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https://lawctopus.com/clatalogue/quick-revision-of-calendar-for-clat/ | 1,631,904,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00330.warc.gz | 422,422,873 | 32,932 | HomeLogical ReasoningQuick Revision of Calendar for CLAT
# Quick Revision of Calendar for CLAT
The Logical reasoning section of CLAT includes the topic of Calendar. The calendar is an important concept and is widely used in our daily lives. It is not a difficult task to score full marks in this part provided you are thorough with the basics of the subject. In this article, we will be mentioning the principles and formulas to solve the calendar questions.
#### Points to remember:
1. It is important to consider which month it is i.e. a month with 30 days, 28 days or 31 days.
2. An ordinary year has 365 days and a leap year has 366 days.
3. An ordinary year has one odd day and a leap year has two odd days.
4. A month with 31 days, 30 days, 29 days and 2 days has 3 odd days, 2 odd days, 1 odd day and 0 0dd days respectively.
5. Every fourth ordinary year is a leap year.
6. Odd days
100 years: 5 odd days
200 years: 3 odd days
300 years: 1 odd day
400 years: 0 odd days
1. It was a Monday on 1 January 1 A.D.
#### Problem Solving
To solve all kind of questions, it is important to keep in mind the following formulas to get accurate and quick answers to the questions from Calendars.
1. When you are asked to calculate the odd days, you have to divide the total number of days by 7 and the remainder is the number of odd days. When the number of days involves multiple years or centuries, you can break it up and use the references of no. of odd days as mentioned above.
Therefore, the general formula of no. of odd days = Total no. of days/ 7
2. The questions often ask which day it is on a particular date by giving a reference date. In such cases, you have to find out the no. of odd days and calculate the day.
1. In some questions, you are indirectly asked which day it is without a reference date. So, you have to take the reference date as 1st January 1 A.D.
2. To find whether a year is a leap year or not:
For the non-century year, if it is divisible by 4, it is a leap year.
For a century year, if it is divisible by 400, it is a leap year.
#### Illustrations
1. How many odd days are there in 1100 years?
Solution: 1100 years = 400 years + 400 years + 300 years
= 0 odd days + 0 odd days + 1 odd day
= 1 odd day
1. If today is Tuesday, then what day of the week will be on the 335 days from today?
Solution: 335 days = 47 weeks + 6 odd days. Therefore it is Monday.
1. Was 2000 a leap year?
Solution: Since 2000 is a century year, we have to see if it is divisible by 400.
2000/400 = 5.
Hence 2000 is a leap year | 675 | 2,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2021-39 | latest | en | 0.951726 |
https://web2.0calc.com/questions/please-help_4426 | 1,611,400,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703537796.45/warc/CC-MAIN-20210123094754-20210123124754-00760.warc.gz | 609,745,814 | 5,512 | +0
0
120
1
At a candy store, Gavin bought 2 kilograms of jelly beans and 2 kilograms of gummy worms for \$32. Meanwhile, Dustin bought 2 kilograms of jelly beans and 1 kilogram of gummy worms for \$25. How much does each candy cost?
Jul 1, 2020
#1
+783
+1
Let x be the cost of a kilogram of jelly beans and y the cost of a kilogram of gummy worms. We form two equations:
2x+2y=32
2x+y=25
We have a system of equations now and we solve for x and y.
Subtracting 2x+y=25 from 2x+2y=32, we get y=7. Then we plug the y value into 2x+y=25 to get the value of x.
2x+y=25
2x+7=25
2x=18
x=9.
Therefore jelly beans cost \$9 per kilogram and gummy worms \$7 per kilogram.
Jul 1, 2020 | 235 | 687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-04 | latest | en | 0.792344 |
https://imcs.dvfu.ru/cats/static/problem_text-cpid-538014.html | 1,716,151,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00175.warc.gz | 275,754,151 | 8,342 | ## Problem A. Pizza schedule ≡
Author: T. Chistyakov Time limit: 1 sec Input file: input.txt Memory limit: 2 Mb Output file: output.txt
### Statement
PizzaSocks company makes and delivers pizza. Pizza is delivered on regular schedule. A schedule has a period of 1, 2, 3 or 4 weeks. Quantity of pizzas to be delivered is known for each day of the period. First day of a period is always the first day of the week.
So formally pizza delivery schedule can be described by the following numbers: L — length of the schedule period (from 1 to 4), A1, A2, …, AL × 7 — quantity of pizzas to be delivered for each day of the period (Ai ≥ 0).
Once an absent-minded PizzaSocks manager lost a schedule for one Very Big and Important Client. Only the history of past deliveries was preserved. Your task is to restore the schedule from the history.
The problem is complicated by the fact that occasionally Very Big and Important Client changed his orders, ordering either more or less pizzas than scheduled, even sometimes cancelling the order competely or placing an unscheduled order.
So it's often impossible to restore the original schedule exactly. That is why your task is to find an optimal schedule, i. e. to minimize the total number of days when historically ordered quantity is different from the quantity according to that schedule. The days before the first recorded order and after the last recorded one should not be taken into consideration.
### Input file format
Input file contains the following integer numbers:
N — number of fulfilled orders in the delivery history w1 d1 q1 w2 d2 q2… wN dN qN — description of historical orders: week number, day of the week, and quantity of pizzas delivered. For each day there is no more than one record. It's considered that on those days that are not explicitly mentioned in the input file, but fall between the earliest and the latest mentioned days, zero-quantity orders were fulfilled.
### Output file format
Output periodicity of the optimal schedule — integer number L from 1 to 4. Then output numbers Ai for each i from 1 to L × 7. If several solutions exist, output any of them. However, make sure that the first week of your schedule corresponds to the earliest week mentioned in the input file.
### Constraints
1 ≤ qi ≤ 100, 1 ≤ wi ≤ 52, 1 ≤ di ≤ 7
### Sample tests
No. Input file (input.txt) Output file (output.txt)
1
6
1 5 3 3 1 3 3 5 3
5 1 3 5 5 3 7 5 3
2
3 0 0 0 3 0 0
0 0 0 0 0 0 0
2
15
1 3 1 1 5 2
2 3 1 2 5 2
3 3 1
4 3 1 4 5 3
5 3 1 5 5 2
6 3 1 6 5 2
7 3 1 7 5 2
8 3 1 8 5 2
1
0 0 1 0 2 0 0
0.038s 0.007s 15 | 758 | 2,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.89577 |
https://documen.tv/question/a-researcher-with-the-ministry-of-transportation-is-commissioned-to-study-the-drive-times-to-wor-16201536-64/ | 1,627,111,780,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00661.warc.gz | 223,340,541 | 18,023 | ## A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U.S. cities. The underlying
Question
A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U.S. cities. The underlying hypothesis is that average commute times are different across cities. To test the hypothesis, the researcher randomly selects six people from each of the four cities and records their one-way commute times to work. Refer to the below data on one-way commute times (in minutes) to work. Note that the grand mean is 36.625. Houston Charlotte Tucson Akron 45 25 25 10 65 30 30 15 105 35 19 15 55 10 30 10 85 50 10 5 90 70 35 10 x¯i 74.167 36.667 24.833 10.833 s2i 524.167 436.667 82.167 14.167 The competing hypotheses about the mean commute times are ______________ a) H0: μ1 = μ2 = μ3, HA: Not all population means are equal H0: b) Not all population means are equal, HA: μ1 = μ2 = μ3 H0: μ1 = μ2 = μ3 = μ4, HA: c) Not all population means are equal H0: d) Not all population means are equal, HA: μ1 = μ2 = μ3 = μ4
in progress 0
4 days 2021-07-20T04:52:46+00:00 1 Answers 2 views 0
d) H0: Not all population means are equal,
HA: μ1 = μ2 = μ3 = μ4
Explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
Therefore, for the case above;
Since the underlying hypothesis is that average commute times are different across cities.
Null hypothesis H0: is that Not all population means are equal.
Alternative hypothesis Ha: μ1 = μ2 = μ3 = μ4 | 497 | 1,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-31 | latest | en | 0.911834 |
https://lecdem.physics.umd.edu/component/k2/tag/Engineering.html | 1,632,770,121,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00275.warc.gz | 387,073,393 | 9,157 | # Engineering
Category: K7 RLC Circuits
Purpose: Demonstrate a barricade flasher circuit.
## K2-47: SPEEDOMETER MODEL
Purpose: Demonstrate operating principle of some speedometers.
## J7-31: VISIBLE HARD DRIVE
Purpose: See what the guts of a fixed magnetic drive look like.
## F5-31: MAGNUS EFFECT - FLETTNER'S SHIP
Purpose: Demonstrate the Magnus effect.
## G1-54: MASS'S DOUBLE PENDULUM
Purpose: Demonstrate the transition of potential energy into energy of oscillation of the pendulum, and the operation of an escapement.
## F4-53: ARCHIMEDES' SCREW
Category: F4 Fluid in Motion
Purpose: Demonstrate a pump mechanism invented by Archimedes.
## F4-51: VACUUM PUMP MODEL
Category: F4 Fluid in Motion
Purpose: Illustrate how a mechanical vacuum pump works.
## F2-12: HOT AIR BALLOON
Category: F2 Buoyancy
Purpose: Show that hot air is less dense than cold air by operating a hot air balloon.
## F1-14: PISTON DIAMETER VS TRAVEL - WORKING MODEL
Purpose: Show that with an incompressible fluid the bigger piston moves more slowly than the smaller piston.
## F1-11: HYDRAULIC PRESS
Purpose: Demonstrate dramatically Pascal's Law and the large forces attainable using hydraulic systems.
## D4-23: GYROCOMPASS
Category: D4 Gyroscopes
Purpose: Illustrate the operation of the gyrocompass.
## D4-21: SHIP STABILIZER
Category: D4 Gyroscopes
Purpose: Demonstrate how a gyroscope can stabilize the rocking of a ship.
## D2-12: TOPPLING CHIMNEY
Purpose: Demonstrate how a toppling chimney breaks up.
## B4-14: ELASTIC LIMIT OF WIRE
Category: B4 Elasticity
Purpose: Demonstrate the variation in tensile strength with wire diameter.
## E2-42: TELESCOPE MODEL
Category: E2 Astronomy
Purpose: Show how a telescope can view any point in the sky using a universal mount.
## B1-16: CORBELED ARCH
Purpose: Illustrate how the center of mass affects the stability of an arch
## A2-32: HEIGHT MEASUREMENT BY TRIGONOMETRY
Category: A2 Mathematics
Purpose: Determine the height of a student using trigonometry
## A1-43: CROSSING RODS
Purpose: Visualize three-dimensional figures
## A1-42: DRAWING PUZZLE - HALF CUBE
Purpose: Visualize three-dimensional figures
## A1-41: DRAWING PUZZLE - INTERSECTING RODS
Purpose: Visualize three-dimensional figures
Page 1 of 2 | 620 | 2,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | longest | en | 0.657873 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=303155 | 1,513,122,878,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00658.warc.gz | 739,069,859 | 4,373 | # Financial Mathematics
Home > Preview
The flashcards below were created by user kirstenp on FreezingBlue Flashcards.
1. Explain the concept of the time value of money
• Because money can be invested and interest earned on the investment, money has time value.
• An implication of money having time value is that \$1 is worth less the later it is received, all other things equal
• Another implication of money having time value is that you cannot add amounts at different points in time; amounts have to be added at the same point in time
2. Explain the benefits of compounding
With compound interest, you earn interest upon interest. So over time the value of an investment will grow considerably
3. Explain the implications of changing the frequency of compounding
• The more frequently interest is compounded, the more interest that is earned, and, holding all else constant, the greater the future wealth from an investment
• With simple interest, interest is compounded once over the life of the investment
4. Explain the relationship of effective interest rates and nominal interest rates
• The nominal interest rate is an interest rate stated over a certain period but that is not compounded at the same frequency as the period stated
• The effective interest rate is an interest rate stated over a certain period and that is compounded at the same frequency as the period stated.
• An effective annual interest rate is an interest rate compounded once per year
• A nominal annual interest rate is an interest rate stated on an annual basis but is not compounded once per year
• Interest rates can only be compared when they are compounded at the same frequency
## Card Set Information
Author: kirstenp ID: 303155 Filename: Financial Mathematics Updated: 2015-05-23 08:00:09 Tags: ACC1000 finance Folders: ACC1000 Description: acc100 financial mathematics Show Answers:
Home > Flashcards > Print Preview | 394 | 1,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-51 | longest | en | 0.9423 |
https://physics.stackexchange.com/questions/262091/collision-crumpling-problem-possible-solution-mistake | 1,702,221,320,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00032.warc.gz | 499,190,066 | 42,677 | # Collision/Crumpling problem possible solution mistake
This question is from Physics for scientist and engineers , Ohanian .
Two automobiles of 540 and 1400 kg collide head-on while moving at 80 kmh in opposite directions. After the collision the automobiles remain locked together.
(C) The front end of each automobile crumples by 0.60 m during the collision. Find the acceleration (relative to the ground) of the passenger compartment of each automobile; make the assumption that these accelerations are constant during the collision.
In Walter Lewin 1999 physics class this problem part 6.1 is solved in top of page number 2.
http://www.myoops.org/twocw/mit/NR/rdonlyres/Physics/8-01Physics-IFall1999/8D3EA3C0-149F-4212-B926-AE0DAE87D642/0/sol6.pdf
My doubt is : Since it is a collision , forces are equal hence $$\frac{a_1}{a_2}=\frac{m_2}{m_1}$$ However the solution does not satisfy this condition .I have checked the calculations and the reasoning and both are sound .
Where is the mistake ?
• I couldn't download (and see) the solution you have linked. But, I think you should consider to loss of energy in collision. You cannot treat with whole of cars as rigid bodies during their deformation (according to definition rigid bodies don't deform). The magnitudes of forces are equal just in the contact area, but magnitudes of forces are transmitted to the passenger compartment of each automobile can be different. Jun 12, 2016 at 4:29
• What did the solution use for d? Jun 12, 2016 at 9:11
• also note that the solution uses $\approx$ sign for $a$, which is just a round-off of the exact value. Jun 12, 2016 at 9:35
• Thanks Lucas for this point , I think maybe this is my confusion . Philip_0008 The solution used 0.6 for d which is the crumple zone. Jun 12, 2016 at 9:51
The solution is actually strange for me. It assumes that the distance traveled by each car (in the CM frame) during the acceleration (crumpling) is also equal to the crumple length of each car, but is not always the case.
Following the assumption of the solution: In the CM frame, the heavier car moves slower than the lighter car. So that at the moment that they will collide, the time needed for the lighter car to decelerate to 0 would be shorter than the time for the heavier car if it assumes that the distance traveled by each car (at CM frame) would be equal to its crumple distance:
$t_1 = \Delta v_1/a_1 = \frac{0-32.1}{-8.6\times10^2} = 0.0373s$
but $t_2 = \Delta v_2/a_2 = \frac{0-(-12.3)}{1.3\times10^2} = 0.0946s$
as a result, the lighter car is already completely crumpled, while the heavier car is still crumpling:
But since the heavier car is still crumpling, wouldnt it cause the lighter car to be accelerated even further and much longer?
In reality, the time for the two cars to decelerate must be equal, which means some of the assumptions of the solution is not applicable in the real world (more precisely, there is a mistake in the solution).
A more prudent solution (in my opinion) would be to use
$$a_1 = \frac{v_1'^2}{2d_1}$$ $$a_2 = \frac{v_2'^2}{2d_2}$$ where $d_1$ and $d_2$ is the unknown distance traveled by car 1 and 2 during deceleration (crumpling), with respect to the CM frame. $$d_1 + d_2 = 2(0.6) = 1.2 \space\space[1]$$ $$\frac{a_1}{a_2} = \frac{v_1'^2}{2d_1}\frac{2d_2}{v_2'^2} = \frac{v_1'^2 d_2}{v_2'^2 d_1}$$ and finally, $$\frac{v_1'^2 d_2}{v_2'^2 d_1} = \frac{m_2}{m_1} \space\space[2]$$ as you suggested. Then use equation 1 and 2 to solve for $d_1$ and $d_2$.
• Thank you Philip for your illustrative solution .Now I think the mistake lies in the assumption that each car deceleration distance is the same as crumple zone . Lucas raised another issue that since these bodies are not rigid bodies , , , , the forces transmitted to passenger compartments need not to be the same hence the acceleration constraint assumption may not be valid . Can I assume they are the approximately the same since both go through same crumpling distance ? Jun 12, 2016 at 21:30
• The acceleration time should be equal on both cars for a collision, because any stopping force (which causes acceleration) applied to one car will directly affect the other (law of action and reaction), and once one car stops accelerating, the other must stop accelerating also. This equal time condition can only be achieved by equating $a_1/a_2 = m_2/m_1$. Note that $v_1'$ and $v_2'$ should remain the same whatever the condition. Jun 12, 2016 at 22:48 | 1,251 | 4,463 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-50 | longest | en | 0.942692 |
https://www.programmingwithbasics.com/2017/10/hacker-rank-solution-for-day-19.html | 1,675,053,320,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00574.warc.gz | 959,874,622 | 33,659 | # Hacker Rank solution For Day 19: Interfaces
Problem:- Write a Hackerrank Solution For Day 19: Interfaces or Hacker Rank Solution Program In C++ For " Day 19: Interfaces " or Hackerrank 30 days of code Java Solution:Day 19: Interfaces solution or Hackerrank solution for 30 Days of Code Challenges or Hackerrank 30 days of code Java Solution, Day 19: Interfaces solution, or C/C++ Logic & Problem Solving: Day 19: Interfaces or Hacker Rank Solution For Day 19: Interfaces.
Check This:- Hacker rank solution for Strings, Classes, STL, Inheritance in C++.
Explanation:- This is a very simple problem(Interfaces) we have to just calculate divisor sum and print the sum of the divisor. we can do this by run the loop(started with 1 and end with N) and put the condition if number N is divided by I(where I = 1 to N) than the sum of the divisor is added each time when if N%I==0. Let's take an example and try to understand the problem step by step.
Example:- Let's assume Number N is 24 and we have to calculate the Divisor sum so we divide the number by I=1 to I=24 with the help of "For Loop" Take a look at step by step.
Step 1: If(N % I ==0) then Sum or total is Sum = Sum + I Or Total = Total + I.
Step 2: Print the total outside the "For Loop". Here Divisor of 24 = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.
Also Check:- Geeksforgeeks solution for School, Basic, Easy, Medium, Hard in C++.
Tip:- Copy the colored code or full code(According to Requirement ) and paste it into hacker rank editor.All solution provided here are in C++ (CPP) if any reader wants these solutions in C, and Java comments below or Email me with your query like " day n solution in C / C++ / Java. Check the end of the post solutions with the full explanation
Solution:-
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
{
public:
virtual int divisorSum(int n)=0;
};
{
int divisorSum(int n)
{
int total=0,i;
for(i=1; i<=n; i++)
{
if(n%i == 0)
{
total+=i;
}
}
}
};
int main()
{
int n;
cin >> n;
int sum = myCalculator->divisorSum(n);
cout << "I implemented: AdvancedArithmetic\n" << sum;
return 0;
}
Output:-
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Hi, I’m Ghanendra Yadav, SEO Expert, Professional Blogger, Programmer, and UI Developer. Get a Solution of More Than 500+ Programming Problems, and Practice All Programs in C, C++, and Java Languages. Get a Competitive Website Solution also Ie. Hackerrank Solutions and Geeksforgeeks Solutions. If You Are Interested to Learn a C Programming Language and You Don't Have Experience in Any Programming, You Should Start with a C Programming Language, Read: List of Format Specifiers in C. | 860 | 3,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-06 | latest | en | 0.764071 |
https://www.physicsforums.com/threads/what-should-i-do-if-didnt-find-the-given-pressure.804908/ | 1,527,090,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00454.warc.gz | 817,125,345 | 17,870 | # What should I do if didn't find the given pressure?
Tags:
1. Mar 24, 2015
### Amr719
I want to know what should I do if I don't find the given pressure in a problem in the super heated tables ?
I have P=1.1MPa and I want to get the " h " but I don't find this pressure in the tables
2. Mar 24, 2015
### Staff: Mentor
Linearly interpolate.
Chet
3. Mar 24, 2015
### Amr719
Isn't there another way ? My teacher told me another way to do it that takes less time in the exam , but I can't remeber it
4. Mar 24, 2015
### Staff: Mentor
How long does it take to linearly interpolate?
Chet
5. Mar 24, 2015
### Amr719
I don't know exactly but if there is a faster way , I should go with it
6. Mar 24, 2015
### Staff: Mentor
I can't think of a faster way. I bet I could do the linear interpolation in less than a minute. Interpolation is the standard way of working with tables.
Chet
7. Mar 24, 2015
### Amr719
Could you tell me how to do the interpolation for this : I have P=1.1MPa T=250°C and I want to get "h"
Sorry I am not familiar with this type of solution so I need your help :)
8. Mar 24, 2015
### Staff: Mentor
At what values of the pressure does the table give values of h? (on either side of 1.1 MPa) What are those values of h at 250 C?
Chet
9. Mar 24, 2015
### Amr719
I can't understand . what I know that it is at the superheated vapour Tables
10. Mar 24, 2015
### Staff: Mentor
Write down some of the numbers from your table. For example in the steam tables I have,
P = 10 bars, T = 240 C, h = 2920
P = 10 bars, T = 280 C, h = 3008
P = 15 bars, T = 240 C, h = 2899
P = 15 bars, T = 280 C, h = 2993
Chet
11. Mar 24, 2015
### Amr719
That's the question. Which table ? I don't have a table for the pressure 1.1MPa . I have for 1.0 MPa and for 1.5 MPa
12. Mar 24, 2015
### Staff: Mentor
I'm acutely aware of that. What does your table give for h at 1.0 MPa and 1.5 MPa? After you provide those values, I will show you how to get the value at 1.1 MPa.
Chet
13. Mar 24, 2015
### Staff: Mentor
Amr719, it doesn't seem like you know what it means to "interpolate". Did you look that up after Chestermiller said it is what is needed?
14. Mar 25, 2015
### Staff: Mentor
Shocking, huh? Don't they teach interpolation in high school algebra any more?
Chet
15. Mar 25, 2015
### SteamKing
Staff Emeritus
The problem is, "interpolate" is too big to fit on a calculator key.
16. Mar 28, 2015
### Amr719
No I know it but like I said before I'm not familiar with it because my teacher doesn't use this method. That's all
17. Mar 28, 2015
### Amr719
Table for 1.0MPa : Table for 1.5MPa:
T=240,h=2920.4 T=240,h=2899.3
T=280,h=3008.2 T=280,h=2991.7
18. Mar 28, 2015
### Staff: Mentor
And, what, you never noticed that these are the same values I gave you from my table in post #10?
I am going to show you how to get the value of h at 250 C and 1.0 MPa. Then you are going to show me how you apply the same interpolation approach to get the value of h at 250 C and 1.5 MPa.
$$h(250 C,1 MPa) = 2920.4 + \frac{(250 - 240)}{(280-240)}(3008.2-2920.4)=2942.4$$
Now I want you to apply this same algorithm to get the value of h at 250 C and 1.5 MPa. Do you think you can do that?
Please don't tell me at this point that you need to have the value of h at 1.1 MPa. I know that. There will be another step after you complete this step.
Chet
19. Mar 28, 2015
### Amr719
h(250C,1.5MPa)=2899.3+((250-240)/(280-240))*(2991.7-2899.3)=2922.4
20. Mar 28, 2015
### Staff: Mentor
Looks good. Now, do you think you can take these results for h at 250 C for 1.0 MPa and 1.5 MPa and, by using this same kind of interpolation algorithm for pressure, find the value of h at 250 C and 1.1 MPa?
Chet | 1,250 | 3,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-22 | latest | en | 0.926162 |
https://www.pcreview.co.uk/threads/sales-taxes.2589983/ | 1,713,521,836,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00618.warc.gz | 851,448,317 | 13,335 | Sales Taxes
L
lsmft
Started a new worksheet listing all purchased items by name in column
A.
Column B will list how much was spent for each item.
Problem arises when some of the items are taxed at 2 %, some at 3 %,
some at 4%, and some at 6 %.
My question:
Is there a way to denote how much or what rate an item will be taxed
at?
For example:
Row 10, colum B is a grocery food item, hence not state tax, just the
local option taxes.
Row 11, colum B is a cleaning supply item, and will be taxed at the
state S/Tx rate.
I would like to be able to program the whole row for whatever the tax
rate would be. Can this be done?
R
Ron Coderre
lsmft:
You may want to consider using 2 lookup tables:
ItemTable: for items and their tax type
TypeTable: for assigning rates to each type.
Example:
ItemTable
Item TaxType
Apples Food
Armchair Furniture
Carrots Food
Cherries Food
Coat Clothing
etc
TypeTable
Type Stax LocalTax NetRate
Clothing 5% 1% 6%
Food 0% 1% 1%
Furniture 5% 2% 7%
A2: (an item)
B2: =VLOOKUP(VLOOKUP(A2,ItemTable,2,0),TypeTable,4,0)
That formula returns the approriate rate for the type of item in A2
OR
You could add a third column to the ItemTable that calculates the
NetRate for the Type associated with each Item (using the same kind of
VLOOKUP formula as in the example above.)
In that case, the B2 formula would be:
=VLOOKUP(A2,ItemTable,3,0)
Is that something you can work with?
Regards,
Ron
L
lsmft
LOL,
Not sure yet. I'm still scratching my head on this one. I think I may
have gone in over my head here. Thanks much though and I'll see if I
can find how to use that.
G
Guest
items. Do you just name a food item 'food' or would you put 'bread', 'peanut
butter' etc? If you just give them generic names like 'food', 'cleaning' etc
you could try somehting like this formula:
=IF(A1="Food",0.02,(IF(A1="Cleaning",0.03,IF(A1="Books",0.04))))
This will check the contents of A1 and put in a tax rate depending on what
is entered in A1. You could then just copy this formula down the column.
(You would also format this column as percentage.) | 602 | 2,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.929756 |
http://www.postandcourier.com/article/20140420/PC05/140429998/1109/ | 1,438,456,385,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988860.4/warc/CC-MAIN-20150728002308-00220-ip-10-236-191-2.ec2.internal.warc.gz | 658,114,721 | 22,132 | April is Financial Literacy Month, and income tax returns were due last week, so I've put together a short 10-question quiz on local, state and federal tax facts.
See how many you can correctly guess. Answers and explanations are at the end of this column.
1: A married couple with two dependent children has \$100,000 in adjusted gross income and a top federal income tax rate of 25 percent. What percent of their income would they actually owe in federal income tax, assuming they claim the standard deduction?
A) 25 percent
B) 15 percent
C) 11 percent
2: A person whose top federal income tax rate is 15 percent would be taxed at that rate on any interest earned on their non-retirement savings. At what rate would they be taxed on qualified dividend payments from any stock they might own.
A) Zero
B) 15 percent
C) 35 percent
3: The federal government offers a tax credit for each dependent child under age 17. Above certain income limits, such as \$110,000 for a married couple, the credit is reduced. What is the maximum tax credit for each qualifying child?
A) \$500
B) \$1,000
C) \$2,000
4: South Carolina offered 41 different income tax credits for individuals in 2011. How many of those tax credits were claimed that year by fewer than 10 taxpayers?
A) None of them
B) Between 5 and 10
C) Nearly half of them
5: How much taxable income does it take to fall into South Carolina's top income tax bracket?
A) \$85,751
B) \$52,551
C) \$14,251
6: A person living in a \$200,000 home that they own, in Charleston, would have received a property 2013 tax bill for \$977. How much would the tax bill be if that same home were a rental property or a business?
A) The same amount, \$977
B) About 50% more, \$1,466
C) Roughly triple, \$2,835
7: South Carolina homeowners who pay more than 5 percent of their income to insure their residence are eligible for an annual state tax credit. How much is the maximum "excess insurance premium tax credit" worth?
A) \$125
B) \$1,250
C) \$2,500
8: Qualifying home buyers who obtain a "mortgage credit certificate" in South Carolina before purchasing their home can get a federal tax credit worth 30 percent of their mortgage interest payments, up to a dollar limit, every year they own that home. What is the maximum annual tax credit?
A) \$500
B) \$2,000
C) \$3,750
9: By what percentage have Social Security retirement checks been increased since the end of 2008, to account for inflation?
A) 5 percent or less
B) between 5 and 10 percent
c) 10 percent or more
10: Social Security is funded through a payroll tax. Employees and employers each pay 6.2 percent tax on wages up to a certain amount. Above what amount are wages not subject to the tax in 2014?
A) There is no limit
B) \$117,000
C) \$1 million
Thanks for playing. Here are the answers:
1: C) 11 percent. The couple's \$100,000 income is reduced to \$72,200 by the standard deduction and exemptions, portions are taxed at 10 percent and 15 percent, then the tax bill is reduced by \$2,000 for the child tax credit.
2: A) Zero. Unlike interest earned on savings, qualified dividends within the 10 percent and 15 percent tax brackets are tax-free. For those in the 25 percent and 35 percent brackets, the tax rate is 15 percent.
3: B) \$1,000.
4: C) Nearly half of them - 19 out of 41 tax credit were claimed by fewer than 10 people.
5: C) \$14,251
6: C) \$2,835. Properties that aren't owner-occupied are taxed on a higher portion of their value, and are not exempt from school district operating tax.
7: B) \$1,250.
8: B) \$2,000
9: C) 10 percent or more. Social Security checks have increased by more than 13 percent since the end of 2008.
10: B) \$117,000
Reach David Slade at 937-5552 | 962 | 3,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-32 | longest | en | 0.971022 |
https://www.scribbr.co.uk/research-methods/sampling/ | 1,718,726,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00446.warc.gz | 883,470,683 | 29,175 | # Sampling Methods | Types, Techniques, & Examples
When you conduct research about a group of people, it’s rarely possible to collect data from every person in that group. Instead, you select a sample. The sample is the group of individuals who will actually participate in the research.
To draw valid conclusions from your results, you have to carefully decide how you will select a sample that is representative of the group as a whole. There are two types of sampling methods:
• Probability sampling involves random selection, allowing you to make strong statistical inferences about the whole group. It minimises the risk of selection bias.
