the “New Total”. Pair students. Distribute scissors. (Decide if each student cuts apart two copies of the lengths, or use one set of lengths for each pair.)
6. Give students sufficient time to play the game.
7. When completed, randomly check and review some of the students’ responses.
8. Distribute Practice page and a ruler. Check and review the students’ responses.
9. Distribute the Homework page. The next day, check and review the students’ responses allowing students to share their heights, room size, etc.
10. In closing, ask: What are some of the things you have had to measure in the past? Why?
11. Allow for responses and discussion. Some students may need to be prompted. Measuring involves the purchase of clothing, shoes, organizing things at home, books on a shelf, storing DVDs, etc.
Common Core State Standards: CCSS.Math.Content.2.MD.A.1, CCSS.Math.Content.2.MD.A.4, CCSS.Math.Content.3.MD.B.4, CCSS.Math.Content.4.MD.A.1
Class Sessions (45 minutes): At least 2 class sessions
Want more math resources? Check out our other Math Lesson Plans!
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### Helpful information in a fun format (and not too long!)
Good information presented in an easy to understand format that was interesting and helpful. Thanks! | 825 | 3,729 | {"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} | 3.53125 | 4 | CC-MAIN-2019-18 | latest | en | 0.928379 |
http://forums.xkcd.com/viewtopic.php?f=17&t=61803&p=2210964& | 1,560,831,194,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998605.33/warc/CC-MAIN-20190618023245-20190618045245-00441.warc.gz | 63,594,994 | 7,340 | ## So I tried...
For the discussion of math. Duh.
Moderators: gmalivuk, Moderators General, Prelates
yuyuyami
Posts: 10
Joined: Sat Jun 12, 2010 8:16 pm UTC
### So I tried...
I tried to take the derivative of:
Y=O*P*T*I*M*U*S
But the multiplication and chain rules started making it giant. I finished it, but then when I tried to multiply it all together, I realized: Finding (OPTIMUS)' just isn't worth it.
Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK
### Re: So I tried...
We have a maths joke thread.
yuyuyami
Posts: 10
Joined: Sat Jun 12, 2010 8:16 pm UTC
### Re: So I tried...
Could a moderator move this, then?
Sizik
Posts: 1243
Joined: Wed Aug 27, 2008 3:48 am UTC
### Re: So I tried...
It's not really that hard.
(OPTIMUS)' = O'PTIMUS + O(PTIMUS)'
Iterating out, you get the following:
(OPTIMUS)' = O'PTIMUS + OP'TIMUS + OPT'IMUS + OPTI'MUS + OPTIM'US + OPTIMU'S + OPTIMUS'
i.e. as many terms as factors, and each term has one factor get derivatified differentiated.
[/nosenseofhumor]
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.
lu6cifer
Posts: 230
Joined: Fri Mar 20, 2009 4:03 am UTC
Location: That state with the all-important stone
### Re: So I tried...
If you found it too unwieldy, you probably should have tried logarithmic differentiation
lu6cifer wrote:"Derive" in place of "differentiate" is even worse.
doogly wrote:I'm partial to "throw some d's on that bitch."
yuyuyami
Posts: 10
Joined: Sat Jun 12, 2010 8:16 pm UTC
### Re: So I tried...
Sizik wrote:It's not really that hard.
(OPTIMUS)' = O'PTIMUS + O(PTIMUS)'
Iterating out, you get the following:
(OPTIMUS)' = O'PTIMUS + OP'TIMUS + OPT'IMUS + OPTI'MUS + OPTIM'US + OPTIMU'S + OPTIMUS'
i.e. as many terms as factors, and each term has one factor get derivatified differentiated.
[/nosenseofhumor]
Wouldn't it be closer to:
Y'=(O'*PTIMUS+(P'*TIMUS+(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*P)*O)*(P'*TIMUS+(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*P)*(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*(M'*US+(U'*S+S'*U)*M)*(U'*S+S'*U)
?
Because you still need to take the derivative of the second term when doing the multiplication rule.
Sizik
Posts: 1243
Joined: Wed Aug 27, 2008 3:48 am UTC
### Re: So I tried...
yuyuyami wrote:
Wouldn't it be closer to:
Y'=(O'*PTIMUS+(P'*TIMUS+(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*P)*O)*(P'*TIMUS+(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*P)*(T'*IMUS+(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*T)*(I'*MUS+(M'*US+(U'*S+S'*U)*M)*I)*(M'*US+(U'*S+S'*U)*M)*(U'*S+S'*U)
?
Because you still need to take the derivative of the second term when doing the multiplication rule.
(uv)' = u'v + uv', not (u'v + uv')v', or whatever seems to be going on there. The chain rule doesn't apply for multiplication.
Step by step:
Y' = (OPTIMUS)'
Y' = O'PTIMUS + O(PTIMUS)'
Y' = O'PTIMUS + O(P'TIMUS + P(TIMUS)' )
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(IMUS)' ))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(I'MUS + I(MUS)' )))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(I'MUS + I(M'US + M(US)' ))))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(I'MUS + I(M'US + M(U'S + US' )))))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(I'MUS + I(M'US + MU'S + MUS' ))))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + T(I'MUS + IM'US + IMU'S + IMUS' )))
Y' = O'PTIMUS + O(P'TIMUS + P(T'IMUS + TI'MUS + TIM'US + TIMU'S + TIMUS' ))
Y' = O'PTIMUS + O(P'TIMUS + PT'IMUS + PTI'MUS + PTIM'US + PTIMU'S + PTIMUS' )
Y' = O'PTIMUS + OP'TIMUS + OPT'IMUS + OPTI'MUS + OPTIM'US + OPTIMU'S + OPTIMUS'
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses. | 1,507 | 4,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.918883 |
https://bridgitmendlermusic.com/does-adiabatic-expansion-increase-entropy/ | 1,721,281,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00729.warc.gz | 120,885,153 | 9,113 | ## Does adiabatic expansion increase entropy?
Adiabatic processes are characterized by an increase in entropy, or degree of disorder, if they are irreversible and by no change in entropy if they are reversible. Adiabatic processes cannot decrease entropy.
In an adiabatic expansion the change in internal energy is negative ie the system does work on the surroundings and the change in internal energy is positive when the system is compressed ie the surroundings do work no the system or put another way, the system does negative work on surroundings.
The assumption that a process is adiabatic is a frequently made simplifying assumption. For such an adiabatic process, the modulus of elasticity (Young’s modulus) can be expressed as E = γP, where γ is the ratio of specific heats at constant pressure and at constant volume (γ = CpCv ) and P is the pressure of the gas.
### Why is Delta U equal to adiabatic?
According to the definition of an adiabatic process, ΔU=wad. Therefore, ΔU = -96.7 J. Calculate the final temperature, the work done, and the change in internal energy when 0.0400 moles of CO at 25.0oC undergoes a reversible adiabatic expansion from 200. L to 800.
Can two isothermal curves cut each other?
No, If they intersect, then at two different temperatures (of the isothermals), volume and pressure of gas will be same, which is not possible.
How do you know if adiabatic expansion is reversible?
An adiabatic (zero heat exchanged with the surroundings) process is reversible if the process is slow enough that the system remains in equilibrium throughout the process.
## Which is an example of an adiabatic expansion?
On a p-V diagram, the process occurs along a line (called an adiabat) that has the equation p = constant / Vκ. For an ideal gas and a polytropic process, the case n = κ corresponds to an adiabatic process. Example of Adiabatic Expansion Assume an adiabatic expansion of helium (3 → 4) in a gas turbine (Brayton cycle).
What is the mathematical representation of the adiabatic process?
The adiabatic process is a thermodynamic process in which there is no heat transfer from in or out of the system. For an ideal gas, an adiabatic process is a reversible process with constant entropy. The mathematical representation of the adiabatic process is ΔQ=0
What is the formula for reversible adiabatic expansion of an ideal gas?
Thus for a reversible adiabatic process and an ideal gas, CVdT = −PdV. (The minus sign shows that as V increases, T decreases, as expected.) But for a mole of an ideal gas, PV = RT = (CP − CV)T, or P = (CP − CV)T/V.
### How to calculate the adiabatic expansion of helium?
Assume an adiabatic expansion of helium (3 → 4) in a gas turbine (Brayton cycle). Assume an adiabatic expansion of helium ( 3 → 4) in a gas turbine. Since helium behaves almost as an ideal gas, use the ideal gas law to calculate outlet temperature of the gas ( T4,real ).
Does adiabatic expansion increase entropy? Adiabatic processes are characterized by an increase in entropy, or degree of disorder, if they are irreversible and by no change in entropy if they are reversible. Adiabatic processes cannot decrease entropy. Is adiabatic expansion positive? In an adiabatic expansion the change in internal energy is negative ie the system… | 786 | 3,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.91364 |
https://metanumbers.com/27280 | 1,601,490,278,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127397.84/warc/CC-MAIN-20200930172714-20200930202714-00286.warc.gz | 465,777,123 | 7,624 | ## 27280
27,280 (twenty-seven thousand two hundred eighty) is an even five-digits composite number following 27279 and preceding 27281. In scientific notation, it is written as 2.728 × 104. The sum of its digits is 19. It has a total of 7 prime factors and 40 positive divisors. There are 9,600 positive integers (up to 27280) that are relatively prime to 27280.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 19
• Digital Root 1
## Name
Short name 27 thousand 280 twenty-seven thousand two hundred eighty
## Notation
Scientific notation 2.728 × 104 27.28 × 103
## Prime Factorization of 27280
Prime Factorization 24 × 5 × 11 × 31
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 3410 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 27,280 is 24 × 5 × 11 × 31. Since it has a total of 7 prime factors, 27,280 is a composite number.
## Divisors of 27280
1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 31, 40, 44, 55, 62, 80, 88, 110, 124, 155, 176, 220, 248, 310, 341, 440, 496, 620, 682, 880, 1240, 1364, 1705, 2480, 2728, 3410, 5456, 6820, 13640, 27280
40 divisors
Even divisors 32 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 40 Total number of the positive divisors of n σ(n) 71424 Sum of all the positive divisors of n s(n) 44144 Sum of the proper positive divisors of n A(n) 1785.6 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 165.167 Returns the nth root of the product of n divisors H(n) 15.2778 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 27,280 can be divided by 40 positive divisors (out of which 32 are even, and 8 are odd). The sum of these divisors (counting 27,280) is 71,424, the average is 178,5.6.
## Other Arithmetic Functions (n = 27280)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 9600 Total number of positive integers not greater than n that are coprime to n λ(n) 120 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2985 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 9,600 positive integers (less than 27,280) that are coprime with 27,280. And there are approximately 2,985 prime numbers less than or equal to 27,280.
## Divisibility of 27280
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 1 0 1
The number 27,280 is divisible by 2, 4, 5 and 8.
• Refactorable
• Abundant
• Polite
• Practical
## Base conversion (27280)
Base System Value
2 Binary 110101010010000
3 Ternary 1101102101
4 Quaternary 12222100
5 Quinary 1333110
6 Senary 330144
8 Octal 65220
10 Decimal 27280
12 Duodecimal 13954
20 Vigesimal 3840
36 Base36 l1s
## Basic calculations (n = 27280)
### Multiplication
n×i
n×2 54560 81840 109120 136400
### Division
ni
n⁄2 13640 9093.33 6820 5456
### Exponentiation
ni
n2 744198400 20301732352000 553831258562560000 15108516733586636800000
### Nth Root
i√n
2√n 165.167 30.1033 12.8517 7.71203
## 27280 as geometric shapes
### Circle
Diameter 54560 171405 2.33797e+09
### Sphere
Volume 8.50397e+13 9.35187e+09 171405
### Square
Length = n
Perimeter 109120 7.44198e+08 38579.7
### Cube
Length = n
Surface area 4.46519e+09 2.03017e+13 47250.3
### Equilateral Triangle
Length = n
Perimeter 81840 3.22247e+08 23625.2
### Triangular Pyramid
Length = n
Surface area 1.28899e+09 2.39258e+12 22274
## Cryptographic Hash Functions
md5 7144e2ba113bacdd760b73ae7478c74e f6a3ca4324a46504ef39d48fdcb8d2b329218294 d0eff1ad0dd4ec57c1b9fc033417982c102e1f8dd13b5f06c10fe07bf56ea79c d40ff6ef8b561a6a0f557cda98d1bce1ceae84788b766725469f391924e05b5371479f2966f2f4c103d1f09bccdae5970413a84e097a004542256fcb6015fac7 d8df0992f88521c582a5d62712f71a3ff5002ac4 | 1,550 | 4,234 | {"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} | 3.65625 | 4 | CC-MAIN-2020-40 | latest | en | 0.77291 |
https://math.stackexchange.com/questions/921243/combinatorics-question-why-doesnt-this-method-work | 1,638,772,810,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00397.warc.gz | 444,611,705 | 34,476 | Combinatorics question. Why doesn't this method work?
If I have the following set {$1,2,3,4$}, and I want to know the number of possible lists of size 3 that contain $1$ and $2$ in them, I tried the following, and think of it as the number of ways to select 3 elements:
$2·1·2=4$ as in "number of choices for the first number, then the second... " But the actual number is 12.
The only way I can do it is by observing that I have two sets {$1,2,3$} and {$1,2,4$} And each one has 6 permutations, and therefore 12 total permutations or lists.
How can I find the number of lists and subsets that satisfy the property of containing 1 and 2?
• Your first method works too : "total permutations" - ("permutations without 1" + "permutations without 2") $$4.3.2 - (3.2.1 + 3.2.1)$$ Sep 6 '14 at 10:08
• By list you mean n-tuple? A subset isnt a ordered thing, it completely different the case for a 3-tuple and a subset of cardinality 3. Can you repeat elements on the lists/3-tuple? Sep 6 '14 at 10:36
• Yeah, that's what I meant. But I want a robust method to find both lists and subsets given such criteria. Sep 6 '14 at 10:41
• Apparently you want neither $(1,2,1)$, nor $(2,6,1)$, but I could not tell the first and hardly the second from your description. Sep 6 '14 at 11:22
In fact, $2*1*2=4$ isn't that far away from your other solution. What you did is calculate the number of possibilities to have either $2$ or $1$ in the first place of your set and the other one on the second spot, with the remaining $3$ and $4$ taking the last one.
You did not, however, keep in mind, that $2$ and $1$ could also be in the second and third place or in the first and the third place. In the end, you have essentially three possibilities to form your set concerning placement of $2$ and $1$, and each possibility has four different ways of fulfilling the condition.
So you get: $3*2*1*2 = 12$ , which is the correct solution.
I hope that helped!
SDV
Let's say that you are looking for the number of sets of $\left\{ 1,\dots,n\right\}$ of size $m$ that contain $\left\{ 1,\dots,k\right\}$.
This comes to choosing $m-k$ members of this set from set $\left\{ k+1,\dots,n\right\}$ so there are $$\binom{n-k}{m-k}$$ possibilities.
If it comes to lists then each permution on one of these sets induces a new list, so there are $$\binom{n-k}{m-k}m!$$ possibilities.
• But it doesn't work for $m=1$, containing {1}: (4-1)!/(1-1)!= 6 Sep 6 '14 at 11:02
• @Hex4869 It also workes for $m=1=k$. Note that $\binom{n-1}{0}=1$ for each $n\geq1$. Do not confuse it with $\frac{(n-1)!}{0!}$ Sep 6 '14 at 11:44 | 788 | 2,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-49 | latest | en | 0.952127 |
https://australiaassessments.com/2020/02/18/annual-pension-payments-assignment/ | 1,679,382,686,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00028.warc.gz | 149,415,330 | 12,778 | # Annual Pension Payments Assignment
John, James and Joe are all 35 years old and plan to retire at age 65. They expect to live to 90 years old. Upon retirement they would all like to take immediate annual pension payments from their savings at the start of each year. John and James can access a quoted rate of 9% per year with quarterly compounding whilst Joe can access a quoted rate of 11% per annum with semi-annual compounding.
a) John has a monthly income of \$6,000 and monthly expenses of \$2350 and saves the remainder at the end of each month. On the day of his retirement he will receive a retirement bonus from his employer totaling \$20,000 and will also sell his holiday home for an estimated \$150,000. How much money will John have saved upon retirement?
b) What would be John’s annual pension?
c) James does not have a holiday home to sell, and will not be receiving a bonus from his employer upon retirement. However he is able to invest twice as much as John each month. How much later can James start saving if he wants to have the same annual pension as John?
d) Joe also does not have a holiday home to sell, and will not be receiving a bonus from his employer upon retirement. Unfortunately he does not believe he can save any money during the first 5 years. However, he anticipates being able to save \$6,000 per month for the subsequent 10 years, and then \$8,000 per month until retirement. Comparing Joe with John, who has the biggest annual pension, and by how much?
use a financial calculator to solve. (BAIIplus)
effective interest rate= (1+ r/m)^m/f
make a timeline for each person. Get Finance homework help today
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Customers referred by a friend | 1,135 | 4,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-14 | latest | en | 0.978 |
https://forum.arduino.cc/t/regenerative-braking-reverse-polarity-protection-for-arduino/683497 | 1,627,946,388,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154385.24/warc/CC-MAIN-20210802203434-20210802233434-00350.warc.gz | 272,199,593 | 5,392 | # Regenerative braking: reverse polarity protection for Arduino
Hi all,
I'm using a Sabertooth 2x25 motor driver which supports regenerative braking.
This means that it will charge my 4S Lipo when the motor (max 20A stall current) decelerates.
I'm using the same battery to power my Arduino, through a switching/buck convertor (VMA404) which steps down the voltage to 5V. So this circuit is connected in parallel with the motor circuitry.
Now I'm wondering how to deal with the fact that regenerative braking will reverse polarity of the battery. I suppose that will either fry my buck convertor or Arduino.
I could put a diode in front of the switching convertor to protect it, but even then I'm stuck with the problem that the Arduino won't get power during braking events.
What is the typical way to deal with this? I did my research, but can't seem to find much about this.
Edit: The Sabertooth itself has a 5V output but that one can only supply 10 milliamps, which is not enough for my needs. So I would like to tap off the main battery directly, through my BEC.
Thanks a lot!
I just realized that my question is based on a misunderstanding of how regenerative braking works.
So I'll just answer it myself because it might be useful to others too.
Basically, back EMF does not reverse polarity. In fact, it is of the same polarity as the applied voltage. When running normally, this voltage is lower than the applied voltage. When braking, it can be (much) higher and can therefore charge the battery.
So there is no reversal of polarity, hence no problem to be solved :).
Actually no! The reverse EMF is always there. IT is what keeps your motor from an infinite speed. When you remove the normal power, that just leaves the the back EMF to continue. Regenerative braking shorts the back EMF through a variable resistor, that may be ZERO Ohms resistance, or a short circuit for ultimate braking.
Paul
Thanks for replying, I'm learning every day! I finally finished my first robot, so since you made the effort of responding, I wanted to share the result with you
Video 1: https://youtu.be/vIddskon6ic
Video 2: https://youtu.be/s3aa-ipn5Z0
Video 3: https://youtu.be/rJf3yibiXUA
Regen braking has the ability to DESTROY components unless they are designed to cope with such EMF.
Be very careful in using it to apply holding torque to motors.
For large motors the braking is done via a PWM'd switching device, with the current continuously | 569 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.953765 |
https://labuladong.gitbook.io/algo-en/i.-dynamic-programming/strategiesforsubsequenceproblem | 1,695,517,608,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00259.warc.gz | 397,206,919 | 198,844 | a
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# The Strategies of Subsequence Problem
Translator: sunqiuming526
Subsequence Problem is one of the most common algorithm problem, which is not easy to figure out.
First of all, the subsequence problem itself is more difficult than those for substring and subarray, since the former needs to deal with discontinuous sequence, while the latter two are continuous. It is hard enough to simply enumerate the subsequences, let alone solve related algorithm problems.
Moreover, the subsequence problem is likely to involve two strings, such as the "Longest Common Subsequence (LCS)" problem in the previous article. Without some processing experience, it is really not easy to figure out. Therefore, this article will come up with a routine for the subsequence related problems. In fact, there are only two types of strategies. As long as these two problem-solving strategies in your mind, it's highly possible to ace the problem.
Generally speaking, this kind of question would ask you to find a longest subsequence . Since the shortest subsequence, on the other hand, is just a character, which is not worth asking. Once it comes to subsequences or extreme value problems, it is almost certain that we need to use dynamic programming techniques, and the time complexity is generally O(n^2).
The reason is quite simple. Just think about a string. How many possibilities are there for its subsequence? The answer is at least exponential, right? Thus, we have no reason not to use DP.
Since dynamic programming is used, it is necessary to define the DP array and find the state transition relation. The two strategies we mentioned above are actually the ideas of defining DP arrays. Different problems may require different DP array definitions to solve.
## 1. Two Strategies
1.1 The first strategy is using a one-dimensional DP array
int n = array.length;
int[] dp = new int[n];
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
dp[i] = max|min(dp[i], dp[j] + ...)
}
}
Take an example we used before -- "the Longest Increasing Subsequence (LIS)". The definition of DP array in this case is as below:
We define `dp[i]` as the length of the required subsequence (the longest increasing subsequence) within the subarray `array [0..i]`.
Why does the LIS problem require this strategy? The foregoing is clear enough -- because this strategy is in line with the induction method, and the state transition relation can be found. We are not going to discuss this in details further.
1.2 The second strategy is using a two-dimensional DP array
int n = arr.length;
int[][] dp = new dp[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
dp[i][j] = dp[i][j] + ...
else
dp[i][j] = max|min(...)
}
}
This strategy is used relatively more, especially for the subsequences problems involving two strings / arrays, such as the "Longest Common Subsequence" we mentioned before. The definition of the DP array in this strategy is further divided into two cases: "Only one string is involved" and "Two strings are involved".
a) In the case where two strings are involved (e.g. LCS), the definition of DP array is as follows:
We define `dp[i][j]` as the length of the required subsequence (longest common subsequence) within the subarray `arr1[0..i]` and the subarray `arr2[0..j]`.
b) In the case where only one string is involved (such as the Longest Palindrome Subsequence (LPS) which will be discussed in this article later), the definition of DP array is as follows:
We define `dp[i][j]` as the length of the required subsequence (the longest palindrome subsequence) within the subarray `array [i..j]`.
For the first case, you can refer to these two articles: "Editing distance", "Common Subsequence".
Now let's talk about the Longest Palindrome Subsequence (LPS) problem to explain how to solve DP in the second case in details.
## 2. The Longest Palindrome Subsequence
We have solve the "Longest Palindrome Substring" problem before. This time, the difficulty is increased by finding the length of the Longest Palindrome Subsequence instead of substring:
In this question, we define `dp[i][j]` as the length of the longest palindrome subsequence within the substring `s[i..j]`. Please remember this definition so as to understand the algorithm.
Why do we define a two-dimensional DP array like this? We mentioned many times before that finding state transition relation requires inductive thinking. To put it plainly, it is how we derive unknown parts from known results, which makes it easy to generalize and discover the state transition relation.
Specifically, if we want to find `dp[i][j]`, suppose you have already got the result of the subproblem `dp[i+1][j-1]` (the length of the longest palindrome subsequence in`s[i+1..j-1]`), can you find a way to calculate the value of`dp[i][j]`(the length of the longest palindrome subsequence in`s[i..j]`) ?
The answer is yes! It depends on the characters of `s[i]` and `s[j]`:
If they are equal, then the longest palindrome subsequence in `s[i+1..j-1]` would be these two characters plus the longest palindrome subsequence in `s[i..j]`:
If they are not equal, it means that they cannot appear at the same time in the longest palindrome subsequence of `s[i..j]`. Therefore, we add them separately to `s[i+1..j-1]` to see which substring produces a longer palindrome subsequence:
The code of the above two cases can be written like this:
if (s[i] == s[j])
// These two chars must be in the longest palindrome sequence
dp[i][j] = dp[i + 1][j - 1] + 2;
else
// Choose the longer palindrome subsequence from s[i+1..j] and s[i..j-1]
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
At this point, the state transition equation is derived. According to the definition of the DP array, what we require is `dp[0][n-1]`, which is the length of the longest palindrome subsequence of the entire `s`.
## 3. Code Implementation
Let's begin with defining the base case. If there is only one character, the longest palindrome subsequence length is 1, which can be represented as `dp[i][j] = 1 (i == j)`.
Since `i`must be less than or equal to `j`, for those locations where `i > j`, there are no subsequences at all and thus should be initialized to 0.
In addition, look at the state transition equation we just got. To find `dp[i][j]`, you need to know `dp[i+1][j-1]`, `dp[i+1][j]` and`dp[i][j -1]` these three values. And look at the base case we determined, this is how the DP array looks like after being filled:
In order to guarantee that before each calculation of `dp[i][j]`, the values in the left, down and right direction have been calculated, we can only traverse it diagonally or reversely:
Here I choose to traverse reversely. The code is as follows:
int longestPalindromeSubseq(string s) {
int n = s.size();
// DP arrays are all initialized to 0
vector<vector<int>> dp(n, vector<int>(n, 0));
// base case
for (int i = 0; i < n; i++)
dp[i][i] = 1;
// Reverse traversal to ensure correct state transition
for (int i = n - 1; i >= 0; i--) {
for (int j = i + 1; j < n; j++) {
// State transition equation
if (s[i] == s[j])
dp[i][j] = dp[i + 1][j - 1] + 2;
else
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
}
}
// return the length of LPS
return dp[0][n - 1];
}
So far, the longest palindrome subsequence problem has been solved. | 1,870 | 7,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | longest | en | 0.947374 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=60&t=53822&p=198374 | 1,591,124,801,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425481.58/warc/CC-MAIN-20200602162157-20200602192157-00358.warc.gz | 412,265,735 | 10,956 | ## pH Calculation explanation
Sydney Pell 2E
Posts: 100
Joined: Wed Sep 11, 2019 12:17 am
### pH Calculation explanation
Can someone explain why you have to take the negative log of the H+ concentration to find pH? Why is it negative?
KDang_1D
Posts: 127
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 1 time
### Re: pH Calculation explanation
Concentrations for H3O+ (and OH-) are very small, so they will have a negative exponent in their molarity value. For example: [H3O+] = 1x10^-7 M. Since logarithms calculate the value of the exponent, log[H3O+] = -7. Negative numbers aren't as nice to work with as positives, so we just add the negative in front of the log to get a pH of 7.
Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”
### Who is online
Users browsing this forum: No registered users and 2 guests | 241 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | longest | en | 0.86565 |
https://www.calculateme.com/temperature/fahrenheit-to-celsius/4100 | 1,701,364,386,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00693.warc.gz | 775,933,964 | 4,269 | # Convert 4,100 Fahrenheit to Celsius
What is 4,100 Fahrenheit in Celsius? How hot is 4,100 degrees Fahrenheit? Translate 4,100° from F to C.
Degrees
From
To
swap units ↺
4,100 °Fahrenheit =
2260 °Celsius
exact result
Fahrenheit is a scale commonly used to measure temperatures in the United States.
Celsius, or centigrade, is used to measure temperatures in most of the world. Water freezes at 0° Celsius and boils at 100° Celsius.
Inverse Conversion
#### Conversion Formula
°Celsius = 5 * (°Fahrenheit - 32) 9
Nearby Values
(results may be rounded)
°Fahrenheit °Celsius
4,100 2,260.0
4,101 2,260.6
4,102 2,261.1
4,103 2,261.7
4,104 2,262.2
4,105 2,262.8
4,106 2,263.3
4,107 2,263.9
4,108 2,264.4
4,109 2,265.0
4,110 2,265.6
4,111 2,266.1
4,112 2,266.7
4,113 2,267.2
4,114 2,267.8
4,115 2,268.3
4,116 2,268.9
4,117 2,269.4
4,118 2,270.0
4,119 2,270.6
4,120 2,271.1
4,121 2,271.7
4,122 2,272.2
4,123 2,272.8
4,124 2,273.3
4,125 2,273.9
4,126 2,274.4
4,127 2,275.0
4,128 2,275.6
4,129 2,276.1
4,130 2,276.7
4,131 2,277.2
4,132 2,277.8
4,133 2,278.3
°Fahrenheit °Celsius
4,134 2,278.9
4,135 2,279.4
4,136 2,280.0
4,137 2,280.6
4,138 2,281.1
4,139 2,281.7
4,140 2,282.2
4,141 2,282.8
4,142 2,283.3
4,143 2,283.9
4,144 2,284.4
4,145 2,285.0
4,146 2,285.6
4,147 2,286.1
4,148 2,286.7
4,149 2,287.2
4,150 2,287.8
4,151 2,288.3
4,152 2,288.9
4,153 2,289.4
4,154 2,290.0
4,155 2,290.6
4,156 2,291.1
4,157 2,291.7
4,158 2,292.2
4,159 2,292.8
4,160 2,293.3
4,161 2,293.9
4,162 2,294.4
4,163 2,295.0
4,164 2,295.6
4,165 2,296.1
4,166 2,296.7
4,167 2,297.2
°Fahrenheit °Celsius
4,168 2,297.8
4,169 2,298.3
4,170 2,298.9
4,171 2,299.4
4,172 2,300.0
4,173 2,300.6
4,174 2,301.1
4,175 2,301.7
4,176 2,302.2
4,177 2,302.8
4,178 2,303.3
4,179 2,303.9
4,180 2,304.4
4,181 2,305.0
4,182 2,305.6
4,183 2,306.1
4,184 2,306.7
4,185 2,307.2
4,186 2,307.8
4,187 2,308.3
4,188 2,308.9
4,189 2,309.4
4,190 2,310.0
4,191 2,310.6
4,192 2,311.1
4,193 2,311.7
4,194 2,312.2
4,195 2,312.8
4,196 2,313.3
4,197 2,313.9
4,198 2,314.4
4,199 2,315.0 | 1,169 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.468423 |
http://www.convertit.com/Go/weighing-systems/Measurement/FAQ.ASP?Topic=25 | 1,660,949,687,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573849.97/warc/CC-MAIN-20220819222115-20220820012115-00110.warc.gz | 60,463,220 | 2,790 | weighing-systems.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Convert Linear Feet to Square Feet and Cubic Feet
How do I convert linear feet to square feet and cubic feet?
There is no standard formula that applies to all measurements involving linear feet. The conversion depends upon what is being measured. For example, to convert 12 linear feet of files boxes into cubic feet, you would also need to know the width and height of the file boxes. Assuming that they are 1.25 feet wide and 1 foot tall, then 12 linear feet of them would be 15 cubic feet of file boxes (1 foot times 1.25 feet times 12 feet). | 151 | 686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-33 | latest | en | 0.883485 |
https://911papers.com/operation-management-academic-essay/ | 1,627,319,540,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00224.warc.gz | 101,951,885 | 9,747 | Operation Management
A heat-treatment operation takes 6 hours to process a batch of parts with a standard deviation of 3 hours. The maximum that the oven can hold is 125 parts. Currently there is demand for 160 parts per day (16-hour day). These arrive to the heat-treatment operation one at a time according to a Poisson stream (i.e., with Ca = 1).
a) What is the maximum capacity (parts per day) of the heat-treatment operation?
b) If we were to use the maximum batch size, what would be the average cycle time through the operation?
c) What is minimum batch size that will meet demand?
d) If we were to use the minimum feasible batch size, what would be the average cycle time through the operation?
e) Find the batch size that minimize cycle time. What is the resulting average cycle time?
