url
stringlengths 14
5.47k
| tag
stringclasses 1
value | text
stringlengths 60
624k
| file_path
stringlengths 110
155
| dump
stringclasses 96
values | file_size_in_byte
int64 60
631k
| line_count
int64 1
6.84k
|
|---|---|---|---|---|---|---|
https://digitalroom7.com/scientific-calculator-plus-991/
|
math
|
Scientific calculator plus 991
Scientific calculator 300 plus, 991 is the calculator for students and engineering student. Calculator 991 plus provides powerful functions in a real calculator 991 300 . It makes advanced calculator 991 become the most useful calculator for university and school.
Application includes some calculators inside:
– Basic calculator 300 plus
– Advanced calculator 115 plus
– Scientific calculator 991
– Scientific calculator 991 plus
– Graphing calculator 84 plus
Let’s see some special features:
◉ Basic calculator 300 plus and 115 plus: supports from basic mathematics to advanced math functions: percentage, powers, roots, trigonometric, logarithms calculator. Fraction calculation 991 supports convert fraction to decimal, mixed fraction and solve fractions problems.
◉ Advanced calculator 115 plus: combines of hyper calculator and simple scientific calc. This calculator is a multi-functional advanced calculator 991 with all features in one app. Such as linear algebra, calculus, complex numbers, display result in rectangular and polar coordinates, matrix and vector.
◉ Scientific calculator 991 plus: includes smart equation solver. Equation solver can solve quadratic, cubic equations, systems of equations. Caluculator solves any polynomial. Smart scientific calculator 991 has exponent calculator contains scientific calculations such as derivative, integral, square root calculation, factorial calculation, pi calculation, equation solver.
◉ Scientific calculator 991: supports calculate with infinity number of digits, prime factors, random numbers, combinations, permutations, GCD and LCM. This calculator 991ex is a real engineering simulator. Scientific calculator has features of 500 es, 500 ms, 300 es plus, 991 es plus. Calculator keyboard layout same as calculator 300 es plus, 991 ex and es plus.
◉ Graphing calculator 84 plus supports function graph, polar, parametric and implicit function. Graphing calculator 84 can draw tangent, tracing, derivative, root, min and max. It simulates for graphing caluculator 83 and t1 84, 84 plus.
◉ Integral calculator 991 plus provide powerful calculator to calculate integrate, derivative, differentiation and integration.
◉ This scientific calculator 991 includes CAS (computer algebra system), which can perform symbolic calculation.
◉ Some other features: math formula, physics formula, unit conversion, theme, font and programming.
We appreciate all valuable feedback.
In this update, we fixed so many problems and improved the math engine.
Thanks for using our app.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00306.warc.gz
|
CC-MAIN-2023-50
| 2,588 | 20 |
http://spru.pt/main.cfm?mnid=3&langID=EN
|
math
|
• I am not a student. Can I apply for a room at the Residence?
• When do applications start?
• I don’t live in Portugal and I will not be able to make the admission interview before my arrival.
• I am in the waiting list. Do I have any chances of getting a room in the Residence?
• I have been told that there are available rooms and I showed interest. Is this enough to have a room reservation?
• Are there schedules to enter and exit the Residence?
• Are my visitors allowed to sleep in my room?
• Is it allowed to smoke in the rooms?
• Can I receive phone calls directly to my room?
• Do you serve meals?
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424090.53/warc/CC-MAIN-20170722182647-20170722202647-00053.warc.gz
|
CC-MAIN-2017-30
| 629 | 10 |
https://classnotes.ng/lesson/linear-graphs-in-two-variable-mathematics-jss2/
|
math
|
Welcome to Class !!
We are eager to have you join us !!
In today’s Mathematics class, We will be discussing Linear Graphs in two Variables. We hope you enjoy the class!
- Continuous graphs
- Graph of real-life situations
- Choosing scales.
Continuous Graph: Graphs are used to show the relationship between two quantities. A continuous graph is in the form of a continuous line and shows the relationship between the two quantities.
1a. A student walked along a road at a speed of 120m per minute.
Make a table of values showing how far the students have walked after 0, 1, 2,3,4,5, minutes
- using the scale of 1cm to 1min on the horizontal axis and 1cm to 100m on the vertical axis, draw a graph of the information. use the graph to find:
- how far the students have walked after 2.6min.
- how long it takes the students to walk 500m
- The student has walked 310cm after 2.6min.
- The student takes about 4.5 in walk 500m
We use distance-and-time graphs to show journeys. It is always very important that you read all the information shown on these type of graphs.
A graph showing one vehicle’s journey
If we look at the graph shown below, you can see that the time in hours is along the horizontal, and the distance in miles is on the vertical axis. This graph represents a journey that Jan took, in travelling to Glasgow and back, from Aberdeen.
Important points to note are:
It took half an hour to travel a distance of miles
Between 1 pm and 3 pm, there was no distance travelled. This means that the car had stopped.
The journey back, after 3 pm, took one hour.
A graph showing two different journeys in the same direction
The next graph shows two different journeys. You can see that there is a difference with the steepness of the lines drawn. Remember that, the steeper the line, the faster the average speed. We can calculate the average speeds, by reading distances from the graph and dividing by the time taken.
Line A: How long does journey A last, and what distance is travelled?
The journey takes 2 hours, and the distance travelled is 15km.
Line B: How long does journey B last, and what distance is travelled?
The journey takes 1 hour, and the distance travelled is also 15 km.
This means that the average speeds are:
A: 15/2 = 7.5 km per hour
B: 15/1 km per hour
You will also notice from the graph that the two lines cross. This means that, if the two vehicles were travelling along the same route, they would have met at that point, which was just after 10 am. The vehicle on journey B overtook vehicle A.
A graph showing two journeys in the opposite direction
A different pair of journeys is shown below. It is important to note that one journey begins at a distance of 0 miles, and the other at a distance of 40 miles, from Perth. In fact, what is happening is that one journey is travelling away from, and the other is travelling towards Perth. Again, the two journeys meet. This time it is miles from Perth.
You will also see that the two journeys contain stops.
If we were to calculate the average speeds for each total journey we would have to include this time as well.
A graph showing a journey (or journeys) should have time on the horizontal axis, and distance from somewhere on the vertical axis.
A line moving up, as it goes from left to right, shows a journey moving away from a place, and a line moving down, as it goes from left to right, represents a journey towards a place.
A horizontal line is a break or rest.
Two lines, sloping the same way, cut: then an overtaking has taken place.
Two lines, sloping opposite ways, cut: a meeting has taken place.
A speed-time graph ( also called velocity-time graph), shows how the speed of an object varies with time during a journey.
There are two very important things to remember about velocity-time graphs.
The distance travelled is the area under the graph.
The gradient or slope of the graph is equal to the acceleration. If the gradient is negative, then there is a deceleration. We may use the equations(1) or some rearrangement of this equation.
Example. A car starts on a journey. It accelerates for 10 seconds at It then travels at a constant speed for 50 seconds before coming to rest in a further 4 seconds.
- Sketch a velocity-time graph.
- Find the total distance travelled.
- Find the deceleration when the car is coming to a stop at the end.
- Find the average speed.
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Plane Figures and Shapes. We are very much eager to meet you there.
For more class notes, homework help, exam practice, download our App HERE
Join ClassNotes.ng Telegram Community for exclusive content and support HERE
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00718.warc.gz
|
CC-MAIN-2024-10
| 4,832 | 54 |
https://demonstrations.wolfram.com/RubiksSnakePuzzle/
|
math
|
Rubik's Snake Puzzle
Requires a Wolfram Notebook System
Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.
Rubik's snake is a mechanical puzzle made of 24 isosceles triangular prisms as links. The toy can be twisted to form objects, animals, or geometric shapes.[more]
This Demonstration builds a forward kinematics model of Rubik's snake with 23 degrees of freedom. You can use slider bars for the left and right columns to control the rotation angle of each joint. Joint 0 in the left frame and joint 0 in the right frame are the same joint, but they drive different links.[less]
Contributed by: Frederick Wu (September 2014)
Open content licensed under CC BY-NC-SA
This Demonstration was inspired by Professor Oussama Khatib's Introduction to Robotics course, available free online through Stanford's Engineering Everywhere.
J. J. Craig, Introduction to Robotics, Mechanics and Control, 3rd ed., New York: Pearson–Prentice Hall, 2005 pp. 62–82.
J. Denavit and R. S. Hartenberg, "A Kinematic Notation for Lower-Pair Mechanisms Based on Matrices," Journal of Applied Mechanics, 77, 1955 pp. 215–221.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00781.warc.gz
|
CC-MAIN-2023-14
| 1,160 | 10 |
https://plus.maths.org/content/tags/news
|
math
|
We talk to Ivan Smith, invited lecturer at the ICM 2018, about his work and what he likes about the Congress.
We talk to Nalini Joshi about her mathematics, the International Mathematical Union, and her work in supporting women in mathematics.
We talk to two of the organisers of the ICM 2022 in Saint Petersburg.
Ali Nesin has been awarded the 2018 Leelavati Prize for creating a mathematical paradise for Turkish students and the world's mathematicians.
In this podcast we try to capture a flavour of Fields medallist Akshay Venkatesh's work.
In these two videos Akshay Venkatesh tells us about winning the Fields Medal, what he enjoys about maths, and introduces us to the ideas in his work.
We talk to Maria Esteban, mathematician and President of the International Council for Industrial and Applied Mathematics.
Nearly 18 million students took part in Brazil's maths olympiad for schools!
Masaki Kashiwara wins the Chern medal for his "outstanding and foundational
contributions to algebraic analysis and representation theory sustained over a period of
almost 50 years."
Akshay Venkatesh has been awarded the Fields medal 2018 for his work exploring the boundaries of number theory.
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00445.warc.gz
|
CC-MAIN-2024-10
| 1,189 | 12 |
https://math.answers.com/other-math/Which_number_added_to_100_and_164_will_make_them_both_perfect_square_numbers
|
math
|
Because numbers are infinite, there is an infinite number of answers. e.g. What number should be added to 2 to make a perfect square? 2+2=4 (a perfect square) 2+7=9 (a perfect square) 2+14=16 (a perfect square) 2+23=25 (a perfect square) etc... Did you have a specific number in mind.
5607 + 18 = 5625, a perfect square. The perfect square of a square root is the number you started with.
can I have the awnser
Hi, do you mean what square number is closest to 4321? 66 squared is 4356 which is 35 from 4321 if that's what you mean? it mean that 35 is least no. whih should be added to 4321 to make it perfect square
2 and 2
A perfect number is only classified as a perfect number because all of its proper divisors add up to itself. For example, the proper divisors of six are one, two and three. Those numbers added together equal six. Therefore, six is a perfect number.
It is: 64 because the square root of 1089 is 33
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00701.warc.gz
|
CC-MAIN-2022-27
| 920 | 7 |
https://www.onlinemathlearning.com/max-min-polynomials.html
|
math
|
More Lessons for PreCalculus
Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to find the maximum and minimum values of polynomial functions and word problems.
Finding Maximum and Minimum Values of Polynomial Functions
Polynomial functions are useful when solving problems that ask us to find things like maximum income, revenue or production quantities. Finding maximum and minimum values of polynomial functions help us solve these types of problems. When setting up these functions, we first determine what the problems are asking us to maximize and then set up the function accordingly.
How we define optimization problems, and what it means to solve them?
Solving Optimization Problems using Derivatives
Step 1: Write a function for the item to be optimized.
Step 2: Write constraint equations as needed to relate the variables. Will be used for substitution into the optimization function.
Step 3: Using substitution make the optimization function of only 1 variable.
Step 4: Find first derivative critical values and analyze to find appropriate relative max or min.
Step 5: Use the selected critical value to answer the question in the problem.
This tutorial demonstrates the solutions to 5 typical optimization problems using the first derivative to identify relative max or min values for a problem.
Example 1: Find a pair of non-negative numbers that have a product of 196 and minimize the sum of four times the first number and the second number.
Example 2: An open topped serving box will be made by cutting squares out of each corner of a 12" by 18" sheet of cardboard and folding the tabs up to form a box. What size squares should be cut out to maximize the volume of the box?
Example 3: The combined perimeter of a circle and rectangle is 100 inches. The length of the rectangle is twice its width. Find the dimensions of each in order to minimize the total area.
Example 4: What point on the function y = x2
- 6x + 10 is closest to the origin?
Example 5: The demand function for a product is p = $1000 - √x. The cost to produce x units is C = $50000 + $100x. What price should be set to maximize the profit of the product?
Find the approximate maximum and minimum points of a polynomial function by graphing
Graph f(x) = x3
Estimate the x-coordinates at which the relative maxima and relative minima occurs.
Optimization Problem - Maximizing the Area of Rectangular Fence Using Calculus / Derivatives
This video shows how a farmer can find the maximum area of a rectangular pen that he can construct given 500 feet of fencing. We can actually solve this quite easily using algebra but this video tries to show the overall process that we use on maximization / minimization problems.
A man has a farm that is adjacent to a river. Suppose he wants to build a rectangular pen for his cows with 500 ft. of fencing. If one side of the pen is the river, what is the area of the largest pen he can build?
Optimization Problem - Max Volume of a Box Made From Square of Material
How to find the maximum volume of a box made from a 2ft x 2ft piece of metal when corners of equal size are removed and then the sides of the box are folded up?
Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00826.warc.gz
|
CC-MAIN-2023-40
| 3,540 | 30 |
https://excelypedia.com/calculate-percentage-in-excel/
|
math
|
Percentages are commonly used when you’re working with data. But they can often be difficult to manually calculate. Fortunately, spreadsheets do the work for you. Let’s learn how to calculate percentage in Microsoft Excel.
How to Calculate Percentage in Excel
Some percentages are easy to see: 50 is 50% of 100, for example. But 73 is what percentage of 384? As you can see, the math gets tricky. Not so when you use Excel to calculate percentage.
The typical formula for percentage is as follows:
Part/Whole x 100
Excel makes things even simpler. When you use Percent Style cell formatting, you don’t have to perform the step of multiplying by 100. Let’s learn how.
To begin, enter the part value into any cell in your worksheet. For this example, we’ll use 73 in cell E1. Simply type that number in, and then click into a second blank cell like E2.
There, place your total number. In this case, your total is 384. You have your numbers; now it’s time to build the percentage formula.
Click into any blank cell, then type an = sign. This tells Excel that you’re inputting a formula. Now, remember how percentages work. You’re dividing the part by the whole. With this in mind, click on cell E1. Then, type /, which is the division sign in Excel. So far, your formula is:
Finally, click on cell E2. This is your complete formula:
Hit Enter on your keyboard, and Excel will return a value: 0.190104167. Be careful here – that isn’t your actual percentage quite yet. Obviously, 73 is more than 0.19% of 384!
With the cell containing 0.19 selected, go to the Home tab on Excel’s ribbon. Near the middle, click the % button. This is called Percentage Style. Your cell transforms to an easy, rounded percentage: 19%. In essence, Percentage Style applies the step of multiplying by 100.
As you can see, it’s a breeze to calculate percentage in Excel.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00219.warc.gz
|
CC-MAIN-2023-40
| 1,868 | 13 |
https://opus.lib.uts.edu.au/handle/10453/27139
|
math
|
A bayesian hierarchical factorization model for vector fields
- Publication Type:
- Journal Article
- IEEE Transactions on Image Processing, 2013, 22 (11), pp. 4510 - 4521
- Issue Date:
Factorization-based techniques explain arrays of observations using a relatively small number of factors and provide an essential arsenal for multi-dimensional data analysis. Most factorization models are, however, developed on general arrays of scalar values. For a class of practical data arising from observing spatial signals including images, it is desirable for a model to consider general observations, e.g., handling a vector field and non-exchangeable factors, e.g., handling spatial connections between the columns and the rows of the data. In this paper, a probabilistic model for factorization is proposed. We adopt Bayesian hierarchical modeling and treat the factors as latent random variables. A Markov structure is imposed on the distribution of factors to account for the spatial connections. The model is designed to represent vector arrays sampled from fields of continuous domains. Therefore, a tailored observation model is developed to represent the link between the factor product and the data. The proposed technique has been shown effective in analyzing optical flow fields computed on both synthetic images and real-life videoclips. © 2013 IEEE.
Please use this identifier to cite or link to this item:
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00096.warc.gz
|
CC-MAIN-2019-13
| 1,415 | 7 |
http://www.book.artbeeweb.nl/?p=3571
|
math
|
Light from any direction hits a surface, penetrates it for a very slight amount – which will give some absorption, and then it gets re-radiated in a color-filtered way equally in all directions. The path back to the surface will give absorption too, proportional to the distance D traveled through the object surface.
As this distance D is inversely proportional to the cosine of the exiting angle (D=D0/cos(a)), the intensity of the diffuse light in that direction will be proportional: I = I0 cos(a). This is the angular distribution of outgoing, diffuse light, according to the mathematician J.H. Lambert (about 1750). At perpendicular scattering, angle a=0 so cos(a)=1 and the response is maximal, while at parallel scattering cos(a)=0 and there is no response at all.
And since cosine calculations are hardwired into modern CPU electronics, this is a speedy rendering approach by any means. Therefore, Poser includes Lambert shading into Diffuse (see the basic and intermediate articles on this). This offers a resource-friendly first step towards more realistically looking renders. And, for people who want more steps in that direction, Poser offers alternatives like Clay.
Now, look what will happen to the render result. A specific area on the render plane (say: a pixel), marked green in the illustration, gets its light from an area on the object surface (marked green as well). At skewer angles a between surface normal and camera, this area on the object gets larger: A = A0 / cos(a).
At skewer angles such an area emits less light per unit of surface (cm2 or alike) and the resulting amount of light onto the pixel in the render plane is I * A = I0 cos(a) * A0 / cos(a) = I0 * A0 is a constant
So the Lambert shading not only matches a nice explanation on diffuse lighting, it also makes that the intensity of light in the render result does not depend on the camera angle to the surface. Because at skewer angles the pixel in the render represents a larger area on the object surface, which diffuses less light towards the camera, and both effects cancel out.
Look at the light
This leaves the effect of the incoming light itself. At skewer angles, the same amount of light will hit larger and larger areas of the object surface, so the intensity per unit of surface decreases accordingly: L = L0 cos(b) . Besides the math, this means that the extent to which shading reveals the shape of an object, depends on the ‘directivity’ of the light only. Point lights are quite directive, even a flat surface it lit under varying angles. An infinite light is less directional, a flat surface is evenly lit but a ball is not. IDL lighting is hardly directive at all, the light comes from all directions and all surface areas are equally lit whatever the shape of the object. Hence the shape of objects is less revealed, and objects will look flat.
Left: point light nearby, the edges get darker faster. Mid: Infinite light. Right: IDL, the ball looks like a disk.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00301.warc.gz
|
CC-MAIN-2023-50
| 2,976 | 9 |
https://repository.nie.edu.sg/handle/10497/23969
|
math
|
Publication: Non-symbolic ratio reasoning in kindergarteners: Underlying unidimensional heuristics and relations with math abilities
Although it is thought that young children focus on the magnitude of the target dimension across ratio sets during binary comparison of ratios, it is unknown whether this is the default approach to ratio reasoning, or if such approach varies across representation formats (discrete entities and continuous amounts) that naturally afford different opportunities to process the dimensions in each ratio set. In the current study, 132 kindergarteners (Mage = 68 months, SD = 3.5, range = 62–75 months) performed binary comparisons of ratios with discrete and continuous representations. Results from a linear mixed model revealed that children followed an additive strategy to ratio reasoning—i.e., they focused on the magnitude of the target dimension across ratio sets as well as on the absolute magnitude of the ratio set. This approach did not vary substantially across representation formats. Results also showed an association between ratio reasoning and children’s math problem-solving abilities; children with better math abilities performed better on ratio reasoning tasks and processed additional dimensions across ratio sets. Findings are discussed in terms of the processes that underlie ratio reasoning and add to the extant debate on whether true ratio reasoning is observed in young children.
Muñez, D., Bull, R., Cheung, P., & Orrantia, J. (2022). Non-symbolic ratio reasoning in kindergarteners: Underlying unidimensional heuristics and relations with math abilities. Frontiers in Psychology, 13, Article 800977. https://doi.org/10.3389/fpsyg.2022.800977
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00283.warc.gz
|
CC-MAIN-2024-18
| 1,708 | 3 |
http://cms.math.ca/10.4153/CJM-2010-030-4
|
math
|
Canad. J. Math. 62(2010), 563-581
Printed: Jun 2010
We give explicit formulas for Whittaker functions on real semisimple Lie groups of real rank two belonging to the class one principal series representations. By using these formulas we compute certain archimedean zeta integrals.
11F70 - Representation-theoretic methods; automorphic representations over local and global fields
22E30 - Analysis on real and complex Lie groups [See also 33C80, 43-XX]
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398466178.24/warc/CC-MAIN-20151124205426-00261-ip-10-71-132-137.ec2.internal.warc.gz
|
CC-MAIN-2015-48
| 451 | 5 |
https://prisoner-dilemma.herokuapp.com/report.html
|
math
|
The Prisoner's Dilemma is a standard example of a game analyzed in game theory that shows why two completely rational individuals might not cooperate, even if it appears that it is in their best interests to do so. It was originally framed by Merrill Flood and Melvin Dresher while working at RAND in 1950. [Source: Wikipedia]
This survey was a modified simulation of the same. It is a part of my Philosophy Project at college. This report shows the results and observations for the same.
This survey received around 150 responses in the first week of being live. A big thank you to all those who contributed! :)
In this case, the penalties were set to resemble the original Prisoner's Dilemma (which however assumes the other person to be intelligent only).
Though, we do get to see rather mixed opinions with only a little cooperation/betrayal margin, opposed to the original paradox, according to which number of betrayals would be a lot more than number of cooperations.
The additional thing here is the assumption of the other person being unintelligent. We can observe a deviation in the distribution in this case. People tend to Betray when they assume that the other person is not intelligent.
Putting it differently, we may say that people do tend to cooperate if they assume that the other person is intelligent, probably expecting for a similar action on the other end.
In this case, the penalties involving at least one of the two people cooperating was reduced by a great extent.
It is very evident from the responses that people prefer cooperating in this case. This shows that people are willing to cooperate as the risk factor reduces. It is also worth noting the order in which the cases are put up. That is, people tend to cooperate in this case after they faced the original situation in which the penalties were higher.
Also, it can be observed that the number of betrayals increases when the other person is assumed to be unintelligent. More on this in the last section.
In this case, mutual cooperation was favoured, by releasing all charges when both cooperate. Also, in case of a conflict, where one betrays and the other cooperates, even the one who betrays gets some punishment (in contrast to the original case where the person who betrays is set free in such a situation).
A very good thing to note is that this simply works as expected :) There are very less betrayals compared to cooperations.
Again, the ratio of number of betrayals to cooperations is higher in the case where the other person is assumed to be unintelligent.
In this case, the punishment for common responses (either both cooperate or both betray) were kept ambiguous. This case was intentionally put up in the end, in an attempt to build up a mindset that generally mutual cooperation leads to lesser punishment. In case of a conflict (one cooperates and other betrays), the one who cooperates suffers larger punishment.
The motive here was to see whether people cooperate based on an instinct built up from the previous cases or they betray just seeing the lesser punishment (and assuming the other would cooperate).
One interesting thing to note is that many people avoided this case since there were considerably lesser (around 20 less) responses to this case when compared to other cases.
For the responses received, we can observe that people tend to cooperate when the other person is assumed to be intelligent. However, in the other situation, the difference between number of cooperations and betrayals decreased.
That is, people tend to go with the instinct instead of the apparent benefit (which however involves some amount of risk in case the other betrays).
This is a comparison for total number cooperations or betrayals when the other person is assumed to be intelligent versus when the other person is assumed to be unintelligent, summed up over all cases.
It can be easily observed that number of betrayals get increased by a lot when the other person is assumed to be unintelligent.
This is an interesting point. Let's say we fix that person A betrays. We have two cases. One, A thinks that the other unintelligent person (B) would cooperate, since B is not intelligent, and hence A would betray for getting released. Two, A thinks that B would betray. But in this case, A should prefer to cooperate in order to get lesser penalty (unless A doesn't want a tit-for-tat). So, A won't betray.
Hence, people tend to betray more under the assumption that the other person is not intelligent, thinking that the other person would cooperate.
As I mentioned before, Prisoner's Dilemma is a well-known paradox and has many use-cases. This modified simulation of the same had two main objectives:
One, to see if the present responses are in line with prior observations.
Two, to observe and analyse the affect of modification in parameters like rationality and reward (rather punishment in this case).
The observations in prior sections catered to these objectives.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00438.warc.gz
|
CC-MAIN-2022-27
| 4,972 | 26 |
https://tasiilaq.net/what-is-the-most-abundant-isotope-of-potassium/
|
math
|
You are watching: What is the most abundant isotope of potassium
The percent herbal abundances of $ce^39K$ and $ce^41K$ room $93.2581\%$ and also $6.7302\%$ respectively.
a) What is the natural abundance of $ce^40K$?
b) determine the isotopic mass of $ce^41K$.
So I started to use this equation:
$mathrmatomic:mass = (fractional:abundance:isotope:1 cdot mass) + (fractional:abundance:isotope:2 cdot mass)$
But then ns realized that i did not have some values. $ce^39K$ is the only isotope which has two values. $ce^40K$ only has actually the mass weight, however, if i were to multiply $ce^40K$"s mass with ($1-0.932581$) would that help me arrive at one answer?
Also, would that answer be accurate due to the fact that it only takes right into account the there are only to values that add up come 100%.
improve this question
edited Oct 30 "15 at 16:55
10k2222 yellow badges4646 silver badges5858 bronze title
request Sep 29 "14 at 1:01
35922 yellow badges55 silver badges1818 bronze badges
| display 2 more comments
1 answer 1
active oldest Votes
A). If you know the percentages the the diversity of $frac23$s the the isotopes then recognize the percentage of the $3^rd$ is as simple as
$ 100\% - (\%_ce^39K+\%_ce^41K) = \%_ce^40K $
B). Detect the mass of the isotope should be pretty simple. It would certainly be
$(mathrmMass of proton cdot mathrm# that protons )+(mathrmMass of a neutron cdot mathrm# the neutrons ) approx mathrmMass of isotope$
You disregard the load of electrons due to the fact that the fixed is tiny relative come the others.
boost this price
reply Sep 29 "14 in ~ 6:14
john SnowJohn eye
4,39944 gold badges2424 silver- badges6464 bronze badges
add a comment |
Thanks for contributing response to tasiilaq.net ridge Exchange!Please be certain to answer the question. Carry out details and share your research!
But avoid …Asking because that help, clarification, or responding to various other answers.Making statements based upon opinion; earlier them up with references or an individual experience.
Use MathJax to format equations. MathJax reference.
To learn more, view our tips on writing great answers.
See more: What Episode Does Naruto Go 9 Tails ? Naruto Nine
Sign increase or log in in
sign up making use of Google
authorize up utilizing Facebook
authorize up making use of Email and also Password
Post together a guest
email Required, however never shown
Post together a guest
Required, however never shown
article Your answer Discard
Not the answer you're spring for? Browse other questions tagged everyday-tasiilaq.net isotope or asking your own question.
The Overflow Blog
Featured ~ above Meta
What does $mathrmA_r(mathrmH)$ represent when calculating the average natural weight of an elemental atom?
Questions about combustion and also isotopic diversity
Biological consequences of Asteroid Mining—Death by Isotope?
go the natural abundance the radioactive elements change?
exactly how come uranium's relative atomic massive is 238.03 when it only includes isotopes with a mass number of 238 or less?
Why room isotopes an issue in reading mass spectra?
just how do we understand the organic abundance of isotope on Earth?
hot Network inquiries much more hot questions
i ordered it to RSS
concern feed To i ordered it to this RSS feed, copy and paste this URL right into your RSS reader.
ridge Exchange Network
site design / logo design © 2021 stack Exchange Inc; user contributions licensed under cc by-sa. Rev2021.10.28.40586
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00034.warc.gz
|
CC-MAIN-2021-49
| 3,471 | 55 |
https://studydaddy.com/question/qnt-273-week-2-dq-1
|
math
|
QNT 273 Week 2 DQ 1
This paperwork of QNT 273 Week 2 Discussion Question 1 includes:
How are the mean, median, and mode similar? How are the mean, median, and mode different? For a given set of data, might the mean, median, and mode be the same? Give specific examples to illustrate your point.
*** *** Week 2 ** *Attached: QNT 273 Week 2 DQ 1.zip
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00448-ip-10-31-129-80.ec2.internal.warc.gz
|
CC-MAIN-2016-50
| 347 | 4 |
http://inftech.webservis.ru/it/conference/isanditc/2000/section5/eng/areng5.html
|
math
|
Russia, Petrozavodsk, Petrozavodsk State University (PetrSU)
INFORMATION-MATHEMATICAL ASPECTS OF THE RENEWAL SYSTEMS CONTROL FOR FINITE TIME INTERVAL
The paper presents some approaches to solving the problems that arise in investigation of the problem of optimal control of a renewal system for finite time interval from the stage of statement and analysis of the problem to the stage of software programming.
Россия, Петрозаводск, Петрозаводский государственный университет (ПетрГУ)
ИНФОРМАЦИОННО-МАТЕМАТИЧЕСКИЕ АСПЕКТЫ ПРОБЛЕМЫ УПРАВЛЕНИЯ ВОССТАНАВЛИВАЕМЫМИ СИСТЕМАМИ НА КОНЕЧНОМ ИНТЕРВАЛЕ ВРЕМЕНИ
В докладе рассказывается о некоторых подходах к решению задач, возникающих при полном цикле исследования (от постановки задачи и ее анализа до создания соответствующего программного обеспечения) по организации оптимального управления восстанавливаемыми системами, которые функционируют на конечном интервале времени.
The problem of renewal system control for finite time interval [1, 5, 6 and 8] is attracting increased interest due to its practical applications. Usually the renewal processes, alternating and semi-Markovian processes are used as a model of renewal systems operation. Some models are based on embedded processes [12, 14]. After a model has been developed, its adequacy has to be checked. In order to determine control parameter values a corresponding optimization problem has to be stated and solved. The solution found is analyzed and if this analysis proves that this solution is practically acceptable then the corresponding computer software should be developed. This software must be convenient enough for the managerial personnel who do not possess specialized mathematical and computer skills.
2. Renewal system simulation
Some approaches to simulation of renewal system operation on the example of paper-machines (PM) with the use of embedded alternating processes as a model are presented in [1, 12 and 14]. In those papers five stages have been distinguished for describing the operation efficiency of a paper-machine. These are working, accident by mechanical cause, accident by technological cause, stopping by administrative and economical cause and planned preventive maintenance. The model is based on three embedded alternating processes.
Unfortunately, the data received from this model is censored. It satisfies the following testing plans [n, B, T], [n, T1, …, Tn], [n, A, (F1, T1, …, Tn)], [n, A, (F1, F2, T1, …, Tn)] etc. This fact complicates the checking of model adequacy and requires development of special statistical methods.
In author’s opinion the use of embedded processes as models is more prospective than the use of semi-Markovian processes.
3. Statistical analysis of mathematical model parameters
We now present (perhaps incomplete) list of statistical problems, which arise when renewal processes are used as a model of a renewal system :
Statistical checking of the model described in the previous section and estimation of reliability parameters of the renewal system have required the development of special statistical methods. The results obtained by the author are presented in [4-6, 9-15].
4. Control system simulation
The following problems arise during control system simulation:
The real practice leads to a series of optimization problems for renewal systems functioning. Therefore, synthesis of optimization mathematical model for functioning of a renewal system is a very complicated problem. To get an idea of how to approach this problem, the author suggests taking a look at the results of development and investigation of optimization models for planning preventive measures (PPM) for a complex system of paper-machines on finite time interval [0, T]. These results can be found in [1-3, 6 and 8]. Optimization problems described there are nonlinear problems of separable continuous-discrete programming with constraints of “or-or” type. Besides that, these are multi-criterion problems. As a rule, the number of variables in these problems is too large to be able to solve them by straight item-by-item examination. The dynamic programming method often does not apply, too. Application of some heuristic methods with local optimization gives acceptable results.
5. Requirements to the information system of control support
The following problems arise when the information system of control support is being developed:
Generalizing the results of development and applying the information and mathematical support for solving different optimization problems of control for some enterprises of North-Western region of Russia (Karelia, Arkhangelsk area), we come to the conclusion, that now there is a need in developing special software environment. It should be oriented on solution of practical problems in the following areas: information systems and networks, service of operative repair of technological equipment, robot-technical complex, transport systems, problems of calendar network planning etc .
Let formulate the main requirements of this environment. It must consist of the following units:
Information unit is intended for statistical data entry, lexicographic data entry and database entry. This information about control support system will be used to solve optimization problems.
Mathematical unit will
Mathematical unit has to have modular structure. Every user can use only modules he/she needs.
Service unit is intended to specify characteristics and parameters of the control support system. This unit is used with specialists having mathematical knowledge and experience of mathematical model development. Such specialists also apply user interface, as users are not supposed to have special mathematical and mathematical programming knowledge. If application is correct then specialist-mathematician does not take part in the environment exploitation. Thus the service includes choice of mathematical model of control support system (possibly, several models) and choice of user interface.
Printing unit is intended to prepare and print different reports. Unit has to have graphic means.
Existing software, which can be used for solving the optimization problems, may be divided into two groups. The first, such as MathLab, MathCad, Mathematica, are used by specialist-mathematician in laboratory conditions. The second type – Excel, Quattro Pro – is intended for economical calculation. It has limited mathematical means and can not be used for solving optimization problems. Besides, practically all software does not have convenient tools for customizing the environment for non-specialist users.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00215.warc.gz
|
CC-MAIN-2019-35
| 7,053 | 27 |
http://www.amazon.co.uk/product-reviews/0201631970
|
math
|
15 of 15 people found the following review helpful
on 17 May 2006
This was one of my university textbooks in the first year, but it is such a good book that I still read it once in a while.
It is clearly written, with many worked examples. It covers Axioms of arithmetic, Sequences, Series, Differentiation and Integration, but it starts off from first priciples in all cases, building up from logic onwards. Interspersed with the text are short biographies of mathematicians.
There are exercises at the end of each section with answers in the back. All in all, this is an ideal book for an undergraduate doing a course in Analysis or the interested layman.
4 of 5 people found the following review helpful
on 5 March 2011
I have to say that this module has saved my life. The course that I am doing at university is not one of the easiest thing to get your head around, but this book puts things into clear and concise definitions and is definetly a recommendation for anyone studying any form of mathemtical analysis or calculus. It cuts things down to an easy to understand level whilst giving full proofs for a number of theorems. Combined with the examples, problems and exercises it helps develop an understand of the subject in great detail.
1 of 4 people found the following review helpful
on 30 October 2010
syllabus specific for my degree, much cheaper on amazon than in shops. saved 40 quid.
1 of 6 people found the following review helpful
on 23 June 2011
I found this book, unlike the abysmal Gods Of Ruin, to be not only useful but, like the wonderful Atlas Shrugged, also quite entertaining. That said, also unlike the horrid Gods Of Ruin, this book does require at least a 4th grade education to understand and follow easily. Anyone interested in a concise, accurate, and informative approach to basic mathematics will certainly find this book rewarding and, unlike Gods Of Ruin, well worth the purchase price.
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440646312602.99/warc/CC-MAIN-20150827033152-00006-ip-10-171-96-226.ec2.internal.warc.gz
|
CC-MAIN-2015-35
| 1,926 | 14 |
http://rivista.math.unipr.it/vols/2010-1-1/05.html
|
math
|
On the zero-viscosity limit for 3D Navier-Stokes equations under slip boundary conditions
Received: 25 February 2010
Accepted: 7 April 2010
Mathematics Subject Classification (2000): 35Q30, 76D05, 76D09.
Keywords: Navier-Stokes equations, slip boundary conditions, zero-viscosity limit.
Abstract: In this survey article we consider the initial-boundary value problem for the three-dimensional Navier-Stokes equations with Navier boundary conditions and study the problem of convergence of the solutions, as the viscosity goes to zero, to the solution of the Euler equations. We present some strong convergence results, obtained in collaboration with H. Beirão da Veiga (see and ), in the case where the region of motion is a bounded domain with flat boundaries.
H. Beirão da Veiga and F. Crispo, Sharp Inviscid Limit Results under Navier Type Boundary Conditions. An Lp Theory, J. Math. Fluid Mech. 12 (2010) 397–411, DOI: 10.1007/s00021-009-0295-4
H. Beirão da Veiga and F. Crispo, Concerning the Wk,p-Inviscid Limit for 3-D Flows Under a Slip Boundary Condition, J. Math. Fluid Mech. (2009), DOI: 10.1007/s00021-009-0012-3
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690307.45/warc/CC-MAIN-20170925021633-20170925041633-00286.warc.gz
|
CC-MAIN-2017-39
| 1,132 | 8 |
https://www.galbraithgroup.com/news/energy-matters-september-2013
|
math
|
Energy Matters September 2013
Read the new issue of our e-zine to get update with current issues and projects from our Energy Team.
View the current issue our e-zine now.
You can also read previous issues of our e-zine.
Find your nearest office
Copyright © 2018 CKD Galbraith LLP | Letting Agent Registration Number: LARN1810017 |
Privacy and Cookies |
Policy Statements |
Employee Privacy Notice |
Client Money Handling Procedures
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00660.warc.gz
|
CC-MAIN-2022-33
| 450 | 10 |
https://askingthelot.com/what-is-a-proportional-relationship-7th-grade-math/
|
math
|
A proportional relationship between two quantities is a collection of equivalent ratios, related to each other by a constant of proportionality. Proportional relationships can be represented in different, related ways, including a table, equation, graph, and written description.
- 1 What is a proportional relationship in math example?
- 2 How do you know if a relationship is proportional?
- 3 What is 7th grade math?
- 4 What are some examples of proportions?
- 5 What does proportion mean in maths?
- 6 What do u mean by proportional?
- 7 How hard is 7th grade?
- 8 Is 7th grade math hard?
- 9 How do you write a proportional equation?
- 10 What is a proportional linear relationship?
- 11 How do you calculate proportionate?
- 12 Does proportional mean equal?
- 13 How do you find a proportional relationship on a graph?
- 14 Which equation represents proportional relationships?
- 15 Which characteristics of a graph tell you that it represents a proportional relationship?
- 16 How old are u in 7th grade?
- 17 Should 7th graders take algebra 1?
- 18 Can a 7th grader learn calculus?
- 19 Is it normal for a 13 year old to be in 7th grade?
- 20 Is seventh grade the worst year?
- 21 What grade is the hardest in school?
- 22 What IQ is needed to skip a grade?
- 23 Can you go to college at age 13?
- 24 What is the passing grade for 7th grade?
What is a proportional relationship in math example?
This week your student will learn to write equations that represent proportional relationships. For example, if each square foot of carpet costs $1.50, then the cost of the carpet is proportional to the number of square feet. The constant of proportionality in this situation is 1.5.
How do you know if a relationship is proportional?
How Do You Know If Two Ratios are Proportional? Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
What is 7th grade math?In 7th grade, students will fully understand how to interpret and compute all rational numbers. They can add, subtract, multiply, and divide all decimals and fractions, as well as represent percents.
What are some examples of proportions?
- 40m of that rope weighs 2kg.
- 200m of that rope weighs 10kg.
What does proportion mean in maths?
Definitions: A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0. A proportion is an equation in which two ratios are set equal to each other. For example, if there is 1 boy and 3 girls you could write the ratio as: … 0.25 are boys (by dividing 1 by 4)
What do u mean by proportional?
: having a size, number, or amount that is directly related to or appropriate for something. : having parts that are the correct or appropriate size in relation to each other. proportional. adjective.
How hard is 7th grade?Work in 7th grade can be challenging at times. It is known to be the most challenging grade in middle school-but everyone gets through it. In order to succeed, it is crucial to pay close attention in class and take good notes. Studying hard is also very important in order to do well in grade seven.
Is 7th grade math hard?
Seventh grade is a challenging year for many reasons. Socially, students are in the heart of middle school and all of the drama that it brings. In many subjects, the work gets harder too! In math, concepts begin to jump from concrete to more abstract, making 7th grade algebra a challenging course for many students.How do you skip a grade?
- A Written Request. Put your request for skipping a grade in writing to the school principal and keep a copy. …
- Expert Guidance. Make sure that legitimate requirements are being used in considering your request. …
- Academic Achievement. …
- Emotional Readiness. …
- Student Acceptance. …
- Need for Change.
How do you write a proportional equation?
The equation that represents a proportional relationship, or a line, is y=kx, where k is the constant of proportionality. Use k=yx from either a table or a graph to find k and create the equation. Proportional relationships can be represented by tables, graphs and equations.
What is a proportional linear relationship?
Proportional linear relationships have a constant rate of change, or slope. Proportional linear relationships have a constant rate of change, or slope. 2. Proportional linear relationships have a common ratio from each output to each input.
How do you calculate proportionate?
The Formula for Percent Proportion is Parts /whole = percent/100. This formula can be used to find the percent of a given ratio and to find the missing value of a part or a whole.
Does proportional mean equal?
When something is proportional to something else, it does not mean the values are equal, just that they change with respect to eachother. The constant of proportionality serves as a multiplier.
How do you find a proportional relationship on a graph?
The best way to show and explain direct proportional relationships is by graphing two sets of related quantities. If the relation is proportional, the graph will form a straight line that passes through the origin.
Which equation represents proportional relationships?
A proportional relationship between a quantity y and a quantity x that has a constant of proportionality k is represented by the equation y = kx. If an equation in a different form can be rewritten as above, then it is a proportional relationship.
Which characteristics of a graph tell you that it represents a proportional relationship?
A graph represents a proportional relationship if it is a line which goes through the origin. The ‘y’-coordinate when ‘x’ equals 1 is the constant of proportionality ‘k,’ and the equation from the graph is given by y=kx.
How old are u in 7th grade?
Student Age (as of September 1, 2021)American Grade Equivalent12 years oldGrade 711 years oldGrade 610 years oldGrade 59 years oldGrade 4
Should 7th graders take algebra 1?
Seventh graders are capable of Algebra 1 or even Geometry, depending on how well they have prepared. It’s not the age, but how well you have prepared them. If the child is going to take a College Major related to Math or Math skills required, then try to take Algebra in 7th. grade at least.
Can a 7th grader learn calculus?
Students who take Algebra 1 in 7th grade can complete Calculus in the 11th grade and take an even more advanced math class, such as college-level Linear Algebra, in grade 12. On the other hand, students who want to jump off the Calculus track have other course options, such as Trigonometry or Statistics.
Is it normal for a 13 year old to be in 7th grade?
Well, most people are 12–13 years old when they are in 7th grade. Heck, I’m 12 and I’m in 7th grade. However, there could be various reasons for why you are still in 7th grade. It is okay for you to be in 7th grade at this age.
Is seventh grade the worst year?
“Seventh grade really is the worst year ever,” agrees Jennifer Powell-Lunder, a psychologist at Pace University who specializes in tween development. Once self-assured, happy kids become encumbered by new feelings of embarrassment, isolation, depression, and, for girls in particular, a loss of self-esteem.
What grade is the hardest in school?
While junior year is often the hardest year of high school, the transition from middle school to 9th grade can also be tough. To make it easier, don’t feel afraid to reach out to your teachers and counselors, and take advantage of the support resources that are available.
What IQ is needed to skip a grade?
To advance successfully, some educators indicate that children should have a measured IQ in at least the 98th+ percentiles (IQ measurements vary depending on the test, but 125-130 is a minimum) and should already work at the average level of the desired grade placement.
Can you go to college at age 13?
Mary Baldwin is one of a handful of colleges in the United States that have programs that accept students as young as 13, according to a study done by Johns Hopkins University.
What is the passing grade for 7th grade?
C – this is a grade that rests right in the middle. C is anywhere between 70% and 79% D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00025.warc.gz
|
CC-MAIN-2022-27
| 8,369 | 78 |
https://www.shaalaa.com/question-bank-solutions/a-gum-bottle-rests-on-its-base-if-it-is-placed-upside-down-how-does-the-thrust-pressure_30519
|
math
|
A gum bottle rests on its base. If it is placed upside down, how does the thrust .
A gum bottle has narrow neck and wider base when placed upside down
Surface area is less, larger is the effect of thrust.
Is there an error in this question or solution?