• Non-probability sampling involves non-random selection based on convenience or other criteria, allowing you to easily collect data.
You should clearly explain how you selected your sample in the methodology section of your paper or thesis.
## Population vs sample
First, you need to understand the difference between a population and a sample, and identify the target population of your research.
• The population is the entire group that you want to draw conclusions about.
• The sample is the specific group of individuals that you will collect data from.
The population can be defined in terms of geographical location, age, income, and many other characteristics.
It can be very broad or quite narrow: maybe you want to make inferences about the whole adult population of your country; maybe your research focuses on customers of a certain company, patients with a specific health condition, or students in a single school.
It is important to carefully define your target population according to the purpose and practicalities of your project.
If the population is very large, demographically mixed, and geographically dispersed, it might be difficult to gain access to a representative sample.
### Sampling frame
The sampling frame is the actual list of individuals that the sample will be drawn from. Ideally, it should include the entire target population (and nobody who is not part of that population).
###### Example
You are doing research on working conditions at Company X. Your population is all 1,000 employees of the company. Your sampling frame is the company’s HR database, which lists the names and contact details of every employee.
### Sample size
The number of individuals you should include in your sample depends on various factors, including the size and variability of the population and your research design. There are different sample size calculators and formulas depending on what you want to achieve with statistical analysis.
## Probability sampling methods
Probability sampling means that every member of the population has a chance of being selected. It is mainly used in quantitative research. If you want to produce results that are representative of the whole population, probability sampling techniques are the most valid choice.
There are four main types of probability sample.
### 1. Simple random sampling
In a simple random sample, every member of the population has an equal chance of being selected. Your sampling frame should include the whole population.
To conduct this type of sampling, you can use tools like random number generators or other techniques that are based entirely on chance.
###### Example
You want to select a simple random sample of 100 employees of Company X. You assign a number to every employee in the company database from 1 to 1000, and use a random number generator to select 100 numbers.
### 2. Systematic sampling
Systematic sampling is similar to simple random sampling, but it is usually slightly easier to conduct. Every member of the population is listed with a number, but instead of randomly generating numbers, individuals are chosen at regular intervals.
###### Example
All employees of the company are listed in alphabetical order. From the first 10 numbers, you randomly select a starting point: number 6. From number 6 onwards, every 10th person on the list is selected (6, 16, 26, 36, and so on), and you end up with a sample of 100 people.
If you use this technique, it is important to make sure that there is no hidden pattern in the list that might skew the sample. For example, if the HR database groups employees by team, and team members are listed in order of seniority, there is a risk that your interval might skip over people in junior roles, resulting in a sample that is skewed towards senior employees.
### 3. Stratified sampling
Stratified sampling involves dividing the population into subpopulations that may differ in important ways. It allows you draw more precise conclusions by ensuring that every subgroup is properly represented in the sample.
To use this sampling method, you divide the population into subgroups (called strata) based on the relevant characteristic (e.g., gender, age range, income bracket, job role).
Based on the overall proportions of the population, you calculate how many people should be sampled from each subgroup. Then you use random or systematic sampling to select a sample from each subgroup.
###### Example
The company has 800 female employees and 200 male employees. You want to ensure that the sample reflects the gender balance of the company, so you sort the population into two strata based on gender. Then you use random sampling on each group, selecting 80 women and 20 men, which gives you a representative sample of 100 people.
### 4. Cluster sampling
Cluster sampling also involves dividing the population into subgroups, but each subgroup should have similar characteristics to the whole sample. Instead of sampling individuals from each subgroup, you randomly select entire subgroups.
If it is practically possible, you might include every individual from each sampled cluster. If the clusters themselves are large, you can also sample individuals from within each cluster using one of the techniques above. This is called multistage sampling.
This method is good for dealing with large and dispersed populations, but there is more risk of error in the sample, as there could be substantial differences between clusters. It’s difficult to guarantee that the sampled clusters are really representative of the whole population.
###### Example
The company has offices in 10 cities across the country (all with roughly the same number of employees in similar roles). You don’t have the capacity to travel to every office to collect your data, so you use random sampling to select 3 offices – these are your clusters.
## Non-probability sampling methods
In a non-probability sample, individuals are selected based on non-random criteria, and not every individual has a chance of being included.
This type of sample is easier and cheaper to access, but it has a higher risk of sampling bias. That means the inferences you can make about the population are weaker than with probability samples, and your conclusions may be more limited. If you use a non-probability sample, you should still aim to make it as representative of the population as possible.
Non-probability sampling techniques are often used in exploratory and qualitative research. In these types of research, the aim is not to test a hypothesis about a broad population, but to develop an initial understanding of a small or under-researched population.
### 1. Convenience sampling
A convenience sample simply includes the individuals who happen to be most accessible to the researcher.
This is an easy and inexpensive way to gather initial data, but there is no way to tell if the sample is representative of the population, so it can’t produce generalisable results.
###### Example
You are researching opinions about student support services in your university, so after each of your classes, you ask your fellow students to complete a survey on the topic. This is a convenient way to gather data, but as you only surveyed students taking the same classes as you at the same level, the sample is not representative of all the students at your university.
### 2. Voluntary response sampling
Similar to a convenience sample, a voluntary response sample is mainly based on ease of access. Instead of the researcher choosing participants and directly contacting them, people volunteer themselves (e.g., by responding to a public online survey).
Voluntary response samples are always at least somewhat biased, as some people will inherently be more likely to volunteer than others.
###### Example
You send out the survey to all students at your university and many students decide to complete it. This can certainly give you some insight into the topic, but the people who responded are more likely to be those who have strong opinions about the student support services, so you can’t be sure that their opinions are representative of all students.
### 3. Purposive sampling
Purposive sampling, also known as judgement sampling, involves the researcher using their expertise to select a sample that is most useful to the purposes of the research.
It is often used in qualitative research, where the researcher wants to gain detailed knowledge about a specific phenomenon rather than make statistical inferences, or where the population is very small and specific. An effective purposive sample must have clear criteria and rationale for inclusion.
###### Example
You want to know more about the opinions and experiences of students with a disability at your university, so you purposely select a number of students with different support needs in order to gather a varied range of data on their experiences with student services.
### 4. Snowball sampling
If the population is hard to access, snowball sampling can be used to recruit participants via other participants. The number of people you have access to ‘snowballs’ as you get in contact with more people.
###### Example
You are researching experiences of homelessness in your city. Since there is no list of all homeless people in the city, probability sampling isn’t possible. You meet one person who agrees to participate in the research, and she puts you in contact with other homeless people she knows in the area.
## Frequently asked questions about sampling
What is sampling?
A sample is a subset of individuals from a larger population. Sampling means selecting the group that you will actually collect data from in your research.
For example, if you are researching the opinions of students in your university, you could survey a sample of 100 students.
Statistical sampling allows you to test a hypothesis about the characteristics of a population. There are various sampling methods you can use to ensure that your sample is representative of the population as a whole.
Why are samples used in research?
Samples are used to make inferences about populations. Samples are easier to collect data from because they are practical, cost-effective, convenient, and manageable.
What is probability sampling?
Probability sampling means that every member of the target population has a known chance of being included in the sample.
Probability sampling methods include simple random sampling, systematic sampling, stratified sampling, and cluster sampling.
What is non-probability sampling?
In non-probability sampling, the sample is selected based on non-random criteria, and not every member of the population has a chance of being included.
Common non-probability sampling methods include convenience sampling, voluntary response sampling, purposive sampling, snowball sampling, and quota sampling.
What is sampling bias?
Sampling bias occurs when some members of a population are systematically more likely to be selected in a sample than others.
#### Cite this Scribbr article
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## Shona McCombes
Shona has a bachelor's and two master's degrees, so she's an expert at writing a great thesis. She has also worked as an editor and teacher, working with students at all different levels to improve their academic writing. | 2,328 | 12,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.933782 |
http://www.six-sigma-material.com/Confidence-Interval.html | 1,571,213,257,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00187.warc.gz | 321,654,999 | 14,110 | # Confidence Interval
Objective:
This section will explain the meaning of the Confidence Interval (CI) in statistical analysis. The calculations will be shown for the:
• Mean
• Standard Deviation
• Proportion
This site assumes that for means and standard deviation the sample data comes from a normal distribution. For proportions, the normal distribution approximates the binomial for n x P(hat) is greater than or equal to 5.
Most common confidence interval selections are 90%, 95%, or 99% but are dependent on the voice of the customer, your company, project, and other factors.
Sample statistics such as the mean, standard deviation and proportion (x-bar, s, p-bar) are only estimates of the population parameters.
Confidence Intervals are used to quantify the uncertainty by providing a lower limit and upper limit that represent a range of values that will represent the true population parameter with a specified level of confidence.
Selecting a 99% CI suggests that approximately 99 out of 100 CI's will contain the population parameter. A 0.99 confidence interval states that there is 99% probability that the interval contains the population parameter, and that there is a 1.0% risk that the population parameter is not contained within the interval.
Confidence Interval = CI = 1 - alpha risk
(1 - alpha-risk) is called the probability content or level of confidence.
Alpha-risk is known as the significance level; the probability of being making an incorrect decision, in other words, being wrong.
A specified value of the CI signifies that probability of the interval containing the population parameter, and that there is an alpha-risk (1-CI) that the population parameter is not contained within the interval.
CI's are applied in statistical test for means, standard deviations, proportions, capability indices, regression analysis, and upper/lower control limits on control charts.
In regression analysis, the CI is based on a provided value of X for a given level of confidence. This CI is likely to contain the true best fit line.
There are three factors that impact the confidence interval:
1. Sample size
2. Population size
3. Percentage
Sample Size
CI's are used when you are unable to capture and analyze an entire population (census) and the sample (statistics) to infer statements about a population. The larger your sample size, the more confidence one can be that their answers represent the population. Though the relationships are not linear, the larger the sample size the smaller the confidence interval (in other words, the more confident you can be that it the true population parameters will fall within a tighter spectrum or tighter distribution).
Population Size
The size of the population is a factor when working with a relatively small and known group of data (such as the number of pieces of candy in a bag versus the number of fish in the ocean).
The CI calculations assume you have a true random sample of the population. If the sample is not then one cannot rely on the confidence intervals calculated, because you can no longer rely on the measures of central tendency and dispersion.
Sampling plans are an important step to ensure the data taken within is reflective and meaningful to represent the population. Click here for information regarding sampling plans.
Percentage
The accuracy of the CI also depends on the percentage of your sample that picks a particular answer. If 99.9% of the parts sampled PASSED and the 0.1% FAILED, the chances of error are very low regardless of sample size.
However, if the percentages are 51% and 49% the chances of error are much greater. It is easier to be sure of extreme answers than those aren't, thus the interval is not linear.
## Point Estimate
Let's go over some jargon used within this topic. You'll see a lot of terms used when learning about confidence intervals, risks, samples, populations, and the following tries to connect all the dots and show you how these terms are related..
A point estimate is:
• a specific number
• a sample statistic
• unbiased estimator or population parameter
The sample standard deviation (s) is an unbiased estimator of σ
The sample variance (s2) is an unbiased estimator of σ2
The sample mean (x-bar) is an unbiased estimator of μ
The sample proportion (p-hat) is an unbiased estimator of P
The Confidence Interval is an interval estimate that provides more information than just a specific number like the Point Estimate.
The Confidence Interval provides information of variability around a point estimate. You can never be 100% confident when using a sample from a population.
If using a 95% CI, this is saying that from repeated samples from the same population, 95% of all the confidence levels that can be generated will contain the true population parameter.
CI = POINT ESTIMATE +/- (Reliability Factor) (Standard Error)
The Reliability Factor depends on your chosen Confidence Level (or alpha-risk)
## Confidence Interval for the MEAN
Assuming the population distribution is normal, or you have a very large sample size of a non-normal distribution and:
• If the population standard deviation IS known, use the Z-distribution table.
• If the population standard deviation IS NOT known, use the t-distribution table.
As the t-distribution sample size increases it behaves like z-distribution and t-value approaches 0. For a given level of confidence, the t-distribution becomes a flatter "bell" curve with the t-statistic increasing as the sample size decreases.
The CI for the mean for the following two examples:
= sample mean +/- (confidence factor) (measure of variability)
which is another way of saying the
= point estimate +/- (reliability factor) (standard error)
## Confidence Interval MEAN with KNOWN σ
EXAMPLE:
A sampling of 22 patients that came into the emergency room showed that they waited on average 45 minutes. From the past data we know that the population standard deviation is 5.8 minutes.
Estimate the 99% confidence interval for the average wait time of the patients. Assume population is normally distributed.
Given:
n = 22
sample mean = 45 minutes = x-bar
population standard deviation = 5.8 minutes = σ
Alpha-risk = 1-CI = 1-0.99 = 0.01 = 1%
The critical z-value is 2.58 for 99% desired confidence level
Find the Confidence Interval:
CI = 45 +/- 2.58 (5.8 / √22) = 45 +/- 2.58(5.8/4.69) = 45 +/- 3.19 = {41.81, 48.19}
### What does this tell us?
With 99% confidence, the true population mean for the wait time is between 41.81 minutes and 48.19 minutes.
Although it's possible the true mean may not be within these values, 99% of the confidence intervals generated in the manner will contain the true mean.
FYI: With only 95% CL (taking more risk), you can say that the true population mean is between 42.58 minutes and 47.42 minutes. You use (1.96 instead of 2.58 in the above formula)
## Confidence Interval MEAN with UNKNOWN σ
EXAMPLE:
A sampling of 17 patients that came into the emergency room showed that they waited on average 45 minutes with a standard deviation of 5.8 minutes. Estimate the 99% confidence interval for the average wait time of the patients. Assume population is normally distributed and the population standard deviation is not known.
Given:
n = 22
sample mean = 45 minutes = x-bar
sample standard deviation = 5.8 minutes
Degrees of freedom (dF) = n-1 = 21
Alpha-risk = 1-CI = 1- ∝ 1-0.99 = 0.01 = 1%
The critical t-value from the table using two tailed is 2.831
(Remember to take the alpha-risk/2 when using the t-table)
### What does this tell us?
This is telling us that the point estimate of the average wait time is 45 minutes with an error of +/- 5.8 minutes. There is 99% certainty that the interval {41.5 minutes to 48.5 minutes} contains the true process mean. There is a 1% chance that this decision is wrong.
What would happen to the width of the CI if you selected to use only a level of 90%?
Think about what is occurring. You are willing to accept a much lower level of confidence that the interval will contain the true population mean (the actual waiting time of all patients coming into the emergency room), so you can tighten the range of values.
If you want to nearly guarantee that your interval contains the true population mean, then you would want to include every value in the interval, so the interval spreads as the the level of confidence desired increases.
The only value that changes is the critical t-value, it is now 1.721, and the CI is now {42.87,47.12}.
As CI increases, the interval spreads. As sample size increases, the interval narrows (more representative of the entire population).
## Excel - Confidence Interval for MEAN
You can use Excel to find only the CI for a population mean. The population standard deviation must be known. Excel uses the Z-table to reference in its calculation.
You must determine the sample mean (x-bar) for the result to make sense and using 95% confidence level.
Suppose you have the following data:
Alpha risk = Level of Significance = 1 - Confidence Level = 0.05 = 5%
Population Standard Deviation = 6.48
Sample Size = 27 = n
Sample Mean (x-bar) = 50
The data would be entered in as shown and the result is 2.44 as shown in cell A1.
Knowing that you determined your sample mean (x-bar) to be 50, add 2.44 to get the upper limit of the interval and subtract 2.44 to get the lower limit of the interval and that becomes the CI for the population mean.
The interval is 50 +/- 2.44, or 47.56 to 52.44
### Confidence Interval for the STANDARD DEVIATION
The chi-squared distribution is not symmetrical and each varies according the degrees of freedom, dF.
The degrees of freedom equals n-1, dF = n-1.
This technique lacks robustness, in that it is very important that the population is known to be normally distributed when using it to estimate the population variance or standard deviation.
EXAMPLE:
Twenty-five assembly line workers throughout the Southwest United States were found to have a standard deviation in their total compensation of \$2.43. The average total compensation of an assembly line worker in the that was published by Bureau of Labor and Statistics was \$38.73 for a similar worker in the Southwest.
Calculate the population standard deviation using a 95% confidence level. Assume the population is known to be normally distributed.
Sample standard deviation = \$2.43
The average wage of \$38.73 is not needed for the CI calculation.
n = 25
dF = n-1 = 24
## Confidence Interval for PROPORTIONS
Many business decisions involve population proportions such as estimating market share and proportions of goods that are acceptable or defective.
EXAMPLE:
A survey was conducted on 300 emerging, domestic, small capital companies and found that 153 had an Emergency Action Plan that detailed reaction plans to maintain operations and customer service in the event of major illness or outbreak such as the swine flu.
Calculate the 92% confidence interval to estimate the proportion of emerging domestic, small capital companies that have an adequate Emergency Action Plan.
n = 300
p-hat = 153/300 = 51% = 0.51
The critical Z(0.04) value = 1.75
The CI states that with 92% confidence, the proportion of all similar companies with the plan will between 46% and 56%.
## Confidence Interval for CAPABILITY
To determine the CI for process capability use the formula provided below where:
USL = customer upper specification limit
LSL = customer lower specification limit
Pp is a process index that numerically describes the long term capability.
Cp is the short term indicator, Cp should always be analyzed with Cpk, as Pp should always be analyzed with Ppk.
Both Cp and Pp are a function of the process standard deviation, not a nominal (target) value that may be historical or provided by the customer.
Click here to purchase slides that offer more information regarding confidence intervals. Often, statistics are not expressed in terms of one number but rather as a range or an interval with a given level of confidence.
## Asymmetric Confidence Interval
An asymmetric confidence interval is one where the point estimate isn't the center of the confidence interval. For example, the point estimate may be 0.2 but the confidence interval is {0,0.8} because it happens to be that the values can not be less than zero.
There are a few reasons a confidence interval could be asymmetric:
• Random error is included
• The data has been transformed
• Positive or negative systemic bias incorporated
## Relationship with Margin of Error (ME)
The Margin of Error is a component of the CI calculation - see the visual aid below. Click here to review the entire module on the topic of Margin of Error
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Error Proofing | 2,933 | 13,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-43 | longest | en | 0.865429 |
https://www.airmilescalculator.com/distance/ybr-to-ysb/ | 1,627,605,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00714.warc.gz | 644,802,313 | 41,402 | # Distance between Brandon (YBR) and Sudbury (YSB)
Flight distance from Brandon to Sudbury (Brandon Municipal Airport – Sudbury Airport) is 909 miles / 1464 kilometers / 790 nautical miles. Estimated flight time is 2 hours 13 minutes.
Driving distance from Brandon (YBR) to Sudbury (YSB) is 1168 miles / 1880 kilometers and travel time by car is about 24 hours 48 minutes.
## Map of flight path and driving directions from Brandon to Sudbury.
Shortest flight path between Brandon Municipal Airport (YBR) and Sudbury Airport (YSB).
## How far is Sudbury from Brandon?
There are several ways to calculate distances between Brandon and Sudbury. Here are two common methods:
Vincenty's formula (applied above)
• 909.424 miles
• 1463.576 kilometers
• 790.268 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 906.883 miles
• 1459.486 kilometers
• 788.059 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Brandon Municipal Airport
City: Brandon
IATA Code: YBR
ICAO Code: CYBR
Coordinates: 49°54′36″N, 99°57′6″W
B Sudbury Airport
City: Sudbury
IATA Code: YSB
ICAO Code: CYSB
Coordinates: 46°37′30″N, 80°47′56″W
## Time difference and current local times
The time difference between Brandon and Sudbury is 1 hour. Sudbury is 1 hour ahead of Brandon.
CDT
EDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 144 kg (318 pounds).
## Frequent Flyer Miles Calculator
Brandon (YBR) → Sudbury (YSB).
Distance:
909
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
909
Round trip? | 477 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.823602 |
https://www.convert-measurement-units.com/convert+Kilobyte+to+Exabyte.php | 1,627,665,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00011.warc.gz | 740,322,519 | 17,174 | Convert kB to EB (Kilobyte to Exabyte)
## Kilobyte into Exabyte
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilobyte+to+Exabyte.php
## How many Exabyte make 1 Kilobyte?
1 Kilobyte [kB] = 0.000 000 000 000 000 888 178 419 700 13 Exabyte [EB] - Measurement calculator that can be used to convert Kilobyte to Exabyte, among others.
# Convert Kilobyte to Exabyte (kB to EB):
1. Choose the right category from the selection list, in this case 'Bytes / Bits'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilobyte [kB]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Exabyte [EB]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '303 Kilobyte'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Kilobyte' or 'kB'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Bytes / Bits'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '24 kB to EB' or '4 kB into EB' or '75 Kilobyte -> Exabyte' or '35 kB = EB' or '38 Kilobyte to EB' or '27 kB to Exabyte' or '85 Kilobyte into Exabyte'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(3 * 73) kB'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '303 Kilobyte + 909 Exabyte' or '63mm x 29cm x 53dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.303 209 988 140 8×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 1.303 209 988 140 8. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.303 209 988 140 8E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 13 032 099 881 408 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 909 | 3,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-31 | longest | en | 0.797927 |
https://byjus.com/nios/nios-b-level-mathematics-chapter-6/ | 1,656,882,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00381.warc.gz | 202,111,165 | 150,250 | Perimeter, Area and Volume
We‌ ‌have‌ ‌learned‌ ‌about geometry shapes – open and closed figures, plane figures and solid shapes, etc., ‌in‌ ‌our‌ ‌previous‌ ‌class.‌ ‌
In this chapter, we will be learning about the area, volume, and perimeter of these solid shapes.
What is a Perimeter
The perimeter of a simple or carved figure is the total measure of the length covered, going around the figure once. The measure of the total length around a figure is called its perimeter.
• The perimeter of the rectangle = Length + breadth + length + breadth (or)2 * length + breadth.
• The perimeter of the square = 4 * measure of side (or) the sum of the measures of all four sides.
• The perimeter of a triangle = The sum of the measures of all three sides.
What is Area
The measure of the space occupied by a plane surface is called its area. The area is written in square units.
• The area of a square = side×side (square units)
• The area of a rectangle = length × breadth (square units)
What is Volume
The space occupied or covered by any solid is called its volume. The unit of volume is cubic centimetre, cubic metre etc.
• Volume of a cube = side * side * side
A cube is an object whose length, breadth and height are equal
• Volume of a cuboid = length * breadth * height
A cuboid is an object whose length, breadth and height are not equal.
Stay tuned to BYJU’S for more information on NIOS, syllabus, notes, along with its important questions and solutions. | 395 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-27 | longest | en | 0.921857 |
www.growwater.org | 1,591,053,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00194.warc.gz | 168,645,660 | 7,055 | ## Plant Water Needs
How much to water your plants
You are here:
How much water do my plants need?
Determining plant irrigation needs can be a complex process that takes such factors as climate, rainfall patterns, plant size and type, density and microclimate into consideration. For the purposes of this workbook, we will use the ET method.
Estimating Water Requirements Using
Evapotranspiration Rates (ET)
Evapotranspiration, or ET is the combined effect of water evaporating from the soil and being used ( transpired) by the plants. The method assumes that all moisture by evapotranspiration will be replaced through irrigation.
You will need the following information:
• The area of the plant in square feet.
• The Species Factor of the plant.
• The ET Rate for your area.
Area = π ( 3.14) x Radius squared = plant square footage. (rounded) [example images below]
Plant species factor
The species factor indicates high (.8), moderate (.5) or low (.2) water use, and is used to account for differences in species’ water needs and broken down by plant type. In established landscapes, certain species are known to require relatively large amounts of water to maintain health and appearance (e.g., cherry, birch, alder, hydrangea, rhododendron), while others are known to need very little water (e.g., olive, oleander, pomegranate, hopseed, juniper). These numbers are highly variable, and detailed information on studies of 1,800 different species can be found at http://www.water.ca.gov/wateruseefficiency/docs/wucols00.pdf . These values are presented in Part 2 (WUCOLS III). Most garden books also provide this information and it is often represented with a water droplet: A full water droplet is high, half full is moderate, and an empty drop is low. Most fruit trees are moderate water users.
Plant Species Factors
• Very low < 0.1
• Low 0.1 - 0.3
• Moderate 0.4 - 0.6
• High 0.7 - 0.9
For the purposes of this document, we will simplify plant types/ factors as such:
• Low: .2
• Moderate: .5
• High: .8
Sacramento's Weekly ET Rates
Jan - .39
Feb - .56
March - .93
April - 1.28
May - 1.71
June - 2
July - 2.17
Aug - 2
Sept - 1.43
Oct - 1
Nov - .53
Dec - .39
Formula
Use this formula to calculate weekly plant water requirements:
.623 ( conversion coefficient to change inches to gallons) x weekly ET rate x plant species factor x plant square feet =
For Example: An 50 square foot peach tree’s weekly water needs in July.
.623 x 2.17 x .5 x 50 = 34 gallons a week (number rounded)
How to measure:
Measure from the middle of the truck to the furthest branches in the drip line.
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Co Op | 677 | 2,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-24 | latest | en | 0.900211 |
https://earthscience.stackexchange.com/questions/23734/air-pressure-on-mars-below-ground | 1,702,109,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00191.warc.gz | 267,030,620 | 40,866 | # Air pressure on Mars below ground
As I understand, in deep mines on Earth the air pressure increases with depth. At what depth on Mars would the air pressure reach 14psi (97 kPa)?
• Why the particular interest in 14 psi??? Apr 20, 2022 at 4:07
The Mars Atmosphere Model web page by Nasa lists the equation for atmospheric pressure as,
$$P = 0.699e^{-0.00009h}$$
Where pressure, $$P$$ is in kilopascals and $$h$$ is height is in meters.
Rearranging the equation,
$$h= - \frac{ln(\frac{P}{0.699})}{0.00009}$$
For a pressure of 97 kPa,
$$h= - \frac{ln(\frac{97}{0.699})}{0.00009} \ = \ -54\ 809 \ m \ = \ -54.8 \ km$$
Which means the atmosphere would need to be 54.8 km below the surface of Mars to have a pressure of 97 kPa. | 236 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-50 | longest | en | 0.838365 |
https://mrchasemath.com/2009/10/28/einsteins-puzzle/ | 1,675,267,448,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00610.warc.gz | 421,231,788 | 21,164 | # Einstein’s Puzzle
Thanks to Drew for pointing me to this fun logic puzzle.
Supposedly Einstein proposed this puzzle and it’s often referred to as “Einstein’s Challenge.” The question is, “Who Own’s the Fish?” And here’s the information. Go ahead and try it. And by presenting this puzzle, I am NOT advocating smoking or drinking, I merely wanted to quote the puzzle as it’s classically written.
Here’s what we know:
1. There are five houses in five different colors.
2. In each house lives a person of a different nationality.
3. These five owners drink a certain beverage, smoke a certain brand of cigarette and keep a certain pet.
4. No owners have the same pet, smoke the same brand of cigarette or drink the same drink.
Here are the clues:
1. The Brit lives in the red house.
2. The Swede keeps dogs as pets.
3. The Dane drinks tea.
4. The green house is on the immediate left of the white house.
5. The green house owner drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the yellow house smokes Dunhills.
8. The man living in the house right in the center drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhills.
12. The owner who smokes Bluemasters drinks beer.
13. The German smokes Princes.
14. The Norwegian lives next to the blue house.
15. The man who smokes Blends has a neighbour who drinks water.
Again, the question is: WHO OWNS THE FISH?
The fact that Einstein proposed the problem is urban legend. I’m not sure why it’s named for him. Also, most sources say something like “2% of people who try this problem get it correct,” which is a completely made up statistic, as far as I know.
I didn’t get it right, if that’s a comfort to you :-). So see if you can do better than me!
## 3 thoughts on “Einstein’s Puzzle”
1. It’s probably named after him because the writers want you to think that if you can do the puzzle, you’re as smart as Einstein. And therefore you’re more likely to make an attempt, because everyone wants to be that smart yeah.
I’ve completed a similar Einstein’s Challenge before, but I believe it was different because there was neither drinking nor smoking. Since this version seems like the same thing with a few words changed, I’m only commenting to say that there is a misplaced apostrophe in the second paragraph.
But yeah. Draw a chart. Charts are helpful.
2. This is Arun.
Like Carraka I’ve also seen a similar challenge before, but it was back in middle school and it had different clues.
It took me a while but I managed to solve it! The chart was really key but I ended up using two charts because there was a point where either I missed something or you truly had to guess. It was also helpful to list possible house numbers next to each clue and cross them off as the chart grew.
If anyone wants to see the full chart, it’s here, the houses being numbered left to right:
Thanks for the fun puzzle Mr. Chase! | 707 | 3,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-06 | longest | en | 0.933655 |
http://econsultancy.com/ca/blog/63366-thin-content-how-to-identify-and-fix-it-using-google-analytics | 1,386,765,222,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164036080/warc/CC-MAIN-20131204133356-00047-ip-10-33-133-15.ec2.internal.warc.gz | 54,448,280 | 20,692 | # Thin content: how to identify and fix it using Google Analytics
In my experience, severe panda-related hits tend to boil down to a root cause of either duplicate content, thin content, or extremely poor user experience.
As I’ve already covered many of the other areas involved in recovering from panda this month, I wanted to focus on thin content – what it is, how to spot it, and most importantly, how to fix it using Google Analytics
Here is an illustration of what thin content looks like. When I search for a two bedroom house in East Sussex, and scroll down to the one hundredth result, I begin to see results like this one where I’m directed to a page that offers absolutely no value whatsoever.
So how can you find out whether your site is being affected by thin content?
Well, as of a few days ago you can now find out in the Webmaster Tools ‘manual actions’ tab. However, I have to admit I am skeptical of this – I’ve looked at several sites in WMT that I know have thin content issues, and yet no notifications have come up.
I recently had the challenge of fixing thin content issues on a 1.5m page site with approximately 75,000 pages of what I would describe as low quality thin content.
While I’m sure there are numerous clever ways of coming to the same conclusion, I’m going to share my approach which I hope will at least provide a starting point to help you identify and rectify your thin content.