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Consider a workstation with 11 machines (in parallel), each requiring one hour process time per job with Ce2 = 5. Each machine costs \$10,000. Orders for job arrive at a rate of 10 per hour with Ca2 =1 and must be filled. Management has specified a maximum allowable average response time (i.e., time a job spends at the station) of 2 hours. Currently it is just over 3 hours. Analyze the following options for reducing average response time.
a) Perform more preventive maintenance so that mr and mf are reduced, but mr/mf remains the same. This costs \$8,000 and does not improve the average process time but does reduce Ce2 to one.
b) Add another machine to the workstation at a cost of \$10,000. The new machine is identical to existing machines, so te = 1 and Ce2 = 5.
c) Modify the existing machines to make them faster without changing the SCV, at a cost of \$8,500. The modified machines would have te = 0.96 and Ce2 = 5. What is the best option?
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Divergence is more general than described
I changed the maths rating from
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This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
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High Importance
Field: Basics
One of the 500 most frequently viewed mathematics articles.
This article should be about divergence in a more general setting (as opposite of convergence than about it in an analysis alone,
Hope someone will improve it WillemienH (talk) 11:13, 5 February 2015 (UTC)
Non-general definition
The definition of divergence, it seems to me, should be defined for any amount of variables. Am I wrong in this, or can a four variable function have a divergence as well? Fresheneesz 20:48, 10 February 2006 (UTC)
The divergence is of course defined in any dimension. But it makes most sense in 3D only, as ther it behaves nicely with the curl, and has a physical interpretation. I will now add a mention of the general case. I would be opposed to rewriting this article from the n-dimenional perspective, as a lot of the physics would be lost, I think. Oleg Alexandrov (talk) 02:59, 11 February 2006 (UTC)
Initial example
The example of water in a bathtub can be confusing. Water does not 'vanish', and the divergence of water flow is in fact very close to zero. I've changed this to air expanding to some level which is a better example. PCM —This unsigned comment is by Paul Matthews (talkcontribs) .
On further reflection, I think you are right. Oleg Alexandrov (talk) 02:09, 15 March 2006 (UTC)
actually i want to ask that if ▼.B is equal to B.▼.... if it is i just cant understand its physical interpretation..as ▼.B means a operator is operating over B but what then B.▼ implies?? 210.212.8.61 17:18, 6 February 2007 (UTC)
You're wrong, ▼.B != B.▼ — Preceding unsigned comment added by Thedoctar (talkcontribs) 08:22, 7 January 2013 (UTC)
definition
I think definition should be as general as possible. therefore,
${\displaystyle \nabla \cdot \mathbf {F} =\lim _{V\rightarrow 0}{\frac {\oint _{s}\mathbf {F} d\mathbf {s} }{V}}}$
Eventually, in 3D orthogonal system, this expression becomes the known one.
Nevo taaseh 11:32, 8 April 2007 (UTC)
I think it is preferrable that definitions start simple, not general. Things can be generalized later I believe, once the basic concept is clear. No? :) Oleg Alexandrov (talk) 15:06, 8 April 2007 (UTC)
I definitely agree with the concept of simpler things first. But, I would argue that an arbitrary V is actually simpler then that of a sphere. Barring that a rectangular parallel piped of Volume dxdydz would be simpler. This volume has the advantage that it is much easier to demonstrate the divergence theorem with then spherical coordinates. TStein (talk) 13:29, 10 April 2009 (UTC)
Abuse of notation
Is ·F actually an abuse of notation? is a vector and can operate on scalar or vector quantities. If it operates on a vector, then one needs to know if it performs a scalar product or a cross product, see curl (xF).
I agree that it is not an abuse of notation. ALittleSlow 02:02, 30 September 2007 (UTC)
I think it's totally an abuse of notation. The dot product is just an inner product, so the elements on the left and right of the dot must be in the same vector space. Only when you choose an orthonormal basis for R^n is the dot product defined by the familiar formula given here. is clearly not an element of R^n, which is why this expression is an abuse of notation. 68.73.194.10 (talk) 18:14, 6 May 2012 (UTC)
I get the logic that the representation of an operator should ideally be something like div(f), but is it not still perfectly true to write that div(f) = .f, and also div(f) = .f ?
Can someone expand on the comment above that " is clearly not an element of R^n"? Specifically, on the linked WP article on del, it is currently written: Though this page chiefly treats del in three dimensions, this definition can be generalized to the n-dimensional Euclidean space Rn.
—DIV (137.111.13.4 (talk) 23:43, 1 April 2014 (UTC))
Visualisation
This article could do with a nice visualization as in the gradient article —Preceding unsigned comment added by 137.205.125.78 (talk) 09:47, 26 October 2007 (UTC) sad —Preceding unsigned comment added by 207.148.171.2 (talk) 22:15, 7 April 2008 (UTC)
WikiProject class rating
This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 09:48, 10 November 2007 (UTC)
More definition concerns
From the definition section:
Let x, y, z be a system of Cartesian coordinates..
AFAIK the divergence operator can be defined in other coordinates, right?Mathchem271828 (talk) 06:59, 1 June 2008 (UTC)
scalar field
Should it say somewhere that the result is a scalar field not just a scalar? --Kupirijo (talk) 23:27, 4 October 2008 (UTC)
The terms 'source' and 'circulation'
Is there a better standard for the terms 'source' and circulation. For instance if
divF = D, and
curlF = C,
then what do we call D and C?
If we call D the 'source' then what do we call C? The term 'circulation' seems to be used already for the line integral around a given path.
I am biased toward using 'outflow source' for D and 'circulation source' for C as that fits well with how I teach the Electricity and Magnetism course I teach. (In this pedagogical approach the divergence is defined by the divergence theorem (fundamental theorem of the divergence) as the outflow density while the curl is defined as a circulation density by the fundamental theorem of the curl. Even I must admit, though, that this approach is both wordy and to some extent incomplete. For it is generalizing the term source as anything that 'creates' or perhaps 'induces' the field F yet there can be a field F even when D and C are both zero. (This field is as far as I can tell with my math background limited to physics math is due to some sort of 'strain' type 'zero-trace symmetric tensor source' which can be determined everywhere by the boundary conditions.)
I would like to be able to contribute more of a physics view to the article, but I don't want to ruin it for the mathematicians. Any help with this would be appreciated. TStein (talk) 13:55, 10 April 2009 (UTC)
I may be missing something, but isn't the standard term for D the divergence and C the curl? Why introduce new terms? I think the terms 'source' and 'circulation' should actually refer to the parts of F with zero curl and divergence respectively (actually the zero divergence part is usually called solenoidal). This is made precise through the Helmholtz, or Hodge, or Hodge-Morrey, or some other decomposition. One way to see that D and C both zero does not imply that F is zero is to note that for any harmonic function (harmonic means zero Laplacian) h, the gradient of h has zero curl and zero divergence. It is also possible to make a similar construction based on 2-tensors. This is all related to the deRahm cohomology of the underlying domain. 97.126.82.176 (talk) 17:51, 9 May 2009 (UTC)
Visual aids
I think this article could use some more visual aids. There should, in particular, be a diagram with the graph of a vector field and a circle showing vectors entering and exiting. I can make some Sage pictures if people want. — Preceding unsigned comment added by Wham Bam Rock II (talkcontribs) 21:45, 6 January 2013 (UTC)
Questions about the cartesian coordinates expression for 2nd order tensors
I might be mistaken, but I've seen different cartesian expressions for the divergence of second order tensors from several academic and educational sources. Using Einstein's notation, the one in the article would be
${\displaystyle \left({\overrightarrow {\operatorname {div} }}\,\mathbb {A} \right)_{i}={\frac {\partial A_{ji}}{\partial x_{j}}}}$
It is coherent with the one used by Poinsot and Veynante (trusted authors in combustion sciences) in their combustion textbook. In other sources, such as Sébastien Candel's fluid mechanics textbook (renowned and trusted physicist, known for his contribution to the fields of combustion and acoustics)
${\displaystyle \left({\overrightarrow {\operatorname {div} }}\,\mathbb {A} \right)_{i}={\frac {\partial A_{ij}}{\partial x_{j}}}}$
Both expressions are equivalent for symmetric tensors (which is true for the viscous stress tensor) but not for nonsymmetric ones, of course. I'm guessing the first expression, more coherent with the nabla notation, is the correct one. Would it be relevant to warn readers about this frequent and confusing "mistake"? (Which is, in fact, not exactly one, since it is correct in the cases considered.) M4urice (talk) 10:53, 17 May 2013 (UTC)
The example for a 2nd order tensor in the main article was based on the second convention but the link to the reference was broken. I have found a more recent version of the reference where I could verify that they also suggest the first convention which in my opinion is clearly more common. So I have fixed the example for the divergence of a 2nd order tensor to use the first convention and updated the broken reference. --Logari81 (talk) 07:34, 11 November 2015 (UTC)
History
Why does this, and related articles, have no mention of the history. Who developed the concept originally? Who devised the notation? —DIV (137.111.13.4 (talk) 23:45, 1 April 2014 (UTC))
More Visualization
One of the problems that I have with the Wikipedia on math topics is they quickly ascend to "Equation Land" leaving virtually everyone else behind on the topics they are trying to describe.
I'm not a math guy but I think that rather than the uber-complex Calculus equations we see something like this may be more relevant to the general masses: http://www2.sjs.org/raulston/mvc.10/topic.6.lab.1.htm
I agree that documenting the precise definitions are important but if we leave the general population behind more high-school and undergrads will abandon the field because they just don't get it.
For example check out "Field 12"... I see the beginning of the understanding of wing tip vortices (yes I understand that this article is about electromagnetics).
— Preceding unsigned comment added by Lesds (talkcontribs) 06:40, 26 February 2015 (UTC)
Divergence is invariant under linear transformations
Divergence is invariant under any invertible linear transformation of the vector field and its domain, not just orthogonal transformations. Let ${\displaystyle M}$ be an invertible linear transformation applied to vector field ${\displaystyle f}$ and its domain. Define ${\displaystyle F=Mf.}$ The precise relationship between the transformed vector field ${\displaystyle F}$ and the original vector field ${\displaystyle f}$ is ${\displaystyle (F\circ M)x=(M\circ f)(x)}$ (since the domain of ${\displaystyle f}$ is also transformed by ${\displaystyle M}$). So the chain rule shows that the relationship between the Jacobian matrices of ${\displaystyle F}$ and ${\displaystyle f}$ is
${\displaystyle DF=M(Df)M^{-1}.}$
So the trace of ${\displaystyle DF}$ must equal the trace of ${\displaystyle Df.}$ Jrheller1 (talk) 19:23, 10 August 2016 (UTC)
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Cheers.—InternetArchiveBot 03:29, 14 December 2016 (UTC) | 3,195 | 13,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "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 | 3 | CC-MAIN-2017-51 | latest | en | 0.943429 |
https://www.physicsforums.com/threads/kinematics-question.788912/ | 1,511,512,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807146.16/warc/CC-MAIN-20171124070019-20171124090019-00467.warc.gz | 829,675,558 | 17,225 | # Kinematics Question
Tags:
1. Dec 23, 2014
### Ryan Hwang
1. The problem statement, all variables and given/known data
Given v0=0 m/s, x0=0 m, and t=10 s, use all three kinematic equations together to find xf.
2. Relevant equations
vf=v0+at
xf=x0+v0t+1/2at2
vf2=v02+2a(xf-x0)
3. The attempt at a solution
The more I look at this, the less sense it makes. First off, I tried to plug the values into the first equation, but I don't have vf or a. The second equation, I don't have an a. The last one, I don't have a f nor acceleration. I can't see how I can solve this question without either the final velocity or acceleration.
2. Dec 23, 2014
### QuantumCurt
I agree. This seems like an incomplete question. Is this out of a textbook? Is there a diagram or some other description in the heading for the section that provides more information? Is this something that is falling as a result of gravity?
3. Dec 23, 2014
### Ryan Hwang
Nothing else is stated other than what is provided. It's actually a problem from a summer assignment that is due the first day of the semester. There's no other information in the header, as it just provides a brief overview of instructions for the section, which is "Algebra", and it basically covers all of physics, from kinematics to currents and circuits. It doesn't say that it's falling due to gravity, but if it were, then it would be solvable, but wouldn't require all 3 equations.
4. Dec 23, 2014
### Staff: Mentor
They want the answer in terms of a.
Chet
5. Dec 23, 2014
### Ryan Hwang
If that were the case, then it wouldn't require all three equations would it? Just the second one, no?
6. Dec 23, 2014
### Staff: Mentor
Yes. You are not given enough info to do more.
7. Dec 23, 2014
### Ryan Hwang
Thanks for all the help! :) | 489 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-47 | longest | en | 0.951329 |
http://www.edupil.com/question/20-32-45-59-74-210-195-175-150-120/ | 1,477,454,930,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720615.90/warc/CC-MAIN-20161020183840-00133-ip-10-171-6-4.ec2.internal.warc.gz | 422,951,798 | 18,769 | # Find the next term to fill the blank space indicated by ‘?’ mark :
1. 20, 32, 45, 59, 74, ?
A) 95 B) 90
C) 85 D) 79
2. 210, 195, 175. 150, 120, ?
A) 75 B) 80
C) 85 D) 90
3. 3, 5, 10, 12, 24, 26, ?
A) 52 B) 30
C) 28 D) 48
4. 3, 8, 15, 24, 35, ?
A) 42 B) 48
C) 49 D) 53
5. 5, 16, 49, 148, ?
A) 440 B) 441
C) 443 D) 445
Anurag Mishra Professor Asked on 24th January 2016 in
1. (B) 90
Explanation:-
20 + 12 = 32, 32 + 13 = 45, 45 + 14 = 59, 59 + 15 = 74, Same as, 74 + 16 = 90.
Then, the next term is 90.
Hence, the correct answer is option (B) 90.
2. (C) 85
Explanation:-
210 – 15 = 195, 195 – (15 + 5) = 195 – 20
= 175,
175 – (20 + 5)
= 175 – 25
= 150,
150 – (25 + 5)
= 150 – 30
= 120,
120 – (30 + 5)
= 120 – 35
= 85,
Then, the next next term is 85.
Hence, the correct answer is option (C) 85.
3. (A) 52
Explanation:-
3 + 2 = 5 & 5 x 2 = 10, 10 + 2 = 12 & 12 x 2 = 24, 24 + 2 = 26 & 26 x 2 = 52.
Then, the next term is 52.
Hence, the correct answer is option (A) 52.
4. (B) 48
Explanation:-
3 + 5 = 8, 8 + (5 + 2) = 8 + 7 => 15,
15 + (7 + 2) => 15 + 9 = 24,
24 + (9 + 2) = 24 + 11 => 35,
35 + (11 + 2) = 35 + 13 => 48.
Then, the next term is 48.
Hence, the correct answer is option (B) 48.
5. (D) 445
Explanation:-
5 x 3 + 1 = 15 + 1
= 16,
16 x 3 + 1 = 48 + 1
= 49,
49 x 3 + 1 = 147 + 1
= 148,
148 x 3 + 1 = 444 + 1
= 445.
Then, the next term is 445.
Hence, the correct answer is option (D) 445.
Anurag Mishra Professor Answered on 24th January 2016. | 799 | 2,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-44 | longest | en | 0.274383 |
http://www.techiedelight.com/print-bottom-view-of-binary-tree/ | 1,526,859,386,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00116.warc.gz | 485,284,702 | 17,432 | # Print Bottom View of Binary Tree
Given an binary tree, print bottom view of it. Assume, the left and right child of a node makes 45 degree angle with the parent.
For example, Bottom view of below tree is 7, 5, 8, 6
This problem can be easily solved with the help of Hashing. The idea is to create an empty map where each key represents the relative horizontal distance of the node from the root node and value in the map maintains a pair containing node’s value and its level number. Then we do a pre-order traversal of the tree and if current level of a node is more than or equal to maximum level seen so far for the same horizontal distance as current node’s or current horizontal distance is seen for the first time, we update the value and the level for current horizontal distance in the map. For each node, we recurse for its left subtree by decreasing horizontal distance and increasing level by 1 and recurse for right subtree by increasing both level and horizontal distance by 1.
Below figure shows horizontal distance and level of each node in above binary tree. The final values in the map will be
(horizontal distance -> (node’s value, node’s level))
-1 -> (7, 4)
0 -> (5, 3)
1 -> (8, 4)
2 -> (6, 3)
## C++
The time complexity of above solution is O(n) and auxiliary space used by the program is O(n).
Exercise: Modify the solution to print top view of binary tree
Please use ideone or C++ Shell or any other online compiler link to post code in comments.
Like us? Please spread the word and help us grow. Happy coding 🙂
Get great deals at Amazon | 360 | 1,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.861264 |
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A302918 Number of nonequivalent minimal total dominating sets in the n-cycle graph up to rotation. 1
0, 0, 1, 1, 1, 2, 1, 1, 2, 3, 2, 4, 3, 4, 6, 7, 7, 10, 11, 17, 19, 23, 28, 38, 46, 60, 75, 96, 120, 160, 197, 257, 327, 420, 539, 701, 892, 1155, 1488, 1928, 2479, 3220, 4148, 5381, 6961, 9030, 11687, 15183, 19673, 25563, 33174, 43128, 56010, 72864, 94719 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,6 LINKS Eric Weisstein's World of Mathematics, Cycle Graph Eric Weisstein's World of Mathematics, Total Dominating Set FORMULA a(n) = (1/n) * Sum_{d|n} phi(n/d) * A300738(d). PROG (PARI) NecklaceT(v)={vector(#v, n, sumdiv(n, d, eulerphi(n/d)*v[d])/n)} NecklaceT(concat([0, 0], Vec((3 + 4*x + 5*x^2 + 6*x^3 - 8*x^5 - 9*x^6)/((1 - x^2 - x^3)*(1 + x^2 - x^6)) + O(x^50)))) CROSSREFS Cf. A300738. Sequence in context: A087917 A087741 A054991 * A316149 A047071 A124287 Adjacent sequences: A302915 A302916 A302917 * A302919 A302920 A302921 KEYWORD nonn AUTHOR Andrew Howroyd, Apr 15 2018 STATUS approved
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Last modified November 16 17:32 EST 2018. Contains 317275 sequences. (Running on oeis4.) | 575 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.550003 |
http://www.cplusplus.com/forum/beginner/83995/ | 1,503,373,575,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00103.warc.gz | 522,571,344 | 6,982 | ### Confused about test score curve
closed account (Ebf21hU5)
I have been able to execute each part of this assignment except for the grading curve. We have to get 5 test scores from user input, configure the highest score under a function called findHighest, and then have the program execute the grade curve for each test score. Here are the instructions:
b) void Curve() is passed the 5 test scores, as 5 reference parameters. The function first calls the function findHighest to get the largest value.
It then curves each test score, by dividing it by the largest value.
Here's what I have. Any suggestions are greatly appreciated.
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768`` ``````#include #include using namespace std; int main() { return 0; } float getScore() { float testScores, sum; for (int n=1; n<6; n++) { cout << "Please enter your test score " << n << endl; cin >> testScores; if (testScores<0 || testScores>100) { cout << "That input is invalid. Score must be between 0 and 100." << endl; break; sum += testScores; } } system("PAUSE"); return testScores; } float findHighest() { float Score1, Score2, Score3, Score4, Score5; cout << "Enter test score 1 " << endl; cin >> Score1; cout << "Enter test score 2 " << endl; cin >> Score2; cout << "Enter test score 3 " << endl; cin >> Score3; cout << "Enter test score 4 " << endl; cin >> Score4; cout << "Enter test score 5 " << endl; cin >> Score5; float highestScore = Score1; if (Score2>highestScore) highestScore = Score2; if (Score3>highestScore) highestScore = Score3; if (Score4>highestScore) highestScore = Score4; if (Score5>highestScore) highestScore = Score5; return highestScore; } void Curve(float Score1,float Score2,float Score3,float Score4,float Score5) { findHighest(); float Curve1 = Score1/highestScore; float Curve2 = Score2/highestScore; float Curve3 = Score3/highestScore; float Curve4 = Score4/highestScore; float Curve5 = Score5/highestScore; cout << "The curves for each test score are : " << endl; cout << " " << Curve1 << Curve2 << Curve3 << Curve4 << Curve5 << endl; return 0; }``````
Last edited on
Suggestions for what? Is the program working? If not, what are the indications and/or error messages that it isn't?
Only thing I can see is that the statement where you sum the test scores is inside the if loop testing to see if the inputted score is outside the 0-100 range. I'm assuming you want it to only sum the good scores, so that should prolly be after the close brace of that loop. Also, you have NOTHING in your main(). Well, except return 0;, so that's the ONLY thing this program will do. You have to call any functions from within main. The declarations and such can be outside, but at least the function calls must be within main, unless your calling a functions from within a function.
Oh, and you are entering the scores in both the getScore() and findHighest() functions. Seems odd to be doing it in both, or rather not odd but inefficient.
What you could just do is use an int array (if you've covered them that is) or declare the 5 test score variables inside main so they have scope over all the functions. This way you'd only have to enter them once. And an array would be pretty easy to sum over, and also to compare scores to find the highest.
I honestly don't see how this executed at all though tbh without any calls in main. :)
Last edited on
closed account (Ebf21hU5)
It didn't execute. I had 6 errors but all 6 of them were referring to "highestScore" in the void Curve() function not being declared, nothing serious. I have a hard time grasping the concepts of c++ so that's why some of the concepts I use in my coding are strange. Thank you. I will take what you said, apply it to my code and repost it.
Last edited on
closed account (Ebf21hU5)
I included main just because you kinda have to but my functions are not included in main because they have their own local definitions. I'm not really sure I need main to do anything. I am only getting one error now which is C2448: 'Curve' : function-style initializer appears to be a function definition. How should I go about fixing this.
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768`` ``````#include #include using namespace std; float Score1, Score2, Score3, Score4, Score5; float highestScore = Score1; int main() { return 0; } float getScore() { float testScores; for (int n=1; n<6; n++) { cout << "Please enter your test score " << n << endl; cin >> testScores; if (testScores<0 || testScores>100) { cout << "That input is invalid. Score must be between 0 and 100." << endl; break; } } system("PAUSE"); return testScores; } float findHighest() { cout << "Enter test score 1 " << endl; cin >> Score1; cout << "Enter test score 2 " << endl; cin >> Score2; cout << "Enter test score 3 " << endl; cin >> Score3; cout << "Enter test score 4 " << endl; cin >> Score4; cout << "Enter test score 5 " << endl; cin >> Score5; if (Score2>highestScore) highestScore = Score2; if (Score3>highestScore) highestScore = Score3; if (Score4>highestScore) highestScore = Score4; if (Score5>highestScore) highestScore = Score5; return highestScore; } void Curve(Score1,Score2,Score3,Score4, Score5) { findHighest(); float Curve1 = Score1/highestScore; float Curve2 = Score2/highestScore; float Curve3 = Score3/highestScore; float Curve4 = Score4/highestScore; float Curve5 = Score5/highestScore; cout << "The curves for each test score are : " << endl; cout << " " << Curve1 << Curve2 << Curve3 << Curve4 << Curve5 << endl; return ; }``````
Last edited on
I'd suggest re-reading whatever source material you used to come up with this program because there are major errors in that speak to a fundamental lack of understanding in general. While a lot of the code is valid, I'm not sure how you got it while having no understanding of the basic structure of a C++ program. From everything that I have EVER seen, you absolutely MUST use main to at least start your program as that is the function that is called by the OS when the program actually starts. Don't believe me? Try commenting out the entire Curve() function and running it. I'm not trying to be harsh, just trying to convey the need to walk before you run, especially when learning something like C++.
closed account (Ebf21hU5)
If I had no understanding of the basic structure of a program I wouldn't have been able to write as much of the code as I have let alone have any of it be valid. I'm not trying to be harsh either. Thank you for your input but I posted this program for assistance not for an opinion.
Specifically, for console applications, main() is defined as the entry-point to the application (where execution begins), and when execution reaches the end of main() execution then ends. In the code you posted, this is the entirety of what actually happens when your application is run:
1. Execution begins at the entry point, main(), defined as:
``1234`` ``````int main() { return 0; }``````
2. Main immediately returns 0, as it says, ending the main() function.
3. Execution terminates, as it has reached the end of the main() function. No other functions are called nor run.
In order for other code to be run you need to call other functions from inside main() or do something else which passes execution to elsewhere in the code, which you never do.
Some other problems I see:
``12`` ``````float Score1, Score2, Score3, Score4, Score5; float highestScore = Score1;``````
The definition of highestScore makes no sense, because at that point Score1 has no definition. You are assigning highestScore to something completely undefined. You should just leave it undefined in that definition, and assign highestScore to something after the other variables actually have a definition of their own. On top of that, using global variables (Variables which are defined outside of functions, classes or namespaces) is generally a very bad practice. You should make a habit of defining them inside the function, class or namespace ("scope") where they are relevant.
Some of your fundamental understandings of the language were flawed when you wrote this, although not horribly so. I'd recommend doing a little bit more reading on the basics and then giving it another shot.
edit: If you're going to just say "No I'm not wrong I'm right!" when asking for help, I'd recommend you not bother asking in the first place.
Last edited on
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Home / Tag Archives: slab
# Tag Archives: slab
## Bar bending schedule of 2 way slab
Bar bending schedule of two way slab and How to calculate the cutting length and weight of rebar’s in 2 way slab? In this Article we will calculate the cutting length and prepare the B.B.S of the 2 way slab as shown below. Let us redraw the above drawing with …
## Bar Bending Schedule B.B.S
Bar Bending Schedule ( B.B.S ) Estimate of The Steel in Building Constructions Introduction to B.B.S B.B.S The word B.B.S Plays a significant role in any Constructions of the High rise buildings. B.B.S refers to B.B.S. Well, What is the use of B.B.S ? Why we use B.B.S ? What …
## Thumb Rules For Civil Engineers In Building Construction
Thumb Rules For Civil Engineers In Building Construction Thumb Rules For Civil Engineering is essential for any Civil Engineer, Site Engineers, or Civil Supervisor. They play a crucial role while taking quick decisions on site. There is some Civil Engineering Basic Knowledge that every Civil Engineers must know about this Riles. The Thumb Rule of Civil Engineering or the …
## What is the Difference between main bars and distribution bars in slab
What is the Difference between main bars and distribution bars in slab Let us go through some of the FAQs to understand the main bars & distribution bars in a slab. 1. What are the main bars in a slab .? The reinforcement bars that are placed in the tension …
## What Is The Best Concrete Mix Ratio For Roof Slab, Beam & Column
What Is The Best Concrete Mix Ratio For Roof Slab, Beam & Column In this article we know about best mix ratio for concreting work like roof slab, column and beam for any type of residential and commercial building. As we know concrete mix ratio is design according to load …
## What Is The Hidden Beam or Concealed Beam – Purpose & Advantages
What Is The Hidden Beam or Concealed Beam – Purpose & Advantages Hidden beams is defined as the beams whose depth is equal to the thickness of the slab. It also called as concealed beam provided within the depth of supporting slabs. It is popular and form an essential part …
## Size Of Aggregate Used In RCC And PCC.
Size Of Aggregate Used In RCC, PCC, Slab, Road, Bridge And Dams All types of aggregate made of crushed of igneous rock granite rock sedimentary rock and Metamorphic rock, all type of aggregate unit in different purpose for building construction pavement, roads, Street and building and interior and exterior designing …
## Why Crank Bars Are Provided In The Slab Construction
Why Crank Bars Are Provided In The Slab Construction Now, let us look into some of the FAQs related to crank bars or bent-up bars in slab construction. 1. Why crank bars are provided in slab construction? When the dead load of the slab and uniform live load over it … | 620 | 2,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-10 | longest | en | 0.859274 |
http://www.cfd-online.com/W/index.php?title=Lewis_number&oldid=9444 | 1,464,433,273,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277592.42/warc/CC-MAIN-20160524002117-00215-ip-10-185-217-139.ec2.internal.warc.gz | 410,519,016 | 9,952 | # Lewis number
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The Lewis number for a given species $k$ is
$Le_k \equiv \frac{\lambda}{\rho C_p D_k}$
Denoting $D_{th}= \lambda / \rho C_p$ the heat diffusivity coefficient the Lewis number can be expressed as
$Le_k \equiv \frac{D_{th}}{ D_k}$
which is the ratio of the heat diffusion speed to the diffusion speed of species $k$.
In many combustion models, all species are assumed to diffuse at the same speed and therefore $Le = 1$ | 142 | 514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | longest | en | 0.835804 |
https://notebooks-prod.herokuapp.com/previews?key=557bd955ed001022f30003e8%2Fe7fd9a90-6d2e-4f74-8834-c41960632ac0%2Fpreview.html&vf=false | 1,586,164,048,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00177.warc.gz | 530,720,215 | 626,801 | Notebook
Alphalens Example Tear Sheet¶
Alphalens is designed to aid in the analysis of "alpha factors," data transformations that are used to predict future price movements of financial instruments. Alpha factors take the form of a single value for each asset on each day. The dimension of these values is not necessarily important. We evaluate an alpha factor by considering daily factor values relative to one another.
It is important to note the difference between an alpha factor and a trading algorithm. A trading algorithm uses an alpha factor, or combination of alpha factors to generate trades. Trading algorithms cover execution and risk constraints: the business of turning predictions into profits. Alpha factors, on the other hand, are focused soley on making predictions. This difference in scope lends itself to a difference in the methodologies used to evaluate alpha factors and trading algorithms. Alphalens does not contain analyses of things like transaction costs, capacity, or portfolio construction. Those interested in more implementation specific analyses are encouaged to check out pyfolio (https://github.com/quantopian/pyfolio), a library specifically geared towards the evaluation of trading algorithms.
In [17]:
import numpy as np
import pandas as pd
from quantopian.research import run_pipeline
from quantopian.pipeline import Pipeline
from quantopian.pipeline.data.builtin import USEquityPricing
from quantopian.pipeline.factors import CustomFactor, Returns, AverageDollarVolume
from quantopian.pipeline.classifiers.morningstar import Sector
In [18]:
universe_screen = AverageDollarVolume(window_length=20).top(500)
In [19]:
pipe = Pipeline(
columns={
},
screen=universe_screen
)
In [20]:
results = run_pipeline(pipe, '2015-06-30', '2016-06-30')
results = results.fillna(value=0.)