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178349708.2/warc/CC-MAIN-20210224223004-20210225013004-00272.warc.gz
|
CC-MAIN-2021-10
| 252 | 4 |
http://wiki.irises.org/TbPthruT/TbTomcat
|
math
|
, R. 2005). Seedling# 746-1. TB, height 37" (94 cm), Midseason to late bloom. Standards white, neyron rose (RHS 56C) at midrib flushing toward rim; style arms white; falls honeysuckle (159C) blending to magnolia purple (70C), Honeysuckle rim (161C) and lighter honeysuckle (161D) outer rim; beards indian orange (32A) tipped ethyl blue (112D), ethyl blue end; ruffled, lightly laced; pronounced sweet fragrance. Seedling# 420-1: (seedling# 91-231-11, 'It's Getting Better'
sibling, x (Hager seedling# T5587: (seedling# 5300BrGdT: (seedling# 4292FlrY: (seedling# 3673LmY: ('Dream Affair'
x seedling# 3196Cr: ('Ice Sculpture'
)) x 'Catalyst'
) x seedling# 4558LcY: ('Fringe Benefits'
x seedling# 4112BrtOrLc: (seedling# 3689RfOr: (seedling# 3112BrtOr: ('Flaming Light'
x seedling# 2735Or) x 'Fresno Calypso'
) x seedling# 3690BchOr: (seedling# 3480BchOr: (seedling# 2995OrBch: ('Hayride'
x seedling# 2735Or: (seedling# 2022B: (seedling# 1572: (seedling# 666cream: ('Norah'
) x 'Glittering Amber'
) x seedling# 1652C: (seedling# 983A: (seedling# 536A: ((Hall seedling# 44-05 x T. Craig seedling# 7669) x 'Happy Birthday'
) x Crosby 5425) x 'Glittering Amber'
)) x 'Picture Perfect'
)) x Keppel seedling# 72-3E, 'Marmalade'
sibling) x 'Fresno Calypso'
)))) x seedling# 5292Br/Gd, 'Earthborn'
sibling))) X Hager seedling# T5527, 'Passing Clouds'
sibling. Lauer's Flowers 2005. Honorable Mention 2007; Award of Merit 2009
Please do not enter images that are not your own without owners' permission, this is against Wiki policy
"Although the Encyclopedia is free to all, it is supported by Emembership in AIS, If you would like to help sustain this reference, for $15 you can become an Emember, click here
Interested in Tall Bearded Iris?
Please visit the: Tall Bearded Iris Society
Your Observations Are Valued.
Please make note of bud count, branching, purple based foliage and bloom time, etc. Because these are affected by climate, note date, year and geographic location and write these and other comments in the comment box below.
- TB 'Tomcat' bloomed at the Minnesota Landscape Arboretum, zone 4b, approximately June 2-11, in 2017 (2 rhizomes). 'Tomcat' has purple based foliage, varying from very light to medium color at the base of the leaves, at least during and after flowering. -- HollyJohnson2017-04-01 - 03 Oct 2017
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247481766.50/warc/CC-MAIN-20190217071448-20190217093448-00603.warc.gz
|
CC-MAIN-2019-09
| 2,324 | 25 |
https://legal-dictionary.thefreedictionary.com/generalization
|
math
|
can be wrong even on less important things such as, let's say, dogs.
PE teachers, teachers of students who are visually impaired, and O&M specialists can promote physical activity participation in environments outside of PE by embedding some of these generalization
tactics into the school day.
The exact form of the generalization
gradient is dependent, in part, on how well participants can distinguish among the various stimuli constituting the stimulus set.
Both conditions produced higher ORF rates on training passages compared to the control group; however, only the multiple exemplar intervention produced generalization
gains for oral reading fluency compared to the control group but no differences were noted between the two experimental conditions.
Many researches have studied fully automated generalization
of topographic maps and mapping agencies in many countries have introduced automated generalization
So far, research on map generalization
has almost exclusively focused on maps and spatial data at large scales (i.
The benefit of methods for the generalization
of spatial data was recognized in the 20th century, and in a few decades various algorithms were developed to carry out different generalization
Stokes and Osnes (1989) presented a comprehensive model of generalization
that included 12 strategies that support generalization
This procedure often produces an asymmetrical generalization
gradient, with the gradient displaced away from S+ and towards stimuli on the opposite end of the stimulus dimension from S-.
discrete trial instruction) when used alone result in poor generalization
of language skills (e.
This column proposes the following empirical generalization
, called the law of annual halving, for nonprofits: the probability that a donor will donate again in the future is cut in half for every additional year that donor is inactive (recency).
In that literature, what is usually meant by a "law" is either a "lawlike" generalization
or (less often) a "nomic" regularity described by such a generalization
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945668.34/warc/CC-MAIN-20180422232447-20180423012447-00345.warc.gz
|
CC-MAIN-2018-17
| 2,050 | 23 |
https://ems.press/journals/rmi/articles/12657
|
math
|
We examine a certain class of trilinear integral operators which incorporate oscillatory factors , where is a real-valued polynomial, and prove smallness of such integrals in the presence of rapid oscillations.
Cite this article
Michael Christ, Diogo Oliveira e Silva, On trilinear oscillatory integrals. Rev. Mat. Iberoam. 30 (2014), no. 2, pp. 667–684DOI 10.4171/RMI/795
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00050.warc.gz
|
CC-MAIN-2023-50
| 374 | 3 |
http://www.e-booksdirectory.com/details.php?ebook=5060
|
math
|
Geometry and Group Theory
by Christopher Pope
Publisher: Texas A&M University 2008
Number of pages: 181
Lecture notes on Geometry and Group Theory. In this course, we develop the basic notions of Manifolds and Geometry, with applications in physics, and also we develop the basic notions of the theory of Lie Groups, and their applications in physics.
Home page url
Download or read it online for free here:
by Keith Ball, Vitali Milman - Cambridge University Press
Convex bodies are at once simple and amazingly rich in structure. This collection involves researchers in classical convex geometry, geometric functional analysis, computational geometry, and related areas of harmonic analysis.
by Jozsef Sandor - American Research Press
Contents: on Smarandache's Podaire theorem, Diophantine equation, the least common multiple of the first positive integers, limits related to prime numbers, a generalized bisector theorem, values of arithmetical functions and factorials, and more.
by Robert Sharpley - University of South Carolina
This course is a study of modern geometry as a logical system based upon postulates and undefined terms. Projective geometry, theorems of Desargues and Pappus, transformation theory, affine geometry, Euclidean, non-Euclidean geometries, topology.
by L. Henkin, P. Suppes, A. Tarski - North Holland Publishing Company
The volume naturally divides into three parts. Part I consists of 14 papers on the foundations of geometry, Part II of 14 papers on the foundations of physics, and Part III of five papers on general problems and applications of the axiomatic method.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00192.warc.gz
|
CC-MAIN-2019-04
| 1,601 | 15 |
https://maisaustralia.com.au/australia/which-exactly-are-much-like-requirements-in-math/
|
math
|
Which Exactly Are Much Like Requirements in Math?
Perhaps one among the absolute most widely used questions that people inquire whenever they get stuck into a math problem is”what are like provisions in mathematics?”
The easy response for the question is”we do not really know.”
We have more confounded, as we know more about these, although there are many things which can be understood to people. This informative article may allow you to understand the way it relates to a ability and exactly what exactly are similar provisions of mathematics.
Z http://shunchengfloor.com/2020/01/21/ama-mathematics-features/ is all about realizing the relationships between what exactly which you’re studying. Your job is to break down what is being studied into easy to comprehend hard to memorize chunks of advice. This is one of math’s primary goals: to help you learn by the mistakes and not to produce them all the moment.
You may possibly have seen there are unique systems which can be used to find mathematics. In some instances, we’ve got a publication where you will find these steps all at once, however also for the large part, you’re authorized to http://www.omeam.org.br/truth-about-mathematics-how-to-boost-a-kid-with-poems-about-mathematics/ just take every step one at one moment. This provides you the chance to know without having to memorize everything, and become knowledgeable about each step.
You will find a number of matters which you can do to help your self learn mathematics. These can be from undertaking things such as finding a graphing calculator into moving any time.
It can likewise be handy to accomplish what is referred to as”refreshing” your mind using your own learning strategies. Some people today learn best by considering the problem and getting rid of the idea they’ve initial. Then consider of what this means, they try to look at what is actually being analyzed and try to figure out the implications of what it is that they have heard.
Term in math may be employed to describe the manner that mathematics will work and the way that it pertains for the surface world. Since you may observe, in the event that you are just starting to learn, you will find different options available for your requirements.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305317.17/warc/CC-MAIN-20220127223432-20220128013432-00507.warc.gz
|
CC-MAIN-2022-05
| 2,261 | 9 |
http://newsgroups.derkeiler.com/Archive/Comp/comp.soft-sys.matlab/2008-09/msg00842.html
|
math
|
Re: Cross Power Spectrum and Image Registration
- From: "Jin Liu" <jliu@xxxxxxxxxxxxxxx>
- Date: Mon, 8 Sep 2008 16:27:02 +0000 (UTC)
"Jameson" <rtgroups@xxxxxxxxxxx> wrote in message <fuag2t$mvq$1$8302bc10@xxxxxxxxxxxxxxxx>...
I've managed to recover the correct parameters for a translated image and in
a separate test, for a rotated image, using phase correlation. When I
combine the transforms on the input image applying both a translation and a
rotation, the results seem to be wrong.
Here is my algorithm:
(1) Given image A and image to register with B
(2) Recover (x,y) as ArgMaxX(F) and ArgMaxY(F) of the cross power spectrum
of A and B
(3) Convert original images A and B to Log Polar Space, A' and B'
(4) Recover (s, a) as ArgMaxX(F) and ArgMaxY(F) of the cross power spectrum
of A' and B'
When the image to register is only translated, or only rotated, the correct
values are found. When I apply both a rotation and a translation to the
original image, incorrect values result (or perhaps my interpretation of
ArgMaxX and ArgMaxY needs to be modified).
(A) Translate by 10, 10 with a rotation of 0, results in RegX = 10, RegY =
10, RegAngle= 357.1875
(B) Translate by 0, 10 with a rotation of 0, results in RegX = 0, RegY =
10, RegAngle= 0
(C) Translate by 0, 0 with a rotation of 45, results in RegX = -52, RegY
= -93, RegAngle= 45
(D) Translate by 20, 20 with a rotation of 22.5, results in RegX = 0, RegY =
12, RegAngle= 0.75
So my questions are:
in the above results, would the angle in (A) of 357 [it should be zero] be
likely caused by not applying a filter on my log polar transform? Given the
results of (C), a simple rotation of 45 degrees is recovered, but what
should I do with RegX and RegY? They should be zero. Given the results of
(D), where translation is applied alongside rotation, how should I interpret
this result (x = 0, y = 12, angle = 0.75)?
And one final question, am I supposed to recover the angle of rotation,
rotate/scale the image and *then* apply the phase correlation to this new,
rotated image, in order to recover the correct translation? Or visa-versa?
Or simply do this in two passes on the original images.
I'm so close to getting this working!
Thanks for any assistance you can give me.
Can you please send me your code of transform and inverse transform of log polar code?
Thanks -- Jin
- Prev by Date: Re: Slash / Backslash path problem PC/MAC
- Next by Date: Re: Slash / Backslash path problem PC/MAC
- Previous by thread: Slash / Backslash path problem PC/MAC
- Next by thread: Simulink stops collecting data
|
s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368708546926/warc/CC-MAIN-20130516124906-00064-ip-10-60-113-184.ec2.internal.warc.gz
|
CC-MAIN-2013-20
| 2,562 | 46 |
http://jaynesbooks.blogspot.com/2012/12/musing-mondays-dec17.html
|
math
|
Musing Mondays (Dec.17)
Is there a particular book that is your nemesis - the book you are determined to one day finish?
If this had been asked a year ago, I would have given you a totally different answer, but I am still determined to finish Les Miserables by the end of this calendar year.
After that, I would suppose it would have to be War & Peace. I started reading it a couple years ago, but didn't and I keep kicking myself because I didn't finish it and its one that I would really like to finish. And maybe that will be my goal for next year.
I don't know how much I will be on in the next couple of weeks, but if I have some time, I will answer that weeks question.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221211719.12/warc/CC-MAIN-20180817045508-20180817065508-00442.warc.gz
|
CC-MAIN-2018-34
| 675 | 5 |
https://myhomeworkwriters.com/501mid-psychology-homework-help/
|
math
|
501mid | Psychology homework help
1. Identify and briefly describe the three criteria that help to define science.
2. Explain what a negative correlation between depression and self-esteem means.
3. Explain what deception and debriefing are. When it is acceptable to use deception in a research study?
4. Provide an example of a variable measured on a(n): nominal scale, ordinal scale, interval scale, and ratio scale.
5. If observers in a study disagree 35 times out of 90, what is the interrater reliability? Interpret this interrater reliability coefficient.
6. Explain the difference between naturalistic observation and laboratory observation, noting advantages and disadvantages of each.
7. Identify and briefly describe probability sampling versus nonprobability sampling.
8. Imagine that you are writing a survey on student perceptions of the food in the dining hall. Write one open-ended, one closed-ended, and one partially open-ended question concerning quality of the food in the dining hall.
9. What are the problems with the following survey questions?
• Do you agree that Americans should be more concerned with conserving fuel and reducing pollution from auto emissions?
• Do you favor reducing the outrageous number of administrators in the federal government?
• Most people believe that politicians are overpaid. Do you agree?
10. Calculate the mean, median, and mode for the following distribution of scores: 2, 2, 6, 9, 10, 11, 15, 17, 18, 20.
11. Explain the difference between qualitative and quantitative variables, noting the relationship of nominal, ordinal, interval, and ratio data to these terms.
12. For a hypothetical normal distribution of test scores, approximately 95% fall between 38 and 62, 2.5% are below 38, and 2.5% are above 62. Given this information, (a) the mode=_____ and (b) the standard deviation=_____.
13. Calculate s (standard deviation) and A.D. (average deviation) for the following sample: 2, 2, 6, 9, 10, 10, 15, 18, 18, 20.
14. Students in the psychology department consume an average of 5 cups of coffee per day with a standard deviation of 1.75 cups. The number of cups of coffee consumed is normally distributed.
• What proportion of students consume an amount equal to or less than 6 cups a day?
• How many cups would an individual at the 80th percentile drink?
15. In a recent study it was found that the correlation between self-esteem and depression was -.64. Tom interprets this to mean that low levels of self-esteem lead to high levels of depression and vice verse. How is top misinterpreting this data?
16. What is the third variable problem, and how does the partial correlation technique help with this problem? Does it completely solve the problem? Why or why not?
17. Calculate the Pearson product-moment correlation for the data below. X 3 4 2 1 Y 5 5 3 4
18. Explain when each of the correlation coefficients listed below should be used.
• Pearson product-moment correlation coefficient.
• Spearman rank-order correlation coefficient.
• point-biserial correlation coefficient
• phi coefficient
19. Assume that the regression equation for the relationship between SAT scores and IQ scores is y = 9 +.105x. What would you expect the IQ score to be for the following individuals, given their SAT scores? Individual SAT Score IQ Score Susan 850 Sally 1175 Sam 1225 Sean 1050
20. If a researcher decided to use the .01 level of significance rather that using the more conventional .05 level of significance, what type of error is more likely to be made? Why?
21. Identify when it would be appropriate to use a parametric versus a nonparametric test.
22. Explain how a one-tailed hypothesis differs from a two-tailed hypothesis.
23. Explain how Type I and Type II errors are related to false alarm and misses.
24. What are inferential statistics and how do they differ from descriptive statistics?
25. If weight in the general population is normally distributed with an average of 160 and a standard deviation of 20 pounds, what is the probability of selecting someone who weights 120 or less or 170 or more pounds?
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00242.warc.gz
|
CC-MAIN-2023-40
| 4,097 | 35 |
http://www.ems-ph.org/journals/show_abstract.php?issn=1463-9963&vol=8&iss=4&rank=1
|
math
|
Interfaces and Free Boundaries
Full-Text PDF (324 KB) | Table of Contents | IFB summary
Optimal transportation networks as flat chains
Emanuele Paolini (1) and Eugene Stepanov (2)(1) Dipartimento Matematica 'U.Dini', Università degli Studi di Firenze, Viale Morgagni, 67/a, 50134, FIRENZE, ITALY
(2) Dipartimento di Matematica, Università di Pisa, via Buonarroti 2, 56127, PISA, ITALY
We provide a model of optimization of transportation networks (e.g.\ urban traffic lines, subway or railway networks) in a geografical area (e.g. a city) with given density of population and that of services and/or workplaces, the latter being the destinations of everyday movements of the former. The model is formulated in terms of Federer-Fleming theory of currents, and allows to get both the position and the necessary capacity of the optimal network. Existence and some qualitative properties of solutions to the respective optimization problem are studied. Also, in an important particular case it is shown that the model proposed is equivalent to another known model of optimization of optimal transportation network, the latter not using the language of currents.
No keywords available for this article.
|
s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368701233842/warc/CC-MAIN-20130516104713-00057-ip-10-60-113-184.ec2.internal.warc.gz
|
CC-MAIN-2013-20
| 1,199 | 7 |
https://slideplayer.com/slide/4098948/
|
math
|
Presentation on theme: "Barry Latham Bloom High School Conceptual Physics, Hewitt, 1999."— Presentation transcript:
Barry Latham Bloom High School Conceptual Physics, Hewitt, 1999.
12.1: The Falling Apple Newton is credited with the idea of the falling apple to prove gravity. Probably never happened. If no outside forces are present, an object continues on a straight line path forever. Inertia! What keeps the Moon in orbit then? RQ 1-2
12.2: The Falling Moon The Moon really is falling toward the Earth! We just keep getting out of the way! If we fire a cannonball from Earth, we get the same results. Image: Fire a cannonball with faster and faster velocities. Faster velocities mean that the cannonball will farther and farther. Eventually it will go all the way around!
Tangential Velocity Tangent- “perpendicular to” Gravity tries to pull the Moon toward the Earth, the tangential velocity keeps it from crashing into us The only difference that matters for the apple vs. Moon is the distance from the Earth. If the Moon is 60 times further away than an apple at 1s… (Transparency 19, p. 170) The apple will fall 4.9m in the first second The Moon will only fall 1.4mm RQ 3-6
12.3: The Falling Earth The Moon is “falling” toward the Earth The Earth is then “falling” toward the Sun Why don’t we crash into the Sun? RQ 7
12.4: Newton’s Law of Universal Gravitation Universal Gravitation: Everything is gravitationally attracted to everything else! F=ma Only for local objects or those in an independent frame of reference F=Gm 1 m 2 /r 2 F=force of attraction (N) m 1 =mass of first object (kg) m 2 =mass of second object (kg) r=distance separating centers of m 1 and m 2 (m) G=gravitational constant (6.67x10 -11 Nm 2 /kg 2 ) Makes the units cancel out correctly
Measurement of G “G” was measured by Cavendish 150 years AFTER Newton “discovered” gravity Device measured a small twist in a quartz wire due to attraction between two Pb spheres A small value of G means that gravity is very weak!
Scientific Notation Review Scientific notation is needed when working with F because the numbers are SO big! 6.67x10 -11 is way easier to write than 0.0000000000667 every single time. The equatorial radius of the Earth is 6,370,000 m Keep dividing by 10 until you get to the one’s place Use the number of 10’s as your exponent 6.37x10 6 m
One billion examples Meters: Earth-Moon distance Kilograms: mass of Earth’s oceans Seconds: 31.7 years (Mr. Latham in 09/2008) Minutes: 1903 years Years ago: no Humans on Earth People: Population of China Atoms: enough to make the dot on a printed “i”
Weigh the Earth (without a scale or balance) Using F g =mg and F=Gm 1 m 2 /r 2 we can find the mass of the Earth! mg=(m object )(g on Earth) Gm 1 m 2 /r 2 =(G)(m object )(m Earth )/(r Earth ) 2 Set them equal to each other (m object )(g on Earth)=(G)(m object )(m Earth )/(r Earth ) 2 Solve for (m Earth ) (r Earth ) 2 (g on Earth)/(G)=(m Earth ) Plug & Chug (scientific calculator needed!) (r Earth ) 2 (g on Earth)/(G)/=(m Earth ) (6.4x10 6 m) 2 (9.80 m/s 2 )/(G)=(m Earth ) (m Earth )=6.02x10 24 kg RQ 8-10
12.5: Gravity & Distance: The Inverse Square Law The quantity varies as the inverse square of the distance, keeping masses constant (p. 175) F≈1/d 2 If distance doubles, Force decreases by 1/2 2 (or 1/4) If distance triples, Force decreases by 1/3 2 (or 1/9)
Inverse Square & Weight As distance increases, F decreases P. 176 (Transparency 20)
CD 12-1 worksheet Inverse Square Law 1. Complete the areas and thicknesses 2. Complete the areas 3. How does depth perception help us? Mislead us? RQ 11-12
12.6: Universal Gravitation Applications Most celestial objects are a sphere because gravity pulls equally in all direction. Because all objects also pull on each other, the planets change their orbits when they come close enough to each other Perturbation Uranus displayed perturbations Neptune was calculated to exist before it was seen Pluto was also calculated to exist before it was seen RQ 13-14
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00005.warc.gz
|
CC-MAIN-2021-31
| 4,042 | 15 |
https://mathschoolinternational.com/Math-Books/Geometry-Theorems/18-Theorems-of-Geometry-by-William-Smith.aspx
|
math
|
Math shortcuts, Articles, worksheets, Exam tips, Question, Answers, FSc, BSc, MSc
MathSchoolinternational.com contain houndreds of Free Math e-Books. Which cover almost all topics of mathematics. To see an extisive list of
Geometry Theorems eBooks . We hope mathematician or person who’s interested in mathematics like these books.
18 Theorems of Geometry written by William Smithy
This is an other great mathematics book cover the following topics.
Read Online 18 Theorems of Geometry by William Smith
Download Similar Books
Other Famous Books Topics:-
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00483.warc.gz
|
CC-MAIN-2023-14
| 555 | 8 |
https://www.studyelectronics.in/integrator-and-differentiator-using-opamp/
|
math
|
Integrator simulates mathematical integration of a function and differentiator simulates mathematical operation differentiation of a function. Here we are discussing about Integrator and Differentiator using opamp. Integration is basically a summing process that determines the total area under the curve of a function. Differentiation is determining
the instantaneous rate of change of a function.
Input signal is connected to inverting terminal and non inverting terminal is grounded. Negative feedback technique is used where in feedback path a capacitor is used. In practical integrator this capacitor in the feedback path have a parallel resistor.
How a capacitor charges ?
To know the working of integrator, recall charging of a capacitor, When capacitor in RC circuit is charging, charging current decreases continuously thus capacitor voltage also decreases continuously (exponentially). In Opamp integrator charging current is made constant, thus a rate of change of voltage decreases as a straight line instead of exponential curve.
Consider above figure, Since the non inverting terminal of opamp is grounded. So inverting terminal is in virtual ground. Thus constant voltage supply generate only constant input current. Because of the high input impedance of opamp only negligible amount of current passes through the inverting terminal. In other words, input current completely passes through the capacitor in the feedback path.
Since the input current is constant and it is completely flowing through the capacitor.voltage across the capacitor decreases linearly.
Positive side of the capacitor remains zero volt, Because of the virtual ground. Negative side’s voltage linearly decreases from zero and form a negative ramp.
Output voltage is same as that on the negative side of the capacitor. When a constant positive input voltage in the form of a pulse is applied, the output will be a ramp decreases negatively until opamps maximum negative saturation.
Rate of Change of the Output Voltage
Rate of Change of the Output Voltage is equal to the rate of change of the capacitor voltage. It is equal to
Integrator can be used to generate triangular wave oscillator.
An ideal opamp integrator have only capacitor in the feedback path. But capacitor conducts only AC but not DC. When DC current passes is connected across the capacitor capacitor act as an open loop. This offest error can be avoided by connecting a feedback resistor parallel to this capacitor. This resistor should be larger than input resistance. And to balance this bias another resistor is connected to the non inverting terminal.
op-amp differentiator circuit have slight differences from opamp integrator. That is feedback capacitor is replaced by a resistor and input resistor is replaced by a capacitor. Differentiator produces an output which is proportional to the rate of change of the input voltage.
Working of opamp differentiator
When a positive going ramp function is applied across the opamp differentiator. Voltage across the capacitor in the inverting terminal is also same as the ramp function. Because of the presence of virtual ground. Due to the high input impedance of opamp only negligible current is feed into the opamp, so input current completely flows through the current.
Since current across capacitor is constant (because of constantly increasing capacitor voltage ), current across resistor and output is constant. Since current in out put is constant, output voltage is constant (by ohms law).
During the ramp with positive slope, output will be negative, because input signal is inverting terminal. Similarly when the input ramp is with negative slope, output will be positive . Thus When a triangular wave is applied as input, a square wave is generated in the output terminal.
Ideal differentiator have only a capacitor in inverting terminal. Capacitor conducts high frequency signals and combination of this capacitor and feedback resistor act as a high frequency amplifier. Electrical noise mostly have high frequency. So this combination of capacitor and feedback resistor amplify noise. This can be rectified by using a resistor in series with capacitor as shown in the figure. This resistor make the input section as a low pass filter, So output can be protected from electrical noise. A bias compensating resistor may also connected to non inverting terminal.
|Introduction to Op-amp|
|Linear and Non Linear Applications of Op-Amp|
|Summing Amplifier using Op-Amp|
|Comparator Circuit using Op-Amp|
|Integrator and Differentiator using op-amp|
|Current Voltage Converters using Op-Amp|
We prepared this list from the suggestions of Junior Engineers who got selection in 2014.
RRB JE PREMIUM STUDY KIT from Bookhive
RRB JE Recruitment exam from GK Publications
COMBO PACK OF RRB JE Study Guide 2019 from ARIHANT
Shortcut in Reasoning from Disha Publications
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00497.warc.gz
|
CC-MAIN-2021-31
| 4,879 | 30 |
https://jacopretorius.net/2011/02/microsoft-interview-questions-and-answers-2.html
|
math
|
One train leaves Los Angeles at 15mph heading for New York. Another train leaves from New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide, how far will the bird have traveled?
Interestingly enough we’re not given the distance between LA and New York – maybe we’re supposed to just know it? Anyways, the key is to figure out how long the trains will take to collide and then calculate how far the bird flies in that time. The 2 trains are travelling at 15mph and 20mph respectively so they will cover the distance between the 2 cities at 35mph. So if the distance between the 2 cities is x, the 2 trains will cover the distance in x/35 hours. The bird will cover (x/35) * 25 = (5/7)x miles in this time.
You have two jars, 50 red marbles and 50 blue marbles. A jar will be picked at random, and then a marble will be picked from the jar. Placing all of the marbles in the jars, how can you maximize the chances of a red marble being picked? What are the exact odds of getting a red marble using your scheme?
Very cool question. Most people will try to use statistics to solve it (and fail), but the answer is actually very simple. Put one red marble in the one jar and put all the other marbles (49 red and 50 blue) in the other jar. So calculating the odds of a red marble being picked would be..
0.5 * p(picking red from first jar) + 0.5 * p(picking red from second jar) = 0.5 + 0.5 (49/99)
So it’s almost 75%. Awesome question. Next time you get asked in an interview ‘tell us about a time you had a difficult problem to solve’ you can say ‘well, this one time I had 2 jars of marbles…’
Imagine you are standing in front of a mirror, facing it. Raise your left hand. Raise your right hand. Look at your reflection. When you raise your left hand your reflection raises what appears to be his right hand. But when you tilt your head up, your reflection does too, and does not appear to tilt his/her head down. Why is it that the mirror appears to reverse left and right, but not up and down?
This one messes with your head a little. The trick is that left and right are relative to your body, where up and down are relative to the earth. To understand what I’m saying look at what’s on your left and then turn around. Now what’s on your left is actually the complete opposite. But if you do the same for what’s above you it doesn’t change. So the answer here is – it’s due to the definition of left and right and up and down.
You have 5 jars of pills. Each pill weighs 10 gram, except for contaminated pills contained in one jar, where each pill weighs 9 gm. Given a scale, how could you tell which jar had the contaminated pills in just one measurement?
Another very cool question, with strong links to programming. To make the explanation a little easier let’s name the 5 jars from A to E. Now we need to measure 5 pills from A, 4 from B, 3 from C, 2 from D and one from E. Now the weights will tell us exactly which jar contains the contaminated pills. For example, if the weight is 145 grams that means there were 5 contaminated pills so the contaminated jar is A. 146 Grams means it’s B, 147 means it’s C, 148 means it’s D and 149 means it’s E.
If you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?
I think this is actually the same puzzle presented to Bruce Willis in the movie Die Hard. Fill the 3 quart pail and pour it into the 5 quart pail. Fill the 3 quart pail and use it to fill the 5 quart pail (which contains 3 quarts at this stage). There will be 1 quart left in the 3 quart pail. Empty the 5 quart pail and pour the 1 quart left in the 3 quart pail into the 5 quart pail. Fill the 3 quart pail and pour it into the 5 quart pail (which contains 1 quart at this stage). The 5 quart pail will now contain 4 quarts.
You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color. How many do you have to grab to be sure you have 2 of the same?
I couldn’t get this one at all and then I explained it to a friend who managed to figure it out. As with any good riddle the answer is ridiculously easy when you have it. You have to pick 4 jelly beans. If you pick 3 you could possibly pick 3 different colors, but once you have 4 you must have at least 2 of the same color. The wording makes it tricky because it seems to imply that you must be holding exactly 2 beans of the same color in your hand, but actually you need to have at least 2 of the same color.
A colleague sent me this one (which he was once asked in an interview) – it’s probably my favourite.
*A pirate ship captures a treasure of 1000 golden coins. The treasure has to be split among the 5 pirates: 1, 2, 3, 4, and 5 in order of rank. The pirates have the following important characteristics:
Infinitely smart. Bloodthirsty. Greedy.
Starting with pirate 5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the pirate is thrown overboard. A proposal is accepted if and only if a majority of the pirates agrees on it.
The Question: What proposal should pirate 5 make?
As with many puzzles you can’t really solve it unless you work backwards. Let’s imagine the first 3 proposals have been rejected – pirates 5, 4 and 3 have been thrown overboard and only 1 and 2 are left. Pirate 2 now has to make a proposal. No matter what proposal he makes, pirate 1 will reject it, pirate 2 will be thrown overboard and pirate 1 will get all the money. Even if pirate 2 proposes that pirate 1 gets all the money he will still reject it because he’s bloodthirsty – he would prefer killing 2 and getting all the money to simply getting all the money.
Right, now let’s back up one step. Pirates 1, 2 and 3 are left and pirate 3 has to make a proposal. He knows that if he gets thrown overboard pirate 2 is screwed, so he should be really easy to convince. He will propose pirate 3 gets 999 coins, pirate 2 gets 1 coin and pirate 1 gets nothing. Pirate 2 will accept this since it’s better than him rejecting it and getting nothing – pirate 3 has his majority vote.
Let’s back up another step. Pirates 1, 2, 3 and 4 are left and pirate 4 has to make a proposal. He knows that if he gets thrown overboard pirate 1 gets nothing, pirate 2 gets 1 coin and pirate 3 gets 999 coins. He needs 3 votes for a majority so he will propose pirate 4 gets 997 coins, pirate 3 gets nothing, pirate 2 gets 2 coins and pirate 1 gets 1 coin. (If he proposes that pirate 2 gets only 1 coin pirate 2 will reject it since he’s bloodthirsty and he can get the same amount of coins in the next round)
Let’s back up another step – we’re back to the initial scenario. Knowing what will happen in the next round, pirate 5 will propose: pirate 5 gets 997 coins, pirate 4 gets nothing, pirate 3 gets 1 coin, pirate 2 gets nothing and pirate 1 gets 2 coins.
Awesome question. Happy coding.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00284.warc.gz
|
CC-MAIN-2022-05
| 7,119 | 24 |
http://freehitcountercode.com/what-number-has-31-tens-and-no-ones/
|
math
|
What Number Has 31 Tens and No Ones?
Numbers are an integral part of our everyday lives. They serve as a foundation for mathematical calculations, measurements, and various other aspects of our existence. Sometimes, we come across intriguing number puzzles that pique our curiosity. One such puzzle is the question, “What number has 31 tens and no ones?” In this article, we will explore this fascinating puzzle and provide a detailed explanation.
Explaining the Puzzle:
To find the number that has 31 tens and no ones, we need to understand place value and how it relates to the digits in a number. In the decimal system, each digit’s position determines its value based on powers of ten. The rightmost digit represents ones, the digit to its left represents tens, the one after that represents hundreds, and so on.
To solve this puzzle, we need to build a number that has 31 tens and no ones. Since the digit representing tens is the second digit from the right, we can conclude that the number must have a zero in the ones place. This is because if there were any digit other than zero in the ones place, the number would have ones, which contradicts the given condition.
Hence, the number that has 31 tens and no ones is 310.
Understanding the Concept:
To fully comprehend this puzzle, let’s break down the number 310. The digit 3 represents three hundred, the digit 1 represents ten, and the digit 0 represents zero ones. Therefore, 310 is the number that has 31 tens and no ones.
This concept is crucial in mathematics as it helps us understand the value and significance of each digit in a number. By analyzing the position of digits, we can determine the overall value of a number accurately.
Q: Are there any other numbers that have 31 tens and no ones?
A: No, there is only one number that satisfies this condition, and that is 310. This is because, in the decimal system, the tens digit is always followed by the ones digit, and any non-zero digit in the ones place would create ones in the number.
Q: Can we have a number with 31 tens and a different number of ones?
A: No, if we have 31 tens in a number, it implies that the tens digit is fixed. Changing the number of ones would require altering the ones digit, which contradicts the given condition of having no ones.
Q: How does this puzzle relate to real-world applications?
A: While this puzzle may seem purely theoretical, it enhances our understanding of numbers and place value, which forms the basis of various mathematical calculations. This knowledge is essential in fields like finance, physics, engineering, and computer science, where precise numerical understanding is vital.
Q: Can this puzzle be solved using other number systems?
A: No, this particular puzzle is specific to the decimal system, where the base is ten. In other number systems, such as binary or hexadecimal, the positions of digits and their values are different, so the solution to this puzzle would not be applicable.
The puzzle of finding a number that has 31 tens and no ones has been successfully solved. By understanding the concept of place value, we determined that the number is 310. This puzzle not only challenges our mathematical abilities but also provides a deeper understanding of numbers and their significance. So, the next time you come across a number puzzle, take a moment to explore it further, as it might offer an intriguing journey into the world of mathematics.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00219.warc.gz
|
CC-MAIN-2023-40
| 3,441 | 18 |
https://nicklaffin.weebly.com/technology-in-mathematics.html
|
math
|
Wolfram Mathematica is a powerful computer calculator, capable of performing complex mathematics and creating 3D models. For my Coordinating Seminar class I chose to cover Mathematica and matrix operations to show my fellow students how useful the program could be. I created a short hand out covering the many fundamentals of matrices and how to input the commands into Mathematica. To learn more about Wolfram Mathematica click this link.
To view the Mathematica handout on matrix operations, click the download below.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00013.warc.gz
|
CC-MAIN-2019-04
| 520 | 2 |
https://justaaa.com/mechanical-engineering/14397-05-kilograms-of-air-are-compressed-from-100-kpa
|
math
|
0.5 kilograms of air are compressed from 100 kPa and 300 K in a polytropic process, n = 1.3, to a state where V2 = 0.5 V1. The air is further compressed at constant pressure until the final volume is 0.2 V1 . Draw a sketch of the processes on a p-V diagram. Determine the work for each process.
In this problem, first of all draw the process on the P-V diagram then calculate pressure and Volume at each state and then calculate the work for both process as below:
Note:- The negative sign shows that the work done on the system.
Get Answers For Free
Most questions answered within 1 hours.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00591.warc.gz
|
CC-MAIN-2023-06
| 590 | 5 |
https://www.genealogy.math.ndsu.nodak.edu/id.php?id=174251
|
math
|
Karl Wilhelm Borchardt
Dr. phil. Universität Königsberg 1843
Dissertation: Gewisse Systeme nichtlinearer Differential-Gleichungen
Mathematics Subject Classification: 34—Ordinary differential equations
Advisor 1: Carl Gustav Jacob Jacobi
No students known.
If you have additional information or corrections regarding this mathematician, please use the update form. To submit students of this mathematician, please use the new data form, noting this mathematician's MGP ID of 174251 for the advisor ID.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00070.warc.gz
|
CC-MAIN-2023-14
| 504 | 7 |
http://www.animeforum.com/showthread.php?85252-Please-explain-to-me...&p=2251086&viewfull=1
|
math
|
I could never really get into Death Note.They talk too much for me and the concept doesn't really make
sense to me either. Like... I would probably get more into it if I wasn't so confused. Now, although I am
anticipating a bashing for coming at the "boring-ness" of it but if you could find it in you to give me an
explanation to my question...that'd be nice too. A brief...overview (for the lack of a better word) would
be even nice(r). :] thanks in advance.
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042987662.63/warc/CC-MAIN-20150728002307-00259-ip-10-236-191-2.ec2.internal.warc.gz
|
CC-MAIN-2015-32
| 460 | 5 |
http://lpassignmentbiwx.laluzsiuna.info/analysis-of-cost-profit-and-total.html
|
math
|
Limitations of cost-volume-profit (cvp) analysis: cost volume profit (cvp) is a short run, marginal analysis: it assumes that unit variable costs and unit revenues are constant, which is appropriate for small deviations from current production and sales, and assumes a neat division. Understanding the cost-volume-profit (cvp) analysis in business can be useful, especially to entrepreneurs, as it can help them determine the when the entrepreneur knows the total number of units sold to reach their break-even point, it will help the entrepreneur avoid sustaining any financial. This type of analysis is known as 'cost-volume-profit analysis' (cvp analysis) and the purpose of this article is to cover some of the straight forward calculations and graphs methods for calculating the break-even point the break-even point is when total revenues and total costs are equal, that is. Cost-volume-profit (cvp) analysis expands the use of information provided by breakeven analysis a critical part of cvp analysis is the point where total revenues equal total costs (both fixed and variable costs) at this breakeven point (bep), a company will experience no income or loss. Cost-volume-profit (cvp) analysis helps managers understand the relationships that exist break-even analysis: an important element of cvp the break-even point is the level of sales the firm break-even point in total sales dollars = fixed expenses/cm ratio where, cm ratio = cm per.
Cost-volume-profit (cvp) analysis is one of the most powerful tools that managers at their command it helps them understand the interrelationship between cost, volume again choose some volume of sales and plot the point representing total sales tk at the at the activity level we have selected. 2 total variable costs( tvc) - these are costs that vary with output 3 total costs (tc) - it is the sum of total fixed costs and total variable costs 12 a competitive firm takes the market price as constant if it wants to maximize profits, the optimum level of production in the short run is when its. Assumptions of cost-volume-profit analysis the profit-volume and cost-volume-profit graphs just illustrated rely on some important assumptions 2 the sales revenue and total costs are not always linear in as normally assumed in the theory 3 two or more break-even points may exist for a.
Total profit equals total revenue minus total cost in order to maximize total profit, you must maximize the difference between total revenue and total cost the first thing to do is determine the profit-maximizing quantity. Accounting cost - tend to be retrospective they recognize costs only when these are made and properly recorded short-run cost analysis short run for the firm is a time horizon when one input is held constant point of maximum profit by definition, profit equals total revenue(tr) less total cost. Cost volume profit analysis graph is a simple model that represents these expenses and the revenues that are generated may represent the financial structure thus, the amount of income or loss at each level of sales can be derived from the total sales and total cost lines uses of cost volume profit. 3 and low profitabilitybanks in china: an analysis of cost and profit efficiencies 1 its banking businesses are underdeveloped and its client groups input prices p1 and p 2 negatively contribute to the total cost and y1 and y 2 positively affect the dependent variable p 2 and p 2 p 2 are significant. Cost volume profit (cvp analysis), also commonly referred to as break even analysis, is a way for companies to determine how changes in the contribution margin is the product's selling price less the variable costs associated with producing that product this value can be given in total or per unit.
The cvp (cost-volume-profit analysis) expresses the relationships among costs, volume, and profit [or loss] it answers questions such as how many the breakeven point is the sales level at which operating income is zero (in other words, no profit and no loss since total costs / expenses are equal. Definition of 'cost-volume profit analysis': cost-volume profit analysis is a simplified model, useful for elementary instruction and for short-run decisions case 4-33 cost structure target profit and break-even analysis [lo4, lo5, lo6] pittman company is a small but growing manufacturer of. Cost-volume-profit (cvp) analysis is used to determine how changes in costs and volume affect a company's operating income and net income the break‐even point represents the level of sales where net income equals zero in other words, the point where sales revenue equals total variable. Cost-volume-profit (cvp), in managerial economics, is a form of cost accounting it is a simplified model, useful for elementary instruction and for short-run decisions a critical part of cvp analysis is the point where total revenues equal total costs (both fixed and variable costs.
Cost accounting for cost-volume-profit analysis: understanding & calculating the breakeven point (total costs = total revenues), breakeven revenue in this video i explain the costs of production including fixed costs, variable costs, total cost, and marginal cost make sure that you know how to. This lesson introduces cost-volume-profit analysis cvp analysis is a way to quickly answer a number of important questions about the profitability of a total fixed costs: stay essentially the same month to month, regardless of the number of units produced unit fixed costs: goes down as. About this quiz chapter: cost, volume and profit relationships (cvp analysis) total points: 26. Cost-volume-profit cvp analysis is based entirely on unit costs the cvp analysis determines the changes in costs and volume thataffects a company's operating income and net income however itassumes that the sales price, variable costs and the total fixedcosts per unit remain constant. 2total variable cost (tvc) costs that vary with output 3total costs (tc) the sum of total fixed costs and total variable costs cost-volume profit (cvp) analysis this is a technique used for planning short-term run profits by finding the relationship between profits and factors that influence.
Cost-volume profit (cvp) analysis is a method of cost accounting that looks at the impact that varying levels of costs and volume have on operating profit contribution margin is the difference between total sales and total variable costs for a business to be profitable, the contribution margin must. Cost volume profit analysis thinks like a number line wherein it starts with negatives, then comes 0 and then positives similarly, with the increasing level of sales, first will see a phase of losses, second a breakeven and third where we make profits the first priority of any businessman is to safeguard its.
Total cost calculations provide a method for entrepreneurs to expense the opportunity costs associated with long-term ventures total revenue profit calculations provide a means for entrepreneurs to measure the time invested in the business. In general, cost volume profit analysis is designed to show how changes in product margins, prices, and unit volumes impact the profitability of a business cost volume profit analysis is one of the fundamental financial analysis tools for ascertaining the underlying profitability of a business.
Cost-volume profit analysis: a cost volume profit analysis is a cost accounting method in the managerial economics use to determine the breakeven point of to calculate the break-even point we have first calculate the number of units, sales, total variable cost and total fixed cost (investopedia. Cost-volume-profit analysis (cvp analysis) deals with how profit and costs change with a change in volume contribution margin refers to sales revenue minus total variable costs the determination of the break-even point is one of the applications of cost-volume-profit (cvp) analysis.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039748315.98/warc/CC-MAIN-20181121112832-20181121134832-00294.warc.gz
|
CC-MAIN-2018-47
| 7,909 | 8 |
http://www.google.com/patents/US3216001?dq=ascentive
|
math
|
US 3216001 A
Abstract available in
Claims available in
Description (OCR text may contain errors)
Nov. 2, 1965 K. HINRICHS ANALOG-T0DI GI TAL CONVERTER Filed Oct. 15, 1960 DECIMAL. DECADES III 6 Sheets-Sheet 1 /.000o 0/000 fQ0/00 0. 00/0 0.9000 0.0900 0.000 0.000 fila. 0.8000 0.0800 0.0080 0.0008
0/000 00 00 dog 0 QOOO/ REFERENCE 20 o k 0 O x 0 73 020 21 DECIMAL DECADES 2: 11 I I In: REFERENCE STANDARD (0.033 00/25 0/300 00/30 Si-fiffi :1'
I0. 0/200 0.0/20 SEQUENCER 4- 0//00 0-0/0 comm. 27 /.0000 0./000 00/00 DIGITAL CODE. 0.8000 Q0a00 Q0000 CONVERTER H 0/000 00/00 GCU/O JNVENTOR. 'Q0/00 'Q00/0 h MA/IQICHS- Nov. 2, 1965 K. HINRICHS ANALOG-TO-DIGITAL CONVERTER 6 Sheets-Sheet 2 Filed Oct. 13. 1960 M GU L.
B (D Z IK r INVENTOR. 15224 ,M/VE/CY/S Mrraeflz Y Nov. 2, 1965 K. HINRICHS 3,216,001
ANALOG-TO-DIGITAL CONVERTER Filed Oct. 13. 1960 6 Sheets-Sheet s D AND AND Pea/ 7 62 06% Nov.- 2, 1965 K. HINRICHS ANALOG-TO-DIGITAL CONVERTER 6 Sheets-Sheet 4 Filed Oct. 15. 1960 wasp QQMQQ QM N300 Q&
Nov. 2, 1965 K. HINRICHS ANALOG-TO-DIGITAL CONVERTER 6 Sheets-Sheet 5 Filed Oct; 13. 1960 INVENTOR.