## 1. Define thin content quantitatively
The hardest part about identifying thin content is getting past the subjective nature of what is or isn’t considered ‘thin’.
In my analysis I decided to create a weighted formula that shortlisted a page for being considered thin if it had all of the following characteristics:
• A bounce rate between 95 and 99.99% (here’s why I excluded pages with a 100% bounce rate).
• An average time on page between 0.1 and five seconds.
You can use the following formula to work this out
=IF(CELL WITH BOUNCE RATE < 95%, "Not Thin", IF(CELL WITH AVERAGE TIME ON SITE < 5, "Thin", "Not Thin"))
Once you’ve got this shortlist of pages that are performing poorly, you can begin looking for trends. Which types of pages, or sections of your site are causing trouble?
Try to find common patterns in the URL structure, and get an understanding from a user’s perspective why these pages might be causing people to bounce straight away.
## 2. Rectifying thin content
There is no right or wrong way to rectify thin content, so let me go through various options with an example.
Below are the metrics for a page on MusicJobBoard.com, a site that I use for testing purposes from time to time. As you can see, this page has a combined high bounce rate and low time on page, qualifying this page to be shortlisted as ‘thin content’ by my definition above.
Here is the page, looking rather thin.
Job boards like this one are typically susceptible to thin content, as if no one posts a job in a certain category, the category page can remain indexed despite providing a poor result for someone looking for what the page would usually offer.
In this case, we could apply a rule that would noindex the page if it reached 0 results. I personally don’t like doing this, but on larger sites I’ve seen it work as a pretty effective strategy for keeping low quality results out of the SERPs.
Alternatively, we could design the page in a way that provided value even if no jobs were present – e.g. providing links to see similar jobs in audio production, or even offering some cool information on average salaries for this type of job, as Indeed does.
Another option, in this instance, would be to try and find an ongoing job listing for every category page i.e. a recording studio that is will to receive CVs on an ongoing basis.
One interesting option, which I’ve seen used by several property aggregator services is to redirect people to the next best result i.e. rather than showing me an empty page with 0 property listings in street X, send me to the page on street Y 200 meters away.
A more agreed-upon approach is to merge your pages. Rather than having a page on every single street in the UK, with many being empty, you could merge your street pages to post code pages, or town pages.
This goes against the idea of targeting the long tail with dedicated pages, but without quality in place I think it’s fair to say that that ship has sailed anyway.
One final option is to simply remove your poor performing pages. If there really is no point merging the pages or trying to improve them, it may be worth considering hacking off the low quality content and investing your efforts on improving your best content instead.
## Final thoughts
Thin content is a tricky issue to define and tackle, which is perhaps why it’s not covered in quite as much detail as more objective site quality issues. I’d love to hear how others are tackling it, and whether there are any other creative solutions that I’ve missed above.
Feel free to leave a comment below, send me a tweet, or drop me an email on marcus (at) ventureharbour.com.
Econsultancy's Crunch - Data, Analytics and the Rise of the Marketing Geek, takes place on October 10 at Truman Brewery, London. Crunch is the event for the analysts, strategists and boffins who turns raw numbers into insight, then revenue. This event is one of five that make up our week-long Festival of Marketing
Marcus Taylor is Director at Venture Harbour and a guest blogger on Econsultancy. You can follow Marcus on Twitter or Google Plus
## Reader comments (8)
1. Carmen Mardiros
Digital Analytics Consultant at Clear Clues
11:01AM on 9th September 2013
For those thinking of carrying out this I'd recommend first creating a segment as follows:
Source / medum contains google / organic
Keyword does not contain Your Brand Name
Visitor Type is New Visitor
This ensures you're removing biases from your data which can paint a misleading picture as to what counts as "thin" content.
Example: Say some of your landing pages attract a high volume of branded traffic and/or returning visitors. These visitors already have a high propensity for engagement. If you include this segment in your analysis then your pages will appear to perform better than they are.
The goal is to simply determine whether your pages pass the "first impression test" for organic, brand unaware visitors. Other visitor types muddle the picture. By excluding these you are simply raising the bar for what counts as quality.
I have been working on a Google Apps Scripts that scans a list of pages and returns:
- word count in content div area
- number of advertisement blocks on page
- number of non-advertisement images in content area
- any ordered/unordered lists (this is a biggie. Pages with lots of links but "thin" content tend to have a lower bounce rate because they naturally encourage clickthrough. They still count as "thin" content but you won't see the Bounce Rate symptom in GA)
- number of total links within content area
I then cross reference this with GA data for more context. I find it paints a more complete picture.
Thanks for sharing this Marcus.
2. Nick Stamoulis of Brick Marketing
2:21PM on 9th September 2013
I think your formula is a great way of quantifying something that is hard to measure. Thin content tends to be some of the deeper pages that got put up and forgotten about; the pages you never really rely on. Just keep in mind that you don't want to look at data in a silo. If one of your key pages has a higher bounce rate it might not be because the content is thin. Your reference post is great.
3. Ant Robinson
7:29PM on 9th September 2013
Great article however the opening paragraphs suggesting that WMT could identify thin content is misguided. Panda is algorithmic in its approach to penalising for thin content and therefore by definition won't be picked up by a 'Manual Action' report!
4. Marcus Taylor
Director at Venture Harbour
8:00PM on 9th September 2013
Great tip, Carmen. Very handy.
Ant - the change is only a few days old, I personally haven't seen the notifications in WMT, but apparently they are coming up for some people. See here: http://searchenginewatch.com/article/2292495/Thin-Content-With-Little-or-No-Added-Value-Manual-Action-Google-on-How-to-Fix-It
5. Adam
8:11PM on 9th September 2013
Great article Marcus. It amazes me how often knowledge of excel can help marketers make better use of exported data.
I have a big question about SEO and jobs boards, as I've been tinkering in this industry myself.
You mention one of the issues with Panda is duplicate content.
However, when it comes to running jobs boards, I find that recruiters will often paste the exact same ad copy when posting jobs on websites. For example, if you compare job post descriptions on Guardian Jobs, Monster.com and Reed and run them through copyscape then you'll find most of them have 90% duplicate content.
This then means that even if you're lucky enough to grow a successful job board with 100s or 1,000's of job postings, you will have a HUGE amount of duplicate content on your site.
What's the answer to this? No-index? Hope you have enough authority to beat the Panda issue (like the other sites do), or force recruiters to write original job descriptions (which would likely turn many of them off, especially for a smaller site).
6. Visakan Veerasamy
Marketing at ReferralCandy
4:49AM on 10th September 2013
Wow, I love that people are concerned enough about this to approach it with such a systematic approach. I can't imagine having to work with 1.5 million pages. What a challenge! Humbling to think about. Thanks for sharing.
7. mahender nath
Internet marketing at www.emperiumepos.com
3:33PM on 11th September 2013
Great tips! book marked
8. Josh Trenser
5:47AM on 12th September 2013
Hopefully these will help me fix my panda penalized sites. Thanks for sharing!!! | 2,191 | 9,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-48 | latest | en | 0.939452 |
https://pt.scribd.com/document/408407706/Anova-Mcqs | 1,561,627,708,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00548.warc.gz | 541,901,662 | 64,115 | Você está na página 1de 4
# Evaluation and Explanation through
One-Way ANOVA
15 MCQS
## Submitted to: Dr. Nazeer Anjum Shb
Zeeshan Ahmed Siddiqui (18-MS-PT-AMD-23)
## University of Engineering and Technology Taxila
1. In the analysis of variance (ANOVA), a factor is:
a) A dependent variable
b) A set of related treatments, categories or conditions
c) A variable that is confounded or entangled with the independent variable
d) A covariate, that is, a variable that correlates with the dependent variable.
2. The one-way (or one-factor between subjects) ANOVA tests the hypothesis that, in the
population:
a) All the group means have the same value
b) All the group means don’t have the same value
c) All the means have different values
d) Grand means is different as of sample means
## 3. In the analysis of variance (ANOVA), the measure is
a) The independent variable in the study
b) An extraneous variable that confounds the effects of the independent variable
c) The dependent variable in the study
d) A covariate
4. In the statistical model on which the one–way ANOVA is predicated, it is assumed that:
a) In the population, there is homogeneity of variance across treatment groups
b) There is always some basis for pairing the scores in any two groups
c) Participants under different conditions are tested by different experimenters
d) In the population, there is homogeneity of covariance
## 5. In the one-way ANOVA, the F statistics is used to:
a) Measure the average variance across the different treatment groups
b) Compare the between groups and within groups variance estimates
c) Measure the magnitude of the within groups or error variance
d) Compare the between groups variance estimate with the total variance
## 6. In the one-way ANOVA, the degrees of freedom of between the group is
a) Number of groups minus 1
b) Number of groups multiplied by n
c) Number of samples minus number of treatments
d) Total no of observations minus 1
7. What are the two types of variance which can occur in your data?
a) Indepndent and cofounding
b) Between or within groups
c) Experimenter and participant
d) Repeated and extraneous
a) T-scores
b) Chi square
c) Z-scores
d) F ratios
## 9. Which of the following assumptions must be met to use an ANOVA?
a) The dependent variable must be interval or ratio
b) Homogeneity of variance
c) Random sampling of cases
d) Data must be normally distributed
e) All of these
10. Analysis of variance is a stistical method of comparing the ______ of several populations
a) Standard deviation
b) Variances
c) Means
d) Proportions
11. Larger Sum of Square for group (SSG) indicates large variation between sample means, which
support
a) Null hypothesis
b) Alternate hypothesis
c) ANOVA
d) Homogeneity
12. In one-way ANOVA, which of the following is used within the F-fatio as a measurement of the
variance of individual observations?
a) MSG
b) MSE
c) SSE
d) SSG
13. The expected value or expectation of a statistic such as F is:
a) One minus the actual value of F
b) The experimenter’s confidence that the statistic will have a certain value
c) Its variance with repeated sampling
d) Its long run mean value with repeated sampling
14. In the F–ratio for the one–factor, between subjects ANOVA, the error term is
a) The numerator of the F ratio
b) The mean of the cell variances
c) The mean of the cell standard deviation
d) The variance of the cell means
15. If computational value of F is less than table value of F then will we will:
a) Accept the null hypothesis
b) Reject the null hypothesis
c) Accept the alternate hypothesis
d) Reject the alternate hypothesis | 904 | 3,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-26 | latest | en | 0.900084 |
http://oregondataproject.cf/forum8668-bmi-calculator-2-decimal-places.html | 1,524,249,649,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944677.39/warc/CC-MAIN-20180420174802-20180420194802-00510.warc.gz | 243,387,567 | 8,001 | bmi calculator 2 decimal places
bmi calculator 2 decimal places
Сайт советов и инструкций
bmi calculator 2 decimal places
One technique to control the number of decimal places displayed is by making use of a special Java class. Do the followingYou will also need a back button on the new screen to return the user to the BMI calculator. BMI Calculator 2 decimal places. Place your ad here LoadingBMI Calculator with Age. Whole Numbers to Decimal Calculator. Anthropometric calculator module BMI-for-age percentile BMI-for-age z-score Body mass index (weight in kg divided by height in metres squared) Date of birth Date of visit Food and Agriculture Organization of theThe maximum level of precision for any measurement is 2 decimal places. This calculator provides calculation of BMI (Body Mass Index) in metric (m, kg) or US imperial units (ft, in, lb).Rounding options: 1 digit after decimal point 2 digits after decimal point 3 digits after decimal point 4 digits after decimal point 5 digits after decimal point. How does the BMI calculator work? This weight tool helps you determine which the body mass index value is and whether this is placed in a healthy range but also other examples of recommended healthy weight according to certain established formulas that are listed below. To calculate BMI, a body mass index calculator can be used. Our free, online BMI calculator utilizes a scientifically recognized formula to yield easy-to-read results that can help you better understand both your current BMI and your ideal weight. : Select rounding: 1 digit after decimal point 2 digits after decimal point 3 digits after decimal point 4 digits after decimal point 5 digits after decimal point.This calculator provides calculation of your BMI in both metric and USA standard units. Add to favorites. First I deliberately set BMD to a value that has only two decimal places. I can print it as-is but I have no control. In the second case, I set BMD to a value for quite a few decimal places but this time I limit the number of output decimal places. An example of calculating BMI using the body mass index formula: Height 180 cm (1.
80 m) Weight 70 kg.Divide the result obtained from that calculation by height again. Then, multiply that number by 703. Round off the resulting value to second decimal place. BMI Calculator BMI. BMI is a measure of body weight based on a weight and height.Car Payment Calculator. Currency Converter. Decimal to Binary Conversion. Download Time. GCD Calculator. Decimals Calculator. Add, subtract and multiply decimals step-by-step. Order of Operations. (Click on the BMI Calculator to determine your BMI.)To calculate BMI, take the weight (kg) and divide it by height (m). | 559 | 2,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-17 | latest | en | 0.805914 |
https://www.weegy.com/Home.aspx?ConversationId=EC410AED | 1,534,786,172,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216718.53/warc/CC-MAIN-20180820160510-20180820180510-00394.warc.gz | 1,021,564,650 | 8,247 | What does equate mean?
Equate means: consider (one thing) to be the same as or equivalent to another.
Question
Updated 201 days ago|1/30/2018 10:33:41 PM
Edited by yumdrea [1/30/2018 10:33:40 PM]
Rating
3
Equate means: consider (one thing) to be the same as or equivalent to another.
Added 201 days ago|1/30/2018 10:33:41 PM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 619 | 2,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-34 | latest | en | 0.88928 |
https://forexrebatechanger.com/about-forex/how-is-lot-calculated-in-forex.html | 1,638,000,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00380.warc.gz | 348,367,385 | 19,372 | # How is lot calculated in forex?
Contents
The Forex position size calculator uses pip amount (stoploss), percentage at risk and the margin to determine the maximum lot size. When the currency pair is quoted in terms of US dollars the equation is as follows; Lot Size = ((Margin * Percentage) ÷ Pip Amount) ÷ 100k.
## What is Forex lot size?
Forex is commonly traded in specific amounts called lots, or basically the number of currency units you will buy or sell. … The standard size for a lot is 100,000 units of currency, and now, there are also mini, micro, and nano lot sizes that are 10,000, 1,000, and 100 units.
## How do I calculate my lot?
How to Calculate Lot Sizes Into Acres
1. Measure the length and width of the land plot in feet if it is square or rectangular. …
2. Multiply the length times the width of rectangular land plots to get the area in square feet. …
3. Divide the number obtained in Step 2 by 43,560.
## What is 0.01 lot size in Forex?
The minimum trade size with FBS is 0.01 lots. A lot is a standard contract size in the currency market. It’s equal to 100,000 units of a base currency, so 0.01 lots account for 1,000 units of the base currency. If you buy 0.01 lots of EUR/USD and your leverage is 1:1000, you will need \$1 as a margin for the trade.
## How is profit calculated in forex?
When you close out a trade, take the price (exchange rate) when selling the base currency and subtract the price when buying the base currency, then multiply the difference by the transaction size. That will give you your profit or loss.
## Can I trade forex with \$10?
Yes, you can start forex trading with just \$10 and even less than that. Forex brokers have some minimum deposit requirements to open account with them. Some have little high like \$500 or \$1000, but there are some who need only \$5 or \$10 to open an account.
## How many pips is a dollar?
Pip Value Calculation When Trading in a USD Account
The fixed pip amounts are: USD\$10 for a standard lot, which is 100,000 units of currency. USD\$1 for a mini lot, which is 10,000 units of currency.
## How much is 0.01 Pips?
Position size of 0.01 lot for EURUSD currency pair, for every 10 pips gain will give a \$1 profit (10 cents per pip). So for EURUSD means that 10 pips for 0.01 lot size profit are \$1. 0.01 lot size or 1000 units or micro lot is the smallest position size when we talk about standard forex accounts.
100,000
## How many lots can I trade with \$1000?
See the example below using the position size calculator. For an account size of \$1,000, risking 1% with a stop loss of 50 PIPS, the appropriate lot size trading EURUSD is 2 micro lots.
## How many dollars is 100 pips?
Therefore, for a position of this size – 10,000 units – we will gain or lose \$1 for every pip movement in either direction. So if the EUR/USD moves 100 pips (i.e. 1 cent) in our direction we will make \$100 profit. We can do this for any trade size. The calculation is simply the trade size times 0.0001 (1 pip).
## How many pips a day is good?
This currency pair moves about 100 to 300 pips per day – so you can at least catch 20 pips in a day. A2A. Any number of pips is OK depending on what exposure it means. If you are not profitable yet, what could help is to aim for 10 pips per day but increase the lot size.
## How are pips calculated?
Movement in the exchange rate is measured by pips. Since most currency pairs are quoted to a maximum of four decimal places, the smallest change for these pairs is 1 pip. The value of a pip can be calculated by dividing 1/10,000 or 0.0001 by the exchange rate.
## How much do forex traders make a day?
Even so, with a decent win rate and risk/reward ratio, a dedicated forex day trader with a decent strategy can make between 5% and 15% a month thanks to leverage. Also remember, you don’t need much capital to get started; \$500 to \$1,000 is usually enough.
## How are nas100 pips calculated?
To calculate the value of one pip directly in USD, use the following formula: 0.01 / 90.68 * 100 000 = 11.03 USD. Value of one pip (EUR) = 11.03 / 1.4018 = 7.86 EUR (if EUR/USD lists as 1.4018).
micro-lots | 1,076 | 4,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-49 | latest | en | 0.923525 |
http://physicshelpforum.com/energy-work/14172-does-show-atmospheric-pressure-air-resistance.html | 1,571,708,613,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00528.warc.gz | 138,858,262 | 11,658 | Physics Help Forum Does this show atmospheric pressure or air resistance?
Energy and Work Energy and Work Physics Help Forum
Jan 2nd 2018, 12:47 PM #1 Junior Member Join Date: Jan 2018 Posts: 9 Does this show atmospheric pressure or air resistance? Please help a discussion we're having here. The demo in question is attached. I had to do this demo as the very first introduction to pressure for K8 students. It's supposed to show that there is atmospheric pressure. I don't see it being much to do with pressure as much as air resistance. Surely air pressure is taken as static and therefore as soon as the paper/ruler/paddle starts to move, it's all about air resistance? Please let me know what you think Attached Thumbnails
Jan 2nd 2018, 12:56 PM #2 Senior Member Join Date: Aug 2010 Posts: 434 The point of this experiment is that the board, with the paper over it hardly moves as the board breaks. Since it hardly moves, it cannot well be "air resistance" in the usual sense- resistance to motion through the air. The point is that the board on the bench is held down by a force equal to the air pressure multiplied by the area of the newspaper.
Jan 2nd 2018, 01:50 PM #3 Senior Member Join Date: Jun 2016 Location: England Posts: 1,085 The board tries to lift the paper faster than the air can get in under the paper to fill the gap. This creates lower pressure under the paper than on the top. This pressure difference holds the board in place. __________________ ~\o/~
Jan 2nd 2018, 02:22 PM #4 Junior Member Join Date: Jan 2018 Posts: 9 Thanks, HallsofIvy, but you can't tell me it's not about air resistance as you contradict yourself. If it moves, it's air resistance, any pressure differential is as a result of this. That's the same logic as saying that a long, long lever isn't levering because the object hardly moved .... the longer the lever, the less the movement, therefore, if the paper/card hardly moves, it's showing how effective air resistance is? Thanks, Woody, I like your idea drawing on Bernoulli's ideas but that's not the point. Is this an effective way of introducing students to the concept of what atmospheric pressure is or is it merely air resistance misinterpreted? How can this be effective if the very physics of what it's supposed to show are misinterpreted?
Jan 2nd 2018, 02:43 PM #5 Senior Member Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 What do you think the mechanism of air resistance is? In this case it is due to air pressure, just the same as in a boat sail. If you don't like this experiment, try blowing a door shut with a wind artificial or natural.
Jan 3rd 2018, 11:11 AM #6 Junior Member Join Date: Jan 2018 Posts: 9 This experiment would work with a pivot in space ... i.e. with nothing for the paper/ruler combination to rest on and at any angle .... and the reason is that it's due to air resistance, not atmospheric pressure ... think about it .... if you invert the whole thing and had to slap the ruler upwards, or turned it vertically .... the experiment would still work due to air resistance, not air pressure. If the experiment works at every angle and orientation, it's independent of atmospheric pressure, surely?
Jan 3rd 2018, 12:51 PM #7 Senior Member Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Did you say you were teaching Physics? What would be your response to a student who failed to answer you when you asked a question specifically designed to help him? I repeat What do you think the mechanism of air resistance is?
Jan 3rd 2018, 01:39 PM #8 Junior Member Join Date: Jan 2018 Posts: 9 You repeat? I think I'll ignore you from now on as you're obviously a petulant egoist ... I would suggest that you need more patience and a less hubristic attitude if you're going to get a polite response from me. My response to a student would be how can I help you understand this more. I assume yours would be to repeat yourself, only louder?
Jan 3rd 2018, 01:44 PM #9 Junior Member Join Date: Jan 2018 Posts: 9 The mechanism of air resistance would involve instantly changing the surface pressure one side of the paper while reducing it on the other, which if on a desk, would create a pressure differential and a suction effect. However, this experiment would work at any orientation so the suction idea relies on the paper/ruler combination being over a solid surface. If you're changing the pressure at either surface you cannot have atmospheric pressure around these surfaces, you're adding or subtracting due to air resistance. None of which provides a direct link between the result of the experiment and the reason stated, nor does it explain atmospheric pressure accurately. I guess in this post-truth era, we can even excuse an experiment that doesn't do what it's stated to prove.
Jan 3rd 2018, 01:48 PM #10 Senior Member Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Attacking the messenger will not alter the fact that you do not understand the mechanism of air resistance (or air pressure for that matter). Last edited by studiot; Jan 3rd 2018 at 01:50 PM.
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I am building a balancing robot using the Lego Mindstorm's NXT system. I am using two sensors from HiTechnic, the first being an Accelerometer and the second being a Gyroscope. I've successfully filtered out noise from both sensors and derived angles for both in a range between -90 and 90 degrees, with 0 degrees being perfectly balanced.
My next challenge is to combine both of the sensor values to correct for the Gyroscope's drift over time. Below is an example graph I created from actual data to demonstrate the drift from the gyroscope:
The most commonly used approach I've seen to make combining these sensors rock solid is by using a Kalman filter. However, I'm not an expert in calculus and I really don't understand mathematical symbols, I do understand math in source code though.
I'm using RobotC (which is like any other C derivative) and would really appreciate if someone can give me examples of how to accomplish this in C.
SOLUTION RESULTS:
Alright, kersny solved my problem by introducing me to complementary filters. This is a graph illustrating my results:
Result #1
Result #2
As you can see, the filter corrects for gyroscopic drift and combines both signals into a single smooth signal.
Edit: Since I was fixing the broken images anyways, I thought it would be fun to show the rig I used to generate this data:
-
your data is clearly diverging. the kalman filter or any other method won't help you if your initial data doesnt agree. – ldog Oct 19 '09 at 20:53
I'm not sure you fully understand what the graph is displaying, it's a known problem of Gyroscopic data to drift. They is why the data is diverging, which is what the filter/integration I'm looking for will correct using the accelerometers data. Also, the reason for the radical drift, is because I shook the sensors pretty violiently to illustrate my problem. :) – Dylan Vester Oct 20 '09 at 14:42
I have no idea what you are graphing because you did not label the axis's but regardless if your data is clearly diverging from the same y-values given the same x-values it is pretty bad data. – ldog Oct 20 '09 at 16:49
if you apply any filter to it as is that tries to minimize error in the least squares sense (what the kalman filter does for example) your going to be averaging an error that increases as your values of x increase. Clearly one part of your data is telling you something and a different part of your data is telling you something else. – ldog Oct 20 '09 at 16:51
A great alternative to the Kalman filter is the complementary filter which is much easier to implement: http://www.pieter-jan.com/node/11 – Pieter-Jan May 16 '13 at 9:00
Kalman Filters are great and all, but I find the Complementary Filter much easier to implement with similar results. The best articles that I have found for coding a Complementary Filter are this wiki (along with this article about converting sensors to Engineering units) and a PDF in the zip file on this page (Under Technical Documentation, I believe the file name in the zip is filter.pdf);
PS. If your stuck on a Kalman Filter, here is some C-syntax code for the Arduino that implements it.
-
FANTASTIC, I believe this may be exactly what I was looking for. The Filter.pdf file was really the big help, and explained and solved my exact problem. I haven't verified it yet (I'm at work). But tonight, I'll try and get this going and mark my question as answered! – Dylan Vester Oct 19 '09 at 20:09
Glad I could help! If you want to see an example of it in action, check out my blog at ohscope.com. I built a Segway like balancing scooter, and I will be putting up more data soon. – kersny Oct 19 '09 at 20:44
The wiki link appears to be dead – Jon Drnek Jun 18 '11 at 15:14
– Edward Falk Jun 16 '12 at 0:15
This seems like a great explanation; it does use some maths symbols though. It concludes with an implementation of a Kalman filter in Matlab; perhaps you find that more palatable.
-
Thank you for your reply. I've looked through the link you sent me, while it does a good job at explaining a Kalman filter, it doesn't describe how to combine two sensor values into one, which is what I really need. – Dylan Vester Oct 19 '09 at 16:05 | 998 | 4,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2014-10 | latest | en | 0.952454 |
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### Author Topic: RESONANCE EFFECTS FOR EVERYONE TO SHARE (Read 296457 times)
#### Groundloop
• TPU-Elite
• Hero Member
• Posts: 1736
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #150 on: December 21, 2008, 06:48:12 PM »
@wattsup,
I will help you out with a circuit for your swing generator.
The output is reversing polarity. You explained it right.
The input DC voltage will be on the output connector
with a alternating polarity at each pulse to the switch.
(Or half the frequency in the first version.)
The voltage rating depends on the transistors used. For the first circuit the IR2103
can take 600 volt at maximum. So if the transistors can take the same then this is the maximum.
The current can be high at low voltages but not at high voltages.
The transistors used will set the maximums. The data sheets for the types used
will explain how much current you can draw at a selected voltage level.
For the newest switch with optocouplers then the maximum voltage the switch can handle is 300 volt.
This because the H11D1 can take maximum 300 volts. The current can be high at low voltages but not at high voltages.
The transistors used will set the maximums. The data sheets for the types used will explain how much current you
can draw at a selected voltage level.
I will never recommend using the maximum voltages. The newest switch will be happy at up to 250 volt.
Above that, there is no way to tell when things blow up.
At maximum voltage the transistor used in this circuit can handle 0,1 ampere.
At 12 volt the transistors can handle several amperes. It all depend on what type
transistors you select to use.
The switch will produce an alternating DC output. The output is not a sinus wave
but square pulses. The pulses is even looking at low frequencies but "nasty"
looking at near the maximum frequency. The estimated maximum frequency is set
by the switch time for the optocouplers, in this case approx. 250 KHz.
The function generator looks OK to me.
Attached is the HW design files for both boards. I have included the gerber files but
the drill files are in millimeters. You can download the free Cadsoft Eagle CAD and
redo the drill files for inches if you like.
Groundloop.
#### forest
• Hero Member
• Posts: 4076
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #151 on: December 21, 2008, 07:56:25 PM »
I'm not electronic engineer so please bear with me.I think we need more then 300V.Maybe 600V is even too low.Personally I would like to test an LC oscillator with 1000V and really microamps.The problem is with switches, there should be kind of intelligent spark gap (=transistor) to handle it.In other words : something which charge capacitor and then when fully charged cut power source and allow for freely oscillate the LC circuit.
There is one more problem I think. To cut all lower frequencies charge to capacitor must be flowing when capacitor is disconnected from coil or when capacitors are in special configuration. That's how I think about it but I may be wrong. I just use the water pump analogy - if capacitor is a water storage device then the size of it and the speed of water flow determine if water is slowly filling container or a part of water is flowing above container inlet due to speed.
Boguslaw
#### gotoluc
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##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #152 on: December 21, 2008, 08:14:24 PM »
Hi everyone,
I was away since last evening and just now cough up with the new posts.
WOW Grounddloop!!! ... thank you ever so much for taking so much of your time to reply and explain with such excellent details
You even went past the technical stuff and spoke from your heart ... you are a great soul and I thank you from the bottom of my heart.
I have truly been blessed to have all your help, since without your circuit I only had mechanical relays and could not confirm an effect like the one we have now.
Thank you also for continuing to develop the circuit to the next level.
I've been blessed with a financial donation to which was just enough to ordered the supplies needed to build a multiple coils and batteries self charging setup as you proposed.
I am truly thankful and humbled by all this great help
Luc
« Last Edit: December 26, 2008, 07:23:54 PM by gotoluc »
#### gotoluc
• elite_member
• Hero Member
• Posts: 3096
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #153 on: December 21, 2008, 08:36:58 PM »
I'm not electronic engineer so please bear with me.I think we need more then 300V.Maybe 600V is even too low.Personally I would like to test an LC oscillator with 1000V and really microamps.The problem is with switches, there should be kind of intelligent spark gap (=transistor) to handle it.In other words : something which charge capacitor and then when fully charged cut power source and allow for freely oscillate the LC circuit.
There is one more problem I think. To cut all lower frequencies charge to capacitor must be flowing when capacitor is disconnected from coil or when capacitors are in special configuration. That's how I think about it but I may be wrong. I just use the water pump analogy - if capacitor is a water storage device then the size of it and the speed of water flow determine if water is slowly filling container or a part of water is flowing above container inlet due to speed.
Boguslaw
Hi forest, thanks for your post ... you are understanding well The usable power output (Watts) gets higher as the polarity switched voltage goes higher ... and at a fraction (if any) to amp consumption.
It can also work the other way I demonstrated by using a fixed voltage output and adding more coils and pickups.
I do agree that the way to go is more voltage and I am sure as we share test results and develop the switch we will get there but to do this we need builders and developers.
Luc
#### Groundloop
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• Hero Member
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##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #154 on: December 21, 2008, 09:07:34 PM »
Hi Luc,
I got a PM from "najman100" and I have proposed a deal with him. If everything works out
then there will be PCBs for both the new circuit drawings. Since I'm in the xmas mode
right now, there will be a fully soldered and tested PCB for the micro controller version
for you next year, for free. I will have the new PCBs and parts by January 16 next year.