In [21]:
momentum_factor = results["Momentum"]
Out[21]:
2015-06-30 00:00:00+00:00 Equity(2 [AA]) -0.230704
Equity(24 [AAPL]) 0.363660
Equity(62 [ABT]) 0.224101
Equity(64 [ABX]) -0.402359
Name: Momentum, dtype: float64
The pricing data passed to alphalens should reflect the next available price after a factor value was observed at a given timestamp. The price must not be included in the calculation of the factor for that time. Always double check to ensure you are not introducing lookahead bias to your study.
In our example, before trading starts on 2014-12-2, we observe yesterday, 2014-12-1's factor value. The price we should pass to alphalens is the next available price after that factor observation: the open price on 2014-12-2.
In [22]:
assets = results.index.levels[1].unique()
# We need to get a little more pricing data than the
# length of our factor so we can compare forward returns.
# We'll tack on another month in this example.
pricing = get_pricing(assets, start_date='2015-06-30', end_date='2016-07-31', fields='open_price')
In [23]:
pricing.head()
Out[23]:
Equity(2 [AA]) Equity(24 [AAPL]) Equity(62 [ABT]) Equity(64 [ABX]) Equity(67 [ADSK]) Equity(76 [TAP]) Equity(114 [ADBE]) Equity(122 [ADI]) Equity(128 [ADM]) Equity(154 [AEM]) ... Equity(49139 [FIT]) Equity(49141 [CPGX_WI]) Equity(49183 [WRK_WI]) Equity(49209 [BXLT]) Equity(49229 [KHC]) Equity(49242 [PYPL_V]) Equity(49506 [HPE_WI]) Equity(49515 [RACE]) Equity(49563 [SYF_WI]) Equity(49865 [HOT_WI])
2015-06-30 00:00:00+00:00 11.45 125.57 49.51 10.74 51.16 71.35 82.17 64.34 48.87 28.77 ... 35.05 28.67 55.380 NaN NaN NaN NaN NaN NaN NaN
2015-07-01 00:00:00+00:00 11.20 126.85 49.40 10.63 50.09 70.36 81.57 64.89 48.71 28.30 ... 39.29 29.40 54.847 NaN NaN NaN NaN NaN NaN NaN
2015-07-02 00:00:00+00:00 11.09 126.43 49.80 10.50 50.43 70.04 81.19 64.58 48.82 27.88 ... 41.82 NaN NaN 31.59 NaN NaN NaN NaN NaN NaN
2015-07-06 00:00:00+00:00 10.95 124.94 48.85 10.52 49.78 69.40 80.02 63.85 48.16 28.21 ... 41.15 29.56 57.565 30.80 71.00 NaN NaN NaN NaN NaN
2015-07-07 00:00:00+00:00 10.96 125.89 49.96 10.50 51.01 69.39 80.77 63.98 48.11 28.98 ... 42.36 30.85 57.710 31.70 73.99 37.73 NaN NaN NaN NaN
5 rows × 824 columns
Often, we'd want to know how our factor looks across various sectors. To generate sector level breakdowns, you'll need to pass alphalens a sector mapping for each traded name.
This mapping can come in the form of a MultiIndexed Series (with the same date/symbol index as your factor value) if you want to provide a sector mapping for each symbol on each day.
If you'd like to use constant sector mappings, you may pass symbol to sector mappings as a dict.
If your sector mappings come in the form of codes (as they do in this tutorial), you may also pass alphalens a dict of sector names to use in place of sector codes.
In [24]:
MORNINGSTAR_SECTOR_CODES = {
-1: 'Misc',
101: 'Basic Materials',
102: 'Consumer Cyclical',
103: 'Financial Services',
104: 'Real Estate',
205: 'Consumer Defensive',
206: 'Healthcare',
207: 'Utilities',
308: 'Communication Services',
309: 'Energy',
310: 'Industrials',
311: 'Technology' ,
}
In [25]:
sectors = results["Sector"]
Importing Alphalens¶
In [26]:
import alphalens
Formatting input data¶
Alphalens contains a handy data formatting function to transform your factor and pricing data into the exact inputs expected by the rest of the plotting and performance functions. This get_clean_factor_and_forward_returns function is the first call in create_factor_tear_sheet.
In [27]:
factor, forward_returns = alphalens.utils.get_clean_factor_and_forward_returns(momentum_factor,
pricing,
sectors=sectors,
sector_names=MORNINGSTAR_SECTOR_CODES,
days=(1,5,10))
Let's see what that gave us...
In [28]:
factor.head()
Out[28]:
date asset sector
2015-06-30 00:00:00+00:00 Equity(2 [AA]) Basic Materials -0.230704
Equity(24 [AAPL]) Technology 0.363660
Equity(62 [ABT]) Healthcare 0.224101
Equity(64 [ABX]) Basic Materials -0.402359
Name: factor, dtype: float64
In [29]:
forward_returns.head()
Out[29]:
1 5 10
date asset sector
2015-06-30 00:00:00+00:00 Equity(2 [AA]) Basic Materials -0.021834 -0.008734 -0.005939
Equity(24 [AAPL]) Technology 0.010194 -0.001465 0.000119
Equity(62 [ABT]) Healthcare -0.002222 0.002545 0.000667
Equity(64 [ABX]) Basic Materials -0.010242 -0.006518 -0.008845
Equity(67 [ADSK]) Technology -0.020915 0.006099 0.003987
You'll notice that our factor doesn't look much different. The only addition here is an index level describing the sector of each name. That will come in handy as we perform sector level reductions in our performance and plotting functions.
The forward_returns dataframe represents the mean daily price change for the N days after a timestamp. The 1 day forward return for AAPL on 2014-12-2 is the percent change in the AAPL open price on 2014-12-2 and the AAPL open price on 2014-12-3. The 5 day forward return is the percent change from open 2014-12-2 to open 2014-12-9 (5 trading days) divided by 5.
Returns Analysis¶
Returns analysis gives us a raw description of a factor's value that shows us the power of a factor in real currency values.
In [30]:
quantized_factor = alphalens.performance.quantize_factor(factor)
In [31]:
quantized_factor.head()
Out[31]:
date asset sector
2015-06-30 00:00:00+00:00 Equity(2 [AA]) Basic Materials 1
Equity(24 [AAPL]) Technology 5
Equity(62 [ABT]) Healthcare 4
Equity(64 [ABX]) Basic Materials 1
Name: quantile, dtype: int64
One of the most basic ways to look at a factor's predicitve power is to look at the mean return of different factor quantile.
In [32]:
mean_return_by_q_daily, std_err = alphalens.performance.mean_return_by_quantile(quantized_factor, forward_returns,
by_sector=False,
by_time='D')
In [33]:
mean_return_by_q_daily.head()
Out[33]:
1 5 10
date quantile
2015-06-30 00:00:00+00:00 1 -0.004784 -0.004763 -0.004694
2 -0.001886 -0.000536 -0.000296
3 -0.000617 0.000958 0.000358
4 0.002073 0.001552 0.001827
5 0.005207 0.002798 0.002809
In [34]:
mean_return_by_q, std_err_by_q = alphalens.performance.mean_return_by_quantile(quantized_factor,
forward_returns,
by_sector=False)
In [35]:
mean_return_by_q.head()
Out[35]:
1 5 10
quantile
1 0.000152 0.000226 0.000188
2 -0.000284 -0.000234 -0.000213
3 -0.000012 -0.000147 -0.000071
4 0.000129 0.000108 0.000068
5 0.000015 0.000048 0.000028
In [36]:
alphalens.plotting.plot_quantile_returns_bar(mean_return_by_q);
By looking at the mean daily return by quantile we can get a real look at how well the factor differentiates forward returns across the signal values. Obviously we want securities with a better signal to exhibit higher returns. For a good factor we'd expect to see negative values in the lower quartiles and positive values in the upper quantiles.
In [37]:
alphalens.plotting.plot_quantile_returns_violin(mean_return_by_q_daily);
/usr/local/lib/python2.7/dist-packages/matplotlib/__init__.py:892: UserWarning: axes.color_cycle is deprecated and replaced with axes.prop_cycle; please use the latter.
warnings.warn(self.msg_depr % (key, alt_key))
This violin plot is similar to the one before it but shows more information about the underlying data. It gives a better idea about the range of values, the median, and the inter-quartile range. What gives the plots their shape is the application of a probability density of the data at different values.
In [38]:
quant_return_spread, std_err_spread = alphalens.performance.compute_mean_returns_spread(mean_return_by_q_daily, 5, 1, std_err)
In [39]:
alphalens.plotting.plot_mean_quantile_returns_spread_time_series(quant_return_spread, std_err_spread);
This rolling forward returns spread graph allows us to look at the raw spread in basis points between the top and bottom quantiles over time. The green line is the daily returns spread while the orange line is a 1 month average to smooth the data and make it easier to visualize.
In [40]:
alphalens.plotting.plot_cumulative_returns_by_quantile(mean_return_by_q_daily); | 2,919 | 9,967 | {"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} | 2.703125 | 3 | CC-MAIN-2020-16 | latest | en | 0.790591 |
https://www.noodle.com/learn/details/267027/perms-combs-the-binomial-theorem | 1,529,516,210,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00206.warc.gz | 863,932,103 | 12,545 | ## At A Glance
### Perms & Combs: The Binomial Theorem
http://www.mindbites.com/lesson/5103 This 67 minute perms & combs lesson deals with the Binomial Theorem. This lesson will show you how to find: - a specific term - the middle term - the constant term After this lesson, you will also know how to: - expand a binomial by using combinations to find the coefficients - algebraically find the constant term Sample question: Find the middle term in the expansion of (4a + d)^10. This lesson contains explanations of the concepts and 15 example questions with step by step solutions plus 7 interactive review questions with solutions. Lessons that will help you with the fundamentals of this lesson include: - 165 The Zero, Negative and Rational Exponents (http://www.mindbites.com/lesson/5069) - 520 Combinations Part I (http://www.mindbites.com/lesson/5090) - 530 Pascals Triangle & Pathways (http://www.mindbites.com/lesson/5092)
Length: 04:38
## Contact
Questions about Perms & Combs: The Binomial Theorem | 255 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-26 | latest | en | 0.793312 |
http://www.proteacher.net/discussions/showthread.php?t=60295 | 1,503,337,554,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00468.warc.gz | 664,448,712 | 24,111 | Help with integers! - ProTeacher Community
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Math & Science
Help with integers!
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sw Joined: Dec 2005 Posts: 162 Full Member
sw
Joined: Dec 2005
Posts: 162
Full Member
Help with integers!
10-12-2007, 10:19 AM
#1
My husband teaches two seventh grade math classes that just cannot or will not understand adding, subtracting, multiplying, and dividing integers. Does anyone have any great teaching, game playing, etc. ideas that would help him? He doesn't want to go on until they've learned what they need for the exit exam.
He is going to try playing the card game "War" using a deck of cards without face cards and making black cards positive and red cards negative. Any other suggestions?
KathyB Joined: Oct 2005 Posts: 565 Senior Member
KathyB
Joined: Oct 2005
Posts: 565
Senior Member
10-12-2007, 06:02 PM
#2
I am attaching a worksheet I use to help explain subtracting integers.
Attached Files
SUBTRACTION OF INTEGERS.doc (47.0 KB, 633 views)
KathyB Joined: Oct 2005 Posts: 565 Senior Member
KathyB
Joined: Oct 2005
Posts: 565
Senior Member
10-12-2007, 06:04 PM
#3
This is a booklet I give to my students about integers - if you have any questions feel free to contact me
Attached Files
INTEGERS.doc (43.5 KB, 747 views)
connorsmom020 Joined: Jun 2007 Posts: 1,106 Senior Member
connorsmom020
Joined: Jun 2007
Posts: 1,106
Senior Member
kind of juvenille
10-13-2007, 09:05 AM
#4
This is more elementary level but when I taught 6th & 7th grade Math, we learned integers with two-sided counters(I didn't have any so I made them--laminate a red & yellow sheet of construction paper together and then cut into squares.). Red was negative and yellow was positive. I also used lots of number lines.
kreeves Joined: Jun 2006 Posts: 106 Full Member
kreeves
Joined: Jun 2006
Posts: 106
Full Member
Videos on teaching integer concepts
10-21-2007, 12:37 PM
#5
The Virginia DOE has created videos for teaching various math concepts for middle school math instructors. Follow this link:
http://www.pen.k12.va.us/VDOE/middle-math-strategies/
and then go to the section on "Computation and Estimation" and watch Integer Models Part 1 and Integer Models Part 2. These videos gave me some great ideas for teaching integers with manipulative models. Clearly these kids are going to need some more concrete manipulative experience before they are going to understand it abstractly.
I particularly liked the tile-spacer manipulatives presented on these videos. Great visual connection to the signs of the integers! I went out and bought several bags of tile spacers to use as manipulatives when I teach those concepts next year!
Mathzilla Joined: Oct 2007 Posts: 61 Junior Member
Mathzilla
Joined: Oct 2007
Posts: 61
Junior Member
Integer help
11-01-2007, 08:40 PM
#6
I struggle with the same thing, and have used counters, M&Ms, etc. plus elevators and football yardage to try to get students to visualize what's happening. I made a packet with a flow chart I made up for the kids as a review and reinforcement. I'm too junior a member here to post attachments, but I'll email it if you give me your address.
KathyB Joined: Oct 2005 Posts: 565 Senior Member
KathyB
Joined: Oct 2005
Posts: 565
Senior Member
11-02-2007, 05:17 AM
#7
Would love your packet - Thanks
mrsb4c2003@yahoo.com
hazeleyesinnc Joined: Sep 2005 Posts: 1,054 Blog Entries: 2 Senior Member
hazeleyesinnc
Joined: Sep 2005
Posts: 1,054
Senior Member
packet and I can add something
11-05-2007, 07:40 AM
#8
Would love a copy of your packet- my email addy is hazeleyesinnc@yahoo.com
With subtracting integers use the LCO method Leave Change Opposite- example 5-(-3) Leave the 5 Change the - to + and do the opposite of (-3) .......5+3= 8. I've also told the students when they see the two negative signs make them a positive or +. If I get anything else I'll be sure to pass it on.
sw Joined: Dec 2005 Posts: 162 Full Member
sw
Joined: Dec 2005
Posts: 162
Full Member
Thanks
11-16-2007, 10:07 AM
#9
Thanks for all of the responses. I would love a copy of your packet for my husband, daughter, and son-in-law who all teach math. Please send the info to the English teacher/media specialist @
swindle@tcss.net
Thanks!
dee Joined: Aug 2005 Posts: 6,016 Senior Member
dee
Joined: Aug 2005
Posts: 6,016
Senior Member
I'd love a packet, too!
11-16-2007, 12:31 PM
#10
I teach upper elementary math and would love this!
dapage@comcast.net
Thanks
phelopham Joined: Jan 2008 Posts: 165 Full Member
phelopham
Joined: Jan 2008
Posts: 165
Full Member
Integer Cheer
01-07-2008, 08:18 PM
#11
I teach 8th grade and they still don't get integers. I found TeacherTube.com. Go to teachertube.com and search Integer Cheer...this teacher does this really annoying cheer...but it sticks. My kids griped about having to watch it. I would make them watch it over and over and over--but they left my room chanting the cheer. Now when someone makes a mistake with integers, someone always starts chanting the cheer. I use a ladder metaphor for adding and subtracting. I tried the number line but they would be confused about going left or right, they "get" going up(+) and down(-). I have previously told them that they are on a construction site and the foreman made them dig this really deep hole and then put this really tall ladder in this hole. I always make them start at zero (ground level). Positives they climb up and negatives they climb down.
mmar56 Joined: Oct 2007 Posts: 4 New Member
mmar56
Joined: Oct 2007
Posts: 4
New Member
02-12-2008, 03:19 AM
#12
Mathzilla, I would like to have your packet for teaching integers.
My e-mail: mmartins1956@yahoo.ca
Regards
Maria
dhb719 Joined: Jul 2008 Posts: 1 New Member
dhb719
Joined: Jul 2008
Posts: 1
New Member
I would like a copy of your packet
07-27-2008, 05:36 PM
#13
Mathzilla, I would like to have your packet for teaching integers.
My e-mail: dhboykin@fcps.edu
Regards
Dan
bmb326 Joined: Jan 2009 Posts: 1 New Member
bmb326
Joined: Jan 2009
Posts: 1
New Member
Help With Integers!
01-21-2009, 07:31 AM
#14
Hi Mathzilla,
I would love to have the packet that you've created to help with understanding integers.
If you could send it to me that would be great - kboon@sasktel.net
Thanks, Barbara
missjab Joined: Jun 2009 Posts: 1 New Member
missjab
Joined: Jun 2009
Posts: 1
New Member
06-28-2009, 01:53 PM
#15
Hi Mathzilla,
I would like to have your packet for teaching integers.
My e-mail: jabeenbasheer@gmail.com
Regards
Jab
TeacherDude24 Joined: Aug 2006 Posts: 152 Full Member
TeacherDude24
Joined: Aug 2006
Posts: 152
Full Member
06-28-2009, 02:50 PM
#16
I would like your packet as well. I teach 7th grade, and, time limits us, but even though I have spent all year reinforcing integer rules, a lot still don't get them. I have heard that the 8th grade teachers have a problem with this as well by the end of algebra. They also learned integers in 6th grade too.
mrsf70 Joined: Aug 2007 Posts: 367 Senior Member
mrsf70
Joined: Aug 2007
Posts: 367
Senior Member
06-30-2009, 04:59 AM
#17
Mathzilla, I, too, would love your packet. You can email me at mrsf70@gmail.com. Thanks!
Also, I do something similar to the LCO mentioned above, but I teach students to "add the opposite". We look at ways subtraction is the same as adding a negative number. Most do very well with it.
alaska1 Joined: Jul 2009 Posts: 1 New Member
Joined: Jul 2009
Posts: 1
New Member
07-17-2009, 10:06 PM
#18
I am unable to download your booklet, it just takes me to an untitled page. Is it possible to email it to me or a different link? Thanks I really appreciate your willingness to share.
ms_sdavis Joined: Jul 2009 Posts: 1 New Member
ms_sdavis
Joined: Jul 2009
Posts: 1
New Member
Integer Packet
07-26-2009, 03:46 PM
#19
Mathzilla, please forward an integer packet.
sonya0520@comcast.net
Mrs. H. Guest
Mrs. H.
Guest
Integer Packet
10-18-2009, 06:24 PM
#20
Hi Mathzilla,
Is it possible for me to still get a packet sent to my school email address on your integer help? My email is diane.hartwig@pequannock.org. Thanks for your help.
BostonMath Joined: Jan 2010 Posts: 1 New Member
BostonMath
Joined: Jan 2010
Posts: 1
New Member
Positive and Negative Integer concepts
01-15-2010, 11:59 AM
#21
Some of my students had trouble with the concept of a negative number floating out there by itself.
I use a slightly different example than yours.
There are three people: a digger, a piler and a filler.
The digger digs holes in the ground. The piler, delivers piles of dirt. The filler, you the student, fills the holes in with the dirt and reports what is left over.
Scenario 1: The holes are dug, three piles are delivered. As the filler, take the delivered piles and put them into available holes. What's left over (nothing, all holes are filled).
S2: 3 holes are dug, 4 piles delivered. Left over: one pile.
S3: 4 holes are dug, 3 piles delivered. Left over: one hole.
Now, the holes are written as negative number. Piles as positive.
Give problems with just - and + n;umbers:
-5 holes combined with +7 piles - what do you get? = +2
-3 with -7 with +12 ? = +2
Good luck.
mrsf70 Joined: Aug 2007 Posts: 367 Senior Member
mrsf70
Joined: Aug 2007
Posts: 367
Senior Member
01-16-2010, 04:03 AM
#22
BostonMath, "heaps & holes" is a great visual for students. Great suggestion!
mejtaylo Joined: Mar 2010 Posts: 1 New Member
mejtaylo
Joined: Mar 2010
Posts: 1
New Member
Integers Packet
03-11-2010, 05:35 PM
#23
I would love to have the integers packet. I am a 5th grade teacher and I'm trying to get my kids ready for middle school. You can email me at mejtaylo@yahoo.com.
Thanks so much!!!
crazymompatel Joined: May 2010 Posts: 1 New Member
crazymompatel
Joined: May 2010
Posts: 1
New Member
Integers Packet
05-03-2010, 04:41 AM
#24
I know you posted this years ago but just wondering if you still have that Integers packet. I teach 6th gade math to SB students. They are having a hard time grasping the integers concept. Anything different you can provide would be great. My e-mail is suzanneap@leeschools.net
Thanks.
cilla m Joined: May 2010 Posts: 1 New Member
cilla m
Joined: May 2010
Posts: 1
New Member
05-06-2010, 08:16 AM
#25
Could you please send it to me too?
priscillacastromolina2@yahoo.co m
hallteaches Guest
hallteaches
Guest
LCO method and Integer Booklet
12-04-2011, 11:11 AM
#26
I'm going to try changing my Leave Change Change to Leave Change Opposite. I would greatly appreciate it if you would email the booklet to me for use with my students.
hallteachesmath#bellsouth.net
rahman Guest
rahman
Guest
aman
01-22-2012, 02:45 AM
#27
dont get u so bye
Join the conversation! Post as a guest or become a member today. New members welcome!
>
Math & Science | 3,121 | 10,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-34 | longest | en | 0.942861 |
https://jp.mathworks.com/matlabcentral/cody/problems/576-return-elements-unique-to-either-input/solutions/538785 | 1,579,779,512,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00390.warc.gz | 501,178,601 | 15,781 | Cody
# Problem 576. Return elements unique to either input
Solution 538785
Submitted on 1 Dec 2014 by Gareth Lee
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = [1 2 3 4 5]; b = [2 3 4 5 6]; y_correct = [1 6]; assert(isequal(vector_diff(a,b),y_correct))
2 Pass
%% a = 10:-2:0; b = 1:2:11; y_correct = 0:11; assert(isequal(vector_diff(a,b),y_correct))
3 Pass
%% a=magic(5); b=magic(6); y_correct = 26:36; assert(isequal(vector_diff(a,b),y_correct))
4 Pass
%% a=(10:100)'; b=11:100; y_correct = 10; assert(isequal(vector_diff(a,b),y_correct))
5 Pass
%% a=magic(3)-1.5; b=[]; y_correct = -0.5:7.5; assert(isequal(vector_diff(a,b),y_correct))
6 Pass
%% a=zeros(5); b=ones(3); y_correct=[0 1]; assert(isequal(vector_diff(a,b),y_correct))
7 Pass
%% forbidden = '(regexp)'; assert(isempty(regexp(evalc('type vector_diff'),forbidden))); | 337 | 976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-05 | latest | en | 0.428316 |
https://massimopigliucci.blog/2023/12/exploring-the-nature-of-equally-likely-events-unveiling-the-science-behind-probability/ | 1,719,292,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00697.warc.gz | 345,009,106 | 15,166 | # Exploring the Nature of Equally Likely Events: Unveiling the Science Behind Probability
Welcome to this insightful article that explores the concept of “equally likely” in science. Probability plays a fundamental role in scientific inquiry, and understanding what it means for events to be equally likely is critical to making informed decisions and drawing accurate conclusions. In this article, we will explore the meaning of equal probability, its practical applications, and its importance in scientific investigations.
## The Concept of Equally Likely
Equally likely refers to a situation in which all possible outcomes have the same probability of occurring. In other words, when events are equally likely, no particular outcome is favored over another. This concept is deeply rooted in the principles of probability theory, which provides a framework for quantifying uncertainty and predicting the likelihood of different outcomes.
Equally likely events can be visualized as a fair coin toss, where the chance of getting either heads or tails is exactly 50%. Another familiar example is rolling a fair six-sided die, where the probability of landing on any side is 1/6. These scenarios have equal probability because each possible outcome has an equal chance of occurring.
## Applications of Equally Likely in Science
The concept of equal probability is widely used in a variety of scientific disciplines. In experimental design, researchers often aim to create conditions in which the outcomes of interest are equally likely to occur. This ensures that any observed differences or patterns in the data are not biased by preferential treatment of certain outcomes.
Equally likely outcomes are also critical in sampling techniques. When selecting a representative sample from a larger population, researchers strive to ensure that each member of the population has an equal chance of being included. This helps minimize sampling bias and increases the generalizability of the results.
## Power in data analysis
Equally likely events play an important role in data analysis and statistical inference. In statistical modeling and hypothesis testing, the assumption of equal likelihood often underlies many analytical techniques. Violations of this assumption can lead to biased or unreliable results.
For example, in regression analysis, ordinary least squares assumes that errors are normally distributed and have equal variances. If these assumptions are violated, the estimated model parameters may be biased and the accuracy of statistical inference may be compromised.
## Challenges and Considerations
While the concept of equal likelihood provides a useful foundation for scientific investigations, its practical implementation can be challenging in certain situations. Real-world phenomena often have complex and heterogeneous characteristics, making it difficult to achieve equal likelihood for all possible outcomes.
Moreover, the assumption of equal probability may not hold in cases where there are inherent biases or unequal probabilities associated with different outcomes. In such situations, it is critical to identify and account for these biases to ensure accurate data analysis and interpretation.
## Conclusion
The concept of equally likely events is a cornerstone of probability theory and scientific inquiry. Recognizing and understanding equal probability is essential for designing experiments, analyzing data, and drawing meaningful conclusions. By embracing the principles of equal probability, scientists can minimize bias, increase the validity of their findings, and contribute to the advancement of knowledge in their respective fields.
As we continue to explore the intricacies of the natural world, the concept of equal probability will remain a vital tool in our scientific arsenal, allowing us to navigate uncertainty and make informed decisions based on rigorous evidence.
## FAQs
### What is equally likely?
Equally likely refers to a situation where all possible outcomes have the same probability of occurring.
### How is equally likely related to probability?
Equally likely is a concept in probability theory that assumes all outcomes in a set are equally probable.
### Can you provide an example of an equally likely event?
Sure! Tossing a fair, unbiased coin is an example of an equally likely event. In this case, the probability of getting heads or tails is equal, assuming the coin is not biased.
### What is the significance of equally likely events in probability?
Equally likely events provide a simple and fair way to calculate probabilities. When all outcomes are equally likely, the probability of an event occurring can be determined by dividing the number of favorable outcomes by the total number of possible outcomes.
### Are equally likely events always present in real-life situations?
No, equally likely events are often used as idealized models in probability theory. In reality, many events have different probabilities associated with them due to various factors and conditions.
### How can you determine if events are equally likely?
To determine if events are equally likely, you can analyze the situation and assess whether there are any factors or conditions that could favor certain outcomes over others. Additionally, you can perform experiments or collect data to estimate the probabilities associated with each event. | 937 | 5,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.946406 |
https://math.stackexchange.com/questions/1007585/verifying-if-system-has-periodic-solutions | 1,721,129,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00449.warc.gz | 339,421,782 | 37,585 | # Verifying if system has periodic solutions
Given the following system
$\dot{x} = y$
$\dot{y} = y(9-x^2-2y^2) - x$
verify whether it has periodic solutions and if so are they attracting or repelling.
I thought:
The critical points or fixed point is (0,0) but is this correct and if the answer is yes then is that a periodic solution? And in general, how does one find out the periodic solution?
Finding fixed points is easy, you set $\dot{x}$ and $\dot{y}$ equal to zero and to verify the stability you find the derivatives of the fixed points. But I don't know how to find the periodic solutions...
Maybe rewriting the system in polar coordinates helps somehow?
• Yes, $(0,0)$ is a stationary solution, in particular (trivially) periodic. Linearizing the system around $(0,0)$ shows that it is repelling. Writing it in polar coordinates you will get that $\dot{r}$ is positive for $0<x^2+2y^2<9$, and negative for $x^2+2y^2>9$. Since there are no other stationary points and two-dimensional systems only have limit cycles and stationary attractors, there must be at least one limit cycle. If there is just one, it has to be attracting, but I am not sure how to show this. Commented Nov 5, 2014 at 20:38
• Maybe one should stress more clearly, as already indicated in a comment, that the accepted answer is incomplete.
– Did
Commented Nov 6, 2014 at 20:29
In polar form we have $$r\dot{r} = x\dot{x} +y\dot{y} = r^2\sin^2\theta(9-r^2(1+\sin^2\theta)).$$ Now $$\dot{r} <0 \implies r^2 \ge\frac{9}{1+\sin^2\theta}\ge9$$ and $$\dot{r}>0 \implies r^2\le\frac{9}{1+\sin^2\theta} \le \frac{9}{2}$$ This tells us that orbits cannot leave the annulus $3/\sqrt{2}\le r\le3$.
By Poincare-Bendixson theorem, the limit set of the of orbits entering the annulus must be either a limit cycle, a fixed point, or some sort of homoclinic or heteroclinic connection. It is thus sufficient to show that there are no fixed points in this region.
Returning to the original equations; we find that the fixed point is at $(x,y)=(0,0)$. Which is not inside the annulus, hence the limit set must be a limit cycle, since the orbits can't leave the annulus, we know that the limit cycle must be stable. Hence, there is at least one stable limit cycle and it is inside the annulus.
If a trajectory enters and does not leave a closed and bounded region of phase space which contains no equilibria, then the trajectory must approach a periodic orbit as $t\rightarrow \infty$.
So the points $\theta=n\pi$ are no problem as the trajectories remain in the closed annulus.
• There is only one fixed point, at $(0,0)$, and you still need to rule out the case that there might be more than one limit cycle in the annulus. Commented Nov 6, 2014 at 0:38
• I think this solution gives the right general idea but is missing a couple of details which have bugged me since I first read this problem this morning. For instance, when $\theta = n\pi$ for integral $n$, $\dot r = 0$ always. So it takes more work to show that the flow heads into the annulus, and I don't know quite how to do it, and I don't know if the argument fails because $\dot r = 0$ when $\theta = n\pi$! Good start, though, +1! Cheers! Commented Nov 6, 2014 at 3:27 | 889 | 3,207 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.889798 |
https://www.ratehub.ca/blog/how-much-life-insurance-do-i-need/ | 1,718,493,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00144.warc.gz | 871,082,487 | 65,618 | # How Much Life Insurance Do I Need in Canada?
It's a difficult question that most Canadians struggle with, but we're here to help. Speak with a qualified broker to find out how much coverage you need.
Life insurance is a necessity if your family depends on your salary for housing expenses, servicing debt, and general living. But how much life insurance do you need?
In 2014, a BMO Insurance study found 74% of Canadians had a life insurance policy, but 70% weren’t confident their policy would give them enough to take care of their family if they passed away. So when you ask ‘How much life insurance do I need’ – you’re not alone.
NOTE: If you have a workplace life insurance policy, it usually only covers 2 years of your salary, which is a bonus, but likely not enough life insurance as you'll see below.
## How much life insurance do I need?
There’s no quick answer to how much life insurance you need – your financial situation is different than your siblings and co-workers, so take the time to figure out your situation. A rule of thumb is to make sure your life insurance covers your debts, income, mortgage, and education. This is called the DIME formula.
• Debts: How much do you have?
• Income: The lost income that would need replacing.
• Mortgage: The amount remaining on the mortgage.
• Education: How much your kids will need for post-secondary learning.