BY L IGi ENTOR.
6 Sheets-Sheet 6 JVAEL ,H/v/e/o/s Nov. 2, 1965 Filed Oct. 15, 1960 United States Patent 3,216,001 ANALOG-TO-DIGITAL CONVERTER Karl Hinrichs, Anaheim, Calif., assignor to Beckman Instruments, Inc., a corporation of California Filed Oct. 13, 1960, Ser. No. 62,379 22 Claims. (Cl. 340-347) My invention relates to analog-to-digital converters, and more particularly to analog-to-digital converters of the electronic type in which an analog signal is converted to a coded digital signal, one digital symbol at a time.
A variable is deemed to be in analog form when the magnitude that represents its value is a continuous function of time. Often it is desirable to convert from this form of information to a series of numerical samples at selected time intervals for convenience, accuracy and speed in data handling computations. My invention converts such analog signals to digital signals at selected time intervals, the digital samples being a convenient form for input to digital computers or similar information processing devices. A variable is deemed to be in digital form when each time sample of the signal is represented by a series of magnitudes, each of which is permitted only two quantum states. Such digital samples may be encoded in any of a wide variety of electronic computer codes chosen on a basis of convenience, accuracy and reliability.
A very common and widely used analog-to-digital converter may conveniently be referred to as a digit-ata-time converter, in which the analog signal during selected time intervals is quantized to a set of digital symbols such a 1 and 0,. one symbol at a time. Ordinarily a converter of this type uses selected subtraction. Commencing at a selected time, the analog signal is compared with assembled reference values or standards weighted according to a predetermined number system. The analog signal magnitude at that time is compared with the largest reference standard, and the comparison continues with respect to smaller standards individually in sequence until a standard smaller than the analog signal at that instant is found. At the time of the next succeeding comparison, the value of that smaller reference standard is subtracted from the analog signal. The resulting difference signal is now compared with succeedingly smaller reference standards and new differences formed whenever a reference standard is found smaller than the signal difference with which it had been compared. This process is continued until the selected reference standards, when added, are equal to a value which differs from the instantaneous analog input by less than the smallest reference standard. The number of reference standards used is determined by the resolution required, the system noise level, and the inherent limitations of the conversion equipment.
Illustrative examples of this type of a converter are easily shown. If a pure binary code is desired, the reference values are weighted 0.5000, 0.2500, 0.1250, etc., when the chosen full scale is normalized to unity. If, for example, the analog signal during the conversion time interval has a constant amplitude of 0.5001, and the converter accuracy and sensitivity is approximately one part in ten thousand (0.0001), upon comparison with the first reference standard of 0.5000, an accurate comparison shows that the standard is of a smaller magnitude than the signal; therefore, this standard will be retained and a digital symbol for 1 registered by the converter for this digit. The 0.5000 standard is now subtracted from the analog signal, and the resulting difference signal of 0.0001 used in the next comparison step. Upon comparison with the 0.2500 standard, it will be found that the standard is larger, so that the difference signal remains unaltered for the succeeding comparison and the converter records a symbol 0 for this digit. The next standard 0.1250 is similarly rejected. Succeeding comparisons all result in rejection of their reference standards until the fourteenth and last comparison, whose reference is 0.00006, somewhat less than 0.0001. The converter holds the results of the conversion in a code which represents the retained reference standards by using such a device as a series of binary fiip-flops or relays which can be read, for the above conversion, as 10000000000001 (which may be reconstituted as desired to be l 0.5000 plus 1 0.00006=0.500l
within one part in ten thousand, the instantaneous analog signal magnitude).
Another convenient and commonly used number system is that termed binary-coded-decimal. In this system the chosen full scale range is described by decimal digits, each decimal digit being individually coded in four or more binary digits. Such a system is often used because of its ready conversion to the familiar decimal system. One binary-coded-decimal system in common use is the 8421 code which assigns the same weights to each of the hits as in ordinary binary notation. An encoder adapted to convert two decimal-digit numbers to binary requires eight reference values in two decades, four of the reference standards corresponding to the tens decimal digit and the remaining four reference standards corresponding to the units decimal digit. An illustrative example utilizing again the instantaneous analog signal magnitude 0.5001 would involve a comparison series starting with the reference standard 0.8000 and continuing with the values 0.4000, 0.2000, 0.1000, 0.0800, etc., if the precision of the instrument is again chosen as one part in ten thousand. It is readily seen that the chosen analog signal is less than 0.8000, but larger than 0.4000. After discarding the 0.8000 reference standard and retaining the 0.4000 reference standard, the resulting difference is smaller than the 0.2000 but larger than the 0.1000 standard, resulting in a binary decimal code of 0101 for the first decimal digit. A continuation of this process results in a binary-coded-decimal code of 0000 for the second and third decimal digits, and 0001 for the last decimal digit. By processes analogous to the preceding pure binary code case, the instantaneous analog signal magnitude can be reconstituted as 1 0.4000 plus 1 0.1000 plus l 0.000l=0.5001
again to the precision of one part in ten thousand.
A primary disadvantage of encoders known in the prior art which utilize the digit-at-a-time technique is that the accuracy and resolution of each and every comparison step must equal or surpass that of the final step. Thus, in the illustrative example above for the pure binary system, the instantaneous analog signal is read as larger than the 0.5000 reference standards. Equivalently, it may be said that the signal is initially classified in a class having a lower bound of 0.5000. This bound must be known to the same accuracy as desired for the ultimate comparison step because succeeding comparisons do not extend the class below 0.5000. Similarly, in the binary-coded-decimal example, after four classifications have been made and the maximum bound of 0.6000 and the minimum bound of 0.5000 are selected, these classifications are irrevocable, and will require the same accuracy as for the final classification if all possible analog signals are to be accurately encoded.
The severe problems associated with this limitation of prior art converters may be shown by example. Taking first the pure binary encoder, assume that although the analog signal at the time of the final comparison will have the value of 0.5001, its value at the first comparison step might easily be less than 0.5000 due to system transients, equipment errors, or fluctuations 1n the signal. Accurate comparison at the first step, resulting in rejection of the 0.5000 value, will now incorrectly limit the possible encoded values at the final step to 0.4999. Such devices are prone to SlIllllElI' and even larger errors whenever the difference s gnal is close to the reference standard with which it is be ing compared. Such devices are therefore seriously limited 'in speed and accuracy by this erroneous classification process.
Prior art binary-coded-decimal, or BOD dev1ces are similarly susceptible to erroneous classification. In the foregoing example a standard BCD machine would 'be extremely susceptible to error at the time of the fourth or 0.1000 comparison, and system disturb- 'ances and fluctuations might readily result in the irrevocable rejection of this standard, with the result that the correct magnitude of 0.5001 would again be er roneously encoded as 0.4999. Other examples which illustrate this detrimental sensitivity to error are readily demonstrable.
Because the accuracy requirement for each step of prior art converters has been maximally severe, these devices have been forced to operate at speeds slow enough to permit the transients introduced by the converter itself to settle down to noise level prior to the establishment of each class. These transients are particularly severe in the first few comparison steps since these involve switching of the largest reference standards. The high accuracy requirement [for these large standards conflicts with high switching speeds. The comparator, which is normally an amplifier, can never respond immediately to step voltage changes on its input and will require long settling times to dissipate the effects of large input signals before it is capable of correct comparison of small signals in succeedlng comparisons. For example, even if the input signal of 0.5001 in the preceding example were present at the fourth comparison step of the illustrated BCD converter, the third comparison step of 0.2000 would have presented the comparator with an error signal of 02000-01001 -or 0.0999, and this relatively large signal could easily have disturbed the comparator so that it would be incapable of responding correctly at the fourth Step to the small 0.0001 difference when even mod-est encoding speeds are attempted.
Slower encoding speeds to permit thorough settling of such transients not only lower the informationhandling speed, but further increase the possibility and magnitude of errors when encoding a signal wh1ch 1s changing. This is a primary limitation, since all signals of value fluctuate, there is no information in an absolutely stationary signal since its value is already known. Since the signal may cross classification bounda ries during encoding, conventional systems have often found it necessary to freeze the analog signal at a given time by some type of analog holding or memory technique. Such techniques, however, are sources of errors in themselves and result in frequency limitations on the signal, accuracy degradation, and additional hardware complexity and cost.
- Accordingly, it is a principal object of my invention to provide an analog-to-digital converter which can commence digitizing prior to complete settling of the analog signal fed into the converter.
' A further object of my invention is to provide an analog-to-digital converter in which the resolution of the initial digitizing steps may be much l accurate than that of the final digitizing step.
It is another object of my invention to provide an analog-to-digital converter in which it is possible to make decisions in the early steps at a rate which does not allow for transient settling of the signal to within the required final accuracy.
It is still another object of my invention to provide an analog-to-digital converter which enables corrections to be made in the latter stages of the weighted decision sequence to correct errors made in earlier de- 01810113.
A further object of my invention is to provide an analog-to-digital converter which may accurately accept analog signals that appear initially either higher or lower in magnitude than their value at the end of the comparison cycle.
Other and further objects, features and advantages of the invention will become apparent as the description proceeds.
Briefly, my invention provides correct encoding at high speeds for fluctuating signals in the presence of system transients by utilizing an increased classification span of succeeding steps beyond that of prior art encoders so that allowance may be made for signal and system uncertainties in preceding comparisons. The number of comparison steps utilized and the degree of uncertainty tolerance incorporated for each comparison step in my invention is adjusted to maximize information handling capacity as restricted by the signal characteristics, comparator characteristics, decoding ease and system cost.
A more thorough understanding of my invention may be obtained by a study of the following detailed description taken in connection with the accompanying drawings in which:
FIG. 1a is a graph useful in explaining the operation of prior art analog-to-digital converters to afford a better understanding of my invention;
. FIG. lb is a graph illustrating the operation of my invention;
FIG. 2 is a block diagram of a preferred embodiment of my invention;
FIG. 3 is a partially schematic, partially block diagram of one embodiment of my invention;
FIG. 4 is a schematic illustration of the sequencer portion of the sequencer and control of my invention;
FIGS. 51: and 5b are schematic illustrations of the control portion of the sequencer and control and the circuitry of the digital code converter of my invention; and
FIG. 6 is a schematic illustration of the output register of my invention.
RELATIONSHIP BETWEEN PRIOR ART CON- VERTERS AND THE PRESENT CONVERTER A graphic illustration of a specific embodiment of my invention and its prior art counterpart are illustrated in FIGS. 1b and 101, respectively. Referring now to FIG. 1a, the reference standards for a prior art four decade, BCD encoder are illustrated. For the most significant decimal digit, the first decade includes a series of ten reference values extending from 0 to 1.0000, each value bemg separated by 0.1000 and the chosen full scale being normalized to unity. These reference values may be considered as constituting the boundaries between primary order classes which divide into ten classes the range in which the analog signal can occur. The second decimal decade comprises a secondary class which subdivides the selected primary class into ten subclasses separated by 0.0100; this decade determines the next most significant decimal digit. The third decade comprises ten subclasses each separated by 0.0010 and the fourth decade comprises ten subclasses each separated by 0.0001. Any decimal number lying within the chosen full scale of unity may be encoded by selecting an appropriate class in each of the decades. For example, the encoding of the number 0.5155 is illustrated in FIG. 1a. Thus, the fifth class in the first decade; the first class in the second decade and the fifth class in the third and fourth decades are selected (i.e., 0.5000+0.01o0+0.005o+0.000 s The disadvantage of prior art converters as exemplified by the illustration of FIG 1a is that the accuracy and resolution of each and every comparison step must equal or surpass that of the final comparison step. Thus, in the foregoing example, the analog input signal was read as 0.5155, its true instantaneous value. Assume however, that the signal, due to signal fluctuation or transients within the system, was initially incorrectly read as less than 0.5000 but greater than 0.4000. Then, in the first decade classification which digitizes the most significant decimal digit, the 0.4000 primary class would have been selected rather than the correct 0.5000 primary class. The fact that the correct signal reading was made later in the classification would not serve to completely correct the error since the maximum classes in the second, third and fourth decades are 0.0900, 0.0090 and 0.0009. The digitized output would he, therefore, 0.4999 instead of the true value of 0.5155. As heretofore noted, such errors in reading the true analog value require a longer encoding time since the most significant decimal digit must be read to the same accuracy as the least significant decimal digit. Such extended conversion times place increased limits on the frequency and amplitude of the converter, lowering its value and accuracy in contemporary high speed data handling systems.
My invention provides a converter of improved accuracy and increased operating speed by utilizing an increased classification span in succeeding comparison steps. In a specific embodiment operating as illustrated in FIG. 1b, the increased classification span is provided in the second and third decades, each of which includes additional secondary classes equally distributed positively and negatively. Thus, the second decade extends from -0.0333 to 0.1333, as contrasted to the prior art conversion span of 0.0100 to 0.1000. In like manner, the third decade of the embodiment of my invention shown in FIG. 1b extends from -0.0033 to 0.0133, whereas the prior art conversion span is 0.0010 to 0.0100.
This error tolerance afforded by my converter may be illustrated by encoding an analog signal having an instantaneous true value of 0.5155 If no errors occur in reading the analog signal in any of the comparison steps, each of the classifications will be made as shown by the solid arrows of FIG. 1b. However, if for example, an error is made in the initial decimal decade and the signal is read as less than 0.5000, the 0.4000 primary class will be selected instead of the correct 0.5000 primary class. This error may be compressed for by the error tolerance provided by the succeeding decades, since the 0.1100 subclass may be selected instead of the 0.0100 subclass had the initial reading been correct. Classification in the third and fourth decades would then be made as before with a resultant encoding of 0.4000+0.1100+0.0050+0.0005 or a correct total of 0.5155.
The additional subclasses illustrated in FIG. 1b may be derived from utilizing a code in which certain digit combinations fall outside the selected code, i.e., a number system having the characteristic that certain combinations of quantities exceed a selected total number greater than any individual number in said system; by the addition of certain error tolerance distributing weights which are added during the comparison sequence; or a combination of these two. An illustration of a number system having certain combinations falling outside the code is the 8-4-2-1 BCD system in which all digital combinations greater than nine are outside the system. Other BCD codes and an adaptation of the standard binary system which may be employed in my invention are described hereinafter.
As shown in FIG. 1b, the additional secondary classes are normally equally distributed between positive and negative values of error tolerance. The negative error tolerances provided in the second and third decades permit a number to be read initially higher than its actual value. Thus, if a number having a true instantaneous value of 0.4500 was erroneously read as being greater than 0.5000, a negative error tolerance may be selected in the second and third decades so as to obtain corect encoding. In certain applications, however,. it may be desirable to distribute unevenly the error tolerance and in some instances have the entire error tolerance as either a positive or negative value. An important example of this principle is found in systems wherein the converter is used to synchronize encoding with commutation of analog signals. If the sampling i such as to provide a true zero converter input between channels, and the system is adequately damped, probable error is always negative. A considerable increase in system accuracy and/ or speed can now be obtained if all of the built-in error tolerance of my invention is applied in one direction. The desiderata involved in selection of appropriate error tolerance distribution for specific applications in accordance with my invention will become more apparent as the description proceeds.
MATHEMATICAL DESCRIPTION OF THE PRESENT CONVERTER A few mathematical definitions and relations help to clarify the fundamental features of my invention. The following Table I lists some of these useful concepts.
Table I (a) N =number of comparison steps.
(b) j=designator for a typical comparison step between 1 and N.
(c) T =value of the incoming signal at step 1'. T is the true value of the signal at the last comparison step.
(d) W =reference standard added at step 1' and subsequently retained or rejected.
(e) X total error between apparent signal at step 1' and its final value T (f) K '=built-in tolerance for error at step i.
(g) Y =added tolerance-distributing weight at step i, not subsequently retained.
(h) C -'=comparator input at step 1'.
(i) 6 =comparator output at step j equals 1 for retained standard, equals 0 for rejected standard.
(j) F=fineness or resolution of the comparison process.
The basic operation of the converter of my invention is readily established by noting that the converter output (the sum of all retained reference standards) should equal the true signal input at the last step within the required fineness. Mathematically,
2 1 i i Ni where the retained )1 rejected signal tiifijifi and the comparator input of my invention is i=1 O,-=T +X,;IWX x"Wi Yi the difference between (a) the instantaneous signal-pluserror at step 1 minus the sum of all previously retained reference standards, and (b) the reference standard and error distributing weights for this step.
My invention provides excess counting ability in succeeding steps, i.e., on one or more of the j comparison steps the sum of the succeeding reference standards is greater than the true value of the signal minus the re- 7 tained reference standards up to but not including step i. Expressed mathematically as an inequality:
'so that any given comparison except possibly the last or last fewmay have its accuracy requirement reduced by an error tolerance,
i=2 W. W.
My invention also provides that this error tolerance may be distributed in a manner to suit the circuitry and signal nature by adjusting the error distributing weight for each step as follows;
where m is the chosen proportion. As noted above, normally equal tolerance to plus or minus values of X, will be desired and m will be /2 v The comparison process in my invention, denoted by Equation 3 above, permits large values of error X, to exist at early steps without the subsequent irrevocable errors of prior art converters. An example readily illustrates'the process. If a coding system is chosen which has a first comparison with a reference standard of full scale, or 0.5000 on a normalized-full-scale basis, but the sum of the succeeding smaller comparison steps is made equal to 0.6000,
N W,=0.5000;2W,=0.6000 (7) then from Equation 5 I N 2W -W1=K1=0.1000 x=2 The total error between the apparent input signal at step 1 and its final value (X may then be as much as plus or minus 0.0500 without endangering the comparison which is just the value of the retained reference standard, so that the remaining Weights are all rejected, and
(b) Noise at a negative maximum; W rejected, i.e.,
C1=TN+X1"W1-Y1 0 or T o.0500-o.50o0 0.0s00 0 14 gr T 0.6000 (15) which is just the stated ability of the remaining steps to count to within the required fineness F.
The proof of adequacy of the conversion process illustrated above may be similarly demonstrated for the general case at any step i, with any required error tolerance K;, for any convenient coding array W to any required resol-ution' F. The number of steps required can then be obtained, or the relations inverted to obtain F, K,-, or the necessary counting ability N Z x=j+1 at any step, by use of Relations 1 through 6 above.
8 One-preferred form-of my invention applies the process to the standard binary system. If the system and signal errors likely at any step are considered proportional to th reference standard value of that step,
where p is a chosen per-unit value.
If equal tolerance to plus and minus errors is desired,
Y Z /ZK 'II /ZPW Substituting Equation 16 into Equation'5 gives p Z) i (18 or p . N i +P)= 2 W x=j+1 In a standard binary system with full scale normalized to unity,
j1 I r N 2W W,+ 2 Wx 1 x=1 x=j+1 j-l N l-2W,W,= W (21) x=1 I x=j 1 Combining Equations 19 and 21 provides:
1-1 .(1+p Zj WW, (22
' In a normal binary system J-1 .T-2 E X= .1.+E X (2 X=1 x=1 v From Equation 24, W may be calculated as +Pi: x =1 r (25) Combining Equations 22, 23 and 24 W 1 1 1 1 W W 26 Ml Mi 2% H1 1 1 W I 7 El Repeating the above steps and successively calculating for j-l from 1' gives the equation Since V 1 E F x =1 the optimum reference standard of my invention used at any given step in a regular binary system is Utilizing Equations 17, 30, and 31, a detailed example of the reference standards error-distributing weights and permissible error may be derived as tabulated in Table II, below, for a twenty-one-step standard binary converter with fineness of 0.01% and an error tolerance of 100% (plus or minus 50% distribution of this tolerance is chosen for the example). For convenience, rounding off is made within the fineness tolerance. Note that the weights do not have the same ratios as in the classical pure binary distribution, since they are derived in accordance with my invention. This distribution shares with pure binary, however, stepwise uniformity in con trast to the bunching or grouping distribution of the BCD systems hereinafter described.
Table 11 Reference Error Dis- Permitted Step No Standard tributing Error Weight 0. 3333 0. 1667 i0. 1667 0. 2222 0. 1111 :J;0.1111 0. 1481 0. 0741 $0. 0741 0. 0988 0. 0494 $0. 0494 0. 0659 0. 0329 i0. 0329 0. 0439 0. 0219 =|;0. 0219 0. 0292 0. 0146 $0. 0146 0.0195 0. 0098 :0. 0098 0.0130 0.0065 i0. 0065 0. 0087 0. 0044 i0. 0044 0. 0058 0. 0029 =l:0. 0029 0. 0039 0. 0019 5:0. 0019 0. 0026 0. 0013 $0. 0013 0. 0017 0. 0009 =l 0. 0009 0.0011 0. 0006 10. 0006 0. 0008 0. 0004 i0. 0004 0.0005 0. 0003 i0. 0003 0. 0004 0. 00015 #10. 00015 0. 0003 0. 0001 :110. 0001 0. 0002 0. 00005 i0. 00005 0. 0001 None None Extension of this invention to include systems arranged for an optimum error tolerance distribution derived from multiple time-constant systems or those with predictable systematic disturbances is readily apparent.
Table III illustrates the different error tolerances, number of steps, and degree of resolution obtainable with stepwise-uniform coding, as derived from Equations 16, 30 and 31.
It can be seen from the above table that an enormous return in tolerance to errors (signal fluctuations, amplifier settling, switch transients, etc.) is obtained for the use of only a few steps beyond the minimum required (pure binary). It is also obvious that the return for a greatly increased number of steps, although large, probably does not justify the length and complexity of this more elaborate process. The limiting case is readily seen to be that Where all reference standards have the same size and the number of steps taken is equal to the reciprocal of the fineness necessary.
Although the foregoing mathematical analysis is based upon the normal binary code, it will be apparent as the description proceeds that my invention is equally app-licable for analog-to-digital converters operating with other electronic computer codes. Thus, any of the many codes in which each of a series of decimal digits is represented by a group of two-state weights may be used as the background weighting distribution for my invention. Useful examples of such codes are the 8-4-42-1, 5-3-2-1, and the 95-32-1 binary coded decimal systems (hereinafter referred to as BCD systems). The bunched or grouped forms of my invention do not permit free choice or smooth distribution of error tolerance at each step, but have practical advantage in the simplicity of conversion of the retained reference standards to the conventional and convenient decimal readout. Reference standard simplicity and switch simplifications often result When certain simplified choices of error-distributing weights are used, for example, a single error-distributing weight for each decade of a BCD code such as 8-4-2-1.
Table IV below illustrates some of the BCD forms of my invention.
Table IV 8-4-2-1 0.01% 5-3-2-1 0.01% 9-5-3-2-1 0. 01% 6-321 0.01% Step N 0. Reference Total and Reference Total and Reference Total and Reference Total and Standard Error Tol- Standard Error Tol- Standard Error Tol- Standard Error Tol- W; era-nee K,- i erance K, W, erance K,- W; erance K;
. 8000 0666 5000 2222 9000 4221 6000 1333 4000 0666 3000 1222 5000 3221 3000 1333 2000 0666 2000 .0222 .3000 2221 2000 0333 1000 0666 1000 0222 2000 1221 1000 0333 .0800 0066 0500 0222 1000 1221 0600 0133 0400 .0066 0300 0122 .0900 0421 0300 0133 0200 0066 0200 0022 0500 .0321 0200 .0033 0100 0066 0100 0022 0300 .0221 0100 0033 0080 0006 0050 0022 0200 0121 0060 0013 .0040 0006 0030 0012 0100 0121 0030 0013 .0020 .0006 0020 0002 0090 0010 0006 0010 0002 0050 0008 None 0005' 0002 0030 0004 None 0003 0001 0020 .0002 None 0002 None 0010 0001 None 0001 None .0009 0005 0003 It is readily apparent that the above BCD classifications of my invention may be extended to a wide variety of other grouped codes with a base of or some other convenient number. The code chosen, it is shown, depends strongly on the error-tolerance distribution appropriate to the application.
By a slight increase in nonuniformity and complexity an improved error tolerance for BCD or grouped" codes may be obtained by altering the coding of the most significant decimal decade only, from that shown in Table IV. It is noted for all of the codes shown in Table IV that the counting capability of the accumulated weights of each series exceeds the normalized full scale of 1.0000. Altering the first decade reference standards so that overcounting is not possible permits increased error tolerances in the important first few comparison steps. Table V, below, demonstrates the modified BCD systems of Table IV if their most significant decades are altered to take full advantage of their counting ability. If this process, at some cost in logic and equipment complexity, is carried further to fractional Weights in the first decade, an even smoother gradation of error tolerances is possible. The limiting case when fractional weights are admitted is found to be the stepwise uniform embodiment typified by Tables II and 1H.
These modified codes of the reference standards in the most significant decade, if only Whole multiples of 0.1 are admitted, may be superimposed on any chosen grouped code provided the final sum of the reference standards equals or exceeds 1.0000. Such mixed codes can be designated by the symbols shown in Table V, such as 4221/6321. This signifies that the first decade is coded .4000, .2000, .2000, .1000, and that the remainder of the reference standards are coded in the ratio 6 to 3 to 2 to l, as detailed in Table IV.
Table V In operation,a reference voltage or current source 20 supplies a reference voltage or current to the reference standard generator 13 through a connecting line 21. The reference standard generator 13 proceeds to provide electrical increments or quantities of stepped magnitude which are the analog equivalents of the reference standards W A portion of the reference standard generator denoted as an error-tolerance distributor 23 provides stepped increments which are the analog equivalents to the tolerance-distributing weights Yj- The reference standard W and tolerance-distributing weight Y are subtracted from the analog signal T; in accordance with Equation 3 above, and an output signal 5 is generated by comparator 16 to indicate whether or not the reference standard W should be retained or rejected. The difference (provided at the error junction 12 and applied to the input of the comparator 16) between the analog input (T and a reference signal (W +Y is compared to ground by the comparator 16 to determine if the difference (between the analog input and the reference) is less than or greater than ground. The comparator 16 provides one output when the reference is larger than the analog input, and another output when the analog input is larger than the reference. The output of the comparator 16 is supplied to an input of the sequencer and control device 17. Signals are generated by the sequencer and control device 17 and supplied to the reference standard generator 13 to sequentially select each increment step.
A clock pulse generator 22 connected to the sequencer and control device 17 by connecting line 24 generates periodic spaced pulses for controlling the system operation. The output sequencing pulses of the sequencer and EXAMPLES OF NONUNIFORM JIDBEJAIEIJDIECODING MOST SIGNIFICANT *Note number of steps out by one.
THE SYSTEM AS A WHOLE In order to more fully describe my invention, a complete converter is illustrated in block diagram form in FIG. 2. As shown therein, the analog input T,- is connected to terminal 10. Connecting line 11 couples termihal 10 to a summing point 12. A reference standard generator 13 is likewise connected to summing point 12 through connecting line 14. Connecting line 15 couples summing point 12 as the input 0 of a comparator 16. The output 6,- of comparator 16 is connected to the input 3321/8421 "4221/5321 3221/95321" 22211/95321 Step No.
Wg K,- W] K; W3 K, W; K;
1 3000 4666 4000 2222 3000 4221 2000 6221 2 3000 1666 2000 2222 2000 3221 2000 4221 3 2000 0666 2000 0222 2000 1221 2000 2221 4 1000 0666 1000 0222 1000 1221 1000 2221 5 1000 1221 6 7 8 9 "8421 as in 5321 as in 10 Table IV Table IV 11 95321 as in 12 Table IV 13 "95321 as in 14 Table IV of a sequencer and control device 17 by connecting line control device are initiated by the clock pulses. After each of the increment steps have been generated and compared with the input signal in the comparator 16, an accumulation of increments will be manifested by the device 17. As heretofore noted, if there has been an error in the initial reading of the analog input, the reference standard generator may generate a suitable signal for compensating for this error, although this signal will not correspond to the selected code. Accordingly, a digital code converter 30 is connected to another output of the sequencer and control device 17 so that the digital information may be read out in the desired code.
13 COMPARATOR, REFERENCE AND INPUT CIR CUIT OF SEQUENCER AND CONTROL DEVICE A11 analog-to-digital converter constructed in accordance with my invention is shown in more detail in FIG. 3 wherein like reference numerals are employed to designate elements referred to previously. The analog input T is connected between terminals a and 10b, terminal 10a being connected to summing point 12 through a resistor 40 by connecting line 11 and terminal 10b being connected to a common terminal which may be at ground potential as illustrated. The output of reference standard generator 13 is carried by connecting line 14 to summing point 12. Connecting line 14 includes series resistor 41.
Comparator 16 comprises a high gain feedback amplifier 42 having a feedback resistor 43 connected between its output and input. As is well known in the art, the output voltage of an operational feedback amplifier such as amplifier 42 is very precisely equal to the sum of the currents at its input or summing point 12 multiplied by the resistance of feedback resistor 43. In most feedback amplifiers, the sign of the output voltage will be opposite that of the input voltage.
Reference source 20 may comprise a direct current source illustrated as a battery 44 having its positive pole connected to the common terminal and its negative pole connected to the input of the reference standard generator 13 by connecting line 21.
The input analog signal is assumed to be positive in the system shown in FIG. 3. If the input analog signals vary in polarity, polarity sensing means (not shown) are well known in the art and may be connected to detect polarity changes. By simply changing the polarity of the reference source 20, a quantity of appropriate polarity will be generated by the reference standard generator 13.
Reference standard generator 13 may conveniently comprise a conductance adder providing at its output either voltage or current increments. For the system shown and described hereinafter, the reference standards Wj and tolerance distributing weights Y,- are current values which are compared with the current magnitude of the input analog signal T,-.
For positive analog inputs to the system of FIG. 3, the current increments representing the reference standards W must be negative. Thus, the input analog current and conductance adder output current will be subtracted by the amplifier 42. When the output of the amplifier is negative, it means that the analog input has a higher value than the current output from the reference standard generator (remembering that amplifier 42 acts as a phase inverter). The amplifier output is positive when the output of the reference standard generator is larger than the analog input.
As previously noted, a reference standard is to be retained and subtracted from the analog input when the analog input is larger than the reference standard; conversely, a reference standard is to be rejected when the analog input is smaller than the reference standard. For providing this function, the output portion of the comparator 16 includes and gate 36 and one-shot multivibrator 37. Specific circuitry for these devices is not shown since the details thereof are well known in the art. The amplifier output and a series of A pulses from clock generator 22 are supplied as input signals to the an gate. When both are positive, i.e., when the analog input is smaller than the output of the conductance adder, the and gate receives a positive signal at both of its inputs and passes a pulse which turns over the one-shot multivibrator. When the amplifier output is negative, i.e., when the analog input is larger than the conductance adder output, the and gate 36 does not receive a positive signal at both of its inputs and so will not pass a pulse to the one-shot multivibrator. The multivibrator output, designated by 5,, is connected to the control elements of the sequencer and control device by connecting line 38 only when the reference standard is too large and must be rejected, i.e., 6, is 1 in Equation 1 above when it is present at the output of one-shot multivibrator 37 and 0 in Equationl when it is not so present.
REFERENCE STANDARD GENERATOR The reference standard generator 13 is shown to comprise a conductance adder having a plurality of precision resistors and a plurality of switches for connecting the resistors in particular arrangements. Although for purposes of illustration mechanical single-pole double-throw switches are illustrated, it will be understood that preferably high speed electronic switches known to those skilled in the art will be utilized. The 8-4-2-1 BCD system has been selected for implementing the generator in this embodiment, and the conductances are weighted according to the reference standards (W,-) and error tolerances (K as tabulated above in Table IV.
The first and most significant decimal decade includes single-pole double-throw switches 45, 46, 47 and 48. Each of these switches have two input terminals, one of which is connected to the reference source 20. The output terminal of each of these switches is connected to the respective movable contact; each of these switch output terminals is in turn connected to a respective one of the resistors 50, 51, 52 and 53. These resistors have resistance values which will give the reciprocal conductance values for the desired reference standards, or 0.8000, 0.4000, 0.2000, and 0.1000. As is well known in the art, the actual conductance values may be at any desired level so long as they retain these relationships herein normalized to unity for convenience It will be understood these these resistors may also be valued to give the reciprocal conductance values for a nonuniform BCD system as tabulated in Table V above. As noted supra, altering the coding of the most significant decimal decade permits increased error tolerances in the first few comparison steps. Depending upon the particular code selected, the digital code converter 30 is suitably constructed to decode the particular nonuniform BCD system selected.
The second, third and fourth decimal decades of the reference standard generator 13 are arranged similarly to the first decade, in each case comprising the first four resistors and associated switches of each decade. Thus, resistors 55, 56, 57 and 58 and associated switches 59, 60, 61 and 62 comprise the second decade; resistors 65, 66, 67 and 68 and associated switches 69, 70, 71 and 72 comprise the third decade; and resistors 75, 76, 77 and 78 and associated switches 79, 80, 81 and 82 comprise the fourth and least significant decimal decade. The resistors 55, 56, 57 and 58 provide conductance values of .0800, .0400, .0200 and .0100; the resistors 65, 66, 67 and 68 provide conductance values of .0080, .0040, .0020 and .0010; and the resistors 75, 76, 77 and 78 provide conductance values of .0008, .0004, .0002, and .0001. 4
The decimal decades described above function in a manner identical with that of prior art conductance adders in analog-to-digital converts. When none of the switches in the reference standard generator is actuated, each of the resistors is connected by the switches to the common terminal and thus no reference standard current is generated by the generator 13. On actuation of one or more switches, the output current increment of the reference standard generator 13 is equal to the voltage of reference source 20 (V multiplied by the sum of the conductances of the resistors that are connected to line 21 which is in turn connected to the negative terminal of voltage source 44. Thus, the current output from the conductance adder is given by the equation:
out= reI 1 where G is the total conductance connected to the ref erence source. According to Equation 32 the output current is proportional to the sum of the conductances connected to conductor 21. Therefore, if the resistors are Valued to give the conductances discussed, a series of selectably addible, stepped currents may be obtained for comparison with the analog input.
ERROR-TOLERANCE DISTRIBUTOR CIRCUIT The error-tolerance distributor circuit 23 comprises resistors 54, 85, 87 and respectively associated switches 49, 86, 88. Resistors 54, 85 and 87 are valued to give respective conductance values of 0.0300, 0.0030 and 0.0003. As with the resistors in the reference standard generator, these values are normalized to unity for convenience. With all conductances in circuit, a total tolerance-distributing conductance of 0.0333 is inserted; with resistors 85 and 87 only in circuit, a total tolerancedistributing conductance of 0.0033 is inserted; and with only resistor 87 in circuit, a total tolerance-distributing conductance of 0.0003 is inserted. In so valuing these resistance values the m of E nation 6, su ra, has been selected as /2 so that the error tolerance tabulated in Table IV is evenly distributed between positive and negative values. It will be apparent as the description proceeds that m may be varied simply by changing the values of these resistors; thus, if the particular data handling system in which the analog-to-digital converter of the invention is incorporated encounters only positive errors, the resistors 54, 85 and 87 and associated switches can be deleted, thereby giving a total error tolerance of 0.0666 in the first decimal decade for errors of unidirectional positive polarity. If, on the other hand, the errors are entirely negative, the conductance of resistor 54 may be selected as 0.0600, the conductance of resistor 85 selected as.0.0060, and the conductance of resistor 87 as 0.0006 so as to give a total error tolerance of 0.0666 in the first decimal decade for errors of unidirectional negative polarity.
Referring again to the particular embodiment of my invention illustrated in FIG. 3, a total tolerance-distributing conductance of 0.0333 is added to the reference standards generated during the first decade. Stated in another way, an additional current of 0.0333 magnitude is subtracted from the analog input signal so that the analog signal appears 0.0333 lower in value throughout the comparison of the first decade. The tolerancedistributor circuit in the first decade has the effect of obtaining coded values for a signal which is 0.0333 lower in value than the actual analog input. Thus, if the analog input signal is erroneously read higher than it actually is, the system will be able to compensate for the incorrect reading since it has already subtracted an error tolerancedistributing weight of 0.0333 prior to the first decade comparison.
A major source of erroneous classification lies in the finite response time of the comparator to changes in its input. The largest input changes occur when the most significant digits are switched in the conductance adder. The'possibility that the output of the comparator amplifier has not yet reached its final value is especially present under these circumstances. Therefore, the total tolerance-distributing conductance of 0.0333 is inserted during the first decimal decade comparison. During the comparison of the second decimal decade with the analog input, the resistance 54 is disconnected from the reference 20 by deactuating switch 49 thereby leaving a tolerance-distributing conductance of 0.0033 during the second decimal decade comparison. However, the magnitude changes occurring in Cj, the comparator during this decade comparison, are considerably lower valued than during the first decade comparison and the maximum error tolerance of :0.0033 during this second decade comparison substantially improves the accuracy of the converter. In like manner, the maximum allowable error during the comparison of third decimal decade is :0.0003. Accordingly, switch 86 is deactuated during the third decimal decade comparison and only tolerance distributor switch 88 remains actuated during this decade comparison, During the fourth or units comparison, the final accuracy of the system is determined and therefore no error-tolerance-distributor values are permitted at this time. However, since this comparison is made at the end of the conversion period, signal fluctuations and system transients will have become substantially zero at this time.
By way of illustration, a numerical example may be given to further illustrate the function of the error-tolerance-distributor circuitry. Assume that the true value of the analog signal at the last comparison step (T is 0.4888 but that initially it is erroneously read as greater than 0.5000 but less than 0.5333. As heretofore observed, in the prior art devices the signal would be erroneously classified as greater than 0.5000. In the prior art devices such a decision is irrevocable, resulting in the erroneous encoding of 0.5000 rather than the required 0.4888. In accordance with my invention, however, the tolerance-distributing value of 0.0333 is initially subtracted from the input signal, with the result that a signal level of less than 0.5000 but greater than 0.4000 is compared instead of the measured signal above 0.5000. Thus, the signal is correctly classified as greater than 0.4000 and less than 0.5000 although the signal actually appears to be larger than 0.5000. Even though the 0.0300 error-tolerance-distributor is removed after the first decade comparison, a sufiicient period of time will have elapsed by then so that the system transients will have decayed considerably. The signal may nevertheless still appear larger than the true value T For example, assume that during the second decade comparison, the signal level appears to be greater than 0.4900 but less'than 0.4932. In the prior art apparatus the signal would be erroneously and irrevocably encoded as 0.4900 instead of its true instantaneous value of 0.4888. However, upon the subtraction of 0.0033, the signal appears to have a value greater than 0.4800 but less than 0.4900. Again, although the 0.0030 error-tolerance-distributor is removed at the end of the second decade classification, a further period of time has lapsed so that the read input analog signal will be still closer in value to its true value T Thus, in the third decade classification a maximum allowable error of 20.0003 is permitted. If the actual value of the signal is erroneously read as 0.4890 instead of the actual instantaneous value of 0.4888, the signal will be properly classified because of the substraction of 0.0003 during this encoding. During the fourth encoding decade, the final accuracy of the instrument is determined and accordingly no tolerance; distributing weight is substracted. However, "by the time that the fourth decade has been selected for comparison, the time has elapsed for the first, second and third decades so that the system errors and signal fluctuations will have approached zero.
It may be readily observed that in accordance with my invention a quite simple, yet very convenient means is aiforded for adding tolerance-distributing weights to the reference standards during certain of the decade conversions, thereby permitting the analog input signal to be read in the initial converter steps either above or below its true instantaneous value.
A PRELIMINARY DESCRIPTION OF THE CONTROL CIRCUIT Each of the switches in the conductance adder is under the control of an associated flip-flop in the sequencer and control device 17. The control portion of this apparatus is shown in block diagram form in FIG; 3, each of the switches in the first decimal decade being connected to a flip-flop element 90. Although apurely mechanical connection is shown between each of these switches and the flip-flop element 90, it will be understood that an electrical connection would normally be made between the output of flip-flop element to the input of suitable electronic switches serving as the conductance adder switches. Control flip-flop element 92 is connected to drive each of the switches 59-62 in the second decade. Control flip-flop element 93 is similarly connected to drive switches 69-72 in the third decade and control flip-flop element 94 is connected to actuate switches 79-82 in the fourth decade. Control flip-flop elements 91, 95 and 96 are connected respectively to switches 49, 86 and 88 of the error-tolerance-distributor circuitry.
Each of the control flip-flop elements 90, 92, 93 and 94 are actuated in sequence by a plurality of timing pulses T: T1: T2, T3! T4: T5, T6: T7: T81 T99 T10, T11: T12! T13 T T T and T Each of these pulses is generated .in the sequencing portion of the sequencer and control device 17 and this portion of the invention will be described in more detail hereinafter. Briefly, the periodically generated A and B pulses from the clock 20, supplied to the sequencer and control device 17 through connections 21a and 21b, serve to provide the sequence of seventeen pulses followed by a periodic repetition thereof.
Sequencing pulse T is supplied to flip-flop elements 90, 92, 93 and 94 for resetting to zero all control flip-flops except the one which controls the 0.8000 conductance;
this latter conductance is actuated by the T pulse to begin a conversion cycle.
Also supplied to each control flip-flop element is the pulse through connecting line 38. As previously noted, this pulse occurs whenever the reference standard is too large and should be discarded. Accordingly, appropriate circuitry hereinafter described causes each sequencing pulse following T to deactuate the too large reference standard. Thus, sequencing pulse T, will cause control element 90 to deactuate the 0.8000 conductance if a pulse 6 is present when pulse T occurs. The oneshot multivibrator is designed so that the 6 pulses are of sutficient duration to last until after a succeeding sequencing pulse.
Sequencing pulse T will also cause control element 90 to actuate the 0.4000 conductance. In similar manner, pulses T and T effect a comparison between the analog input and the 0.2000 and 0.1000 conductances.
The T pulse is also supplied to control elements 91,
'95 and 96 for actuating the tolerance-distributing conductances 54, 85 and 87. All of these conductances remain in circuit until pulse T This pulse is supplied to control element 91 for disconnecting conductance 54 and to control element 92 for initially energizing the second decade. Pulse T is also supplied to control element 90 for deactuating the 0.1000 conductance if a 5 pulse is present.
In sequence, pulses T T T and T actuate the 0.0800, 0.0400, 0.0200 and 0.0100 conductances in the second decade. In similar manner, pulse T disconnects conductance 85 at the beginning of the third encoding decade so as to remove the 0.0030 tolerancedistributing conductance. Pulse T also disconnects the 0.0100 conductance if a 5 pulse is present simultaneously. Pulses Tg-"Tn serve to actuate in sequence the reference standard conductances in the third encoding decade, and pulses T T serve to sequentially actuate the reference standard conductances in the fourth encoding deca-de. Pulses T and T deactuate the 0.0010 and 0.0001 conductances if 6 and 6 pulses are respect-ively present. At the beginning of the units decade,
. pulse T also turns off flip-flop 96 for disconnecting the 0.0003 tolerance-distributing conductance.
A PRELIMINARY DESCRIPTION OF THE DIGITAL CODE CONVERTER As hereinabove described, my invention provides a predetermined error tolerance f-or each comparison step or group of comparison steps by utilizing an increased classification span of succeeding steps beyond that of prior art encoders. In the specific embodiment illustrated in FIG. 3, the increased classification span is derived from digit combinations of the 84-2-1 BCD system which fall outside this code. Accordingly, at the end of an encoding sequence, the retained reference standard conductances do not necessarily correspond to the selected 8-4-2-1 BCD system. Therefore, one form or another of code conversion will be required before the output of the analog-to-digital converter is in useful form.
Numerous code conversion techniques known in the art may be employed to convert in an appropriate manner, the information contained in the converter. In some instances, the converter will be connected to the control elements 90, 92, 93 and 94 and convert therefrom directly into a decimal readout device. In the specific embodirnent illustrated in FIG. 3, an alternative type of converter is shown wherein the particular flip-flops of the control elements are monitored by a digital code converter 30 and controlled thereby to assume a true 842-1 BCD relationship.
A preliminary description of the operation of the digital code converter 30 is as follows. As shown in FIG. 3, this converter comprises the four elements 99, 100, 101 and 102. Element 100 is connected to the output of control element 92 by lines 110, 111, 112 and 113. Element 101 is connected to the output of control element 93 by lines 114, 115, 116 and 117 and element 102 is connected to the output of control element 94 by lines 118, 119, 120 and 121. One of the functions of the elements 100-102 in the digital code converter 30 is to monitor the state of the flip-flops in the sequencer and control device 17 and to determine when the decades total greater than nine digits, e.g., whenever an 8 and a 2 reference standard conductance are simultaneously connected. In the circuitry shown, the status of each element in the Sequencer and control device 17 is detected immediately after a sequence of control signals has been completed. Thus, after the second decade conversion has been completed, pulse T is applied to element 100 to determine whether the flip-flops in element 92 total greater than nine digits. If a combination outside the BCD 8-4-2-1 system is shown, a control signal is transmitted back to the sequencer and control device 17 for rearranging the state of the flip-flops to conform to the 8421 BCD system, In like manner, the status of the element 93 is monitored by applying sequencing pulse T to element 101 of the digital code converter and the status of the element 94 is monitored by applying sequencing pulse T to element 102 of the digital code converter.