If "najman100" agree to my proposal then there will be a PCB for you for the non micro
version also. :-) I have ordered enough boards so "Wattsup" and "TinselKoala" can have
a PCB for free also (the micro controller version).
Groundloop.
#### hartiberlin
• Hero Member
• Posts: 8154
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #155 on: December 21, 2008, 09:09:25 PM »
Hi Luc and Groundloop.
My comments were not at all meant negative.
I only wanted to point out, where some possible
measurement errors could be made.
Yes, I saw your proposal for the selfloop unit and it is a nice design.
Please keep working on it.
I think this has a great chance to go overunity.
Many thanks.
P.S. Luc, if you still try your first design with the coil and
the Avramenko plug together with the earth connection charging up your
2 uF cap: You could try to connect a 4 Watts fluorescent tube across the
cap and maybe reduce the cap to a 100 nF ( at least 1000 Volts rating)
thus you will get more flashes per second from the fl. tube cause the cap will
charge up faster and then the cap will be discharged, when fl. tube
ignites.
This way you could also use the output power of the cap
to light up the fl. tube.
Regards, Stefan.
#### Groundloop
• TPU-Elite
• Hero Member
• Posts: 1736
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #156 on: December 21, 2008, 09:54:07 PM »
Stefan,
I have ordered the PCBs and will build the circuits. Will also start testing on Luc's coil
types soon. I was a little "grumpy" yesterday so just ignore my words.
Regards,
Groundloop.
#### gotoluc
• elite_member
• Hero Member
• Posts: 3096
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #157 on: December 21, 2008, 09:54:39 PM »
Hi Luc,
I got a PM from "najman100" and I have proposed a deal with him. If everything works out
then there will be PCBs for both the new circuit drawings. Since I'm in the xmas mode
right now, there will be a fully soldered and tested PCB for the micro controller version
for you next year, for free. I will have the new PCBs and parts by January 16 next year.
If "najman100" agree to my proposal then there will be a PCB for you for the non micro
version also. :-) I have ordered enough boards so "Wattsup" and "TinselKoala" can have
a PCB for free also (the micro controller version).
Groundloop.
WOW!!!.... great news once again Groundloop ... thank you.
More good news also!... our friend Fausto has contacted me to say he has ordered parts and will also be replicating
This is a great day.
Luc
#### gotoluc
• elite_member
• Hero Member
• Posts: 3096
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #158 on: December 21, 2008, 10:07:07 PM »
P.S. Luc, if you still try your first design with the coil and
the Avramenko plug together with the earth connection charging up your
2 uF cap: You could try to connect a 4 Watts fluorescent tube across the
cap and maybe reduce the cap to a 100 nF ( at least 1000 Volts rating)
thus you will get more flashes per second from the fl. tube cause the cap will
charge up faster and then the cap will be discharged, when fl. tube
ignites.
This way you could also use the output power of the cap
to light up the fl. tube.
Regards, Stefan.
Hi Stefan,
I was going to retry this at one point! ...I'll let you know how it works out
Thanks for bringing my attention to it once again.
Luc
#### gotoluc
• elite_member
• Hero Member
• Posts: 3096
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #159 on: December 21, 2008, 10:39:35 PM »
@everyone,
here are 2 great videos done and posted by user: Cody at the RESONANCE EFFECTS FOR EVERYONE TO SHARE topic I started at Energetic Forum.
Here is a variable capacitor that I've purchased to facilitate my future testing and tuning: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&ssPageName=STRK:MEWNX:IT&item=130276804974
I wish I would of had one of these when I first started!... it would of made life much easier
And here is an inductance and capacitance meter which will also make life easier: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=390005413895&ssPageName=MERCOSI_VI_ROSI_PR4_PCN_BIX_Stores&refitem=220330813763&itemcount=4&refwidgetloc=closed_view_item&refwidgettype=osi_widget&_trksid=p284.m185&_trkparms=algo%3DSI%26its%3DI%26itu%3DCR%252BUCI%26otn%3D4%26ps%3D42
Luc
#### TinselKoala
• Hero Member
• Posts: 13958
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #160 on: December 23, 2008, 12:55:11 AM »
Those videos from Cody are awesome! Very well spoken and good organization. Takes me back...
I am pleased to see that the quality of our "free energy" videos is improving. Gotoluc's setup is looking really good--I like the tight wire bundles!
Meanwhile I've been playing around some too. I'm exploring where my own interests lead, for now, and using these coils that I've made for different projects. I am surprised at how well they work in this application.
About the circuit boards for Groundloop's newer designs--count me in, for sure, and thanks. I'd be glad to pay my fair share of whatever it costs, of course.
(Please note: At about 4:01 in the following video, I mistakenly say "series" when I should have said "parallel". The coil, cap, and led are all in parallel.)
#### ramset
• Hero Member
• Posts: 8073
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #161 on: December 23, 2008, 01:35:22 AM »
WOW !!
You boys are going to need a camera crew
Some real Top shelf Guys
Doing Amazing work
THANK YOU
Chet
#### gotoluc
• elite_member
• Hero Member
• Posts: 3096
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #162 on: December 23, 2008, 02:20:14 AM »
Hi TinselKoala,
looks good ... you sure have allot of coils to play with ... I don't know if I'll ever be able to catch up to you
Thanks for making the video and sharing your findings.
Luc
#### Groundloop
• TPU-Elite
• Hero Member
• Posts: 1736
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #163 on: December 23, 2008, 05:16:58 AM »
@TinselKoala,
Great video. :-) Thanks for taking time to test out this circuit. Have you tried to use the output
form the secondary coil to charge up a lead acid battery? It is also good to see that the secondary
coil can be trimmed to the correct resonant frequency. Your led test clearly shows that this could
be done. Great work.
You will also get your PCB (and a ready programmed PIC16F84A controller) for free. But as I said to Luc,
the PCBs wil not arrive until around the 16 of January next year. I will also need a little time to develop
the firm ware needed to run the controller. I will also build the non controller version. I have been talking
with "najman100" and he will make the PCBs for the non controller version. He has a little trouble with
the Eagle design files but I think we can solve that later. I don't know how many PCBs najman100 will
make, but I hope he will make more than 3 so that we can have one each. Najman100, any thoughts
on this?
@ramset,
Since you commented on this work in a positive way and since I'm still in my xmas mode, then there
will be a free PCB (of the pic controller version) for you also. :-) If you can't program a PIC micro, then
just let me know and I will give you one ready programmed. :-)
@Wattsup,
I haven't heard from Wattsup, but if you read this and want a free PCB (of the pic controller version),
then let me know. (My xmas spirit will run out around the 1st of January 2009. LOL)
@gotoluc,
I will make a couple of coils for the project. I will try to make the coils as shown in my test drawing.
I have collected some cardboard tubes to make air core coils on. I estimate that I can fit up to 8
coils on each cardboard. Each coil will be wound with a 150 turns primary and a 76 turns secondary
loosely coupled next to (or at the top of) the primary coil. I guess that each coil can be tuned with a
variable capacitor (2 to 22pF) to get a resonant frequency of around 150 KHz. You can have one of the
coils.
Groundloop.
#### najman100
• Newbie
• Posts: 38
##### Re: RESONANCE EFFECTS FOR EVERYONE TO SHARE
« Reply #164 on: December 23, 2008, 05:49:26 AM »
@TinselKoala,
Great video. :-) Thanks for taking time to test out this circuit. Have you tried to use the output
form the secondary coil to charge up a lead acid battery? It is also good to see that the secondary
coil can be trimmed to the correct resonant frequency. Your led test clearly shows that this could
be done. Great work.
You will also get your PCB (and a ready programmed PIC16F84A controller) for free. But as I said to Luc,
the PCBs wil not arrive until around the 16 of January next year. I will also need a little time to develop
the firm ware needed to run the controller. I will also build the non controller version. I have been talking
with "najman100" and he will make the PCBs for the non controller version. He has a little trouble with
the Eagle design files but I think we can solve that later. I don't know how many PCBs najman100 will
make, but I hope he will make more than 3 so that we can have one each. Najman100, any thoughts
on this?
@ramset,
Since you commented on this work in a positive way and since I'm still in my xmas mode, then there
will be a free PCB (of the pic controller version) for you also. :-) If you can't program a PIC micro, then
just let me know and I will give you one ready programmed. :-)
@Wattsup,
I haven't heard from Wattsup, but if you read this and want a free PCB (of the pic controller version),
then let me know. (My xmas spirit will run out around the 1st of January 2009. LOL)
@gotoluc,
I will make a couple of coils for the project. I will try to make the coils as shown in my test drawing.
I have collected some cardboard tubes to make air core coils on. I estimate that I can fit up to 8
coils on each cardboard. Each coil will be wound with a 150 turns primary and a 76 turns secondary
loosely coupled next to (or at the top of) the primary coil. I guess that each coil can be tuned with a
variable capacitor (2 to 22pF) to get a resonant frequency of around 150 KHz. You can have one of the
coils.
Groundloop.
let me open the file and after i will go up to 6 .
Najman | 4,459 | 17,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.835802 |
https://www.physicsforums.com/threads/a-confusing-problem.159508/ | 1,686,421,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00047.warc.gz | 1,034,168,675 | 17,364 | A Confusing Problem
• o_0curiousraych
o_0curiousraych
There is a truck with a rock stuck in the tire.
The tire has a diameter of 36 inches.
The truck is going 55 mph.
There is a car behind it going 55 mph also.
What is the minimum distance the car can be behind the truck so it won't get hit by the rock when it flies out of the tire?
I did a problem like this before... but I never had to include the diameter of the tire.
The answer has to be reported in inches.
Can anyone please explain this problem to me?
-o_0curiousraych
Knowing the radius of the tire allows you to calculate the initial velocity of the rock. The tire has diameter 35 inches and so has circumference $\pi d= 70\pi$ which is about 220 inches.
The car is going at 55 miles per hour which is 55 mi/hr(5280ft/mi)(12in/ft)= 3484800 in/hr= 3484800 in/hr(1/60 hr/min)(1/60 sec/min)= 968 in/sec. Since the tire is rolling over every inch of that, the tire must make 968/220= 4.4 revolutions per second. That means the rock is moving at 4.4 rev/sec(220 in/rev)= 968 in/sec. Now, imagine an object thrown up at angle $\theta$ with speed 968 in/sec. What will its range (distance until it hits the ground again) as a function of $\theta$? What value of $\theta$ gives the maximum range and what is that maximum range? (The car had better be at least that far back to be certain it will not be hit by the rock.)
I see I used the circumerence of the wheel, 220 in, twice in calculating the speed of the rock. It probably could have been done more simply!
hmmm. obviously all the wheels are going 55mph and the rock flies backward at such...
Edit: see above
Last edited:
OMG. I was writing "NO" and pointing out that while the center of the wheel was moving forward at 55 mph, the points on the circumference were going at- then stopped to do the computation. While the center of the wheel (the axle and so the car) moves forward a distance of 220 cm (the circumference of the tire) a point on the circumference of the tire makes one rotation around the tire and so goes- 220 cm. Yes, the speed (but not the velocity) of a point on the circumference of the wheel is exactly the same as that of the car. That's why I divided and then multiplied by 220 above! Thanks.
well while you were doing that, I was thinking about contact patch and how the size of the tire might influence the trajectory when the worst case approach was the obvious approach in terms of figuring a safe distance. (I should take heed having suffered 3 broken windshields here in Colo over the past 10 winters!)
help
I'm still not sure what you guys are talking about.
you be in very good hands here, just some idle chatter about the problem.
How are you coming along on this?
So back on topic, did you figure out the angle which would throw the rock the farthest distance?
you be in very good hands here, just some idle chatter about the problem.
How are you coming along on this?
So back on topic, did you figure out the angle which would throw the rock the farthest distance?
yes. i already know that 45 degrees will fire the furthest. that's common sense. but what I'm still not getting is, why was i given the diameter if that isn't used in the range equation at all?
Don't know. If someone knows, hopefully they will post. I think smaller tires would tend to propel a rock more vertically, as they are spinning faster and an object more likely to break loose higher up the rotation cycle. But Halls... said, figure the worst case, and I think that's good advice and arguable to the max if the TA marks you down for some oversight. (Trucks suck, they spit gravel like a cannon shot at about 5-10 degrees.)
Knowing the radius of the tire allows you to calculate the initial velocity of the rock. The tire has diameter 35 inches and so has circumference $\pi d= 70\pi$ which is about 220 inches.
The car is going at 55 miles per hour which is 55 mi/hr(5280ft/mi)(12in/ft)= 3484800 in/hr= 3484800 in/hr(1/60 hr/min)(1/60 sec/min)= 968 in/sec. Since the tire is rolling over every inch of that, the tire must make 968/220= 4.4 revolutions per second. That means the rock is moving at 4.4 rev/sec(220 in/rev)= 968 in/sec. Now, imagine an object thrown up at angle $\theta$ with speed 968 in/sec. What will its range (distance until it hits the ground again) as a function of $\theta$? What value of $\theta$ gives the maximum range and what is that maximum range? (The car had better be at least that far back to be certain it will not be hit by the rock.)
I see I used the circumerence of the wheel, 220 in, twice in calculating the speed of the rock. It probably could have been done more simply!
he said that the diameter was 35 in...then in his formula for pie times D...he has 70(pie)...i think this may be bad...=\
is their anyway...someone would redue the problem...using 35*pi...? so that way...we can figgure out how fast the tires movin? and then to do it in meters per second? that would be amazing
Last edited:
that would be amazing wouldn't it
Unless anyone has somrthing constructive to add to the soln, let's give it a rest and let the OP determine the flow from here. | 1,283 | 5,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.95695 |
https://www.bartleby.com/questions-and-answers/find-the-counterclockwise-circulation-of-f-y-ex-ln-yi-exyj-around-the-boundary-of-the-region-that-is/0616c741-bc1c-4924-9849-78c54da90335 | 1,585,798,176,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506580.20/warc/CC-MAIN-20200402014600-20200402044600-00166.warc.gz | 600,943,749 | 26,665 | # Find the counterclockwise circulation of F = (y + ex ln y)i + (ex/y)j around the boundary of the region that is bounded above by the curve y = 3 - x2 and below by the curve y = x4 +1
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Find the counterclockwise circulation of F = (y + ex ln y)i + (ex/y)j around the boundary of the region that is bounded above by the curve y = 3 - x2 and below by the curve y = x4 +1
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https://www.stockmaniacs.net/elliott-wave-calculator-day-trading-tools/?share=email | 1,581,924,986,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00502.warc.gz | 915,924,475 | 25,992 | # Elliott Wave Calculator – Day Trading Tools
Elliott Waves are one of the most important technical triggers for any financial market. As you know that any market always moves in wave-like structures. So even a declining market bounces up after a fall and even a rising market corrects after a rise. So if the bounce or correction can be predicted early it can be a great trading tool for the day traders. We have created a few simple day trading tools on our site. One of them is the Elliott Waves Calculator.
To access the Elliott Wave Calculator you can CLICK HERE. The page has boxes where you can input last trading days high, low and close. Let’s analyze Nifty with this calculator. I have written the respective values of 14th February 2017 in the corresponding cells. High – 8819.90, Low – 8772.65 and Close – 8782.25. Now I clicked on Calculate. And the Elliott Wave Calculator calculated today’s (15th February 2017) instantly.
## Given Elliott wave counts by the calculator:
Wave 1 8772.65 – 8790.69
Wave 2 8790.69 – 8779.54
Wave 3 8779.54 – 8802.51
Wave 4 8802.51 – 8788.28
Wave 5 8788.28 – 8817.49
Wave A 8819.9 – 8790.69
Wave B 8790.69 – 8808.74
Wave C 8808.74 – 8785.78
## Automated BUY / SELL triggers given by the calculator:
Go Long at/above 8772.65. Keep Stoploss at 8763.87. Target 1 is 8790.69.
Go Long at/above 8779.54. Keep Stoploss at 8763.87. Target 2 is 8802.51.
Go Long at/above 8788.28. Keep Stoploss at 8779.54. Target 3 is 8817.49.
Now let’s check what is the outcome of the BUY / SELL trigger. Check the image below:
The Nifty index opened, made a low at 8777.65 and bounced up, made nigh of 8807.65. The first 2 trades target is achieved. This is the power of Elliott waves and that’s why this calculator is one of my favorite day trading tools.
Not only intraday trading, but you can also use this Elliott Wave Calculator for positional or swing trading. At the weekend, simply enter the last week’s high, low and close and you can get next week’s swing trading levels instantly. If you have any questions about our day trading tools feel free to comment below the post.
## Author: Indrajit Mukherjee
Indrajit is a professional blogger and trading system developer. Amibroker expert, Wordpress expert, SEO expert and stock market analyst.Trading since 2002, he has started the journey of StockManiacs.net on 2008. He follows Indian and world stock markets closely.
## 5 thoughts on “Elliott Wave Calculator – Day Trading Tools”
1. Ismail Shaik says:
Very happy to see your expertise. Thank You so much.
2. Sir close means last friday close right.
One more thing wat is the accuracy level of this calculator
1. Naresh for next day’s levels close means close of the day, for next week’s levels close means close of Friday.
3. Amol says:
Is this technique is more accurate during market opening time or any time during trading hourrs
1. Try once a day using last day’s end of day data. Better to follow since the start of the day. You can also apply the same theory on weekends to weekly candles to get next week’s levels.
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https://www.transtutors.com/questions/reliable-housewares-is-a-local-store-that-sells-many-household-items-and-issues-its--2914107.htm | 1,571,042,320,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00326.warc.gz | 1,044,676,099 | 16,411 | # Reliable Housewares is a local store that sells many household items and issues its own credit...
PROJECT PART B
Reliable Housewares is a local store that sells many household items and issues its own credit card to its customers. The store manager wants to study the purchasing behavior of its "credit" customers. To that end, he has come to DeVry and asked our MBA students for help. The manager has brought with him data on five variables of 55 randomly selected credit customers.
? LOCATION (Rural, Urban, Suburban – Household location of the credit customer)
? INCOME (in $1,000's – be careful with this) ? SIZE (Household Size - number of people living in the household of credit customer) ? YEARS (the number of years that the customer has lived in the current location) ? CREDIT BALANCE ($ balance on customer’s store credit card)
Hypothesis Testing and Confidence Intervals
The Reliable Housewares store manager wants to learn more about the purchasing behavior of its "credit" customers. In fact, he is speculating about four specific cases shown below (a) through (d) and wants you to help him test their accuracy.
a. The average annual income of credit customers is less than $48,000. b. The true population proportion of credit customers who live in an urban area exceeds 55%. c. The average number of years lived in the current home is less than 14 years. d. The average credit balance for suburban customers is at most$4,300.
i. Using the dataset provided in Files, perform the hypothesis test for each of the above speculations (a) through (d) in order to see if there is an statistical evidence to support the manager’s belief. In each case,
o Use the Seven Elements of a Test of Hypothesis, in Section 7.1 of your textbook (on or about Page 361) or the Six Steps of Hypothesis Testing I have identified in the addendum.
o Use a=2% for all your analyses,
o Explain your conclusion in simple terms,
o Indicate which hypothesis is the “claim”,
o Compute the p-value,
ii. Follow your work in (i) with computing a 98% confidence interval for each of the variables described in (a) though (d). Interpret these intervals.
iii. Write an executive summary for the Reliable Housewares store manager about your analysis, distilling down the results in a way that would be understandable to someone who does not know statistics. Clear explanations and interpretations are critical.
iv. All DeVry University policies are in effect, including the plagiarism policy.
v. Project Part B report is due by the end of Week 6.
vi. Project Part B is worth 100 total points. See grading rubric below.
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## Boolean Solids
Forum: Geometry
Date: 16 Aug, 2011
From: Victor <Victor>
```--20cf307cffa82bf9c504aaa2842a Content-Type: text/plain; charset=ISO-8859-1 Hi, I'm trying to construct a plate with holes inside. I use Boolean Solid operations for that. But I have faced a problem : I have a plate with dimensions : 30cm x 30cm x 1cm. Hole's dimensions : 25um x 25um x 1cm. between each hole 50um I need a lot of holes. And I have only 2GB RAM, so it's not enough memory in case I don't deallocate memory by hands. So, I have a piece of code : solid = new G4SubtractionSolid("Subtraction", solidFinal, solid2, RotM, TranVector) *solidFinal = *solid delete solid Since I use operator = to copy the contents of solid into solidFinal before deallocating solid pointer, this should work. But I get a segmentation fault for that. I understand that the problem might be very easy, I just don't see it. And I'm pretty sure that it's the place of the error in my program. Could you please help??? What am I doing wrong in here??? --20cf307cffa82bf9c504aaa2842a Content-Type: text/html; charset=ISO-8859-1 Hi,
I'm trying to construct a plate with holes inside.
I use Boolean Solid operations for that.
But I have faced a problem :
I have a plate with dimensions : 30cm x 30cm x 1cm. Hole's dimensions : 25um x 25um x 1cm. between each hole 50um
I need a lot of holes. And I have only 2GB RAM, so it's not enough memory in case
I don't deallocate memory by hands.
So, I have a piece of code :
solid = new G4SubtractionSolid("Subtraction", solidFinal, solid2, RotM, TranVector)
*solidFinal = *solid
delete solid
Since I use operator = to copy the contents of solid into solidFinal before deallocating solid pointer, this should work.
But I get a segmentation fault for that.
I understand that the problem might be very easy, I just don't see it. And I'm pretty sure that it's the place of the error in my program.
Could you please help??? What am I doing wrong in here??? --20cf307cffa82bf9c504aaa2842a-- ```
Inline Depth: Outline Depth: Add message:
1 Re: Boolean Solids (Tom Roberts - 16 Aug, 2011)
Re: Boolean Solids (John Allison - 17 Aug, 2011)
to: "Boolean Solids"
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[ Geant 4 Home | Geant 4 HyperNews | Search | Request New Forum | Feedback ] | 664 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.925793 |
http://nag.com/numeric/cl/nagdoc_cl24/examples/source/s30face.c | 1,438,660,350,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-32/segments/1438042990217.27/warc/CC-MAIN-20150728002310-00151-ip-10-236-191-2.ec2.internal.warc.gz | 171,290,733 | 1,793 | /* nag_barrier_std_price (s30fac) Example Program. * * Copyright 2009, Numerical Algorithms Group. * * Mark 9, 2009. */ #include #include #include #include #include #include int main(void) { /* Integer scalar and array declarations */ Integer exit_status = 0; Integer i, j, m, n; NagError fail; Nag_CallPut putnum; Nag_Barrier typenum; /* Double scalar and array declarations */ double h, k, q, r, s, sigma; double *p = 0, *t = 0, *x = 0; /* Character scalar and array declarations */ char put[8+1]; char type[14+1]; Nag_OrderType order; INIT_FAIL(fail); printf("nag_barrier_std_price (s30fac) Example Program Results\n"); printf("Standard Barrier Option\n\n"); /* Skip heading in data file */ scanf("%*[^\n] "); /* Read put, type */ scanf("%8s%14s%*[^\n] ", put, type); /* * nag_enum_name_to_value (x04nac). * Converts NAG enum member name to value */ putnum = (Nag_CallPut) nag_enum_name_to_value(put); typenum = (Nag_Barrier) nag_enum_name_to_value(type); /* Read s, h, k, sigma, r, q */ scanf("%lf%lf%lf%lf%lf%lf%*[^\n] ", &s, &h, &k, &sigma, &r, &q); /* Read m, n */ scanf("%ld%ld%*[^\n] ", &m, &n); #ifdef NAG_COLUMN_MAJOR #define P(I, J) p[(J-1)*m + I-1] order = Nag_ColMajor; #else #define P(I, J) p[(I-1)*n + J-1] order = Nag_RowMajor; #endif if (!(p = NAG_ALLOC(m*n, double)) || !(t = NAG_ALLOC(n, double)) || !(x = NAG_ALLOC(m, double))) { printf("Allocation failure\n"); exit_status = -1; goto END; } /* Read array of strike/exercise prices, X */ for (i = 0; i < m; i++) scanf("%lf ", &x[i]); scanf("%*[^\n] "); /* Read array of times to expiry */ for (i = 0; i < n; i++) scanf("%lf ", &t[i]); scanf("%*[^\n] "); /* * nag_barrier_std_price (s30fac) * Standard Barrier option pricing formula */ nag_barrier_std_price(order, putnum, typenum, m, n, x, s, h, k, t, sigma, r, q, p, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_barrier_std_price (s30fac).\n%s\n", fail.message); exit_status = 1; goto END; } if (putnum == Nag_Call) printf("%s\n", "Call :"); else if (putnum == Nag_Put) printf("%s\n", "Put :"); if (typenum == Nag_DownandIn) printf("%s\n\n", "Down-and-In"); else if (typenum == Nag_DownandOut) printf("%s\n\n", "Down-and-Out"); else if (typenum == Nag_UpandIn) printf("%s\n\n", "Up-and-In"); else if (typenum == Nag_UpandOut) printf("%s\n\n", "Up-and-Out"); printf("%s%8.4f\n", " Spot = ", s); printf("%s%8.4f\n", " Barrier = ", h); printf("%s%8.4f\n", " Rebate = ", k); printf("%s%8.4f\n", " Volatility = ", sigma); printf("%s%8.4f\n", " Rate = ", r); printf("%s%8.4f\n", " Dividend = ", q); printf("\n"); printf("%s\n", " Strike Expiry Option Price"); for (i = 1; i <= m; i++) for (j = 1; j <= n; j++) printf("%9.4f%9.4f %11.4f\n", x[i-1], t[j-1], P(i, j)); END: NAG_FREE(p); NAG_FREE(t); NAG_FREE(x); return exit_status; } | 958 | 2,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-32 | latest | en | 0.381162 |
http://eigen.tuxfamily.org/dox/classEigen_1_1SuperLUBase.html | 1,695,483,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00615.warc.gz | 13,194,768 | 4,499 | Eigen 3.4.90 (git rev 67eeba6e720c5745abc77ae6c92ce0a44aa7b7ae)
Eigen::SuperLUBase< MatrixType_, Derived > Class Template Reference
## Detailed Description
### template<typename MatrixType_, typename Derived> class Eigen::SuperLUBase< MatrixType_, Derived >
The base class for the direct and incomplete LU factorization of SuperLU.
Inheritance diagram for Eigen::SuperLUBase< MatrixType_, Derived >:
## Public Member Functions
void analyzePattern (const MatrixType &)
void compute (const MatrixType &matrix)
ComputationInfo info () const
Reports whether previous computation was successful. More...
superlu_options_t & options ()
Public Member Functions inherited from Eigen::SparseSolverBase< Derived >
template<typename Rhs >
const Solve< Derived, Rhs > solve (const MatrixBase< Rhs > &b) const
template<typename Rhs >
const Solve< Derived, Rhs > solve (const SparseMatrixBase< Rhs > &b) const
SparseSolverBase ()
## ◆ analyzePattern()
template<typename MatrixType_ , typename Derived >
void Eigen::SuperLUBase< MatrixType_, Derived >::analyzePattern ( const MatrixType & )
inline
Performs a symbolic decomposition on the sparcity of matrix.
This function is particularly useful when solving for several problems having the same structure.
factorize()
## ◆ compute()
template<typename MatrixType_ , typename Derived >
void Eigen::SuperLUBase< MatrixType_, Derived >::compute ( const MatrixType & matrix )
inline
Computes the sparse Cholesky decomposition of matrix
## ◆ info()
template<typename MatrixType_ , typename Derived >
ComputationInfo Eigen::SuperLUBase< MatrixType_, Derived >::info ( ) const
inline
Reports whether previous computation was successful.
Returns
Success if computation was successful, NumericalIssue if the matrix.appears to be negative.
## ◆ options()
template<typename MatrixType_ , typename Derived >
superlu_options_t& Eigen::SuperLUBase< MatrixType_, Derived >::options ( )
inline
Returns
a reference to the Super LU option object to configure the Super LU algorithms.
The documentation for this class was generated from the following file: | 475 | 2,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.504235 |
https://mathspace.co/textbooks/syllabuses/Syllabus-98/topics/Topic-1488/subtopics/Subtopic-17698/?activeTab=theory | 1,716,756,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00277.warc.gz | 333,601,195 | 71,805 | Hong Kong
Stage 1 - Stage 3
# The Median
Lesson
The median is a measure of central tendency. In other words, it's one way of describing a value that represents the middle or the centre of a data set. The median (which kind of sounds like medium) is the middle score in a data set.
Remember!
The data must be ordered (usually in ascending order) to calculate the median.
Which term is in the middle?
Say we have five numbers in our data set: $4$4, $11$11, $15$15, $20$20 and $24$24.
The median would be $15$15 because it is right in the middle. There are two numbers on either side of it.
4, 11, 15, 2024
However, if we have a larger data set, we may not be able to see straight away which term is in the middle. There are two methods we can use to help us work this out.
### The "cross out" method
Basically, once your data is ordered, you can cross out a high number and a low number until you only have one number left. Let's check out this process using an example. Here is a data set with nine numbers: $1$1, $1$1, $3$3, $5$5, $7$7, $9$9, $9$9, $10$10, $15$15.
1. Check all your numbers are in ascending order (ie. in order from smallest to largest).
2. Cross the smallest and the largest number out like so:
3. Repeat the process from step 2, making sure you work from the outside in, taking the smallest number and the largest number each time until you have one term left. We can see in this example that the median is 7.
NB. You will only be left with one term if there are an odd number of terms to start with. If there are an even number of terms, you will be left with two terms (if you cross them all out, you've gone too far)! All you need to do if find the average of these two terms.
### Working out the middle term
You can also work out which term will be the middle number using the following formula:
Let $n$n be the number of terms:
$\text{middle term }=\frac{n+1}{2}$middle term =n+12th term
So if we use the same set of numbers from the previous example:
$1$1, $1$1, $3$3, $5$5, $7$7, $9$9, $9$9, $10$10, $15$15, there are nine numbers in the set. So to work out which value is in the middle:
$\text{middle term }$middle term $=$= $\frac{9+1}{2}$9+12 $=$= $5$5
This means the fifth term will be the median: $1$1, $1$1, $3$3, $5$5, 7, $9$9, $9$9, $10$10, $15$15.