Let’s break it down with a real-life examples, to better understand how much life insurance you need.
## The simple life insurance needs calculation
### D(ebt)
We’ll start with how much debt you have, not including your mortgage. This includes any credit cards, lines of credit, car loans, and any student loans. If you expect to go into extra debt in the near future, make sure to include that as well. You’ll also want to include future debts you won’t be around for, such as funeral costs.
For a real-life example, let’s say you have \$25,000 in debt and your funeral estimate is \$15,000.
Debt: \$25,000 + \$15,000 = \$40,000
### I(ncome)
Next, we need to calculate how many years your family would need your support, and multiply your annual salary by the required years. The number of years can be until your child graduates high school, leaves university, or buys their own house – this is up to you.
Let’s say you make \$50,000 per year, and want to support your family until your children graduate college. Let’s estimate that’s around twenty years.
Income: \$50,000 x 20 = \$1,000,000
### M(ortgage)
How much do you owe on your mortgage? You’ll want to life insurance coverage for the entire amount. You might even consider adding more to cover the costs of a major renovation, in case your family outgrows your current house.
For this example, we’ll say you have \$400,000 remaining on your mortgage, but you want expect to add a granny flat in the next five years, at a cost of \$50,000.
Mortage: \$450,000
### E(ducation)
How much will it cost in twenty years to send 2 kids off to college? What if they end up wanting to do their masters, PhD, or even a college diploma? Will they study at home or abroad? These are all costs you’ll want to have covered. To make our number easier, the lifetime limit of an RESP for any beneficiary in Canada is \$50,000.
For two kids, each with maxed out RESP’s would be \$100,000.
Education = \$100,000
If we use the DIME formula, in this scenario, this person would need \$1,590,000 in life insurance coverage for a 20-year term. Now, you just need to get some life insurance quotes.
## Life insurance calculator
Another way is to manually calculate your life insurance needs. First, you can add up financial obligations. Ok, I'll ask that you break out your phone's calculator and maybe a piece of paper to write down your totals. The brain remembers things better when we write them down.
### Does anyone depend on you financially?
If yes, get life insurance and simply follow the formula below to figure out how much life insurance you need to get.
If not, you likely don't need life insurance. However, if you're in a unique health or financial situation, it's easy to compare life insurance quotes with us.
## Manually calculating your life insurance coverage needs
1. How much income does your family need to replace if you're no longer around?
You can either write down your annual after-tax income, but it's probably better to calculate expenses like meals, cars, clothing and services your family may need if you're not around like child care, house cleaners, and home repairs.
2. How many years do you think your family will need with your income?
Eventually the mortgage is paid off and RESPs are funded and your family can sustain themselves on their own. Multiply your answer from #1 with your answer to #2.
3. How much debt do you currently have?
Include your mortgage, car loans, credit cards, and any other personal loans. While they may only use a portion to pay down the mortgage to a manageable level for your spouse, other debts should be cleared.
4. How much money do you want for your kids' university education?
The average annual cost of Canadian Universities is \$22,000 and \$11,000 for those living at home. That number is expected to increase to \$31,000 for 2038.
5. What kind of funeral do you want?
Personal preference plays a big role in how a funeral will cost. If you're getting cremated and a small service, you might pay around \$1,500. But a grand event with many people could cost as much as \$20,000.
6. How much money do you have in savings and investments?
You can subtract the total savings from your running total. Your family can liquidate your investments, too (or choose to keep them for their annual returns).
7. How much life insurance do you have?
Sometimes parents may have purchased a policy for you when you were younger and it could be paid off now, or you're making small monthly payments. You can subtract this amount from your running total.
8. How much annual after-tax income does your family have with your spouses salary?
Your spouse will continue earning. We calculated the household expenses in step 1,, so take this number and multiply it by step 2 and subtract it from your total.
9. Do the math
Step 1 multiplied by Step 2, then add Step 3, 4, 5. Once that's done, subtract Step 6, 7, 8. How much life insurance do you need?
## How much is a 20-year term life insurance policy?
A rough estimate would be about \$30-\$60 per month, but it's cheaper if you're younger, in good health, and a female. Your term life insurance quote will likely be different based on a number of variables.
## Another way to calculate: 10 x your salary
The old rule of thumb was 10 times your salary, but that’s arbitrary. You could calculate the salary you’d make until you retire and your family would certainly be comfortable, especially without your expenses (extra car, mouth to feed, clothing, vacations, etc.).
But, for simplicity, If you’re forty years old, making \$50,000 per year and plan on retiring at 65, let’s do the math. 25 x \$50,000 = \$1,250,000. Your spouse could likely even retire with that money, especially if it's in the market and they're safely withdrawing at a rate below market appreciation.
Of course, we haven’t taken into account any high-interest savings or investment accounts like RRSPs that you would subtract from that total.
With that in mind, follow this 2 step process; essentially, financial obligations minus liquid assets.
Obligations: Annual income multiplied by the years you want to replace that income, then add in your remaining mortgage, debts, college funds and funeral expenses. If you’re a stay-at-home parent, include how much it might cost to replace all the work you do every day from daycare to cooking and cleaning.
Liquid assets: What’s your obligated amount? Now, subtract your savings, investments RESPs, and any other life insurance you might currently have that maybe your parents bought for you when you were a child.
That number is probably not bad an estimation for how much life insurance you need.
There’s also an argument to buy multiple term life policies. Here’s how that would work. Let’s say you buy a 30-year term policy to only cover what your spouse needs until your retirement. Then, for your children, buy a smaller term say for 20 years to cover your children’s college education. If your life insurance quotes come to less, use it.
## Workplace life insurance
If you have life insurance through work, the amount you’re covered for is usually one or two times your annual earnings. However, that might not be enough – the Canadian Life and Health Insurance Association says a common amount of coverage is often between five and seven times your current net income. What happens if you get fired or move to another company that doesn’t have life insurance? It’s better to have too much life insurance than not enough.
## Quick tips
• Use the DIME method
• Speak with your spouse – would they need your full income or be fine with a portion? | 2,014 | 9,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.963218 |
http://www.epa.gov/oppt/dfe/pubs/lithography/bulletins/bullet04/worksheet.html | 1,419,296,713,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802777438.76/warc/CC-MAIN-20141217075257-00087-ip-10-231-17-201.ec2.internal.warc.gz | 502,715,355 | 6,853 | You are here: EPA Home » DfE » Blanket Wash Comparison Worksheet
# Blanket Wash Comparison Worksheet
DfE Lithography Bulletin 4 Blanket Wash Comparison Worksheet: Could a substitute blanket wash be a better choice for your facility? Use the scorecard below to figure out whether a substitute blanket wash may be a better choice for your facility. Add the scores to see if the substitute wash is better, worse, or the same as your current wash. Test Wash: Supplier Name and Phone: Compare the test wash to the blanket wash you normally use for the following questions and enter the score of your answer in the column on the right:
Compared to your normal blanket wash, -2 -1 Scores 0 1 2 Enter Score Below Is the price per gallon of the test blanket wash: Much more More Same Less Much less ___ Is the amount of test wash used to clean each blanket: Much more More Same Less Much less ___ Is the time required to clean a blanket with the test wash: Much more More Same Less Much less ___ What does the press operator think of the test blanket wash? Is it: Much worse Worse Same Better Much better ___ Consult MSDS forms and contact the blanket wash supplier to answer questions 5-9: Does the test wash contain hazardous chemicals as defined by Federal/State environmental regulations or OSHA? Y = -2 N = 2 ___ Compared to your normal blanket wash, -2 -1 0 1 2 Is the vapor pressure of the test wash: Much higher Higher Same Lower Much lower ___ Is the percentage VOCs of the test wash: Much higher Higher Same Lower Much lower ___ Is the flash point of the test wash: Much lower Lower Same Higher Much higher ___ How does the test wash compare on any other factors: Much worse Worse Same Better Much better ___ Add all nine scores on this worksheet to get the total score for this test blanket wash. Remember, when adding negative numbers -2 + 2 = 0. Total ___
What does the score mean? A score greater than zero means the test wash may be a better choice than your facility's regular wash, a score of zero indicates that the test wash is approximately the same, and a score of less than zero indicates that the test wash may not be a better choice for your facility. If the test wash appears to be approximately the same as your normal wash overall, look at each individual category. Which is most important to you? Different scores in that category may still help you decide which blanket wash is best for your shop. Use the Worksheet to Choose a Better Roller Wash Too!
[ Back to Bulletin 4 ] | 540 | 2,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-52 | latest | en | 0.82427 |
https://nickadamsinamerica.com/tag/reading-comprehension-passages-5th-grade/ | 1,618,949,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00257.warc.gz | 542,918,399 | 22,022 | ## Rhyming Words For KindergartenRhyming Words For Kindergarten
Published at Saturday, April 04th 2020, 00:06:20 AM by Natuche Chauvin. Worksheet. Children who struggle with this lesson may benefit from a review of simple addition and subtraction problems without exchanging or regrouping. They may also benefit from reviewing place value concepts.
## Worksheet Kindergarten is Great for ChildrenWorksheet Kindergarten is Great for Children
Published at Saturday, March 21st 2020, 09:37:14 AM by Patrice Bailly. Worksheet. Children who struggle with this lesson may benefit from a review of simple addition and subtraction problems without exchanging or regrouping. They may also benefit from reviewing place value concepts.
## Fun Sheets For KindergartenFun Sheets For Kindergarten
Published at Friday, March 27th 2020, 07:44:45 AM by Roial Adam. Worksheet. Counting money is one of the most practical early math skills. Our grade 2 counting money worksheets help kids learn to recognize common coins and bills and to count money. U.S. and Canadian currencies are used. 2nd grade counting money worksheet
## Kindergarten Curriculum GuideKindergarten Curriculum Guide
Published at Friday, March 27th 2020, 15:24:17 PM. Worksheet By Pensee Perrier. These Math Flash Cards are great for classroom practice recognizing different shapes. The set will produce thirteen different shapes. These Math Flash Cards are appropriate for Kindergarten, 1st Grade, and 2nd Grade.
## Kindergarten Assessment WorksheetsKindergarten Assessment Worksheets
Published at Friday, March 27th 2020, 14:39:14 PM. Worksheet By Celestyna Martinez. A complete guide to teaching your child the addition facts, including the best practice resources, free printable games, and the missing piece of the puzzle that makes memorizing the facts faster and easier.
### Free Learning Materials For KindergartenFree Learning Materials For Kindergarten
Published at Friday, March 27th 2020, 14:26:34 PM. Worksheet By Sydnee Auger. These fractions worksheets are great for practicing how to subtract measurement you would find on a tape measure. The problems will use 1/2’s, 1/4’s, 1/8’s. 1/16’s and there is an option to select 1/32’s and 1/64’s. These worksheets will generate 10 tape measurement fraction subtraction problems per worksheet.
#### Kindergarten Math ProblemsKindergarten Math Problems
Published at Friday, March 27th 2020, 13:18:53 PM. Worksheet By Stephanie Paul. This section contains all of the graphic previews for the Circle Worksheets. We have identifying radius and diameter for circles worksheets, calculating circumference, area, radius, and diameters worksheets, arcs and central angles for circles worksheets, arcs and chords worksheets, inscribed angles worksheets, graphing of circles worksheets and much more circle worksheets for your use. These geometry worksheets are a good resource for children in the 5th Grade through the 10th Grade.
Published at Friday, March 27th 2020, 12:35:43 PM. Worksheet By Roxane Roussel. These fractions worksheets are great for practicing how to subtract measurement you would find on a tape measure. The problems will use 1/2’s, 1/4’s, 1/8’s. 1/16’s and there is an option to select 1/32’s and 1/64’s. These worksheets will generate 10 tape measurement fraction subtraction problems per worksheet.
###### Best Kindergarten WebsitesBest Kindergarten Websites
Published at Friday, March 27th 2020, 12:15:35 PM. Worksheet By Carressa Moulin. These division worksheets are a good introduction for algebra concepts. You may select various types of characters to replace the missing numbers of the division problems on the division worksheets. The format of the division worksheets are horizontal and the answers range from 0 to 99. These division worksheets can be configured to layout the division problems using the division sign or a slash (/) format. You may select between 12 and 30 problems for these division worksheets.
## Teaching Aids For KindergartenTeaching Aids For Kindergarten
Published at Friday, March 27th 2020, 12:04:36 PM. Worksheet By Tiffney Maillard. This section contains all of the graphic previews for the Differentiation Applications Worksheets. We have rate of change, graphing, graph properties, differentials, optimization, Newton’s Method, and related rates worksheets for your use. These Calculus worksheets are a good resource for students in high school.
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Any content, trademark/s, or other material that might be found on this site that is not this site property remains the copyright of its respective owner/s. | 1,046 | 4,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-17 | latest | en | 0.892297 |
https://www.mathgenie.com/blog/addition-and-subtraction-on-the-abacus | 1,675,275,080,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00713.warc.gz | 904,747,169 | 15,031 | Addition and subtraction are some of the first math skills that your child learns in school. Before teaching your child addition and subtraction, you should make sure that they are familiar with numbers and counting. They need to understand how to count forward and backward in order to understand how addition and subtraction works.
With constant practice, your child will improve their number sense. This understanding of numbers is important for laying the groundwork for basic math problems. Once your child has a solid understanding for numbers, you can begin to teach addition and subtraction.
In American schools, it isn’t very common for the abacus to be utilized and taught in the typical classroom. As a teacher at Math Genie, I had never even heard of an abacus until I applied to work there. After fighting through a math phobia that I had developed as a young child, I learned to love the abacus and taught the system to my adult friends. I encouraged my friends and family to adopt this new style of solving math because I noticed that it made my memory sharper and melted away my long-standing phobia. In retrospect, I never imagined that I would be the type of person who actually enjoyed math. While I admit that there are students of mine who solve their mental math faster than me, I know it’s because I wasn’t exposed to the abacus at a prime age like they are.
Addition- combining the values of two or more numbers.
Subtraction- taking away value from a larger number.
Practice makes perfect
The more your child practices addition and subtraction, the better they will get at mastering these skills. Try to find ways to help them practice and apply these concepts to their everyday lives. Your child will be more motivated to learn something that they find relevant and useful.
How We Teach Addition and Subtraction on the Abacus
Whether it be online or in-person, we teach our students how to solve basic math problems using the abacus. With foundational skills, they advance in time and are eventually able to solve complex problems utilizing the abacus, whether it be in their head or physically touching the beads.
When doing math on the abacus, you must remember to always start on the magic dot. If you don’t, your answers will be skewed.
Under the line and magic dot is the value of one. On top of the line and magic dot, the bead has a value of five. The abacus is a very intuitive tool to use once you get the hang of it. We teach your child how to navigate the abacus in their free class, and the majority of kids grasp it more quickly than their parents do!
We would love to teach you more about the abacus. Come in today to one of our many locations to have a free class and assessment. We can’t wait to meet you.
-The Math Genie team
Sources
https://www.huffpost.com/entry/number-sense-the-most-important-mathematical-concept_b_58695887e4b068764965c2e0
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https://www.proprofs.com/quiz-school/story.php?title=aircraft-drawings | 1,708,664,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00488.warc.gz | 972,730,743 | 106,804 | # The Ultimate Aircraft Drawing Quiz! Trivia
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This is the ultimate aircraft drawing quiz. How good are you when envisioning and drawing out what an aircraft looks like its components and features to a certain scale? Can you draw how the installation should be established? This quiz is designed to see if you have the skills to fill the drawing position based on your knowledge, check it out and up your skills.
• 1.
### 1) A detailed drawing is a description of a single part. 2) An assembly drawing is a description of an object made up of two or more parts. Regarding the above statements,
• A.
Only No. 1 is true.
• B.
Neither No. 1 nor No. 2 is true.
• C.
Both No. 1 and No. 2 are true.
C. Both No. 1 and No. 2 are true.
Explanation
Both statements 1 and 2 are true because they accurately describe the nature of detailed drawings and assembly drawings. A detailed drawing provides a description of a single part, while an assembly drawing provides a description of an object that is made up of multiple parts. Therefore, both statements are correct.
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• 2.
### Which statement is true regarding an orthographic projection?
• A.
There are always at least two views.
• B.
It could have as many as eight views.
• C.
One-view, two-view, and three-view drawings are the most common
C. One-view, two-view, and three-view drawings are the most common
Explanation
An orthographic projection is a method of representing a three-dimensional object in two dimensions. It typically involves creating multiple views of the object from different angles. The statement that one-view, two-view, and three-view drawings are the most common is true because these are the most basic and widely used techniques in orthographic projection. While it is possible to have more than three views, such as four, five, six, or even eight views, these are less common and usually only used for complex objects. Therefore, the most common practice is to create one, two, or three views to accurately represent an object in orthographic projection.
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• 3.
### 1) Schematic diagrams indicate the location of individual components in the aircraft. 2) Schematic diagrams indicate the location of components with respect to each other within the system. Regarding the above statements,
• A.
Only No. 1 is true.
• B.
Both No. 1 and No. 2 are true.
• C.
Only No. 2 is true.
C. Only No. 2 is true.
• 4.
• A.
1.
• B.
2.
• C.
3.
B. 2.
• 5.
• A.
1.
• B.
2.
• C.
3.
C. 3.
• 6.
• A.
1.
• B.
2.
• C.
3.
A. 1.
• 7.
### For the sketching purposes, almost all objects are composed of one or some combination of six basic shapes; these include the
• A.
Angle, arc, line, plane, square, and circle.
• B.
Triangle, circle, cube, cylinder, cone, and sphere.
• C.
Triangle, plane, circle, line, square, and sphere.
B. Triangle, circle, cube, cylinder, cone, and sphere.
• 8.
### What is the class of working drawing that is the description/depiction of a single part?
• A.
Installation drawing.
• B.
Assembly drawing.
• C.
Detail drawing.
C. Detail drawing.
Explanation
A detail drawing is a class of working drawing that provides a detailed description and depiction of a single part. It includes all the necessary information such as dimensions, materials, and manufacturing processes required to produce that specific part. Installation drawings are used to show how different parts fit together in a larger structure or system, while assembly drawings provide an overview of how multiple parts come together to form an assembly. Therefore, the correct answer is detail drawing, as it focuses on the specific details of a single part.
Rate this question:
• 9.
### One purpose for schematic diagrams is to show the
• A.
Functional location of components within a system.
• B.
Physical location of components within a system.
• C.
Size and shape of components within a system.
A. Functional location of components within a system.
Explanation
Schematic diagrams are used to illustrate the functional location of components within a system. These diagrams provide a visual representation of how different components are interconnected and work together to achieve a specific function. They focus on the logical relationships and connections between components rather than their physical location or size and shape.
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• 10.
### A hydraulic system schematic drawing would indicate the
• A.
Specific location of the individual components within the aircraft.
• B.
Direction of fluid flow through the system.
• C.
Type and quantity of the hydraulic fluid.
B. Direction of fluid flow through the system.
Explanation
A hydraulic system schematic drawing is a visual representation of the hydraulic system in an aircraft. It shows the arrangement and interconnection of the various components in the system. One important aspect that a schematic drawing indicates is the direction of fluid flow through the system. This is crucial information as it helps in understanding how the hydraulic fluid moves and circulates within the system, ensuring that the system functions correctly.
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• 11.
### 1) Measurement should not be scaled from an aircraft print because the paper shrinks or stretches when the print is made. 2) When a detailed drawing is made, it is carefully and accurately drawn to scale and is dimensioned. Regarding the above statements,
• A.
Only No. 2 is true.
• B.
Both No. 1 and No. 2 are true.
• C.
Neither No. 1 nor No. 2 is true.
B. Both No. 1 and No. 2 are true.
Explanation
The first statement states that measurement should not be scaled from an aircraft print because the paper used for the print can shrink or stretch, which would affect the accuracy of the measurements. This is true because paper can indeed undergo changes in size due to various factors. The second statement says that when a detailed drawing is made, it is carefully and accurately drawn to scale and dimensions are provided. This is also true because detailed drawings are typically created with precision and accuracy, and include dimensions to provide accurate measurements. Therefore, both statements are true.
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• 12.
### The drawings often used in illustrated parts manuals are
• A.
Exploded view drawings.
• B.
Black drawings.
• C.
Detail drawings.
A. Exploded view drawings.
Explanation
Exploded view drawings are commonly used in illustrated parts manuals as they provide a visual representation of how the different components of a product fit together. These drawings show the individual parts of the product separated from each other, allowing the viewer to understand the assembly process and the relationship between the parts. This type of drawing is particularly helpful for troubleshooting, maintenance, and repair purposes, as it provides a clear and detailed depiction of the product's structure and arrangement.
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• 13.
### A drawing in which the subassemblies or parts are shown as brought together on the aircraft is called
• A.
An assembly drawing.
• B.
An installation drawing.
• C.
A detail drawing.
B. An installation drawing.
Explanation
An installation drawing is a type of drawing that shows how subassemblies or parts are brought together and installed on the aircraft. This drawing provides a clear visual representation of the arrangement and placement of components during the assembly process. It helps in understanding the correct installation procedure and ensures that all parts are properly fitted and connected.
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• 14.
### What type of diagram is used to explain a principle of operation, rather than show the parts as they actually appear?
• A.
A pictorial diagram.
• B.
A schematic diagram.
• C.
A block diagram.
B. A schematic diagram.
Explanation
A schematic diagram is used to explain a principle of operation, rather than show the parts as they actually appear. It uses symbols and lines to represent the different components and connections in a system, making it easier to understand the overall functioning of the system. Unlike a pictorial diagram, which shows the physical appearance of the parts, a schematic diagram focuses on the conceptual representation of the system. A block diagram, on the other hand, is used to show the overall structure of a system by using blocks to represent different components and their interconnections.
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• 15.
### In what type of electrical diagram are images of components used instead of conventional electrical symbols?
• A.
A pictorial diagram.
• B.
A schematic diagram.
• C.
A block diagram.
A. A pictorial diagram.
Explanation
A pictorial diagram is a type of electrical diagram that uses images of components instead of conventional electrical symbols. This type of diagram provides a visual representation of the components and their connections, making it easier to understand for individuals who are not familiar with electrical symbols. Pictorial diagrams are commonly used in technical manuals, user guides, and educational materials to simplify complex electrical systems.
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• 16.
### What type of line is normally used in a mechanical drawing or blueprint to represent an edge or object not visible to the viewer?
• A.
Medium-weight dashed line.
• B.
Medium solid line.
• C.
Alternate short and long light dashes.
A. Medium-weight dashed line.
Explanation
A medium-weight dashed line is typically used in a mechanical drawing or blueprint to represent an edge or object that is not visible to the viewer. This type of line helps to differentiate between visible and hidden elements, providing a clear representation of the object's structure and dimensions. The dashed line allows the viewer to understand the hidden features of the object without cluttering the drawing with unnecessary details.
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• 17.
### Which of the following terms is/are used to indicate specific measured distances from the datum and/or other points identified by the manufacturer to points in or on the aircraft? 1. Zone numbers. 2. Reference numbers. 3. Station numbers.
• A.
1 and 3.
• B.
3.
• C.
2.
B. 3.
Explanation
Station numbers are used to indicate specific measured distances from the datum and/or other points identified by the manufacturer to points in or on the aircraft. Zone numbers and reference numbers are not used for this purpose.
Rate this question:
• 18.
• A.
1.
• B.
3.
• C.
2.
B. 3.
• 19.
### Which statement is applicable when using a sketch or making a part?
• A.
The sketch may be used only if supplemented with three-view orthographic projection drawings.
• B.
The sketch must show all information to manufacture the part.
• C.
The sketch need not show all necessary construction details.
B. The sketch must show all information to manufacture the part.
Explanation
When using a sketch or making a part, it is necessary for the sketch to show all the information required to manufacture the part. This means that the sketch should include all the necessary dimensions, features, and construction details that are needed to accurately produce the part. By providing all the relevant information in the sketch, it ensures that the manufacturer has a clear understanding of how the part should be made and reduces the chances of errors or misunderstandings during the manufacturing process.
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• 20.
### What should be the first step of the procedure in sketching an aircraft wing skin repair?
• A.
Draw heavy guidelines.
• B.
Lay out the repair.
• C.
Block in the views.
C. Block in the views.
Explanation
The first step in sketching an aircraft wing skin repair should be to block in the views. This means creating a rough outline or shape of the repair on the drawing, without going into too much detail. By blocking in the views, the overall structure and dimensions of the repair can be established, providing a framework for the rest of the sketching process. This step helps to ensure that the repair is accurately represented and allows for easier visualization and planning of the subsequent steps.
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• 21.
### 1) According to FAR Part 91, repairs to an aircraft skin should have a detailed dimensional sketch included in the permanent records. 2) On occasion, a mechanic may need to make a simple sketch of a proposed repair to an aircraft, a new design, or modification. Regarding the above statements,
• A.
Only No. 1 is true.
• B.
Only No. 2 is true.
• C.
Both No. 1 and No. 2 are true.
B. Only No. 2 is true.
Explanation
The correct answer is only No. 2 is true. This is because according to FAR Part 43, repairs to an aircraft skin should have a detailed dimensional sketch included in the permanent records, not Part 91 as stated in statement No. 1. However, statement No. 2 is true as mechanics may need to make a simple sketch of a proposed repair, new design, or modification.
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• 22.
### What material symbol is frequently used in drawings to represent all metals?
• A.
Steel.
• B.
Cast iron.
• C.
Aluminum.
B. Cast iron.
Explanation
Cast iron is frequently used in drawings to represent all metals because it is a common and easily recognizable material. It has a distinct appearance with its dark gray color and rough texture, making it a suitable symbol for metals in general. Steel and aluminum, although also commonly used metals, may not be as widely recognized or easily distinguishable in drawings compared to cast iron. Therefore, cast iron is the preferred choice for representing metals in drawings.
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• 23.
### A simple way to find the center of a circle on a sketch or drawing, or a circular piece of material is to
• A.
Draw two non-parallel chord lines across the circle and then a corresponding perpendicular bisector line across each chord line.
• B.
Draw two parallel chord lines across the circle and then a corresponding perpendicular bisector line across each chord line.
• C.
Draw a single chord line across the circle and then a corresponding perpendicular bisector line across the chord line.
A. Draw two non-parallel chord lines across the circle and then a corresponding perpendicular bisector line across each chord line.
Explanation
To find the center of a circle, drawing two non-parallel chord lines across the circle and then a corresponding perpendicular bisector line across each chord line is an effective method. This is because the perpendicular bisector of a chord passes through the center of the circle. By drawing two non-parallel chord lines and their perpendicular bisectors, the intersection point of the bisectors will give the center of the circle. This method utilizes the properties of chords and perpendicular bisectors to accurately locate the center of the circle.
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• 24.
### What are the proper procedural steps for sketching repairs and alterations?
• A.
3, 1, 4, and 2.
• B.
4, 2, 3, and 1.
• C.
1, 3, 4, and 2.
A. 3, 1, 4, and 2.
Explanation
The proper procedural steps for sketching repairs and alterations are to first identify and analyze the existing conditions (step 3), then create a preliminary sketch of the proposed repairs and alterations (step 1), followed by verifying the accuracy of the preliminary sketch (step 4), and finally producing a final sketch of the repairs and alterations (step 2).
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• 25.
### What is the next step required for a working sketch of the illustration?
• A.
Darken the object outlines.
• B.
Sketch extension and dimension lines.
• C.
Add notes, dimensions, title, and date.
B. Sketch extension and dimension lines.
Explanation
The next step required for a working sketch of the illustration is to sketch extension and dimension lines. This step is necessary to provide additional information about the size and placement of objects in the illustration, allowing for accurate measurements and understanding of the overall design. Darkening the object outlines and adding notes, dimensions, title, and date may be important steps as well, but they are not the immediate next step required for a working sketch.
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• 26.
• A.
1.
• B.
2.
• C.
3.
C. 3.
• 27.
### The measurements showing the ideal or "perfect sizes of parts on drawings are called
• A.
Tolerances.
• B.
Allowances.
• C.
Dimensions.
C. Dimensions.
Explanation
Tolerances refer to the allowable variation in the dimensions of a part, while allowances are the intentional extra material added to a part for machining or assembly purposes. Therefore, the correct answer is dimensions, as it refers to the specific sizes or measurements indicated on a drawing for each part.
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• 28.
### Zone numbers on aircraft blueprints are used to
• A.
Locate parts, sections, and views on large drawings.
• B.
Indicate different sections of the aircraft.
• C.
Locate parts in the aircraft.
A. Locate parts, sections, and views on large drawings.
Explanation
Zone numbers on aircraft blueprints are used to locate parts, sections, and views on large drawings. These numbers help in identifying specific areas or regions within the blueprint, making it easier for engineers, technicians, and other professionals to navigate and understand the various components and structures of the aircraft. By using zone numbers, individuals can quickly locate the desired information or section on the blueprint, facilitating efficient analysis, troubleshooting, and maintenance processes.
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• 29.
### When reading a blueprint, a dimension is given as 4.387 inches +.005 -.002. Which statement is true?
• A.
The maximum acceptable size is 4.390 inches.
• B.
The maximum acceptable size is 4.385 inches.
• C.
The maximum acceptable size is 4.382 inches.
B. The maximum acceptable size is 4.385 inches.
Explanation
The given dimension is 4.387 inches +.005 -.002. This means that the actual size can vary within the given range. The maximum acceptable size is determined by adding the maximum allowable variation (+.005) to the nominal dimension (4.387). Therefore, the maximum acceptable size is 4.387 + .005 = 4.392 inches. However, since the given dimension is rounded to three decimal places, the correct answer is 4.385 inches, which is the closest value to the maximum acceptable size.
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• 30.
### What is the allowable manufacturing tolerance for a bushing where the outside dimensions shown on the blueprint are: 1.0625 +.0025 -.0003
• A.
.0028
• B.
1.0650
• C.
1.0647
A. .0028
Explanation
The allowable manufacturing tolerance for the bushing is .0028. This means that the actual outside dimensions of the bushing can vary within the range of +.0028 and -.0028 from the specified dimensions on the blueprint.
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• 31.
### In the reading of aircraft blueprints, the term "tolerance," used in association with aircraft parts or components,
• A.
Is the tightest permissible fit for proper construction and operation of mating parts.
• B.
Is the difference between extreme permissible dimensions that a part may have and still be acceptable.
• C.
Represents the limit of galvanic compatibility between different adjoining material types in aircraft parts.
B. Is the difference between extreme permissible dimensions that a part may have and still be acceptable.
Explanation
The term "tolerance" in the reading of aircraft blueprints refers to the difference between extreme permissible dimensions that a part may have and still be acceptable. This means that there is a range within which the dimensions of the part can vary while still meeting the requirements for proper construction and operation. This allows for some flexibility in the manufacturing process while ensuring that the parts will fit together correctly and function as intended.
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• 32.
### What is the dimension of the chamfer?
• A.
1/16 x 37degrees
• B.
0.3125 +.005 - 0
• C.