In order to generate a BCD 8-4-2-1 code, a control element such as 92 which indicates a total greater than nine digits must have subtracted therefrom ten digits. This subtraction is afforded by proper control signals transmitted from converter element 100 to control element 92 by connecting line 125, from converter element 101 to control element 93 by connecting line 127, and from converter element 102 to control element 94 by connecting line 129.
The subtracted value must be made up by adding or carrying over one digit to the next higher decimal decade. For this purpose each control element is connected to appropriate circuitry in the digital code converted for adding the required one digit upon receipt of a carry pulse. For control elements 92, 93 and 94 this function is provided by converter elements 100, 101 and 102. For control element 90, this function is performed by converter element 99 coupled to control element through lines 107, 108 and 109. The carry pulse from converter element 100 is coupled to control element 90 through lines 107, 108 and 109. The carry pulse from converter element 100 is coupled to control element 90 and converter element 99 by connecting line 126. The carry pulse from converter element 101 is coupled to control element 92 and converter element 100 by connecting line 128. The carry pulse from converter element 102 is coupled to control element 93 and converter element 101 by connecting line 130.
The operation of the digital code converter may be illustrated by a numerical example. In FIG. lb and the description thereof, a typical example was assumed, Wherein an analog signal having -a true value T of 0.5155 was initially incorrectly read as less than 0.5000 but greater than 0.4000. As heretofore noted, the first decimal decade selected the 0.4000 class, the second decimal decade the 0.1100 class, the third decimal decade the 0.0050 class and the fourth decimal decade the 0.0005 class. At the end of the encoding of the second decade, pulse T is applied to element 100 of the digital code converter. This converter element then detects that the second decade is greater than 0.0900 since the 8, 2 and 1 digits have been simultaneously energized to provide the 0.1100 class. A signal is transmitted from element 100 through connecting line 125 to control element 92 for reversing the state of the 8 and the 2 flip-flops to subtract ten digits. The second decade then encodes only the 1 digit or 0.0100 class. Simultaneously, a carry signal is sent from element 100 through connecting line 126 to control element 90 and converter element 99 whereupon the control element 90 is advanced one digit, i.e., from the 0.4000 to the 0.5000 class. The final encoded value is thus the same, except that it is in the desired BCD 8421 code with the first decade selecting the 0.5000 class, the second decade the 0.0100 class, the third decade the 0.0050 class and the fourth decade the 0.0005 class.
A PRELIMINARY DESCRIPTION OF THE OUTPUT REGISTER The output register comprises four buffer registers, respectively labeled 135, 136, 137 and 138 in FIG. 3. Buffer register 135 is connected to the output of control element 90 in the sequencer and control device 17 by connecting lines 140, 141, 142 and 143. Buffer register 136 is similarly connected to control element 92 by lines 144, 145, 146 and 147. Lines 148, 149, 150 and 151 connect control element 93 to the input buffer register 137 and lines 152, 153, 154 and 155 connect control element 94 THE SEQUENCING CIRCUITRY The sequencer portion of the sequencer and control device 17 is shown in detail in FIG. 4. The sequencer comprises five flip-flops 160, 161, 162, 163 and 164. Each flip-flop is depicted with a single input and two outputs. A number of vacuum tube and transistor embodiments of flip-flops suitable for the sequencer are known in the art;
typically each of the flip-flop outputs may be either at ground potential or at a positive potential. Each of the flip-flops will change to the respectively opposite state upon application of a positive input pulse. The sequencer is so designed that in the quiescent state in which all flip- :flops are reset to their zero or off state, the left-hand output is at ground or zero and the right-hand output is at a positive voltage as shown for illustrative purposes. A logic is represented by said zerooutput and a logic 1 is represented by said positive voltage.
. 20 Periodically spaced A pulses are supplied from the clock to flip-flop through connecting line 21a. The normally 1 output of this flip-flop is connected through capacitor to the inputs of and gates 166 and 167. The second input of and gate 166 is formed by the normally 1 output of flip-flop 164. A second input of and gate 167 is formed by the normally 0 output of flip-flop 164. The output of an gate 166 is connected to the input of flip-flop 161. The output of and gate 167 is connected to an or gate 173, the output of which is connected to the input of flip-flop 164. Capacitors 180, 181 and 182 are respectively connected between the normally 1 output of flip-flops 161, 162 and 163 and the inputs of flip-flops 162 and 163 and the or gate 173.
The operation of the circuit of FIG. 4 thus far described is as follows. Prior to the first A pulse from the clock through connecting line 21a, each of the flipflops 160-164 are in their quiescent or off state, and accordingly, the left-hand output of flip-flop 160 is 0 and the right-hand output is 1. After the first A pulse, the state of flip-flop 160 is reversed so that the righthand output will be zero, becoming positive or in its on state delivering a logic 1 again on the second A pulse.
The change in voltage from zero to positive occurring at the right-hand output of flip-flop 160 upon the second A pulse is diiferentiated by capacitor 165, resulting in a positive voltage pulse which is simultaneously applied to and gates 166 and 167. At this time, the second input of an gate 166 is also positive because flip-flop 164 is in its olf state. Accordingly, the and gate permits the pulse output from flip-flop 160 to be connected through an gate 166 to the input of flip-flop 161. Thus, so long as flip-flop 164 is in its off state, a positive input pulse will be applied to flip-flop 161 after each two A pulses. Consequently, flip-flop 161 will change state with every two A pulses. Four A pulses into flipflop 160 are required to change flip-flop 161 from its 011 state to its on state and back to its off state. Accordingly, a positive pulse will enter flip-flop 162 from differentiating capacitor only after four pulses. In similar manner, flip-flop 163 will receive a positive input pulse only after eight A pulses. On the sixteenth input A pulse, flip-flop 163 will return to its 01f state and apply a positive pulse through capacitor 182 to or gate 173. Or gate 173 transmits this positive pulse to the input of flip-flop 164 whereupon flip-flop 164 is changed from its off state to its on state. Since flip-flop 164 is no longer in its quiescent or off state, positive pulses are no longer admitted through and gate 166 to the input of flip-flop 161. However, the pulses originating from capacitor 165 are now allowed to pass through and gate 167 because the second input of this and gate is now positive since it is connected to the left-hand output of flip-flop 164 which is now at a positive potential. Such a positive pulse will be supplied from capacitor 165 at the eighteenth A pulse. At this time, flip-flop.
160 is returned to its 011 state, flip-flops 161, 162, 163 have been and remain in their oif state and flip-flop 164 is returned to its off state because of the positive pulse applied to its input through and gate 167 and or gate 173. Flip-flops 160-164 are thus in their original or quiescent off state and accordingly the eighteenth A pulse is the first pulse of another sequence cycle.
It will thus been seen that the sequencer shown in FIG. 4 is in effect a decade counter having five counting states in which the normally full capacity of thirty-two is attenuated to seventeen by the circuitry hereinbefore shown and described.
21 The states of each of the flip-flops at each of the A pulses is tabulated as follows:
Flip-Flop No. A Pulse No.
0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 O 0 0 1 1 0 1 0 1 1 0 0 l 1 1 0 1 1 l 1 0 0 0 O 0 1 1 0 0 0 1 0 0 0 0 0 Noting the tabulation above it will be apparent that 17 different combinations are afforded by the flip-flop states. And gates 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201 and 202 are connected to the outputs of flip-flops 160 and 164 so that a pulse is gated through the individual and gates at a particular A pulse from the clock. Because the change in state of the flip-flops takes a finite time, one input signal for each of said and gates 185202 is supplied by periodically spaced B pulses which are also generated by the clock and directed to the an gates through line 2112. As shown in the figure, the B pulses are slightly delayed with respect to the A pulses so as to guarantee that no change of state initiated by a preceding A pulse takes place simultaneously with or after the beginning of a B pulse. Consequently, the pulses indicated at the output of the n g as T0: T1 T2: T3 T4: T5: T6: T7 T8: T99 10 11 12, 13, 14, 15 16, 17 are generated With substantially equal time intervals.
It may be noted in the table immediately above that all of the flip-flop states are distinctly defined by the first four flip-flops 160, 161, 162 and 163 with the exception of the pulse outputs T T T and T which are distinguishable only by means of noting the status-of flip-flop 164. Consequently, the associated and circuits have an output of this flip-flop as an additional input.
THE CONTROL CIRCUITRY FIGS. 5a and 5b are detailed illustrations of the control circuitry of the sequencer and control device 17 and the circuitry of the digital code convested 30. Singles generated by the sequencer and control device 17 are coupled to respective switches in the reference standard generator and the error-tolerance distributor for controlling the operation thereof. For example, control element 90 controls the switches of the first decimal decade. As shown in FIG. 5:: this control element comprises flip-flops 210, 211, 212 and 213 each having a pair of stable states producing a ground potential at one output (0) and a positive potential at the other (1). In the quiscent or off state each of the flip-fiops has a 0 left-hand output and a 1 right-hand input.
Each of the left-hand outputs is connected to a respective one of the reference standard switches 4548. Thus, flip-flop 210 will actuate switch 45; flip-flop 211 will actuate switch 46, etc. When the respective fiip-flop is in its off state, no output potential appears at its left-hand output and no potential will be applied to the associated switch; contrariwise, when a respective flip-flop is in its on state, a positive potential appears at its left-hand output for actuating the associated switch. At the beginning of an analog-to-digital conversion (T it is necessary that all conductance adder switches previously actuated be returned to their off position with the exception of switch 45 connected to the initial reference (0.8000) conductance. Accordingly, timing pulse T from the sequencer portion of the sequencer and control 17 (FIG. 4) is connected to each of .the flip-flops 211, 212 and 213 and serves to reset each of these flip-flops to their off state. For achieving this reset, respective and gates 214, 215 and 216 have an input connected to respective lefthand outputs of flip-flops 211, 212 and 213; other inputs of these an gates are connected to T The outputs of the and gates are connected to the inputs of flip-flops 211213 through respective or gates 217, 218 and 219. The pulse T will be gated through an and gate only whenever a positive voltage is applied to the and gate input from the left-hand output from a flip-flop; i.e., only whenever a flip flop is in the on state. For those flipflops in an on state, the pulse T will appear at their inputs and cause the flip-flop states to reverse to the desired off or reset state.
Since pulse T is also the initial pulse of analog-to-digital conversion cycle, this pulse should caus fiip-fiop 210 to remain in the on state, or to change to the on state. And gate 220 has one input connected to T and another input connected to the right-hand output of flip-flop 210. The output of an gate 220 is connected to the input of flip-flop 210 through or gate 225. The pulse T will be gated through and gate 220 only when a positive potential appears at the right-hand output of flip-flop 210; i.e., only when flip-flop 210 is in its off state. Thus, if flipflop 210 is in its 011 state, the pulse T will properly reverse the state of this flipflop so as to actuate the 0.8000 conductance; if flip-flop 210 is in its on state, the pulse T will not be allowed to reverse the flip-flop state so that the 0.8000 conductance remains in circuit.
After a comparison has been made between the 0.8000 reference value and the analog input, a pulse 5 will appear on line 38 if the reference value is greater in magnitude than the analog input. The pulse 6 is connected to inputs of and gates 226, 227, 228 and 229 of element 90. As hereinabove noted, the pulse 6,- is of sufficient duration to continue until after a succeeding sequencing pulse. Thus, if both the pulse 5 and T are at the input of and gate 226, this gate will open and allow the pulse T to disconnect the 0.8000 reference standard. If 6 is not present (i.e., a comparison has shown the 0.8000 reference to be smaller than the analog input) and gate 226 remains closed and does not pass pulse T When 6 is present, an output pulse from and gate 226 is applied to the input of flip-flop 210 through or gate 225. Therefore, the flip-flop 210 will be caused to reverse its state by pulse T when the comparator supplied pulse 6 indicating that the 0.8000 reference was to be discarded.
Timing pulses T T and T are respectively connected to flip-flops 211, 212 and 213 through respective or gates 217, 218 and 219 for actuating in sequence the 0.4000, 0.2000 and 0.1000 reference standards.
Pulses T T and T are respectively connected to the inputs of and gates 227, 228 and 229. Outputs of these and gates are respectively connected to flip-flops 211, 212 and 213 through or gates 217, 218 and. 219. And gates 227-229 functions in a manner similar to that of and gate 226, allowing the timing pulses to deactuate the previous reference whenever a 5 pulse indicates that it should be discarded. For example, when pulse T is supplied by the sequencer, flip-flop 212 is actuated to connect the 0.2000 reference standard. Simultaneously, the 0.4000 reference standard will be discarded if the 5 pulse is present at the input of and gate 227. If the 5 pulse is not present, it indicates that the 0.4000 reference standard should remain in circuit .and be subtracted from the analog input.
23' left-hand output of flip-flop 231 drives switch 49 which actuates the 0.0300 tolerance-distributing conductance. Element 95 includes or gate 255 and flip-flop 256. Or gate 255 has two inputs, one which is connected to the T sequencing pulse and the other of which is connected to the T sequencing pulse. The left-hand output of flipfiop 256 drives switch 49 which actuates the 0.0030 tolerance-distributing conductance. Control element 96 includes or gate 257 and flip-flop 258. Or gate 257 has two inputs one of which is connected to the T sequencing pulse and the other which is connected to the T sequencing pulse. The left-hand output of flip-flop 258 drive switch 88 which actuates the 0.0003 tolerancedistributing conductance. The operation of the control elements driving the errortolerance distributor is as heretofore described. Ssquencing pulses T changes flip-flops 231, 256 and 258 to their onfstate whereby a positive voltage is applied to errortolerance distributor switches 49, 86 and 88 for actuating same and inserting a total tolerance-distributing conductance of 0.0333 in circuit. Flip-flop 231 and associated switch 49 remain on until sequencing pulse T This latter pulse returns flip-flop 231 to its ofi state thus disconnecting the 0.0300 tolerance-distributing conductance. Flip-flop 231 therefore remains on during the first decimal decade and off during the remainder of the encoding cycle. Flip-flop 256 and associated switch 86 remain on until sequencing pulse T This latter pulse returns flip-flop 256 to its off state and thus disconnects the 0.0030 tolerance-distributing conductance. Flipfiop 256 therefore remains on during the first and second decimal decades and off during the remainder of the encoding cycle. Flip-flop 258 and associated switch 88 remain on until a. sequencing pulse T This latter pulse returns flip-flop 258 to its off state and thus disconnects the 0.0030 tolerance-distributing conductance. Flip-flop 258 therefore remains on during the first, second and third decimal decades and off during the fourth or last decimal decade. The second, third and fourth decades of the control portion of the sequencer and control device 17 are similar to the first decade. For example, element 92 comprises flip-flops 235, 236, 237 and 238 having their left-hand outputs respectively connected to conductance adder switches 59, 60, 61 and 62 for actuating same when a respective flip-flop is in the on state.
And gates 240, 241, 242 and 243 each have one input connected to sequencing pulse T and another input respectively connected to the left-hand outputs of flipflops 235-238. The outputs of each of these and gates are respectively connected to the inputs of flip-flops 235- 238 through respective or gates 245, 246, 247 and 248. As in the first decade these and gates allow the T sequencing pulse to reset each of the flip-flops to their oif state.
Flip-flops 235238 are sequentially turned on by respective sequencing pulses T T T and T, which are respectively connected to the inputs of the flip-flops through or gates 245-248. Upon the comparison between the analog input signal and a reference value a pulse 6 denotes whether the reference value was too large and therefore should be disconnected. Pulse 6, is therefore connected to inputs of and gates 250, 251, 252 and 253. Other inputs of these and gates are respectively connected to sequencing pulses T T T and T The outputs of these and gates are respectively connected to flip-flops 235, 236, 237 and 238 through respective or gate-s 245, 246,247 and 248. Thus, upon occurrence of, for example, the pulse T flip-flop 236 and the 0.0400 reference standard conductance will be turned on. Preceding flip-fiop 235 and conductance 0.0800 Will remain on or be turned olf depending upon whether a pulse 8 is applied to and gate 250 when pulse T occurs.
24 THE DIGITAL CODE CONVERTER CIRCUITRY In FIG 5a is shown in detail the proportion of the digital code converter 30 connected to the outputs of the first and second decimal decades. This converter monitors the control elements of the sequencer and control device so as to provide output signals encoded in an 8-4-2-1 BCD relationship. For convenience, element of the digital code converter will be first described. This element is connected to the output of element 92 of the sequencer and control 17. This element controls the encoding cycle during the second decimal decade.
Element 100 includes and gates 260 and 261. The normally 0 outputs of flip-flops 235 and 237 are connected by respective connecting lines and 112a to the input of and gate 260. Pulse T is also connected to the input of and gate 260. The output of and gate 260 is connected to an input of or gate 262. The output of this or gate is connected to the inputs of flip-flops 235 and 237 through connecting line a and respective or gates 245 and 247. The output of or gate 262 is also connected to the input of or gate 263.
The normally 0 output of flip-flops 235 and 236 and the normally 1- output of flip-flop 237 are connected by respective connecting lines 110, 111 and 11217 to the input of and gate 261. Pulse T is also connected to the input of and gate 261. The output of and gate 261 is connected to another input of or gate 262 and to the input of flip-flop 236 through connecting line 1251) and or gate 246.
The operation of this portion of the digital code converter is as follows: As noted above, the digital code converter 100 must detect whenever the second decimal decade encodes a digital combination of ten or greater. These combinations are 8+4+2+1, 8+4+2+0, 8+4+0+l, 8+4+0+0, 8+0+2+1 and'8+0+2+0. For proper conversion, the 1 digit need not be converted. All combinations of the 8 and 2 digits may be con verted by reversing the state of the 8 and 2 flip-flops (in the second decade these are flip-flops 235 and 237) to their oil states and applying a carry pulse to the preceding, or first decimal decade. Thus, 8+4+2 becomes 0+4+0+carry and 8+0+2 becomes 0+0+0+ carry. A combination of the 8 and 4 digits without a 2 digit requires reversing the state of the 8, 4 and 2 flip-flops (in the second decade these are flip-flops 235, 236 and 237) and applying a carry pulse. Thus, the 8 and 4 flip-flops are changed to their off state, while the 2 flip-flop is changed to its on state; i.e., 8+4+0 becomes 0+0+2+carry.
And gate 260 connected to the 0.0800 flip-flop 235 of the 0.0200 flip-flop 237 detects a combination of the 8 and 2 digits at the time of the T sequencing pulse (which terminates the second decade conversion). A pulse is then gated through and gate 260 and or gate 262 through connecting line 125a to reverse the states of the 0.0800 flip-fiop 235 and the 0.0200 flip-flop 237. A carry pulse is also applied to or gate 263 for application through connecting line 126 to the preceding first decade element 90 and preceding digital code converter element 99.
And gate 261 detects the combination of the 0.0800 and 0.0400 flip-flops on and the 0.0200 flip-flop off. A pulse is then gated through and gate 261 and or gate 262 through connecting lines 125a to reverse the states of the 0.0800 flip-flop 235 and the 0.0200 flip-flop 237, and to or gate 263 as a carry pulse to the first decade. The pulse gated through and gate 261 is also conducted through connecting line 125k to reverse the state of the 0.0400 flip-flop.
The carry pulse from the second decimal decade is applied to the first decimal decade through connecting line 126. This pulse indicates to the higher decade that an additional digit must be added; e.g., an additional 0.0100 conductance in the first decade, so as to compensate for the 0.0100 no longer encoded in the second decade because of the action of the digital code converter element 100. If the 1 flip-flop 213 is off in the first decade, the additional digit may be added by simply changing the 1 flip-flop to its on state. Thus, connecting line 126 is connected directly to flip-flop 213 through or gate 219 so that a carry pulse will change the state of flip-flop 21 3.
However, the 1 flip-flop may already have been in the on state so that a carry pulse will cause it to change to the off state. In this case, a digit will not be added; rather, a digit will be subtracted whereby two digits must then be added. In like manner, the 4 flip-flop must be turned on when the carry pulse will change the state of flip-flop 213.
Element 99 of the digital code converter includes capacitors 266, 267 and 268. One plate of capacitor 266 is connected to the normally 0 output of flip-flop 211 through resistor 271 and connecting line 107; one plate of capacitor 267 is connected to the normally 0 output of flip-flop 212 through resistor 272 and connecting line 108; and one plate of capacitor 268 is connected to the normally 0 output of flip-flop 213 through resistor 273 and connecting line 109.
The purpose of the aforesaid capacitors 265-268 is to hold the voltage of the flip-flop outputs for some time after a flip-flop has changed state. When, for example, a flip-flop changes from zero to a positive voltage, charging current flows through the series resistor to the capacitor and it will take an interval of time before the voltage at the previously grounded capacitor plate approaches the final output value of the flip-flop. As carry pulses may appear only at intervals at least as long as four sequencing pulses, the time constant of a resistor capacitor combination may be fairly long. Although capacitors are shown as the storage devices, it will be apparent to those skilled in the art that other devices having similar characteristics may be substituted therefor.
The purpose of these R-C circuits will become clear in the following paragraphs.
Element 99 of the digital code converter further includes and gates 275, 276 and 277. And gate 275 is adapted for detecting the condition wherein a carry pulse is received when the 1 flip-flop 213 is already in the on condition. Then, the left-hand output of flipflop 213 was positive shortly before or at the time of the carry pulse so that the upper plate of capacitor 268 remains positive for some time after flip-flop 213 changes state. This positive voltage is connected to the input of and gate 275 through resistor 280; the positive carry pulse is also connected as an input to this and gate so that a positive pulse is gated through and gate 275 to the input of flip-flop 212, through or gate 218, to add two digits. Thus, although flip-flop 213 changes state immediately upon the occurrence of a carry pulse, the R-C circuit associated with this flip-flop (resistor 273 and capacitor 268) affords a means for reading the state of this flip-flop just prior to the carry pulse.
If the 1 and 2 flip-flops are both on when a carry pulse is received, it was noted heretofore that the 4 flip-flop must then be actuated and the 1 and 2 flipfiops turned off. For this purpose, and gate 276 is connected to the upper plate of capacitor 267 through resistor 281. Another input of and gate 276 is connected to the output of and gate 275. Since the upper plate of capacitor 267 will be positive when the 2 flip-flop was on just prior to the carry pulse, and the output of and gate 275 is positive when the 1 flipflop was on just prior to the carry pulse, the output of and gate 276 indicates that the 4 flip-flop should be actuated. Therefore, the output of an gate 276 is connected to the input of flip-flop 211 through or gate 217. The carry pulse directly turns off the flipflop 275 and the 2 flip-flop is turned off by the output of and gate 275.
If the 1, 2 and 4 flip-flops are all on when a carry pulse is received, the 8 flip-flop must be actuated and the 1, 2 and 4 flip-flops turned off. Accordingly, and gate 277 has an input connected to the upper plate of capacitor 266 through resistor 282 and another input connected to the output of and gate 276. The upper plate of capacitor 266 is positive when the 4 flip-flop 211 was on just prior to the carry pulse. The output of and gate 276 is positive. when the 2 and 1 flip-flops 212 and 213 were on just prior to the carry pulse. The output of and gate is therefore positive when the 1, 2 and 4 flip-flops were on just prior to the carry pulse. This output is connected to flipflop 210 through or gate 225, for turning this flip-flop on. The 1 flip-flop is turned off directly by the carry pulse; the 2 flip-flop is turned off by the output of and gate 275 and the 4 flip-flop is turned off by the output of and gate 276.
In the first decade the "8 flip-flop will never be on simultaneously with either or both the 2 or 4 flipfiops since this indicates a number above 0.9999 and above the range of a four decade, binary decimal coded converter.
Similarly, if the 8 and 1 flip-flops in the first decade are on simultaneously, a carry pulse will not be received over connecting line 126 since this also would indicate an analog input beyond the range of a f our decade analog to binary decimal coded converter. However, in the second decade, a carry pulse may be received from the third decade over connecting line 128 when the 8 and 1 flip-flops are both on since a carry may be sent to the next highest or first decimal decade. Thus, element of the digital code converter has an additional and gate 285 Which detects when the 8 and l flip-flops of the second decade are both on when a carry pulse is received.
And gate 285 has one input connected to the upper plate of capacitor 286 through resistor 287. The upper plate of this capacitor is also connected to the output of the 8 flip-flop 235 through resistor 288. The lower plate of capacitor is grounded.
A second input of an gate 285 is connected to the upper plate of capacitor 289 through resistor 290. The upper plate of this capacitor is connected to the output of the 1 flip-flop 238 through resistor 291. The R-C circuits 288, 286 and 291, 289 function in an identical manner to the R-C circuits in the digital code con verter element 99, allowing the state of the 8 and 1 flip-flops to be read just prior to a carry pulse. The upper plates of respective capacitors 286 and 289 are positive when the flip-flops 235 and 238 are on just prior to a carry pulse. Thus, a carry pulse from line 128 is applied to a third input of and gate 285 resulting in a positive pulse at the output of and gate 285 which turns off the 8 flip-flop 235 and applies a carry pulse to connecting line 126 through or gate 263. The l flip-flop 238 is turned off directly by the carry pulse on connecting line 128. A carry pulse from the third decimal decade therefore properly results in a carry pulse to the first decade when the second decade is already filled, i.e., already encodes a total of nine digits. The aforementioned circuitry subtracts nine digits in the second decade and adds one digit in the first decade.
The remaining circuitry in digital code converter element 100 is similar to previously described circuitry in the digital code converter element 99 and performs analogous functions.
Although a carry pulse may be received in the second decade when it totals nine digits (0.0900), a higher total will never be encoded just prior to receipt of a carry pulse. This operation is obviated by the additional circuitry in the element 100 which detects a total of greater than 0.0900 immediately upon termination of the encoding of the second decade (sequencing pulse T and immediately converts the encoded bits in the second decade to conform to the true 8-4-2-1 binary coded decimal code. This conversion is therefore entirely completed prior to the receipt at sequencing pulse T or T of a carry pulse from the third or fourth decades.
It may further be noted that a carry pulse from the third decade over connecting line 128 is connected immediately to the first decade over connecting line 126 when the second decade is filled; i.e., only an and gate is in series with the carry pulse and passage of the pulse does not wait any flip-flop actuation.
As shown in FIG. b, element 101 of the digital code converter 30 which is connected to the output of control element 93 is identical to element 100 heretofore described. Element 102 of the digital code converter which is connected to the output of control element 94 is also similar to element 100. Code conversion element 102 is connected to the lowest order decade. Since there are no succeeding decades, no carry signals are applied to element 102 with the result that, as shown, no storage and related and and or gates are required in this converter element.
THE OUTPUT REGISTER CIRCUITRY After an encoding cycle has been completed, the states of the control flip-flops in the sequencer and control device 17 are indicative of the magnitude of the analog input in true 8-4-2-1 binary coded decimal code. If this output could be read in a very brief period of time, the control flip-flops 210-213 in element 90 and analogous control flip-flops in elements 92, 93 and 94 could themselves be utilized as the output indicating means.
However, the output information may often be used in such a manner that a very quick reading is either impossible, or quite diflicult. Since a complete encoding requires 17 sequencing pulses (T -T it will be apparent that if an additional output storage means were provided, the output could be read over a period as long as 17 sequencing pulses. This additional time may be used, for example, to feed the information to a digital computer in serial form, one bit at a time.
In order to provide this storage function, an output register 33 is provided. This register has previously been shown in block diagram form in FIG. 3. FIG. 6 shows a detailed illustration of this register.
As shown in FIG. 6, element 135 of the output register includes and gates 300, 301, 302 and 303 each having an input connected to sequencing pulse T The output of each of these and gates is connected to the input of respective flip-flops 305, 306, 307 and 308 through respective or gates 310, 311, 312 and 313.
The flip-flops in each of the output register elements serve as a buffer register into which the information from the control portion of the sequencer and control is transferred at the end of an encoding cycle. Prior to a transfer, each of these registers must be cleared to erase the previously registered output. This clearing function may be performed at any convenient time during an encoding cycle; for example, in the register shown in FIG. 6, the sequencing pulse T is connected to each register for this purpose. In order that each flip-flop shall be in a zero state after the occurrence of this pulse, an output from the normally zero outputs of each of the flip-flops 305-308 is returned to a respective an gate 300-303 as an input. Therefore, the pulse T will be gated through only those and gates having an input connected to an on flip-flop; therefore, only the on flip-flops will change state upon application of pulse T The information in the control portion of the sequencer and control device is transferred to flip-flops 305-308 by lines 140-143. Thus, line 140 connects the normally zero or left-hand output of flip-flop 210 to the input of and gate 315 output register element 135. In like manner, connecting line 141 connects flip-flop 211 to and gate 316, connecting line 142 connects flip-flop 212 to 28 and gate 317, and connecting line 143 connects flipflop 213 to and gate 318.
Each of the and gates 315-318 serve to gate the information from the sequencer and control device 17 into the output register '33 at the end of the encoding cycle. For this purpose each of these and gates has an additional input to which sequencing pulse T is supplied, and an output connected to a respective buffer register flip-flop 305-308 through respective or gates 310-313.
Pulse T occurs in sequence just after all encoder decades have been actuated. This pulse transfers the information in the control element 90 by applying a positive pulse through each of the and gates 315-318 which are connected to on flip-flops in the control element. The positive pulses allowed through one or more of the and gates 315-318 cause the associated buffer register flip-flops to change to an on state thereby exactly duplicating in flip-flops 305-308 the state of the control flipfiops 210-213, i.e., the flip-flops in output register element 135 are in the identical 0 and 1 states as the control flip-flops in the first decade of the sequencer and control 17. This information will remain stored in the output register until sequencing pulse T of the immediately following analog-to-digital conversion cycle.
Elements 136, 137 and 138 of the output register 33 are shown in detail in FIG. 6. These elements are identical in structure and function to previously described element 135, element 136 affording a storage of the information in the second decade, element 137 affording an information storage for the third decade and element 138 affording an information storage for the fourth decade.
My invention therefore provides an analog-to-digital converter characterized by improved accuracy and high digitizing speeds. This unique invention recognizes that electronic systems have finite response times and accordingly, permits substantial errors to be made in the initial digitizing steps without affecting the accuracy of the final digitized output. Converters constructed according to my invention can therefore be operated at substantially higher operating speeds with improved accuracy.
Although the specific embodiment of my invention which has been shown and described in detail utilized the 8-4-2-1 BCD relationship, it is to be clearly understood that the same is by way of illustration and example only since it is apparent that a wide variety of digital codes may be utilized. In particular, several additional representative codes in which my invention may be encoded are tabulated in Tables II, IV and V hereinabove. Likewise, the particular structural details in the specific embodiment are by way of illustration and example only, the spirit and scope of my invention being limited only by the terms of the appended claims.
1. An analo-g-to-digital converter comprising a source of selectively addible, digitally weighted electn'cal quantities; control means connected to said source for selecting said digitally weighted quantities and providing an indication thereof; means for selectively generating error tolerance distributing electrical quantities; summing means for selectively adding said error tolerance distributing electrical quantities and said digitally weighted electrical quantities; comparator means for comparing the analog input signal with the output of said summing means to provide an output signal indicative of whether the analog input signal is larger or smaller than the output of said summing means; means connecting the output of said comparator means to said control means so that said control means selects those digitally weighted electrical quantities which cause the difference between said analog input signal and the output of said summing means to approach zero, said control means upon completion of a conversion cycle being adapted by its electrical quantity indication to manifest digital values corresponding in combination to the value of said analog input signal.
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131298228.32/warc/CC-MAIN-20150323172138-00261-ip-10-168-14-71.ec2.internal.warc.gz
|
CC-MAIN-2015-14
| 96,169 | 171 |
http://nrich.maths.org/public/leg.php?code=12&cl=2&cldcmpid=7237
|
math
|
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
What is the remainder when 2^2002 is divided by 7? What happens
with different powers of 2?
An investigation that gives you the opportunity to make and justify
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
Find a cuboid (with edges of integer values) that has a surface
area of exactly 100 square units. Is there more than one? Can you
find them all?
Can you work out the arrangement of the digits in the square so
that the given products are correct? The numbers 1 - 9 may be used
once and once only.
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
An environment which simulates working with Cuisenaire rods.
The clues for this Sudoku are the product of the numbers in adjacent squares.
Can you make square numbers by adding two prime numbers together?
Ben passed a third of his counters to Jack, Jack passed a quarter
of his counters to Emma and Emma passed a fifth of her counters to
Ben. After this they all had the same number of counters.
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
Investigate the smallest number of moves it takes to turn these
mats upside-down if you can only turn exactly three at a time.
The discs for this game are kept in a flat square box with a square
hole for each disc. Use the information to find out how many discs
of each colour there are in the box.
Given the products of adjacent cells, can you complete this Sudoku?
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
How many different shaped boxes can you design for 36 sweets in one
layer? Can you arrange the sweets so that no sweets of the same
colour are next to each other in any direction?
What happens if you join every second point on this circle? How
about every third point? Try with different steps and see if you
can predict what will happen.
The planet of Vuvv has seven moons. Can you work out how long it is
between each super-eclipse?
How many different sets of numbers with at least four members can
you find in the numbers in this box?
There are ten children in Becky's group. Can you find a set of
numbers for each of them? Are there any other sets?
Can you work out some different ways to balance this equation?
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
Complete the magic square using the numbers 1 to 25 once each. Each
row, column and diagonal adds up to 65.
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
Suppose we allow ourselves to use three numbers less than 10 and
multiply them together. How many different products can you find?
How do you know you've got them all?
Follow the clues to find the mystery number.
Have a go at balancing this equation. Can you find different ways of doing it?
When Charlie asked his grandmother how old she is, he didn't get a
straightforward reply! Can you work out how old she is?
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
A student in a maths class was trying to get some information from
her teacher. She was given some clues and then the teacher ended by
saying, "Well, how old are they?"
Number problems at primary level to work on with others.
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
Number problems at primary level that may require determination.
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
Can you find a way to identify times tables after they have been shifted up?
If you have only four weights, where could you place them in order
to balance this equaliser?
Can you complete this jigsaw of the multiplication square?
What is the lowest number which always leaves a remainder of 1 when
divided by each of the numbers from 2 to 10?
A mathematician goes into a supermarket and buys four items. Using
a calculator she multiplies the cost instead of adding them. How
can her answer be the same as the total at the till?
Got It game for an adult and child. How can you play so that you know you will always win?
The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed?
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
A game that tests your understanding of remainders.
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280763.38/warc/CC-MAIN-20170116095120-00217-ip-10-171-10-70.ec2.internal.warc.gz
|
CC-MAIN-2017-04
| 5,773 | 77 |
http://iagod.trickip.org/hepworth/What-Are-Mathematical-Problem-Solving-Skills.html
|
math
|
Mathematics as a Complex Problem-Solving Activity.
Successful mathematical problem solving depends upon many factors and skills with different characteristics. In fact one of the main difficulties in learning problem solving is the fact that many skills are needed for a learner to be an effective problem solver. Also, these factors and skills make the teaching of problem solving one of the most complex topic to teach. This paper will discuss.
Problem Solving Skills: Definition, Steps, and Examples.
The first article Mathematical Problem Solving in the Early Years pointed out that young children are natural problem setters and solvers: that is how they learn. This article suggests ways to develop children’s problem solving strategies and confidence. Problem solving is an important way of learning, because it motivates children to connect previous knowledge with new situations and to.
Evaluation of Students’ Mathematical Problem Solving.
A problem-solving approach is not only a way for developing students’ thinking, but it also provides a context for learning mathematical concepts. Problem-solving allows students to transfer what they have already learned to unfamiliar situations. A problem-solving approach provides a way for students to actively construct their ideas about mathematics and to take responsibility for their.
Enhancing Students ’ Mathematical Problem -Solving Skills.
Such motivation gives problem solving special value as a vehicle for learning new concepts and skills or the reinforcement of skills already acquired (Stanic and Kilpatrick, 1989, NCTM, 1989). Approaching mathematics through problem solving can create a context which simulates real life and therefore justifies the mathematics rather than treating it as an end in itself. The National Council of.
Mathematical Fluency and Problem-Solving and Reasoning.
Keywords: problem-solving strategies, visualisation techniques, bar model, mathematical problem-solving skills. INTRODUCTION Problem-solving is an important skill that one must have. Problem-solving in mathematics helps students to experience on how to solve daily life problems by applying their mathematical knowledge and skill. Word problem solving is one of the important components of.
Resources tagged with: Problem solving - NRICH.
In summary: Mathematical fluency skills help students think faster and more clearly, giving them the energy, attention and focus to tackle complex problem-solving and reasoning questions. The future needs problem-solvers with reasoning skills. But as education shifts its focus to the critical and creative angle of mathematics problems, we can’t lose sight of the abilities and skills that.
Problem Solving in Mathematics - ThoughtCo.
Mathematical Problem Solving in the Early Years: Developing Opportunities, Strategies and Confidence Age 3 to 7. In this article for Early Years practitioners, Dr Sue Gifford outlines ways to develop children's problem-solving strategies and confidence in problem solving. Using National Young Mathematicians' Award Tasks to Develop Problem-solving and Group-working Skills Age 7 to 11. This.
Problem Solving: Teaching and Learning Strategies.
Although underpinning knowledge is tested in its own right, problem solving is a core element of Functional Skills mathematics yet should not obscure or add additional mathematical complexity beyond the level of the qualification. Defining problem solving is a challenge but the attributes below are helpful. Not all (in fact often just one) of the listed attributes must be present in a single.
Mathematical Reasoning - Problem Solving Materials.
Problem-solving tasks develop mathematical skills and problem-solving tactics. These open-ended investigations for Reception or Early Years settings are designed to take advantage of outdoor learning environments, but many of them can be adapted to run inside. Nick's Guidance Session 1 Shape Session plan Download all files. Objectives. Open-ended investigative tasks provide fun, stimulating.
Transferable Skills in the Mathematical Tripos.
The Mathematical Problem Solving course is an elective course in the elementary education major curriculum. Its syllabus was developed based on the future teachers’ mathematical skills and the needs of their pupils with the goal of developing their skills in the areas of problem solving and problem posing. The focused set of course materials combines relevant concepts in a pedagogically rich.
On Teaching Mathematical Problem-Solving and Problem Posing.
Mathematical Problem Solving. The goal of any mathematics course is to introduce you to a variety of methods used to solve real-world problems. In this way, you are able to practice solving problems and, along the way, improve your critical thinking skills. It may seem that solving a mathematical problem involves nothing more than knowing the right formula and following the correct steps.
How to promote problem solving in the early years.
Developing problem solving skills is an area of development early years’ practitioners are familiar with, and the importance of developing these problem solving skills is well known, but what exactly are problem solving skills? And how do we encourage children to develop these? Within the Early Years Foundation Stage, problem solving comes under the category Mathematical Development, however.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587854.13/warc/CC-MAIN-20211026072759-20211026102759-00491.warc.gz
|
CC-MAIN-2021-43
| 5,375 | 24 |
https://trendyanswers.com/category/mathmatics/
|
math
|
What is the difference between $, $$ and $x symbols? This question is a common question and people often ask for clarification because they want to know if there is a big difference in using these two symbols. In this post, we’ll explore the history behind these two symbols and explain how they differ, so ..
Category : Mathmatics
What is cos30 degree? In trigonometry, the cosine function can be defined by the proportion of the adjacent side in relation to the hypotenuse. When the angle of the right triangle is 30 degrees, the cosine value for that angle, i.e., that of Cos 30 degree is given in the form of a fraction, 3/2. ..
An exact form is usually a sign that we can see some irrational element with the solution. Therefore, we should radically keep answers…. or make the answer appear as a number multiple of pi, for instance. Read more: Is 0870 A Premium Number? What is the exact form in simple words? A fraction can be ..
XV in Numbers: “XV” Roman Numerals could be read as numbers combing those transformed numerals i.e.XV = X + VXV= 10 + 5XV= 15The higher roman numerals precede the lower ones, which results in the proper translation of the XV Roman numerals. Read More: How many electrons valence in the SF4 Lewis structure? What is ..
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00304.warc.gz
|
CC-MAIN-2023-14
| 1,261 | 5 |
http://www.wyzant.com/resources/answers/12313/what_is_distributive_property
|
math
|
6x+3y+x i need help with this problem. i am studying for an exam tomorrow please help !!
What is distributive property?
The distributive property is easy to show in a simple example which, if used, allows people to do reasonably complicated computations in their head without any need to write them down.
Let's say someone asks you to compute 15*22. Well you can treat this as
15*22 = 15*(20+2) = 15*20+15*2 = 300+30 = 330
What I put in bold is what you would call the distributive property of multiplication with respect to addition.
More in general, using letterals,
a*(b+c) = ab+ac
In your problem, did you mean (6x+3y)*x ? In that case it would be 6x*x + 3y*x = 6x2 + 3xy
6(x-y) - 3(2x-3y) (distributive property)
= 6x-6y-6x+9y ( 6x and -6x cancel each other out)
= -6y+9y ( add)
The above relation is algebraic way of stating the distributive law or distributive property. In words, this can be stated as follows:
The result of adding two numbers and multiplying the result by a third number will be the same value as multiplying each of the first two numbers with the third number and adding them together. This law sometimes provides an easy way to mentally multiply two numbers. Given the distributive law, I can do the following multiplication by distribution:
Using the distributive law again on the first term (27*40), the above expression becomes
27*41=20*40+7*40+27*1=800+280+27=1107. Distributive law is applicable to division as well
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011335666/warc/CC-MAIN-20140305092215-00046-ip-10-183-142-35.ec2.internal.warc.gz
|
CC-MAIN-2014-10
| 1,448 | 16 |
https://www.intechopen.com/books/numerical-simulation-from-theory-to-industry/on-new-high-order-iterative-schemes-for-solving-initial-value-problems-in-epidemiology
|
math
|
Most problems arising from mathematical epidemiology are often described in terms of differential equations. However, it is often very difficult to obtain closed form solutions of such equations, especially those that are nonlinear. In most cases, attempts are made to obtain only approximate or numerical solutions. In this work, we revisit the SIR epidemic model with constant vaccination strategy that was considered in , where the Adomian decomposition method was used to solve the governing system of nonlinear initial value differential equations.
In this work we develop new accurate iterative schemes which are based on extending Taylor series based linearization method to obtain accurate and fast converging sequence of hybrid iteration schemes. At first order, the hybrid iteration scheme reduces to quasilinearization method (QLM) which was originally developed in . More recently Mandelzweig and his co-workers [8–10] have extended the application of the QLM to a wide variety of nonlinear BVPs and established that the method converges quadratically. In this work we demonstrate that the proposed hybrid iteration schemes are more accurate and converge faster than the QLM approach.
To implement the method we consider the SIR model that describes the temporal dynamics of a childhood disease in the presence of a preventive vaccine. In SIR models the population is assumed to be divided into the standard three classes namely, the susceptibles (S), who can catch the infection but are so far uninfected, the infectives (I), those who have the disease and can transmit it to the susceptibles, and the removed (R), who have either died or who have recovered and are therefore immune.
The governing equations for the problem are described by
whereS(t), I(t) and R(t) denote the susceptibles, infectives and the removed classes respectively.
The total population is denoted by N = S + I + R, μ is the death rate, P is the fraction of citizens vaccinated at birth each year, β is the average contact rate, π is the constant birth rate, and κ is the rate at which an individual recovers from the disease and enters the removed group which also contains vaccinated individuals. Equations (1 - 3) are solved using the new hybrid iteration schemes and the results are compared with results from the Runge-Kutta MATLAB in-built solver ode45.
2. Numerical solution
To simplify the formulation of the solution, equations (1) - (3) are scaled by dividing by N. We define new variables z1 = S/N, z2 = I/N and z3 = R/N. This leads to z1 + z2 + z3 = 1 and if we assume that π = μ, the scaled new system becomes
wheres0 and i0 are given constants. The solution for z3(t) can be obtained from z3 = 1−z1 −z2.
Previous studies [4–7, 12] have shown that the long term behaviour of systems like (4) - (5) can be classified into two categories namely, endemic or eradication. From the long term behaviour of z1(t) and z2(t) it holds that the solution asymptotically approaches a disease free equilibrium (DFE) or the endemic equilibrium (EE) where
Here Rv, the vaccination reproduction number, is the threshold that determines the stability of the equilibria and is defined by
It was shown in that the DFE is locally stable if Rv<1 and the EE is locally stable provided 1 <Rv≤ 4(κ + π)/π. In this work, we use develop new iteration schemes to solve the system (4) - (5) using parameters that yield both the DFE and EE.