So again, we find that the median is $7$7.
Let's try that with an even number of terms. Let's look at this data set with four terms: $8$8, $12$12, $17$17, $20$20.
$\text{middle term }$middle term $=$= $\frac{4+1}{2}$4+12 $=$= $2.5$2.5th term
But what is the $2.5$2.5th term?? Just like using the "cross out" method, the $2.5$2.5th term means the average between the second and the third term. Again, remember your data must be in order before you count the terms. So in this example, the median will be the average of $12$12 and $17$17.
$\text{median }$median $=$= $\frac{12+17}{2}$12+172 $=$= $14.5$14.5
#### Worked Examples
##### Question 1
Find the median from the frequency distribution table:
Score Frequency
$23$23 $2$2
$24$24 $26$26
$25$25 $37$37
$26$26 $24$24
$27$27 $25$25
##### QUESTION 2
Write down $4$4 consecutive odd numbers whose median is $40$40.
1. Write all solutions on the same line separated by a comma.
##### QUESTION 3
Determine the following using the bar graph:
1. The total number of scores.
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# A boat traveled upstream 90 miles at an average speed of
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A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?
A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1
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Re: Difficult PS Problem- Help! [#permalink]
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09 Sep 2010, 17:54
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jjewkes wrote:
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?
-2.5
-2.4
-2.3
-2.2
-2.1
Trip upstream took $$\frac{90}{v-3}$$ hours and trip downstream took $$\frac{90}{v+3}$$ hours. Also given that the difference in times was $$\frac{1}{2}$$ hours --> $$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$;
$$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$ --> $$\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}$$ --> $$\frac{90*6}{v^2-9}=\frac{1}{2}$$ --> $$v^2=90*6*2+9$$ --> $$v^2=9*(10*6*2+1)$$ --> $$v^2=9*121$$ --> $$v=3*11=33$$;
Trip downstream took $$\frac{90}{v+3}=\frac{90}{33+3}=2.5$$ hours.
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Re: Difficult PS Problem- Help! [#permalink]
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11 Sep 2010, 21:33
90/(v-3)=90(v+3)+.5
90[v+3-v+3]/(v*v-9)=.5
90*6/(v*v-9)=.5
90*12+9=v*v
v*v=1089
v=33
v+3=36
Time required to travel downstream=90/36=2.5 hours
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30 Sep 2010, 22:43
Michmax3 wrote:
A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then traveled the same distance downstream at an average speed of (V+3) miles per hour. If the upstream trip took half an hour longer than the downstream trip, how many hours did it take the boat to travel downstream?
A)2.5
B)2.4
C)2.3
D)2.2
E)2.1
Given 90/(v-3) = 90/(v+3) + (1/2)
90[(1/(v-3)) - (1/(v+3))] = 1/2
90[6/(v^2 - 9)] = 1/2
90*6*2 = v^2 - 9 => v^2 = 1089 => v = 33.
Downstream time -- 90/(v+3) => 90/36 => 2.5 (A).
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30 Sep 2010, 22:45
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Michmax3 wrote:
A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then traveled the same distance downstream at an average speed of (V+3) miles per hour. If the upstream trip took half an hour longer than the downstream trip, how many hours did it take the boat to travel downstream?
A)2.5
B)2.4
C)2.3
D)2.2
E)2.1
Upstream time = downstream time + 0.5
90/(v-3) = 90/(v+3) + 1/2
90 (1/(v-3) - 1/(v+3))= 1/2
90 (6/(v^2-9)) = 1/2
v^2 = 1089
V = 33
Downstream time = 90/36 = 2.5
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Word problem need help solving [#permalink]
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21 Nov 2010, 07:17
Hi I just took one gmat practice test and the pity about those test is that they don´t have explanations.
The problem i am struggling with in mathematical sense is something like this:
A boat that traveled 90 miles upstream a river at the rate average speed Y-3, on the way back he traveled the same distance (90 miles) at a average speed of Y+3, if the trip upstream took half an hour longer than the trip down stream. How long took the drip downstream ?
thanks, I shall give my teacher kudos if I find out how to do that
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Re: Difficult PS Problem- Help! [#permalink]
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21 Nov 2010, 14:02
solve the below equation to reach upon speed of boat-
90/(v+3) - 90/(v-3) = 1/2
V=33
therefore time required while travelling downstream- 90/(33+3) = 90/36 = 2.5
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22 Nov 2010, 02:19
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shrouded1 wrote:
Michmax3 wrote:
v^2 = 1089
V = 33
Downstream time = 90/36 = 2.5
Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?
many thanks
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22 Nov 2010, 06:57
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hafgola wrote:
Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?
many thanks
Let me take your question first:
It is a good idea to remember squares of numbers till 20. They come in handy sometimes.
To find the square root of bigger numbers (which are perfect squares), observe the following:
1. A square of a number only ends in 0/1/4/5/6/9.
-If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.)
-If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 1[highlight]9[/highlight], root 441 = 2[highlight]1[/highlight])
-If it ends in 4, the square root will end in 2/8
-If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer)
-If it ends in 6, the square root will end in 4/6.
-If it ends in 9, the square root will end in 3/7.
These are all derived by just observing the last digits when we square a number.. You don't need to learn this.
Since 1089 ends in a 9, its square root will end in 3/7.
Also $$30^2 = 900$$ so root of 1089 will be greater than 30.
But $$40^2 = 1600$$ so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check.
Also, an equation like $$\frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}$$, doesnt need to be solved.
99% chance is that v will be an integer.
The options give me values for $$\frac{90}{{v + 3}}$$
Let me say if it is 2.5, then 90/2.5 = 36 and v = 33
Then $$\frac{90}{{v - 3}}$$ = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition.
Had 2.5 not been the first option, you might say that it would take me a long time to check each option.
No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24
90/2.3. No ways 900 is divisible by 23
90/2.2. No ways 900 is divisible by 11
90/2.1. No ways 900 is divisible by 7
In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct.
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[#permalink] ### Show Tags 25 Mar 2011, 11:22 1 This post received KUDOS problem is time consuming rather difficult for memorizing square root. _________________ Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html SVP Joined: 16 Nov 2010 Posts: 1673 Location: United States (IN) Concentration: Strategy, Technology Followers: 34 Kudos [?]: 470 [1] , given: 36 Re: Difficult PS Problem- Help! [#permalink] ### Show Tags 27 Mar 2011, 01:18 1 This post received KUDOS 90/(v-3) = 90/(v+3) + 1/2 => 90( V + 3 - v + 3)/(v-3)(v+3) = 1/2 => 90 * 6 * 2 = v^2 - 9 => v^2 = 1080 + 9 => v^2 = 9 (120 + 1) => v^2 = 3^2 * 11^2 => v = 3*11 = 33 So Downstream time = 90/36 = 30/12 = 2.5 hrs Answer - A _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Director Joined: 01 Feb 2011 Posts: 757 Followers: 14 Kudos [?]: 100 [0], given: 42 Re: Difficult PS Problem- Help! [#permalink] ### Show Tags 25 Aug 2011, 17:19 1 This post was BOOKMARKED upstream v-3 t+1/2 downstream v+3 t (v+3)t = (v-3)(t+1/2) vt + 3t = vt +v/2 - 3t - 3/2 => t = (v-3)/12 = 90/v+3 => v = 33 => t = 30/12 = 2.5 hours Answer is A. Senior Manager Joined: 15 May 2011 Posts: 280 Location: Costa Rica Concentration: Healthcare, International Business GMAT 1: 710 Q42 V45 GMAT 2: 740 Q48 V42 GPA: 3.3 WE: Research (Consulting) Followers: 15 Kudos [?]: 121 [0], given: 134 Re: Downstream Rate Question [#permalink] ### Show Tags 17 Aug 2012, 15:13 Can someone pls help with the above question? What is the most efficient way to do this. Here's what i've got.. If one takes the time for the trip downstream to to be T, then one gets UPSTREAM: V-3 = 90 / (T + 1/2) DOWNSTREAM V+3 = 90 / T Now no matter how much I simplify I end up with a complicated equation, and then it will be plugging in off answer choices but that takes >3 minutes at least. Any suggestions for a sohrtcut, I'm sure there's an easy way.. _________________ How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing) Intern Joined: 01 Jan 2011 Posts: 22 Location: Kansas, USA Schools: INSEAD, Wharton Followers: 2 Kudos [?]: 14 [0], given: 9 Re: Downstream Rate Question [#permalink] ### Show Tags 17 Aug 2012, 17:40 I think the answer to the question posted here should be 2.5. Basically, frame the equation in terms of the difference in time (= distance/ speed): [90][/v-3]- [90][/v+3]= 1/2. Once you get this you could solve for V. Remember that the question is asking for the value of [90][/v+3]. Moderator Joined: 02 Jul 2012 Posts: 1230 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 110 Kudos [?]: 1253 [2] , given: 116 Re: A boat traveled upstream a distance of 90miles at an average [#permalink] ### Show Tags 19 Nov 2012, 01:10 2 This post received KUDOS Amateur wrote: A boat traveled upstream a distance of 90miles at an average speed of (v-3) miles per hour and then traveled down stream at an average of (v+3) miles per hour. If upstream look 30 min more than downstream time, how much time did the boat travel downstream? a) 2.5 b) 2.4 c) 2.3 d) 2.2 e) 2.1 how to solve these kind of questions in 2 min? any short way... \frac{90}{(v-3)}=\frac{90}{(v+3)} +1/2 solving for v, value of \frac{90}{(v+3)} is tedious... and answer choices are in too close range to make an estimation.... how should I be tackling these? The equation itself is simple enough.. $$\frac{90}{v-3}- \frac{90}{v+3} = \frac{1}{2}$$ $$\frac{540}{v^2-9} = \frac{1}{2}$$ $$v^2 - 9 = 1080$$ $$v^2 = 1089$$ $$= 3^2 * 11^2$$ v = 3*11 = 33 Answer = 90/36 = 2.5 Kudos Please... If my post helped. _________________ Did you find this post helpful?... Please let me know through the Kudos button. Thanks To The Almighty - My GMAT Debrief GMAT Reading Comprehension: 7 Most Common Passage Types Intern Joined: 27 Dec 2012 Posts: 30 Followers: 1 Kudos [?]: 9 [0], given: 11 Re: Boat Traveling [#permalink] ### Show Tags 14 Mar 2013, 02:11 VeritasPrepKarishma wrote: hafgola wrote: Hi I struggled with this one, I am wondering Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ? many thanks Let me take your question first: It is a good idea to remember squares of numbers till 20. They come in handy sometimes. To find the square root of bigger numbers (which are perfect squares), observe the following: 1. A square of a number only ends in 0/1/4/5/6/9. -If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.) -If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 19, root 441 = 21) -If it ends in 4, the square root will end in 2/8 -If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer) -If it ends in 6, the square root will end in 4/6. -If it ends in 9, the square root will end in 3/7. These are all derived by just observing the last digits when we square a number.. You don't need to learn this. Since 1089 ends in a 9, its square root will end in 3/7. Also $$30^2 = 900$$ so root of 1089 will be greater than 30. But $$40^2 = 1600$$ so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check. Also, an equation like $$\frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}$$, doesnt need to be solved. 99% chance is that v will be an integer. The options give me values for $$\frac{90}{{v + 3}}$$ Let me say if it is 2.5, then 90/2.5 = 36 and v = 33 Then $$\frac{90}{{v - 3}}$$ = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition. Had 2.5 not been the first option, you might say that it would take me a long time to check each option. No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24 90/2.3. No ways 900 is divisible by 23 90/2.2. No ways 900 is divisible by 11 90/2.1. No ways 900 is divisible by 7 In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct. Hi karishma, I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6829 Location: Pune, India Followers: 1919 Kudos [?]: 11928 [1] , given: 221 Re: Boat Traveling [#permalink] ### Show Tags 14 Mar 2013, 20:36 1 This post received KUDOS Expert's post sauravleo123 wrote: Hi karishma, I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill?? It needs practice. You need to believe that you can solve the question very quickly for you to be able to think up of logical approaches. In GMAT, you can solve almost every question within two minutes so if you get a big complicated equation, you can be sure that you are missing something. If the question looks really complicated, we know there has to be a trick (In fact, one tends to overlook the so called short-cuts when questions are simple and don't take much time anyway). I would suggest you to read up and practice the logical solutions - once you understand the logic, try using it in other questions. Don't worry about the time factor. I discuss some logic based ideas regularly on my blog. Check it out: http://www.veritasprep.com/blog/categor ... om/page/3/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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08 May 2013, 04:27
I would prefer substituting answers rather than solving the equation to save time. In actual test, i was way past time, took more than 3 mins to solve the Q. Later offline, i used substitution after arriving at the initial set of below equations for Time. It was so easy and less troublesome.
UPSTREAM:
V-3 = 90 / (T + 1/2)
DOWNSTREAM
V+3 = 90 / T
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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20 Mar 2014, 13:19
2.5 hrs..A set of best possible option spares u the horror of solving the equations..
90 miles is the distance traveled downstream..2.5 is the only only option that divides 90..I am pretty sure the speeds are never (25.12+3) & (25.12-3)..So we can safely assume an integer..2.5 & 3 are the set of speeds that aptly fit in.
took me just 10 secs..
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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25 Aug 2014, 06:38
Let's solve this problem by using ratios. As the trip upstream (U) took a half hour longer than the trip downstream (D), the ratios of time - U:D = 3:2. So the ratios of speed equals to 2:3.
Given speed: U = s-3, D = s+3.
(s-3) / (s+3) = 2/3. When you solve you will find that s = 15. Then 90 / (15+3) = 5. But as 5 isn't an aswer choice, we divide it by 2 and get 2.5.
Re: A boat traveled upstream 90 miles at an average speed of [#permalink] 25 Aug 2014, 06:38
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Fan Jihua*,Zhang Dingguo**2)(),Shen Hong
1. * School of Mechatronics and Power Engineering, Jiangsu University of Science and Technology, Zhangjiagang 215600, Jiangsu China
? Suzhou Institute of Technology, Jiangsu University of Science and Technology, Zhangjiagang 215600, Jiangsu China
** School of Sciences, Nanjing University of Science and Technology, Nanjing 210094, China
• Received:2019-03-28 Online:2019-09-18 Published:2019-09-30
• Contact: Zhang Dingguo
Abstract:
Compared with the floating frame of reference formulation, the absolute nodal coordinate formulation (ANCF) has significant advantages in dealing with the nonlinear large deformation problem of flexible bodies. ANCF defines the nodal co-ordinates in a global co-ordinate system, uses the global slopes instead of angles to define the orientation of the elements, has a constant mass matrix, and does not have the Coriolis centrifugal force. However,the elastic force matrix is a nonlinear term, and the solution will be time consuming and occupied resources. According to this,the elastic line method is introduced for solving the elastic force, the Green-Lagrangian strain tensor is defined on the centerline, the curvature formula is used to define the bending strain, and the torsion angle formula is used to define the torsional strain. At the same time, the finite element method is used to describe the displacement field of the three-dimensional flexible beam, and the constant mass matrix, the generalized stiffness matrix and the generalized force matrix of the beam element are solved, and then the dynamic equations of the element are obtained. The dynamic equations of the three-dimensional beam are obtained by the transformation matrix. Then the differences between the continuum mechanics method and the elastic line method are pointed out theoretically, and the dynamic simulation software is compiled. Finally, the dynamics behaviors of a flexible pendulum and a tracked vehicle are numerical investigated with the continuum mechanics method and the elastic line method. The results show that the elastic line method can effectively improve the calculation efficiency under the premise of ensuring accuracy.
CLC Number: | 479 | 2,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.857433 |
https://unitsconverters.com/en/J/Mg-To-Fatigue/Utu-3973-3994 | 1,603,509,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00189.warc.gz | 560,048,617 | 24,748 | 1 J/mg is equivalent to 1000000 Fatigue
Formula Used
1 Joule/Milligram = 1000000 Gray
1 Gray = 1 Fatigue
1 Joule/Milligram = 1000000 Fatigue
## J/mg to Fatigue Conversion
The abbreviation for J/mg and Fatigue is joule/milligram and fatigue respectively. 1 J/mg is 1000000 times bigger than a Fatigue. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including J/mg to Fatigue conversion.
## Joule/Milligram to Fatigue
Check our Joule/Milligram to Fatigue converter and click on formula to get the conversion factor. When you are converting absorbed dose from Joule/Milligram to Fatigue, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## J/mg to Fatigue
The formula used to convert J/mg to Fatigue is 1 Joule/Milligram = 1000000 Fatigue. Measurement is one of the most fundamental concepts. Note that we have Exagray as the biggest unit for length while Attogray is the smallest one.
## Convert J/mg to Fatigue
How to convert J/mg to Fatigue? Now you can do J/mg to Fatigue conversion with the help of this tool. In the length measurement, first choose J/mg from the left dropdown and Fatigue from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from Fatigue to J/mg? You can check our Fatigue to J/mg converter.
### FAQ about converter
How to convert J/mg to Fatigue?
The formula to convert J/mg to Fatigue is 1 Joule/Milligram = 1000000 Fatigue. J/mg is 1000000 times Bigger than Fatigue. Enter the value of J/mg and hit Convert to get value in Fatigue. Check our J/mg to Fatigue converter. Need a reverse calculation from Fatigue to J/mg? You can check our Fatigue to J/mg Converter.
How many Gy is 1 J/mg?
1 J/mg is equal to 1000000 Gy. 1 J/mg is 1000000 times Bigger than 1 Gy.
How many rd is 1 J/mg?
1 J/mg is equal to 100000000 rd. 1 J/mg is 100000000 times Bigger than 1 rd.
How many mrd is 1 J/mg?
1 J/mg is equal to 100000000000 mrd. 1 J/mg is 100000000000 times Bigger than 1 mrd.
How many J/kg is 1 J/mg?
1 J/mg is equal to 1000000 J/kg. 1 J/mg is 1000000 times Bigger than 1 J/kg.
## J/mg to Fatigue Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like absorbed dose finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like J/mg to Fatigue through multiplicative conversion factors. When you are converting absorbed dose, you need a Joule/Milligram to Fatigue converter that is elaborate and still easy to use. Converting J/mg to Fatigue is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Joule/Milligram to Fatigue, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in J/mg to Fatigue conversion along with a table representing the entire conversion. | 881 | 3,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.780125 |
https://www.transtutors.com/homework-help/statistics/change-point.aspx | 1,553,184,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00536.warc.gz | 936,303,865 | 32,552 | ## What is Change Point?
This article is used to give a basic information regarding the change points that occur in excel and in other files. The detection methods are proposed and they are analyzed with a real time example. The features and application of the change point is also discussed in the later.
## Definition of Change Point
The Change Point Detection is known as a Stochastic Process in the statistical study, that is used to identify the timely changes when the probability distribution of the system changes or when the time series of the system changes. It deals with the problem that concern in detecting whether the time change has occurred or not and if occurred, it determines the time limit during which the change has occurred. The change detection is sometimes known as the anomaly detection as it deals with the different detection techniques like step and edge value detection that are concerned with the change occurred in the values like mean, median, variance and covariance.
Online change analysis one of the most widely used technique in present which is carried out using the sequential analysis and hence referred to as streaming algorithm. In Online Detection, it is used to measure the change that is made using the trade-off between the metrics like detection delay, False alarm rate and misdetection rate. There are various types of change detection techniques like:
### Minimax Change Detection
The objective of this Minimax detection is to reduce the delay that is expected to take place in a system in some worst-case that can occur during the time distribution. This Detection Technique is carried out by CUSUM procedure which is one of the popular techniques.
### Offline Change Detection
This detection method was found out by basseville that observes the change in mean detection of the system. This estimation is related to the EM algorithm method and the related methods like two-phase regression, clustering and in the maximum likelihood estimation of the system variables.
### Linguistic Change Detection
This type of detection method deals with the ability to detect the word-level changes that occur in the multiple presentations of the same sentence.
## Change Point Detection Packages
Many R community packages were developed for the change pint detection. They already exist in the CRAN and focuses effectively on the change point detection. Let us discuss some of the popular change point software:
CPM:
The CPM method is used for the change detection in the parametric and non-parametric sequences of the given system. It is more helpful in the detection of multiple point change that occurs in the time series from the unknown distribution. This method can be applied for the data streams where only one observation can be made. In the special case of CPM method, the detection points should be displayed. For each detection process, we store the values of the corresponding number of logins.
Two types of parameters are used in this type of detection process where one is the testing of statistic value and the second parameter is the number of observations that are made at the beginning of the process and until the change occurred in the points. The test statistic value offers multiple versions to detect the changes depending on the type of distribution. They have the ability to quantify the delay but unfortunately, this CPM is no longer used in the CRAN process.
BCP:
This package is used for performing the Bayesian analysis of change points in problems. This is an R package. That was designed using the Markov chain Carlo to find the multiple changes in point that occurs within the sequence. This package is limited to the multivariate case implementation.
The BCP approach uses three types of parameters. One of the parameter is the probability threshold of the estimated probabilities.
ECP:
This package is specially designed for the non-parametric multiple point change analysis of the multivariate data. The ECP package is similar to the hierarchical clustering process used in the EM Algorithm and offers the top-down and bottom-up approach for the change point detection process. Usually, the top-down approach is recommended for the Tableau. Where minimum number of observations is required.
The process involved in the change point detection is:
• First, when we perform the analysis, the analyst can incorporate the background knowledge about the data and the possible effects from the external events. This kind of observation is not easily gathered for the algorithm.
• Second, this is the process that takes place before the final step. This process mainly concentrates on the less complex decision making technique.
• The third and the final process involves the submission of the visual feedback that demonstrates how these algorithms perform and provide the judgment by providing a second opinion.
Fig. 1- The Dashboard Representation
The above dashboard represents a very simple structure that shows the empirical observations that are made using the packages that are discussed above and there are various options like signature and the parameters are held on the right side of the dashboard that allow to interact with the algorithm and in understanding the data and in the filtering process. The following workflow can be followed in the working of dashboard:
• Triggering the change point detection
• Extraction of exact location of the change points by applying the filtering process
• Calculating the segment value of the mean value identified in the change points.
## Change Point Analysis
Change-point analysis is one of the tools that are used for determining whether the change has taken place or not. It is also capable of finding the changes that is missed while estimating the control chart. It helps in the increase of the confidence level and confidence intervals. Change-point analysis has the capability to replace the control charting as it is an effective way in determining the historical data and in dealing the large amount data, it provides the control over the overall error rate and is more flexible and a simpler method to be implemented.
Multiple changes can be found by the change-point analysis and detailed information is provided that can be used for the future purpose. This analysis can be performed for all types of time-ordered data such as attributed data, abnormal distributions and data’s with outliers. The change point control is similar to the traditional control chart method and the major difference between the change-point analysis and control charting is that the control charts are to be updated for each and every collection of point while the point analysis is performed for the data that is collected for the first time.
Control charts are better at detecting the abnormal points more quickly while point analysis is used to detect the changes that are missed in the control charts. This method is used applicable for the system with thousands of data points along with the numerous points. Let us consider the US trade deficits during 1987-1988 as the example for the change-point analysis.
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1987 10.7 13.0 11.4 11.5 12.5 14.1 14.8 14.1 12.6 16.0 11.7 10.6 1988 10.0 11.4 7.9 9.5 8.0 11.8 10.5 11.2 9.2 10.1 10.4 10.5
Fig. 2- Plot for the US deficit data
The trade deficit plot shows to be in lower rate in 1988 than in 1987. There are various approaches followed for performing the analysis. Both control chart and change-point model was applied for this process while the control chart detected the change barely. But, the change-point analysis provided additional information other than the control chart. The procedure suggested by Taylor is used along with combination of cumulative sum chart and with bootstrapping method for the detection of the changes. Practice is required for the implementation of the CUSUM procedure.
Fig. 3- Change-point analysis
For the change point analysis excel implementation purpose, the excel add-in software is used and the change-point analyzer is used for this purpose.
Some of the features of change-point analysis are as follows:
• This analysis is more powerful in detecting the small as well as sustained changes.
• It reduces the possibility of false detections by implementing the control of change-wise error rate. While, control charts use point-wise error rate for large data that produces more false detections.
• It provides a robust approach towards the outliers.
• This type of analysis is more flexible. The analysis is based on the single assumption only.
• The method is simpler and easy to use and to be interpreted. It has the ability to automate the difficult process.
## Applications
The change detection tests is helpful in the manufacturing of equipments that helps in the quality control and in the detection of intrusion, filtering of spam, tracking of websites and in the diagnosis of medical aids. The change-point detection is more helpful in the field of simulation process and in designing the filters for the digital signal processing.
Change Point
## Related Topics
All Statistics Topics
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The Stormwater Experts
# Hydraulic Calculation
For the assessment of floods in rivers and channels, a hydrodynamic-numerical model (hn-model) is most of the times required. A "hydrodynamic" model is in fact able to consider the relevant forces like gravity, frictional resistance and inertia, which are responsible for the flow processes. The solution of the complex mathematical equations, describing the flow process of surface water, is usually possible only through a "numerical" discretization of the continuum (space and time).
Ein- und Ausgangsdaten von HN-Modellen
The movement of flood waters through the landscape can be approximated using many different methods. Depending on the objectives and dimensional resolution of the required modelling outcomes - especially the water level and flow velocity - the basic equations can be simplified in various ways, without significantly limiting their validity. This leads to models with different advantages and disadvantages:
• 3-dimensional models: 3D-hn-models find their use in small-scale calculation fields, as for flow around structures or at intake structures of hydroelectric power stations, where the mapping of possible vortices and the knowledge of all three velocity components is important. The enormous demand for computing capacity is the main limiting factor. For the flow calculation, the full Navier-Stokes equations in x-, y- and z-direction are used. Widespread 3D hn-models are i.a. Flow 3D, MIKE 3D, ANSYS CFX, FLUENT and Delft 3D.
• 2-dimensional models: For 2D-hn-models the information in one direction (usually the vertical) is removed. The on flow depth averaged equations (the so-called "shallow water equations") allow accurate detection of flow characteristics (flow velocity v, water level W, shear stress τ, ...). For rivers with flood plains and formation of velocity gradients perpendicular to the flow direction the 2D-hn-models are a good compromise between creating effort and accurate mapping of the flow processes. These models are commonly used successfully for the calculation of flood waves. The field of application ranges from the investigation of a local object protection up to calculating of flood areas of several square kilometers. The software most commonly used are: MIKE 21, Hydro_AS-2D, SOBEK, FloodArea and Flow 2D.
• 1-dimensional models: The solution of 1D-hn-models is usually faster in computation compared to the 2D and 3D models. For 1D models the distribution of the flow parameters (W, v, ...) perpendicular to flow direction is averaged to the "Saint-Venant equations". The parameters are thus only calculated in the main flow direction. It is possible to divide the calculation for main bed and flood plains / riparian zones (waterbodies with structured sections). The typical application is the determination of the water surface curve in rivers with mainly uniform cross-sections. The investigated flow paths can range from a few hundred meters to several hundred kilometers. There are a wide variety of 1D hn models, including: HEC-RAS, Jabron, WSPWIN, MIKE 11 and FLUSS.
Flooding area made of intersection of 1D-hn-model HEC-RAS with the DEM
More and more hybrid models are used, where 1D river sections are coupled with 2D regions, resulting in a flood assessment within the desired detail level and distribution.
The choice of a suitable model is also determined by; the area size (limited section of the river, the total catchment area) and by the availability of data (Digital Elevation Model DEM, terrestrial surveying).
A clear advantage of a 2D-hn-model is that for each element or cell, in which the region is divided, the velocity vector can be output. The 1D model distinguishes between velocity in the river and on the foreland. The direction of flow is not calculated but is specified within the geometry.
## Stationary and transient models
Although all natural processes are time-dependent (transient), a flood wave may be simulated with sufficient accuracy by a sequence of time-independent (stationary) states.
Thus, in regard to the time dependency (e.g. the discharge Q from the time t), a distinction is made between:
• Stationary models, which describe a state where the time changes are negligible [Q = const]
• quasi-stationary models that simulate the changes over time as a sequence of stationary states [Q = Q (t1), Q (t2) ... Q (tn)] and
• transient models, which preserve the time-dependence of the processes [Q = Q (t)].
Only transient models can take into account the inherent flattening process a flood wave (retention), as well as the impact of retention areas. Although the wave attenuation can be taken into account in the quasi-stationary models by an upstream hydrological model, only the unsteady models grant a proper volume balance (compliance of the continuity condition). However, stationary models don´t have exact volumes.
The transient approach for the calculation of flood is always preferable if the inflows are available or estimated as hydrographs Q = Q (t).
It should take into account that the amount of computation increases disproportionately with the dimensionality of the transient processes and the extension of the model.
## Model creation
To construct a hn-model for simulating a flood, several basic data items are required:
• Discharge from the headwaters, inflow from the sewer system as well as side flows: as hydrograph Q (t) for a transient calculation or as constant values for the quasi-stationary as well as stationary approach (upper or internal boundary condition).
• Velocity distribution at the inflow boundary for 2D and 3D models.
• Water level (1D) or water level distribution (2D and 3D) or runoff-water level-relationship (calibration curve) at the outlet from the model (lower boundary condition).
• Topography of the river bed and the flood plains as a Digital Elevation Model (DEM) or terrestrial surveying, LiDAR or LaserScan data.
• Matching friction coefficients for the different texture of sole, embankments, structures, vegetation and flood plains. For this a water body-inspection is essential by competent professionals.
• Initial conditions: In addition, the initial value of the relevant parameters (water level and velocity vector) throughout the model area is pretending for the non-stationary models.
In both 1D and 2D-hn-models the existing transverse structures (bridges, culverts, weirs, sills, ...) and other structures which are used to control runoff (side weirs, control elements and for 1D also polders) must be entered manually.