0.0625 x 45degrees
C. 0.0625 x 45degrees
Explanation
The chamfer dimension is given as 0.0625 x 45 degrees. This means that the chamfer has a width of 0.0625 units and an angle of 45 degrees.
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• 33.
### What is the maximum diameter of the hole for the clevis pin?
• A.
0.3175
• B.
0.3130
• C.
0.31255
A. 0.3175
Explanation
The maximum diameter of the hole for the clevis pin is 0.3175. This means that the hole should be able to accommodate a pin with a diameter of 0.3175 units.
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• 34.
### What would be the minimum diameter of 4130 round stock required for the construction of the clevis that would produce a machined surface?
• A.
55/64 inch.
• B.
1 inch.
• C.
7/8 inch.
B. 1 inch.
Explanation
The minimum diameter of the round stock required for the construction of the clevis should be 1 inch. This is because a machined surface requires enough material to be removed during the machining process in order to achieve the desired dimensions and smoothness. A larger diameter provides more material to work with, allowing for greater precision and a smoother finish. Therefore, a 1-inch diameter is the minimum necessary for producing a machined surface on the clevis.
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• 35.
### Using the information, what size drill would be required to drill the clevis bolt hole?
• A.
5/16 inch.
• B.
21/64 inch.
• C.
1/2 inch.
A. 5/16 inch.
Explanation
The correct answer is 5/16 inch. This is because the question asks for the size of the drill required to drill the clevis bolt hole, and 5/16 inch is the only option provided that matches this requirement.
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• 36.
• A.
3.
• B.
1.
• C.
4.
A. 3.
• 37.
### The diameter of the holes in the finished object is
• A.
3/4 inch.
• B.
31/64 inch.
• C.
1/2 inch.
C. 1/2 inch.
Explanation
The correct answer is 1/2 inch because it is the only option that matches the given information. The question states that the diameter of the holes in the finished object is 1/2 inch, and none of the other options (3/4 inch and 31/64 inch) match this measurement.
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• 38.
### The vertical distance between the top of the plate and the bottom of the lowest 15/64-inch hole is
• A.
2.250
• B.
2.242
• C.
2.367
C. 2.367
Explanation
The correct answer is 2.367. This means that the vertical distance between the top of the plate and the bottom of the lowest 15/64-inch hole is 2.367 inches.
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• 39.
### The -100 in the title block (Area 1) is applicable to which doubler part number(s)?
• A.
-101.
• B.
-102.
• C.
Both
A. -101.
Explanation
The -100 in the title block (Area 1) is applicable to the doubler part number -101. This means that the doubler part number -101 has a specific designation or characteristic indicated by the -100 in the title block. The -100 does not apply to doubler part number -102, as it is not mentioned in the given options. Therefore, the correct answer is -101.
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• 40.
### Which doubler(s) require(s) heat treatment before installation?
• A.
-101.
• B.
-102.
• C.
Both.
B. -102.
Explanation
The correct answer is -102. This means that only doubler -102 requires heat treatment before installation. The other doubler, -101, does not require heat treatment.
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• 41.
### Using only the information is given (when bend allowance, set back, etc., have been calculated) which doubler is it possible to construct and install?
• A.
-101.
• B.
-102.
• C.
Both.
A. -101.
Explanation
Based on the given information, it is possible to construct and install doubler -101. There is no mention of any specific criteria or conditions that would restrict the use of doubler -101 or favor doubler -102. Therefore, both doublers could potentially be used, but since the question asks for the doubler that is possible to construct and install, -101 is the correct answer.
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• 42.
### How many parts will need to be fabricated by the mechanic in the construction and installation of one doubler?
• A.
2.
• B.
3.
• C.
4.
B. 3.
Explanation
In the construction and installation of one doubler, the mechanic will need to fabricate three parts. This suggests that the doubler consists of multiple components or sections that need to be fabricated separately and then assembled together.
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• 43.
### An aircraft reciprocating engine has a 1,830 cubic inch displacement and develops 1,250 brake-horsepower at 2,500 RPM. What is the brake mean effective pressure?
• A.
217.
• B.
205.
• C.
225.
A. 217.
Explanation
The brake mean effective pressure (BMEP) is a measure of the average pressure exerted on the piston during the power stroke of an engine. It is calculated by dividing the brake horsepower (BHP) by the displacement volume of the engine. In this case, the BHP is given as 1,250 and the displacement volume is 1,830 cubic inches. Dividing the BHP by the displacement volume gives us a BMEP of approximately 217. Therefore, the correct answer is 217.
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• 44.
### An aircraft reciprocating engine has a 2,800 cubic-inch displacement, develops 2,000 brake-horsepower, and indicates 270 brakes mean effective pressure. What is the engine speed (RPM)?
• A.
2,200.
• B.
2,100.
• C.
2,300.
B. 2,100.
Explanation
The engine speed (RPM) can be calculated using the formula: RPM = (2 * BHP * 33,000) / (mean effective pressure * displacement). Plugging in the given values, we get RPM = (2 * 2000 * 33,000) / (270 * 2800) = 2100.
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• 45.
### An aircraft reciprocating engine has a 2,800 cubic-inch displacement and develops 2,00 brake-horsepower at 2,200 RPM. What is the brake mean effective pressure?
• A.
257.5
• B.
242.5
• C.
275.0
A. 257.5
Explanation
The brake mean effective pressure (BMEP) is a measure of the average pressure exerted on the piston during the power stroke of an engine. It is calculated by dividing the brake horsepower by the displacement volume. In this case, the engine has a displacement of 2,800 cubic inches and develops 200 brake horsepower. Dividing 200 by 2,800 gives a BMEP of 0.0714 psi. Multiplying this by 3,600 (to convert from psi to lb/ft^2) gives a BMEP of 257.5 lb/ft^2. Therefore, the correct answer is 257.5.
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• 46.
### Determine the cable size of a 40-foot length of single cable in free air, with a continuous rating, running from a bus to the equipment in a 28-volt system with a 15-ampere load and a 1-volt drop.
• A.
No. 10.
• B.
No. 11.
• C.
No. 18.
A. No. 10.
Explanation
The correct answer is No. 10. The cable size is determined based on the length of the cable, the voltage of the system, the load, and the allowable voltage drop. In this case, the length of the cable is 40 feet, the system voltage is 28 volts, the load is 15 amperes, and the allowable voltage drop is 1 volt. Based on these parameters, a No. 10 cable is the appropriate size to ensure that the voltage drop does not exceed 1 volt.
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• 47.
### Determine the maximum length of a No. 16 cable to be installed from a bus to the equipment in a 28-volt system with a 25-ampere intermittent load and 1-volt drop.
• A.
8 feet.
• B.
10 feet.
• C.
12 feet.
A. 8 feet.
Explanation
In a 28-volt system with a 25-ampere intermittent load and a 1-volt drop, the maximum length of a No. 16 cable can be determined using the formula: maximum cable length = (voltage drop / (load current x cable resistance per unit length)). In this case, the voltage drop is 1 volt, the load current is 25 amperes, and the cable resistance per unit length for a No. 16 cable is known. By plugging in these values into the formula, it is determined that the maximum length of the cable is 8 feet.
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• 48.
### Determine the minimum wire size of a single cable in a bundle carrying a continuous current of 20 amperes 10 feet from the bus to the equipment in a 28-colt system with an allowable 1-volt drop.
• A.
No. 12.
• B.
No. 14.
• C.
No. 16.
A. No. 12.
Explanation
The minimum wire size of a single cable in a bundle carrying a continuous current of 20 amperes 10 feet from the bus to the equipment in a 28-volt system with an allowable 1-volt drop is No. 12. This is determined by considering the current, distance, voltage, and allowable voltage drop. No. 12 wire is capable of carrying a continuous current of 20 amperes without exceeding the allowable voltage drop of 1 volt.
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• 49.
### Determine the maximum length of a No. 12 single cable that can be used between a 28-volt bus and a component utilizing 20 amperes continuous load in free air with a maximum acceptable 1-volt drop.
• A.
22.5 feet.
• B.
26.5 feet.
• C.
12.5 feet.
B. 26.5 feet.
Explanation
The maximum length of a cable can be determined by considering the voltage drop and current carrying capacity. In this case, the cable needs to carry a continuous load of 20 amperes with a maximum acceptable 1-volt drop. By using Ohm's law (V = IR), we can calculate the resistance of the cable. Since the voltage drop is 1 volt and the current is 20 amperes, the resistance is 0.05 ohms. Next, we need to find the maximum length of the cable that will result in a voltage drop of 1 volt. By using the formula V = IR, where V is the voltage drop, I is the current, and R is the resistance, we can rearrange the formula to solve for length. Plugging in the values, we get length = voltage drop / (current * resistance) = 1 volt / (20 amperes * 0.05 ohms) = 1/1 = 26.5 feet. Therefore, the maximum length of the cable is 26.5 feet.
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• 50.
### Determine how much fuel would be required for a 30-minute reserve operating at 2,300 RPM.
• A.
25.3 pounds.
• B.
35.5 pounds.
• C.
49.8 pounds.
A. 25.3 pounds.
Explanation
The amount of fuel required for a 30-minute reserve operating at 2,300 RPM is determined by the fuel consumption rate at that RPM. The correct answer of 25.3 pounds suggests that the engine consumes fuel at a rate of approximately 0.84 pounds per minute (25.3 pounds divided by 30 minutes). This means that for every minute of operation at 2,300 RPM, the engine consumes 0.84 pounds of fuel.
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# Tutorial 3 - Scaling, Inverse Kinematics, and Inverse Dynamics
## I. Objectives
### Purpose
The purpose of this tutorial is to demonstrate how OpenSim solves an inverse kinematics problem and an inverse dynamics problem using experimental data. To diagnose movement disorders and study human movement, biomechanists frequently ask human subjects to perform movements in a motion capture laboratory and use computational tools to analyze these movements. A common step in analyzing a movement is to compute the joint angles and joint moments of the subject during movement. OpenSim has tools for computing these quantities:
1. Inverse kinematics is used to compute joint angles.
2. Inverse dynamics is used to compute net joint reaction forces and net joint moments.
Inverse kinematics computes the joint angles for a musculoskeletal model that best reproduce the motion of a subject. Inverse dynamics then uses joint angles, angular velocities, and angular accelerations of the model, together with the experimental ground reaction forces and moments, to solve for the net reaction forces and net moments at each of the joints. The schematic below shows an overview of the inverse kinematics and inverse dynamics problems.
In this tutorial, you will:
• Become familiar with OpenSim's Scale, Inverse Kinematics and Inverse Dynamics tools
• Solve an inverse kinematics and inverse dynamics problem using experimental data
• Interpret the results of the inverse dynamics solution
• Investigate the dynamic inconsistencies that arise during inverse dynamics
### Format
Each section of the tutorial guides you in using certain tools within and asks you to answer a few questions. The menu titles and option names you must select and any commands you must type to run OpenSim will appear in bold face. The questions can be answered based on information from OpenSim and basic knowledge of the human musculoskeletal system. After you complete the tutorial, feel free to explore OpenSim and the other analysis tools further on your own. Depending on the amount of exploration you do, this tutorial should take 1-2 hours to complete.
## II. Generic Musculoskeletal Model
In this tutorial, you will be using a generic musculoskeletal model with 23 degrees of freedom and actuated by 54 muscles entitled 3DGaitModel2354. It is a simplified version of the lower-extremity model of Delp et al. [1], modified to include a torso and back joint based on the model of Anderson and Pandy [2].
To load the generic musculoskeletal model into OpenSim:
• Click the File menu and select Open Model.
• Find the Models folder, which is located under your OpenSim installation directory, e.g., C:\OpenSim3.2. (The actual directory name depends on version used and can be changed during installation).
• Open the Gait2354_Simbody folder, select the file gait2354_simbody.osim, and click Open.
The experimental gait data were collected by Jill Higginson and Chand John in the Neuromuscular Biomechanics Lab at the University of Delaware as part of the study cited below. The data include marker trajectories and ground reaction forces for an adult male walking at a self-selected speed on an instrumented split-belt treadmill. Please note that the data distributed with OpenSim is from a different subject than the one described in the paper. Data collection protocols were the same for both subjects.
Chand T. John, Frank C. Anderson, Jill S. Higginson & Scott L. Delp (2012): Stabilisation of walking by intrinsic muscle properties revealed in a three-dimensional muscle-driven simulation, Computer Methods in Biomechanics and Biomedical Engineering.
## III. Scaling A Musculoskeletal Model
The purpose of scaling a generic musculoskeletal model is to modify the anthropometry, or physical dimensions, of the generic model so that it matches the anthropometry of a particular subject. Scaling is one of the most important steps in solving inverse kinematics and inverse dynamics problems because these solutions are sensitive to the accuracy of the scaling step. In OpenSim, the scaling step adjusts both the mass properties (mass and inertia tensor), as well as the dimensions of the body segments. Scaling can be performed using a combination of two methods:
(1) Measurement-based Scaling: This type of scaling determines scale factors for a body segment by comparing distance measurements between specified landmarks on the model, known as virtual markers, and the corresponding experimental marker positions.
(2) Manual Scaling: This type of scaling allows the user to scale a segment based on some predetermined scale factor. Manual scaling is sometimes necessary when suitable marker data are not available, or if the scale factors were determined using an alternative algorithm.
To scale the generic model:
• Click the Tools menu and select Scale Model.
• At the bottom of the Scale Tool dialog, click Load to input a settings file.
• In the file browser, ensure that you are in the Gait2354_Simbody folder, select the file subject01_Setup_Scale.xml and click Open.
This xml file contains pre-configured settings to scale the generic musculoskeletal model to the dimensions of the subject. Notice all of the textboxes in the dialog were filled in appropriately.
#### Questions
1. Based on information in the Scale Tool dialog, what is the mass of the generic musculoskeletal model? What was the mass of the subject?
2. To see the loaded scale factors, click on the Scale Factors tab. Which body segments were scaled manually?
3. Based on information in the Inverse Kinematics Tool dialog, at what frequency was the experimental motion data captured? Hint: Look for the box titled Marker Data.
To complete the scale step:
• In the Scale Tool dialog, click Run. Then click Close.
• To save the scaled model, click File and select Save Model.
• Ensure that you are in the Gait2354_Simbody folder, type gait2354_scaled.osim into the File name textbox, and click Save.
When complete, a new, scaled model entitled subject01 will appear in View window. Notice the pink markers around the new model. These blue spheres graphically represent the experimental markers from the motion capture data used in the measurement-based scaling.
At this point, you may close the generic model (right-click the model name in the Navigator window, and select "close") or hide the model (right-click the model name, and select Display >> Hide).
## IV. Inverse Kinematics
Kinematics is the study of motion without considering the forces and moments that produce that motion. Thus, to perform kinematical analyses, such as inverse kinematics, mass and inertia properties are not needed. The purpose of inverse kinematics is to find the joint angles of the model that best reproduce the experimental kinematics of a particular subject. In this tutorial, the experimental kinematics used by the inverse kinematics tool are based on experimental marker positions.
The inverse kinematics tool goes through each time step, or frame, of recorded motion and computes the set of joint angles that put the model in a configuration that "best match" the experimental kinematics. OpenSim determines this "best match" by solving a weighted least squares optimization problem with the goal of minimizing marker error. For more information about optimization and least-squares problems, see Chapter 1 of Convex Optimization by Stephen Boyd and Lieven Vandenberghe (http://www.stanford.edu/~boyd/cvxbook/).
Marker error is defined as the distance between an experimental marker and the corresponding virtual marker. Each marker has an associated weighting value, specifying how strongly that marker's error term should be minimized in the least squares problem. In each frame, the inverse kinematics tool solves for a vector of generalized coordinates (e.g., joint angles), q, that minimizes the weighted sum of marker errors, which is expressed as
$\min_q \left[ \sum_{i \in \mathrm{markers}} w_i \left\| \mathbf{x}_i^{\mathrm{exp}} - \mathbf{x}_i(\mathbf{q}) \right\|^2 \right]$
where q is the vector of generalized coordinates (e.g., joint angles), xiexp is the position of experimental marker i, xi(q) is the position of the corresponding virtual marker i (which depends on q), and wi is the weight associated with marker i.
To solve the inverse kinematics problem:
• Click the Tools menu and select Inverse Kinematics.
• In the Inverse Kinematics Tool dialog, click Load to load an Inverse Kinematics setup file.
• In the file browser, ensure that you are in the Gait2354_Simbody folder, select the file subject01_Setup_IK.xml and click Open.
This xml file contains pre-configured settings to solve the inverse kinematics problem for the scaled model. Notice all of the textboxes in the dialog were filled in appropriately.
#### Questions
4. Click the Weights tab and scroll through the list of markers in the top half of the dialog. Which markers have weighting values less than one? Why?
Hint: Think about joints that have not been modeled.
To complete inverse kinematics step:
• In the Inverse Kinematics Tool, click Run. Then click Close.
Note: Even though you closed the dialog, the Inverse Kinematics tool is still running. Notice the progress bar in the lower right-hand corner of the program. Wait until the bar disappears before proceeding.
The model will begin to move slowly, as the inverse kinematics problem is being solved for each frame of the experimental data.
To compare experimental marker data with inverse kinematics results, in the Navigator panel, go to Motions and right-click on Results (which are what the Inverse Kinematics Tool just generated). Then choose Associate Motion Data... from the drop down menu. Choose subject_01_walk1.trc and click Open. Model markers are virtual markers shown in pink and experimental markers are shown in blue. Hit play in the Motion Toolbar. The virtual markers should correspond closely to the experimental marker locations as the animation proceeds.
*Note: If using a Virtual Machine on a Mac, Command + Ctrl + Left Click on each motion.
When completed, examine the accuracy of the inverse kinematics solution:
• Click the Window menu and select Messages.
• The Messages window records details of the all steps you have performed. Take a minute to explore the Messages window. Then, scroll to the very bottom.
• The next to last line provides the markers errors and model coordinate errors (e.g., joint angle errors) associated with the last frame of the motion.
Note: All marker errors have units in meters, and all coordinate errors have units in radians.
#### Questions
5. Based on information in the Messages window, what is the root-mean-squared (RMS) error of all the markers in the last frame of the motion? Include units. Does this seem reasonable? Explain.
6. What was the value of the maximum marker error in the last frame? Include units. Which marker had this maximum error, and why?
Hint: Think about the weighted least squares problem.
To visualize the inverse kinematics solution, animate the model by using the motion slider and video controls. The model should walk through one full gait cycle.
Remember: You can loop and control the speed of the animation.
The inverse kinematics solution is saved to subject01_walk1_ik.mot as specified in the setup file.
Note: Be sure to use the exact file name given, as this file is used to complete the next step.
## V. Inverse Dynamics
Dynamics is the study of motion and the forces and moments that produce that motion. Thus, to perform dynamical analyses, such as inverse dynamics, estimation of mass and inertia is required. The purpose of inverse dynamics is to estimate the forces and moments that cause a particular motion, and its results can be used to infer how muscles are utilized for that motion. To determine these forces and moments, equations of motion for the system are solved iteratively [3]. The equations of motion are derived using the kinematic description and mass properties of a musculoskeletal model. Then, using the joint angles from inverse kinematics and experimental ground reaction force data, the net reaction forces and net moments at each of the joints are calculated such that the dynamic equilibrium conditions and boundary conditions are satisfied [3].
Note: Joint reaction, or inter-segmental, force is the total force acting across a particular joint in a model. This should not be confused with joint bone-on-bone force, which is the actual force seen across the articulating surfaces of the joint and include the effect of muscle activity. For a thorough discussion on this topic see pp 77-79 in [4].
To solve the inverse dynamics problem:
• Click the Tools menu and select Inverse Dynamics.
• In the Inverse Dynamics Tool dialog, click Load to load an Inverse Dynamics setup file.
• In the file browser, ensure that you are in the Gait2354_Simbody folder, select the file subject01_Setup_InverseDynamics.xml and click Open.
Note: If the Motion From File textbox appears red, this means the textbox was filled with an inappropriate file name. Make sure the motion file was saved with the correct file name in the Inverse Kinematics section.
• Note the folder listed in the Directory textbox, located in the Output section of the dialog. The storage file containing the inverse dynamics results will be saved in this folder: examples\Gait2354_Simbody\ResultsInverseDynamics
• Click Run at the bottom of the dialog. Then click Close.
When completed, examine the results of the inverse dynamics solution by plotting the net moments at the left and right ankles:
• Click Tools and select Plot.
• In the Plotter window, click the Y-Quantity button and select Load File.
• In the file browser, go to the ResultsInverseDynamics folder, select the file inverse_dynamics.sto, and click Open.
• In the menu, select ankle_angle_r_moment and ankle_angle_l_moment by clicking the corresponding checkboxes, then click OK.
Note: To quickly find these quantities, type ankle into the pattern text box.
• Click the X-Quantity button, select time, and click OK.
• Back in the Plotter window, click Add to add the moment curves to the plot.
• Print your plot by right clicking on the plot and selecting Print.
Note: To export the plot as an image by right-clicking the plot and selecting Export Image.
• After printing the plot and answering the following questions, close the Plotter window.
#### Questions
7. On your plot of the ankle moments, identify when heel strike, stance phase, toe off, and swing phase occur for each curve (i.e., left leg and right leg).
8. Based on your plot and the angle convention for the ankle, give an explanation of what is happening at the ankle just before toe-off.
Hint: It may be useful to use the Coordinate sliders to understand the angle convention for the ankle.
In solving the inverse dynamics problem, both kinematical data and force plate data were used, making this an over-determined problem. In other words, the problem has more equations than unknowns (i.e., degrees of freedom). Due to errors in the experimental motion data and inaccuracies in the musculoskeletal model, it turns out that Newton's second law is violated, or [3]. One method to handle this dynamical inconsistency is to compute and apply residual forces and moments to a particular body segment in the model, such that Newton's second law becomes:
† An analogous equation relates the ground reaction moment, to the residual moment, .
In this musculoskeletal model, the residuals are applied to the pelvis segment.
To see the residuals from the inverse dynamics solution:
• Click Tools and select Plot.
• In the Plotter window, click the Y-Quantity button and select Load File.
• In the file browser, go to the ResultsInverseDynamics folder, select the file inverse_dynamics.sto, and click Open.
• In the menu, select pelvis_tx_force, pelvis_ty_force, and pelvis_tz_force by clicking the corresponding checkboxes, then click OK.
• Click the X-Quantity button, select time, and click OK.
#### Questions
9. What are the maximum magnitudes of the residual forces? Using the mass of the subject from Question 1, what fraction of body weight are the maximum residual forces?
While computing and applying residual forces and moments makes the model's motion dynamically consistent with the external forces , this strategy is undesirable because the residuals can be large. More advanced strategies have been developed to deal with the problem of residuals and dynamic inconsistencies, such as least-squares optimization [3], the Residual Elimination Algorithm (REA) [5], and the Residual Reduction Algorithm (RRA) [6]. Additionally, OpenSim implements a Residual Reduction Algorithm as part of its workflow for generating muscle-actuated simulations [7]. Reference [6] is attached at the end of this tutorial. For additional information on these strategies, please also refer to [3], [5], and [7].
##### References
1. Delp, S.L., Loan, J.P., Hoy, M.G., Zajac, F.E., Topp E.L., Rosen, J.M. An interactive graphics-based model of the lower extremity to study orthopaedic surgical procedures. IEEE Transactions on Biomedical Engineering, vol. 37, pp. 757-767, 1990.
2. Anderson, F.C., Pandy, M.G. A dynamic optimization solution for vertical jumping in three dimensions. Computer Methods in Biomechanical and Biomedical Engineering, vol. 2, pp. 201-231, 1999.
3. Kuo, A.D. A least squares estimation approach to improving the precision of inverse dynamics computations, Journal of Biomechanical Engineering, vol. 120, pp. 148-159, 1998.
4. Winter, D.A. Biomechanics and Motor Control of Human Movement, Wiley and Sons, pp. 77-79, 1990.
5. Thelen, D.G., Anderson, F.C. Using computed muscle control to generate forward dynamic simulations of human walking from experimental data, Journal of Biomechanics, vol. 39, pp. 1107-1115, 2006.
6. John, C.T., Anderson, F.C., Guendelman, E., Arnold, A.S., Delp, S.L. An algorithm for generating muscle-actuated simulations of long-duration movements, Biomedical Computation at Stanford (BCATS) Symposium, Stanford University, 21 October 2006, Poster Presentation.
7. Delp, S.L., Anderson, F.C., Arnold, A.S., Loan, P., Habib, A., John, C.T., Guendelman, E., Thelen, D.G. OpenSim: Open-source software to create and analyze dynamic simulations of movement. IEEE Transactions on Biomedical Engineering, vol. 55, pp. 1940-1950, 2007. | 3,941 | 18,471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-26 | latest | en | 0.868221 |
http://www.superteacherworksheets.com/measurement-milliliters-liters.html | 1,435,708,878,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094629.80/warc/CC-MAIN-20150627031814-00104-ip-10-179-60-89.ec2.internal.warc.gz | 631,544,003 | 8,570 | # Measuring Capacity (Milliliters and Liters)
## Measurement - Millimeters, Litres
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Use these printables to help students learn about measuring capacity, or volume, using metric units. Skills include estimating capacity in milliliters and liters, as well as converting units.
Estimating- Liters and Milliliters Free
Determine whether each is best measured in milliliters or liters
Milliliters and Liters - Conversions Member
In & out boxes and word problems for converting to and from milliliters and liters
Converting Milliliters and Liters- Basic Member
Convert from milliliters to liters and vice-versa; Easy conversions do not include decimal numbers
Converting Milliliters and Liters- Advanced Member
Convert to and from milliliters and liters; Advanced conversions include decimal numbers
Which is Greater? - Liters and Milliliters Member
Determine which of the amounts given is the larger one. (example: 8,000 mL or 80 L?)
Capacity Questions: Liters and Milliliters Member
Four critical thinking questions about measurement with liters and milliliters. | 426 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-27 | latest | en | 0.876041 |
https://acedmypaper.com/1003043/ | 1,632,724,933,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00367.warc.gz | 137,164,970 | 10,407 | ## [answered] A car rounds a curve with a radius of 25 m at a speed of 20
A car rounds a curve with a radius of 25 m at a speed of 20 m/s. What is the centripetal acceleration of the car?
ac = 16 m/s2
a
Speed = 20m/s
R =25m
a=(20*20)/25 =16 m/s(square)
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https://dokumen.tips/documents/basic-principle-of-strain-gauge-accelerometer-basic-principle-of-strain-gauge.html | 1,653,806,415,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00103.warc.gz | 269,502,980 | 23,345 | # Basic Principle of Strain Gauge Accelerometer ... Basic Principle of Strain Gauge Accelerometer When a cantilever beam attached with a mass at its free end is subjected to vibration,
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• Basic Principle of Strain Gauge Accelerometer
When a cantilever beam attached with a mass at its free end is subjected to vibration,
vibrational displacement of the mass takes place. Depending on the displacement of the
mass, the beam deflects and hence the beam is strained. The resulting strain is proportional
to the vibration displacement of the mass and hence the vibration/acceleration being
measured when calibrated.
Description of Strain Gauge Accelerometer
The main parts of a strain gauge accelerometer are as follows:
A cantilever beam fixed to the housing of the instrument.
A mass is fixed to the free end of the cantilever beam.
Two bounded strain gauges are mounted on the cantilever beam as shown in diagram.
Damping is provided by a viscous fluid filled inside the housing.
Operation of Strain Gauge Accelerometer
The accelerometer is fitted on to the structure whose acceleration is to be measured.
Due to the vibration, vibrational displacement of the mass occurs, causing the cantilever beam to be strained.
http://1.bp.blogspot.com/-Q47bBsOknBA/UHfFldZrjcI/AAAAAAAAAjg/4I96OsIx5o4/s1600/Strain+gauge+Accelerometer.jpg
• Hence the strain gauges mounted on the cantilever beam are also strained and due to this their resistance change.
Hence a measure of this change in resistance of the strain gauge becomes a measure of the extent to which the cantilever beam is strained.
But the resulting strain of the cantilever beam is proportional to the vibration/acceleration and hence a measure of the change in resistance of the strain gauges becomes a measure of vibration/acceleration.
The leads of the strain gauges are connected to a wheat stone bridge whose output is calibrated in terms of vibration/acceleration.
Introduction
A piezoelectric accelerometer utilizes the piezoelectric effect of certain materials to
measure dynamic changes in mechanical variables, such as mechanical shock,
vibration and acceleration. Like other transducers, piezoelectric accelerometers convert
one form of energy into another and provide an electrical signal in response to the
condition, property or quantity. Acceleration acts upon a seismic mass that is restrained
by a spring or suspended on a cantilever beam, and converts a physical force into an
electrical signal.
There are two types of piezoelectric accelerometers: high and low impedance. High
impedance accelerometers have a charge output that is converted into a voltage using
a charge amplifier or external impedance converter. Low impedance units use the same
piezoelectric sensing element as high-impedance units, and incorporate a miniaturized
built-in charge-to-voltage converter and an external power supply coupler to energize
the electronics and decouple the subsequent DC bias voltage from the output signal.
Working Principle of Piezoelectric Accelerometer
A piezoelectric accelerometer consists of a mass attached to a piezoelectric crystal
which is mounted on a case. When the accelerometer body is subjected to vibration, the
mass on the crystal remains undisturbed in space due to inertia. As a result, the mass
compresses and stretches the piezoelectric crystal. This force is proportional to
acceleration in accordance with Newton’s second law, F = ma, and generates a charge.
• The charge output is then converted into low impedance voltage output with the help of
electronics.
Benefits of Piezoelectric Accelerometer
The key benefits of piezoelectric accelerometers are:
Wide frequency range
No moving parts
Excellent linearity over their dynamic range
Low output noise
Self-generating - no external power required
Acceleration signal can be integrated to provide velocity and displacement
Applications
Major applications of piezoelectric accelerometers include:
Engine testing - Combustion and dynamic stressing
Ballistics - Combustion, explosion, and detonation
Industrial/factory - Machining systems, metal cutting, and machine health monitoring
Original equipment manufacturer - Transportation systems, rockets, machine tools,
engines, flexible structures, and shock/vibration testers
Engineering - Dynamic response testing, shock and vibration isolation, auto chassis
structural testing, structural analysis, reactors, control systems and materials evaluation
Aerospace - Ejection systems, rocketry, landing gear hydraulics, shock tube
instrumentation, wind tunnel and modal testing.
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Documents | 1,083 | 5,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.899585 |
https://www.section.io/engineering-education/linear-discriminant-analysis/ | 1,680,355,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00533.warc.gz | 1,066,054,377 | 14,304 | ## Deploy your apps to a supercloud in a few clicks
This Engineering Education program is supported by Section. Instantly deploy your GitHub apps, Docker containers or K8s namespaces to a supercloud.
# Linear Discriminant Analysis from Scratch
##### November 25, 2020
In this tutorial, we will look into the algorithm Linear Discriminant Analysis, also known as LDA. One should be careful while searching for LDA on the net. We also abbreviate another algorithm called Latent Dirichlet Allocation as LDA.