3. Method of solution
To develop the method of solution, we assume that the true solution of (4 - 5) is zs,α(s = 1, 2) and zs,γare the initial approximations. We introduce the following coupled system,
This idea of introducing the coupled equations of the form (9-10) have previously been used in the construction of Newton-like iteration formulae for the computation of the solutions of nonlinear equations of the form f (x) = 0.
We write equation (9) as
We use the quasilinearization method (QLM) of Bellman and Kalaba to solve equation (13). The QLM determines the (i+ 1)th iterative approximation zj,i+1 as the solution of the differential equation
which can be written as
We assume that zj,0is obtained as a solution of the linear part of equation (13) given by
which yields the iteration scheme
We note that equation (19) is the standard QLM iteration scheme for solving (4 - 5).
When i= 0 in (16) we can approximate zjas
Thus, setting i= 0 in (16) we obtain
which yields the iteration scheme
whereis the solution of
The general iteration scheme obtained by setting i= m (m ≥ 2) in equation (16), hereinafter referred to as scheme-m is
whereis obtained as the solution of
The iteration schemes (19),(24 - 25) can be solved numerically using standard methods such as finite difference, finite elements, spline collocation methods,etc. In this study we use the Chebyshev spectral collocation method to solve the iteration schemes. For brevity, we omit the details of the spectralmethods, and refer interested readers to ([2, 13]). Before applying the spectral method, it is convenient to transform the domain on which the governing equation is defined to the interval [-1,1] on which the spectral method can be implemented. We use the transformation t = tF(τ + 1)/2 to map the interval [0, tF] to [-1,1], where tFis a finite time. The basic idea behind the spectral collocation method is the introduction of a differentiation matrix D which is used to approximate the derivatives of the unknown variables z at the collocation points as the matrix vector product
whereN + 1 is the number of collocation points (grid points), D = 2D/tF, and Z = [z(τ0), z(τ1),..., z(τN)]Tis the vector function at the collocation points τj.
Applying the Chebyshev spectral method to (19), for instance, gives
andI is an (N + 1) × (N + 1) identity matrix.
4. Results and discussion
In this section we present the results of solving the governing equations (4-5) using the iteration scheme-m. For illustration purposeswe present the results for m = 0, 1, 2 to illustrate the effect of increasing m in the accuracy and convergence of the iteration schemes. The number of collocations points in all the results presented here is N = 40. In order to assess the accuracy of the proposed method, the present numerical results were compared against results generated using the MATLAB initial value solver ode45. In the numerical simulations presented here, following , the governing parameters were carefully selected in order to represent the cases which give rise to both the disease free equilibrium (DFE) and endemic equilibrium (EE).We consider the following cases
1. Case 1: s0 = 1, i0 = 0, β = 0.8, κ = 0.03, π = 0.4, P = 0.9.
In this case we observe that Rv= 0.186 <1, hence we expect the disease to be eradicated from the population after some time.
2. Case 2: s0 = 0.8, i0 = 0.2 β = 0.8, κ = 0.03, π = 0.4, P = 0.9.
In this case we observe that Rv= 0.186 <1 and as expected, using these parameters, the disease should be eradicated as t → ∞.
3. Case 3: s0 = 0.8, i0 = 0.2 β = 0.8, κ = 0.03, π = 0.4, P = 0.
In this caseRv= 1.86 >1 which leads to the endemic equilibrium (no disease eradication).
4. Case 4: s0 = 0.8, i0 = 0.2 β = 0.8, κ = 0.03, π = 0.4, P = 0.3.
In this case Rv= 1.32 >1 which leads to the endemic equilibrium (no disease eradication).
The results for Case 1 are shown on Figs. 1 - 2. In this case, the initial guess and the first few iterations match the numerical solution all the iterative schemes in the plots of s(t), r(t). We observe that s(t) decreases monotonically with time while r(t) increases with time. The graph of the profile for i(t) is not shown because i(t) = 0 in this particular case.
Figs. 3 - 5 show the numerical approximation of the profiles of the different classes for Case 2. Again, all the iterative schemes rapidly converge to the numerical solution. The population of the susceptibles decreases with time and that of the removed (those recovered with immunity) increases with time. The infected population initially increases and reaches a maximum, then gradually decreases to zero as t → ∞.
Figs. 6 - 8 show the numerical approximation of the profiles of the different classes for Case 3. It can be noted from the graphs that the Scheme-2 converges fastest towards the numerical results. Only 10 iterations are required for full convergence in Scheme-2 compared to 14 iterations in Scheme-1 and 28 iterations in Scheme-1.
Figs. 8 - 11 shows the variation all the population groups with time for Case 4. Again, we observe that Scheme-2 converges fastest towards the numerical results. Only 5 iterations are required for full convergence in Scheme-2 compared to 6 iterations in Scheme-1 and 12 iterations in Scheme-1.
In this work, a sequence of new iteration schemes for solving nonlinear differential equations is used to solve the SIR epidemic model with constant vaccination strategy. The proposed iteration schemes are derived as an extension to the quasi-linearization method to obtain hybrid iteration schemes which converge very rapidly. The accuracy and validity of the proposed schemes is confirmed by comparing with the ode45 MATLAB routine for solving initial value problems. It is hoped that the proposed method of solution will spawn further interest in computational analysis of differential equations in epidemiology and other areas of science.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655889877.72/warc/CC-MAIN-20200705215728-20200706005728-00048.warc.gz
|
CC-MAIN-2020-29
| 9,357 | 46 |
http://www.edurite.com/kbase/answers-to-study-island-math-questions
|
math
|
answers to study island math questions
Best Results From
Yahoo Answers Youtube
From Yahoo Answers
Question:I am an idiot at math and my daughter needs help... What are the answers to these math questions?
1.What is 3/4 equal to? a)9/12 b)4/7 c)2/1 d)not here
2.What is equal to 8/4? a)32/16 b)1/2 c)16/4 d) not here
3.What is equal to 12/20? a)3/20 b)2/12 c) 6/10 d) not here
4.What is equal to 22/88? a) 88/22 b)2/11 c)8/2 d)not here
5.Which is both 6/12 & 10/20 equal to? a)20/240 b) 12/32
c)1/2 d)not here
6)Which is both 3/4 & 9/2 equal to? a)12/48 b) 75/100 c) 3/16
d) not here
So, please help me with these questions, she is having a hard time understanding. Haha, I a just horrible at math aren't I? Thanks you guys. The ONLY thing that kept me in the Gifted program in 6th grade for was for English, Reading, Science, and Social Studies. So I was and am not completely dumb! :-]
Answers:Theses are cake questions. I sucked at high school math
Question:I answered some of my math study guide stuff, i just need help with the rest of these questions:
2. The anaconda was 288 inches long. How many feet is 288 inches?
3. What time is 2&1/2 hours after 10:15 a.m.?
4. Compare 5/8 (>, <, =) 2/3
5. 6&1/4 - 5&5/8
6. 3/4 * 2/5
7. 3/4 \ 2/5
8. How much money is 60% of $45?
9. Find the perimeter of this rectangle: 1&1/8 in. + 3/4 in.
10. How many vertices does a triangular prism have?
11. A ton is 2000 pounds. How many pounds is 2&1/2 tons?
12. The paper cup would not roll straight. One end was 7 cm in diameter, and the other end was 5 cm in diameter. In one roll of the cup,
(a) how far would the larger end roll?
(b) how far would the smaller end roll?
(Round each answer to the nearest centimeter)
13. Jefferson got a hit 30% of the 240 times he went to bat during the season. Write 30% as a reduced fraction. Then find the number of hits Jefferson got during the season.
14. Jena has run 11.5 miles of a 26/2-mile race. Find the remaining distance Jena has to run by solving this equeation:
11.5 mi + d = 26.2 mi
15. The sales-tax rate was 7%. The two CDs cost $15.49 each. What was the total cost of the two CDs including tax?
Answers:2. 1 foot = 12 in so 288 /12
3. 10 + 2 = 12 o'clock 15 mins + 30 mins is 45 min so answer: 12:45 pm
4. (5/8) < (2/3)
6. 6/20 or 3/10
8. 45 * .6 = $27
11. 5000 pounds
Question:17. Expressions, equations, and inequalities use similar algebraic symbols, but they have differences. Compare and contrast the meaning and uses of expressions, equations, and inequalities.
19. How can you determine when and where two things (people, trains, rabbits, etc) in a "race" will meet if they have different starting points and different rates? What happens if they have the same rate (m)? What happens if they have the same starting point (b)?
20. What are some of the basic rules and procedures used to solve algebraic equations that involve one variable?
21. What does a variable have one value? Many values? Give examples.
These are questions for a FRESHMAN (they are extremely hard, aren't they?) summer math project and I would really appreciate it if you could help me answer these last few. THANK YOU SO MUCH!!
Answers:21. A variable will have one value for a linear equation (i.e. no squared terms or higher). Something like x + 9 = 11 will only give one value. A variable will have more than one value for any other equation or inequality (x^2 - 5x + 4 = 0, or x + 3 > 99 will give more than one value for x).
Question:what is a site that you can type any math question in and it gives you the answer? i want to check my paper.
Answers:there is none but there is this website called study island and its only a website for pre-k through eighth grade. You get to pick a subject like math as you infer.
Study Island :Ms. Moore and Mr. Marsh help 7th grade math students use Study Island, December 15, 2009 . Visit our website www.dmscougar-roarrr.com
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00411.warc.gz
|
CC-MAIN-2017-39
| 3,888 | 49 |
https://saylordotorg.github.io/text_introduction-to-economic-analysis/s12-investment.html
|
math
|
The distinction between the short-run supply and the long-run supply is governed by the time that investment takes. Part of the difference between the short-run demand and the long-run demand arises because we don’t scrap capital goods—cars, fridges, and air conditioners—in response to price changes. In both cases, investment is an important component of the responsiveness of supply and demand. In this section, we take a first look at investment. We will take a second look at investment from a somewhat different perspective later when we consider basic finance tools near the end of the book. Investment goods require expenditures today to produce future value, so we begin the analysis by examining the value of future payments.
The promise of $1 in the future is not worth $1 today. There are a variety of reasons why a promise of future payments is not worth the face value today, some of which involve risk that the money may not be paid. Let’s set aside such risk for the moment. Even when the future payment is perceived to occur with negligible risk, most people prefer $1 today to $1 payable a year hence. One way to express this is by the present valueValue in today’s dollars of a stream of future payments.: The value today of a future payment of a dollar is less than a dollar. From a present value perspective, future payments are discountedThe practice of adjusting a sequence of payments to account for the interest these payments would earn in financial markets..
From an individual perspective, one reason that one should value a future payment less than a current payment is due to arbitrageBuying and reselling to make a profit..Arbitrage is the process of buying and selling in such a way as to make a profit. For example, if wheat is selling for $3 per bushel in New York but $2.50 per bushel in Chicago, one can buy in Chicago and sell in New York, profiting by $0.50 per bushel minus any transaction and transportation costs. Such arbitrage tends to force prices to differ by no more than transaction costs. When these transaction costs are small, as with gold, prices will be about the same worldwide. Suppose you are going to need $10,000 one year from now to put a down payment on a house. One way of producing $10,000 is to buy a government bond that pays $10,000 a year from now. What will that bond cost you? At current interest rates, a secure bondEconomists tend to consider U.S. federal government securities secure, because the probability of such a default is very, very low. will cost around $9,700. This means that no one should be willing to pay $10,000 for a future payment of $10,000 because, instead, one can have the future $10,000 by buying the bond, and will have $300 left over to spend on cappuccinos or economics textbooks. In other words, if you will pay $10,000 for a secure promise to repay the $10,000 a year hence, then I can make a successful business by selling you the secure promise for $10,000 and pocketing $300.
This arbitrage consideration also suggests how to value future payments: discount them by the relevant interest rate.
Example (Auto loan): You are buying a $20,000 car, and you are offered the choice to pay it all today in cash, or to pay $21,000 in one year. Should you pay cash (assuming you have that much in cash) or take the loan? The loan is at a 5% annual interest rate because the repayment is 5% higher than the loan amount. This is a good deal for you if your alternative is to borrow money at a higher interest rate; for example, on (most) credit cards. It is also a good deal if you have savings that pay more than 5%—if buying the car with cash entails cashing in a certificate of deposit that pays more than 5%, then you would be losing the difference. If, on the other hand, you are currently saving money that pays less than 5% interest, paying off the car is a better deal.
The formula for present value is to discount by the amount of interest. Let’s denote the interest rate for the next year as r1, the second year’s rate as r2, and so on. In this notation, $1 invested would pay $(1 + r1) next year, or $(1 + r1) × (1 + r2) after 2 years, or $(1 + r1) × (1 + r2) × (1 + r3) after 3 years. That is, ri is the interest rate that determines the value, at the end of year i, of $1 invested at the start of year i. Then if we obtain a stream of payments A0 immediately, A1 at the end of year one, A2 at the end of year two, and so on, the present value of that stream is
Example (Consolidated annuities or consolsBonds that pay a fixed amount in perpetuity (forever).): What is the value of $1 paid at the end of each year forever, with a fixed interest rate r? Suppose the value is v. ThenThis development uses the formula that, for -1 < a < 1, …, which is readily verified. Note that this formula involves an infinite series.
At a 5% interest rate, $1 million per year paid forever is worth $20 million today. Bonds that pay a fixed amount every year forever are known as consols; no current government issues consols.
Example (Mortgages): Again, fix an interest rate r, but this time let r be the monthly interest rate. A mortgage implies a fixed payment per month for a large number of months (e.g., 360 for a 30-year mortgage). What is the present value of these payments over n months? A simple way to compute this is to use the consol value, because
Thus, at a monthly interest rate of 0.5%, paying $1 per month for 360 months produces a present value M of . Therefore, to borrow $100,000, one would have to pay per month. It is important to remember that a different loan amount just changes the scale: Borrowing $150,000 requires a payment of per month, because $1 per month generates $166.79 in present value.
Example (Simple and compound interest): In the days before calculators, it was a challenge to actually solve interest-rate formulas, so certain simplifications were made. One of these was simple interestAn approximate interest rate that is calculated by dividing the annual interest rate by the number of periods in one year., which meant that daily or monthly rates were translated into annual rates by incorrect formulas. For example, with an annual rate of 5%, the simple interest daily rate is The fact that this is incorrect can be seen from the calculation which is the compound interestThe correct rate of interest that is calculated by adding together the cumulative interest payments earned during one year. calculation. Simple interest increases the annual rate, so it benefits lenders and harms borrowers. (Consequently, banks advertise the accurate annual rate on savings accounts—when consumers like the number to be larger—but not on mortgages, although banks are required by law to disclose, but not to advertise widely, actual annual interest rates on mortgages.)
Example (Obligatory lottery): You win the lottery, and the paper reports that you’ve won $20 million. You’re going to be paid $20 million, but is it worth $20 million? In fact, you get $1 million per year for 20 years. However, in contrast to our formula, you get the first million right off the bat, so the value is
Table 11.1 "Present value of $20 million" computes the present value of our $20 million dollar lottery, listing the results in thousands of dollars, at various interest rates. At 10% interest, the value of the lottery is less than half the “number of dollars” paid; and even at 5%, the value of the stream of payments is 65% of the face value.
Table 11.1 Present value of $20 million
The lottery example shows that interest rates have a dramatic impact on the value of payments made in the distant future. Present value analysis is the number one tool used in MBA programs, where it is known as net present value, or NPVThe present value of a stream of net payments., analysis. It is accurate to say that the majority of corporate investment decisions are guided by an NPV analysis.
Example (Bond prices): A standard Treasury billA bond that has a fixed redemption value. has a fixed future value. For example, it may pay $10,000 in one year. It is sold at a discount off the face value, so that a one-year, $10,000 bond might sell for $9,615.39, producing a 4% interest rate. To compute the effective interest rate r, the formula relating the future value FV, the number of years n, and the price is
We can see from either formula that Treasury bill prices move inversely to interest rates—an increase in interest rates reduces Treasury prices. Bonds are a bit more complicated. Bonds pay a fixed interest rate set at the time of issue during the life of the bond, generally collected semiannually, and the face value is paid at the end of the term. These bonds were often sold on long terms, as much as 30 years. Thus, a three-year, $10,000 bond at 5% with semiannual payments would pay $250 at the end of each half year for 3 years, and pay $10,000 at the end of the 3 years. The net present value, with an annual interest rate r, is
The net present value will be the price of the bond. Initially, the price of the bond should be the face value, since the interest rate is set as a market rate. The U.S. Treasury quit issuing such bonds in 2001, replacing them with bonds in which the face value is paid and then interest is paid semiannually.
A simple investment project requires an investment, I, followed by a return over time. If you dig a mine, drill an oil well, build an apartment building or a factory, or buy a share of stock, you spend money now, hoping to earn a return in the future. We will set aside the very important issue of risk until the next subsection, and ask how one makes the decision to invest.
The NPV approach involves assigning a rate of return r that is reasonable for the specific project and then computing the corresponding present value of the expected stream of payments. Since the investment is initially expended, it is counted as negative revenue. This yields an expression that looks like
where R1 represents first-year revenues, R2 represents second-year revenues, and so on.The most common approach treats revenues within a year as if they are received at the midpoint, and then discounts appropriately for that mid-year point. The present discussion abstracts from this practice. The investment is then made when NPV is positive—because this would add to the net value of the firm.
Carrying out an NPV analysis requires two things. First, investment and revenues must be estimated. This is challenging, especially for new products where there is no direct way of estimating demand, or with uncertain outcomes like oil wells or technological research.The building of the famed Sydney Opera House, which looks like billowing sails over Sydney Harbor in Australia, was estimated to cost $7 million and actually cost $105 million. A portion of the cost overrun was due to the fact that the original design neglected to install air conditioning. When this oversight was discovered, it was too late to install a standard unit, which would interfere with the excellent acoustics, so instead an ice hockey floor was installed as a means of cooling the building. Second, an appropriate rate of return must be identified. The rate of return is difficult to estimate, mostly because of the risk associated with the investment payoffs. Another difficulty is recognizing that project managers have an incentive to inflate the payoffs and minimize the costs to make the project appear more attractive to upper management. In addition, most corporate investment is financed through retained earnings, so that a company that undertakes one investment is unable to make other investments, so the interest rate used to evaluate the investment should account for opportunity cost of corporate funds. As a result of these factors, interest rates of 15%–20% are common for evaluating the NPV of projects of major corporations.
Example (Silver mine): A company is considering developing a silver mine in Mexico. The company estimates that developing the mine requires building roads and opening a large hole in the ground, which would cost $4 million per year for 4 years, during which time the mines generates zero revenue. Starting in year 5, the expenses would fall to $2 million per year, and $6 million in net revenue would accrue from the silver that is mined for the next 40 years. If the company cost of funds is 18%, should it develop the mine?
The earnings from the mine are calculated in the table below. First, the NPV of the investment phase during years 0, 1, 2, and 3 is
A dollar earned in each of years 4 through 43 has a present value of
The mine is just profitable at 18%, in spite of the fact that its $4 million payments are made in 4 years, after which point $4 million in revenue is earned for 40 years. The problem in the economics of mining is that 18% makes the future revenue have quite modest present values.
|Year||Earnings ($M)/yr||PV ($M)|
There are other approaches for deciding to take an investment. In particular, the internal rate of return (IRR)Method of analyzing an investment that solves the equation NPV = 0 for the interest rate. approach solves the equation NPV = 0 for the interest rate. Then the project is undertaken if the rate of return is sufficiently high. This approach is flawed because the equation may have more than one solution—or no solutions—and the right thing to do in these events is not transparent. Indeed, the IRR approach gets the profit-maximizing answer only if it agrees with NPV. A second approach is the payback period, which asks calculates the number of years a project must be run before profitability is reached. The problem with the payback period is deciding between projects—if I can only choose one of two projects, the one with the higher NPV makes the most money for the company. The one with the faster payback may make a quite small amount of money very quickly, but it isn’t apparent that this is a good choice. When a company is in risk of bankruptcy, a short payback period might be valuable, although this would ordinarily be handled by employing a higher interest rate in an NPV analysis. NPV does a good job when the question is whether or not to undertake a project, and it does better than other approaches to investment decisions. For this reason, NPV has become the most common approach to investment decisions. Indeed, NPV analysis is more common than all other approaches combined. NPV does a poor job, however, when the question is whether to undertake a project or to delay the project. That is, NPV answers “yes or no” to investment, but when the choice is “yes or wait,” NPV requires an amendment.
Risk has a cost, and people and corporations buy insurance against financial risk.For example, NBC spent $6 million to buy an insurance policy against U.S. nonparticipation in the 1980 Moscow Summer Olympic Games—and the United States didn’t participate (because of the Soviet invasion of Afghanistan)—and NBC was paid $94 million from the policy. The standard approach to investment under uncertainty is to compute an NPV, using a “risk-adjusted” interest rate to discount the expected values received over time. The interest rate is increased or lowered depending on how risky the project is.
For example, consider a project like oil exploration. The risks are enormous. Half of all underwater tracts in the Gulf Coast near Louisiana and Texas that are leased are never drilled, because they turn out to be a bad bet. Half of all the tracts that are drilled are dry. Hence, three quarters of the tracts that are sold produce zero or negative revenue. To see how the economics of such a risky investment might be developed, suppose that the relevant rate of return for such investments is 18%. Suppose further that the tract can be leased for $500,000 and the initial exploration costs $1 million. If the tract has oil (with a 25% probability), it produces $1 million per year for 20 years and then runs dry. This gives an expected revenue of $250,000 per year. To compute the expected net present value, we first compute the returns:
Table 11.2 Oil tract return
At 18%, the investment is a loss—the risk is too great given the average returns.
A very important consideration for investment under uncertainty is the choice of interest rate. It is crucial to understand that the interest rate is specific to the project and not to the investor. This is perhaps the most important insight of corporate finance generally: The interest rate should adjust for the risk associated with the project and not for the investor. For example, suppose hamburger retailer McDonald’s is considering investing in a cattle ranch in Peru. McDonald’s is overall a very low-risk firm, but this particular project is quite risky because of local conditions. McDonald’s still needs to adjust for the market value of the risk it is undertaking, and that value is a function of the project risk, not the risk of McDonald’s other investments.
This basic insight of corporate financeThe study of funding of operations of companies. (the study of funding of operations of companies)—the appropriate interest rate is determined by the project, not by the investor—is counterintuitive to most of us because it doesn’t apply to our personal circumstances. For individuals, the cost of borrowing money is mostly a function of our own personal circumstances, and thus the decision of whether to pay cash for a car or borrow the money is not so much a function of the car that is being purchased but of the wealth of the borrower. Even so, personal investors borrow money at distinct interest rates. Mortgage rates on houses are lower than interest rates on automobiles, and interest rates on automobiles are lower than on credit cards. This is because the “project” of buying a house has less risk associated with it: The percentage loss to the lender in the event of borrower default is lower on a house than on a car. Credit cards carry the highest interest rates because they are unsecured by any asset.
One way of understanding why the interest rate is project-specific, but not investor-specific, is to think about undertaking the project by creating a separate firm to make the investment. The creation of subsidiary units is a common strategy, in fact. This subsidiary firm created to operate a project has a value equal to the NPV of the project using the interest rate specific to the subsidiary, which is the interest rate for the project, independent of the parent. For the parent company, owning such a firm is a good thing if the firm has positive value, but not otherwise.It may seem that synergies between parent and subsidiary are being neglected here, but synergies should be accounted for at the time they produce value—that is, as part of the stream of revenues of the subsidiary.
Investments in oil are subject to another kind of uncertainty: price risk. The price of oil fluctuates. Moreover, oil pumped and sold today is not available for the future. Should you develop and pump the oil you have today, or should you hold out and sell in the future? This question, known as the option value of investment, is generally somewhat challenging and arcane, but a simple example provides a useful insight. An optionThe right to buy or sell at a price determined in advance. is the right to buy or sell at a price determined in advance.
To develop this example, let’s set aside some extraneous issues first. Consider a very simple investment, in which either C is invested or not.This theory is developed in striking generality by Avinash Dixit and Robert Pindyck, Investment Under Uncertainty, Princeton University Press, 1994. If C is invested, a value V is generated. The cost C is a constant; it could correspond to drilling or exploration costs or, in the case of a stock option, the strike priceThe amount one pays to obtain a share of stock. of the option, which is the amount one pays to obtain the share of stock. The value V, in contrast, varies from time to time in a random fashion. To simplify the analysis, we assume that V is uniformly distributed on the interval [0, 1], so that the probability of V falling in an interval [a, b] is (b – a) if 0 ≤a ≤ b ≤ 1. The option only has value if C < 1, which we assume for the rest of this section.
The first thing to note is that the optimal rule to make the investment is a cutoff value—that is, to set a level V0 and exercise the option if, and only if, V ≥ V0. This is because—if you are willing to exercise the option and generate value V—you should be willing to exercise the option and obtain even more value. The NPV rule simply says V0 = C; that is, invest whenever it is profitable. The purpose of the example developed below is to provide some insight into how far wrong the NPV rule will be when option values are potentially significant.
Now consider the value of option to invest, given that the investment rule V ≥ V0 is followed. Call this option value J(V0). If the realized value V exceeds V0, one obtains V – C. Otherwise, one delays the investment, producing a discounted level of the same value. This logic says,
This expression for J(V0) is explained as follows. First, the hypothesized distribution of V is uniform on [0, 1]. Consequently, the value of V will exceed V0 with probability 1 – V0. In this event, the expected value of V is the midpoint of the interval [V0, 1], which is ½(V0 + 1). The value ½(V0 + 1) – C is the average payoff from the strategy of investing whenever V ≥ V0, which is obtained with probability 1 – V0. Second, with probability V0, the value falls below the cutoff level V0. In this case, no investment is made and, instead, we wait until the next period. The expected profits of the next period are J(V0), and these profits are discounted in the standard way.
The expression for J is straightforward to solve:
Rudimentary calculus shows
First, note that and which together imply the existence of a maximum at a value V0 between C and 1, satisfying Second, the solution occurs at
The positive root of the quadratic has V0 > 1, which entails never investing, and hence is not a maximum. The profit-maximizing investment strategy is to invest whenever the value exceeds V0 given by the negative root in the formula. There are a couple of notable features about this solution. First, at r = 0, V0 = 1. This is because r = 0 corresponds to no discounting, so there is no loss in holding out for the highest possible value. Second, as r → ∞, V0 → C. As r → ∞, the future is valueless, so it is worth investing if the return is anything over costs. These are not surprising findings, but rather quite the opposite: They should hold in any reasonable formulation of such an investment strategy. Moreover, they show that the NPV rule, which requires V0 = C, is correct only if the future is valueless.
How does this solution behave? The solution is plotted as a function of r, for C = 0, 0.25, and 0.5, in Figure 11.1 "Investment strike price given interest rate ".
The horizontal axis represents interest rates, so this figure shows very high interest rates by current standards, up to 200%. Even so, V0 remains substantially above C. That is, even when the future has very little value, because two-thirds of the value is destroyed by discounting each period, the optimal strategy deviates significantly from the NPV strategy. Figure 11.2 "Investment strike price given interest rate " shows a close-up of this graph for a more reasonable range of interest rates, for interest rates of 0%–10%.
Figure 11.1 Investment strike price given interest rate r in percent
Figure 11.2 "Investment strike price given interest rate " shows the cutoff values of investment for three values of C, the cost of the investment. These three values are 0 (lowest curve), 0.25 (the middle, dashed curve), and 0.5 (the highest, dotted line). Consider the lowest curve, with C = 0. The NPV of this project is always positive—there are no costs, and revenues are positive. Nevertheless, because the investment can only be made once, it pays to hold out for a higher level of payoff; indeed, for 65% or more of the maximum payoff. The economics at an interest rate of 10% is as follows. By waiting, there is a 65% chance that 10% of the potential value of the investment is lost. However, there is a 35% chance of an even higher value. The optimum value of V0 trades these considerations off against each other.
For C = 0.25, at 10% the cutoff value for taking an investment is 0.7, nearly three times the actual cost of the investment. Indeed, the cutoff value incorporates two separate costs: the actual expenditure on the investment C and the lost opportunity to invest in the future. The latter cost is much larger than the expenditure on the investment in many circumstances and, in this example, can be quantitatively much larger than the actual expenditure on the investment.
Some investments can be replicated. There are over 13,000 McDonald’s restaurants in the United States, and building another one doesn’t foreclose building even more. For such investments, NPV analysis gets the right answer, provided that appropriate interest rates and expectations are used. Other investments are difficult to replicate or logically impossible to replicate—having pumped and sold the oil from a tract, that tract is now dry. For such investments, NPV is consistently wrong because it neglects the value of the option to delay the investment. A correct analysis adds a lost value for the option to delay the cost of the investment—a value that can be quantitatively large—as we have seen.
Figure 11.2 Investment strike price given interest rate r in percent
Example: When should you refinance a mortgage? Suppose you are paying 10% interest on a $100,000 mortgage, and it costs $5,000 to refinance; but refinancing permits you to lock in a lower interest rate, and hence pay less. When is it a good idea? To answer this question, we assume that the $5,000 cost of refinancing is built into the loan so that, in essence, you borrow $105,000 at a lower interest rate when you refinance. This is actually the most common method of refinancing a mortgage.
To simplify the calculations, we will consider a mortgage that is never paid off; that is, one pays the same amount per year forever. If the mortgage isn’t refinanced, one pays 10% of the $100,000 face value of the mortgage each year, or $10,000 per year. If one refinances at interest rate r, one pays r × $105,000 per year, so the NPV of refinancing isNPV = $10,000 – r × $105,000.
Thus, NPV is positive whenever
Should you refinance when the interest rate drops to this level? No. At this level, you would exactly break even, but you would also be carrying a $105,000 mortgage rather than a $100,000 mortgage, making it harder to benefit from any further interest-rate decreases. The only circumstance in which refinancing at 9.52% is sensible is if interest rates can’t possibly fall further.
When should you refinance? That depends on the nature and magnitude of the randomness governing interest rates, preferences over money today versus money in the future, and attitudes to risk. The model developed in this section is not a good guide to answering this question, primarily because the interest rates are strongly correlated over time. However, an approximate guide to implementing the option theory of investment is to seek an NPV of twice the investment, which would translate into a refinance point of around 8.5%.
You are searching for a job. The net value of jobs that arise is uniformly distributed on the interval [0, 1]. When you accept a job, you must stop looking at subsequent jobs. If you can interview with one employer per week, what jobs should you accept? Use a 7% annual interest rate.
(Hint: Relate the job-search problem to the investment problem, where accepting a job is equivalent to making the investment. What is c in the job-search problem? What is the appropriate interest rate?)
For the past 60 years, the world has been “running out of oil.” There are news stories about the end of the reserves being only 10, 15, or 20 years away. The tone of these stories is that, at this time, we will run out of oil completely and prices will be extraordinarily high. Industry studies counter that more oil continues to be found and that the world is in no danger of running out of oil.
If you believe that the world will run out of oil, what should you do? You should buy and hold. That is, if the price of oil in 20 years is going to be $1,000 per barrel, then you can buy oil at $40 per barrel, hold it for 20 years, and sell it at $1,000 per barrel. The rate of return from this behavior is the solution to
This equation solves for r = 17.46%, which represents a healthy rate of return on investment. This substitution is part of a general conclusion known as the Ramsey ruleFor resources in fixed supply, prices rise at the interest rate.:The solution to this problem is known as Ramsey pricing, after the discoverer Frank Ramsey (1903–1930). For resources in fixed supply, prices rise at the interest rate. With a resource in fixed supply, owners of the resource will sell at the point maximizing the present value of the resource. Even if they do not, others can buy the resource at the low present value of price point, resell at the high present value, and make money.
The Ramsey rule implies that prices of resources in fixed supply rise at the interest rate. An example of the Ramsey rule in action concerns commodities that are temporarily fixed in supply, such as grains after the harvest. During the period between harvests, these products rise in price on average at the interest rate, where the interest rate includes storage and insurance costs, as well as the cost of funds.
Example: Let time be t = 0, 1, … , and suppose the demand for a resource in fixed supply has constant elasticity: Suppose that there is a total stock R of the resource, and the interest rate is fixed at r. What is the price and consumption of the resource at each time?
Solution: Let Qt represent the quantity consumed at time t. Then the arbitrage condition requires
Thus, Finally, the resource constraint implies
This solves for the initial consumption Q0. Consumption in future periods declines geometrically, thanks to the constant elasticity assumption.
Market arbitrage ensures the availability of the resource in the future and drives up the price to ration the good. The world runs out slowly, and the price of a resource in fixed supply rises on average at the interest rate.
Resources like oil and minerals are ostensibly in fixed supply—there is only so much oil, gold, bauxite, or palladium in the earth. Markets, however, behave as if there is an unlimited supply, and with good reason. People are inventive and find substitutes. England’s wood shortage of 1651 didn’t result in England being cold permanently, nor was England limited to the wood it could grow as a source of heat. Instead, coal was discovered. The shortage of whale oil in the mid-19th century led to the development of oil resources as a replacement. If markets expect that price increases will lead to substitutes, then we rationally should use more today, trusting that technological developments will provide substitutes.Unlike oil and trees, whales were overfished and there was no mechanism for arbitraging them into the future—that is, no mechanism for capturing and saving whales for later use. This problem, known as the tragedy of the commons, results in too much use (Garett Hardin, Science, 1968, Tragedy of the Commons). Trees have also been overcut, most notably on Easter Island. Thus, while some believe that we are running out of oil, most investors are betting that we are not, and that energy will not be very expensive in the future—either because of continued discovery of oil or because of the creation of alternative energy sources. If you disagree, why not invest and take the bet? If you bet on future price increases, that will tend to increase the price today, encouraging conservation today, and increase the supply in the future.
A tree grows slowly but is renewable, so the analysis of Section 11.4 "Resource Extraction" doesn’t help us to understand when it is most profitable to cut down the tree. Consider harvesting for pulp and paper use. In this case, the amount of wood chips is what matters to the profitability of cutting down the tree, and the biomass of the tree provides a direct indication of this. Suppose the biomass sells for a net price p, which has the costs of harvesting and replanting deducted from it, and the biomass of the tree is b(t) when the tree is t years old. It simplifies the analysis slightly to use continuous time discounting where
Consider the policy of cutting down trees when they are T years old. This induces a cutting cycle of length T. A brand new tree will produce a present value of profits of
This profit arises because the first cut occurs at time T, with discounting e-ρT, and produces a net gain of pb(T). The process then starts over, with a second tree cut down at time 2T, and so on.
Profit maximization gives a first-order condition on the optimal cycle length T of
This can be rearranged to yield
The left-hand side of this equation is the growth rate of the tree. The right-hand side is approximately the continuous-time discount factor, at least when T is large, as it tends to be for trees, which are usually on a 20- to 80-year cycle, depending on the species. This is the basis for a conclusion: Cut down the tree slightly before it is growing at the interest rate. The higher that interest rates are, the shorter the cycle for which the trees should be cut down.
The pulp and paper use of trees is special, because the tree is going to be ground up into wood chips. What happens when the object is to get boards from the tree, and larger boards sell for more? In particular, it is more profitable to get a 4 × 4 than two 2 × 4s. Doubling the diameter of the tree, which approximately raises the biomass by a factor of six to eight, more than increases the value of the timber by the increase in the biomass.
It turns out that our theory is already capable of handling this case. The only adaptation is a change in the interpretation of the function b. Now, rather than representing the biomass, b(t) must represent the value in boards of a tree that is t years old. (The parameter p may be set to one.) The only amendment to the rule for cutting down trees is as follows: The most profitable point in time to cut down the tree occurs slightly before the time when the value (in boards) of the tree is growing at the interest rate.
For example, lobsters become more valuable as they grow. The profit-maximizing time to harvest lobsters is governed by the same equation, where b(T) is the value of a lobster at age T. Prohibiting the harvest of lobsters under age T is a means of insuring the profit-maximizing capture of lobsters and preventing overfishing.
The implementation of the formula is illustrated in Figure 11.3 "Optimal solution for T". The dashed line represents the growth rate while the solid line represents the discount rate, which was set at 5%. Note that the best time to cut down the trees is when they are approximately 28.7 years old and, at that time, they are growing at 6.5%. Figure 11.3 "Optimal solution for T" also illustrates another feature of the optimization—there may be multiple solutions to the optimization problem, and the profit-maximizing solution involves cutting from above.
Figure 11.3 Optimal solution for T
The U.S. Department of the Interior is in charge of selling timber rights on federal lands. The department uses the policy of maximum sustainable yieldPolicy that maximizes the long-run average value of a sustainable resource. to determine the specific time that the tree is cut down. Maximum sustainable yield maximizes the long-run average value of the trees cut down; that is, it maximizes
Maximum sustainable yield is actually a special case of the policies considered here, and arises for a discount factor of 0. It turns out (thanks to a formula known variously as l’Hôpital’s or l’Hospital’s Rule) that the
Thus, the rule as ρ → 0, and this is precisely the same rule that arises under maximum sustainable yield.
Thus, the Department of the Interior acts as if the interest rate is zero when it is not. The justification given is that the department is valuing future generations at the same level as current generations—that is, increasing the supply for future generations while slightly harming the current generation of buyers. The major consequence of the department’s policy of maximum sustainable yield is to force cutting of timber even when prices are low during recessions.
Many people purchase durable goods as investments, including Porsche Speedsters (see Figure 11.4 "The Porsche Speedster"), Tiffany lamps, antique telephones, postage stamps and coins, baseball cards, original Barbie dolls, antique credenzas, autographs, original rayon Hawaiian shirts, old postcards, political campaign buttons, old clocks, and even Pez dispensers. How is the value of, say, a 1961 Porsche Speedster or a $500 bill from the Confederacy, which currently sells for over $500, determined?
The theory of resource prices can be adapted to cover these items, which are in fixed supply. There are four major differences that are relevant. First, using the item doesn’t consume it: The goods are durable. I can own an “I Like Ike” campaign button for years, and then sell the same button. Second, these items may depreciate. Cars wear out even when they aren’t driven, and the brilliant color of Pez dispensers fades. Every time that a standard 27.5-pound gold bar, like the kind in the Fort Knox depository, is moved, approximately $5 in gold wears off the bar. Third, the goods may cost something to store. Fourth, the population grows, and some of the potential buyers are not yet born.
Figure 11.4 The Porsche Speedster
To understand the determinants of the prices of collectibles, it is necessary to create a major simplification to perform the analysis in continuous time. Let t, ranging from zero to infinity, be the continuous-time variable. If the good depreciates at rate δ, and q0 is the amount available at time 0, the quantity available at time t is .
For simplicity, assume that there is constant elasticity of demand ε. If g is the population growth rate, the quantity demanded, for any price p, is given by for a constant a that represents the demand at time 0. This represents demand for the good for direct use, but neglects the investment value of the good—the fact that the good can be resold for a higher price later. In other words, xd captures the demand for looking at Pez dispensers or driving Porsche Speedsters, but does not incorporate the value of being able to resell these items.
The demand equation can be used to generate the lowest use value to a person owning the good at time t. That marginal use value v arises from the equality of supply and demand:
Thus, the use value to the marginal owner of the good at time t satisfies
An important aspect of this development is that the value to the owner is found without reference to the price of the good. The reason this calculation is possible is that the individuals with high values will own the good, and the number of goods and the values of people are assumptions of the theory. Essentially, we already know that the price will ration the good to the individuals with high values, so computing the lowest value individual who holds a good at time t is a straightforward “supply equals demand” calculation. Two factors increase the marginal value to the owner—there are fewer units available because of depreciation, and there are more high-value people demanding them because of population growth. Together, these factors make the marginal use value grow at the rate
Assume that s is the cost of storage per unit of time and per unit of the good, so that storing x units for a period of length Δ costs sxΔ. This is about the simplest possible storage cost technology.
The final assumption we make is that all potential buyers use a common discount rate r, so that the discount of money or value received Δ units of time in the future is e-rΔ. It is worth a brief digression to explain why it is sensible to assume a common discount rate, when it is evident that many people have different discount rates. Different discount rates induce gains from trade in borrowing and lending, and create an incentive to have banks. While banking is an interesting topic to study, this section is concerned with collectibles, not banks. If we have different discount factors, then we must also introduce banks, which would complicate the model substantially. Otherwise, we would intermingle the theory of banking and the theory of collectibles. It is probably a good idea to develop a joint theory of banking and collectibles, given the investment potential of collectibles, but it is better to start with the pure theory of either one before developing the joint theory.
Consider a person who values the collectible at v. Is it a good thing for this person to own a unit of the good at time t? Let p be the function that gives the price across time, so that p(t) is the price at time t. Buying the good at time t and then selling what remains (recall that the good depreciates at rate δ) at time t + Δ gives a net value of
For the marginal person—that is, the person who is just indifferent to buying or not buying at time t—this must be zero at every moment in time, for Δ = 0. If v represents the value to a marginal buyer (indifferent to holding or selling) holding the good at time t, then this expression should come out to be zero. Thus, dividing by Δ,
Recall that the marginal value is , which gives The general solution to this differential equation is
It turns out that this equation only makes sense if for otherwise the present value of the marginal value goes to infinity, so there is no possible finite initial price. Provided that demand is elastic and discounting is larger than growth rates (which is an implication of equilibrium in the credit market), this condition will be met.
What is the initial price? It must be the case that the present value of the price is finite, for otherwise the good would always be a good investment for everyone at time 0, using the “buy and hold for resale” strategy. That is,
This condition implies that and thus
This equation may take on two different forms. First, it may be solvable for a nonnegative price, which happens if
Second, it may require destruction of some of the endowment of the good. Destruction must happen if the quantity of the good q0 at time 0 satisfies
In this case, there is too much of the good, and an amount must be destroyed to make the initial price zero. Since the initial price is zero, the good is valueless at time zero, and destruction of the good makes sense—at the current quantity, the good is too costly to store for future profits. Enough is destroyed to ensure indifference between holding the good as a collectible and destroying it. Consider, for example, the $500 Confederate bill shown in Figure 11.5 "$500 Confederate States bill". Many of these bills were destroyed at the end of the U.S. Civil War, when the currency became valueless and was burned as a source of heat. Now, an uncirculated version retails for $900.
Figure 11.5 $500 Confederate States bill
The amount of the good that must be destroyed is such that the initial price is zero. As q0 is the initial (predestruction) quantity, the amount at time zero after the destruction is the quantity q(0) satisfying
Given this construction, we have that
where either q(0) = q0 and p(0) ≥ 0, or q(0) < q0 and p(0) = 0.
Destruction of a portion of the stock of a collectible, followed by price increases, is actually a quite common phenomenon. In particular, consider the “Model 500” telephone by Western Electric illustrated in Figure 11.6 "Western Electric Model 500 telephone". This ubiquitous classic phone was retired as the United States switched to tone dialing and push-button phones in the 1970s, and millions of phones—perhaps over 100 million—wound up in landfills. Now the phone is a collectible, and rotary phone enthusiasts work to keep them operational.
Figure 11.6 Western Electric Model 500 telephone
The solution for p(0) dramatically simplifies the expression for p(t):
This formula enables one to compare different collectibles. The first insight is that storage costs enter linearly into prices, so that growth rates are approximately unaffected by storage costs. The fact that gold is easy to store—while stamps and art require control of humidity and temperature in order to preserve value, and are hence more expensive to store—affects the level of prices but not the growth rate. However, depreciation and the growth of population affect the growth rate, and they do so in combination with the demand elasticity. With more elastic demand, prices grow more slowly and start at a lower level.
Typically, wheat harvested in the fall has to last until the following harvest. How should prices evolve over the season? If I know that I need wheat in January, should I buy it at harvest time and store it myself, or wait and buy it in January? We can use a theory analogous to the theory of collectibles developed in Section 11.6 "Collectibles" to determine the evolution of prices for commodities like wheat, corn, orange juice, and canola oil.
Unlike collectibles, buyers need not hold commodities for their personal use, since there is no value in admiring the wheat in your home. Let p(t) be the price at time t, and suppose that the year has length T. Generally there is a substantial amount of uncertainty regarding the size of wheat harvests, and most countries maintain an excess inventory as a precaution. However, if the harvest were not uncertain, there would be no need for a precautionary holding. Instead, we would consume the entire harvest over the course of a year, at which point the new harvest would come in. It is this such model that is investigated in this section.