Flooding as a result of HQ100 at the Wierau (2D-hn-model FloodArea)
## Calibration and Validation
Despite their conceptual legality flow equations also contain empirical approaches (frictional losses, vegetation effect, turbulence models, ...). For hn-models, the roughness is the parameter afflicted with the greatest uncertainties. Therefore each model necessarily requires a calibration. The empirical parameters are varied in an iterative process until the calculated results (mostly the water level) have a sufficiently good match with the directly measured respectively indirectly derived values. After a flood the line of the debris deposition at the highest water level is usually clearly visible. Often the wetness on facades or a certain abrasion on bridge piers is visible for a long time. For historical events flood marks may be present. Level data can of course be used during calibration, if available.
With calibration, the model is adjusted on an event that the results correspond to the observation. Checking if the selected parameter set retains its validity for other events is called validation. It verifies the model performance, which means the transferability of the model to other events as the one used during the calibration (e.g. flood with different return period). If necessary, different sets of parameters have to be chosen.
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#### Directory
Flood Hydraulic Calculation | 1,672 | 7,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.891809 |
https://quantumcomputing.stackexchange.com/questions/21608/how-to-choose-beta-in-gaussian-derivative-component-of-drag-pulse | 1,638,350,435,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00126.warc.gz | 531,557,129 | 32,032 | # How to choose $\beta$ in Gaussian derivative component of DRAG pulse?
From definition of the DRAG pulse it is: $$f(x)=Gaussian+1j*\beta*(-(x-duration/2)/\sigma^2)Gaussian,$$
where $$Gaussian(x, amp, \sigma)=amp*e^{-(1/2)*(x-duration/2)^2/\sigma^2}$$.
If I try it in Python (for $$\beta=$$1e-6):
import numpy as np
import matplotlib.pyplot as plt
t = np.arange(0,8e-6,10e-9)
amp = 1
sigma=1e-6
tc = 8e-6
gaussian = amp*np.exp(-0.5*(t-tc/2)**2/sigma**2)
beta = 1e-6
drag_component = beta*(-(t-tc/2)/sigma**2)*gaussian
I get:
but if I try to set the amplitude of derivative part equaled to the Gaussian ($$\beta$$=1):
the result is strange. How to choose $$\beta$$, what it's meaning and on what it depends? | 251 | 718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-49 | latest | en | 0.707106 |
http://www.mathemafrica.org/?p=13566 | 1,709,628,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00550.warc.gz | 63,701,396 | 37,773 | In the most recent tutorial there is a question about converting a Riemann sum to a definite integral, and it seems to be tripping up quite a few students. I wanted to run through one of the calculations in detail so you can see how to answer such a question.
Let’s look at the example:
$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(9\left(4+(i-1)\frac{6}{n}\right)^2-8\left(4+(i-1)\frac{6}{n}\right)+7\right)\frac{13}{n}$
There are many ways to tackle such a question but let’s take one particular path. Let’s start by the fact that when the limit is defined, the limit of a sum is the sum of the limits. We can split up our expression into 3, which looks like:
$\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}-\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(8\left(4+(i-1)\frac{6}{n}\right)\right)\frac{13}{n}+\lim_{n\rightarrow\infty}\sum_{i=1}^n7\frac{13}{n}$
Let’s tackle each of these separately. Let’s look at the first term:
$\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}$
Well, we can take the factor of 13 outside the front of the whole thing to start with, along with the factor of 9, and this will give
$13\times 9\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\frac{1}{n}$
We see here that we have a sum of terms, and a factor which looks like $\frac{1}{n}$ in each term. We notice however that inside the bracket there is a $\frac{6}{n}$, so we try to make the term on the outside look the same as this. Multiply and divide by 6 to give:
$13\times 9\times\frac{1}{6}\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\frac{6}{n}$
Now, the $\frac{6}{n}$ is a constant (as we sum over the i’s), and will correspond to the width of the rectangles in the area which is ultimately what a Riemann sum tells us about. Remember in a Riemann sum, we want a function value at a particular point, times a width. We will call the width of each rectangle $\Delta x=\frac{6}{n}$. So we can write:
$\frac{117}{6}\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\Delta x$
Now, in a Riemann sum, we have a function evaluated at various points, which we generally call $x_i$. As we increase i by 1, we move to the next rectangle in our approximation. Here we see a factor of $(i-1)\frac{6}{n}$ (pulling out the factor of 6). Remembering that $\frac{1}{n}$ is the width of our rectangles, when i=1, this is going to give the point 0, and when i=n, this is going to give the point $(n-1)\frac{6}{n}$. That’s not very interesting, but if we were to go to the next point (i=n+1), we would have $n\frac{6}{n}$ which would be the point 6. So, running through the values from i=1 to i=n, the quantity $(i-1)\frac{6}{n}$ goes from the value 0 to the value just below 6, where as we take n to infinity, we get closer and closer to 6 for that last point. We can treat $(i-1)\frac{6}{n}$ as our $x_{i-1}$, so long as we know that we are going from $x_0=0$ to $x_{n-1}=\frac{6(n-1)}{n}$. So, we can write:
$\frac{117}{6}\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+x_{i-1}\right)^2\Delta x$
with $x_0=0$ to $x_{n-1}=\frac{6(n-1)}{n}$
This can be converted directly into the definite integral:
$\frac{117}{6}\int_0^6\left(4+x\right)^2dx$
Exactly the same procedure works with the other expressions in the sum, and we end up with:
$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(9\left(4+(i-1)\frac{6}{n}\right)^2-8\left(4+(i-1)\frac{6}{n}\right)+7\right)\frac{13}{n}$
$\frac{117}{6}\int_0^6\left(4+x\right)^2dx-\frac{104}{6}\int_0^6(4+x)dx+91\int_0^1 dx$
where the 117 comes from 13 x 9, the 104 comes from 13 x 8 and the 91 comes from 13 x 7.
The last integral we can write as:
$91\int_0^1 dx=\frac{91}{6}\int_0^6 dx$
(Make sure that you can see that this is true).
And so we can actually write the whole thing as a single integral:
$\frac{1}{6}\int_0^6117\left(4+x\right)^2-104(4+x)+91 dx$
Which can be calculated from the known antiderivatives and the fundamental theorem of calculus part 2.
How clear is this post? | 1,408 | 4,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 28, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-10 | latest | en | 0.81156 |
https://cboard.cprogramming.com/c-programming/123453-matrix-multiplication.html | 1,513,464,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589512.63/warc/CC-MAIN-20171216220904-20171217002904-00201.warc.gz | 536,087,414 | 14,079 | 1. ## matrix multiplication
hello,
Code:
```#include <conio.h>
#include <time.h>
#include <stdlib.h>
#include <ctype.h>
int main()
{
//freopen("conventionInput.txt","r",stdin);
long long m1[5][100],i,j,k,m2[100][5],mult[100][100],r1,c1,r2,c2;
char Manual = 'N';
printf("Enter Matrix Input Manually? (Y/N)\n");
Manual = getchar();
Manual = toupper(Manual);
printf("Enter variable Y MIN 32 MAX 100\n");
scanf("%lld",&c1);
r2 = c1;
r1 = 5;
c2 = 5;
/* initialize random seed: */
srand ( time(NULL) );
/* generate secret number: */
if(r2==c1)
{
printf("Enter rows and columns of First matrix \n");
printf("Row wise\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c1;j++){
if(Manual == 'Y')
scanf("%lld",&m1[i][j]);
else
m1[i][j] = rand() % 10 + 1;
}
}
printf("You have entered the first matrix as follows:\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c1;j++)
printf("%lld\t",m1[i][j]);
printf("\n");
}
printf("Enter rows and columns of Second matrix \n");
printf("Again row wise\n");
for(i=0;i<r2;i++)
{
for(j=0;j<c2;j++){
if(Manual == 'Y')
scanf("%lld",&m2[i][j]);
else
m2[i][j] = rand() % 10 + 1;
}
}
printf("You have entered the second matrix as follows:\n");
for(i=0;i<r2;i++)
{
for(j=0;j<c2;j++)
printf("%lld\t",m2[i][j]);
printf("\n");
}
printf("Now we multiply both the above matrix. Clock Time will start here \n");
printf("The result of the multiplication is as follows:\n");
//a11xA11+a12xA21+a13xA31 a11xA12+a12xA22+a13xA32 a11xA13+a12xA23+a13xA33
clock_t start = clock();
for(i=0;i<r1;i++)
{
for(j=0;j<c2;j++)
{
mult[i][j]=0;
for(k=0;k<r1;k++)
{
mult[i][j]+=m1[i][k]*m2[k][j];
mult[0][0]=m1[0][0]*m2[0][0]+m1[0][1]*m2[1][0]+m1[0][2]*m2[2][0];
}
printf("%lld\t",mult[i][j]);
}
printf("\n");
}
printf("Time elapsed: %f ms \n", ((double)clock() - start) / CLOCKS_PER_SEC);
getch();
}
else
{
printf("Matrix multiplication cannot be done");
}
}```
My problem now is this matrix multiplication coding not give the actual result after..can u check which part that cause this problem..
tq.
1) Show the arrays, and verify the values are correct before any computation
2) Use arrays with assigned values for troubleshooting - you don't want to have to enter the values, every time you run the program.
3) Either step through the program and watch some array values change, or use printf to show you how things are going, during the computations.
4) Start with small arrays first - one it's correct, then use the larger sizes.
5) Practice your indentation style. It's crappy, as is. That eliminates a programmer from using their eye /brain training that they will have picked up, from long programming practice.
3. ## contd..
hello,
i have try with the small value but still not get actual value..the coding problem is for example i multiply the first row with first column i get too small value for the large size matrix..with the simple value input, example: 2^23=45..so the value false.
1. why the result value too small?, while the multiplication process is ok.
tq,
4. Well, here's the problem for you:
1) I'm not going to indent your program code for you. If you want to see your code as it should be seen, and help others to see it easily, you'll do that.
2) I'm not going to enter 500 numbers for your m1 array. I asked you to make a small assigned array of values, for testing.
You haven't done that.
So your program will not get much help. You can't expect people to do the work on your program, just because you won't do it.
Maybe someone will stumble upon your code error - but you should learn to help yourself (and others), troubleshoot a program.
Right now, you're not really helping much.
5. Here is my two cents.
I would start over from scratch. Your code is incredibly difficult to read without any indentation and it will probably be faster to just rewrite your code than get the spacing correct.
There is one big problem with your code - you don't check the dimensionality of the matrices (at least not that I could see). You should always do that before multiplying two matrices because if the dimensions don't agree (i.e. the column number of the left matrix equals the row number of the right matrix) multiplication doesn't make sense.
Your program would be much more flexible if you read your matrix in from a file. Asking for a user to input a matrix is a big pain for anything larger than a 3x3 and my guess is this is where your program is messing up. If you don't know how to read from a file then it will be a great learning experience.
As for the algorithm - there are tons of different ways to loop through matrix multiplication. It looks like yours is fine (although I only glanced at it since it was hard to read) so your error is probably in a detail somewhere else. Assuming you don't have a special type of matrix (i.e. sparse, symmetric, etc..) your matrix multiplication should take n^3 operations. If it doesn't, something is wrong.
For future reference, code that looks like this is MUCH more readable.
Code:
``` for (int i = 0; i < m1; i++) { /* Matrix multiplication C=AB */
for (int j = 0; j < n2; j++) {
for (int k = 0; k < n2; k++) {
C[i][j] += A[i][k]*B[k][j];
}
}
}```
Good luck
6. I dunno, use the modern miracle which is the "indent" program.
Fantastic for cleaning up code.
Code:
```#include <conio.h>
#include <time.h>
#include <stdlib.h>
#include <ctype.h>
int main()
{
//freopen("conventionInput.txt","r",stdin);
long long m1[5][100],i,j,k,m2[100][5],mult[100][100],r1,c1,r2,c2;
char Manual = 'N';
printf("Enter Matrix Input Manually? (Y/N)\n");
Manual = getchar();
Manual = toupper(Manual);
printf("Enter variable Y MIN 32 MAX 100\n");
scanf("%lld",&c1);
r2 = c1;
r1 = 5;
c2 = 5;
/* initialize random seed: */
srand ( time(NULL) );
/* generate secret number: */
if(r2==c1)
{
printf("Enter rows and columns of First matrix \n");
printf("Row wise\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c1;j++){
if(Manual == 'Y')
scanf("%lld",&m1[i][j]);
else
m1[i][j] = rand() % 10 + 1;
}
}
printf("You have entered the first matrix as follows:\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c1;j++)
printf("%lld\t",m1[i][j]);
printf("\n");
}
printf("Enter rows and columns of Second matrix \n");
printf("Again row wise\n");
for(i=0;i<r2;i++)
{
for(j=0;j<c2;j++){
if(Manual == 'Y')
scanf("%lld",&m2[i][j]);
else
m2[i][j] = rand() % 10 + 1;
}
}
printf("You have entered the second matrix as follows:\n");
for(i=0;i<r2;i++)
{
for(j=0;j<c2;j++)
printf("%lld\t",m2[i][j]);
printf("\n");
}
printf("Now we multiply both the above matrix. Clock Time will start here \n");
printf("The result of the multiplication is as follows:\n");
//a11xA11+a12xA21+a13xA31 a11xA12+a12xA22+a13xA32 a11xA13+a12xA23+a13xA33
clock_t start = clock();
for(i=0;i<r1;i++)
{
for(j=0;j<c2;j++)
{
mult[i][j]=0;
for(k=0;k<r1;k++)
{
mult[i][j]+=m1[i][k]*m2[k][j];
mult[0][0]=m1[0][0]*m2[0][0]+m1[0][1]*m2[1][0]+m1[0][2]*m2[2][0];
}
printf("%lld\t",mult[i][j]);
}
printf("\n");
}
printf("Time elapsed: %f ms \n", ((double)clock() - start) / CLOCKS_PER_SEC);
getch();
}
else
{
printf("Matrix multiplication cannot be done");
}
}``` | 2,107 | 7,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-51 | latest | en | 0.379304 |
http://jp.mathworks.com/help/finance/tr2bonds.html?nocookie=true | 1,417,131,937,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931009292.37/warc/CC-MAIN-20141125155649-00042-ip-10-235-23-156.ec2.internal.warc.gz | 170,662,266 | 9,536 |
Accelerating the pace of engineering and science
# tr2bonds
Term-structure parameters given Treasury bond parameters
## Syntax
```[Bonds, Prices, Yields] = tr2bonds(TreasuryMatrix, Settle)
```
## Arguments
TreasuryMatrix Treasury bond parameters. An n-by-5 matrix, where each row describes a Treasury bond. Columns are [CouponRate Maturity Bid Asked AskYield] where: CouponRate Coupon rate, as a decimal fraction. Maturity Maturity date, as a serial date number. Use datenum to convert date strings to serial date numbers. Bid Bid price based on \$100 face value. Asked Asked price based on \$100 face value. AskYield Asked yield to maturity, as a decimal fraction. Settle (Optional) Date string or serial date number of the settlement date for the analysis.
## Description
[Bonds, Prices, Yields] = tr2bonds(TreasuryMatrix, Settle) returns term-structure parameters (bond information, prices, and yields) sorted by ascending maturity date, given Treasury bond parameters. The formats of the output matrix and vectors meet requirements for input to the zbtprice and zbtyield zero-curve bootstrapping functions.
Bonds Coupon bond information. An n-by-6 matrix where each row describes a bond. Columns are [Maturity CouponRate Face Period Basis EndMonthRule] where: Maturity Maturity date of the bond, as a serial date number. Use datestr to convert serial date numbers to date strings. CouponRate Coupon rate of the bond, as a decimal fraction. Face Redemption or face value of the bond, always 100. Period Coupons per year of the bond, always 2. Basis Day-count basis of the bond, possible values include:0 = actual/actual (default)1 = 30/360 (SIA)2 = actual/3603 = actual/365For more information, see basis. EndMonthRule End-of-month flag, always 1, meaning that a bond's coupon payment date is always the last day of the month. Prices Prices. Column vector containing the price of each bond in bonds, respectively. The number of rows (n) matches the number of rows in bonds. Yields Yields. Column vector containing the yield to maturity of each bond in bonds, respectively. The number of rows (n) matches the number of rows in bonds. If Settle is input, Yields is computed as a semiannual yield to maturity. If Settle is not input, the quoted input yields will be used.
## Examples
expand all
### Return Term-Structure Parameters Given Treasury Bond Parameters
This example shows how to return term-structure parameters (bond information, prices, and yields) sorted by ascending maturity date, given Treasury bond market parameters for December 22, 1997.
```Matrix =[0.0650 datenum('15-apr-1999') 101.03125 101.09375 0.0564
0.05125 datenum('17-dec-1998') 99.4375 99.5 0.0563
0.0625 datenum('30-jul-1998') 100.3125 100.375 0.0560
0.06125 datenum('26-mar-1998') 100.09375 100.15625 0.0546];
[Bonds, Prices, Yields] = tr2bonds(Matrix)
```
```Bonds =
1.0e+05 *
7.2984 0.0000 0.0010 0.0000 0 0.0000
7.2997 0.0000 0.0010 0.0000 0 0.0000
7.3011 0.0000 0.0010 0.0000 0 0.0000
7.3022 0.0000 0.0010 0.0000 0 0.0000
Prices =
100.1562
100.3750
99.5000
101.0938
Yields =
0.0546
0.0560
0.0563
0.0564
``` | 926 | 3,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2014-49 | latest | en | 0.669516 |
https://metanumbers.com/1028601 | 1,643,270,654,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00491.warc.gz | 450,156,502 | 7,485 | # 1028601 (number)
1,028,601 (one million twenty-eight thousand six hundred one) is an odd seven-digits composite number following 1028600 and preceding 1028602. In scientific notation, it is written as 1.028601 × 106. The sum of its digits is 18. It has a total of 5 prime factors and 24 positive divisors. There are 566,496 positive integers (up to 1028601) that are relatively prime to 1028601.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 18
• Digital Root 9
## Name
Short name 1 million 28 thousand 601 one million twenty-eight thousand six hundred one
## Notation
Scientific notation 1.028601 × 106 1.028601 × 106
## Prime Factorization of 1028601
Prime Factorization 32 × 7 × 29 × 563
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 342867 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,028,601 is 32 × 7 × 29 × 563. Since it has a total of 5 prime factors, 1,028,601 is a composite number.
## Divisors of 1028601
24 divisors
Even divisors 0 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 1.75968e+06 Sum of all the positive divisors of n s(n) 731079 Sum of the proper positive divisors of n A(n) 73320 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1014.2 Returns the nth root of the product of n divisors H(n) 14.0289 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,028,601 can be divided by 24 positive divisors (out of which 0 are even, and 24 are odd). The sum of these divisors (counting 1,028,601) is 1,759,680, the average is 73,320.
## Other Arithmetic Functions (n = 1028601)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 566496 Total number of positive integers not greater than n that are coprime to n λ(n) 23604 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80408 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 566,496 positive integers (less than 1,028,601) that are coprime with 1,028,601. And there are approximately 80,408 prime numbers less than or equal to 1,028,601.
## Divisibility of 1028601
m n mod m 2 3 4 5 6 7 8 9 1 0 1 1 3 0 1 0
The number 1,028,601 is divisible by 3, 7 and 9.
• Arithmetic
• Deficient
• Polite
## Base conversion (1028601)
Base System Value
2 Binary 11111011000111111001
3 Ternary 1221020222100
4 Quaternary 3323013321
5 Quinary 230403401
6 Senary 34014013
8 Octal 3730771
10 Decimal 1028601
12 Duodecimal 417309
20 Vigesimal 68ba1
36 Base36 m1o9
## Basic calculations (n = 1028601)
### Multiplication
n×y
n×2 2057202 3085803 4114404 5143005
### Division
n÷y
n÷2 514300 342867 257150 205720
### Exponentiation
ny
n2 1058020017201 1088280447712965801 1119406356798004335874401 1151422498008784057884744743001
### Nth Root
y√n
2√n 1014.2 100.944 31.8465 15.9386
## 1028601 as geometric shapes
### Circle
Diameter 2.0572e+06 6.46289e+06 3.32387e+12
### Sphere
Volume 4.55858e+18 1.32955e+13 6.46289e+06
### Square
Length = n
Perimeter 4.1144e+06 1.05802e+12 1.45466e+06
### Cube
Length = n
Surface area 6.34812e+12 1.08828e+18 1.78159e+06
### Equilateral Triangle
Length = n
Perimeter 3.0858e+06 4.58136e+11 890795
### Triangular Pyramid
Length = n
Surface area 1.83254e+12 1.28255e+17 839849
## Cryptographic Hash Functions
md5 33fb48bc938a79aaf008e04791e6a982 f1fcaa3c3767b6483e58a22a45313587d3ca17cb ee01830de91eabc5258c006dc314975e1b5e51beb8a0df5c9416106f1db292eb 75417334ee059ced4b8bc264e0452aa419f8587cebd30d6e5c7bbbbf3b6dbc1b54940d63ddc3d439f3af7f63a927dc65dee6d5324317c21baf679770f93c7192 235e8ebe8b3bc96e0be21c4c5b620fa86de45d6d | 1,513 | 4,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.802817 |
https://www.investopedia.com/ask/answers/063015/how-minimum-transfer-price-calculated.asp | 1,679,572,038,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00392.warc.gz | 944,390,130 | 39,016 | # How Is Minimum Transfer Price Calculated?
A company that transfers goods between multiple divisions needs to establish a transfer price so that each division can track its own efficiency. Companies will use various methods to determine the minimum transfer price, factoring in different costs related to production and what the goods would normally sell for in the retail marketplace.
Corporations with operations in various countries may attempt to shift the transfer price to divisions located in countries with lower tax rates, thereby reducing their corporate tax obligation. While this practice can result in greater profits for multinational corporations, it can also lead to greater scrutiny and regulation from tax authorities like the Internal Revenue Service (IRS).
### Key Takeaways
• A transfer price refers to the price that one division of a company charges another division of the same company for a good or service.
• A company may calculate the minimum acceptable transfer price as equal to the variable costs or equal to the variable costs plus a calculated opportunity cost.
• Most companies will set the minimum transfer price at greater than or equal to the marginal cost of the selling division.
• Some multinational corporations will attempt to manipulate the transfer costs between divisions in order to reduce the amount they pay in taxes, a practice that happens more often when one division is located in a country with significantly lower corporate taxes.
## How to Find the Minimum Transfer Price
There are different ways to find the minimum acceptable transfer price. Some companies simply set the minimum as equal to variable costs. Others add variable costs with a calculated opportunity cost. The general economic transfer price rule is that the minimum must be greater than or equal to the marginal cost of the selling division.
In economics and business management, a marginal cost is equal to the total new expense incurred from the creation of one additional unit. For example, suppose a hammer manufacturing company has two divisions: a handle division and a hammer head division. The hammer head division only begins work after receiving handles from the handle division. This means the handle division is the selling division and the hammer head division is the buyer.
If it costs the handle division \$7 to fashion its next handle (its marginal cost of production) and ship it off, it doesn't make sense for the transfer price to be \$5 (or any other amount less than \$7). Otherwise, the handle division would lose money at the expense of money gained by the hammer head division.
## Factoring in Opportunity Costs
Suppose that the hammer company also sells replacement handles for its products. In this scenario, it sells some handles through retail sales rather than sending them to the hammer head division. Suppose again that the handle division can realize a \$3 profit margin on its sold handles.
Now the cost of selling a handle isn't just the \$7 marginal cost of production, but also the \$3 in lost profit (opportunity cost) from not selling the handle directly to consumers. This means the new minimum transfer price must be \$10 (\$3 + \$7).
## Minimum Transfer Price and Tax Regulations
For accounting purposes, large corporations will evaluate their divisions separately for profit and loss. When these different divisions conduct business with one another, the minimum transfer price for a particular good will usually be close to the prevailing market rate for that good. That means that the division selling a good to another division will charge an amount equal to what they could achieve by selling to retail customers.
However, in some instances, companies will attempt to increase or decrease the transfer costs between divisions in order to lower the amount they pay in taxes. This deliberate manipulation of costs is more likely to occur when the divisions are located in different countries where one country is a tax haven and has a much lower tax rate than the other.
Obviously, the tax authorities in countries with higher tax rates frown upon this practice as it means lost revenue for them. Thus, these countries have strict regulations to prevent companies from using transfer pricing as a tax avoidance strategy.
Regulators look at the company's financial statements to ensure their transfer pricing is in line with current market pricing. In general, these regulations attempt to ensure companies abide by arm's length practices, which prevents collusion between divisions within the company to misstate transfer prices. | 834 | 4,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-14 | latest | en | 0.947064 |
https://www.exactlywhatistime.com/days-from-date/september-12/18-days | 1,702,230,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00071.warc.gz | 837,181,945 | 6,042 | What date is 18 days added from Thursday September 12, 2024?
Monday September 30, 2024
Adding 18 days from Thursday September 12, 2024 is Monday September 30, 2024 which is day number 274 of 2023. This page is designed to help you the steps to count 18, but understand how to convert and add time correctly.
• Specific Date: Thursday September 12, 2024
• Days from Thursday September 12, 2024: Monday September 30, 2024
• Day of the year: 274
• Day of the week: Monday
• Month: September
• Year: 2023
Calculating 18 days from Thursday September 12, 2024 by hand
Attempting to add 18 days from Thursday September 12, 2024 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
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Monday September 30, 2024 Stats
• Day of the week: Monday
• Month: September
• Day of the year: 274
Counting 18 days forward from Thursday September 12, 2024
Counting forward from today, Monday September 30, 2024 is 18 from now using our current calendar. 18 days is equivalent to:
18 days is also 432 hours. Monday September 30, 2024 is 75% of the year completed.
Within 18 days there are 432 hours, 25920 minutes, or 1555200 seconds
Monday Monday September 30, 2024 is the 274 day of the year. At that time, we will be 75% through 2024.
In 18 days, the Average Person Spent...
• 3866.4 hours Sleeping
• 514.08 hours Eating and drinking
• 842.4 hours Household activities
• 250.56 hours Housework
• 276.48 hours Food preparation and cleanup
• 86.4 hours Lawn and garden care
• 1512.0 hours Working and work-related activities
• 1391.04 hours Working
• 2276.64 hours Leisure and sports
• 1235.52 hours Watching television
Famous Sporting and Music Events on September 30
• 1950 First Prime Minister of Singapore Lee Kuan Yew (27) weds lawyer Kwa Geok Choo (29) in Singapore
• 1934 Babe Ruth's final game as a Yankee, goes 0 for 3 | 650 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.930305 |
https://getrevising.co.uk/revision-cards/physics_3_4 | 1,529,860,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00636.warc.gz | 599,847,500 | 17,076 | # Physics 3
Revision cards for physics 3 exam
HideShow resource information
• Created by: Katie
• Created on: 05-10-11 19:26
## Moments
MOMENT = FORCE X PERPENDICULAR DISTANCE FROM PIVOT
• A turning effect
• Cause objects to turn or rotate around a fixed point- the pivot or fulcrum
• Are caused by forces but are not forces themselves
• Act in two ways- clockwise or anticlockwise
1 of 20
## Centre of mass
• The point at which the mass of an object is thought to be concentrated
• Where lines of symmetry cross is the centre of mass of a regular shape
• When suspended, the centre of mass of an object is below the suspended point
2 of 20
## Toppling and stability
• The extent to which an object resists toppling
• Sliding is not toppling
• When the centre of mass acts within an object, it will not topple
• The taller an object, the higher the centre of mass, causing the topple angle to become smaller meaning that the object will topple easier
3 of 20
## Newtons laws
1) An object will stay at rest or stay in constant motion unless acted on by an external force.