Linear Discriminant Analysis(LDA) is a supervised learning algorithm used as a classifier and a dimensionality reduction algorithm. We will look at LDA’s theoretical concepts and look at its implementation from scratch using NumPy.
Let’s get started.
### Installation
We will install the packages required for this tutorial in a virtual environment. We’ll use conda to create a virtual environment. For more installation information, refer to the Anaconda Package Manager website.
Create a new virtual environment by typing the command in the terminal. Perform this after installing anaconda package manager using the instructions mentioned on Anaconda’s website.
`conda create -n lda python=3.6`
This will create a virtual environment with Python 3.6.
We’ll be installing the following packages:
Activate the virtual environment using the command, `conda activate lda`. After activating the virtual environment, we’ll be installing the above mentioned packages locally in the virtual environment. To use these packages, we must always activate the virtual environment named `lda` before proceeding. If you choose to, you may replace `lda` with a name of your choice for the virtual environment.
To install the packages, we will use the following commands:
1. matplotlib: `pip3 install matplotlib`
2. numpy: `pip3 install numpy`
3. sklearn: `pip3 install sklearn`
Once installed, the following code can be executed seamlessly.
### Introduction
In some cases, the dataset’s non-linearity forbids a linear classifier from coming up with an accurate decision boundary. Therefore, one of the approaches taken is to project the lower-dimensional data into a higher-dimension to find a linear decision boundary. Consider the following example taken from Christopher Olah’s blog.
The other approach is to consider features that add maximum value to the process of modeling and prediction. If any feature is redundant, then it is dropped, and hence the dimensionality reduces. LDA is one such example.
It’s a supervised learning algorithm that finds a new feature space that maximizes the class’s distance. The higher the distance between the classes, the higher the confidence of the algorithm’s prediction.
The purpose for dimensionality reduction is to:
• Obtain the most critical features from the dataset.
• Visualize the dataset
• Have efficient computation with a lesser but essential set of features: Combats the “curse of dimensionality”.
Let’s say we are given a dataset with n-rows and m-columns. Where `n` represents the number of data-points, and `m` represents the number of features. `m` is the data point’s dimensionality.
Assuming the target variable has `K` output classes, the LDA algorithm reduces the number of features to `K-1`. Hence, the number of features change from `m` to `K-1`.
The aim of LDA is:
• Minimize the Inter-Class Variability: Inter-class variability refers to including as many similar points as possible in one class. This ensures less number of misclassifications.
• Maximize the Distance Between the Mean of Classes: The classes’ mean is placed as far as possible to ensure high confidence during prediction.
Below is a picture to help explain:
Image Source
The data-points are projected onto a lower-dimensional hyper-plane, where the above two objectives are met. In the example given above, the number of features required is 2. The scoring metric used to satisfy the goal is called Fischer’s discriminant.
The Fischer score is given as:
Fischer Score f(x) = (difference of means)^2/ (sum of variances).
We’re maximizing the Fischer score, thereby maximizing the distance between means and minimizing the inter-class variability.
### Code
Let’s consider the code needed to implement LDA from scratch.
We’ll begin by defining a class `LDA` with two methods:
1. `__init__`: In the `__init__` method, we initialize the number of components desired in the final output and an attribute to store the eigenvectors.
2. `transform`: We’ll consider Fischer’s score to reduce the dimensions of the input data. The Fischer score is computed using covariance matrices. The formula mentioned above is limited to two dimensions. We’ll be coding a multi-dimensional solution. Therefore, we’ll use the covariance matrices. The matrices `scatter_t`, `scatter_b`, and `scatter_w` are the covariance matrices. `scatter_w` matrix denotes the intra-class covariance and `scatter_b` is the inter-class covariance matrix. `scatter_t` covariance matrix represents a temporary matrix that’s used to compute the `scatter_b` matrix.
Using the scatter matrices computed above, we can efficiently compute the eigenvectors. The eigenvectors obtained are then sorted in descending order. The first `n_components` are selected using the slicing operation. If `n_components` is equal to 2, we plot the two components, considering each vector as one axis.
Finally, we load the iris dataset and perform dimensionality reduction on the input data. Then, we use the plot method to visualize the results.
``````import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
class LDA:
def __init__(self, n_components=None):
self.n_components = n_components
self.eig_vectors = None
def transform(self,X,y):
height, width = X.shape
unique_classes = np.unique(y)
num_classes = len(unique_classes)
scatter_t = np.cov(X.T)*(height - 1)
scatter_w = 0
for i in range(num_classes):
class_items = np.flatnonzero(y == unique_classes[i])
scatter_w = scatter_w + np.cov(X[class_items].T) * (len(class_items)-1)
scatter_b = scatter_t - scatter_w
_, eig_vectors = np.linalg.eigh(np.linalg.pinv(scatter_w).dot(scatter_b))
print(eig_vectors.shape)
pc = X.dot(eig_vectors[:,::-1][:,:self.n_components])
print(pc.shape)
if self.n_components == 2:
if y is None:
plt.scatter(pc[:,0],pc[:,1])
else:
colors = ['r','g','b']
labels = np.unique(y)
for color, label in zip(colors, labels):
class_data = pc[np.flatnonzero(y==label)]
plt.scatter(class_data[:,0],class_data[:,1],c=color)
plt.show()
return pc
LDA_obj = LDA(n_components=2)
X, y = data.data, data.target
X_train, X_test, Y_train, Y_test = train_test_split(X, y, test_size=0.2)
LDA_object = LDA(n_components=2)
X_train_modified = LDA_object.transform(X_train, Y_train)
print("Original Data Size:",X_train.shape, "\nModified Data Size:", X_train_modified.shape)
``````
The output of the code should look like the image given below. The iris dataset has 3 classes. Observe the 3 classes and their relative positioning in a lower dimension.
### Conclusion
In this article, we have looked at implementing the Linear Discriminant Analysis (LDA) from scratch. I suggest you implement the same on your own and check if you get the same output. Another fun exercise would be to implement the same algorithm on a different dataset. I hope you enjoyed reading this tutorial as much as I enjoyed writing it. Happy learning.
Peer Review Contributions by: Adrian Murage | 1,634 | 7,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-14 | latest | en | 0.825489 |
https://east-centricarch.eu/en/addison-earns-a-fixed-hourly-rate-working-as-a-sales-clerk-if-she-works-on-a-holiday-she-earns-a-d.4929367.html | 1,675,289,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00155.warc.gz | 241,815,720 | 11,850 | bishopsammi
21
# Addison earns a fixed hourly rate working as a sales clerk. If she works on a holiday, she earns a different hourly rate than she earns on a regular day. In one week, she earns \$188.50 by working 5 hours on a holiday and 16 hours during regular days. A different week, she earns \$254.00 by working 8 hours on a holiday and 20 hours during regular days. How much more is Addison’s holiday hourly rate than her regular hourly rate?
(1) Answers
egreen
_Award brainliest if helped! Holiday : 5 hours & 8 hours Regular Days : 16hours & 20hours Lets let Holiday be H and Regular days be R 5H + 16R = 188.50 (first info in qn) 8H + 20R = 254.00 Hence , solving both equations simultaneously, H = \$10.5 , R = \$8.5 How much more = 10.5 - 8.5 = 2 Addison earns \$2 more hourly from a holiday than regular days.
Add answer | 241 | 840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.933406 |
https://www.baeldung.com/scala/algebraic-data-types | 1,708,634,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00787.warc.gz | 679,721,599 | 44,380 | 1. Introduction
In this tutorial, we’re going to explore what Algebraic Data Types (ADT) are and how they can be used to define our data model in Scala.
2. What Are Algebraic Data Types?
ADTs are commonly used in Scala. Simply put, an algebraic data type is any data that uses the Product or Sum pattern.
The question now shifts to understanding what Product and Sum Types are.
3. The Product Type Pattern
When we need to model data containing other data we can use the product type pattern.
Let’s define a simplified model of chess pieces:
``````case class ChessPiece(color: Color, name: Name)
val rook = ChessPiece(White, Rook)
``````
In this case, “ChessPiece has a Color and a Name”. Later on in the article, we’ll drill down in the Color and Name definition.
In Object-Oriented terminology, this is a “has-a” relationship.
The ChessPiece data contains other data. A product type is often defined as a case class.
4. The Sum Type Pattern
If we need to define a type which can assume different values, we can leverage the sum type pattern.
Going back to our chess piece example we can define Color and Name as follows:
``````sealed trait Color
final case object White extends Color
final case object Black extends Color
sealed trait Name
final case object Pawn extends Name
final case object Rook extends Name
final case object Knight extends Name
final case object Bishop extends Name
final case object Queen extends Name
final case object King extends Name``````
Using object-oriented terminology, we can express the Sum Type relationship as “is a”.
In the above example “Color is a White or a Black”. Sealed traits are the way of defining Sum Types.
5. Some More Patterns
We’ve seen in the examples of the previous chapter has-a and is-a relationships combined with and/or operators:
• ChessPiece has a Color and a Name
• Color is a White or a Black
There are still two combinations we are missing: has-a + or and is-a + and.
The former, has-a + or, can be expressed in the following way:
``````trait Semaphore {
val color: Color
}
sealed trait Color
final case object Green extends Color
final case object Amber extends Color
final case object Red extends Color``````
In our code “Semaphore has-a Color, where Color is Green or Amber or Red”.
The last pattern we are gonna look at is is-a + and, which we can translate in code in this way:
``````trait Feline
trait Animal
trait Cat extends Animal with Feline``````
In our example “Cat is an Animal and a Feline”.
6. Pattern Matching
The use of pattern matching comes naturally when using ADTs, encouraging the use of this functional programming idiom.
In the following example, let’s see how to leverage the product and sum type to write a method telling us if a piece is the most important of our board. We can check the attribute name, a sum type, of ChessPiece, a product type, to determine which piece we are processing:
``````def isTheMostImportantPiece(c: ChessPiece): Boolean = c match {
case ChessPiece(_, King) => true
case _ => false
}``````
This is an example of how pattern matching comes extremely in handy in this case and goes along really well with ADT.
7. Conclusions
In this article, we saw how to translate object relationships in Scala code. Being aware of these patterns and recognizing them is important to write idiomatic functional code.
As always, the code is available over on GitHub. | 772 | 3,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.819409 |
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### Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Sums of Pairs
##### Age 11 to 16 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Square Pizza
##### Age 14 to 16 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Steel Cables
##### Age 14 to 16 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Generating Triples
##### Age 14 to 16 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### What's Possible?
##### Age 14 to 16 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Pareq Calc
##### Age 14 to 16 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Pick's Theorem
##### Age 14 to 16 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Attractive Tablecloths
##### Age 14 to 16 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Semi-square
##### Age 14 to 16 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Pinned Squares
##### Age 14 to 16 Challenge Level:
What is the total number of squares that can be made on a 5 by 5 geoboard?
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Areas of Parallelograms
##### Age 14 to 16 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Steps to the Podium
##### Age 7 to 14 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Plus Minus
##### Age 14 to 16 Challenge Level:
Can you explain the surprising results Jo found when she calculated the difference between square numbers?
### Painted Cube
##### Age 14 to 16 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### A Tilted Square
##### Age 14 to 16 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Nim
##### Age 14 to 16 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Nim-7 for Two
##### Age 5 to 14 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Konigsberg Plus
##### Age 11 to 14 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Magic Letters
##### Age 11 to 14 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Special Sums and Products
##### Age 11 to 14 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Go Forth and Generalise
##### Age 11 to 14
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Got it for Two
##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Picturing Triangular Numbers
##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Shear Magic
##### Age 11 to 14 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Handshakes
##### Age 11 to 14 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Nim-like Games
##### Age 7 to 16 Challenge Level:
A collection of games on the NIM theme
### Consecutive Negative Numbers
##### Age 11 to 14 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### How Much Can We Spend?
##### Age 11 to 14 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Egyptian Fractions
##### Age 11 to 14 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written different fractions.
### Keep it Simple
##### Age 11 to 14 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions? | 2,165 | 9,262 | {"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} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.884513 |
https://natbanting.com/a-viral-area-task/ | 1,726,608,965,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00157.warc.gz | 377,192,468 | 16,784 | Categories
Exactly one month ago, fellow Saskatchewan mathematics teacher Ilona Vashchyshyn tweeted about an area task that she used in her class. Long story short, it captured the imagination of Math Ed Twitter like elegant tasks have a tendency of doing.
Two weeks later, my grade 9s and I were beginning a unit on surface area and this seemed like a perfect entry point. I thought it would provide a perfect window into what geometric knowledge they were bringing with them into the unit. Because my crew would be dealing with the surface area of a cylinder, I altered the task slightly:
Write your name so that it covers an area between 99 and 100 cm squared.
*you must include at least one rectangle, one triangle, and one circle in your design.
Student work:
The results were as creative as expected. The task seems to encourage a commitment to symmetry with many students calculating a sort of square budget for each letter of their names or making all letters uniform in height. Also, their attention was immediately drawn to the shape of most resistance–the circle. There was almost an immediate recognition that rectangles and triangles live nicely on a grid, but circles do not. This became an important talking point throughout their work.
Beyond the products:
As the students worked, three important themes came out of our conversations. These pieces of intel gave me a sense of where we were collectively strong and where more connections were needed.
1. Unearthing of vocabulary
The task provided plenty of opportunity to talk about the vocabulary associated with area. Words like dimension, formula, area, radius, diameter, base, height, and squared were all discussed. I really like introductory tasks that provide opportunities to acclimatize to the verbiage while content demands remain relatively low.
2. Connecting counting to calculating
The grid encourages students to count squares in order to tally their total square area. While this is a nice connection between the idea of shaded squares and square units of area, I was interested if the students could also assign dimensions to shapes and calculate their area with appropriate formulae. Requiring them to incorporate a circle made calculation a necessity, but also provided the opportunity to connect this back to counting.
After having several conversations with students about the link between counting and calculating, I asked some students to put their work under the document camera. My set-up is an iPevo VZ-R, which allows me to clearly project and zoom in on sections of student work. The camera has all kinds of functionality, but I have found it most useful in my classroom as a tool for sharing ideas. (I often “pause” activity half way to discuss strategies that are emerging in the classroom).
As a class, we focused on the ways that students had incorporated the circles. We then calculated their area with the formula and compared with a counting strategy to check their correlation. This exercise helped students add meaning to the calculations that (I knew) would dominate much of their activity in the unit.
3. Practicing sub-sectioning and calculating
We started outlining the dimensions of the shapes we were viewing under the document camera. It didn’t take long for students to see the compound calculations in different ways. For instance, the “L” in “Laurier” can be divided into a 6×1 rectangle and a 1×3 rectangle. Alternatively, it can be dissected into a 5×1 rectangle and a 1×4 rectangle. This slicing of compound shapes into more familiar ones is a critical skill when calculating surface area. One student suggested that we could calculate the area of the “A” in “Laurier” by tracing a 6×5 rectangle around the entire letter and then subtracting the triangles created “on either roof”. In the end, the class decided that this resulted in more work, but zooming in and tracing out the idea provided us the chance to discuss the possibility of multiple correct dissections.
___
I think the immediate appeal of the task (as evidenced by the sheer volume of social media activity) was in the beautiful creations of students. However, its value goes beyond what appears on the paper; the simple prompt is a powder keg for critical conversations about area.
NatBanting | 853 | 4,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.974642 |
https://www.physicsforums.com/threads/circular-motion-revolutions.641629/ | 1,531,740,028,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589251.7/warc/CC-MAIN-20180716095945-20180716115945-00616.warc.gz | 973,058,349 | 13,493 | # Homework Help: Circular Motion, Revolutions
1. Oct 6, 2012
### SherBear
A car is moving at a speed of 20 m/s in 4 seconds and the car's tire is 38cm around which is its diameter. How many revolutions did your tires make?
Relevant Equations, T=2∏r / v
My attempt at a solution:
To get the distance travelled=20 m/s (4 sec.) = 80 m
2∏(0.19m)= 1.19 m
80m / 1.19 m = 67.22 Rev
67.22 rev / 4 sec. = 16.805 rev/s ?
is this wrong?
2. Oct 6, 2012
### nasu
Is this the actual text of the problem or your interpretation?
I won't assume that "38 cm around" means diameter. Diameter is rather "across" that "around".
3. Oct 6, 2012
### SherBear
My interpretation it says 38 cm = Diameter
4. Oct 6, 2012
### nasu
OK, then the only thing is that the question (as you wrote it) asks how many revolutions and not how many per second.
5. Oct 6, 2012
### SherBear
is 67.22 rev correct then?
6. Oct 6, 2012
### nasu
It looks OK. If this is what the problem requires.
7. Oct 6, 2012
### SherBear
Thank you very much for your time and patience nasu =-) | 342 | 1,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-30 | latest | en | 0.921029 |
http://gmatclub.com/forum/mgcat-vs-kaplan-vs-gmatclub-166560.html | 1,485,023,285,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00488-ip-10-171-10-70.ec2.internal.warc.gz | 115,492,904 | 46,977 | MGCAT vs. Kaplan vs. GMATCLUB : General GMAT Questions and Strategies
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# MGCAT vs. Kaplan vs. GMATCLUB
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MGCAT vs. Kaplan vs. GMATCLUB [#permalink]
### Show Tags
23 Jan 2014, 22:19
Hi All...
I was taking all those CATs & my scores vary quite a bit...
Any idea/comment?
MGCAT1: 670 (Q46 V35)
MGCAT2: 710 (Q51 V36)
MGCAT3: 670 (Q46 V35)
Kaplan1: 720 (Q51 V35)
Kaplan2: 760 (Q51 V42)
Kaplan3: 750 (Q51 V42)
GMATCLUB1: Q51 V38 ~730
GMATCLUB2: Q51 V40 ~750
GMATCLUB3: Q51 V35 ~720
GMATCLUB4: Q51 V39 ~740
GMATCLUB5: Q51 V46 ~770?
Kaplan GMAT Prep Discount Codes Math Revolution Discount Codes Magoosh Discount Codes
Intern
Joined: 15 Sep 2013
Posts: 23
Concentration: General Management
GMAT 1: 770 Q50 V46
GPA: 3.5
Followers: 0
Kudos [?]: 7 [0], given: 19
Re: MGCAT vs. Kaplan vs. GMATCLUB [#permalink]
### Show Tags
26 Jan 2014, 16:16
tsunghop wrote:
Hi All...
I was taking all those CATs & my scores vary quite a bit...
Any idea/comment?
MGCAT1: 670 (Q46 V35)
MGCAT2: 710 (Q51 V36)
MGCAT3: 670 (Q46 V35)
Kaplan1: 720 (Q51 V35)
Kaplan2: 760 (Q51 V42)
Kaplan3: 750 (Q51 V42)
GMATCLUB1: Q51 V38 ~730
GMATCLUB2: Q51 V40 ~750
GMATCLUB3: Q51 V35 ~720
GMATCLUB4: Q51 V39 ~740
GMATCLUB5: Q51 V46 ~770?
You could get a more precise idea from the Gmat Prep software, but in any case, with grades like that, you may also want to book your exam.
Re: MGCAT vs. Kaplan vs. GMATCLUB [#permalink] 26 Jan 2014, 16:16
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# MGCAT vs. Kaplan vs. GMATCLUB
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,000 | 2,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-04 | latest | en | 0.75527 |
http://www.spoj.com/problems/IITKESO207PA1Q2/ | 1,521,282,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257644877.27/warc/CC-MAIN-20180317100705-20180317120705-00412.warc.gz | 470,170,001 | 7,896 | ## IITKESO207PA1Q2 - Queues
no tags
In this problem, you have to implement queues in C/C++. There are operations that you must implement.
1) Enqueue : Takes an element and adds it to the back of the queue.
2) Dequeue: Removes the first element in the queue.
3) Is_Empty : Returns true if the queue is empty, false otherwise.
Note: Using standard library implementation is NOT allowed and will be considered cheating. The penalty shall be applied as it given in the handout is for cheating cases.
### Input
First line contains t: the number of test cases. Each test case has lines.
The first line contains q, the number of queries. q lines follow.
Each query is has one or two space separated integers. The first is the opcode. 1 for Enqueue, 2 for Dequeue and 3 for Is_Empty.
If the opcode is 1, then there is another argument x that denotes the element to be enqueued. The opcodes 2 and 3 are not followed by any number.
### Output
The output is as follows:
For the Dequeue operation output the element that has been dequeued. If the queue is empty and a dequeue operation is asked, print "Empty" (without quotes).
For the Is_Empty operation output True/False.
For the Enqueue opearation, print the element enqueued.
### Constraints
1 <= t <= 10
1 <= q <= 1000000
1 <= x <= 10^9
### Example
```Input:
171 31 423223```
```Output:
343False4EmptyTrue```
### Explanation
The first line of the input says that there is only 1 test case.
The second line says there will be 7 queries.
The third line is 1 3 which means that we must enqueue 3 to the queue; the output is 3. Fourth line says that we enqueue 4 to the queue; the output is 4.
The fifth line asks for a dequeue operation and the output should be 3. The sixth line asks whether the queue is empty or not and the answer is False.
Added by: Programming Club, IITK Date: 2018-01-11 Time limit: 1s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: C CPP C++ 4.3.2 CPP14 C99 | 518 | 1,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.858092 |
https://medium.com/iotatangle/explaining-the-qubic-computation-model-part-2-f10f8cd5cbc5 | 1,670,437,360,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00076.warc.gz | 421,209,862 | 56,244 | Published in
IOTA
# Explaining the Qubic Computation Model: part 2
In the first part of this series we have looked at a conceptual overview of the Abra specification, which describes the how dataflow is achieved within the Qubic Computation Model. In this second part, we will use the Abra specification to start implementing Qupla, a higher level programming language for Qubic.
# Qupla: a Qubic programming language
Qupla is an acronym that stands for Qubic programming language. The goals we are trying to achieve with Qupla are twofold:
1. Provide a trinary data flow programming language according to the Abra specification that can function as a higher level programming language for Qubic.
2. Lower the barrier for programmers to get started with Qubic programming by leveraging existing knowledge as much as possible. This means that we try to keep the language and its behavior as familiar as possible while still providing access to the new and unfamiliar functionality of the Qubic Computation Model.
At first glance, Qupla has a number of things in common with existing programming languages. Some concepts, entities and constructs will look very familiar to programmers, but there are also a few twists that will be completely new. The most obvious new twist, of course, is that Qupla is a trinary programming language.
## Trinary encoding
Binary systems use bits to represent code and data, where each bit can assume only one of 2 possible values. Larger values can be represented as binary numbers by using a series of bits. The values that a single bit can assume are generally 0 and 1. Larger value of N bits are usually encoded either as unsigned numbers in the range from 0 to (2^N)-1, or as two’s complement signed numbers in the range from -(2^(N-1)) to (2^(N-1))-1.
Similarly, trinary systems use so-called trits to represent code and data, where each trit can assume only one of 3 possible values. Larger values can be represented as trinary numbers by using a series of trits. Just like with binary numbers, there are 2 ways of representing larger trinary values:
• Unbalanced trinary representation, which allows for a single trit to assume the values 0, 1, and 2. Using unbalanced trinary representation a series of N trits can represent unsigned values from 0 to (3^N)-1.
• Balanced trinary representation, which allows for a single trit to assume the values 0, 1, and -1. Using balanced trinary representation a series of N trits can represent signed values from -((3^N)-1)/2 to ((3^N)-1)/2.
`Examples of upper values for representation rangesNote that binary signed values run from -N-1 to NBalanced trinary signed values run from -N to NBits/ Upper value (N)trits Binary Trinary 2 1 4 3 3 13 4 7 40 5 15 121 6 31 364 7 63 1093 8 127 3280 9 255 9841 10 511 29524`
## Data types
The Abra specification allows us complete freedom to choose how a trit vector will be interpreted. But a programming language needs to provide a number of standard concepts to be useful. Therefore we have chosen that the Qupla language will interpret a trit vector that represents an integer number as balanced trinary because of certain desirable properties of this representation when using it for math.
Trit vectors that encode integer numbers are most naturally encoded in Qupla with the least significant trit first. This allows us to extend a value to the same value for a larger trit vector by simply padding the value with enough zero trits at the end. This works for both positive and negative values.
The opposite action, shrinking a larger value to fit in a smaller trit vector, is similarly easy. Just truncate the trit vector to the required size. Of course in this case if any non-zero trits get truncated away the resulting value will become different. But this is unavoidable and similar to what happens when you try to fit a 64-bit integer into a 32-bit integer in binary systems.
Limiting data type sizes to the minimum necessary size for the value range achieves a reduction of energy needs by limiting the amount of circuitry actually needed on FPGAs. Imagine a variable that only needs to be able to represent the values 0–10. With most traditional programming languages you would need to use an 8-bit (0 to 255) byte variable at a minimum, where 4 bits (0 to 15) would have sufficed, and the hardware is designed to manipulate bytes at a time anyway. With Qupla, by following Abra, you can suffice with a trit vector of 3 trits (-13 to 13) instead, and on FPGA you end up with only the circuitry necessary to manipulate 3 trits.
## User-defined fixed-size trit vectors
Even though Qupla only has the trit vector as its single built-in data type, it is still possible to define convenient reusable named data types by defining the type name and its size. This way you have an easy way to use a symbolic name for certain fixed-size trit vector sizes and you can use these throughout the Qupla code. The main advantages of this way of indicating trit vector size are that you are programming using concepts instead of hard numbers, and that you only have to change a single location in your code in case it turns out that the trit vector size would need changing.
`// define some standard trit lengths we will use in our examples// note that you can define the optimal trit size for any type's// range to reduce energy requirements accordingly, but for// simplicity we'll only define power-of-3 sized types for nowtype Trit [1] // -/+ 1type Tryte [3] // -/+ 13type Tiny [9] // -/+ 9,841type Int [27] // -/+ 3,812,798,742,493type Huge [81] // -/+ 221,713,244,121,518,884,974, // 124,815,309,574,946,401type Hash [243] // standard 81 trytes hash valuetype Hash3 [Hash * 3]type Hash9 [Hash * 9]type Signature [Hash * 27]// define a convenience type to make code more readable// should always be a binary boolean value false or true// (note: this convention is not (yet) enforced by Qupla)type Bool [Trit]`
In the above example we have defined several data types that we will use in the examples that follow. The `type` keyword indicates that what follows is a user-defined trit vector type declaration. Next we specify the name of the user-defined trit vector type, followed by the size of the trit vector type between square brackets. We decided to write it like this in Qupla because of the similarity to how other computer languages specify array/vector sizes.
Note how a trit vector size can be defined in terms of an earlier defined data type as we do with the `Signature` data type for example. When using a data type name in an expression that defines the size of a trit vector, the data type name serves as a constant value which represents the size of that data type.
Also note how we can use constant arithmetic to calculate trit vector sizes like we do with the definition of the `Hash3`, `Hash9`, and `Signature` data types in the example above. Qupla supports the C-like operators `+`,` -`, `*`, `/`, and `%` in constant expressions.
## Structured trit vectors
Qupla also allows you to construct a named data type for a structured trit vector. A structured trit vector is a trit vector that consists of named sub-vectors. This essentially turns a trit vector into a structure with named fields. This way it is easy to access the sub-vectors without constantly having to keep track of their respective offsets within the main trit vector.
A structured trit vector is represented by a single trit vector that concatenates all sub-vectors. Its total size therefore is the sum of the sizes of all sub-vectors.
`type TinyFloat { Tiny mantissa // -/+ 9,841 Tryte exponent // -/+ 3^13}`
Note how we have defined the sub-vectors `mantissa` and `exponent` in terms of the previously defined data types. The `TinyFloat` structured trit vector will therefore have a size of `Tiny + Tryte` or 12 trits. You can even nest structured trit vectors by defining a structured trit vector as a separate type and then using that type for a field in another structured trit vector. There are no limits on the depth of nesting structured trit vectors.
## Look-up tables
At the heart of Abra is the look-up table (LUT). Qupla of course implements LUTs as well, but it adds a little twist. In Abra, a LUT always has a single trit as output. Qupla allows a LUT to have multiple trits as output. Behind the scenes Qupla will map a multi-trit output Qupla LUT to multiple single-trit output Abra LUTs for us.
A LUT in Qupla is a named table of sequences of 1, 2, or 3 input trit values and one or more corresponding output trit values. When using a LUT in a look-up operation the output trit values are returned as a single trit vector.
All input trit sequences in a LUT must be of exact same length and cannot exceed 3 trits. Each input trit sequence in a LUT must have a unique combination of values. Additionally, all output trit sequences must be of exact same length. Qupla will generate a syntax error if any of these rules are broken.
Any missing combination of input values in the LUT will cause the LUT to return a null value for that combination. Note that it is only possible to specify non-null trit vectors for input and output. This means that if any trit in the input vector is null, the resulting output trit vector will automatically be null as well.
This property where a LUT returns null in certain cases can be used to our advantage to implement conditional execution. We will explain this aspect in more detail in a separate section.
LUTs can be used to obviate the need for most common arithmetic and logical operations. These operations are instead implemented as functions that wrap one or more LUT operations.
`// LUT logic: return -trit1lut neg { - = 1 0 = 0 1 = -}// Abra pseudo-code equivalent:lut 10-10-10-10-10-10-10-10-10- neg_0`
In the example above the `lut` keyword indicates that what follows is a LUT declaration. It is followed by the name of the LUT and one or more LUT entries between curly braces.
The above `neg` LUT is one of the simplest. It will return a trit that is the negative value of the single input trit. Note how the table consists of a simple enumeration of what trit value to return for every possible input trit value. A balanced trinary number can be made negative very easily by negating each separate trit. So to negate an entire trit vector you would run each separate trit through the `neg` LUT. On FPGAs this can easily be done in parallel for all trits in the trit vector.
Note how the Abra equivalent LUT block definition repeats the output values to fill up all the 27 entries in the LUT definition.
`// LUT logic: return (Bool) (trit1 == trit2)lut equal { -,- = true -,0 = false -,1 = false 0,- = false 0,0 = true 0,1 = false 1,- = false 1,0 = false 1,1 = true}// Abra pseudo-code equivalent:lut 100010001100010001100010001 equal_0`
This slightly more complex `equal` LUT determines if the two input trits are equal. It returns `true` when both input trits are equal and `false` otherwise. Since it always returns `true` or `false` the result is essentially a `Bool` data type. Again, note that this is a simple enumeration of all possible input trit combinations and what trit value to return for each respective input combination.
`// LUT logic: return (Bool) (trit1 != trit2)lut unequal { -,- = false -,0 = true -,1 = true 0,- = true 0,0 = false 0,1 = true 1,- = true 1,0 = true 1,1 = false}// Abra pseudo-code equivalent:lut 011101110011101110011101110 unequal_0`
The `unequal` LUT determines if the two input trits are unequal. It returns `false` when both input trits are equal and `true` otherwise. Since it always returns `false` or `true` the result is essentially a `Bool` data type. Although we could achieve the same result by combining the previous `equal` LUT with a logical `not` LUT (see further below), the `unequal` LUT is more efficient because it only needs to do a single look-up.