Let δ represent the depreciation rate (which, for wheat, includes the quantity eaten by rodents), and let s be the storage cost. Buying at time t and reselling at t + Δ should be a break-even proposition. If one purchases at time t, it costs p(t) to buy the good. Reselling at t + Δ, the storage cost is about sΔ. (This is not the precisely relevant cost; but rather it is the present value of the storage cost, and hence the restriction to small values of Δ.) The good depreciates to only have left to sell, and discounting reduces the value of that amount by the factor For this to be a break-even proposition, for small Δ,
taking the limit as Δ → 0,
This arbitrage condition ensures that it is a break-even proposition to invest in the good; the profits from the price appreciation are exactly balanced by depreciation, interest, and storage costs. We can solve the differential equation to obtain
The unknown is p(0). The constraint on p(0), however, is like the resource extraction problem—p(0) is determined by the need to use up the harvest over the course of the year.
Suppose demand has constant elasticity ε. Then the quantity used comes in the form Let z(t) represent the stock at time t. Then the equation for the evolution of the stock is This equation is obtained by noting that the flow out of stock is composed of two elements: depreciation, δz, and consumption, x. The stock evolution equation solves for
Thus, the quantity of wheat is consumed exactly if
But this equation determines the initial price through
This equation doesn’t lead to a closed form for p(0) but is readily estimated, which provides a practical means of computing expected prices for commodities in temporarily fixed supply.
Figure 11.7 Prices over a cycle for seasonal commodities
Generally, the price equation produces a “sawtooth” pattern, which is illustrated in Figure 11.7 "Prices over a cycle for seasonal commodities". The increasing portion is actually an exponential, but of such a small degree that it looks linear. When the new harvest comes in, prices drop abruptly as the inventory grows dramatically, and the same pattern is repeated.
Figure 11.8 Log of price of gold over time
How well does the theory work? Figure 11.8 "Log of price of gold over time" shows the log of the future price of gold over time. The relevant data come from a futures market that establishes, at one moment in time, the price of gold for future delivery, and thus represents today’s estimate of the future price of gold. These data, then, represent the expected future price at a particular moment in time (the afternoon of October 11, 2005), and thus correspond to the prices in the theory, since perceived risks are fixed. (Usually, in the real world, risk plays a salient role.) We can observe that prices are approximately an exponential, because the log of prices is approximately linear. However, the estimate of r + δ is surprisingly low, at an annual level of less than 0.03, or 3% for both discounting and depreciation. Depreciation of gold is low, but this still represents a very low interest rate.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00382.warc.gz
|
CC-MAIN-2021-49
| 49,563 | 124 |
http://www.novell.com/documentation/gw8/gw8_admin/data/a7q51ga.html
|
math
|
Section 62.0, Understanding the Monitor Agent Consoles
Section 63.0, Configuring the Monitor Agent
Section 64.0, Configuring the Monitor Application
Section 65.0, Using GroupWise Monitor
Section 66.0, Comparing the Monitor Consoles
Section 67.0, Using Monitor Agent Startup Switches
For a complete list of port numbers used by Monitor, see Section A.8, Monitor Agent Port Numbers and Section A.9, Monitor Application Port Numbers.
For detailed Linux-specific Monitor information, see Section C.0, Linux Commands, Directories, and Files for GroupWise Administration.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00182.warc.gz
|
CC-MAIN-2023-50
| 565 | 8 |
https://link.springer.com/chapter/10.1007%2F11558958_102
|
math
|
Aggregation-Based Multilevel Preconditioning of Non-conforming FEM Elasticity Problems
Preconditioning techniques based on various multilevel extensions of two-level splittings of finite element (FE) spaces lead to iterative methods which have an optimal rate of convergence and computational complexity with respect to the number of degrees of freedom. This article deals with the construction of algebraic two-level and multilevel preconditioning algorithms for the Lamé equations of elasticity, which are discretized by Crouzeix-Raviart non-conforming linear finite elements on triangles. An important point to note is that in the non-conforming case the FE spaces corresponding to two successive levels of mesh refinements are not nested. To handle this, a proper aggregation-based two-level basis is considered, which enables us to fit the general framework of the two-level preconditioners and to generalize the method to the multilevel case. The derived estimate of the constant in the strengthened Cauchy-Bunyakowski-Schwarz (CBS) inequality is uniform with respect to both, mesh anisotropy and Poisson ratio, including the almost incompressible case.
KeywordsNodal Unknown Finite Element Space Numerical Linear Algebra Nonlinear Elliptic Problem Mesh Anisotropy
Unable to display preview. Download preview PDF.
- 9.Blaheta, R., Margenov, S., Neytcheva, M.: Robust optimal multilevel preconditioners for nonconforming finite element systems. To appear in Numerical Linear Algebra with ApplicationsGoogle Scholar
- 13.Farago, I., Karatson, J.: Numerical solution of nonlinear elliptic problems via preconditionionig operators. Theory and applications. NOVA Science (2002)Google Scholar
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00348.warc.gz
|
CC-MAIN-2018-39
| 1,693 | 6 |
https://districtfreshkitchen.com/how-to-cook-brown-rice-for-dogs/
|
math
|
how to cook brown rice for dogs
# How to Cook Brown Rice for Dogs: A Nutritious and Delicious Meal Option for Your Furry Friend
Cooking homemade meals for your furry friend can be a great way to ensure they receive a nutritious and balanced diet. Brown rice is an excellent source of carbohydrates and it is packed with essential vitamins and minerals that can benefit your dog’s overall health. In this article, we will guide you through the step-by-step process of cooking brown rice for dogs, highlighting its benefits and answering some frequently asked questions.
## Benefits of Brown Rice for Dogs
When it comes to your dog’s diet, including brown rice can provide several health benefits. Here are some key advantages:
### 1. Excellent Source of Carbohydrates
Brown rice is a complex carbohydrate that provides a sustainable source of energy for your dog. It is digested slowly, helping to regulate blood sugar levels and prevent spikes and crashes in energy throughout the day.
### 2. Fiber-Rich
Rich in dietary fiber, brown rice promotes healthy digestion in dogs. It aids in maintaining regular bowel movements and can help prevent constipation.
### 3. Nutrient Powerhouse
Brown rice is loaded with essential vitamins and minerals, including manganese, selenium, magnesium, and B vitamins. These nutrients support your dog’s immune system, promote healthy bones and teeth, and contribute to overall wellness.
### 4. Hypoallergenic Option
If your dog has food allergies or sensitivities, brown rice is often well-tolerated and considered a hypoallergenic option. It is free from common allergens such as wheat and corn.
## Step-by-Step Guide: Cooking Brown Rice for Dogs
Now that you are aware of the benefits, let’s dive into how to cook brown rice for your furry friend. Follow these simple steps to prepare a delicious and nutritious meal:
### Step 1: Gather the Ingredients
To cook brown rice for your dog, you will need the following ingredients:
– 1 cup of organic brown rice
– 2 cups of water or low-sodium chicken broth
### Step 2: Rinse the Rice
Before cooking, rinse the brown rice thoroughly with cold water to remove any dust or impurities.
### Step 3: Measure the Water or Broth
Measure two cups of water or low-sodium chicken broth. Using broth adds extra flavor to the rice.
### Step 4: Bring to a Boil
In a medium-sized pot, bring the water or chicken broth to a boil over high heat.
### Step 5: Add the Brown Rice
Once the liquid is boiling, add the rinsed brown rice to the pot and stir well.
### Step 6: Reduce Heat and Simmer
Reduce the heat to low and cover the pot with a lid. Allow the rice to simmer for approximately 45 minutes or until tender. Stir occasionally to prevent sticking.
### Step 7: Let It Rest
After cooking, let the brown rice rest for a few minutes, covered. This allows the rice to absorb any remaining moisture and ensures a fluffy texture.
### Step 8: Serve and Enjoy
Once the rice has cooled down, it is ready to be served to your furry friend. You can mix it with their regular dog food or serve it as a standalone meal.
## Frequently Asked Questions
### 1. Can I feed brown rice to my diabetic dog?
Yes, you can feed brown rice to a diabetic dog. However, it is important to consult with your veterinarian to determine appropriate portion sizes and to ensure a balanced diet for your furry friend.
### 2. Can I freeze cooked brown rice for later use?
Absolutely! You can freeze cooked brown rice in single-serving portions for future use. Make sure to thaw it thoroughly before serving it to your dog.
### 3. Can I substitute brown rice with white rice?
While brown rice is generally a healthier option due to its added fiber content, you can substitute it with white rice occasionally. White rice is easier to digest and can be beneficial in situations where your dog is experiencing an upset stomach.
### 4. How much brown rice should I feed my dog?
The portion size of brown rice for your dog will depend on their size, age, and overall health. As a general guideline, you can start with 1/4 cup to 1/2 cup of cooked brown rice per meal and adjust as needed.
### 5. Are there any risks associated with feeding brown rice to dogs?
When cooked and served in moderation, brown rice is generally safe for dogs. However, every dog is unique, and it is essential to observe any changes in your dog’s digestion or behavior when introducing a new food to their diet. If you notice any adverse effects, consult your veterinarian.
Cooking brown rice for your furry friend can be a simple and rewarding way to enhance their diet with wholesome ingredients. By following the step-by-step guide outlined in this article, you can provide your dog with a nutritious and delicious meal option. Remember to consult your veterinarian for specific dietary recommendations tailored to your dog’s individual needs. Enjoy cooking for your canine companion and watch them thrive on this healthy addition to their diet!
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00479.warc.gz
|
CC-MAIN-2023-40
| 4,970 | 45 |
http://www.ask.com/web?q=What+Does+a+Histogram+Look+like%3F&o=2603&l=dir&qsrc=3139&gc=1
|
math
|
A histogram is a bar graph of raw data that creates a picture of the data
distribution. The bars represent the ... What does it look like? The c-chart shows
as being like a snapshot, while a Run Chart or Control Chart is more like a movie.
(Viewgraph ... construct a Histogram when you want to do the following (
Viewgraph 2): ! Summarize large data sets graphically. When you look at
Histograms. Histogram: a graphical display of data using bars of different heights.
... The horizontal axis is continuous like a number line: ...
The histogram is the best way to tell what you image will look like when you go
back home. It takes ... If you dont have it, next camera you get make sure it does.
Jun 22, 2011 ... But what is that weird thing called a histogram, and what does it mean? ... the
concentration falling sharply, as we look to the slightly dimmer highlights. ... but
you may learn more by learning how to adjust contrast like a pro ...
At first glance, histograms look very similar to bar graphs. ... Another key
difference between bar graphs and histograms has to do with the ordering of the
This is how an ideal histogram might look, evenly distributed, edge to edge, not
up ... for it but you do have a decision to make when you see something like this.
Student: Histograms and bar graphs look very similar. What's ... It doesn't look like
the order matters at all as long as the billions of dollars still matches every
category. ... Mentor: Then how do they decide which scores go into which bars?
A bar graph has nothing to do with pubs, and a histogram is not someone you
hire to ... The bar graph shows what things would look like if we stacked up the ...
Luckily for the math-challenged photographers like me, you do not need to know
.... histogram data, look for clipping and blinkies, and do not judge an exposure ...
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783393997.50/warc/CC-MAIN-20160624154953-00150-ip-10-164-35-72.ec2.internal.warc.gz
|
CC-MAIN-2016-26
| 1,837 | 23 |
https://dekookguide.com/how-many-60-lb-bags-of-concrete-in-a-yard/
|
math
|
You can use this concrete calculator to help you determine the number of bags of QUIKRETE® Concrete Mix, Mortar Mix, or Fast-Setting Concrete you will need for the following projects. (All calculations are rounded up to the next highest whole bag). Calculators:
Quantity and/or coverage calculations for all QUIKRETE® products can be found on all Spec Data sheets.
how many square feet are in a yard of concrete?
The amount of square feet you will fill with a yard of concrete will depend on the thickness.
FOR EXAMPLE: How many bags of concrete do I need for a 10 x 10 slab?
10 x 10 = 100 sq. ft.Â
For the calculations above, I rounded up to get an even number of bags.
How Many Square Feet Are in a Yard of Concrete
A yard of concrete is a 3-dimensional measurement that is 3’x3’x3’. If you multiply 3 x 3 x 3 it equals 27. That means a yard of concrete contains 27 cubic feet (1’x1’x1’ or 12”x12”x12”) of concrete.
For our metric friends, a meter of concrete is different from a yard. A meter is 3.28 feet, so a cubic meter is 3.28’x3.28’x3.28’ or 35.29 cubic feet. That’s 8.29ft³ more concrete or almost a third of a yard larger!
How Many 40 Pound Bags of Concrete in a Yard
A 40-pound bag of concrete mix makes 0.3ft³ of concrete which would cover 3.6ft² to a thickness of 1”. A yard of concrete is 27ft³, so 27 ÷ 0.3ft³ = 90 bags. A 40-pound bag of 4000psi concrete mix costs between $2.78 and $2.88 in my area. So, a yard of concrete mix will cost between $250.20 and $ 259.20 before taxes and weigh 3,600-pounds.
How many 60 pound bags of concrete are in a half yard?
How many bags do I need for 1 yard of concrete?
How many 80lb bags of concrete does it take to pour a yard?
How many cubic feet will 60 lb bag of concrete?
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00497.warc.gz
|
CC-MAIN-2023-14
| 1,772 | 16 |
https://wiki-offline.jakearchibald.com/wiki/Two-ray_ground-reflection_model
|
math
|
The two-rays ground-reflection model is a multipath radio propagation model which predicts the path losses between a transmitting antenna and a receiving antenna when they are in line of sight (LOS). Generally, the two antenna each have different height. The received signal having two components, the LOS component and the reflection component formed predominantly by a single ground reflected wave.
From the figure the received line of sight component may be written as
and the ground reflected component may be written as
where is the transmitted signal, is the length of the direct line-of-sight (LOS) ray, is the length of the ground-reflected ray, is the combined antenna gain along the LOS path, is the combined antenna gain along the ground-reflected path, is the wavelength of the transmission (, where is the speed of light and is the transmission frequency), is ground reflection coefficient and is the delay spread of the model which equals . The ground reflection coefficient is
where or depending if the signal is horizontal or vertical polarized, respectively. is computed as follows.
The constant is the relative permittivity of the ground (or generally speaking, the material where the signal is being reflected), is the angle between the ground and the reflected ray as shown in the figure above.
From the geometry of the figure, yields:
Therefore, the path-length difference between them is
and the phase difference between the waves is
The power of the signal received is
where denotes average (over time) value.
If the signal is narrow band relative to the inverse delay spread , so that , the power equation may be simplified to
where is the transmitted power.
When distance between the antennas is very large relative to the height of the antenna we may expand ,
using the Taylor series of :
and taking the first two terms only,
The phase difference can then be approximated as
When is large, ,
Expanding using Taylor series
and retaining only the first two terms
it follows that
which is accurate in the far field region, i.e. when (angles are measured here in radians, not degrees) or, equivalently,
and where the combined antenna gain is the product of the transmit and receive antenna gains, . This formula was first obtained by B.A. Vvedenskij.
Note that the power decreases with as the inverse fourth power of the distance in the far field, which is explained by the destructive combination of the direct and reflected paths, which are roughly of the same in magnitude and are 180 degrees different in phase. is called "effective isotropic radiated power" (EIRP), which is the transmit power required to produce the same received power if the transmit antenna were isotropic.
In logarithmic units
In logarithmic units :
Path loss :
Power vs. distance characteristics
When the distance between antennas is less than the transmitting antenna height, two waves are added constructively to yield bigger power. As distance increases, these waves add up constructively and destructively, giving regions of up-fade and down-fade. As the distance increases beyond the critical distance or first Fresnel zone, the power drops proportionally to an inverse of fourth power of . An approximation to critical distance may be obtained by setting Δφ to π as the critical distance to a local maximum.
An extension to large antenna heights
The above approximations are valid provided that , which may be not the case in many scenarios, e.g. when antenna heights are not much smaller compared to the distance, or when the ground cannot be modelled as an ideal plane . In this case, one cannot use and more refined analysis is required, see e.g.
As a case of log distance path loss model
The standard expression of Log distance path loss model is
The path loss of 2-ray ground reflected wave is
for the critical distance.
As a case of multi-slope model
The 2-ray ground reflected model may be thought as a case of multi-slope model with break point at critical distance with slope 20 dB/decade before critical distance and slope of 40 dB/decade after the critical distance. Using the free-space and two-ray model above, the propagation path loss can be expressed as
where and are the free-space and 2-ray path losses.
- Radio propagation model
- Free-space path loss
- Friis transmission equation
- ITU-R P.525
- Link budget
- Ray tracing (physics)
- Reflection (physics)
- Specular reflection
- Six-rays model
- Ten-rays model
- Jakes, W.C. (1974). Microwave Mobile Communications. New York: IEEE Press.
- Rappaport, Theodore S. (2002). Wireless Communications: Principles and Practice (2. ed.). Upper Saddle River, NJ: Prentice Hall PTR. ISBN 978-0130422323.
- Vvedenskij, B.A. (December 1928). "On Radio Communications via Ultra-Short Waves". Theoretical and Experimental Electrical Engineering (12): 447–451.
- Loyka, Sergey; Kouki, Ammar (October 2001). "Using Two Ray Multipath Model for Microwave Link Budget Analysis". IEEE Antennas and Propagation Magazine. 43 (5): 31–36.
- S. Salous, Radio Propagation Measurement and Channel Modelling, Wiley, 2013.
- J.S. Seybold, Introduction to RF propagation, Wiley, 2005.
- K. Siwiak, Radiowave Propagation and Antennas for Personal Communications, Artech House, 1998.
- M.P. Doluhanov, Radiowave Propagation, Moscow: Sviaz, 1972.
- V.V. Nikolskij, T.I. Nikolskaja, Electrodynamics and Radiowave Propagation, Moscow: Nauka, 1989.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00577.warc.gz
|
CC-MAIN-2021-31
| 5,396 | 57 |
https://pennstate.pure.elsevier.com/en/publications/hyperelliptic-jacobians-and-simple-groups-usub3sub2supmsup
|
math
|
In a previous paper, the author proved that in characteristic zero the jacobian J(C) of a hyperelliptic curve C : y2 = f(x) has only trivial endomorphisms over an algebraic closure Ka of the ground field K if the Galois group Gal(f) of the irreducible polynomial f(x) ∈ K[x] is either the symmetric group Sn or the alternating group An. Here n > 4 is the degree of f. In another paper by the author this result was extended to the case of certain "smaller" Galois groups. In particular, the infinite series n = 2r + 1, Gal(f) = L2(2r) := PSL2(F2r) and n = 24r+2 + 1, Gal(f) = Sz(22r+1) were treated. In this paper the case of Gal(f) = U3(2m) := PSU3(F2m) and n = 23m + 1 is treated.
All Science Journal Classification (ASJC) codes
- Applied Mathematics
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00010.warc.gz
|
CC-MAIN-2022-49
| 754 | 3 |
https://www.reference.com/math/formula-finding-volume-trapezoidal-prism-258e5af598c5d62f
|
math
|
The formula is the length of the prism times the area of the trapezoid, which is one-half times (a+b) times the height; the area is also called the cross-sectional area. "A" and "B" are the two bases of the trapezoid. The bases are the sides that run parallel to one another, which means they never touch.
The student finds the area of the trapezoid first. If the smaller trapezoid base is 2 and the longer one is 4, she adds those two numbers together to get 6. Then she multiplies that number by the height of the trapezoid, 3, to get 18. Finally, she divides that by 2, which is the same as multiplying by 0.5, to get 9.
To find the volume of the prism, the student multiplies the area of the trapezoid by the length of the prism. If the length of the prism is 10, then she multiplies 9 times 10 to get 90. Therefore, the volume of the trapezoid is 90. The unit depends on the original units the trapezoidal prism was measured in. If it was measured in inches, the resulting unit would be cubic inches, as inches are multiplied by inches three separate times.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00033.warc.gz
|
CC-MAIN-2019-43
| 1,062 | 3 |
http://www.tutorsglobe.com/question/discuss-the-main-functions-of-each-of-these-modules-in-the-51399471.aspx
|
math
|
Discuss the main functions of each of these modules in the
What are the main software modules of a DDBMS? Discuss the main functions of each of these modules in the context of the client-server architecture.
Expected delivery within 24 Hours
once the functions have been determined and the project team is in place the next step is to begin the system
use xdr and htonl to encode a 1000-element array of integers measure and compare the performance of each how do these
ipv6 simplifies arp out of existence by allowing hardware addresses to be part of the ipv6 address how does this
create 2 web pages that does the following1 page 1 html page that contains a form that collect information from user
the average height of 40 students in a class is 155cm recently five students whose average heights is 148 cm left the
the shaft is subjected to an axial force p if the reactive pressure on the conical bearing is uniform determine the
compute the shortest distance from new york to london hint 1 a great circle is the shortest path between two points on
race performance companynbspmonth qty costs000 sales000nbspjan 15000 1250 2340nbspfeb 16400 1620 1952nbspmar 13625 1467
1 list the first four problem-solving steps to solve the following a fluid pump curve can be approximated by
given the above description draw an er diagram for the database state any assumptions you make for your er-diagram to
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171281.53/warc/CC-MAIN-20170219104611-00649-ip-10-171-10-108.ec2.internal.warc.gz
|
CC-MAIN-2017-09
| 1,498 | 14 |
https://web2.0calc.com/questions/hard-counting-pls-help
|
math
|
Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?
A 1 step stair case requirest 4, 2 step 10, so we need 6 more. 2-3 step needs 8 more,so following the pattern we get 10 toothpicsk from 3-4, 12 from 4-5.
10+12 = 22.
If you don't understand anything feel free to ask.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00472.warc.gz
|
CC-MAIN-2020-24
| 361 | 4 |
https://answers.unrealengine.com/questions/861766/view.html
|
math
|
How to repeat simple move to actor using gate?
I'm using blueprints. I want to repeat my simple move to actor blueprint every x seconds so the ai is always following the player, but I don't understand how gates work. Right now I have this and it doesn't work.
I think you want to do this.
Start with with F, then it repeats it every 0.2 seconds.
answered Jan 01 '19 at 08:52 PM
Follow this question
Once you sign in you will be able to subscribe for any updates here
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529737.79/warc/CC-MAIN-20190723215340-20190724001340-00229.warc.gz
|
CC-MAIN-2019-30
| 466 | 7 |
https://billiontrader.com/convexity-adjustment-part-2/
|
math
|
As it was written in the previous article “Futures and forward convexity adjustment”, there is a systematic advantage to being short EuroDollar futures relative to FRAs.
This advantage is characterized as a convexity bias and appropriate methods exist to adjust Eurodollar futures prices to eliminate the difference between the futures and forward LIBOR rates.
The article explained that marking-to-market is one of the sources of convexity bias:
– When interest rates rise, the borrower receives a payment that can be invested at the higher interest rate.
– When interest rates fall, the borrower makes a payment that can be funded at the lower interest rate.
However, the settlement structure of the Eurodollar contract is another reason for convexity bias.
The following examples illustrate the main difference between the settlement structures of FRA and Eurodollar futures.
1. In Novemeber, we buy FRA @1.45% 3-month LIBOR rate in order to guarantee a 1.45% borrowing rate for a $500M loan for 3 months from March to June.
The FRAs payoff is well describe in the article “FRA” and following the same logic we have:
– If interest rate rise up to 1.6%, we make $738 188 to be settled by the seller in March.
– If interest rate is 1.2%, we suffer a loss of -$1 235 177 to be settled by us to the seller in March.
2. Let’s consider again the example in which we wish to guarantee a 1.45% borrowing rate for a $500M loan for 3 months.
Taking into account that if we bought the FRA in the previous example in order to hedge our interest rate exposure against interest rate rise, we need to short 500 Eurodollar futures contracts (the loan is $500M and thus 500 contracts have notional of $500M) in order to make the same hedge.
Suppose, in Novemeber, the March Eurodollar implied 3-month LIBOR is 1.45% and thus the futures price is 94.2, i.e., (100 − 94.2) / (4 * 100) = 1.45% over 3 months.
According to the article “EuroDollars”, the payoff at expiration is:
(94.2 − (100 − LIBOR at expiration)) * 100 * $25.
– If the 3-month LIBOR rate in March is 1.2%, the Eurodollar futures price will be 95.2 (100 – 1.2% * 4)
The payment is: ((94.2 − 95.2) * 100 * $25) * 500= −$1.25M
– If the 3-month LIBOR rate in March is 1.6%, the Eurodollar futures price will be 93.6 (100 – 1.6% * 4)
The payment is: ((94.2 − 93.6) * 100 * $25) * 500= $0.75M
This is like the payment on the FRA paid in June (in-arrears) except that the futures contract settles in March due to its specification, i.e., the FRAs result in June = ED futures result in March as it follows from the table below.
The results are the same but the months are different. If we compare the results in March, we see the FRA result differs from the result of Eurodollar futures. Clearly, if we want to fix the same interest rate of @1.45%, we must adjust the number of Eurodollar futures accordingly.
Put it differently, the rate implied by the Eurodollar contract cannot equal the prevailing FRA rate for the same loan.
In order to have the same @1.45% interest rate (both for FRA and Eurodollar futures), we adjust the number of Eurodollar futures by shorting fewer than 500 contracts. Using the implied 3-month Eurodollar rate of 1.45% as the discount factor, we receive:
500 / (1+ 0.0145) = 492.85 contracts to short.
Now, we recalculate the results in March, taking into account 492.85 contracts instead of 500.
– If the 3-month LIBOR rate in March is 1.6%, the Eurodollar futures price will be 93.6 (100 – 1.6% * 4). The payment in March is:
((94.2 − 93.6) * 100 * $25) * 492.85= $739 275.
This sum is equal $739 275 * 1.016 = $751 103 to be paid in June
– If the 3-month LIBOR rate in March is 1.2%, the Eurodollar futures price will be 95.2 (100 – 1.2% * 4). The payment in March is:
((94.2 − 95.2) * 100 * $25) * 492.85= −$1 232 125.
This sum is equal −$1 232 125 * 1.012 = −$1 246 910 to be paid in June
Now, if we compare the results for each month, we see the difference exists but it is relatively small. One more conclution we can make is that the settlement structure of the Eurodollar contract works systematically in favor of the borrower. Put differently, we can verify that it systematically works against a lender.
As it follows from the table above, Eurdollar futures has an advantage over FRAs, which is called a convexity bias:
In June, if quarterly LIBOR is 1.6%, we pay $8M (500M * 1.6%) in borrowing cost and:
Eurodollar contract earns $751 103 on the , a net borrowing expense is $8M – $751 103= $7 248 897
A net borrowing with the FRA contract will result in $8M – $750 000 = $7 250 000
Therefore, totally, we make a profit, relative to an FRA, of $7 250 000 − $7 248 897 = $1 103
In June, if quarterly LIBOR is 1.2%, we pay $6M (500M * 1.2%) in borrowing cost and:
Eurodollar contract loses -$1 246 910, a net borrowing expense is $6M + $1 246 910= $7 246 910
A net borrowing with the FRA contract will result in $6M + $1 250 000 = $7 250 000
Therefore, totally, we make a profit, relative to an FRA, of $7 250 000 − $7 246 910 = $3 090
The examples above show that Eurodollar futures contracts have an advantage over FRA and thus Eurodollar futures prices should be lower than their so-called fair values. Put differently, the 3-month interest rates implied by Eurodollar futures prices should be higher than the 3-month forward rates to which they are tied.
The difference between the FRA rate and the Eurodollar rate, known as convexity bias, is adjusted by the formula discussed in the article “Futures and forward convexity adjustment”.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00102.warc.gz
|
CC-MAIN-2023-50
| 5,594 | 44 |
https://homework.zookal.com/questions-and-answers/the-pipe-has-a-mass-of-10-kgm-and-is-871396491
|
math
|
the pipe has a mass of 10 kgm and is...
Question: the pipe has a mass of 10 kgm and is...
Solution by an expert tutor
This question has been solved
Subscribe to see this solution
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00014.warc.gz
|
CC-MAIN-2021-17
| 178 | 5 |
https://www.ichec.ir/article_3288.html
|
math
|
Analysis of mean permeate flux for ultrafiltration process by two methods: resistance-in-series model and artificial neural network
عنوان دوره: 1
Chemical engineering dept., Faculty of engineering, University of Guilan, Rasht, Iran.
Two resistance-in-series and artificial neural network (ANN) methods have been studied to compute the amount of mean permeate flux of a hollow-fiber module for ultrafiltration of PVP (360) aqueous solution. The present resistance-in-series model includes two types of pressure distribution equations. In model 1, momentum balance with term of convection has been considered. In model 2, the pressure distribution has been obtained by Hagen Poiseuille equation. The total AAD% and MSE in model 1 are 7.47 and 0.0966. These numbers are a bit lower than model 2 with AAD% 7.53 and MSE equals to 0.0975. The calculated amounts of AAD% and MSE in the lowest velocity, 0.0723 (m/s), are 5.2 and 0.042 and in the highest velocity, 0.2195 (m/s), are 12.5 and 0.4. Thus in both models, the errors and deviations has been arised by increasing feed velocities. In ANN method, input data has been chosen feed velocity, feed concentration and transmembrane pressure and the output data has been selected mean permeate flux. Back-propagation ANN model has been developed with a tree-layer network and Levenberg-Marquardt (LM) as learning algorithm. The transfer function is tansig and six different structures are examined which the number of neurons in hidden layer are varying between 15 and 20. The network with 17 hidden neurons have showed good performance with least deviation and error (AAD%=6.14 and MSE=0.0004). As the amounts of R2 are between 0.87 and 0.99, all structures have got the acceptable proportion of variances.
Ultrafiltration Membrane؛ Resistance-in-Series Model؛ Artificial Neural Network؛ Momentum Balance؛ Hagen Poiseuille Equation
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00564.warc.gz
|
CC-MAIN-2021-25
| 1,889 | 5 |
https://morethingsjapanese.com/how-do-i-calculate-dpmo-in-excel/
|
math
|
How do I calculate DPMO in Excel?
How do I calculate DPMO in Excel?
DPMO = Total Number of Defects found in Sample / (Sample Size * Number of Defects Opportunities per Unit in the Sample) * 1000000
- DPMO = 156 / (80 * 100) * 1000000.
- DPMO = 19500.
How do you calculate DPMO sigma in Excel?
Generally, you would use a table, such as provided in my Six Sigma Demystified book, or you can calculate sigma level directly in Excel using the formula =(NORMSINV(1-$D2))+1.5, where the data in cell D2 is entered as a decimal (for example, 30% error rate = 300,000 DPMO = 0.3 which would calculate Sigma Level as 2.02) …
How do you calculate defects per million opportunities DPMO?
Defects-Per- Million-Opportunities, which is abbreviated as DPMO. It is also called as NPMO or Nonconformities per Million Opportunities. It is defined as the ratio of the number of defects in a sample to the total number of defect opportunities multiplied by 1 million.
How do you calculate 3 sigma in Excel?
In Excel STDEV yeilds one sample standard deviation. To get 3 sigma you need to multiply it by 3. Also, if you need the standard deviation of a population, you should use STDEVP instead.
What is the difference between PPM and DPMO?
DPMO (Defects Per Million Opportunities) is used as an alternative to PPM (Parts Per Million Defective). For customers, defective items or Non- Conforming outcomes is a prime concern and they have penalty clauses based on PPM. Companies sometimes prefer to use DPMO instead of PPM as a measure of process performance.
How do you calculate DPMO and sigma level?
Once the number of products, defects, and opportunities are known, both DPMO and Sigma level can be calculated.
- Defects per opportunity (DPO)= Defect/(Product x Opportunities).
- Defects per million opportunities (DPMO) Six-Sigma is determined by evaluating the DPMO, Multiply the DPO by one million.
How do you calculate process sigma using DPMO?
DPMO is equal to the number of defects times 1,000,000. This number is divided by the number of defect opportunities per unit, times the number of units. Once you have calculated defects per million opportunities you can use a conversion table or a spreadsheet formula to turn DPMO into the Sigma.
What is the formula for 3 sigma?
The three-sigma value is determined by calculating the standard deviation (a complex and tedious calculation on its own) of a series of five breaks. Then multiply that value by three (hence three-sigma) and finally subtract that product from the average of the entire series.
What is 2sd?
Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean.
Who first thought of zero defects?
One of the most influential ideas about this was the notion of “zero defects.” This phrase was coined by Philip Crosby in his 1979 book titled, “Quality is Free.” His position was that where there are zero defects, there are no costs associated with issues of poor quality; and hence, quality becomes free.
What is DPO quality?
A measure of quality that reflects whether a specific product or service has any defects.
How to calculate DPMO and Sigma?
Still you can take down the formula to calculate it, DPMO= (defects/opprtunities) * 1000000. defect = total defects in defective units. opportunities = total unit * number of opportunity per unit. For sigma calculation you can use following formula in MSExcel, =Normsinv (1-defects/opportunities)+1.5. Cheers!
How do you calculate defects per million opportunities?
Use the DPMO formula to calculate the number of defects in the process per million opportunities. The formula is given by: DPMO=Number of Defects x 1,000,000 ((Number of Defect Opportunities/Unit) x Number of Units) For example, consider a cell phone manufacturer who wants to calculate the six sigma level of its manufacturing process.
How do you calculate Sigma in Excel?
Generally, you would use a table, such as provided in my Six Sigma Demystified book, or you can calculate sigma level directly in Excel using the formula = (NORMSINV (1-$D2))+1.5, where the data in cell D2 is entered as a decimal (for example,…
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337490.6/warc/CC-MAIN-20221004085909-20221004115909-00112.warc.gz
|
CC-MAIN-2022-40
| 4,245 | 33 |
http://www.expertsmind.com/questions/factor-30128901.aspx
|
math
|
Customer Service Chat
Get quote & make Payment
Posted Date: 1/20/2013 1:48:53 PM | Location : United States
Ask an Expert
factor, Assignment Help, Ask Question on factor, Get Answer, Expert's Help, factor Discussions
Write discussion on factor
Your posts are moderated
Write your message here..
Truth table-boolean expressions, 1. Construct a truth table for the followi...
1. Construct a truth table for the following Boolean expressions: a) ABC + A'B'C' b) ABC + AB'C' + A'B'C' c) A (BC' + B'C) 2. Simplify the following expressions:
Equations with more than one variable, Here we'll be doing is solving equat...
Here we'll be doing is solving equations which have more than one variable in them. The procedure that we'll be going through here is very alike to solving linear equations that i
Intercepts point, We have to probably do a quick review of intercepts befor...
We have to probably do a quick review of intercepts before going much beyond. Intercepts are the points on which the graph will cross the x or y-axis. Determining intercepts is
Use Substitution to solve each system of equations., 2x-y=32 2x+y=60
Intermediate algebra, f(2)=3 and g(x)=x^2+1 then gof(2)
f(2)=3 and g(x)=x^2+1 then gof(2)
Polynomial satisfy - rational root theorem, Example: prove that the roots ...
Example: prove that the roots of the below given polynomial satisfy the rational root theorem. P ( x ) = 12x 3 - 41x 2 - 38x + 40 = ( x - 4) (3x - 2) ( 4x +5) Solution
Word help, Nel skates at 18 mph and and Christine skates at 22 mph if they ...
Nel skates at 18 mph and and Christine skates at 22 mph if they can keep up that pace for 4.5 hours how far will they be a part at the end of the time
Gauss-jordan elimination, Gauss-Jordan Elimination Next we have to disc...
Gauss-Jordan Elimination Next we have to discuss elementary row operations. There are three of them & we will give both the notation utilized for each one as well as an instanc
Pre-algebra, what is the price per gallon, if it cost 61.80 to fill a 15-ga...
what is the price per gallon, if it cost 61.80 to fill a 15-gallon tank?
Solve the inequality, |3-x\2|-3>2 less then equal to
|3-x\2|-3>2 less then equal to
Accounting Assignment Help
Economics Assignment Help
Finance Assignment Help
Statistics Assignment Help
Physics Assignment Help
Chemistry Assignment Help
Math Assignment Help
Biology Assignment Help
English Assignment Help
Management Assignment Help
Engineering Assignment Help
Programming Assignment Help
Computer Science Assignment Help
IT Courses and Help
Why Us ?
~24x7 hrs Support
~Quality of Work
~Time on Delivery
~Privacy of Work
Human Resource Management
Literature Review Writing Help
Follow Us |
T & C
Copyright by ExpertsMind IT Educational Pvt. Ltd.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170875.20/warc/CC-MAIN-20170219104610-00513-ip-10-171-10-108.ec2.internal.warc.gz
|
CC-MAIN-2017-09
| 2,736 | 51 |
http://c3rcos.com/index.php/kindle/an-introduction-to-bayesian-inference-in-econometrics
|
math
|
By Arnold Zellner
This can be a classical reprint version of the unique 1971 variation of An creation to Bayesian Inference in Economics. This ancient quantity is an early creation to Bayesian inference and method which nonetheless has lasting worth for state-of-the-art statistician and pupil. The assurance levels from the basic suggestions and operations of Bayesian inference to research of purposes in particular econometric difficulties and the trying out of hypotheses and types.
Read Online or Download An introduction to Bayesian inference in econometrics PDF
Similar econometrics books
For a one-year graduate path in Econometrics. this article has goals. the 1st is to introduce scholars to utilized econometrics, together with simple options in regression research and a few of the wealthy number of types which are used while the linear version proves insufficient or irrelevant. the second one is to provide scholars with enough theoretical heritage that they are going to realize new versions of the types realized approximately right here as only normal extensions that healthy inside a typical physique of rules.
Trade, Complexity, and Evolution is an important contribution to the hot paradigm straddling economics, finance, advertising, and administration, which recognizes that advertisement platforms are evolutionary platforms, and needs to consequently be analyzed utilizing evolutionary instruments. Evolutionary structures show complex behaviors which are to an important measure generated endogenously, instead of being exclusively the made of exogenous shocks, for that reason the conjunction of complexity with evolution.
Study and resolve the typical misconceptions and fallacies that non-statisticians carry to their interpretation of statistical effects. discover the various pitfalls that non-statisticians―and additionally statisticians who current statistical stories to non-statisticians―must steer clear of if statistical effects are to be competently used for evidence-based company selection making.
- Conjoint Measurement: Methods and Applications
- Anticipating correlations : a new paradigm for risk management
- Investment Climate, Growth, and Poverty: Berlin Workshop Series 2005
- Games, fixed points and mathematical economics
- Simplicity, Inference and Modelling: Keeping it Sophisticatedly Simple
- New Introduction to Multiple Time Series Analysis
Additional info for An introduction to Bayesian inference in econometrics
Xn) ERn: plxl + PzXz + ... 11) B is convex. Let x, y E B then plxl + ... pnXn ~ m } (0 < f.. )m -. -. 12) (2 13) . 15) and B is convex. Exercise Consider a firm producing output y using inputs Xu ... , xn. Let Y be set of points (y, xl> ... , xn) which the firm can produce; its production set. Show that convexity of Y corresponds to diminishing (or constant) returns to scale. It is now possible to define a concave function. Such a function is simply one which bounds a convex set. In Fig. 12 are depicted two functions, f and g mapping a convex set S to the real line.
If a constraint holds with equality: g;(x) = 0, it is said to be binding, if it holds with strict inequality: g;(x) > 0, it is said to be slack. 1) gives rise to binding or slack constraints. :; 0 be real variables. Denote (A. 11 A.. ;'s are called multipliers or shadow prices. The reason for these mathematical entities to have such an economic sounding name will be discussed later in the chapter. ) with respect to each X; (i = 1, ... 1 (j = 1, ... ). ) We now turn to the central theorems of this chapter.
Baumol calls this satisficing behaviour. 3 provides a means of modelling this type of firm. 1). 3 are satisfied, and suppose that Q, A > 0, for otherwise the firm's behaviour would be uninteresting. 7) Then: A = R"/(1 - R") (from (3. 3, the corresponding constraint must bind. e. the firm's marginal revenue is strictly positive. For an unconstrained revenue maximiser we would have RQ = 0. How does the firm compare with a straightforward profitmaximiser? 9) since A > 0, CQ > 0) A pure profit-maximiser would have nQ = 0 Exercise Show that, with 'normal' marginal revenue and cost curves, the satisficer's output is greater than the profit-maximiser's but less than the unconstrained revenue maximiser's.
An introduction to Bayesian inference in econometrics by Arnold Zellner
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254253.31/warc/CC-MAIN-20190519061520-20190519083520-00140.warc.gz
|
CC-MAIN-2019-22
| 4,337 | 18 |
https://www.cheathappens.com/72588-PC-SD_Gundam_G_Generation_Cross_Rays_cheats
|
math
|
Our SD Gundam G Generation Cross Rays +29 trainer is now available for version 05.30.2020 and supports STEAM. These SD Gundam G Generation Cross Rays cheats are designed to enhance your experience with the game.
1. SD Gundam G Generation Cross RaysTrainer (05.30.2020)
• Unlimited Move and Attack• Super Score• Easy XP• Edit: Current HP• Edit: Current EN• Edit: Current MP• Edit: Unit Has Moved• Edit: Max HP• Edit: Max EN• Edit: ATK• Edit: DEF• Edit: MOB• Edit: MOV range• Edit: Command• Edit: Ranged• Edit: Melee• Edit: Defense• Edit: Reaction• Edit: Awaken• Edit: Auxiliary• Edit: Communications• Edit: Navigation• Edit: Maintenance• Edit: Charisma• Edit: Current Credits Capital
Updated:May 28, 2020
Compability:Win 7 , Win 8.1, Win 10+
This cheat has been scanned and is virus and adware free.
Having trouble getting our trainer to work? These steps resolve 99% of all trainer issues.
• Temporarily disable/uninstall all antivirus, firewall and other security software.
• Right click and choose "Run as Administrator" on trainer and game.
• Make sure trainer version matches game version and distribution.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00153.warc.gz
|
CC-MAIN-2021-17
| 1,166 | 10 |
http://mathforum.org/kb/message.jspa?messageID=8139884
|
math
|
In article <[email protected]>, WM <[email protected]> wrote:
> On 24 Jan., 13:36, "Jesse F. Hughes" <[email protected]> wrote: > > WM <[email protected]> writes: > > > Well, what you present below is *not* a proof of (*). > > That is wrong. You have no reason to believe that your definition of > proof is correct or the only one.
We have lots of reason to beeive that anything WM presents as a proof that is not copied from soemone more competent, is incorrect.
Some of those reasons are the obvious flaws in logic that WM is know for. > > > > > Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in > > N. > > > > Is this what you mean up 'til now? > > Yes. > > > > > > 4) Certainly you agree that, since all t_i = (t_i1, t_i2, ..., t_in) > > > have only a finite, though not limnited, number n of digits, the > > > diagonalization for every t_i yields a finite d_i =/= t_ii. > > > (The i on the left hand side cannot be larger than the i on the right > > > hand side. In other words, "the list" is a square. Up to every i it > > > has same number of lines and columns. ) > > > > No idea what you mean by the parenthetical remark. > > You will have have recognized that here the diagonal argument is > applied. It is obvious that up to every line = column the list is a > square.
Not at all. there is no reason why line n, for any n > 1, must be of length >= n.
In decimal notation, one could start with 10 lines of length 1 followed by 90 lines of length 2, followed by 900 lines of length 3, etc., and never repeat a number.
> > > > I do agree that d_i is defined for every i in N. In particular, (d_i) > > is an infinite sequence of digits. Is this what you're claiming, too? > > You've lost me. I don't know what you mean when you say, "everything > > here happens among FISs." And I'm also puzzled by the meaning of the > > next sentence. > > Every t_i is finite. Hence, in a square, if the width is finite, also > the length must be finite.
But a "diagonal" need not be, and will not be finite. > > > > Here are some obvious things. > > > > d(j) is defined for every j in N. > > d(j) != 0 and d(j) != 9 for any j in N. > > > > Hence the number d does not have a terminating decimal > > representation. > > Neither the set of t_i does have a largest element. Nevertheless there > is no t_i of actually infinite length. > > > > This looks like I do *not* agree with your claim that "d cannot be > > longer than every t_i". > > A sequence of squares will never result in a square such that all > sides are finite but the diagonal d is infinite. The overlap of d and > t_i cannot be larger than t_i. > > In particular, what would be changed in the length of d if we admitted > also non-terminating t_i (of infinite length)?
The diagonal is already required to be infinitely long, so its length need not change. --
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00323.warc.gz
|
CC-MAIN-2018-17
| 2,855 | 9 |
https://auramaine.com/library/college-algebra-graphs-and-models
|
math
|
By Marvin L. Bittinger
The Graphs and Models sequence by means of Bittinger, Beecher, Ellenbogen, and Penna is understood for assisting scholars “see the mathematics” via its specialize in visualization and know-how. those books proceed to keep up the positive factors that experience helped scholars be successful for years: specialise in features, visible emphasis, side-by-side algebraic and graphical recommendations, and real-data applications.