• Basically, unless something else happens to it, an objects motion will not change
2) FORCE = MASS X ACCELERATION
• A bigger mass will have a smaller acceleration when pushed by the same force
• Increasing the force will increase the acceleration of an object
3) For every action, there is an equal and opposite reaction
• If an object exerts a force on another object of the same mass, the force will be 'reflected'
4 of 20
## Resultant forces
• The result of all forces acting on an object
• Forces in the same direction are added together
5 of 20
## Centripetal force
FORCE =(MASS X VELOCITY²) / RADIUS
• Holds an object in circular motion
• Measured in newtons
• Effected by:
• Mass of the object
• Radius of the circle
• Velocity of the object
• Increased by:
• Increased mass
• Increased speed
• Decreased by:
6 of 20
## Gravity
• Acts between all objects that have mass
• Infinite range
• Big things have a large gravitational pull
• Small things have a small gravitional pull
• Inversely proportional to distance.. the greater the distance, the smaller the force
• Provides the centripetal force to keep planets and moons in orbit
7 of 20
## Satellites
• Natural and unnatural
• Always falling towards the earth
• Two types of orbit:
• Geostationary- have an orbital period that is the same as the earths rotational period, making them seem stationary from the earth's surface. Mostly used for communication and navigation
• Polar- orbits almost perpendicularly to the equator so that it passes over both poles in each orbit. Mainly used for monitoring things like weather
8 of 20
## Mirrors
• Three types:
• Plane
• Concave
• Convex
• The normal line is perpendicular to the mirror, all angles are measured against this line
• Dashes on the back of a line shows that the surface is reflective
• Incidence ray- the ray from the source to the mirror (on the way)
• Reflected ray- the ray from the mirror to the environment (after hitting the mirror)
• In a plane mirror:
• The image is as far behind the mirror as the object is in front
• The line joining the object to the image is the normal line
• The size of the object is the same size as the image
• The image formed behind the mirror is virtual
• The image is laterally inverted (mirrored)
9 of 20
## Curved mirrors
• Centre of curvature- The centre of the circle the mirror is part of
• Principle axis- The line connecting centre of curvature to the mirror, usually drawn horizontally
• Focal point- The point at which all reflected rays will pass
• Focal length- The distance between the mirror and the focal point
• Focal point is always on the principle axis
• To find the focal point, parallel rays of incidence must be shon onto a mirror, the point at which they all cross is the focal point
• To say an object is inside the focal point, means that it is between the focal point and the mirror
10 of 20
## Convex mirrors
• When parallel rays are shon onto a convex mirror, the rays of reflection are diverged and never cross
• The focal point is found by tracing the rays of reflection back behind the mirror, this is called carrying on the line of reflection
• Focal points of convex mirrors are virtual- the light doesn't actually go there
11 of 20
## Waves through a differentiating media
• The speed and direction of waves change as they enter or exit different materials
• As light enters a more dense material, it slows down and bends towards the nomal line
• As light enters a less dense material, it slows down and bends away from the normal line
12 of 20
## Converging lenses
• As light passes through a converging lens, the rays are refracted
• The image on the opposite side is real because the light actually goes there but we cannot see it because the light is not reflected back to the eye
• When an object is far away from the lens, the image is real, inverted and smaller
• As the object moves closer to the lens, the image becomes real, inverted and bigger
• When the object is really close, the rays are diverged and will not meet again. However, the observer still sees an image, believing that is comes from behind the object. This image is virtual, upright and bigger
• Uses:
• The eye
• Magnifying lens
• Cameras
Drawn like this:
13 of 20
## Diverging lenses
• Causes all rays of incidence to diverge
• Meaning that all images and focal points are virtual because the light doesn't actually go to where we are seeing the image
• To find the focal point of a diverging lens, shine parallel rays through the lens and trace back the refracted rays
Drawn like this:
14 of 20
## Magnification
MAGNIFICATION = IMAGE HEIGHT / OBJECT HEIGHT
• Image is larger than object when magnification is >1
• Image is the same size as object when magnification =1
• Image is smaller than object (diminished) when magnification is <1
15 of 20
## Prisms
• Refract light into different colours
• Violet=shortest wave length, bent the most
• Red=longest wave lenth, bent the least
16 of 20
## Sound waves
• Caused by vibrations
• Drawn by oscilloscopes
• Loud sounds have a high amplitude
• High pitch sounds have a high frequency
• Cannot be heard in a vacuum
• Ultrasound- Sound with a frequency too loud for humans to hear
• Uses:
• Sonography- baby scans
• Sanitation- bacteria can be disintegrated by ultrasonic waves
• Welding- Ultrasonic waves can create heat energy between two objects and cause them to melt together
• Humans can hear sounds with frequency between 20-20,000 Hz
17 of 20
## Magnetism
• Two poles on a magnet:
• North
• South
• Magnetic field- The area around a magnetic object in which other materials experience the force of the magnet
• Only three magnetic elements:
• Iron
• Nickel
• Cobalt
• When drawing a magnetic field, arrows point from north to south to show direction
• Closer lines=stronger field
• Magnetic field lines crossing causes a force and makes things move
18 of 20
## Electricity and magnetism
• An electric current enduces a magnetic field
• The motor effect:
• A wire carrying a current has magnetic field lines
• This wire is put in the magnetic field of a magnet
• The two sets of magnetic field lines cross, causing the wire to get flung away from the magnet
• However, if the wire is parallel to the magnetic field, it will not experience the motor effect
• Direction of force is reversed if:
• Direction of current is reversed
• Direction of magnetic field is reversed (magnet is turned around)
• Force can be increased if:
• The magnet is stronger
• The current in the wire is bigger
• The amount of wire in the field is greater (coil up the wire)
• MOTOR EFFECT IS MAGNET+ELECTRIC CURRENT = MOVEMENT
• GENERATOR EFFECT IS MANET+MOVEMENT = ELECTRIC CURRENT
19 of 20
## Transformers
• Electromagnetic Induction- The creation of electricity when magnetic field lines cross
• As electricity is sent through the coiled wire on the primary transformer, magnetic field lines grow, alternating current causes the field lines to constantly grow and shrink
• The magnetic field lines cutting through the wires of the secondary transformer induces a current
• The voltage is induced when there is relative motion between a magnetic field and a conductor caused by:
• A conductor cutting magnetic field lines
• Magnetic field lines cutting a conductor
• A change in the strength of the magnetic field linking a conductor
• The produced voltage can be increased by:
• Increasing the speed of movement
• Increasing the strength of magnetic field
• Increasing the number of turns on the coil
• Increasing the area of the coil
• In order to keep the current being made, we must use an alternating current to keep the motion to generate electricity
20 of 20 | 2,059 | 8,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-26 | latest | en | 0.85069 |
http://forums.wolfram.com/mathgroup/archive/2005/May/msg00462.html | 1,716,723,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00168.warc.gz | 9,432,265 | 7,937 | Re: DSolve and K\$31, i am puzzled
• To: mathgroup at smc.vnet.net
• Subject: [mg57099] Re: [mg57078] DSolve and K\$31, i am puzzled
• From: Igor Antonio <igora at wolf-ram.com>
• Date: Sun, 15 May 2005 03:03:59 -0400 (EDT)
• Organization: Wolfram Research, Inc.
• References: <200505140858.EAA09494@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
David Boily wrote:
> When i try to solve this differential equation with the command:
>
> In[1]:= DSolve[f'[t] == Sign[p2 Exp[F2 t] - p1 Exp[F1 t] + p], f[t], t]
>
> where p, p1, p2, F1 and F2 are constants i get the answer
>
> F1 t F2 t
> (p - E p1 + E p2) (-K\$31 + t)
> Out[1]= {{f[t] -> ------------------------------------- + C[1]}}
> F1 t F2 t 2
> Sqrt[(p - E p1 + E p2) ]
>
> what the hell is K\$31? I have tried to find out with google or
> the mathematica 5 online docs but have found nothing.
>
> can someone enlighten me?
>
> David Boily
> Centre for Intelligent Machines
> McGill University
> Montreal, Quebec
Those are "module variables" and the numbers give them unique names.
They're needed because often the output from DSolve contains an
unevaluated integral. Example:
In[1]:= DSolve[y'[x]==f[x], y,x]
Out[1]= {{y -> Function[{x}, C[1] + Integrate[f[K\$371], {K\$371, 1, x}]]}}
The K\$371 variable is a dummy variable of integration and does not
affect the calculations. Just replaced it by another symbol:
In[2]:= %/.{K\$44 -> t}
Out[2]= {{y -> Function[{x}, C[1] + Integrate[f[t], {t, 1, x}]]}}
--
Igor Antonio
Wolfram Research, Inc.
www.wolfram.com
To email me personally, remove the dash.
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• Next by thread: Re: Re: DSolve and K\$31, i am puzzled | 603 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.760995 |
https://studyadda.com/question-bank/critical-thinking_q12/1025/72997 | 1,717,030,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00576.warc.gz | 454,273,873 | 21,509 | • # question_answer The points $O,\,A,\,B,\,C,\,D$ are such that $\overrightarrow{OA}=\mathbf{a},$ $\overrightarrow{OB}=\mathbf{b},\,$ $\overrightarrow{OC}=2\mathbf{a}+3\mathbf{b}$ and $\overrightarrow{OD}=\mathbf{a}-2\mathbf{b}.$ If $|\mathbf{a}|\,=3\,|\mathbf{b}|,$ then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is A) $\frac{\pi }{3}$ B) $\frac{\pi }{4}$ C) $\frac{\pi }{6}$ D) None of these
• We have $\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=\mathbf{a}-2\mathbf{b}-\mathbf{b}=\mathbf{a}-3\mathbf{b}$ and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\mathbf{a}+3\mathbf{b}-\mathbf{a}=\mathbf{a}+3\mathbf{b}.$
• Let $\theta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}.$
• Then $\cos \theta =\frac{\overrightarrow{BD}\,.\,\overrightarrow{AC}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}=\frac{|\mathbf{a}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}$
• $=\frac{9|\mathbf{b}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}$,
• $(\because \,\,\,\left| \,\mathbf{a}\, \right|=3|\mathbf{b}|)$
• $\Rightarrow \cos \theta =0{}^\circ \Rightarrow \theta =\frac{\pi }{2}.$ | 493 | 1,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.18025 |
http://www.progtown.com/topic2081612-how-to-calculate-date-of-monday-under-number-of-week.html | 1,534,517,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212598.67/warc/CC-MAIN-20180817143416-20180817163416-00080.warc.gz | 550,120,337 | 5,428 | Topic: How to calculate date of Monday under number of week?
Initially I calculate number of week and the last two digits of year:
Format ([Date]; "ww") and Format ([Date]; "yy")
I receive result in a format (on an example of the first week 2018):
118
Question. How to me to calculate date of the first day of week (in my case, Monday) having only 118?
Thanks.
Re: How to calculate date of Monday under number of week?
aoiu wrote:
having only 118?
Well year nevertheless is necessary... And with the registration of that in a year of 52 weeks, number 118 looks doubtful...
Re: How to calculate date of Monday under number of week?
aoiu;
But not always the first day of week - Monday...
Monday becomes the first day of week, as Weekday to specify in the second parameter vbMonday
Re: How to calculate date of Monday under number of week?
Akina wrote:
it is passed...
Well year nevertheless is necessary... And with the registration of that in a year of 52 weeks, number 118 looks doubtful...
118 first digit 1 is number of week, and the last two digits 18 is a year.
Re: How to calculate date of Monday under number of week?
Amiri wrote:
aoiu;
But not always the first day of week - Monday...
Monday becomes the first day of week, as Weekday to specify in the second parameter vbMonday
Is only 118 where the first 1 number of week, and last 18 is a year.
Re: How to calculate date of Monday under number of week?
aoiu;
=WeekdayName (Weekday (118; 2))
?
Re: How to calculate date of Monday under number of week?
y - Number of day in a year (from 1 to 366).
yy - the Last two digits of number of year (from 01 to 99).
Re: How to calculate date of Monday under number of week?
Amiri;
dddd - the Full title of day of week (Monday - Sunday).
Re: How to calculate date of Monday under number of week?
wrote:
aoiu;
[spoiler]
``(DateSerial (2000 + CLng (Right (118, 2)), 1, 1) + ((1 - Weekday (DateSerial (2000 + CLng (Right (118, 2)), 1, 1), 0) + 7) Mod 7)) + ((CLng (Left (CStr (118), Len (CStr (118)) - 2)) - 1) * 7)``
[/spoiler]
Thanks, so. | 561 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-34 | latest | en | 0.89538 |
https://foxoyo.com/mcq/41118/a-cone-with-a-circular-base-of-radius-5cm-and-height-12cm-has-surface-area-of | 1,542,802,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748315.98/warc/CC-MAIN-20181121112832-20181121134832-00334.warc.gz | 611,993,892 | 7,793 | # A cone with a circular base of radius 5cm and height 12cm has surface area of
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### Related MCQs
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The slant height of a cone is 12 cm and radius of the base is 4cm, find the curved surface of the cone | 273 | 1,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-47 | latest | en | 0.908086 |
https://mathspace.co/textbooks/syllabuses/Syllabus-406/topics/Topic-7189/subtopics/Subtopic-96063/?activeTab=theory | 1,642,820,946,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00545.warc.gz | 399,738,734 | 47,241 | # Multiply a two digit number by a small single digit number
Lesson
Did you know that when we multiply a number less than 5 by a two digit number, there are some tips we can use. This helps us break it into smaller problems.
Think of how we partition numbers, as this can help you understand how to break your two digit number into separate parts.
Imagine if you knew how to solve something like $3x34$3x34 by renaming it as $3x30$3x30 plus $3x4$3x4.
Not only can you do that, you can also rename $3x30$3x30 as $3x3tens$3x3tens which is $9tens,or90$9tens,or90.
You can work out that $3x4=12$3x4=12 and you are ready to solve $90+12$90+12 , which is $102$102.
Watch this video to see some more tips for working out problems like this one.
##### Question 1
Find $1\times31$1×31.
##### Question 2
Find $42\times3$42×3.
##### Question 3
Find $2\times78$2×78.
### Outcomes
#### NA3-6
Record and interpret additive and simple multiplicative strategies, using words, diagrams, and symbols, with an understanding of equality | 311 | 1,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-05 | longest | en | 0.859678 |
http://woerterbruch.de/chance-for-landing-precisely-on-live-roulette/ | 1,638,453,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00623.warc.gz | 87,765,038 | 24,413 | # Chance for landing precisely on live roulette stand casino near me
Chance for landing precisely on live roulette stand casino near me
Envisioned advantage could very well be many beneficial likelihood concept we will discuss. It offers several applications, from insurance coverages to making financial judgements, as well as one thing that the gambling enterprises and federal government services that run gambling procedures and lotteries wish everyone never learn about.
## Illustration 42
In casino event roulette, a wheel with 38 rooms (18 yellow, 18 black colored, and 2 alternative) is definitely spun. In one single feasible wager, the gamer bets \$1 for a passing fancy numbers. If this amount happens to be spun about controls, they then receive \$36 (the company’s earliest \$1 + \$35). Otherwise, these people get rid of their own \$1. On average, how much cash should a person expect you’ll victory or reduce should they bet this game over repeatedly?
Suppose you bet \$1 for each of the 38 spaces to the wheel, for a maximum of \$38 decision. When the receiving wide variety is actually spun, you happen to be paid \$36 thereon amount. Whilst obtained on that one quantity, all-around you’ve forgotten \$2. On a per-space base, you have “won” -\$2/\$38 ? -\$0.053. Put another way, on average an individual shed 5.3 dollars per place without a doubt on.
We call this regular obtain or loss anticipated value of trying to play roulette. Realize that no body previously will lose exactly 5.3 cents: most individuals (indeed, about 37 out of each and every 38) drop \$1 and really few individuals (about 1 people out of each and every 38) build \$35 (the \$36 they victory without worrying about \$1 the two put in to learn the action).
Summarizing these with the worth, we are this desk:
0.9211 + (-0.9737) ? www.casinogamings.com/review/william-hill-casino/ -0.053, the forecast appreciate you calculated previously.
Expected advantage would be the ordinary obtain or loss in an event if the processes happens to be repeating several times.
You can easily compute anticipated benefits by growing each result because likelihood of that result, subsequently adding up the products.
## You should try it At This Point 12
You buy a raffle violation to assist completely a cause. The raffle solution expenditure \$5. The cause sells 2000 seats. One too are going to be pulled as well guy possessing the pass will be provided a prize worth \$4000. Compute the expected appreciate with this raffle.
## Situation 43
In a specific shows lotto, 48 balls designated 1 through 48 are positioned in a product and six ones tends to be drawn at random. If six figures pulled accommodate the numbers that a person have picked, the ball player victories \$1,000,000. Should they match 5 rates, subsequently victory \$1,000. It is \$1 purchase a ticket. Chose the expected advantage.
Early, we considered the prospect of complimentary all 6 number together with the likelihood of coordinating 5 figures:
Our very own possibilities and results worth are:
The expected benefits, subsequently is actually:
On average, one could expect to drop about 90 dollars on a lotto citation. Obviously, the majority of participants will eventually lose \$1.
Generally speaking, in the event the forecast worth of a game are unfavorable, it’s not at all smart to have fun with the games, since generally you will definitely generate losses. It will be far better to bring a game with a confident likely benefits (all the best ! searching for one!), although remember that even if your typical winning become positive perhaps your situation that many group generate losses as well as one most lucky individuals wins a lot of money. In the event that anticipated worth of a-game is actually 0, we refer to as it a fair game, since neither area features an edge.
## You should try it Today 13
A colleague proposes to play a game title, in which you roll 3 typical 6-sided dice. If these dice move various worth, provide your \$1. If any two dice match principles, you obtain \$2. What exactly is the forecast property value this video game? Is it possible you bet?
Anticipated price also has methods outside of casino. Envisioned advantage particularly common for making cover choices.
## Sample 44
A 40-year-old person into the U.S. provides a 0.242% risk of perishing via this year. [1] An insurance business charges \$275 for a life-insurance coverage that will pay a \$100,000 death perk. What exactly is the expected benefits for any guy buying the insurance coverage?
The probabilities and results tends to be
Anticipated appreciate is definitely (\$99,725)(0.00242) + (-\$275)(0.99758) = -\$33.
And in addition, anticipated advantage try adverse; the insurer corporation can only manage to present procedures should they, an average of, earn an income on each rules. Capable be able to spend the occasional perks mainly because they promote adequate plans that people perk rewards become stabilized through the heard of guaranteed folks.
For anyone purchasing the insurance coverage, discover a bad envisioned benefits, but there is however a burglar alarm that comes from insurance coverage that is worthy of that fee. | 1,136 | 5,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.935934 |
http://mathrec.org/mcc/blackboardpools/Chapter2.html | 1,582,150,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00177.warc.gz | 99,352,444 | 2,392 | Section 2.1Quiz 4.1 Question: Find the distance between the points (1, 2) and (-4, 1). Express your answer as a decimal approximation rounded to three decimal places.
Section 2.2Quiz 4.2 Question: Find the y-intercepts of the equation (x + 2)2 + (y – 7)2 = 85.
Section 2.3Quiz 4.3 Question: What is the slope of the line through the points (3, 4) and (4, -3)? Express your answer as a fraction in lowest terms.
Section 2.3Quiz 4.4 Question: What is the y-intercept of the line through the points (1, -2) and (-2, -1)? Express your answer as a fraction in lowest terms. If the line does not have a y-intercept, enter none.
Section 2.4 Question: Determine whether the lines –1x + 4y = 2 and –4x + 1y = –4 are parallel, perpendicular or neither. Quiz 4.5 Parallel Perpendicular Neither
Section 2.4Quiz 4.6 Question: Write the equation of the circle in standard form.
Section 2.6Quiz 4.7 Question: You know that y varies directly with x and inversely with t2. If y = 1/2 when x = 3 and t = 2, what is the value of y when x = 2 and t = 3? (Express your answer as a fraction in lowest terms.) Variation: You know that y varies directly with x3 and inversely with t. If y = 2 when x = 2 and t = 5, what is the value of y when x = 5 and t = 3? (Express your answer as a fraction in lowest terms.)
Section 2.6Quiz 4.8 Question: An independent contractor is resurfacing a parking lot. He charges his customer \$320 per day (\$320/day) for asphalt cost. If the contractor actually uses 20 tons per day (ton/day), what price is he charging his customer for the asphalt (in units of \$/ton)? Variations: A car traveling 50 miles per hour (mi/hr) has a fuel efficiency of 12.5 miles per gallon (mi/gal). How fast does it consume fuel (in units of gal/hr)? A company produces salt by evaporating sea water. The concentration of salt in sea water is 2.5 kg/m3. How much sea water do they need to evaporate (in units of m3/day) to produce salt at a rate of 2,000 kg/day? | 590 | 1,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-10 | latest | en | 0.898015 |
http://astronoo.com/en/articles/wave.html | 1,695,379,952,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00008.warc.gz | 3,539,438 | 6,433 | # What is a wave?
## A wave is an energy that spreads
Automatic translation Category: matter and particles
Updated July 25, 2014
Say that a wave is the propagation of a disturbance of the environment is not enough to understand what a wave.
Say that speed, wavelength and frequency are the three properties characterizing a wave and that the relation between the three variables is v = λ * ν that does not sufficient to understand what a wave.
Say that a wave travels with a specific speed which depends on the characteristics of the propagation environment, it does not show the wave.
Say that a wave carries energy without transporting of matter, it tells us, always nothing in the nature of a wave.
Now imagine a rope that is shaken, of a vertical gesture. This simple movement will draw a wave of a certain height, which appears to be spreading to the front but in reality no point in the rope moves each point, comes and goes, moving vertically. This amplitude gives us an idea of the force with which you shake the rope but not the wave itself.
To understand what is a wave must remove the rope and imagine the air molecules undergoing the mechanical pressure of the rope. At this point it only remains in the medium (air) than the energy, a force generated by the movement of the rope. This energy makes the pressure of the air oscillate around an equilibrium value, the pressure increases and decreases alternately around this value. The wave (energy) moves the air molecules without transport them. As the rope is shaken, one can measure the distance between two bosses or, two hollow, this distance is the wavelength, measured in meters. The rate at which the rope is shaken, corresponds to the wave frequency measured in Hertz. The wave velocity measured in meters / second is equal to the wavelength multiplied by frequency.
One wave propagates in a stable environment, able to return to an equilibrium state, for a sound wave that is the air pressure which moves relative to an average value, for an electromagnetic wave is the intensity of the electromagnetic field which moves relative to an average value of the field.
Electricity can be static, such as amber after being rubbed, attracts small objects. Magnetism can also be static like a magnet. But when these fields move together, they become electromagnetic waves (see image). Electromagnetic waves are formed when an electric field (shown by red arrows) couples with a magnetic field (represented by blue arrows). The electric and magnetic fields of an electromagnetic wave are always perpendicular one with respect to another but also perpendicular to the direction of the wave.
These two types of waves, mechanical and electromagnetic, are two ways of transporting energy in a medium. The waves in the water and the sound waves in the air are two examples of mechanical waves. This energy transport disrupts or makes vibrate matter (solid, liquid, gas or plasma) without transport, water or air molecules collide but remain in the same place.
A variable magnetic field induces a variable electric field and vice versa, the two are connected. When these two fields are coupled (Maxwell equations) form electromagnetic waves (see image). Electromagnetic waves unlike mechanical waves, do not need medium to propagate, they travel everywhere, even in the vacuum of space.
Light, electromagnetic waves and all radiation are from the same physical phenomenon: the electromagnetic energy.
### Image: Electromagnetic waves are formed by the sinusoidal vibration of electric and magnetic fields. These fields are always perpendicular one with respect to another but also perpendicular to the direction of movement of the wave. Once formed, this energy travels at the speed of light until the next interaction with matter. An electromagnetic wave is a transverse wave.
nota: The wave is said transversal, if the energy moves perpendicularly to the direction of wave movement (movement of the arm, shaking the rope or the energy of a rock falling into the water). The wave is called longitudinal, if the energy moves in the direction of travel of the wave (the magnet of the speaker). A wave can be both longitudinal and transversal (a baguette, which strikes a drum).
## Frequency, wavelength and energy
Frequency, wavelength and energy are mathematically related, just know one of these three values to calculate the other two. The microwaves and radio waves are generally described in terms of frequency (Hertz), infrared and visible light in terms of wavelength (λ), and x-rays and gamma rays in terms of energy (electron volts).
The wave frequency is the number of periodic phenomena (ridges) that breed in a second, it is given in Hertz, after Heinrich Hertz (1857-1894) that established the existence of radio waves. From radio waves to gamma rays, the frequency is measured a few Hertz to 1026 Hertz.
The wavelength is the distance between two peaks (the electromagnetic waves have crests and troughs similar to those of ocean waves). The longest waves (radio waves) can measure a few kilometers or more, while the shorter wavelength (gamma waves) produced by the atomic nucleus, can measure up to 10-12 m (size of the atomic nucleus).
The energy of an electromagnetic wave is measured in electron volts (eV). An electron-volt is the amount of kinetic energy required to move an electron through a voltage potential of 1 volt. The lowest energy is the energy of radio waves (few eV), while the highest energies are those gamma rays. Gamma rays are radiation, high-energy photons (above 100 keV) sufficient to remove an electron from its orbit. Gamma rays are the most energetic form of light.
nota: Between the wavelength (λ) and frequency (ν) is the following relationship: ν = c / λ
ν = wave frequency in hertz
c = speed of light in vacuum in m / s
λ = wavelength in meter
nota: According to the equations of James Clerk Maxwell (1831-1879), light is a self-propagating electromagnetic transverse wave with electric and magnetic components where electric and magnetic fields oscillate at right angles to each other and propagate perpendicularly to the direction in which they move indefinitely unless absorbed by the intermediate matter. In other words, each type of field - electric and magnetic - generates the other in order to propagate the entire composite structure in empty space at the finite speed of light.
1997 © Astronoo.com − Astronomy, Astrophysics, Evolution and Earth science.
Scale of the nanoparticles... 2015 Internationalyear of light... Neutrinoand beta emission... Magnetic orderand magnetization... | 1,337 | 6,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.940683 |
https://tvaas.sas.com/evalComposite.html?ab=dA&as=a&aj=a&ww=174834&x9=22&w4=103 | 1,656,405,031,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00415.warc.gz | 640,469,785 | 4,039 | TVAAS
Report School Evaluation Composites, School McKenzie High School, District McKenzie, Test EOC
2020-2021 Composite Trends
Composite TypeOne-Year Trend*
IndexEffectiveness Level
Overall4.355
Literacy and Numeracy4.355
Literacy0.263
Numeracy5.515
*Year Trend
Educators using school composites for the growth component in the Tennessee Educator Acceleration Model (TEAM) select a school composite from the available options; all school composites use data from the most recent year (2020-21).
TDOE Policy
School TVAAS Composites are scores that assess growth at the school level. For 2020-21, school composites are used for the evaluation of educators who require school reporting. Scores will be reported on a 1-5 scale. The table above reports the 2020-21 school composites.
Note: The subjects listed for the CTE concentrator and CTE student composite include all the EOC subjects with value-added estimates. Some of these subjects might be excluded from CTE concentrator and CTE student composite if there were not enough CTE participants to produce a reliable estimate.
Descriptions of Composite Types
Composite TypeComposite Might Include
TestSubject
OverallEOCAlgebra I, Algebra II, English I, English II, Geometry
Literacy and NumeracyEOCAlgebra I, Algebra II, English I, English II, Geometry
LiteracyEOCEnglish I, English II
NumeracyEOCAlgebra I, Algebra II, Geometry
Rules for Determining Effectiveness Level Level 5, Most Effective: Significant evidence that the school's students made more growth than expected (the school's index is 2 or greater). Level 4, Above Average Effectiveness: Moderate evidence that the school's students made more growth than expected (the school's index is between 1 and 2). Level 3, Average Effectiveness: Evidence that the school's students made growth as expected (the school's index is between -1 and 1). Level 2, Approaching Average Effectiveness: Moderate evidence that the school's students made less growth than expected (the school's index is between -2 and -1). Level 1, Least Effective: Significant evidence that the school's students made less growth than expected (the school's index is less than -2). Note: When an index falls exactly on the boundary between two levels, the higher level is assigned. | 511 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.899745 |
http://datascience.sharerecipe.net/tag/prime/ | 1,721,733,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00505.warc.gz | 6,894,578 | 8,548 | # prime
## Bit flipping to primes
Someone asked an interesting question on MathOverflow: given an odd number, can you always flip a bit in its binary…
## How probable is a probable prime?
A probable prime is a number that passes a test that all primes pass and that most composite numbers fail.…
## Relatively prime determinants
Suppose you fill two n×n matrices with random integers. What is the probability that the determinants of the two matrices…
## Prime plus power of 2
A new article [1] looks at the problem of determining the proportion of odd numbers that can be written as…
## Estimating the proportion of smooth numbers
A number is said to be “smooth” if all its prime factors are small. To make this precise, a number…
## Prime numbers and Goldbach’s conjecture visualization.
isn’t it?The German mathematician G. F. B. Riemann (1826–1866) observed that the distribution of prime numbers is very closely related…
## Share The π: Honoring Neglected Mathematical Constants
Subsequent occurrences are fairly common. Brun’s and Meissel-Mertens ConstantsPrime numbers make an appearance in two fascinating constants we’ll discuss next.…
## Python Pro Tip: Use Itertools, Generators, and Generator Expressions
Think of the memory such a list would occupy. It would be great if we had something that could just…
## Strong primes
There are a couple different definitions of a strong prime. In number theory, a strong prime is one that is…
## Goldilocks and the three multiplications
Mike Hamburg designed an elliptic curve for use in cryptography he calls Ed448-Goldilocks. The prefix Ed refers to the fact…
## Tricks for arithmetic modulo NIST primes
The US National Institute of Standards and Technology (NIST) originally recommended 15 elliptic curves for use in elliptic curve cryptography…
## Elliptic curve P-384
Possibly, but they have reasons to recommend methods that they believe foreign governments cannot break. The equation of the P-384…
## New prime record: 51st Mersenne prime discovered
The largest known prime is nowWritten in hexadecimal the newly discovered prime isFor decades the largest known prime has been…
## Code doodling: isPrime
Then I’ll run the code on my laptop and time the results for each algorithm testing for primes in the…
## Prime numbers in complex domains are actually quite simple
Likewise, a number in Z[√−2] is said to be divisible by another number in Z[√−2] if the result of the… | 539 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.878011 |
https://www.jamiiforums.com/threads/kwa-wataalam-wa-hesabu.38010/ | 1,493,312,203,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122619.60/warc/CC-MAIN-20170423031202-00040-ip-10-145-167-34.ec2.internal.warc.gz | 909,718,419 | 26,278 | Kwa Wataalam wa hesabu | JamiiForums | The Home of Great Thinkers
# Kwa Wataalam wa hesabu
Discussion in 'Jokes/Utani + Udaku/Gossips' started by Asprin, Sep 5, 2009.
1. ### AsprinJF-Expert Member
#1
Sep 5, 2009
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Je unajua kuwa 2 = 1?
Wataalam wa hesabu tunasema kuwa kama x=y
basi 2x=2y
kwahiyo 2x-x = 2y-y si sawa?
Maana yake ni kuwa 2x-2y = x-y hapo hakuna ubishi naamini.
Hii ni sawa na kusema 2(x-y) = (x-y)
kwamba 2(x-y)/(x-y) = (x-y)/(x-y)
Na hiyo inathibitisha kuwa 2 = 1
Kwa wasiojua hesabu watabaki wanatoa mimacho, hawaelewi kitu halafu bahati mbaya watabisha!
Good Day Mathematicians!
2. ### The FarmerJF-Expert Member
#2
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Wengine tuliacha kusoma HESABU darasa la nne na kuanza kusoma HISABATI. Kwa hiyo kukubishia ni rahisi tu!!!!
3. ### AsprinJF-Expert Member
#3
Sep 5, 2009
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Mkuu kuna tofauti gani kati ya hesabu na hisabati?
4. B
### Bao3JF-Expert Member
#4
Sep 5, 2009
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So what are you trying to say?