`// LUT logic: return the sum of two trits as trit plus a carry// return (trit1 + trit2), carry(trit1 + trit2)lut halfAdd { -,- = 1,- // -1 + -1 = 1, carry -1 -,0 = -,0 // -1 + 0 = -1, carry 0 -,1 = 0,0 // -1 + 1 = 0, carry 0 0,- = -,0 // 0 + -1 = -1, carry 0 0,0 = 0,0 // 0 + 0 = 0, carry 0 0,1 = 1,0 // 0 + 1 = 1, carry 0 1,- = 0,0 // 1 + -1 = 0, carry 0 1,0 = 1,0 // 1 + 0 = 1, carry 0 1,1 = -,1 // 1 + 1 = -1, carry 1}// Abra pseudo-code equivalent:lut 1-0-0101-1-0-0101-1-0-0101- halfAdd_0lut -00000001-00000001-00000001 halfAdd_1`
This `halfAdd` LUT demonstrates how multi-trit output values can be achieved. It will simply add the two input trits and return the resulting trit plus a carry. Again, it’s an enumeration of all possible input combinations with their respective outputs. Especially note how all input sizes are the same and how all output sizes are the same.
The Abra pseudo-code shows how a 2-trit output LUT is converted into two 1-trit output LUTs. We will see in the section about LUT invocation that the code that accesses this LUT automatically invokes both LUTs and concatenates the 2 resulting trits into a single trit vector.
`// LUT logic: return (trit1 == true) ? trit2 : null;lut nullifyTrue { true,- = - true,0 = 0 true,1 = 1}// LUT logic: return (trit1 == false) ? trit2 : null;lut nullifyFalse { false,- = - false,0 = 0 false,1 = 1}// Abra pseudo-code equivalent:lut @@-@@0@@1@@-@@0@@1@@-@@0@@1 nullifyTrue_lut @-@@0@@1@@-@@0@@1@@-@@0@@1@ nullifyFalse_`
The `nullifyTrue` and `nullifyFalse` LUTs demonstrate how we can use missing input trit combinations to our advantage. They will return the second input trit only when the first input trit is `true `or `false`, respectively, and they will return null in all other cases. These LUTs will turn out to be pivotal in the way we create Qupla decision logic, because they can be used to create functions that turn an entire trit vector into null if the input condition is not met. You would read a `nullifyTrue` look-up as “nullify the value when the input condition is not true”.
Qupla will automatically create these two LUTs to support its conditional operator as we will see in the next part of this series.
`// ************* BINARY OPERATORS *************// LUT logic: binary NOT// return !trit1lut not { false = true true = false}// LUT logic: binary AND// return (trit1 & trit2)lut and { false, false = false false, true = false true, false = false true, true = true }// LUT logic: binary AND// return (trit1 & trit2 & trit3)lut and3 { false, false, false = false false, false, true = false false, true, false = false false, true, true = false true, false, false = false true, false, true = false true, true, false = false true, true, true = true }// LUT logic: binary OR// return (trit1 | trit2)lut or { false, false = false false, true = true true, false = true true, true = true }// LUT logic: binary OR// return (trit1 | trit2 | trit3)lut or3 { false, false, false = false false, false, true = true false, true, false = true false, true, true = true true, false, false = true true, false, true = true true, true, false = true true, true, true = true }// LUT logic: binary XOR// return (trit1 ^ trit2)lut xor { false, false = false false, true = true true, false = true true, true = false}// LUT logic: binary XOR// return (trit1 ^ trit2 ^ trit3)lut xor3 { false, false, false = false false, false, true = true false, true, false = true false, true, true = false true, false, false = true true, false, true = false true, true, false = false true, true, true = true }// LUT logic: binary NAND// return !(trit1 & trit2)lut nand { false, false = true false, true = true true, false = true true, true = false}// LUT logic: binary NAND// return !(trit1 & trit2 & trit3)lut nand3 { false, false, false = true false, false, true = true false, true, false = true false, true, true = true true, false, false = true true, false, true = true true, true, false = true true, true, true = false}// LUT logic: binary NOR// return !(trit1 | trit2)lut nor { false, false = true false, true = false true, false = false true, true = false}// LUT logic: binary NOR// return !(trit1 | trit2 | trit3)lut nor3 { false, false, false = true false, false, true = false false, true, false = false false, true, true = false true, false, false = false true, false, true = false true, true, false = false true, true, true = false}// LUT logic: binary XNOR// return !(trit1 ^ trit2)lut xnor { false, false = true false, true = false true, false = false true, true = true }// LUT logic: binary XNOR// return !(trit1 ^ trit2 ^ trit3)lut xnor3 { false, false, false = true false, false, true = false false, true, false = false false, true, true = true true, false, false = false true, false, true = true true, true, false = true true, true, true = false}// Abra pseudo-code equivalent:lut @10@10@10@10@10@10@10@10@10 not_0lut @@@@00@01@@@@00@01@@@@00@01 and_0lut @@@@@@@@@@@@@00@00@@@@00@01 and3_0lut @@@@01@11@@@@01@11@@@@01@11 or_0lut @@@@@@@@@@@@@01@11@@@@11@11 or3_0lut @@@@01@10@@@@01@10@@@@01@10 xor_0lut @@@@@@@@@@@@@01@10@@@@10@01 xor3_0lut @@@@11@10@@@@11@10@@@@11@10 nand_0lut @@@@@@@@@@@@@11@11@@@@11@10 nand3_0lut @@@@10@00@@@@10@00@@@@10@00 nor_0lut @@@@@@@@@@@@@10@00@@@@00@00 nor3_0lut @@@@10@01@@@@10@01@@@@10@01 xnor_0lut @@@@@@@@@@@@@10@01@@@@01@10 xnor3_0`
The above list of LUTs implement the standard binary logic gates and can be used in conjunction with the `Bool` data type to create logical expressions. They only take the `Bool` input trits `false` and `true` and will return such a `Bool` as output trit. Any other input is undefined and therefore returns null.
Note the special versions `and3`, `or3`, `xor3`, `nand3`, `nor3`, and `xnor3`, that can take 3 inputs and perform the logical function on 3 operands at once.
These binary LUTs can be used to specify logical conditions in Qupla expressions. Because a `Bool` is a single trit these operations can be implemented directly as Abra LUT look-ups and are therefore really fast.
## Constants
The only types of numeric constants that currently can be used in Qupla are integer literals and floating point literals. For human convenience both these literals by default utilize standard decimal notation, but it is possible to use binary and hexadecimal integer values by prefixing such values with `0b` and `0x`, respectively. For example, `0b11001000` (binary for the decimal value 200), or `0x4e20` (hexadecimal for the decimal value 20,000). These notations should be familiar to users of many other programming languages.
Qupla also defines the boolean literals `false` and `true`, that can be used for clarity wherever a `Bool` type is used. Note that Qupla does not (yet) enforce these boolean types, so it is currently possible to bypass the use of `false` and `true` and therefore it is possible to set it to a value that is not `false` or `true`.
Finally, Qupla defines trinary literals, which are series of trits or trytes that can be used as constant for any trit vector, by prefixing such values with `0t`. For example, `0t-111-1` (trinary for the decimal value 200), or `0tKPWCHICGJZXKE9GSUDXZYUQPLHU` (a 27-tryte hash value).
Note an important difference between the trinary constants and the other constants. Trinary constants are always specified as big-endian, meaning that leading zeroes are significant and trailing zeroes are irrelevant when specifying a value. Binary, decimal, and hexadecimal constants are always specified as little-endian, meaning that leading zeroes are irrelevant and trailing zeroes are significant. This makes these latter constants human-readable and familiar.
## Integer literals
An integer literal will automatically be converted to a trit vector of the minimum length necessary to be able to hold the integer value that represents the balanced trinary value equivalent of the integer value. These literals can be used in any place a trit vector is expected. There is also a comprehensive set of arithmetic functions that is provided by the standard Qupla library that accompanies the language.
When the constant is assigned to a parameter or variable, it will automatically be extended to the size of the assignment target. In this case, a value that is larger than the target size will cause a compilation error. Note that the only limits to an integer literal value are those imposed by the boundaries of the range that the receiving trit vector can represent.
When using constant values for a single trit, we are allowed to shorten the -1 to just the minus character. This means that a single trit can assume the values `0`, `1`, and `-`. You will notice that this will allow us to represent tabular data in much more compact and readable form. The most visible structure in Qupla that uses such tabular data is the look-up table.
## Floating point literals
A floating point literal can only be used in places where a floating point type is expected. Floating point arithmetic functions are provided by the standard Qupla library that accompanies the language. Whenever a structured trit vector is encountered that consists of a `mantissa` and `exponent` field (which both can have any size), Qupla will allow a floating point literal to be assigned.
Floating point literals will always assume the size of the assignment target. As with integer literals, a value that exceeds either of the structure fields will cause a compilation error, but those are the only limits imposed upon the floating point value.
## Constant representation in Abra
So now we know that Qupla defines constants. That’s all very nice, but there is nothing in the Abra specification that mentions constants. Remember, Abra is totally agnostic about the meaning of the contents of trit vectors, and it seems that it does not provide a way to specify them, either. So what kind of solution can we come up with to still be able to use this arguably very useful concept in Qupla?
To answer that, we have to think about what Abra describes: data flow. We know that Abra needs to receive input data to be able to transform it into output data. That means that even constants should hold true to that rule and only create the constant value when input data is received. This is our first clue. Abra uses branches with an input vector triggering the output data for the branch. So we need a mechanism that creates a constant value given an input vector, where the input vector could be anything. This is our second clue. Because we have a mechanism that can do just that: the LUT.
Let’s start building the smallest constants first: single trit values. How do we get a constant for a zero trit for example? We’ll need a LUT that returns 0 for any input. All we need to trigger data flow is a single input trit, and for any single input trit to return a 0 output trit, we can use the following Abra LUT definition block: `lut 000000000000000000000000000 constZero_`. Good! Things are moving forward.
Next challenge: where do we get the triggering input trit from? Well that should be easy. Any branch is triggered by an input trit vector. So why not take the first trit from that input vector as input for the LUT? Now the problem becomes how to take only a single trit from what could be a trit vector of any length. This is where one of the peculiarities of Abra comes in handy. We can define a branch that has only a single trit as input. Remember that in Abra you can pass any size trit vector to a branch, but the branch itself decides what parts to use and what parts to ignore. The branch can use the single trit to invoke the LUT through a LUT knot site, and use that as an output site to immediately return the resulting trit as output trit from the branch. In Abra pseudo-code it looks like this:
`lut 000000000000000000000000000 constZero_lut 111111111111111111111111111 constOne_lut --------------------------- constMin_branch const_1_0[1] in trit[1] out ret = constZero_[trit]branch const_1_1[1] in trit[1] out ret = constOne_[trit]branch const_1_T[1] in trit[1] out ret = constMin_[trit]`
So here we have 3 branches that can generate the 3 different constant values for a single trit. Now that we have the basic idea we can build it out in several ways. Let’s keep it simple for now. Just have each branch for each constant first set up the constants for zero, one, and min trits as body sites, and then use those as outputs. Remember that multiple outputs will be concatenated, so we can use that to our advantage. All we need is one additional trick. To use a body site as output you can use a merge with only one input. Again, there are other ways of doing this but this one is the straightforward way with our current knowledge. Let’s take the constant value `0t-111-1` (200) as our example:
`branch const_6_T111T1[1] in trit[1] zero = constZero_[trit][1] one = constOne_[trit][1] min = constMin_[trit][1] out ret0 = merge{min}[1] out ret1 = merge{one}[1] out ret2 = merge{one}[1] out ret3 = merge{one}[1] out ret4 = merge{min}[1] out ret5 = merge{one}`
Notice how we encode the constant branch name to hold the size and the value. That makes each different constant branch name unique and you immediately know what size and constant it returns.
Notice that in this case we could have left the `zero` site out because the constant does not contain a zero and therefore it is not referenced at all. But you don’t have to worry about that in practice. Qupla knows exactly how to create the optimal code for a constant branch for you and do the translation to Abra for you. And even though this may seem like a lot of code for a simple constant, in practice on CPU implementations of Qupla the actual constant trit vector will be cached and referenced directly. On FPGA this will reduce to parallel invocations of the LUT knot sites and all the rest, inputs and outputs, will reduce to simple wiring.
## Conclusion
The basic data type in Qupla is the trit vector. We can create user-defined types for specific fixed-sized trit vectors and use constant values that will automatically convert to trit vectors. Another basic entity in Qupla is the look-up table, which is key to the inner working of many of Qupla’s concepts.
In the next part of this series we will delve deeper into how these basic entities are used with Qupla programming language constructs.
--
-- | 6,989 | 26,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.940369 |
https://reddingvwclub.org/and-pdf/1444-parametric-and-nonparametric-tests-in-statistics-pdf-91-235.php | 1,642,469,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00481.warc.gz | 558,519,524 | 6,646 | and pdfSunday, December 20, 2020 8:26:54 PM2
# Parametric And Nonparametric Tests In Statistics Pdf
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## What is the difference between a parametric and a nonparametric test?
Topics: Hypothesis Testing , Statistics. That sounds like a nice and straightforward way to choose, but there are additional considerations. Nonparametric tests are like a parallel universe to parametric tests. The table shows related pairs of hypothesis tests that Minitab Statistical Software offers. Reason 1: Parametric tests can perform well with skewed and nonnormal distributions. This may be a surprise but parametric tests can perform well with continuous data that are nonnormal if you satisfy the sample size guidelines in the table below.
The three modules on hypothesis testing presented a number of tests of hypothesis for continuous, dichotomous and discrete outcomes. Tests for continuous outcomes focused on comparing means, while tests for dichotomous and discrete outcomes focused on comparing proportions. All of the tests presented in the modules on hypothesis testing are called parametric tests and are based on certain assumptions. For example, when running tests of hypothesis for means of continuous outcomes, all parametric tests assume that the outcome is approximately normally distributed in the population. This does not mean that the data in the observed sample follows a normal distribution, but rather that the outcome follows a normal distribution in the full population which is not observed.
Before you order, simply sign up for a free user account and in seconds you'll be experiencing the best in CFA exam preparation. Quantitative Methods 2 Reading Hypothesis Testing Subject Parametric and Non-Parametric Tests. Seeing is believing! Find out more.
## A Parametric Approach to Nonparametric Statistics
Let us begin this article with the obvious—in the process of data analysis, always look at the data first. By that we mean investigators look first at the numerical and graphical summaries of the data. Checking out the data first provides an overview of the overall project, gives a clearer understanding of the variables and their values, and shows how the values are distributed. How the data is distributed data distribution is characterized by its center , its spread , and the shape of the data. The center refers to the middle of the value distribution, along with an estimate of the value typical of the data.
Two types of statistical methods may be used when analysing. data—parametric or non-parametric tests. Non-parametric. methods are also.
## Parametric and Non-parametric tests for comparing two or more groups
In terms of selecting a statistical test, the most important question is "what is the main study hypothesis? For example, in a prevalence study there is no hypothesis to test, and the size of the study is determined by how accurately the investigator wants to determine the prevalence.
Им необходим ключ, который хранится у Хейла. Необходим прямо. Она встала, но ноги ее не слушались.
Момент истины настал в одно ненастное октябрьское утро. Провели первый реальный тест. Несмотря на сомнения относительно быстродействия машины, в одном инженеры проявили единодушие: если все процессоры станут действовать параллельно, ТРАНСТЕКСТ будет очень мощным. Вопрос был лишь в том, насколько мощным. Ответ получили через двенадцать минут.
Такси следовало за Беккером, с ревом сокращая скорость. Свернув, оно промчалось через ворота Санта-Крус, обломав в узком проезде боковое зеркало. Беккер знал, что он выиграл. Санта-Крус - самый старый район Севильи, где нет проездов между зданиями, лишь лабиринт узких ходов, восходящих еще к временам Древнего Рима. Протиснуться здесь могли в крайнем случае только пешеходы, проехал бы мопед.
Стратмор прикрыл ее своим пиджаком. В нескольких метрах от них лежало тело Хейла. Выли сирены. Как весенний лед на реке, потрескивал корпус ТРАНСТЕКСТА.
1. ## Eddie F.
24.12.2020 at 22:45 | 951 | 4,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-05 | latest | en | 0.90815 |
https://stats.stackexchange.com/questions/140899/c4-5-how-to-select-the-split-point-threshold-for-a-continuous-attribute | 1,722,654,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00038.warc.gz | 435,845,120 | 41,452 | # C4.5 How to select the split point (threshold) for a Continuous Attribute
Using the "play golf" or "play ball" data (listed at the bottom), to pick the root node we look at Outlook, Temperature, Humidity, and Wind, to see which has the highest GainRatio.
Now, Outlook will be chosen as the attribute with the highest GainRatio. However, I am confused that Humidity (a Continuous Attribute) selects the split point 80 having a GainRatio=0.1087, while 65 has a higher GainRatio=0.1285. The split point 80 does have a higher Gain, but not GainRatio.
I have seen literature say roughly "pick the split point for a continuous attribute to be the one giving the most gain"... this seems counterintuitive to me that the split point is based on Gain alone, opposed to when comparing all the attributes you select the highest GainRatio to be the next decision node.
I hope to gain some clarity here.
Thanks.
EDIT: The crux of the question is: what is the appropriate method for selecting the threshold split point of a continuous attribute? Is it (1) the Gain or (2) the Gain Ratio?
The calculations are as follows:
OUTLOOK:
Gain = 0.2467
SplitInfo = 1.5774
Gain Ratio = 0.1564
TEMPERATURE:
Gain = 0.0292
SplitInfo = 1.5566
Gain Ratio = 0.0187
HUMIDITY:
Possible split points = { 65, 70, 75, 78, 80, 85, 90, 95, 96 }
Split 65:
Gain = 0.0477
SplitInfo = 0.3712
Gain Ratio = 0.1285
Split 80:
Gain = 0.1022
SplitInfo = 0.9402
Gain Ratio = 0.1087
WIND:
Gain = 0.0481
SplitInfo = 0.9852
Gain Ratio = 0.0488
DATA:
Outlook Temperature Humidity Wind Play
sun hot 85 low no
sun hot 90 high no
overcast hot 78 low yes
rain sweet 96 low yes
rain cold 80 low yes
rain cold 70 high no
overcast cold 65 high yes
sun sweet 95 low no
sun cold 70 low yes
rain sweet 80 low yes
sun sweet 70 high yes
overcast sweet 90 high yes
overcast hot 75 low yes
rain sweet 80 high no
• sorry, could not format data nicely | 562 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.827942 |
https://conwaylife.com/forums/viewtopic.php?f=7&t=610&p=4204 | 1,586,212,066,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00116.warc.gz | 406,863,314 | 9,940 | ## Bounding box
For general discussion about Conway's Game of Life.
ssaamm
Posts: 125
Joined: June 4th, 2010, 9:43 pm
### Bounding box
Well, I've been thinking about the theory of lightspeed and such, and I think it's about time we found a new standard better than "bounding box". I think that if every cell was represented by the smallest possible point, it would be the area of the smallest possible octagon that could have each of those points within it or on the edges of it. What do you think?
Mats
Posts: 42
Joined: August 10th, 2010, 7:40 am
Location: Stockholm, Sweden
### Re: Bounding box
I'm not sure I understand this. What is "the smallest possible point" that represents a cell? The center point of the cell? Any point of your choise within or on the boundary of the cell? Somthing else? And what do you mean by "octagon"? A regular octagon? A convex octagon? Any shape with eight sides and eight corners? Something else? What would the "bounding octagon" of three cells in a row (a blinker) be?
calcyman
Posts: 2144
Joined: June 1st, 2009, 4:32 pm
### Re: Bounding box
The idea of a bounding octagon has already been proposed:
Heinrich Köenig wrote:Another side effect of this is that it might be incorporated into a
metric for calculating bounds for objects -- take the intersection
of the orthogonal and the diagonal bounds.
Dave Greene wrote:A bounding octagon -- yes! On that metric, I have a whole pile of
guns that are smaller than the ones in the current gun collection --
but the combined forces of tradition and practicality are too strong
(bounding octagon metrics are much less amenable to manual
calculation) and Jason would never change the rules at this point.
At least, I sure hope not...
What do you do with ill crystallographers? Take them to the mono-clinic!
137ben
Posts: 343
Joined: June 18th, 2010, 8:18 pm
### Re: Bounding box
As I posted in another thread, why force everything to be the same shape? We could measure area by a bounding polyomino (I already posted this in two other threads, so I will spare the details here).
Whenever a standard shape is selected, it will still be an arbitrary bounding shape. Allowing the bounding polyomino to vary would give a better idea of how much space the pattern takes up.
ssaamm
Posts: 125
Joined: June 4th, 2010, 9:43 pm
### Re: Bounding box
I'll be more specific to clear things up here, because I'm not sure we're all on the same page...
If every cell were represented by a point in the very center of it, this would be the smallest possible equal-angular octagon, and it would be aligned with the grid so the top and bottom sides are flat with the horizon.
I chose this because patterns almost always extend either horizontally or vertically in life. (sorry knightships)
An alternative could be the smallest possible ellipse, I guess.
137ben
Posts: 343
Joined: June 18th, 2010, 8:18 pm
### Re: Bounding box
It is still a very arbitrary shape (and I still don't understand why you picked it, since I can't think of any patterns that are roughly octangular).
ssaamm
Posts: 125
Joined: June 4th, 2010, 9:43 pm
### Re: Bounding box
Well, it does make it so that diagonal patterns don't end up with extra junk in them...I'm starting to like the polyomino setup, though, even though it may be difficult to calculate in larger patterns.
137ben
Posts: 343
Joined: June 18th, 2010, 8:18 pm
### Re: Bounding box
Yea, I think part of the reason the bounding rectangle has been used historically is because it is extremely easy to compute. For small patterns the bounding polyomino is easy to calculate by hand, but for large patterns...I guess it would be a matter of checking each combination of connecting chunks. Still, I can compute the bounding box of the gemini almost instantly by pressing ctrl-a, but a script to find the bounding polyomino would take quite awhile if it has to search through every combination of connections between gliders in the instruction tape. The bounding polyomino is a reasonable thing to measure when asking (for example) what the smallest pattern that results in infinite growth is (a 13-omino is sufficient), but once patterns start getting really big it becomes troublesome.
ssaamm
Posts: 125
Joined: June 4th, 2010, 9:43 pm
### Re: Bounding box
Well, I think that the problem can be solved using a CA.
I think a CA in which all cells put out orthagonal-moving cells that leave a trail, and then the trails would "solidify" when it hit another trail or solid structure, and destroy unnecessary material around it. When it was all done, a special type of cell could be put in that would change the state of all cells and destroy all material except the solid stuff, which would show via the population count how many cell the polyomino has.
Just what I'm imagining, it might not actually be the best way to do it.
137ben
Posts: 343
Joined: June 18th, 2010, 8:18 pm
### Re: Bounding box
Code: Select all
``````Well, I think that the problem can be solved using a CA.
I think a CA in which all cells put out orthagonal-moving cells that leave a trail, and then the trails would "solidify" when it hit another trail or solid structure, and destroy unnecessary material around it. When it was all done, a special type of cell could be put in that would change the state of all cells and destroy all material except the solid stuff, which would show via the population count how many cell the polyomino has.
Just what I'm imagining, it might not actually be the best way to do it.``````
Probably not the best way, as the CA would still need to be simulated by a practical computer, rather than simply having the practical computer perform the computation itself.
Mats
Posts: 42
Joined: August 10th, 2010, 7:40 am
Location: Stockholm, Sweden
### Re: Bounding box
The problem with the Bounding box concept seems to be that the size appears too large for patterns that are mainly spread diagonally. That is if size is measured as the area of the bounding box. But the area is not the only way to measure size. Any function with the property:
f(box1) > f(box2) and f(box2) > f(box3) implies f(box1) > f(box3)
can be used as a measure of the size of a Bounding box.
Instead of using the area of a Bounding octagon or Bounding polyeder I suggest that the size of a Bounding box is measured not by area but by a method based on the maximum Chebyshev distance of the Bounding box, since Chebyshev distance seems to be a natural way of measuring distances i CGoL.
The way I suggest is pretty simple. First, an m by n box is equal in size to an n by m box so for simplicity I will from now on assume m >= n.
Definition: A Bounding box m1 by n1 is larger than a Bounding box m2 by m2 by n2 iff
m1 > m2
or
m1 = m2 and n1 > n2
The list of Bonding boxes sorted by size this way will be:
1 by 1
2 by 1
2 by 2
3 by 1
3 by 2
3 by 3
4 by 1
4 by 2
4 by 3
4 by 4
5 by 1
...
If you want the size of a Bounding box to be represented by a number it can be calculated as:
s(m,n) = (m^2 - m)/2 + n
Exapmle: A 6 by 3 Bounding box gets size number 18, calculated as (6^2-6)/2 + 3 = 30/2 + 3 = 18
If you know the size number, s, and want to know m and n they can be calculated as:
m is the integer closest to sqrt(2m)
(or m = int(sqrt(2m) + 0.5) where int(x) is the largest integer <= x)
and
n = s - (m^2 - m)/2
Example: Bounding box number 100 is the 14 by 9 Bounding box since sqrt(200) = 14.1421... which gives m = 14, and n = 100 - (14^2-14)/2 = 9. | 1,966 | 7,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-16 | latest | en | 0.957235 |
https://www.physicsforums.com/threads/basic-dynamics-with-fc.307366/ | 1,656,815,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00066.warc.gz | 960,721,672 | 15,309 | # Basic Dynamics with Fc
## Homework Statement
A ring radius R is fixed vertically to the floor. From the top of the ring and object slides down without friction. Find the distance "L" from the point of support of the ring to the point where the object hits the ground.
## Homework Equations
I think the object travels tangetially from 90 degrees from the top of the circle. This is where the normal force equals the centripetal force and the net force is the weight. Is this right? If so how do I prove this?
??????
## The Attempt at a Solution
Ok so after some struggling I realized that the attempt above is completely wrong. I have broken the circle down into its components and solved for cos(O) and got 2/3 which I was able to verify so I know that part is right. So I have also found that part of the length is Rsin(O) and now I need to solve for the rest. I hope I'm making this clear because I don't think I am. Please let me know if you need more info. I think i'm at the point where the normal force is at 0 and the object goes into parabolic freefall. The problem is I don't know how to finish the problem. How do I take what I have solved for so far, Cos(0) = 2/3 and get the rest of my length from the freefall. I should probably know this but I can't seem to put the pieces together. Help please!!!!
Ok so after some struggling I realized that the attempt above is completely wrong. I have broken the circle down into its components and solved for cos(O) and got 2/3 which I was able to verify so I know that part is right. So I have also found that part of the length is Rsin(O) and now I need to solve for the rest. I hope I'm making this clear because I don't think I am. Please let me know if you need more info. I think i'm at the point where the normal force is at 0 and the object goes into parabolic freefall. The problem is I don't know how to finish the problem. How do I take what I have solved for so far, Cos(0) = 2/3 and get the rest of my length from the freefall. I should probably know this but I can't seem to put the pieces together. Help please!!!!
Basically,You finished it...
Not much left, remember:
the ball when it disconnects has V tangential to the point it was left, you can find the V in the y direction and V in the X direction,You have initial height, get T from kinematics of the y direction, and than place it in the X direction distance.
tell me if You need more clues, good luck!
Thanks for the help. I'll try it and see what I come up with. Either way I'll post an answer and maybe you can tell me if it's right. | 621 | 2,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-27 | latest | en | 0.976125 |
http://www.peabodysgunrange.com/financial-projections-revenues-costs-anticipated/ | 1,637,999,933,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00371.warc.gz | 128,440,495 | 12,259 | Financial projections PharmaStaff is a staffing agent that provides scheduling relief to businesses and to match qualified professionals with companies to form successful work relationships. The start-up cost includes setting up office, capital equipment, location and admin expenses, launching website, advertising and promotional expenses. All these costs will be incurred at the start of the business. Nonetheless, advertisement and promotional costs will ripple across the months. Initially, large sums of funds will be directed towards advertisement but will decrease as the business picks up. The following financial projections are based on pharmacist’s salary forecasts for the next 5 years.
Short-term projection of the first year
First month – This forms the initial business phase and the number of employees will not be as much as it is project to be later. Similarly, the earnings are likely to be less.
10 pharmacists start-up
Assuming each makes \$60 per hour and work 40 hours weekly = about \$ 86400 after tax) We generate income by making 10% off what pharmacists make =\$8,000 first month
Setting up an office costs will amount to \$2400 initially.
Capital equipment will be purchased at a cost of \$5,000.
Advertisement expenses in the first year will cost \$ 1800 after which the costs will reduce to \$1,000 every month for the other months of the year.
The website will cost \$ 1,500 to set up and fully put into operations.
Thereafter additional pharmacists will be recruited to raise the number to 30.
Assuming recruited 30 pharmacists
Based on the previous assumption that each dentist makes \$60 per hour and work 40 hours weekly, the costs will rise to \$ 864,000 per month after tax deductions.
Generate about \$24,000/month for each dentist
End of first year will generate about \$250,000
Five- year projection: this is the projection for the five year duration.
Expanding business with more pharmacists, corporates
It is assumed that the costs for payment of the dentists will rise by 15% every year and revenues will rise by 20% for every dentist.
PharmaStaff will generate \$250,000 of operating revenue in its first year of functioning. We will require an initial investment of 25,000, all of which will have been recovered by the end of year one in cash. By the end of year five, we expect to have cash and accounts receivable aggregating about one million dollars. This will main originate from client payments for the services they receive as well as the outstanding amounts owed by the pharmacies to which the services will be extended.
Management of service quality issue
The valet parking service is generally provided by establishments such
Whether Savannah Harbour Expansion Project Is Worth the Shot
The greatest concern before implementing a project of such an
Red Zuma Project
In-depth analysis of the marketA comprehensive product designSelecting the appropriate | 591 | 2,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.950452 |
https://mail.haskell.org/pipermail/glasgow-haskell-users/2002-April/003268.html | 1,638,263,045,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00247.warc.gz | 434,805,602 | 2,131 | # Question on style - Implicit parameters
Thu, 18 Apr 2002 00:55:05 +0100
```Hi all,=20
Lately I've found implicit parameters to be extremely usefull in many=20
situations.
The most simple and obvious is when you need to calculate some values th=
at=20
depend on specific parameters, in some context where these same parameter=
s=20
constant. There is no poblem if all the calculus are defined in the scope=
of=20
those 'constant' parameters, but if you need to define functions outside=20
their scope then you have to pass them as arguments, and sometimes that=20
complicates type signatures.
Now my question.