With the Fifth Edition, visualization is taken to a brand new point with expertise, and scholars locate extra ongoing review. moreover, ongoing evaluation has been further with new Mid-Chapter combined Review workout units and new Study consultant summaries to assist scholars arrange for exams.
- College Algebra: Graphs and types, 5th Edition
Read or Download College Algebra: Graphs and Models PDF
Similar study & teaching books
In cutting-edge city colleges, a brand new method of educating math and technological know-how is being constructed. academics are studying that by way of tending to scholars' social and emotional health, they've got better luck educating. Edited via a stellar staff, such as either psychologists and lecturers, this important source stocks confirmed suggestions for making improvements to educational functionality between tuition age little ones similar to: Discussions of the influence of "tracking" in colleges at the scholars; the actual wishes of minority scholars; getting ready scholars for a expertise pushed global within the lecture room; the new findings of the 3rd overseas Math and technological know-how learn
This e-book provides the main finished research to this point of the start line of moment language acquisition. With its specialise in the language enter that newcomers obtain and what they do with this enter, the examine sheds gentle on questions nonetheless unanswered in moment language acquisition literature, resembling what wisdom is dropped at the purchase method and the way inexperienced persons use this information to strategy new linguistic details.
The Graphs and versions sequence by means of Bittinger, Beecher, Ellenbogen, and Penna is understood for assisting scholars “see the mathematics” via its specialise in visualization and know-how. those books proceed to take care of the positive aspects that experience helped scholars be successful for years: specialize in services, visible emphasis, side-by-side algebraic and graphical options, and real-data purposes.
- The Ainu language: the morphology and syntax of the Shizunai dialect
- Researching Second Language Classrooms (ESL and Applied Linguistics Professional Series)
- Understanding Language Use in the Classroom: A Linguistic Guide for College Educators
- Educational Judgments
- My Kids Can: Making Math Accessible to All Learners, K-5
- Growing up with Two Languages: A Practical Guide
Additional resources for College Algebra: Graphs and Models
A 125p12q -14r 22 -4 b 25p8q 6r -15 48. a 3x 5y -8 4 b z -2 15 Solve. Write the answer using scientific notation. 79. S. Roadways. S. roadways in a recent year (Source: Keep America Beautiful). On average, how many pieces of trash were on each mile of roadway? Convert to scientific notation. 51. 16,500,000 52. 359,000 53. 000000437 54. 0056 55. 234,600,000,000 56. 8,904,000,000 57. 00104 58. 00000000514 59. Mass of a Neutron. 00000000000000000000000000167 kg. 60. Tech Security Spending. It is estimated that $37,800,000,000 will be spent on information technology security systems in 2013 (Source: IDC).
A polynomial with just one term, like - 9y 6, is a monomial. If a polynomial has two terms, like x 2 + 4, it is a binomial. A polynomial with three terms, like 4x 2 - 4xy + 1, is a trinomial. Expressions like 2x 2 - 5x + 3 , x 9 - 2x, and x + 1 x4 + 5 are not polynomials, because they cannot be written in the form anx n + an - 1x n - 1 + g + a1x + a0, where the exponents are all nonnegative integers and the coefficients are all real numbers. a Addition and Subtraction If two terms of an expression have the same variables raised to the same powers, they are called like terms, or similar terms.
How many miles is it from Earth to Neptune? Express your answer in scientific notation. 01 * 10. 993 + 10. 7993 * 109 mi. a Now Try Exercise 79. Most calculators make use of scientific notation. For example, the number 48,000,000,000,000 might be expressed as shown on the left below. The computation in Example 8 can be performed on a calculator as shown on the right below. 8 E 13 a 9. 3 E 7 * 3. 01 E 1 2. 7993 E 9 Order of Operations Recall that to simplify the expression 3 + 4 и 5, first we multiply 4 and 5 to get 20 and then we add 3 to get 23.
College Algebra: Graphs and Models by Marvin L. Bittinger
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576355.92/warc/CC-MAIN-20190923105314-20190923131314-00102.warc.gz
|
CC-MAIN-2019-39
| 4,769 | 20 |
https://books.google.com.jm/books?id=BMc2AAAAMAAJ&lr=
|
math
|
Other editions - View all
12 cents 15 dollars added amount annexed answer barrels of flour bushels called ciphers column common denominator common fraction composite number compound interest COMPOUND NUMBERS contained cube root cubic decimal fractions denotes difference dividend division dollars apiece dolls equal example expressed factors farthings Federal Money feet fourth gain gallons given fractions given number greatest common divisor Hence hogsheads horse hundred hundredths improper fraction inches interest of $1 lars least common multiple merchant bought miles mills mixed number months multiplicand Multiply Operation oranges ounces paid payable pence present worth principal proceed quantity quarts quotient rate per cent ratio receive Reduce remainder right hand rods shillings simple fraction slate sold Solution square root subtract tens tenths third thousandths Troy Weight units weight whole number Write yards of cloth
Page 316 - The square described on the hypothenuse of a rightangled triangle is equal to the sum of the squares described on the other two sides.
Page 313 - Multiply the divisor, thus increased, by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 223 - If there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above ; and, in like manner, from one payment to another, till all the payments are absorbed ; provided the time between one payment and another be one year or more.
Page 223 - But if any payment be made before one year's interest hath accrued, then compute the interest on the principal sum due on the obligation, for one year, add it to the principal, and compute the interest on the sum paid, from the time it was paid, up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added as above.
Page 71 - The number to be divided is called the dividend. The number by which we divide is called the divisor.
Page 176 - RULE. Divide as in whole numbers, and from the right hand of the quotient point off as many places for decimals as the decimal places in the dividend exceed those in the divisor.
Page 133 - Weight is used in weighing groceries ana all coarse articles ; as, sugar, tea, coffee, butter, cheese, flour, hay; &c., and all metals except gold and silver. 16 drams (dr.) make 1 ounce, marked oz. 16 ounces " 1 pound, " Ib. 25 pounds " 1 quarter, " qr. 4 quarters, or 100 Ibs. " 1 hundred weight, " cwt. 20 hundred weight " 1 ton,
Page 221 - The rule for casting interest, when partial payments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. " If the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent interest is to be computed on the balance of principal remaining due.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00223.warc.gz
|
CC-MAIN-2023-50
| 2,946 | 10 |
http://www.mywordsolution.com/question/draw-demand-marginal-revenue/91794
|
math
|
1. a) Draw demand, marginal-revenue, average-total-cost, and marginal-cost curves for a monopolist. Demonstrate profit-maximizing price and amount of profit.
b) On this graph, demonstrate the deadweight loss.
2. A publisher faces the following demand schedule for next novel from one of its popular authors:
Price Quantity demanded
$100 0 novels
The author is paid $2 million to prepare the book, and the marginal cost of publishing the book is a constant $10 per book.
b. Compute marginal revenue. How does marginal revenue compare to price? Describe.
c. Graph the marginal-revenue, marginal-cost and demand curves. At what quantity do the marginal-revenue and marginal-cost curves cross? What does this signify?
d. In your graph, shade in the deadweight loss. Describe in words what this means.
e. If the author were paid $3 million instead of $2 million for the book, how will this affect the publisher’s decision regarding what price to charge? Describe.
f. Assume that the publisher was not profit-maximizing but was concerned with maximizing economic efficiency. What price will it charge for the book? How much profit will it make at this price?
3.a) Give two exs of price discrimination. Identify what type of price discrimination each ex is (i.e. perfect, 2nd degree or 3rd degree).
b) Name and describe the three conditions that must hold in order for a firm to be able to price discriminate.
Larry and Moe run the only saloon in town. Larry wants to sell as many drinks as possible without losing money. Moe wants to make the largest possible profits. Using the graph below (you can re-draw it on your own paper or print this out), show the price and quantity of favored by each person.
Extra credit: Curly wants to maximize revenue. Show his favored price and quantity.
5. Based on market research, a film production company in Ectenia obtains the following information about the demand and production costs of its new DVD:
Price Quantity Total cost
1000 0 1000
950 5 1125
900 10 1500
850 15 2250
800 20 3500
750 25 5250
700 30 7500
650 35 10250
600 40 13500
550 45 17250
500 50 21500
a. Determine the price and quantity which maximizes the company’s profit.
b. Compute and graph demand, marginal revenue and marginal cost.
c. Illustrate the price and quantity that would maximize social welfare and the deadweight loss.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917125719.13/warc/CC-MAIN-20170423031205-00223-ip-10-145-167-34.ec2.internal.warc.gz
|
CC-MAIN-2017-17
| 2,336 | 31 |
https://flattening.askdefine.com/
|
math
|
- present participle of flatten
- Ellipticity redirects here. For the mathematical topic of ellipticity, see elliptic operator.
The flattening, ellipticity, or oblateness of an oblate spheroid is the "squashing" of the spheroid's pole, down towards its equator.
First and second flatteningThe first, primary flattening, f, is the versine of the spheroid's angular eccentricity ("o\!\varepsilon\,\!"), equalling the relative difference between its equatorial radius, a\,\!, and its polar radius, b\,\!:
- The flattening of the Earth in WGS-84 is 1:298.257223563 (which corresponds to a radius difference of 21.385 km of the Earth radius 6378.137 - 6356.752 km) and would not be realized visually from space, since the difference represents only 0.335 %.
- The flattening of Jupiter (1:16) and Saturn (nearly 1:10), in contrast, can be seen even in a small telescope;
- Conversely, that of the Sun is less than 1:1000 and that of the Moon barely 1:900.
The amount of flattening depends on
There is also a second flattening, f' (sometimes denoted as "n"), that is the squared half-angle tangent of o\!\varepsilon\,\!:
Flattening without pickingFlattening without picking is an efficient full-volume automatic dense-picking method for flattening seismic data. First, local dips (step-outs) are calculated over the entire seismic volume. The dips are then resolved into time shifts (or depth shifts) relative to reference trace using a non-linear Gauss-Newton iterative approach that exploits Discrete Cosine Transforms (DCT's) to minimize computation time. At each point in the image two dips are estimated; one dip in the x direction and one dip in the y direction. Because each point in the image has two dips, each horizon is estimated from an over-determined system of dips in a least-squares sense.
flattening in Danish: Fladtrykthed
flattening in German: Abplattung
flattening in French: Aplatissement
flattening in Dutch: Afplatting
flattening in Japanese: 扁平率
flattening in Slovak: Sploštenie
flattening in Serbian: Елиптицитет
flattening in Swedish: Avplattning
flattening in Chinese: 扁率
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742978.60/warc/CC-MAIN-20181116045735-20181116071735-00488.warc.gz
|
CC-MAIN-2018-47
| 2,114 | 19 |
https://blog.brainscape.com/flashcards/chapter-14-thermal-physics-7218667/packs/11373106
|
math
|
Flashcards in Chapter 14 - Thermal physics Deck (28)
The temp and pressure at which all three phases of matter can co-exist in thermal equilibrium
Another word for the state of matter e.g. solid
Condition in which there is no net transfer of thermal energy between two objects
0th law of thermodynamics namics?
If two objects A and B are separately in thermal equilibrium with object C, then A and B must also be in thermal equilibrium with each other (all same temp)
Celcius temp scale?
Temperature scale where 0 degrees celsius is the freezing point of pure water and 100 degrees Celsius is its boiling point
Absolute temp scale?
Temp scale with two fixed points as the minimum possible temp (0K) - absolute zero, and the triple point of water 273.16 K, making 1k-2k and 1 Degrees celsius-2 degrees Celsius the same size difference
What do the letters stand for in Q/E=mc(change in temp)?
Heat energy = mass x specific heat capacity x change in temperature
Specific heat capacity definition?
The amount of energy needed to heat 1kg of a substance by 1K or 1 degrees Celsius
What does the letters stand for in c=(change in E) / (mass x change in temp)?
Specific heat capacity = (change in thermal energy) divided by (the mass x the change in temp)
What are the units for the specific heat capacity?
J Kg^-1 K^-1
What does the letters stand for in E = m x L?
Thermal energy = mass x latent heat capacity
How do you do the mean root squared?
Square every individual piece of data, add them all up, divide by the total number of pieces of data and then square root it
Lowest attainable temperature, where the substance in question has minimum (but not zero thanks to quantum effects) internal energy
The unit used for absolute temperature, the size of 1K-2K is the same as 1 degree C - 2 degrees C
What is the kinetic model?
A model which describes substances in terms of the arrangement and motion of the atoms or molecules that make up the substance
The erratic random movement of microscopic particles in a fluid, as a result of continuous bombardment from molecules of the surrounding medium
Sum of the randomly distributed kinetic and potential energies of all the particles in a substance
Specific heat capacity?
The energy needed to raise the temp of 1kg of a substance by 1K, whilst the substance is in constant phase (e.g. remains as a liquid during this)
Method of mixtures?
Method of determining the specific heat capacity of a substance by mixing it or bringing it into contact with a fixed amount of a substance (at a diff temp) with known heat capacity and measuring the temp changes
Specific latent heat of vaporisation?
Energy needed to change the state of a fixed mass (1kg) of a substance from liquid to gas at constant temp (it is also the energy released by the opposite process)
The phase change of liquid to gas (evaporate)
The phase change of solid to gas
The phase change of solid to liquid
How do you change degrees Celsius to kelvin?
What effects the density of a substance?
The size of the spacing between particles in the substance
Description of the structure of a solid?
Molecules regularly arranged and packed closely with strong electrostatic forces of attraction between them, holding them in position
Describe the structure of a liquid?
Molecules are fairly close together, with more KE so can change position and flow past each other
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487647232.60/warc/CC-MAIN-20210619081502-20210619111502-00305.warc.gz
|
CC-MAIN-2021-25
| 3,366 | 44 |
http://reference.wolfram.com/language/ref/WheelGraph.html
|
math
|
Revolutionary knowledge-based programming language.
Computation-powered interactive documents.
Semantic framework for real-world data.
Central infrastructure for Wolfram's cloud products & services.
Software engine implementing the Wolfram Language.
Instant deployment across cloud, desktop, mobile, and more.
Technology-enabling science of the computational universe.
Knowledge-based, broadly deployed natural language.
Curated computable knowledge powering Wolfram|Alpha.
gives the wheel graph with n vertices .
The first few wheel graphs :
Directed wheel graphs:
Graph GraphData CycleGraph StarGraph
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814290.11/warc/CC-MAIN-20180222200259-20180222220259-00119.warc.gz
|
CC-MAIN-2018-09
| 602 | 13 |
http://pubmedcentralcanada.ca/pmcc/articles/PMC2647141/?lang=en-ca
|
math
|
The mean age (±SD) of the 123 subjects in this study was 62.2±8.2 years (41-83 years) with a racial distribution as follows: Caucasian: 90.2%; African American: 6.5%; and Other: 3.2%. In the 120 men in whom information was available, 79% of the non-cancer group had a DRE non-suspicious for cancer, while in the cancer group 71% of men had a DRE non-suspicious for cancer. Within the cancer group, 52% of men had a biopsy Gleason score of 6, 38% had a score of 7, and 10% had a score of 8 or 9.
The utility of the isoforms of free PSA, particularly [-2]proPSA was examined in the entire dataset (PSA range 0.48-33.18 ng/mL) and in the clinically important ranges incorporating 4-10 ng/mL total PSA and 2-10 ng/mL total PSA. Overall, the cancer and non-cancer groups were equivalent with respect to age (non-cancer: 61.7±8.6 years, cancer: 62.6±7.8 years) as well as total PSA concentrations (non-cancer: 6.80±5.20 ng/mL, cancer: 6.94±5.1 ng/mL). A comparison between the two groups for the measured PSA derivatives and calculated values as well as testosterone is shown in . %fPSA was significantly lower (p<0.05) in the cancer group while [-2]proPSA and %[-2]proPSA were significantly higher in the cancer group.
Comparison of mean serum values for the non-cancer and cancer groups for all subjects (n=123)
Results for the 4-10 ng/mL and 2-10 ng/mL PSA truncated ranges were similar and therefore only the 2-10 ng/mL data are presented here. In that range, similar to the overall range of data (), %[-2]proPSA was significantly higher (p<0.05) in the cancer group, as was [-2]proPSA alone and the ratio of [-2]proPSA/BPSA. %fPSA was equivalent between the cancer and non-cancer groups.
Comparison of mean serum values for the non-cancer and cancer groups for subjects in the 2-10 ng/mL PSA range (n=89)
ROC analysis of the non-cancer controls and cancer cases in all subjects is shown in and . Of the PSA derivatives, %[-2]proPSA had the greatest area under the curve (AUC) of 0.69 followed by %fPSA with an AUC of 0.61. At a sensitivity of 90 percent, corresponding specificity was 32% for %fPSA and 37% for %[-2]proPSA while at 95% sensitivity, specificity was 30% for %fPSA and 15% for %[-2]proPSA, respectively. At the optimal cutoff point for %[-2]proPSA (1.4%) corresponding to the maximal sum of sensitivity and specificity, sensitivity was 60% with a specificity of 70%. A logistic regression model was constructed combining PSA, BPSA, %fPSA, %[-2]proPSA, [-2]proPSA/BPSA, and testosterone, although only %[-2]proPSA was significant in the final model. The AUC was 0.73 and sensitivity at 90% and 95% specificity was 39% and 28%, respectively ().
ROC analysis for all subjects (n=123)
ROC analysis for all subjects (n=123) comparing PSA (AUC=0.52), %fPSA (AUC=0.61), %[-2]proPSA (AUC=0.69), and a composite logistic regression model combining PSA, BPSA, %fPSA, %[-2]proPSA, %[-2]proPSA/BPSA, and testosterone (AUC=0.73)
In the 2-10 ng/mL PSA range, individually, %[-2]proPSA again had the largest AUC (0.73) while the AUC for %fPSA was 0.53 (, ). When PSA, BPSA, %fPSA, %[-2]proPSA, [-2]proPSA/BPSA, and testosterone were evaluated together using logistic regression (), only %[-2]proPSA was a significant discriminator for prostate cancer (AUC=0.73). The specificities for %fPSA, %[-2]proPSA, and the logistic regression model at 90% sensitivity were 18%, 41%, and 32%, respectively and at 95% sensitivity were 15%, 31%, and 26% respectively.
ROC analysis for subjects in the 2-10 ng/mL PSA range (n=89)
Figure 2 ROC analysis for all subjects with PSA between 2 and 10 ng/mL (n=89) comparing PSA (AUC=0.52), %fPSA (AUC=0.53), %[-2]proPSA (AUC=0.72), and a composite logistic regression model combining PSA, BPSA, %fPSA, %[-2]proPSA, [-2]proPSA/BPSA and testosterone (more ...)
|
s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164746201/warc/CC-MAIN-20131204134546-00018-ip-10-33-133-15.ec2.internal.warc.gz
|
CC-MAIN-2013-48
| 3,792 | 11 |
https://www.wcpss.net/site/default.aspx?PageType=3&DomainID=6311&ModuleInstanceID=29224&ViewID=6446EE88-D30C-497E-9316-3F8874B3E108&RenderLoc=0&FlexDataID=70878&PageID=14108&Comments=true
|
math
|
Alexander the Great said, "I am indebted to my father for living, but to my teacher for living well". All of our teachers at Lockhart work hard every day to be sure that our students are ready for the world ahead of them. Math is something we all use every day and our students need to be prepared for the ever changing future that faces them. One of the best ways to be prepared for all the coming advancements in technology, etc. is by having a strong, fundamental understanding about math. Research shows that students who spend more time with manipulatives and learning number sense have a stronger understanding of math classes in middle and high school. At Lockhart, we are spending time training to be sure that we are focusing on a deeper understanding of math in the classroom so that our students are better prepared.
There are lots of things you can do at home to help your student with math. Since your student is reading every night, make up a math story problem based on the story that you just read. For example, if you just read The Three Little Pigs you could say: if the wolf blew down 12 sticks, 5 bricks and 3 pieces of straw how many did he blow down altogether?
Challenge your student to see the math that is all around us! Another way you can help your student get a strong sense of numbers is simply by asking "I wonder?" questions. For example: "I gave the cashier $5 for a $2.78 charge. I wonder how much money I'm going to get back?" or "Here's a cup of pasta that I need to cook. I wonder how many pieces of pasta that is?". Math learning opportunities are everywhere! Time how long it takes to go through the drive-through. See how many odd pages there are in the book they're reading. Count how many cars go through the intersection. Math can be fun; inspire your student to see the fun that math can be.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00787.warc.gz
|
CC-MAIN-2022-21
| 1,834 | 3 |
https://www.whitwellprimary.co.uk/maths-4/
|
math
|
In mathematics the children will be following the Power Maths scheme. Miss Webster continues to teach the Year 3 children for mathematics and Miss White will teach the Year 4 children for mathematics.
In Year 4 we will focus on fractions for our first topic this term. The children will learn to recognise tenths and hundredths, identify equivalent fractions, simplify fractions and look at fractions greater than one.
The children will continue to look at fractions as the term progresses. They will learn to add and subtract fractions with the same denominator. They will also look at subtracting a fraction from a whole amount. They will then understand how to find a fraction of an amount.
Their final topic for this term will be decimals. The children will explore tenths and hundredths as decimals. They will understand how to divide 1 and 2-digit numbers by 10 and 100. Finally, the children will complete calculations resulting in a decimal answer.
For more information on the mathematics topics taught in Year 3 please visit Miss Webster's class page.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00337.warc.gz
|
CC-MAIN-2022-27
| 1,060 | 5 |
https://forums.gearboxsoftware.com/t/question-regarding-fire-rate/1308463
|
math
|
It seems the list of fire rates is incomplete; as I see it the caps are results of the engine letting the gun fire every N frames. For N = 1, 2, 3, 4, 5, 6 this results in the fire rates 60/1=60, 60/2=30, 60/3=20, 60/4=15, 60/5=12, 60/6=10. The next number on the list is 6 (=60/10), but what about the fire rates for every 7, 8, or 9 frames? They are not whole numbers (the post you linked lists only the whole numbers which divide 60 evenly), but they should exist: 60/7=8.571, 60/8=7.5, 60/9=6.667.
Your Slagga has a listed fire rate of 8.8, which means it would fire every 7 frames (actual fire rate 8.571). A magazine size of 236 means it will take 7236=1652 frames to empty the magazine, that’s 1652/60=27.567 seconds. If you are sure that it took 2 more seconds to empty the magazine, there must have been something else dropping your fire rate.
Let’s calculate the time with a fire rate of 7.5 (which is every 8 frames on 60 fps, or every 4 frames on 30 fps; the fire rate of 8.57 is impossible on 30 fps because that would be every 3.5 frames, which isn’t a whole number of frames). (2368)/60=31.467 seconds, that’s two seconds off as well, but this time in the other direction. If a fluctuating frame rate was what messed up your results, you must have been at 30 fps for about half the time.
Now for the Renegade, the nearest possible fire rate cap below 9.7 is 8.571 (a shot every 7 frames) as well. The magazine size of 60 makes for a really simple formula: 7 frames per shot times 60 shots, divided by 60 frames per second, or (7*60)/60=7 - perfect! On 30 fps you would get the 7.5 fire rate cap, which would take 8 seconds to empty the magazine. I’ll assume your timing is accurate, so no reduced frame rate here.
Disclaimer: I’ve done the math without a calculator, so some numbers may be off. I’m pretty sure the results are accurate, but when in doubt, double-check them.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00606.warc.gz
|
CC-MAIN-2020-45
| 1,903 | 5 |
https://www.nagwa.com/en/videos/926108937473/
|
math
|
Find the missing divisor. 32 divided by what equals eight? Hint: eight times what equals 32?
In this problem, we’re given a division question with a missing number in it. And we’re told to find the missing divisor. The word divisor means the number that we divide by in a division calculation. So the divisor in this particular calculation is the number that we’re dividing by. It’s the second number in a division. The first number in our division is 32. Here’s what 32 looks like. We’re told to divide it by something, that’s our missing divisor, and it will give us an answer, eight.
One way of thinking about our division is as splitting 32 into groups of a certain number, and then we’ll have eight equal groups at the end of it. But eight groups of what? What should we divide 32 by to give us eight groups? We’re given a hint to help us. We can use a multiplication fact to help us find the related division fact. Eight times what equals 32? Eight times one is eight. Eight times two is 16. Eight times three is 24. And eight times four is 32. We know that eight lots of four make 32. And, we can use this multiplication fact to help us find the missing divisor.
If we know that there are eight lots of four in 32, then if we start with 32, we know we want to divide by something to give us eight equal groups. The number that we’re dividing by and the number that’s in each group is four. There are eight lots of four in 32. One, two, three, four, five, six, seven, eight. The missing divisor is four because 32 divided by four equals eight.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358064.34/warc/CC-MAIN-20210227024823-20210227054823-00138.warc.gz
|
CC-MAIN-2021-10
| 1,571 | 4 |
https://www.scribd.com/document/16866602/Eurotherm-TRI-2007
|
math
|
This action might not be possible to undo. Are you sure you want to continue?
C.B. da Silva1, I. Malico2, P.J. Coelho1 and J.C.F. Pereira1
Mechanical Engineering Department, Technical University of Lisbon, Av. Rovisco Pais, 1049-001 Lisboa, Portugal 2 Physics Department, University of Évora, R. Romão Ramalho, 59, 7000-671 Évora, Portugal
An analysis of the interaction between turbulence and radiation in homogeneous and isotropic turbulence has been carried out. A direct numerical simulation code was used to generate instantaneous turbulent scalar fields, and the radiative transfer equation was solved to provide statistical data of the radiation intensity and absorption coefficient, as well as correlations related to radiative emission and absorption. In addition, the time averaged radiative transfer equation was solved and the mean radiation intensity, mean absorption coefficient, and mean emission and absorption terms were computed and compared with those derived from the statistical data. An analysis of the number of samples required to achieve statistically meaningful results is presented. The influence of the optical thickness of the medium, mean and variance of the temperature, and variance of the mean molar fraction of the absorbing species were studied. The moments of the radiation intensity, Planck and incident mean absorption coefficient and emission and absorption correlations relevant to the turbulence – radiation interaction were calculated. It was found that in all cases the correlation between fluctuations of the absorption coefficient and fluctuations of the radiation intensity is small, which supports the optically thin flame approximation, and justifies the good predictions achieved using the time averaged radiative transfer equation.
The interaction between turbulence and radiation (TRI) is a relevant issue in turbulent reactive flows, yielding a significant increase of the radiative heat fluxes in comparison with laminar flows (Li and Modest, 2005, Coelho, 2007). However, our knowledge about such interaction is still limited. Direct numerical simulation (DNS) provides fundamental and reliable insight on turbulent flows, but it can only be applied to simple geometries and low Reynolds number flows, because of the high computational requirements. Recently, DNS has been used to investigate TRI in simple premixed and diffusion combustion systems (Wu et al., 2005, Deshmukh et al., 2005). Silva et al., 2006, have reported statistical data of the radiation intensity field in a homogeneous isotropic turbulent non-reactive flow using a pseudo-spectral code for the DNS. In the DNS calculations, a transport equation for a passive scalar was solved. The instantaneous scalar field computed from DNS was used as input data for the radiative transfer calculations, i.e., the instantaneous fields of temperature and molar fraction of an absorption species were determined from that scalar field, prescribing the mean value and the variance of the temperature and of the molar fraction of the species. Then, these instantaneous scalar fields were used to solve the radiative transfer equation (RTE) in a narrow band, along a large number of lines of sight, in order to collect statistical data. A statistical narrow band model was used to calculate the radiative properties of the medium. However, some features of the radiative calculations reported in that work were not fully consistent with the requirements of the flow simulation, such as the definition of the boundary conditions, and the dependence of the statistical data on the number of samples was not investigated. These drawbacks are eliminated in the present work, which extends the analysis to the
full spectrum rather than just one band, and examines the influence of several parameters on the radiation statistics and correlations. In addition, predictions obtained from the solution of the time averaged form of the RTE are included.
The DNS calculations were carried out using a standard pseudo-spectral code in which the temporal advancement is made with an explicit 3rd order Runge-Kutta scheme (Canuto, 1988). The physical domain is a periodic cubic box of side 2π. A DNS simulation of statistically steady (forced) homogeneous isotropic turbulence was carried out using a uniform mesh with 1923 grid nodes. The instantaneous field of a passive scalar was computed and taken as input data for the radiative transfer simulations. The analysis was performed using up to 40 instantaneous fields after all the turbulence quantities are statistically stationary. Details of the simulation may be found in Silva and Pereira, 2007. The radiative transfer calculations were performed using also a cubic box. The size of the radiation domain is different from the size of the flow domain, and it was defined in a way different from that formerly used in Silva et al., 2006. Here, the length of the side was taken as the ratio of the optical thickness of the medium, which was prescribed, to the Planck mean absorption coefficient in the absence of turbulent fluctuations. The mean and the variance of temperature and of the molar fraction of an absorbing species, taken as CO2, were also prescribed. Data from the flow domain (DNS) were rescaled into the radiation domain as reported in Silva et al., 2006, providing the instantaneous fields of temperature and molar fraction of CO2. These two scalars were assumed to be fully correlated. This assumption is justified by the experimental data in many reactive flows, which reveals a strong correlation between temperature and molar fraction of the species, and by the Burke-Schumann theory for diffusion flames (Kuo, 1986). The radiative properties of the medium were evaluated using the correlated k-distribution (CK) method (Goody et al., 1989). Under the conditions of homogeneous and isotropic turbulence, the statistical data computed from a time series of scalar data along a single optical path parallel to a coordinate axis is identical to the statistical data calculated from all optical paths parallel to the coordinate axes at a given time, as illustrated in figure 1. The statistical data reported below was obtained from the DNS data using all the available optical paths parallel to the coordinate axes, which are statistically indistinguishable. This means that 6×1922×Nt ≈ 2.2×105×Nt samples are used to obtain the results described below, Nt being the number of instantaneous fields considered. The integration of the RTE along a line of sight yields
s s s I i ,)* k (s ) = I i ,)* k (0 ) exp &' ( k i s * ds * # + ( k i s * I b,)* k s * exp &' ( k i s ** ds ** # ds * $ 0 ! $ s* ! 0 % " % "
where ki is the absorption coefficient associated with the ith quadrature point, and Δνk is the kth wavenumber interval length. In the DNS calculations, periodic boundary conditions have been enforced at the boundaries of the computational domain, in order to describe a homogeneous isotropic flow. Therefore, a similar boundary condition should be used in the radiative transfer calculations if the DNS data are taken as an input. This periodic boundary condition requires that the radiation intensity entering the calculation domain at s = 0 is equal to the leaving intensity, i.e., I i ,"! k (s ) = I i ,"! k (0 ). Accordingly, the radiation intensity entering the domain is calculated from this condition. Such a condition is actually found in real radiative transfer problems in the limit of a transparent medium or of a homogeneous and isothermal optically thick medium. The integrals in
equation (1) were numerically evaluated using Simpson’s rule, and the parameters of the CK method were interpolated from the tabulated data from Soufiani and Taine, 1997, using cubic splines in order to keep the order of accuracy of the radiative calculations consistent with the order of accuracy of the DNS solver (see Silva et al., 2006, for details). The statistical results obtained from the solution of the RTE along a sufficiently large number of optical paths, using the instantaneous scalar data from DNS, were compared with the solution of the time averaged form of the RTE. If the optically thin flame approximation (Coelho, 2007) and the CK model are used, the integration along a line of sight yields (Coelho, 2004)
) I = ## % i I i ,"$ (s ) = ## % i I i ,"$ (0 ) exp( k i s )+ k i I b,"$ k k i ( ! exp( k i s ) ! 1 !
where ωi is the quadrature weight and I i ,"! k is the time-averaged spectral radiation intensity integrated over the kth band for the ith quadrature point. The mean values of the absorption coefficient and emission term are evaluated from integration of the instantaneous values of these quantities weighted by a pdf. The blackbody radiation intensity is only a function of temperature, while the absorption coefficient depends on the temperature and on the molar fraction of CO2. Since the mean value and the variance of these scalars is prescribed and they are fully correlated, the probability density function (pdf) of ki and the joint pdf of ki and I b,"! k may be easily determined.
3 Results and Discussion
The standard radiative transfer calculations were carried out assuming that the mean temperature of the medium is 1500 K, and that the medium is a mixture of CO2 and N2, the mean molar fraction of CO2 being 0.10. The rms of temperature and of CO2 molar fraction were taken as 150 K and 0.01, respectively. The optical thickness of the medium is equal to one. In all other calculations where the influence of one variable was studied, only the variable under consideration was changed, while the remaining ones take the standard values mentioned above. Figure 2 shows the normalized values of the mean, root mean square (rms), skewness and flatness of the radiation intensity leaving the computational domain as a function of the number of instantaneous fields. The normalized values are defined as follows:
I Ib ( ) T
& I '2 # $ ! % "
Ib ( ) T
I '3 & I ' 2 # $ ! % "
I '4 & I '2 # $ ! % "
The results plotted in figure 2 show that Nt = 1 is sufficient to accurately determine the normalized mean and rms of the radiation intensity, but not enough to obtain statistically independent results for the skewness and flatness. However, when Nt exceeds 20 to 25 the influence of Nt on the results becomes marginal. The pdfs of the radiation intensity and of the Planck mean absorption coefficient, κP, which are given in figure 3 for Nt = 10, 20, 30 and 40, confirm good convergence of the results and show that the shape of the two pdfs is similar. The discussion below may be more clearly understood if equation (2) is rewritten for the time averaged RTE using total properties and total radiation intensity rather than the CK model. In that case, equation (2) reads as
) I = I (0 ) exp( k G s )+ kI b k G ( ! exp( k G s ) ! 1 !
Figure 1: Equivalence between temporal and spatial statistics under homogeneous and isotropic turbulence.
Figure 2: Radiation intensity as a function of the number of instantaneous fields.
Figure 3: Probability density functions of the radiation intensity and Planck mean absorption coefficient as a function of the number of instantaneous fields, Nt. The Planck mean and the incident mean absorption coefficients, κP and κG, respectively, are defined as follows:
$P = !
" 0 " 0
$# I b# d#
I b# d# I# d# = !
$G = !
$# I# d#
$# G# d#
The second equality in equation (5b) is only valid if the radiation intensity does not change with the direction, which is valid on a statistical basis in the case of a homogeneous and isotropic medium, i.e., different directions are statistically indistinguishable. The following definitions have also been used
$ Ib = $ P Ib = !
$ I = $G I = !
$# I b# d#
$# I# d#
Applying the periodic boundary condition, I = I (0 ) , to equation (4) yields I = kI b k G . This solution may be rewritten as
I I b ( )= " I b " P I b ! (" P " G )! ( b I b ( ) T I T )
Figure 4 shows statistical data for the radiation intensity, normalized according to equation 3, as a function of the optical thickness of the medium, mean temperature, rms of temperature, and rms of CO2 molar fraction. Figure 4 also shows the normalized mean radiation intensity predicted from the solution of the time averaged RTE using the CK model. The change of the optical thickness of the medium was accomplished by modifying the size of the domain, while keeping the Planck mean absorption coefficient of the medium, evaluated at the mean temperature and mean CO2 molar fraction, unchanged. This is the reason why the mean radiation intensity does not change with the optical thickness, as shown in figure 4(a). The statistical data exhibit a small change of I , which may be attributed to statistical errors, but the predictions based on the solution of equation 4 do not. The rms of the radiation intensity increases marginally with the increase of the optical thickness of the medium. The skewness is different from zero, i.e., the pdf is slightly asymmetric, and the flatness is just a little above 3, which means that the pdf is not exactly a Gaussian, but it is close to a Gaussian. These moments of the radiation intensity remain approximately constant when the optical thickness changes. Figure 5 shows the influence of the same variables on several correlations relevant in the emission and absorption terms of the RTE. Figure 5(a) shows that ! I b ! P I b is constant and equal to approximately 0.94 when the optical thickness of the medium varies keeping ! P ( ) constant. T Predictions based on the assumption of Gaussian scalars distribution, which are also shown in figure 5(a), closely match this value. The invariance with the optical thickness is expected, since neither the absorption coefficient nor the temperature have been changed. If the absorption coefficient and the blackbody radiation intensity are expanded as a sum of a mean value and a fluctuation, the following relation holds
" ! Ib ! P Ib = 1 + ! " Ib ! P Ib
! I !G I = 1+ ! " I " !G I
The absorption coefficient of the medium and the blackbody radiation intensity are local quantities, i.e., they depend only on the local temperature and absorption coefficient of the medium. Moreover, the absorption coefficient generally increases with the decrease of the blackbody radiation intensity, ! i.e., they are anti-correlated (Coelho, 2007). Therefore, " ! I b <0, yielding ! I b ! P I b < 1 , as found
! above. Moreover, ! I b is also independent of the optical thickness of the medium, as expected.
3.2 3.1 3.0 1.04 1.03 1.02 1.01 1.00 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.1 1 10 100 Optical thickness 3.5
3.4 3.3 3.2 3.1 3.0 1.10 1.05 1.00 0.5
I I b (T )
0.4 0.3 0.2 0.1 600
I I b (T ) Predictions
Temperature [K] 3.120 3.110 3.100 3.090 3.080 3.070 1.04 1.03 1.02 0.310 0.308 0.306 0.304 0.196 0.194 0.192 0.190 0.000
3.0 2.5 1.05 1.00 0.95 0.3 0.2 0.1 0.0 -0.1 0 50 100 150 rms ( T ) [K] 200
0.010 0.015 rms ( xCO )
Figure 4: Statistics of radiation intensity. In contrast to the blackbody radiation intensity, the radiation intensity is not a local quantity, but depends on the temperature and species concentrations along the optical path. As a consequence of this, the correlation between the absorption coefficient and the radiation intensity is expected to be small, which justifies the so-called optically thin flame approximation (Coelho, 2007). Therefore ! I ! G I is close to unity, i.e., " ! I ! is indeed small. The influence of the temperature on the radiation statistics, correlations and absorption coefficients is illustrated in figures 4(b), 5(b) and 6(a), respectively. The rms of temperature was also changed to keep T !2 T 2 constant. Since the optical thickness of the medium is now maintained constant, and the absorption coefficient depends on the temperature, the absorption coefficient changes, as well as the size of the domain. Figure 6(a) shows that the mean absorption coefficients, ! P and ! G , decrease with the increase of the temperature, as pointed out above. Still, the rms of the Planck mean absorption coefficient does not change significantly. Therefore, the second term on the right
1.00 1.05 0.98 0.96 0.94 0.92 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 0 1 10 100
Ib / Ib /
I I b (Predictions) P
_I _/ b
2 _ Ib2 _ P
2 _ I_
0 -0.20 -0.40 -0.60 -0.80
-1.00 -1.20 600
800 1000 1200 1400 1600 1800
0.90 0.8 0.4 0.0 -0.4 -0.8
0.90 0.20 0.00 -0.20 -0.40 -0.60 -0.80
-1.2 0 50 100 150 200
-1.00 -1.20 0
0.005 0.01 0.015
rms ( T ) [K]
rms ( x CO )
Figure 5: Correlations between the absorption coefficient and the radiation intensity.
rms ( rms (
2.5 2.0 1.5 1.0 0.5
0.96 0.94 0.20 0.16 0.12 0.08
1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.30 0.25 0.20 0.15 0.10 0.05 0.00
0.0 0.00 0 600 800 1000 1200 1400 1600 1800 Temperature [K]
0.04 50 100 rms ( T ) [K] 150
0 0.005 0.01 0.015 0.02 0.025 rms ( xC O 2)
Figure 6: Statistics of the Planck mean and incident mean absorption coefficients.
of equation (8a) increases in modulus with the increase of the temperature, but it is negative, and therefore ! I b ! P I b decreases, as shown in figure 5(b). A similar behaviour is found for ! I ! G I , which departs from 1 if the temperature increases, which implies an increase of the absolute value of " ! I ! , although its normalized value is rather small. Now, ! I b may be approximated by the following expression (see, e.g., Coelho, 2007, for details)
& (' T'# T '2 ( I b = ( P I b ( )$1 + 6 2 + 4 P ! T $ (P T ! T % "
in which correlations of order higher than 2 were neglected. Since T ! 2 T 2 does not change, only the last term on the right of equation (9) varies in the present case. From the previous analysis, we conclude that ! ' T ' ! T increases in modulus when the temperature increases, but it is negative. Therefore, ! I b ! I b ( ) decreases with the increase of the mean temperature. Consequently, it can T
T be concluded from equation (7) that I I b ( ) also decreases with the increase of the mean T temperature, because ! G " ! P and I b I b ( ) does not change. This is in agreement with the
findings from both the statistical analysis of the solution of the RTE based on DNS data, and from the predictions based on the solution of the time averaged RTE along with the assumption of Gaussian scalar distributions (see figure 4b). In addition, figure 4(b) shows that the skewness and the flatness are close to those found for standard conditions, except at T = 1650 K, where both the skewness and the flatness indicate that the pdf of the radiation intensity is not as close to a Gaussian as in the other cases. The influence of the rms of temperature is shown in figures 4(c), 5(c) and 6(b). In this case, the mean temperature is kept equal to 1500 K, and so the intensity of temperature fluctuations is changing. Figure 6(a) shows that the mean values of both the Planck mean and the incident mean absorption coefficients are approximately constant, while the rms increases with the increase of the rms of temperature. Hence, as in the previous case, the second term on the right of equation (8a) increases in modulus with the increase of the rms of temperature, but it is negative, and therefore ! I b ! P I b decreases, as shown in figure 5(c). This means that the increase of the second term on the right of equation (9) is overshadowed by the decrease of the third term. The absorption term, ! I ! G I , exhibits the same behaviour, ! " I " ! G I being again very small. The increase of the temperature fluctuations yields an increase of both the mean and rms of radiation intensity, as expected (Coelho, 2007). The increase is not large, because the turbulence intensity is relatively low. It is about 3% if the turbulence intensity is 10%. In this case, the second term on the right of equation (9) is equal to 0.06. However, the third term is negative, as explained above, and thus the term into parenthesis is only marginally greater than unity. The predictions based on the time averaged RTE reproduce the observed trend. The asymmetry of the radiation intensity increases with the increase of the rms of temperature, and the flatness does not exhibit a monotonous behaviour, but it is close to 3, as in the case of a Gaussian pdf. Finally, figures 4(d), 5(d) and 6(c) show the influence of the rms of the CO2 molar fraction. Its mean value was set to 0.1. Figure 6(c) shows that ! P decreases with the increase of the rms of CO2 concentration. This is due to the anti-correlation between the absorption coefficient and the
blackbody radiation intensity, on the one hand, and to the nonlinearity between the blackbody radiation intensity and the temperature, on the other hand. The calculations based on assumed Gaussian scalars distribution correctly predicted that evolution, but slightly overpredict the statistical data. The rms of the Planck mean absorption coefficient also decreases with the increase of the rms of CO2 concentration, as revealed by figure 6(c). Moreover, the ratio of the rms to the mean value of the Planck mean absorption coefficient also decreases with the increase of the rms of CO2 concentration. Accordingly, the analysis of equation (9) shows that ! I b ! P I b increases, as
T confirmed by the results plotted in figure 5(d). Moreover, I b I b ( ) remains constant while ! P ! G decreases. Therefore, the first and the second terms on the right of equation (7) have T T opposite contributions to I I b ( ). Figure 4(c) shows that I I b ( ) increases slightly with the increase of the rms of the CO2 molar fraction, which means that the contribution of the first term in equation (7) dominates. The normalized correlation ! I ! G I increases also slightly. However, in all studied cases, the correlation between fluctuations of the absorption coefficient and fluctuations of the radiation intensity is small, which supports the optically thin flame approximation, and justifies the good predictions achieved using the time averaged radiative transfer equation.
The interaction between turbulence and radiation in homogeneous and isotropic turbulence was studied. The influence of the optical thickness of the medium, mean and variance of the temperature, and variance of the mean molar fraction of the absorbing species were investigated. It was found that about 2×105 samples, corresponding to a single instantaneous field, provide accurate results for the mean and variance of the radiation intensity leaving the domain, but more than 20 instantaneous fields are needed to obtain accurate values for the higher moments of the radiation intensity, as well as a converged probability density function of the radiation intensity. If the optical thickness of the medium is changed while the Planck mean absorption coefficient for the mean properties is fixed, the moments of the radiation intensity and the emission and absorption correlations remain approximately constant. The increase of the temperature of the medium, keeping the same intensity of temperature fluctuations, causes a decrease of the mean values of the normalized radiation intensity and Planck mean absorption coefficient, as well as the emission and absorption normalized correlations. If the temperature if left constant while its rms increases, then the mean normalized radiation intensity increases, the mean value of the Planck mean absorption coefficient remains approximately constant, and the normalized emission and absorption correlations decrease. The increase of the rms of the molar fraction of the absorbing species, while keeping its mean value constant, yields an increase of the normalized mean radiation intensity, a decrease of the mean value of the Planck mean absorption coefficient and an increase of the normalized emission and absorption correlations. In all cases, the correlation between fluctuations of the absorption coefficient and fluctuations of the radiation intensity is small, which supports the optically thin flame approximation, and justifies the good predictions achieved using the time averaged radiative transfer equation.
This work was developed within the framework of project POCI/EME/59879/2004, which is financially supported by FCT-Fundação para a Ciência e a Tecnologia, programme POCI 2010 (29.82% of the funds from FEDER and 70.18% from OE).
Canuto, C., Hussaini, M.Y., Quarteroni, A., Zang, T.A., 1988, Spectral Methods in Fluid Mechanics, Springer-Verlag, New York, pp. 201-212. Coelho, P.J., 2004, Detailed Numerical Simulation of Radiative Transfer in a Non-Luminous Turbulent Jet Diffusion Flame, Combustion and Flame, 136, 481-492. Coelho, P.J., 2007, Numerical Simulation of the Interaction between Turbulence and Radiation in Reactive Flows, Progress in Energy and Combustion Science, 33, 311-383. Deshmukh, K.V., Haworth, D.C. and Modest, M.F., 2007, Direct Numerical Simulation of Turbulence–Radiation Interactions in Homogeneous Nonpremixed Combustion Systems, Proc. Combustion Institute, 31, 1641-1648. Goody, R.M., West, R., Chen., L and Chrisp, D, 1989, The Correlated-k Method for Radiation Calculations in Nonhomogeneous Atmospheres, J. Quant. Spectrosc. Radiat. Transfer, 42, 539-550. Kuo, K.K., 1986, Principles of Combustion, John Wiley & Sons, New York. Li, G. and Modest, M.F., 2005, Numerical Simulation of Turbulence–Radiation Interactions in Turbulent Reacting Flows, in Modelling and Simulation of Turbulent Hest Transfer, Sundén B. and Faghri, M. (eds.), Southampton, WIT Press, pp. 77-109. Silva, C.B., Malico, I., Coelho, P.J. and Pereira, J.C.F., 2006, An Exploratory Investigation of Radiation Statistics in Homogeneous Isotropic Turbulence, in Computational Thermal Radiation in Participating Media II, Lemonnier, D., Selçuk, N. and Lybaert, P. (eds.), Lavoisier, pp. 215-224. Silva, C.B. and Pereira, J.C.F., 2007, Analysis of the Gradient Diffusion Hypothesis in Large Eddy Simulations Based on Transport Equations, Physics of Fluids, 19, 035106. Soufiani, A, Taine, J., 1997, High Temperature Gas Radiative Property Parameters of Statistical Narrow-Band Model for H2O, CO2 and CO, and Correlated-k Model for H2O and CO2, Int. J. Heat and Mass Transfer, 40, 987-991. Wu, Y., Haworth, D.C., Modest, M.F. and Cuenot, B., 2005, Direct Numerical Simulation of Turbulence/Radiation Interaction in Premixed Combustion Systems, Proc. Combustion Institute, 30, 639-646.
This action might not be possible to undo. Are you sure you want to continue?
We've moved you to where you read on your other device.
Get the full title to continue reading from where you left off, or restart the preview.
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698544358.59/warc/CC-MAIN-20161202170904-00375-ip-10-31-129-80.ec2.internal.warc.gz
|
CC-MAIN-2016-50
| 26,563 | 90 |
https://www.autoitscript.com/forum/tags/max%20character%20length/
|
math
|
Search the Community
Showing results for tags 'max character length'.
Found 1 result
Hi, Just a general question, I'm trying to write a calculator for a specific sum but came across a problem. Question is, what is the max amount of characters in a string of chars that can be assigned to a variable? for example the issues i'm having are as follows. If I have the code... Local $var = 1001011001010100001 msgbox(0,"",$var) the message box displays 1001011001010100001 as expected. But if the string is slightly longer such as... Local $var = 1001011001010100001001001000100100 msgbox(0,"",$var) I get the following in my message box, 9223372036854775807. If I wrap my number in "" then the message box displays correctly.. example Local $var = "1001011001010100001001001000100100" msgbox(0,"",$var) However if my string of characters is too long, autoit returns an error of 'Unterminated string'. Is this a limitation of AutoIt? I'm trying to work out a sum but none of my calculators can display enough digits and they all just give an error.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039747024.85/warc/CC-MAIN-20181121032129-20181121054129-00264.warc.gz
|
CC-MAIN-2018-47
| 1,043 | 4 |
https://www.prep4usmle.com/forum/thread/132295/
|
math
|
|Prep for USMLE|
| Forum | Resources||New Posts | Register | Login||» |
Always take suhagra 100 mg exactly as prescribed by your physician. This medicine is not appropriate for pregnant women or children, or teenagers under 18 years old. age. Before taking this medication, tell your physician if you suffer from serious liver or heart problems or if you've recently suffered a stroke, heart attack, or suffer from an abnormally low level of blood pressure. Inform your doctor that you are suffering from these or other health conditions before taking the medication. Sildenafil is a medication that acts by decreasing PDE-5 while also increasing NO activation in the penis. The mechanism behind the penis being sexually erected is the release of Nitric oxide after sexual stimulation takes place. The no receptor is activated by an enzyme called Guanylate cyclase which triggers an increase in cyclic Guanosine Monophosphate (cGMP) that causes the relaxation of smooth muscle and the flow of blood to the penis.
Always ensure that you take SUHAGRA 100MG precisely as recommended by your doctor. This medication is not suitable for women who are pregnant or children or teenagers who are less than 18 years of. age. Before taking this medicine consult your physician if you are suffering from serious heart or liver problems or if you've had a heart attack or unusually low blood pressure. Inform your doctor if you suffer from any of these issues before beginning to take the drug.
visit also: tadagra | manforce 50mg
Edited by Ninagrace123 on Feb 19, 2022 - 1:33 AM
| Similar forum topics|
Suhagra 100 Mg pills
Suhagra 100 Mg Best Pill
Best ED pill Suhagra 100
| Related resources|
Exam Master USMLE Step 2 CS Online
Archer Online USMLE Step 2
Advertise | Support | Premium | Contact
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337889.44/warc/CC-MAIN-20221006222634-20221007012634-00654.warc.gz
|
CC-MAIN-2022-40
| 1,827 | 14 |
https://plosjournal.deepdyve.com/browse/journals/analysis-and-mathematical-physics/2013/v4/i3
|
math
|
1 - 2 of 2 articles
$$\mathcal R $$
be the restriction of the spherical Radon transform to the set of spheres centered on a hypersurface
$$\mathcal S $$
. We study the construction of a function
$$\mathcal R (f)$$
by a closed-form formula. We approach the...
be a locally compact group,
be a locally compact Abelian (LCA) group,
$$\theta :H\rightarrow Aut(K)$$
be a continuous homomorphism, and let
$$G_\theta =H\ltimes _\theta K$$
be the semi-direct product of
Read and print from thousands of top scholarly journals.
Already have an account? Log in
Bookmark this article. You can see your Bookmarks on your DeepDyve Library.
To save an article, log in first, or sign up for a DeepDyve account if you don’t already have one.
Sign Up Log In
To subscribe to email alerts, please log in first, or sign up for a DeepDyve account if you don’t already have one.
To get new article updates from a journal on your personalized homepage, please log in first, or sign up for a DeepDyve account if you don’t already have one.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00133.warc.gz
|
CC-MAIN-2021-49
| 1,021 | 20 |
http://e-booksdirectory.com/details.php?ebook=4791
|
math
|
by Jesper M. Moller
Number of pages: 75
These notes are intended as an introduction to general topology. They should be sufficient for further studies in geometry or algebraic topology. The text covers: Sets and maps; Topological spaces and continuous maps; Regular and normal spaces; Relations between topological spaces.
Home page url
Download or read it online for free here:
by Alex Kuronya
Contents: Basic concepts; Constructing topologies; Connectedness; Separation axioms and the Hausdorff property; Compactness and its relatives; Quotient spaces; Homotopy; The fundamental group and some applications; Covering spaces; etc.
by T. W. Körner - University of Cambridge
Contents: What is a metric?; Examples of metric spaces; Continuity and open sets for metric spaces; Closed sets for metric spaces; Topological spaces; Interior and closure; More on topological structures; Hausdorff spaces; Compactness; etc.
by Robert B. Ash - Institute of Electrical & Electronics Engineering
A text for a first course in real variables for students of engineering, physics, and economics, who need to know real analysis in order to cope with the professional literature. The subject matter is fundamental for more advanced mathematical work.
by John McCleary - American Mathematical Society
A focused introduction to point-set topology, the fundamental group, and the beginnings of homology theory. The text is intended for advanced undergraduate students. It is suitable for students who have studied real analysis and linear algebra.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00877.warc.gz
|
CC-MAIN-2022-49
| 1,528 | 13 |
https://researchpaperswriter.com/the-accompanying-table-shows-annual-return-data-from-2001-2009-for-vanguards-balanced-index-and/
|
math
|
The accompanying table shows annual return data from 2001-2009 for Vanguard’s Balanced Index and European Stock Index mutual funds.(excel attached) A. )Set up the hypotheses to test whether the mean returns of the two funds differ. (Hint: This is a matched -pairs comparison) B.)At the 5% level of significance, determine if the mean weight gain of women due to pregnancy is more than 35 pounds.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00584.warc.gz
|
CC-MAIN-2021-39
| 397 | 1 |
http://www.netcomputerscience.com/2015/05/ugc-net-computer-science-solved-paper-ii-dec-2004-p3.html
|
math
|
21. What item is at the root after the following sequence of insertions into an empty splay tree:
1, 11, 3, 10, 8, 4, 6, 5, 7, 9, 2 ?
(A) 1 (B) 2
(C) 4 (D) 8
22. Suppose we are implementing quadratic probing with a Hash function, Hash (y)=X mode 100. If an element with key 4594 is inserted and the first three locations attempted are already occupied, then the next cell that will be tried is:
(A) 2 (B) 3
(C) 9 (D) 97
23. Weighted graph:
(A) Is a bi-directional graph
(B) Is directed graph
(C) Is graph in which number associated with arc
(D) Eliminates table method
24. What operation is supported in constant time by the doubly linked list, but not by the singly linked list?
(A) Advance (B) Backup
(C) First (D) Retrieve
25. How much extra space is used by heap sort?
(A) O(1) (B) O(Log n)
(C) O(n) (D) O(n2)
26. Error control is needed at the transport layer because of potential error occurring ..............
(A) from transmission line noise
(B) in router
(C) from out of sequence delivery
(D) from packet losses
27. Making sure that all the data packets of a message are delivered to the destination is ................ control.
(A) Error (B) Loss
(C) Sequence (D) Duplication
28. Which transport class should be used with a perfect network layer?
(A) TP0 and TP2 (B) TP1 and TP3
(C) TP0, TP1, TP3 (D) TP0, TP1, TP2, TP3, TP4
29. Which transport class should be used with residual-error network layer?
(A) TP0, TP2 (B) TP1, TP3
(C) TP1, TP3, TP4 (D) TP0, TP1, TP2, TP3, TP4
30. Virtual circuit is associated with a ..................... service.
(A) Connectionless (B) Error-free
(C) Segmentation (D) Connection-oriented
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218194600.27/warc/CC-MAIN-20170322212954-00169-ip-10-233-31-227.ec2.internal.warc.gz
|
CC-MAIN-2017-13
| 1,629 | 35 |
https://books.google.ie/books?id=9Ns3AAAAMAAJ&pg=PR6&focus=viewport&vq=%22have+an+angle+of+one+equal+to+an+angle+of+the+other,%22&dq=editions:UOM39015065618988&lr=&output=html_text
|
math
|
« PreviousContinue »
The rate of change is introduced informally in connection with the quadratic function, while formal treatment is postponed until later. The difficulties in grasping the concept of a derivative are thus separated, and time for thorough assimilation is afforded.
9. Integration is used to obtain the volumes of a pyramid, cone, and sphere.
10. Tables of squares, square roots etc., that is, tables of functions familiar to the student, are used in advance of other tables. The table of logarithms is introduced as a general tool, and it is not regarded as something to be used primarily with the trigonometric functions. Use is made of various tables throughout a large part of the course, so that the student acquires facility in their use and in the selection of the most suitable table for a given computation.
11. The use of the slide rule, and of logarithmic and semilogarithmic cross section paper, is explained in connection with the logarithmic function.
12. The functions en and bt, which occur frequently in the applications of mathematics, are treated at some length.
13. An effort has been made to render explicit the purpose of the various parts of the course.
14. Formal proofs of a number of theorems are omitted, and some are assumed without proof. The appeal to the intuition underlying most of these assumptions is justified by the belief that the logical presentation of these theorems requires a foundation too abstract for the general student, or too cumbersome for the purpose to be served.
15. The course includes more than a year's work so that teachers have an opportunity for choice of topics, and abundant material is provided for selected sections which progress more rapidly than the average.
16. The course increases in difficulty with a corresponding increase in interest and gain in power on the part of the student.
17. A very wide range of problems, varying in difficulty, makes it possible for the instructor to emphasize different aspects of the subject, to select exercises suitable for students of different abilities, and to assign different sets of exercises in different years.
18. Many of the exercises in the later chapters, while dealing with the newer topics, are constructed to afford review of principles presented earlier in the course, and to correlate various parts of the subject.
We thank several friends for the inspiration of their interest and for their suggestions. We also thank the Trustees of The University of Rochester for making it possible for us to use the text in the classroom throughout our experimentation, in which we have participated equally. Combined courses are still to be regarded as in the experimental stage, but it is our conviction that they are fundamentally sound, and we shall feel well repaid if this volume contributes something of value to their development.
ARTHUR SULLIVAN GALE
CHARLES WILLIAM WATKEYS THE UNIVERSITY OF ROCHESTER May, 1920.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00550.warc.gz
|
CC-MAIN-2022-21
| 2,948 | 15 |
http://quant.stackexchange.com/questions/tagged/credit-ratings+inflation
|
math
|
Quantitative Finance Meta
to customize your list.
more stack exchange communities
Start here for a quick overview of the site
Detailed answers to any questions you might have
Discuss the workings and policies of this site
Do bond credit ratings suffer from “ratings inflation”?
A friend of mine who studies game theory suggested that credit ratings from the bond ratings agencies, such as Moody's, S&P, and Fitch, may suffer from a sort of "ratings inflation" similar to the ...
Aug 22 '11 at 16:01
newest credit-ratings inflation questions feed
Hot Network Questions
Why is router considered to be a layer 3 device while it can handle layer 4 headers too?
Tikz plot polar equation with foreach
Good Textbooks for Real Analysis and Topology.
Birds sitting on electric wires: potential difference between their feet
Why did Microsoft choose the word "Recycle Bin"?
Why did early telephones use a rotary dial instead of 10 individual buttons?
Counting characters in a text file using java
How do Time Lords prove to others that they are the same person as their past regenerations?
Is machine language always binary?
Algebra: What does "is defined for" mean?
Is it OK for an MSc Student to Review a Scientific Paper?
Turn off Windows Phone keys vibration
When to use esc_html and when to use sanitize_text_field?
301 redirect + RewriteRule not working together?
Drawing vertical hierarchical n-ary tree in tikz
How to use special characters like the question mark as a variable in a formula?
Single word for shopping after comparing price and features across multiple shops and product types
How to select the first "lowest" (first valley) number in a list?
What makes it illegal to use the information learned by exploiting a bug?
Why don't I have to install programs?
How can I prove that there is no set containing itself without using axiom of foundation?
Riesz's Theorem of compactness
What does the "phase margin" parameter of an opamp imply?
more hot questions
Life / Arts
Culture / Recreation
TeX - LaTeX
Unix & Linux
Ask Different (Apple)
Geographic Information Systems
Science Fiction & Fantasy
Seasoned Advice (cooking)
Personal Finance & Money
English Language & Usage
Mi Yodeya (Judaism)
Cross Validated (stats)
Theoretical Computer Science
Meta Stack Exchange
Stack Overflow Careers
site design / logo © 2014 stack exchange inc; user contributions licensed under
cc by-sa 3.0
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00181-ip-10-147-4-33.ec2.internal.warc.gz
|
CC-MAIN-2014-15
| 2,392 | 52 |
https://modernmood.me/18-absolutely-unusual-objects-that-are-not-so-unusual-in-alternate-realities/
|
math
|
It would be fun to think about parallel realities, alternate worlds, and universes where everything differs from what is normal and usual for us. For example, the seas could be the skies in other worlds, apples wouldn’t be fruits, and all colors and shapes would be mixed into entirely different things. We can imagine whatever our mind allows us to, but we can also look through some real examples of stuff, objects, and phenomena that look exactly like they are coming from another world. We have prepared 18 photos that will take you on a virtual tour to elsewhere. So prepare yourself to see some really fascinating and unusual stuff.
#1. There was a billiard ball inside of my bocce ball
#2. It’s weird how something this creative can be so creepy
#3. The philippine 1000 peso banknote is quite cool
#4. I feel like I can SMELL the chlorine while sleeping
#5. Need a regular banana for scale.
#6. “Give me a hand.”
#7. Ice cubes or thick glass lenses?
#8. These salt & pepper pots are inseparable
#9. My cat has 26 toes (18 is normal, 28 is WR)
#10. Playing a duet would be uncomfortable, though
#11. Fleshy handsome Squirtle
#12. This log with coins couldn’t bear the heaviness
#13. Justice in Paradise
#14. A stained glass window hanging in a forest in Belgium
#15. This big cat bus is for big cat tours.
#16. Fresh wolf print while lost in Denali, Alaska
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337338.11/warc/CC-MAIN-20221002150039-20221002180039-00209.warc.gz
|
CC-MAIN-2022-40
| 1,372 | 17 |
https://blog.findwhosabi.com/2021/09/16/2021-bbnaija-10-things-emmanuel-said-when-he-met-biggie-in-the-diary-room/
|
math
|
The Diary Session is the period housemates interact with the moderator of Big Brother Nigeria (BBNaija) on things bothering them during their stay in the house although they are still denied the opportunity of seeing his face and knowing his identity.
This is the period housemate pour out their feelings in confidentiality without having to feel other housemates would listen. Biggie on the other hand instructs them on what to do that won’t be known by other housemates.
Each housemate took their turn in the Diary Session and this is a summary of Emmanuel when he met Biggie one on one.
1} I’m feeling okay and totally normal.
2} I’m mimicking Saskay and it’s really tough because most times Saskay is sleeping and I’m not the sleeping type.
3} my relationship with Liquorose is going through trials and tribulations right now.
4} Liquorose attitude came as a surprise to me and I still don’t know what I did wrong.
5} I’ve been asking myself and Liquorose question about what I did wrong to her
6} Liquorose sudden attitude has been disturbing me.
7} it’s not just possible to allow Liquorose go, i will keep trying to get her back because deep down I like her alot.
8} I want me and Liquorose to get back the way we use to be.
9} I feel like I messed up and it’s really hard on me, I’m trying not to loose focus in the game.
10} I really care about Liquorose.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00431.warc.gz
|
CC-MAIN-2022-05
| 1,384 | 13 |
http://mathhelpforum.com/discrete-math/100690-counting.html
|
math
|
10. In a 6-cylinder automobile engine, the even numbered cylinders are on the engine’s left side and the odd numbered cylinders are on the right side. (Drawing a picture will help you to visualize this.) A good firing order is a permutation of the numbers 1 to 6 in which the right and left sides are alternated.
(a) How many possible good firing orders are there? (b) Repeat this problem for an engine with 2n cylinders.
Any thoughts? Thanks.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189734.17/warc/CC-MAIN-20170322212949-00294-ip-10-233-31-227.ec2.internal.warc.gz
|
CC-MAIN-2017-13
| 445 | 3 |
https://svpwiki.com/Propositions-of-Astronomy
|
math
|
Planets (condensed matter) in Space (dispersed matter)
(click to enlarge)
Second Proportion: The square of the diameter of the moon is to the square of the diameter of the earth, as the moon's time around the earth is to the earth's time round the sun, - the time here meant being circular time, as before.
First Proportion: As one primary circumference of a circle is to the moon's time about the earth over the value of a complete circle in space, so is the moon's time round the earth to the earth's time round the sun over the value of a complete circle in space.
Second Proportion: The square of the diameter of the moon is to the square of the diameter of the earth, as the moon's time round the earth over the value of a complete circle in space, is to the earth's time round the sun over the value of a complete circle in space.
The mean distance of the sun's center from the center of the earth, or that at which the earth would revolve, if the area or plane of her elliptical orbit were made the area of a circle, is eleven thousand six hundred and sixty four diameters of the earth, neither more nor less; admitting, therefore, that the earth's diameter is 7,912 English miles (which no doubt is pretty nearly), then the sun's center is distant from the earth's center as above 92,285,568 English miles, and neither one mile more or less.
Axioms as proven herein and as self-evident: First: The circle is the basis or beginning of all magnitude or area. Second: Any expression of numbers in relation to material things is also an expression of magnitude. Third: A point is therefore a magnitude when considered as one. Fourth: A point in reference to space or extension on all sides of it, is therefore a molecule or globe, and in reference to a plane, it is a circle.
Algebraic Values of Trigonometric Functions
Kepler Music of the Spheres
Kepler Music Theory
Kepler Theory of Harmony
Kepler's First Law
Kepler's Second Law
Kepler's Third Law
Part 22 - Solar Rings and Planetary Formation
Propositions Demonstrating the Relative Properties of Straight and Curved Lines
Quadrature of the Circle
Table 1 - Relations of Thirds
The Ox and the Chamois
This Three Dimensional Cube Universe of Nine
We Now Build the Nine Equators of Cube-Sphere Wave-Fields
Wheelwork of Nature
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00447.warc.gz
|
CC-MAIN-2023-50
| 2,281 | 22 |
https://openstax.org/books/contemporary-mathematics/pages/10-6-area
|
math
|
After completing this section, you should be able to:
- Calculate the area of triangles.
- Calculate the area of quadrilaterals.
- Calculate the area of other polygons.
- Calculate the area of circles.
Some areas carry more importance than other areas. Did you know that in a baseball game, when the player hits the ball and runs to first base that he must run within a 6-foot wide path? If he veers off slightly to the right, he is out. In other words, a few inches can be the difference in winning or losing a game. Another example is real estate. On Manhattan Island, one square foot of real estate is worth far more than real estate in practically any other area of the country. In other words, we place a value on area. As the context changes, so does the value.
Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. As opposed to linear measurements like perimeter, which in in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area.
Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as or Note that linear measurements have no exponent above the units or we can say that the exponent is 1.
Area of Triangles
The formula for the area of a triangle is given as follows.
The area of a triangle is given as where represents the base and represents the height.
For example, consider the triangle in Figure 10.133.
The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area:
In Figure 10.134, the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then
Finding the Area of a Triangle
Find the area of this triangle that has a base of 4 cm and the height is 6 cm (Figure 10.135).
Using the formula, we have
Your Turn 10.41
Area of Quadrilaterals
A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle.
In Figure 10.137, a grid is represented with twelve squares across each row, and twelve squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where , is . The area of the square in Figure 10.137 is
The formula for the area of a square is or
Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in Figure 10.138 has width equal to 5 in and length equal to 12 in. The area is
The area of a rectangle is given as
Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?
Finding the Area of a Rectangle
You have a garden with an area of 196 square feet. How wide can the garden be if the length is 28 feet?
The area of a rectangular region is Letting the width equal :
Your Turn 10.42
Determining the Cost of Floor Tile
Jennifer is planning to install vinyl floor tile in her rectangular-shaped basement, which measures 29 ft by 16 ft. How many boxes of floor tile should she buy and how much will it cost if one box costs $50 and covers
The area of the basement floor is We will divide this area by Thus, Therefore, Jennifer will have to buy 24 boxes of tile at a cost $1,200.
Your Turn 10.43
The area of a parallelogram can be found using the formula for the area of a triangle. Notice in Figure 10.139, if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram:
The area of a parallelogram is
For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be
Finding the Area of a Parallelogram
In the parallelogram (Figure 10.140), if find the exact area of the parallelogram.
Using the formula of we have
Your Turn 10.44
Finding the Area of a Parallelogram Park
The boundaries of a city park form a parallelogram (Figure 10.142). The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for $0.50 per square foot, installation is $1.50 per square foot, and the cost of the equipment for the day is $100.
Step 1: As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft.
Step 2: The park has the shape of a parallelogram, and the formula for the area is :
Step 3: The city needs to purchase of sod. The cost will be $0.50 per square foot for the sod and $1.50 per square foot for installation, plus $100 for equipment:
Your Turn 10.45
Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases and and height is given here.
The formula for the area of a trapezoid is given as
For example, find the area of the trapezoid in Figure 10.143 that has base equal to 8 cm, base equal to 6 cm, and height equal to 6 cm.
The area is .
Finding the Area of a Trapezoid
(Figure 10.144) is a regular trapezoid with Find the exact perimeter of , and then find the area.
The perimeter is the measure of the boundary of the shape, so we just add up the lengths of the sides. We have Then, the area of the trapezoid using the formula is .
Your Turn 10.46
The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals.
The area of a rhombus is found using one of these formulas:
- where and are the diagonals.
- where is the base and is the height.
For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is and the measure of find the measure of To solve this problem, we input the known values into the formula and solve for the unknown. See Figure 10.146.
We have that
Finding the Area of a Rhombus
Find the measurement of the diagonal if the area of the rhombus is and the measure of
Use the formula with the known values:
Your Turn 10.47
Finding the Area of a Rhombus
You notice a child flying a rhombus-shaped kite on the beach. When it falls to the ground, it falls on a beach towel measuring by You notice that one of the diagonals of the kite is the same length as the width of the towel. The second diagonal appears to be 2 in longer. What is the area of the kite (Figure 10.147)?
Using the formula, we have:
Your Turn 10.48
Area of Polygons
To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius of a regular polygon is also a line segment that starts at the center but extends to a vertex. See Figure 10.148.
The area of a regular polygon is found with the formula where is the apothem and is the perimeter.
For example, consider the regular hexagon shown in Figure 10.149 with a side length of 4 cm, and the apothem measures
We have the perimeter, We have the apothem as Then, the area is:
Finding the Area of a Regular Octagon
Find the area of a regular octagon with the apothem equal to 18 cm and a side length equal to 13 cm (Figure 10.150).
Using the formula, we have the perimeter Then, the area is
Your Turn 10.49
Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency.
Carpeting comes in units of square yards. Your living room measures 21 ft wide by 24 ft long. How much carpeting do you buy?
We must convert feet to yards. As there are 3 ft in 1 yd, we have and Then,
Your Turn 10.50
Area of Circles
Just as the circumference of a circle includes the number so does the formula for the area of a circle. Recall that is a non-terminating, non-repeating decimal number: . It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area.
The area of a circle is given as where is the radius.
For example, to find the of the circle with radius equal to 3 cm, as shown in Figure 10.152, is found using the formula
Finding the Area of a Circle
Find the area of a circle with diameter of 16 cm.
The formula for the area of a circle is given in terms of the radius, so we cut the diameter in half. Then, the area is
Your Turn 10.51
Determining the Better Value for Pizza
You decide to order a pizza to share with your friend for dinner. The price for an 8-inch diameter pizza is $7.99. The price for 16-inch diameter pizza is $13.99. Which one do you think is the better value?
The area of the 8-inch diameter pizza is The area of the 16-inch diameter pizza is Next, we divide the cost of each pizza by its area in square inches. Thus, per square inch and per square inch. So clearly, the 16-inch pizza is the better value.
Your Turn 10.52
Applying Area to the Real World
You want to purchase a tinted film, sold by the square foot, for the window in Figure 10.153. (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase.
First, the rectangular portion has . For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is
Your Turn 10.53
Area within Area
Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems.
Finding the Area within an Area
The patio in your backyard measures 20 ft by 10 ft (Figure 10.155). On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over?
The hot tub has a radius of 3 ft. That area is then The total square feet taken up with the table and chairs and the hot tub is The area left over is equal to the total area of the patio, minus the area for the table and chairs and the hot tub. Thus, the area left over is
Your Turn 10.54
Finding the Cost of Fertilizing an Area
A sod farmer wants to fertilize a rectangular plot of land 150 ft by 240 ft. A bag of fertilizer covers and costs $200. How much will it cost to fertilize the entire plot of land?
The plot of land is It will take 7.2 bags of fertilizer to cover the land area. Therefore, the farmer will have to purchase 8 bags of fertilizer at $200 a bag, which comes to $1,600.
Your Turn 10.55
People in Mathematics
Heron of Alexandria
Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, Mechanics and Optics. His contribution to the field of mathematics was enormous. Metrica, a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle (Figure 10.159),
Heron’s formula is where is the semi-perimeter calculated as
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00540.warc.gz
|
CC-MAIN-2023-23
| 14,014 | 111 |
https://ask.manytutors.com/my/tags/maths?page=4
|
math
|
These are problems that other students can't solve. Your future exam questions are probably inside so use these to practice!
1) Snap homework photo 2) Upload and wait 3) Solution emailed to you free!
The instruction says find the area...
How to find out the answer
How to find the perimeter?
Im not getting the answer!
Please help me with question 12
This is an science question not mat...
I really need help with question 11...
How ti do these question?
I can't solve this question
how to find the shade area?
It is in Malay.. Pls help me!! Topi...
It is science actually hehe
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00408.warc.gz
|
CC-MAIN-2019-35
| 577 | 14 |
https://www.coursehero.com/file/8495467/Homework-2-solutions/
|
math
|
kanafani (ak22724) – Homework 2 – Sutcliffe – (52440)1Thisprint-outshouldhave20questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.Read the questions carefully.Look forwhether youve been give or asked for moLaR-ity or moLaLity! The last half of the questionsrequire you to consider the vant’Hoff factor i,for solutes that might dissociate into severalions, so read carefully. Assume ideal behaviorof solutions in questions 13 through 18. Thiscovers the remaining material on Exam 1. Wewill cover chapter 6 on Exam 2.00210.0 pointsWhat is the molarity of a solution composedof 9.815 g of HCl in 0.3472 L of solution?M=0.269199 mol HCl0.3472 L soln= 0.775343 M HCl003(part 1 of 2) 10.0 pointsA student investigating the properties of so-lutions containing carbonate ions prepared asolution containing 6.684 g Na2CO3in a flaskof volume 250 mL. Some of the solution wastransferred to a buret.What volume of so-lution should be dispensed from the buret toprovide 7.76 mmol Na2CO3?(part 2 of 2) 10.0 pointsWhat volume of solution should be dispensedfrom the buret to provide 9.059 mmol CO2−3?
5.18mExplanation:00610.0 pointsToluene (C6H5CH3) is a liquid compoundsimilar to benzene (C6H6).Find the molal-ity of toluene in a solution that contains 35.6grams of toluene and 125 grams of benzene.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00072.warc.gz
|
CC-MAIN-2021-17
| 1,356 | 2 |
https://ask.learncbse.in/t/the-u-s-dairy-industry-wants-to-estimate-the-mean-yearly-milk-consumption/52053
|
math
|
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 16 people reveals the mean yearly consumption to be 60 gallons with a standard deviation of 20 gallons.
a. What is the value of the population mean? What is the best estimate of this value?
b. Explain why we need to use the t distribution. What assumption do you need to make?
c. For a 90 percent confidence interval, what is the value of t?
d. Develop the 90 percent confidence interval for the population mean.
e. Would it be reasonable to conclude that the population mean is 63 gallons?
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00575.warc.gz
|
CC-MAIN-2020-45
| 578 | 6 |
https://dspace.sunyconnect.suny.edu/handle/1951/67577
|
math
|
AbstractOn investigating the Weierstrass function: Its applications and impacts to the instruction in calculus Nadia Syeda, State University of New York, University at Buffalo The Weierstrass function is continuous everywhere but differentiable nowhere, is used as an example showing that continuity does not imply differentiability. The "deplorable evil" Weierstrass function ( Poincare,1893), had a revolutionizing impact in the mathematical community as it became an ideal counterexample breaking through the misconception that continuity entails differentiability. Unfortunately, most calculus students do not have much opportunity to investigate the function other than seeing it as an oddly behaved function. This project defies the norm, studying the function in threefold: (1) Research the historical motivation for the project, (2) investigate the non-differentiable behavior of the function, and (3) develop an instructional resource for college calculus in studying this specific function. The threefold approach to this study is unique because it serves as an example of how a mathematical study can thrust the envelope of venturing beyond the realm of formal proofs and analyses. Historical Motivation: Back in 1893 during the midst of the development of calculus ideas, there was a general belief that continuity of a function is sufficient, guaranteeing that the function is differentiable (Hermite's letter to Stieltjes, 1893). Weierstrass developed a function as a counter-example, a function with no derivatives, and it was labeled as "deplorable evil" (Poincaré, YEAR). He established a function that was differentiable at no point of location of the respective function yet was still continuous everywhere. Graphically speaking, Weierstrass function is a curve that has no gradient. To date, the function remains of interest to many, but rarely can it be seen in college calculus. The analysis portion begins with a review of the differentiability entails continuity, whereas the converse is false, such as f(x) = |x| showing that x=0 is continuous but not differentiable. Graphically, a point that is continuous but not differentiable is not "smooth," and this notion extends to the Weierstrass function where the curve is continuous everywhere, but "smooths" nowhere. This study reviews the established proof of non-differentiability, and begs the question, "What happens if one attempts to differentiate the Weierstrass function.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00515.warc.gz
|
CC-MAIN-2023-14
| 2,453 | 1 |
http://sepwww.stanford.edu/data/media/public/docs/sep124/yaxun1/paper_html/node4.html
|
math
|
To compare the migration results we uniformly select three vs, s and s separately for the conventional method; the chosen reference parameters are shown in Figure . For the modified 3D Lloyd's method, we also start with three vs, three s and three s as the initial input and we set the maximum number of anisotropic parameters equal to 27 for each depth level. This guarantees that the number of parameters selected by the modified 3D Lloyd's algorithm will be no larger than that obtained by uniform sampling. In this case, the total number of depth levels is nz=410; hence, for a uniform sampling of reference parameters, we must perform wavefield extrapolations. After selecting by the modified Lloyd's algorithm, however, the total number of extrapolations reduces to 4952, reducing the computation time by about .Figure illustrates the reference parameters selected by the modified 3D Lloyd's algorithm. It is clear in Figure that Lloyd's algorithm has done a very good job of describing the actual model. Figures - show the error maps between the actual model and the selected reference parameters both by the conventional method and the modified 3D Lloyd's method. Obviously, using Lloyd's method, we obtain much smaller differences between the actual model and the reference parameters.
Figure (a) shows the anisotropic prestack migration result of the conventional approach, while Figure (b) shows the result of using the modified 3D Lloyd's algorithm for reference-parameter selections. Though in both cases, all reflectors are nicely imaged, we can still identify the differences between Figure (a) and (b). Using the modified 3D Lloyd's algorithm yields a more focused and continuous salt flank (portion B); the layers, especially A and C, have stronger amplitudes and are better imaged. The angle-domain common image gathers (ADCIGs) for different surface locations computed from images obtained by using both methods are illustrated next to each other in Figure . Figures (a), (c) and (e) are the ADCIGs computed from the image obtained by using the uniform sampling method (Figure (a)) at surface location x=5,875 meters, 11,375 meters and 13,875 meters respectively, while Figures (b), (d) and (f) are the corresponding ADCIGs from the image obtained by using the modified Lloyd's algorithm (Figure (b)) at surface location x=5,875 meters, 11,375 meters and 13,875 meters respectively. The differences between the ADCIGs computed from Figure (a) and those computed from Figure (b) are minor, but we can still see that the ADCIGs shown in Figures (b), (d) and (f) are generally more flat and continuous than those illustrated in Figures (a), (c) and (e), which indicates that the reference anisotropic parameters selected by the modified 3D Lloyd's algorithm are more accurate than those selected by the uniform sampling method. One thing needed for an extra emphasis is that the computational cost by using Lloyd's algorithm is only half of that by using the conventional method.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00367.warc.gz
|
CC-MAIN-2017-43
| 2,995 | 2 |
https://www.nagwa.com/en/explainers/603143879028/
|
math
|
In this explainer, we will learn how to use the kinematics equations of uniform acceleration to model the vertical motion of a body with uniform acceleration due to gravity.
Near the surface of Earth, a body that is not acted on by a net vertical force other than its own weight will accelerate uniformly vertically downward. The acceleration due to gravity is represented by and has a magnitude of approximately 9.8 m/s2. Acceleration due to gravity varies with the distance from the center of mass of Earth, but the variation in with altitude is small near the surface of Earth and is treated as a constant value.
It is important to distinguish between acceleration due to gravity near the surface of Earth and the net acceleration of a body near the surface of Earth. A body near the surface of Earth does not necessarily accelerate vertically downward at 9.8 m/s2. For example, a body at rest, or moving parallel to the surface of Earth (modeling Earth as a sphere), has zero vertical acceleration. This is because the weight of a body in contact with the surface of Earth is not the only force acting on the body.
This explainer exclusively considers bodies on which the only force acting is the force of their weights, so we will neglect resistive forces exerted on bodies by air. The motion of such bodies can be modeled using kinematic equations for motion with uniform acceleration, such as where is the initial velocity of the particle, is the final velocity of the particle, is the acceleration of the particle, and is the displacement of the particle.
Let us look at an example of a body moving freely that is acted on only by its weight.
Example 1: Finding the Initial Velocity of a Particle That Is Projected Vertically Upward
A particle was projected vertically upward from the ground. Given that the maximum height the particle reached was 62.5 m, find the velocity at which it was projected. Take the acceleration due to gravity .
The motion of the particle can be modeled using the equation
The acceleration of the particle is only due to gravity, so the magnitude of the acceleration is 9.8 m/s2. The acceleration is vertically downward. The particle is projected vertically upward, so acceleration due to gravity is in the opposite direction to the initial velocity of the particle. If the direction of the initial velocity of the particle is taken as positive, the acceleration is negative; hence, we have that
This equation can be rearranged to make the subject:
The particle is instantaneously at rest when it reaches its maximum height, so at that instant, it has zero velocity. We see then that
Taking the positive root of , we obtain
The negative root is neglected as this corresponds to vertically downward velocity of the particle, and the particle cannot have initially vertically downward velocity as it is projected vertically upward.
A body that is uniformly accelerated has a continuously varying velocity, but the average value of this velocity is the mean of its velocities before and after acceleration, as given by
Let us look at an example where the average velocity of a body accelerated only by gravity is determined.
Example 2: Finding the Average Velocity of a Falling Body
If a body, which was dropped from a building, took 3 seconds to reach the ground, find its average velocity as it fell. Let the acceleration due to gravity .
The acceleration of a body accelerating uniformly is given by
This expression can be rearranged to determine the average velocity as follows:
Substituting the known values of and , and assuming that is zero as the body was dropped rather than projected vertically, we have that
The average velocity of the object is the mean of its velocities before and after acceleration, given by
We have already seen that is zero, so we have that
Now, let us look at an example where we determine the displacement of a body projected vertically with a known velocity.
Example 3: Finding the Maximum Height Reached by a Ball Projected Vertically Upward
A particle is projected vertically upward at 7 m/s from a point 38.7 m above the ground. Find the maximum height the particle can reach. Consider the acceleration due to gravity to be .
The particle reaches its maximum height at the instant that it is instantaneously at rest and about to start falling back toward the point from which it was projected.
In this example, we are given the initial velocity of the particle, the initial vertical displacement of the particle, and the vertical acceleration of the particle. The velocity of the particle at its greatest vertical displacement will be instantaneously zero.
The displacement of the particle from the point it was projected at this instant can be determined using a formula containing these terms; hence, we can use the formula where , , and . Rearranging the equation to make the subject, we obtain
Substituting known values, we find
The initial vertical displacement of the particle was 38.7 m, so the greatest vertical displacement of the particle is given by
Let us look at another such example.
Example 4: Using the Equations of Motion to Solve a Vertical Projection Question
Fill in the blank: If a body is projected vertically upward with speed to reach maximum height , then the speed that the body should be projected by to reach height is .
The particle reaches its maximum height at the instant that it is instantaneously at rest and about to start falling back toward the point from which it was projected. The displacement of the particle from the point it was projected at this instant can be determined using the formula
Rearranging the equation to make the subject and noting both that is zero and that because the vertically downward direction is taken as negative, and are negative, we therefore obtain
The numerator and denominator of the right side of the expression are negative, so it is equivalent to the expression
For a value of equal to , the value of is equal to . It is important to note that we must not confuse the value of the initial velocity with the symbol used for the final velocity term, .
Using the values of and , we have that
We need to determine how an initial velocity, denoted by , corresponding to a value for of compares to the initial velocity .
We know that a particle launched vertically upward with a velocity has a velocity of zero when its vertical displacement is .
We can use the equation and substitute the values of for initial velocity, for acceleration, and for displacement. Noting that is zero and that is negative when is positive, we have that
Substituting we have that
We can rearrange this expression to obtain an expression that relates and :
Taking square roots of both sides, we find that
We see that has twice the value of .
Let us look at an example where a body is projected vertically with an unknown velocity and the final velocity of the body is also unknown.
Example 5: Finding the Time It Takes for a Body to Reach the Base of a Tower
A body was projected vertically downward from the top of a tower whose height is 80 m. Given that it covered 35.9 m during the 1st second of its motion, find the time taken to reach the ground rounded to the nearest two decimal places. Let the acceleration due to gravity .
Neither the initial velocity of the body nor its velocity at any point during its motion can be determined from the information given in the question, and so the answer cannot be determined by comparing the initial and final velocities of the body.
It might seem that the final velocity of the body is zero, as the body eventually reaches the ground, but at the instant that the body reaches the ground, its instantaneous velocity is not zero. If the body does not rebound, the velocity of the body a short time after it reaches the ground will be zero, but it does not reach the ground with zero velocity.
The velocities of the body are not known, but the initial velocity of the body can be determined using the formula and rearranging to make the subject to get
The initial velocity is in the same direction as the acceleration. We take this as the positive direction.
Substituting known values, we find that
Knowing the initial velocity, the final velocity can be determined using the formula as follows:
We can now rearrange the equation to make the subject, obtaining
To two decimal places, is 1.97 seconds.
Let us summarize what we have learned from these examples.
- Near the surface of Earth, a body that is not acted on by a net vertical force other than its own weight will uniformly accelerate vertically downward at 9.8 m/s2.
- The motion of a body accelerated only by gravity can be modeled using kinematic equations of motion with constant acceleration, where the acceleration has a magnitude of 9.8 m/s2.
- A body moving vertically that is accelerated only by gravity can have an initial velocity either in the same direction as or in the opposite direction to its acceleration.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00253.warc.gz
|
CC-MAIN-2023-23
| 8,948 | 60 |
http://mathforum.org/kb/message.jspa?messageID=7628778
|
math
|
I don't know what problem number 31 of the Rhind Mathematical Papyrus has to do with Mayan astronomy, but if I must I can also shed light on this problem. In my opinion, the Rhind Mathematical Papyrus offers problems that can be solved on several level. On the first level, beginners learn how to handle unit fraction series. On the advanced level they are asked to solve more demanding problems, and on the highest level they are being told about theoretical insights. RMP 31 on the advanced level is about a geometrical problem, it offers a fine example of Egyptian wit, plus a theoretical insight:
note that all 20th century scholars, Gillings, Peet, et al, agree with this phase of the problem ... however, none fairly rport Ahmes' conversion of
28/97 by solving 2/97 + 28/97
with 2/97 solved in the RMP 2/n table manner ... that scaled 2/97 by 56/56 and 26/97 by 4/4 ... scribal steps that you totally ignore by jumping from a false granary problem to a correct quotient and remainder answer.
Please transliterate each of Ahmes' problems as scholars have been doing since 1879. Your granary ring is silly.
Please correct scholarly 20th century translation errors that followed a false additive pattern ... that you oddly 'advocate' thereby 'throwing out the scribal solutions to 2/97 and 26/97 ... with a blind in one-eye failed 'experiment.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00097.warc.gz
|
CC-MAIN-2018-22
| 1,346 | 6 |
https://pldb.pub/languages/omnitab-80.html
|
math
|
OMNITAB 80 is a programming language created in 1980. OMNITAB 80 is a high-level statistical analysis program. OMNITAB, the precursor of Minitab(TM), was developed in the Statistical Engineering Division and is maintained by the Statistical Engineering Division. OMNITAB performs many different statistical analyses including: arithmetic and trigonometric calculations, and matrix and array operations. The software responds to simple instructions and uses reliable computational algorithms.
|#3734on PLDB||42Years Old|
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00582.warc.gz
|
CC-MAIN-2022-27
| 519 | 2 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.