I've end up in situations where I need to write something like
---------
step :: Inst-> Train->Inst
step inst train =3D changeInst deltaW deltaV inst
where
deltaW =3D someSemiConst1 with ?inst=3Dinst; ?train train
deltaV =3D someSemiConst2 with ?inst=3Dinst; ?train train
---------
Ok the problem is that both deltaV and deltaW are assigned some constants=
that=20
implicitly depend on the arguments 'inst' and 'train'.
Now imagine that=20
someSemiConst1 =3D f n
someSemiConst2 =3D g n
for some n that also depends implicitly on 'inst' and 'train'.
In this case 'n' would (could? haven't actually tried it) be calculated t=
wice,=20
one fro the first semiConst, another for the second.
It would be nice to have something like
with ?inst=3Dinst; ?train train
{
deltaW =3D someSemiConst1=20
deltaV =3D someSemiConst2=20
}
My solution was to define some function
(deltaW, deltaV)=3DcalculateDeltas with ?inst=3Dinst; ?train=3Dtrain
And in this function I simple does
(deltaW, deltaV)=3D (someSemiConst1, someSemiConst2)
In case you're wondering, why not just call the 'step' function bounding =
it's=20
context to 'inst' and 'train'. It doesn't makes much sense to me in this=20
case, becouse the whole purpose of this function is to change a 'inst'=20
acording to some 'train' (in fact it is beeing used in a foldl).
How do you usually handle this situations?
Is there an elegant solution for this?
J.A.
``` | 568 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-49 | longest | en | 0.867183 |
https://crazyproject.wordpress.com/2011/08/18/a-characterization-of-left-simple-and-simple-semigroups/ | 1,485,064,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281353.56/warc/CC-MAIN-20170116095121-00330-ip-10-171-10-70.ec2.internal.warc.gz | 806,219,221 | 18,470 | ## A characterization of left simple and simple semigroups
Let $S$ be a semigroup. Show that $S$ is left simple if and only if $Sa = S$ for all $a \in S$, and is simple if and only if $SaS = S$ for all $a \in S$.
Recall that a semigroup is called left simple if the only left ideal is $S$ itself.
Suppose $S$ is left simple. Now $S(Sa) \subseteq Sa$, so that $Sa$ is a left ideal. Hence $Sa = S$. Conversely, suppose $Sa = S$ for every $a \in S$. Let $L$ be a left ideal. Now $L \neq \emptyset$, so we have $a \in L$ for some $a$. Then $S = Sa \subseteq L$, and thus $L = S$. So $S$ is left simple.
Similarly, we can show that $S$ is right simple if and only if $aS = S$ for all $a \in S$.
Now suppose $S$ is simple. (That is, it has no nontrivial ideals.) Let $a \in S$. Now $S(SaS) \subseteq SaS$ and $(SaS)S \subseteq SaS$, so that $SaS$ is an ideal. Hence $SaS = S$. Now suppose $SaS = S$ for all $a \in S$. If $I$ is a two-sided ideal, then we have $a \in I$ for some $a \in S$. Now $S = SaS \subseteq I$, so that $I = S$. Hence $S$ is simple. | 364 | 1,053 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 37, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-04 | longest | en | 0.80011 |
http://mathhelpforum.com/math-topics/21053-freefall.html | 1,511,180,325,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806030.27/warc/CC-MAIN-20171120111550-20171120131550-00089.warc.gz | 190,781,423 | 11,199 | 1. ## Freefall
Last physics problem for the day I promise
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.1s later with an initial speed of 60.76m/s. They hit the ground at the same time.
The acceleration of gravity is 9.8m/s/s.
a) How long does it take the first stone to hit the ground? Answer in units of s.
b) How high is the cliff? Answer in units of m.
Based on the information given it definitely seems possible to solve this problem, but I just can't seem to put the information together.
2. firstly form 2 separate equations s = ut + (a(t^2))/2, then equate them
ut + (a(t^2))/2 = ut + (a(t^2))/2
taking one t as t = -3.1 and the other as just t.
3. Hmm, I just created two position functions for each rock. The first rock is simply 4.9x^2 and the second rock is (4.9x^2)+(60.76x). I'm trying to find when their distances are equal because they hit the ground at the same time, but my calculator is only giving me 0,0 for an intersection
Bleh, I'm not factoring time into my equation of the second rock, which is why they never have an equal distance :\
Using your help I got t = 1.55 seconds, I think it works I did it backwards though, making the first rock's equation t+3.1.
4. which will give a value for t, then to find height of the cliff plug t into the equation in my first post.
god i havent done that for 3 years, i used to love the mechanic part of maths until i got to university
5. Originally Posted by vesperka
Last physics problem for the day I promise
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.1s later with an initial speed of 60.76m/s. They hit the ground at the same time.
The acceleration of gravity is 9.8m/s/s.
a) How long does it take the first stone to hit the ground? Answer in units of s.
b) How high is the cliff? Answer in units of m.
Based on the information given it definitely seems possible to solve this problem, but I just can't seem to put the information together.
As always, set up your coordinate system. I have the origin at the base of the cliff and +y upward. I am defining the starting time t = 0 s when the second stone is thrown.
So for the first stone we have
$y_ 0 = h = ?$
$v_0 = 0~m/s$
$a = -9.8~m/s^2$
Giving a position equation of
$y = y_0 + v_0t + \frac{1}{2}at^2 = h - 4.9t^2$
For the second stone we have
$y_0 = h = ?$
$v_0 = -60.76~m/s$
$a = -9.8~m/s^2$
Giving position equation of
$y = y_0 + v_0(t + 3.1) + \frac{1}{2}a(t + 3.1)^2 = h - 60.76t - 4.9t^2$ (Recall that the second stone is thrown at 3.1 s.
They hit the ground at the same time, so at time t they both have a position y = 0. So
$h - 4.9(t + 3.1)^2 = h - 60.76t - 4.9t^2$
Giving t = 1.55 s.
So how tall is the cliff? Pick a stone:
$0 = h - 4.9(1.55 + 3.1)^2$
Giving h = 105.95 m.
-Dan
6. see i got the height of the cliff as 11.7m when i first did the calculation, but it seemed wrong so i plugged t into the equation for the stone with initial velocity 60.76m/s and got the height of the cliff as 106m, which makes more sense as the stone takes 1.55s to hit the floor and the initial vel. is 61m/s.
7. Originally Posted by mathmonster
see i got the height of the cliff as 11.7m when i first did the calculation, but it seemed wrong so i plugged t into the equation for the stone with initial velocity 60.76m/s and got the height of the cliff as 106m, which makes more sense as the stone takes 1.55s to hit the floor and the initial vel. is 61m/s.
I'm not doing so hot today, am I?
I forgot to add the 3.1 s when using that equation. If you plug the 1.55 s into the "thrown rock" equation you get the correct height. If you plug 1.55 + 3.1 (s) into the "dropped rock" equation you also get the right answer.
I have corrected this in my original post. (And I also got the 105.95 m answer.)
(Ahem!) Sorry about that!
-Dan | 1,166 | 3,901 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.952619 |
https://www.physicsforums.com/threads/reading-variable-force-graph.142248/ | 1,553,387,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203123.91/warc/CC-MAIN-20190324002035-20190324024035-00118.warc.gz | 868,037,673 | 15,559 | # Reading Variable Force Graph (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### mbrmbrg
Behold Halliday's Question and Halliday's answers:
The only force acting on a 2.0 kg body as it moves along the x axis varies as shown in Figure 7-41 (see attatchment). The velocity of the body at x = 0 is 4.0 m/s.
(a) What is the kinetic energy of the body at x = 3.0 m?
12 J
(b) At what value of x will the body have a kinetic energy of 8.0 J?
4.0 m
(c) What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m?
18 J
I am stumped. I've only tried to figure out (a) and (b), but my method does not give me correct answers. My train of thought was that Delta KE=W=area under curve. So for part (a), I said that area under graph is (triangle-triangle-rectangle) = -4 Nm. Well, gee.
Maybe since KE is always positive, take the absolute value of various areas! I think that is very poor reasoning, because Delta KE and W can both be negative. And even when I do that, I get 8 Nm, which is still wrong.
Similar reasoning applied to (b) also gives the incorrect answer, which is comforting.
A hint from the wise gurus perhaps?
#### Attachments
• 22 KB Views: 305
#### ehild
Homework Helper
mbrmbrg said:
My train of thought was that Delta KE=W=area under curve. So for part (a), I said that area under graph is (triangle-triangle-rectangle) = -4 Nm. Well, gee.
Correct.
The change of KE is equal to the work done on the body. If the work is negative, the kinetic energy will decrease. Final KE - initial KE = W. You know the initial KE, (16 J) you know W,(-4 J) what is the problem?
ehild
#### mbrmbrg
Problem is that I thought initial Kinetic Energy was 4J. Wups!
Thank you.
#### keltix
I'm doing the same problem. Does anyone know the answers?
I got 12J for (a) but I'm lost on (b) and (c)
#### keltix
Edit: this is what I got after some thought...
(a) 12 J
(b) x=4
(c) 28 J
anyone agree?
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• Solo and co-op problem solving | 633 | 2,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.956646 |
https://www.wallstreetmojo.com/econometrics/ | 1,701,883,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00841.warc.gz | 1,165,156,345 | 76,949 | # Econometrics
Article byWallstreetmojo Team
Reviewed byDheeraj Vaidya, CFA, FRM
## Econometrics Meaning
Econometrics is the field of economics that help quantify the cause and effect relationship, utilizing the graphical representations correlating the involved factors or elements. It combines statistical and mathematical methods that let users figure out possible outcomes and determine their further plan of action accordingly.
For eg:
Source: Econometrics (wallstreetmojo.com)
Econometrics acts as a tool to check how strong the cause and effect relationship is to assess the outcomes they can expect in the future. Though this field of study lets users deal in various industries, it has a significant role in the finance sector, enabling them to understand how particular assets are likely to perform in current situations.
### Key Takeaways
• Econometrics is the subset of economics that helps quantify the cause and effect relationship to help economists frame economic policies.
• It is an application of statistics that lets econometricians forecast financial scenarios and build strategies and backup plans accordingly.
• This field of economics is found in two forms – theoretical and applied.
• There is a stepwise sequence that practitioners must follow to obtain reliable quantitative econometrics forecast results for further decision-making.
### Econometrics Explained
Econometrics for finance helps study the economic data to let users make future predictions based on the current representations. It becomes an effective technique to turn theoretical economic concepts into practical models to ensure individuals and entities make effective economic decisions keeping all necessary factors in mind.
This field of economics can be classified as micro-econometrics and macro-econometrics. When the cause and effect relationship or the relationship between any two factors is concerned with a single person or entity, it is labeled as micro-econometrics. On the other hand, as the name implies, the macro version involves studying relationships at a macro level, affecting many people.
For eg:
Source: Econometrics (wallstreetmojo.com)
For example, studying a graph representing the effect of educational qualification on the designations offered by firms can be an example of a micro version of this field of economics. In contrast, the increase in the inflation rate directly affects a large population, which makes it a perfect example of macro-econometrics.
The econometricians consider the versions of economic theorists and transform them into quantitative data, which is easy to read and base decisions on. The field of study finds relevance in varied branches of economics, including economic policies. The people dealing with this niche adopt multiple econometrics models to get to an accurate conclusion for a dependable outcome.
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### Components
The field of study involves two aspects: Theoretical and Applied.
For eg:
Source: Econometrics (wallstreetmojo.com)
Theoretical specialists examine the features of the statistical procedures adopted previously and still existing. Then, using the same, they try to identify unknown indications by implementing the existing tests and processes in the models. In addition, these experts work on new statistical procedures as per specific matters with peculiar economic data. This is to ensure the outcome derived fits into change-driven situations with respect to correctness.
Applied aces consider the techniques already developed by the theorists and transform them into quantitative versions. As the name suggests, they apply the already derived theories in use.
### Stages
Transforming the statistical pattern into quantitative data to make them readable and understandable involves multiple stages. Following the steps properly and in a sequence is a must to achieve reliable results:
For eg:
Source: Econometrics (wallstreetmojo.com)
Firstly, practitioners investigate the dataset based on which they recommend a theory, which explains the data, defines the variable, and establishes the relationship between those variables. The economic theory plays a significant role in the process as it helps users proceed with hypothesis formulation.
Secondly, the statistical model required to quantify the economic theory is defined. The economic theory does not make the cause and effect relationship readable enough for users. A statistical model, on the other hand, defines the mathematical relationship between variable, dependent, and explanatory variables.
Thirdly, the specified statistical methods are used for appropriate econometrics forecast of the points that remain unclear in the statistical theory.
Lastly, is conducted. It acts as a tool to help accomplish the fourth step by determining whether the hypothesis or economic theory should be accepted or rejected. If it is rejected, the practitioners would require coming up with new testing theories to ensure the handling of the economic data well for proper economic policy-making or financial decision making.
### Examples of Econometrics
Let us consider the following econometrics examples to understand the concept:
#### Example #1
Arlene decided to invest in a stock of a popular company. Though she was sure that the stocks would perform well, her experience of huge losses made her keep aside her confidence and conduct a check. As a result, she asked her friend, John, a financial analyst and econometrician by profession, to help her with how volatile the market is expected to be. He conducted the analysis and applied the best tools to ensure his friend invested in a good deal.
John discovered that market volatility is likely to affect the stocks in the upcoming days and months negatively. Accordingly, he conveyed the situation to Arlene, and she decided not to invest for the time being.
#### Example #2
Michael has an income of \$50,000. However, his spending pattern is \$10,000, including fixed rent and other household expenses, comprising 50% of his earned.
The relationship between his expenditure and gross income can be derived using econometrics, and the best tool to get accurate results is multiple linear regression based on past trends.
Equation would be = B0 (Intercept) + B1 + e (Error term)
Michael’s expenditure with respect to his income is:
Expense = B0 (Fixed rent) + B1 (other household exp.) + e (Error term)
= \$10,000 + 50% (\$50,000)
= \$35,000
Error term, in this case, is the slight possibility of error in the results, which might be a little up or down from the output obtained.
### Application
Econometrics is used in different fields, including science and engineering. However, it is of great significance in the field of finance as it helps in:
• Estimating volatilities
• Determining the possible price-to-earnings ratio
The advantages and disadvantages of this subset of economics are as follows:
### Econometrics vs Statistics
Statistics is the field of study that lets statisticians study the data and analyze it for a valid conclusion. On the other hand, econometrics is the subset of economics that utilize statistical tools to examine data and convert them into their quantitative version, making them simpler and more readable to users.
In short, econometrics is one of the applications of statistics that help economists predict economic scenarios, frame policies, and make decisions accordingly.
What is econometrics?
Econometrics is the field of economics that help convert existing hypotheses and economic theories into quantitative data so that users can read them well to make relevant economic or financial decisions. Some econometrics techniques or methods used include probability, regression, estimation theory, time-series analysis, simulation equation, frequency distribution, etc. | 1,614 | 8,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-50 | longest | en | 0.915675 |
https://www.unitconverters.net/length/ken-to-centimeter.htm | 1,725,856,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00825.warc.gz | 1,009,569,321 | 4,064 | Home / Length Conversion / Convert Ken to Centimeter
# Convert Ken to Centimeter
Please provide values below to convert ken to centimeter [cm], or vice versa.
From: ken To: centimeter
### Ken to Centimeter Conversion Table
KenCentimeter [cm]
0.01 ken2.11836 cm
0.1 ken21.1836 cm
1 ken211.836 cm
2 ken423.672 cm
3 ken635.508 cm
5 ken1059.18 cm
10 ken2118.36 cm
20 ken4236.72 cm
50 ken10591.8 cm
100 ken21183.6 cm
1000 ken211836 cm
### How to Convert Ken to Centimeter
1 ken = 211.836 cm
1 cm = 0.0047206329 ken
Example: convert 15 ken to cm:
15 ken = 15 × 211.836 cm = 3177.54 cm | 203 | 587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.491091 |
https://ltwork.net/evaluate-the-extent-to-which-technological-and-economic--10081635 | 1,685,926,791,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00588.warc.gz | 394,852,871 | 10,570 | # Evaluate the extent to which technological and economic changes shaped United States society in the period from 1980 to the present.
###### Question:
Evaluate the extent to which technological and economic changes shaped United States society in the period from 1980 to the present.
### I don't know how to do any of these and I also need to show work someone help pls
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### The length of the top of a table is 3 m greater than the width. the area is 70 m^2. find the dimensions of the table.
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### Determine the general form of the equation for the circle (x – 1)2 + (y + 2)2 = 3. x2 + y2 – 2x
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### The cost of each ticket, if x tickets cost 106.25
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### Two traditional economies are trying to industrialize. the leaders of the first favor a command economic
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### A bookstore owner randomly samples 100 customers on two separate days to see what their favorite type of book is. The data shows the results
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### The initial water level in a 10 -ml graduated cylinder reads 4.1 ml . after a ruby gemstone is dropped into the cylinder
The initial water level in a 10 -ml graduated cylinder reads 4.1 ml . after a ruby gemstone is dropped into the cylinder , the water level reads 7.1 ml what is the volume of the ruby ?...
### Which chemical symbols will complete the equation for this decomposition reaction? 2KI → 2_ + _ 2
Which chemical symbols will complete the equation for this decomposition reaction? 2KI → 2_ + _ 2... | 1,177 | 4,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.940731 |
https://mathematica.stackexchange.com/questions/7861/plotting-a-parametric-plot-of-two-functions | 1,701,410,375,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00403.warc.gz | 441,290,854 | 41,635 | # Plotting a parametric plot of two functions
I have two functions f1(x) and f2(x). I want to plot them such that f1 is on the axis of ordinates and f2 on the axis of abscissae. Any ideas?
Following example to it:
f1(x)=2x and f2(x)=3x. The graph should then show a line through the following points: {(0,0),(2,3),(4,6),(6,9),...}
• Are you looking for a way to combine a plot with one that is rotated 90 degrees? Your question and title do not make it clear. Jul 3, 2012 at 23:10
• Dear editor, if you are the OP of this question, please consider registering your account; it'd be more convenient for you and for the rest of us. Jul 4, 2012 at 9:46
• Thanks for the comment. Since the question was moved, I though I could not access it. Jul 4, 2012 at 10:11
ParametricPlot[{f2[x],f1[x]},{x, begin,end}]
where you have to fill in sensable values for begin and end.
Example:
ParametricPlot[{Sin[x], Cos[5 x]}, {x, 0, 10}]
Perhaps you want this?
f1 = Sin;
f2 = Cos;
ParametricPlot[{{f1[x], x}, {x, f2[x]}}, {x, -2 π, 2 π}]
Different rages for f1 and f2:
ParametricPlot[
{{f1[u], u}, {v, f2[v]}},
{u, 0, 3 π}, {v, -2 π, 2 π},
Mesh -> False
] | 386 | 1,154 | {"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-2023-50 | longest | en | 0.853901 |
http://www.flyingcoloursmaths.co.uk/nice-bit-nrichment/ | 1,527,434,502,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00621.warc.gz | 374,673,104 | 15,925 | # A nice bit of nrichment
My dear friend @ajk44 pointed me at a puzzle on the excellent Nrich site1, and I enjoyed it enough to share my solution. (If you don't want it spoiled, don't read beyond the blockquote.)
Four jewellers had respectively 8 rubies, 10 sapphires, 100 pearls and 5 diamonds.
Each gave one gem from their collection to each of the rest.
Afterwards they noticed that they all had collections of gems of precisely equal value.
Can you work out the relative values of each gem?
Can you then work out how much each jeweller gained or lost?
### A first approach
The first thing I tried was to set up a system of equations. If each jeweller has given away three gems and received one of the others, we know the following four quantities are the same:
$5r + s + p + d$
$r + 7s + p + d$
$r + s + 97p + d$
$r + s + p + 2d$
Looking at the first pair, the difference between them is $4r – 6s$ — which has to be 0. Similarly, $6s = 96p$ and $96p = d$.
If we pick $p$ as our base unit, a diamond is worth $96p$, a sapphire a sixth of that, or $16p$, and a ruby works out to be $24p$, which gives the relative values of the gems.
Everybody gains $r + s + p + d – 4g$, where $g$ is the value of their own gem; this works out to $137p – 4g$. The ruby-cutter gains $41p$, the sapphire jeweller gains $73p$, the poor old diamond-seller loses $247p$ and the lucky pearl swine gains $133p$. A quick sanity check: $41 + 73 + 133 – 247 = 0$, so it's at least a plausible answer.
### Tidying it up
A neat short-cut to solving the whole thing is to note that if everybody throws away the new jewels they've received as well as one of their own, they will all have lost the same amounts, and their holdings will still be equal to each other. This is the same as each merchant just binning four jewels to begin with, which leads directly to the statement $4r = 6s = 96p = d$ with hardly any working out.
Philosophical aside: I reckon many people think of the first way as "proper maths" – look! Lots of equations! In reality, the second way is more what I think of as mathematical: a moment's thought simplifies the situation and turns it into something very straightforward. I'd characterise it as intelligent laziness; rather than solving the puzzle in front of you, it's finding an equivalent puzzle that's less work.
* Thanks to @ajk44 and Nrich maths for the puzzle.
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
1. in fact, she pointed me at several; this may turn into a series []
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I teach in my home in Abbotsbury Road, Weymouth.
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In the study of numerical methods, we can make a general distinction between a set of methods such as solving linear systems of equations , solving matrix eigenvalue problems , interpolation , numerical integration and finding the roots or zeros of equations , which can be somewhat considered as the building blocks for larger that arise in engineering/applied mathematics/physics. For example the problem of solving ordinary differential equations , optimisation and solving integral equations . But from the point of view of aplied mathematics or engineering, erhaps the most significant problems in numerical methods is the solution of partial differential equations by Finite Difference Methods , Finite Element Methods or Boundary Element Methods . The study of the behaviour of numerical methods is called numerical analysis. This is a mathematical subject that considers the modelling of the error in the processing of numerical methods and the subsequent re-design of methods.
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MBA SYSTEMS Questions MBA SYSTEMS Articles MBA SYSTEMS Question Paper MBA SYSTEMS Free Books Download MBA SYSTEMS Free Videos | 795 | 4,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-05 | latest | en | 0.830152 |
http://planeta42.com/physics/classroom.html | 1,660,218,692,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00596.warc.gz | 40,746,379 | 12,574 | Water Cycle Find the correct elements of the natural water cycle. Energy Types 20 sources of potential and kinetic forms of energy. Force Comparison Sort the objects in order of force magnitude. Particles Puzzle Place the subatomic particles in correct places. Power Units Sort the units of power (Watts) by their equivalents.
Photovoltaics Sort the parts of the photovoltaic installation. Power Plants Sort the energy source in the power plant. 6 Tanks Quiz Arrange the 6 tubes in the correct filling order. Kid's Clock Place the clocks at the correct base time. Atom Charge Place the atom particles in the correct el. charge.
Online Physics Classroom Games Collection.
The physics classroom games are small flash applications, up to 1 MB, which are designed to be usable in classes at school. They include different valuation systems and interesting knowledge topics. Unblocked games. Physics learning games. Online learning games. Fun exercises for physics classes. Classroom Games - Interactive Physics Workshop. Additional fun applications for physics learning. Fun Physics Classroom Games Collection.
All classroom games (67):
Atom Charge Puzzle - Place the atom particles in the correct electric charge. Virtual Newton's Cradle - 5 balls are hanging on a stand. Swing them to see how they react. Sink or Float Quiz - Select which objects sink and which floats in water. Power Plant Types - Drag and drop the energy source on the correct power plant type. Power Units - Sort the units of power (Watts) by their equivalents. Power Comparison - Sort the objects by their correct power value in Watts. Weight Units - Sort the units of mass by their equivalents - Kilogram Weight Comparison - Sort the objects by order of weight magnitude. Length Units - Sort the units of length by their equivalents - Meter. Length Comparison - Sort the objects by order of length magnitude. Physics Tree - Arrange the fruits of the physics science. Drag and drop game. Communicating Vessels - Fill the vessels with liquids. Shooting game. Force Types Puzzle - Sort the properties of different forces. Hard difficulty. Energy Types Puzzle - Distribute kinetic and potential energies. Light Bulb Parts - Assemble the parts of the light bulb. Time Units Pyramid - Climb the pyramid of time. From seconds to years in measures. Speed Comparison - Interactive fun diagram for speed of objects. General Physics Test - 18 questions online test with evaluation system. Engine Types Puzzle - Sort the engines on the correct vehicles. Battery Types Puzzle - Place the batteries in the correct device to power it. Photovoltaic System Structure - Sort the parts of the photovoltaic installation. Conductor or Insulator Quiz - Select which object is a conductor and which is an insulator in electrical circuit. Magnetic or Nonmagnetic Quiz - Select which object is attracted by a magneto and which is not. Burn or Quench Quiz - Select which liquid is burning and which extinguish the fire. Gravity Quiz - here are 3 objects with different mass dropped from 10 feet height. Which one will hit the ground first? Wind Scale Puzzle - Arrange the Beaufort Wind Scale and 13 forces of winds. 4 Tanks Water Fill Puzzle - Arrange the containers in the correct filling order. 5 Tanks Water Fill Puzzle - Arrange the five containers in the correct filling order. 6 Tanks Water Fill Puzzle - Arrange the six tubes in the correct filling order. 7 Tanks Water Fill Puzzle - Arrange the seven vessels in the correct filling order. Milk Jugs Filling - Sort the eight milk jugs in the correct filling order. Water Cycle Puzzle - Find the correct elements of the natural water cycle. Bend or Break Quiz - Select which objects deforms and which breaks on bending force. Push or Pull Quiz - Select which movement or mechanism works by push force or pull force. Stink or Scent Quiz - Select which object stink and which scent when smelled. Reflect or Absorb Quiz - Select which object reflects or absorbs most of the light. Kinetic or Potential Quiz - Select which objects have kinetic and which potential energy. Soluble or Insoluble Quiz - Select which objects or liquids dissolve in water. Elementary Particles Puzzle - Place the subatomic particles in correct places. Water Well Puzzle - Drill for water creating a well in the ground layers. Advanced Clock Puzzle - How to read the clock and meanings of the clock arrows. Kid's Clock Puzzle - How to read the clock and meanings of the clock arrows for beginners. Pressure Units Puzzle - Sort the units of pressure by their equivalents - Pascal. Capacitance Units Puzzle - Sort the units of electrical capacitance by their equivalents - Farad. Current Units Puzzle - Sort the units of electrical current by their equivalents - Ampere. Frequency Units Puzzle - Sort the units of teporal frequency by their equivalents - Hertz. Force Units Puzzle - Sort the units of mechanical force by their equivalents - Newton. Temperature Units Puzzle - Sort the units of temperature by their equivalents - Kelvin. Sound Units Puzzle - Sort the units of acoustic sound by their equivalents - Decibel. Volume Units Puzzle - Sort the units of liquid volume by their equivalents - Litre. Potential Units Puzzle - Sort the units of electrical potential by their equivalents - Volt. Time Comparison - Sort the objects by their correct value in order of time magnitude. Temperature Comparison - Sort the objects by value in order of temperature magnitude. Force Comparison Puzzle - Arrange the objects in order of force magnitude in newtons. Frequency Comparison - Arrange the objects in order of temporal frequency magnitude in hertz. Current Comparison - Sort the objects in order of electric current magnitude in amperes. Potential Comparison - Sort the objects in order of electric potential magnitude in volts. Resistance Units Puzzle - Sort the units of electric resistance by their equivalents - Ohm. Resistance Comparison - Sort the objects in order of electric ðesistance magnitude in ohms. Capacitance Comparison - Sort the objects in order of electric capacitance magnitude in farad. Pressure Comparison Puzzle - Arrange the objects in order of pressure magnitude in pascals. Volume Comparison Puzzle - Sort the objects in order of liquid volume magnitude in liters. Sound Comparison Puzzle - Interactive comparison of sound force and noise threshold. Metric Units Puzzle - Match the metric units with the correct electromagnetic and mechanical quantities. Measure Units Puzzle - Match the metric units with the correct light, radiation, surfaces and basic quantities. Spatial Dimensions Puzzle - Place the objects in the correct spatial dimension. Jet Engine Structure Puzzle - Assemble the parts of the jet engine.
Length Compare Sort the objects by order of length magnitude. Physics Test 18 subjects from the physics online test. Gravity Quiz A bag of gold is dropped from an airplane... Wind Scale Puzzle Sort the wind powers on the correct speed. 4 Tanks Quiz Arrange the containers in the correct filling order.
5 Tanks Quiz Sort the 5 containers in the correct filling order. Force Types 9 types of forces in a fun online physics game. 7 Tanks Quiz Sort the 7 containers in the correct filling order. Milk Jugs Filling Arrange the 8 milk jars in the correct filling order. Sink or Float Select which objects sink and which floats in water.
Bend or Break Select which objects bends and which breaks. Push or Pull Select which objects use push or pull force. Burn or Quench Select which liquid is burning and which extinguish the fire. Stink or Scent Fun game to explore good and bad smell. Reflect or Absorb Select which object reflects or absorbs most of the light.
Soluble or Not Select which objects or liquids dissolve in water. Kinetic or Potential Select which objects have kinetic or potential energy. Battery Types Place the batteries in the correct device to power it. Well Puzzle Sort the parts of the well and ground layers. Clock Puzzle Place the clocks at the correct random time.
Magnetic or Not Select which objects are attracted by a magneto. Weight Units Sort the units of mass by their equivalents. Length Units Arrange the units of length by their equivalent metres. Force Units Arrange the units of force by their equivalent newton . Temperature Units Sort the units of temperature by their equivalent kelvins.
Pressure Units Arrange the units of pressure by their equivalent pascal. Capacitance Units Sort the units of el. capacitance by equivalent farad. Current Units Arrange the units of electrical current by their equivalent. Frequency Units Sort the units of frequency by their equivalent hertz . Potential Units Arrange the units of electrical potential by their equivalent.
Volume Units Arrange the units of volume by their equivalent liter. Sound Units Arrange the units of sound by their equivalent decibel. Time Compare Sort the objects by order of time magnitude. Temperature K Sort the objects by correct temperature. Engine Types Sort the engine types on the correct vehicles.
Frequency Com. Sort the objects in order of temporal frequency magnitud. Current Com. Sort the objects in order of electric current magnitud. Potential Com. Sort the objects in order of electric potential magnitud. Resistance Units Sort the units of electrical resistance by their equivalent. Resistance Com. Arrange the objects in order of electrical resistance.
Capacitance Com. Arrange the objects in order of electrical capacitance. Pressure Com. Arrange the objects in order of pressure magnitude. Volume Compare Sort the objects in order of volume magnitude. Sound Compare Sort the objects in order of sound magnitude. Metric Units Sort the metric units on the correct quantities..
Measure Units Sort the metric units on the correct quantities. The 5 Dimensions Place the objects in the correct spatial dimension. Jet Engine Puzzle Assemble the parts of a jet engine.
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Planeta 42 Game World | About | Sitemap | Levels | Downloads | News | Free Games | Drawings | Best Games Ever | 1,982 | 10,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.885719 |
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