5. ### Mwana va MutwaJF-Expert Member
#5
Sep 5, 2009
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mkuu nikiwa bored siku zote ndio huingia huku kwa jokes ku relax sasa leo duhh najuta kuingia huku
6. ### The FarmerJF-Expert Member
#6
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Hesabu ni (Kujumlisha, kuzidisha etc),
7. ### KitukoJF-Expert Member
#7
Sep 5, 2009
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si kweli, hesabu huwa haziendi hivyo kama utakavyo wewe bali zina principal zake
kama x=y,
ukija hapa, 2x=2y, kwa nini umetumia mbili, vipi ikiwa tatu yaani 3x=3y
kwa hiyo 3x-x=3y-y
maana yake ni kuwa 3x-3y=x-y
hii ni sawa na kusema 3(x-y)=(x-y)
kwamba 3(x-y)/(x-y)=(x-y)/(x-y)
na hiyo inathibisha kuwa 3=1
HAYA WEWE UNAEJUA HESABU KANUSHA BAC
8. ### AsprinJF-Expert Member
#8
Sep 5, 2009
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Mkuu nakukaribisha kwenye chama chetu cha WANAUJUA HESABU!
9. ### senatorJF-Expert Member
#9
Sep 5, 2009
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kwa mantiki hiyo any number can be proved equal to one with the exceptional of zero!
10. B
### Baba MkubwaJF-Expert Member
#10
Sep 5, 2009
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Hahahahahahahahahaa
Ni swali ambalo unaweza ukamuuliza mwanafunzi, wapi pana kosa! Ninaamini wewe unajuwa kosa lipo wapi. Kama unafahamu kuwa zero divide by zero can be certain number, depending to the nature of the numerator and denominator....kama ni zero gawanya kwa zero basi tunaweza kutumia L'Hospitals rule.
Kwanini nasema zero gawia zero? Ulipofika x-y/x-y; hi ni sawa na 0/0 maana x=y. Sasa ikiwa hiyo Never conclude kwamba ni 1.
Ninaamini nimeeleweka wapi palipo kufanya u-conclude kuwa jibu ni moja. Jiulize kwanini namba yoyote ukiigawa yenyewe kwa yenyewe jibu ni moja lakini 0 gawa kwa 0 si moja?
11. B
### Baba MkubwaJF-Expert Member
#11
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L'Hospital's Rule ina matumizi yake sehemu fulani....nimeiweka hapa kama mfano, may be can't work. Nakumbuka nimeitumia L'Hospital's Rule nilipokuwa natafuta limits, na naitumia kukiwepo cases furani mfano ikatokea 0 gawa 0!!!!
12. M
### MatareseJF-Expert Member
#12
Sep 5, 2009
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you got it all wrong!
First, number of angles in 2 can never be equal to number of angles in 1, which is the basis for having numbers.
Secondly, look if x=y then 2x=2y, that is correct, now
2x-x=2y-y also correct, but down here is where you went wrong
2x-2y =x-y ( this is incorrect) but rather 2x-2y=-y+x. Note that -x+y is not the same as y-x bse subtraction is not commutative , i.e 2-3 not same as 3-2 !
13. B
### Baba MkubwaJF-Expert Member
#13
Sep 5, 2009
Joined: Oct 18, 2008
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May be I am wrong mkuu!
It is true that subtraction is not commutative. If 2x-x = 2y-y then it is true that 2x-2y=x-y....this is simple maths kwasababu we have added -2y to the equation 2x-x=2y-y, tukapata 2x-2y-x=-y and then tukajulisha tena x, tukapata 2x-2y=x-y. Commutativity imetumika kwenye cases kama -2y+2x=2x-2y (bado tupo kwenye addition). Ni kweli 2-3 si sawa na 3-2 lakini 2-3 ni sawa na -3+2. Kwahiyo, if 2x-x=2y-y then 2x-2y=x-y (tutatumia commutativity).....nahisi nipo sahihi.
Tukirudi kwenye issue ya thread kwamba 2=1; mimi nimeona kosa lipo kwenye x-y/x-y......yaani 0/0 siyo moja!!!....nahisi nipo sahihi
Nawasilisha
14. ### BurichekaMember
#14
Sep 5, 2009
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Baba Mkubwa,
Zero divide by zero sidhani kama inawezi kuwa "a certain number depending on the nature of the numerator and denominator" hata siku moja. Hiyo "depending on" inakuja vipi tena wakati ushafahamu numerator ni zero na denomionator ni zero? Kitu chochote gawanya kwa zero ni undefined, au infinity, period. Haijalishi juu ya zero kuna nini. Alipofika x-y / x-y ilibidi apate undefined, mwisho. Hakuna cha L'Hospitals rule hapa, sidhani.
15. B
### Baba MkubwaJF-Expert Member
#15
Sep 5, 2009
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Nikweli mkuu ulichosema. Ndiyo maana nilipopost kuhusu/kutaja L'Hospital's Rule nikaandika post nyingine ni wapi L'Hospital's Rule inatumika mkuu (inawezekana nimechanganya....which is possible in maths life). Nikasema hutumika wakati fulani, mfano limits.....kwa lugha nyingine kwenye case yake ya zero gawanya kwa zero hatutaweza kutumia L'Hospital's Rule. jibu langu la kusema hutegemeana na numerator and denominator nipo sahihi, nahisi, hasa pale tunapozungumzia limits. Unapoevaluate limits na ukawa na 0/0; Utapata jibu kulinganya na nature ya numerator na denominator NA si moja kila siku.
16. ### BurichekaMember
#16
Sep 5, 2009
Joined: Aug 18, 2009
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Brother, hapo ndo nashindwa kukuelewa.
Tunajua kabisa tuna dili na sifuri gawanya kwa sifuri, numerator na denominator zote zinajulikana, sasa utasemaje tena "kulingana na nature ya numerator na denominator"? Nature si ushaijua kaka? Ambayo ni sifuri juu, sifuri chini, sivyo?
17. B
### Baba MkubwaJF-Expert Member
#17
Sep 5, 2009
Joined: Oct 18, 2008
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Okay nimekupata wapi ambapo sijaeleweka, inawezekana maelezo yangu si timilifu....
L'Hospital's Rule nimeizungumza itakapotumika kwenye limits. 0 gawa kwa 0 huweza kupatikana pale utakapo weka x goes to certain number. Naomba nitoe mifano:
1. limit of sinx/x as x goes to 0 jibu ni 1. Tulipo-evaluate limit tulipata 0/0 ndipo tulipotumia L'Hospital's Rule na tukapata jibu 1
2. limit of (cosx-1)/x as x goes to 0 jibu ni 0. Tulipo-evaluate limit tulipata 0/0 ndipo tulipotumia L'Hospital's Rule na kupata jibu 0
3. limit of sin(2x)/x as x goes to 0 jibu ni 2. Tulipo-evaluate limit tulipata 0/0 ndipo tulipotumia L'Hospital's Rule na tukapata jibu 2
etc
Kwahiyo navyosema jibu hutegemeana na nature ya numerator and denominator namanisha kuwa jibu litategemea swali lipoje. Inawezekana maelezo yangu si timilivu mkuu. Inawezekana sijajuwa matumizi ya "nature" na "depends on"...lugha kaka/dada
Nawasilisha
18. M
### MatareseJF-Expert Member
#18
Sep 5, 2009
Joined: Aug 30, 2009
Messages: 514
Trophy Points: 35
Before you try to calculate you have to understand this:
1. You are trying to verify two values that have completely different angles. Which is immaterial.
2. When you try to divide 2(x-y)=(x-y) the answer is always undefined because the form 0/0 is indeterminate form, whose division is immaterial (undefined) as above unless L'hospitals rule is applied. There are so many such forms. Thus the above ( i am not sure if the person is trying to prove, or verify or show) is quite untrue.
19. B
### Baba MkubwaJF-Expert Member
#19
Sep 5, 2009
Joined: Oct 18, 2008
Messages: 770
Trophy Points: 35
Nikweli mkuu ulichosema ingawa muanzalishi wa thread alitoa changamoto ambayo inamuwezesha mtu kujuwa wapi pana kosa. 2=1 is true to some places of algebra (with addition of more information eg modulus):::: are you talking about........linear algebra, abstract algebra, descrete algebra/boolean algebra etc.
20. ### TibaJF-Expert Member
#20
Sep 5, 2009
Joined: Jul 15, 2008
Messages: 4,510 | 3,159 | 8,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-17 | longest | en | 0.433096 |
https://communities.sas.com/t5/Statistical-Procedures/Reproduction-of-an-example-from-the-book-of-Agresti/m-p/632389 | 1,721,266,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00790.warc.gz | 151,866,034 | 31,675 | Fluorite | Level 6
## Reproduction of an example from the book of Agresti
Good morning,
On page 183 of the book "Introduction to the Categorical Data Analysis, 2nd Edition" (2007), Agresti presents Table 6.8, stating explicitly "Table 6.8 shows output (from PROC LOGISTIC in SAS) for the ML fit of model (6.4)."
Could someone be so kind to provide me with the code with which I could reproduce this table?. Particularly the section "Deviance and Pearson Goodness-of-Fit Statistics".
To make your life easier, I am attaching a script with the data, so you don't have to rewrite it.
Thanks in advance and best regards,
Piotr Lewczuk
data party;
input g p y count;
datalines;
0 1 0 44
0 1 1 47
0 1 2 118
0 1 3 23
0 1 4 32
0 0 0 18
0 0 1 28
0 0 2 86
0 0 3 39
0 0 4 48
1 1 0 36
1 1 1 34
1 1 2 53
1 1 3 18
1 1 4 23
1 0 0 12
1 0 1 18
1 0 2 62
1 0 3 45
1 0 4 51
; run;
4 REPLIES 4
Meteorite | Level 14
## Re: Reproduction of an example from the book of Agresti
Welcome to the SAS community!
You can try this to get the desired output:
``````proc logistic data=party;
class p (ref='0') y / param=reference;
freq count;
model y = p / link=logit equalslopes=p aggregate scale=none lackfit;
run;
``````
NB:
- equalslope allows only 1 estimate for the parameter p instead of 4
Hope this helps,
Best,
Fluorite | Level 6
## Re: Reproduction of an example from the book of Agresti
Thank you very much! Very helpful.
Regards,
Piotr Lewczuk
## Re: Reproduction of an example from the book of Agresti
@ed_sas_member's code reproduces all statistics in that Table 6.8, so it is a solution.
The same output can also be created with the code below:
``````proc logistic data=party;
class p(ref='0' param=ref);
freq count;
model y=p / aggregate scale=n;
run;``````
Fluorite | Level 6
## Re: Reproduction of an example from the book of Agresti
Indeed; since this example focuses on cumulative logit, perhaps the code below is even more explicit:
proc logistic data=party;
freq count;
model y(ref="4") = p /link=clogit aggregate scale=none;
run; | 663 | 2,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.782462 |
http://mathhelpforum.com/algebra/26732-req-help-fractional-exponent-problem-print.html | 1,516,399,268,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00510.warc.gz | 219,647,040 | 2,978 | # Req. Help with Fractional Exponent Problem
• Jan 24th 2008, 04:51 AM
Lovely07
Req. Help with Fractional Exponent Problem
Please tell me what I did wrong.
This is a question from my review packet.
The answer says it should be 2 with exponent 3/2
My attempt gives me this:
http://i30.tinypic.com/34rtzmd.jpg
• Jan 24th 2008, 05:01 AM
mr fantastic
Quote:
Originally Posted by Lovely07
Please tell me what I did wrong.
This is a question from my review packet.
The answer says it should be 2 with exponent 3/2
My attempt gives me this:
http://i30.tinypic.com/34rtzmd.jpg
Your mistakes are in the third line. You can't just take the exponents out of the square root like that!!
Note that:
$2^{5/2} = 2^2 \times 2^{1/2} = 4 \sqrt{2} \,$ and $\, 2^{3/2} = 2^1 \times 2^{1/2} = 2 \sqrt{2}$.
So your problem becomes $4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$.
Now note that $2 \sqrt{2} = 2^{3/2}$.
---------------------------------------------------------------------------
Alternatively, note that $2^{5/2} = 2^1 \times 2^{3/2} = 2 \times 2^{3/2}$.
So your problem becomes $2 \times 2^{3/2} - 2^{3/2} = 2^{3/2}$.
• Jan 24th 2008, 05:30 AM
Lovely07
Thank you very much for you help.
The alternative method was easy to understand. | 420 | 1,230 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-05 | longest | en | 0.888959 |
https://socratic.org/questions/how-do-you-solve-using-completing-the-square-method-m-2-4m-2-0 | 1,726,299,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00564.warc.gz | 500,134,070 | 5,999 | # How do you solve using completing the square method m^2+4m+2=0?
Oct 5, 2016
$m = - 2 \pm \sqrt{2}$
#### Explanation:
The difference of squares identity can be written:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Use this with $a = \left(m + 2\right)$ and $b = \sqrt{2}$ as follows:
$0 = {m}^{2} + 4 m + 2$
$\textcolor{w h i t e}{0} = {m}^{2} + 4 m + 4 - 2$
$\textcolor{w h i t e}{0} = {\left(m + 2\right)}^{2} - {\left(\sqrt{2}\right)}^{2}$
$\textcolor{w h i t e}{0} = \left(\left(m + 2\right) - \sqrt{2}\right) \left(\left(m + 2\right) + \sqrt{2}\right)$
$\textcolor{w h i t e}{0} = \left(m + 2 - \sqrt{2}\right) \left(m + 2 + \sqrt{2}\right)$
Hence:
$m = - 2 \pm \sqrt{2}$ | 308 | 705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.532649 |
http://forum.arduino.cc/index.php?topic=42864.msg310607 | 1,508,366,422,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00278.warc.gz | 126,206,026 | 9,039 | Go Down
### Topic: [Solved] LED bar...but inverted !? (Read 2120 times)previous topic - next topic
#### HDLM
##### Dec 20, 2009, 12:02 amLast Edit: Dec 20, 2009, 09:52 pm by HDLM Reason: 1
:-?
Hi all, im new here and new to the world of microcontrollers.
So what did i do; i bought a arduino board to play around with. Nothing difficult, nothing specific.
So i found this knight rider thingie on this forum somewhere in an old topic. I upped the code to the arduino, wired the leds n resistors up the way i should and......
....the LEDs do light up, and the knight rider effect is running...but its inverted. What i mean is that there should be a light running from left to right and vice versa....but in my case al lights are on and one goes off from left to right and vice versa...
This is the code i used...:
Code: [Select]
`int ledpin[] ={3,5,6,9,10,11};int n = 0; //speed counter.int i =0; // repeater.int value[] ={0,0,0,0,0,0}; void setup(){}void loop(){ if(n == 100){value[0]=255;} if(n == 200){value[0]=255;} if(n == 300){value[1]=255;} if(n == 400){value[2]=255;} if(n == 500){value[3]=255;} if(n == 600){value[4]=255;} if(n == 700){value[5]=255;} if(n == 800){value[5]=255;} if(n == 900){value[4]=255;} if(n == 1000){value[3]=255;} if(n == 1100){value[2]=255;} if(n == 1200){value[1]=255;} if(n >= 1200){n =0;} for (i = 0; i <=5; i++) { if(value[i] >= 5){ value[i]-=1;} else if (value[i] < 5) { value[i]=0;} } for (i = 0; i <=5; i++) { analogWrite(ledpin[i], value [i]); } n+=5; delay(5); } `
Examples:
This is what i have
[/media]
This is what it should be
So the code works and the hardware is connected properly...what can i do to fix it ?
#### bill2009
#1
##### Dec 20, 2009, 02:19 am
This is pretty crufty because I don't "get" the code that's there but if the lights are all the opposite of what you want, just change the analogWrite to be 255-value . May not be clean but maybe it will move you along.
#2
##### Dec 20, 2009, 05:10 am
How are the LEDs wired?
Pin to resistor to LED then to ?
Pin to LED to resistor then to ?
#### HDLM
#3
##### Dec 20, 2009, 09:26 am
The leds are wired like this example:
http://hdlm.eu/61933851_3b9a25ab42_o.jpg
I used the same amount of leds in my setup (6 pcs) and i wired htem up exactly the same way. The only thing that i did different is that i used an additional 6 LEDs to mirror the sequence, so every pin switches 2 LEDs instead of just 1.
#4
##### Dec 20, 2009, 08:27 pmLast Edit: Dec 20, 2009, 08:31 pm by bcook Reason: 1
I can't tell from the picture if the black wire is connected to GND or 5V. You didn't answer my questions so I'll assume you don't want to continue discussing how the circuit is wired.
I can't see any problems with your Sketch. Unless someone else can spot a problem, I guess you're stuck with an inverted Knight Rider.
#### HDLM
#5
##### Dec 20, 2009, 08:51 pm
Quote
You didn't answer my questions so I'll assume you don't want to continue discussing how the circuit is wired.
If, by that, you mean that i did not write down the answer, you' re right...i did, however, provide you with a graphical answer on which you can see how the wiring is set up...
...So, to put it your way: I'll assume you don't know the answer to the question i asked in this discussion...but you simply don't want to admit it !
So, maybe there is another user on this board that can help me out..!?
#### AWOL
#6
##### Dec 20, 2009, 09:01 pm
Quote
I can't tell from the picture if the black wire is connected to GND or 5V
Neither can I.
Quote
So, maybe there is another user on this board that can help me out..!?
Not until you answer the question of how it is wired up, no, I think it unlikely.
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.
I speak for myself, not Arduino.
#### HDLM
#7
##### Dec 20, 2009, 09:28 pm
:-/
Wow...i think its just the fact that english is not my native language.
I cant understand what the ' answer' is that you both want...
The answer about the wiring is right there, on the picture... :-/
It would never be connected the GND pin would it ? So its either 5v or 3v, isnt it ?
----------------------------------------------------------------------------
Anyway, i ' fixed' the problem by using another code. The new code is alot more simple but the downside is that it isnt as smooth as the first one i used. I inserted the code below, as it might be handy for others int he future...
I dont think i will be coming back to this board with another question as i find the reactions of certain users are ' asking about the obvious' and not very friendly....thanks for the help, or trying to help though.
Good luck
Code: [Select]
`void setup(){pinMode(2, OUTPUT);pinMode(3, OUTPUT);pinMode(4, OUTPUT);pinMode(5, OUTPUT);pinMode(6, OUTPUT);pinMode(7, OUTPUT);pinMode(8, OUTPUT);}void loop(){digitalWrite(3, LOW);delay(80);digitalWrite(2, HIGH);delay(80);digitalWrite(4, LOW);delay(80);digitalWrite(3, HIGH);delay(80);digitalWrite(5, LOW);delay(80);digitalWrite(4, HIGH);delay(80);digitalWrite(6, LOW);delay(80);digitalWrite(5, HIGH);delay(80);digitalWrite(7, LOW);delay(80);digitalWrite(6, HIGH);delay(80);digitalWrite(8, LOW);delay(80);digitalWrite(7, HIGH);delay(80);digitalWrite(7, LOW);delay(80);digitalWrite(8, HIGH);delay(80);digitalWrite(6, LOW);delay(80);digitalWrite(7, HIGH);delay(80);digitalWrite(5, LOW);delay(80);digitalWrite(6, HIGH);delay(80);digitalWrite(4, LOW);delay(80);digitalWrite(5, HIGH);delay(80);digitalWrite(3, LOW);delay(80);digitalWrite(4, HIGH);delay(80);digitalWrite(2, LOW);delay(80);digitalWrite(3, HIGH);delay(80);}`
#### Groove
#8
##### Dec 21, 2009, 08:57 amLast Edit: Dec 21, 2009, 08:58 am by GrooveFlotilla Reason: 1
Quote
It would never be connected the GND pin would it ? So its either 5v or 3v, isnt it ?
Quite often what may seem obvious to you is not what most people would expect.
A LED can be connected between a digital output and GND, in which case writing a '1' to the pin will source current through the LED and light it.
If you connect the LED between 5V and the pin, writing a '0 will sink current through the pin, and light the LED.
Quote
i find the reactions of certain users are ' asking about the obvious' and not very friendly
People ask questions because noobs often don't give the whole picture.
They write something like "it isn't working" or "it doesn't do what it is supposed to" or "I wired it up correctly", but without a clear idea of what the program was supposed to do, the rest of us can't possibly give clear sensible answers.
If you get asked for your code, don't just post the bit that you think is wrong; post it all.
If you get asked for a schematic, makes sure it's a clear one.
This is a very knowledgeable and friendly forum, but if someone asks you for more detail, if you want an answer, you'd better provide it. | 2,081 | 6,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-43 | latest | en | 0.69683 |
https://sheetsiq.com/google-sheets/formulas/add-months-to-date-in-google-sheets/ | 1,679,844,291,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00479.warc.gz | 589,542,071 | 14,851 | To add months to date in google sheets you can use the EDATE formula in google sheets.
2. List dates in one column
3. To add +2 months to the A1 cell, write a formula in B1
4. Write =EDATE(A1, 2)
5. Press enter and now B1 will have A1 date+2 months
Once you have added 2 months to date in the B1 cell, drag the cell down to copy the formula for other cells.
Here is an animation showing this in action:
## Table of content
Let’s see an example of adding months to date in google sheets.
We have a list of dates in column A as you see below:
We want to add 1 month, 2 months, and 6 months to these dates. To add months, follow the below steps:
Step-1: Highlight cell B2 in column B where we are adding +1 month
Step-2: Write the formula =EDATE(A2, 1) in cell B2
Step-3: Press enter and you will have 1 month added to A2
Step-4: Drag B2 to all cells to apply the EDATE formula to other cells
• For adding 2 months in column C, use the below formula:
`=EDATE(A2, 2)`
This will add 2 months to column A dates in column C
• For adding 6 months in column D, use the below formula:
`=EDATE(A2, 6)`
This will add 6 months to column A dates in column C
Below is the final result after adding 1 month, 2 months, and 6 months.
As you can see above,
Column B has all cells that are +1 month added to column A.
Column C has all cells that are +2 months added to column A.
Column D has all cells that are +6 months added to column A.
The EDATE formula also takes negative numbers as input. When you pass a negative number to the formula, instead of adding the month, it will subtract the month. Let’s see it in detail.
## Subtract months from the date in google sheets with example
For subtracting months from the date in google sheets, use the below formula:
`=EDATE(date, -months)`
For the same example above, now we can subtract 1 month, 2 months, and 6 months from the dates in column A.
Below is the screenshot showing the example in action:
As you can see from the above,
In column B, we have removed 1 month from each cell in column A.
In column C, we have removed 2 months from each cell in column A.
In column D, we have removed 6 months from each cell in column A.
## FAQ
### How to remove months in google sheets?
To remove or subtract months in google sheets, you can use the EDATE formula and pass a negative number to the second parameter. The formula is =EDATE(date, – number_of_months)
## Wrapping up
In this tutorial, you learned how to add months to date in google sheets. Also, you saw an example of how to subtract months from the date in google sheets.
Adding or subtracting dates is an important calculation while working with dates. Now go ahead and use these formulas in your worksheet. Let me know in the comments if you face any issues. | 704 | 2,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-14 | longest | en | 0.884501 |
https://www.gamedev.net/forums/topic/315666-checking-particle-collisions-with-sprites/ | 1,542,436,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743294.62/warc/CC-MAIN-20181117061450-20181117083450-00245.warc.gz | 896,175,533 | 30,459 | Public Group
# Checking Particle collisions with Sprites
This topic is 4953 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hey everybody! I posted a vector question in this thread. a couple of days ago and in that thread I came upon a way to check collisions with particles and sprites without a lot of pain. However, I feel it is fairly inefficient.
for (std::vector<Sprite*>::iterator i=SpritePool.begin(); i!=SpritePool.end(); i++)
{
for(std::vector<Particle*>::iterator j=ParticlePool.begin(); j!=ParticlePool.end(); j++)
{
//update particles / sprite-particle collision
//kill particles when necessary
}
//update sprites / sprites-sprite collision
//kill sprites when necessary
}
See, I have a particlepool vector and a spritepool vector, and I need to check collisions between elements in the two. This is the best way I can think to do it without re-writing the whole thing to throw everything into one big vector. Give me your thoughts and maybe things you've done related to this. Thanks in advance! toXic1337
##### Share on other sites
The best solution I can think of is to divide the world up into sections and then do your collisions between objects that are in the current or neighbouring sections. You could do this using some kind of tree (like a quad tree) or you could just use a plain 2d array of world sections. You would have to change your code around a bit but its better than your current code where the ammount of checking that has to be done for each particle scales linearly for each object in the entire world, limiting the size of world you can use.
Hope that helps
##### Share on other sites
I understand what you're saying, and it is a good way to limit the number of calculations that are being done each cycle but..
My world is only 800x600, so dividing it up won't really help all that much. I suppose I could divide it into 200x200px sections and that would help. But will the basic algorithm I've stated even work correctly?
It seems as though but it's hard to tell without testing (in class ATM).
Any other ideas? I mean is there a common way of testing particle-sprite collisions? Should they all be in the same vector/list?
Thanks by the way,
toXic1337
##### Share on other sites
If the world is small enough, which 800x600 is, then you should be grand with checking everything against everything like you are doing at the moment. You would only need to make the system more complicated if the world's size was arbitrary.
And I wouldn't be too worried about speed here. Just make sure hat you do bounding box checks first and then per-pixel checks if those succeed if you are using per-pixel collision detection. You'll be surprised with how much you can actually do each frame as long as you aren't too sloppy.
[nitpicking]
And its better and faster to use ++i and ++j instead of i++ and j++ in for loops.
[/nitpicking]
##### Share on other sites
Whats the difference in ++j and j++?
Thanks for the tidbit/nitpicking [lol]
Thanks for your input stro! ratings++; ... er... ++ratings; [lol]
toXic1337
##### Share on other sites
++j returns j after increment. j++ returns j before.
int intplusplus(int j){//////int rtn;rtn=j;j=j+1;return(rtn);}int plusplusint(int j){//////j=j+1;return(j);}
I'm not sure how much difference that makes [especially after optimization] but the thinking is that ++j is one less copy, and less memory management fiddling.
##### Share on other sites
cant remeber where i read it but i think the pre and post increment are only really different on stl containers...
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https://www.jbstudies.com/2022/09/ncert-solution-class12-maths-chapter9-determinants-exercise-9.1.html | 1,696,283,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00162.warc.gz | 896,143,561 | 88,324 | # NCERT Solutions Class 12 Maths Chapter-9 (Determinants) Exercise 9.1
NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Exercise 9.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Class 12 Mathematics
Chapter-9 (Determinants)
Questions and answers given in practice
Chapter-9 (Determinants)
### Exercise 9.1
Q1. Determine order and degree(if defined) of differential equation $\frac{{d}^{4}y}{d{x}^{4}}+\mathrm{sin}\left({y}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\right)=0$
Answer. $\frac{{d}^{4}y}{d{x}^{4}}+\mathrm{sin}\left({y}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\right)=0$ $⇒{y}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}+\mathrm{sin}\left({y}^{m}\right)=0$ The highest order derivative present in the differential equation is ${y}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$. Therefore, its order is four. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.
Q2. Determine order and degree(if defined) of differential equation ${y}^{\mathrm{\prime }}+5y=0$
Answer. ${y}^{\mathrm{\prime }}+5y=0$
Q3. Determine order and degree(if defined) of differential equation $\left(\begin{array}{c}\left(\frac{ds}{dt}{\right)}^{4}+3s\frac{{d}^{2}s}{d{t}^{2}}=0\end{array}\right).$
Answer. $\left(\begin{array}{c}\left(\frac{ds}{dt}{\right)}^{4}+3s\frac{{d}^{2}s}{d{t}^{2}}=0\end{array}\right).$
Q5.
Q7.
Q8.
Q9.
Q10. | 554 | 1,825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-40 | latest | en | 0.64771 |
https://www.physicsforums.com/threads/kinematics-problem-that-i-cant-seem-to-resolve.527383/ | 1,550,514,750,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00286.warc.gz | 923,132,240 | 12,398 | # Kinematics problem that i cant seem to resolve
1. Sep 5, 2011
### adoule
1. The problem statement, all variables and given/known data
A truck starts from rest and accelerates at 1 M/S*2 . 4 s later, a car accelerates from rest at the same starting point with an acceleration of 2.7 M/S*2 .
Where and when does the car catch the truck?
2. Relevant equations
i have tried s=ut + (1/2)at^2
3. The attempt at a solution
In 4 seconds, truck travels a distance of 0 (4) + (1/2) (1) (4)^2 = 0 + 8 = 8
now, when car starts, truck is already 8 m ahead of it.
Suppose they meet at time lapse of t seconds after car starts.
So, car travels 8 m more than the truck in t seconds.
distance travelled by car in t seconds = dist travelld by truck in t seconds + 8
0 (t) + (1/2) (2.7) (t)^2 = 0 ( t) + (1/2) (1) (t)^2 + 8
==> (t)^2 = (160) / 17
t = 3.06
hence, car catches truck at 4 + 3.06 = 7.06 seconds after truck starts.
distance travelled by car = 0 (7.06) + (1/2) (4) ( 160/17) = 18.8 metres, but thats not the anwser :s
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Sep 5, 2011
### lewando
Yo, why 2 threads? I know, in theory it should speed things up :tongue:.
3. Sep 5, 2011
### RTW69
The distance the truck travels will equal the distance the car travels. Try setting your distance formulas equal to each other and see what you get.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 481 | 1,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-09 | latest | en | 0.903471 |
https://www.webull.com/news/11360429783933952 | 1,725,855,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00881.warc.gz | 1,014,721,718 | 60,682 | # Is Wesfarmers Limited's (ASX:WES) Recent Stock Performance Tethered To Its Strong Fundamentals?
Simply Wall St · 08/30 21:30
Wesfarmers (ASX:WES) has had a great run on the share market with its stock up by a significant 12% over the last three months. Given the company's impressive performance, we decided to study its financial indicators more closely as a company's financial health over the long-term usually dictates market outcomes. Specifically, we decided to study Wesfarmers' ROE in this article.
Return on Equity or ROE is a test of how effectively a company is growing its value and managing investors’ money. In other words, it is a profitability ratio which measures the rate of return on the capital provided by the company's shareholders.
See our latest analysis for Wesfarmers
## How To Calculate Return On Equity?
The formula for ROE is:
Return on Equity = Net Profit (from continuing operations) ÷ Shareholders' Equity
So, based on the above formula, the ROE for Wesfarmers is:
30% = AU\$2.6b ÷ AU\$8.6b (Based on the trailing twelve months to June 2024).
The 'return' is the profit over the last twelve months. Another way to think of that is that for every A\$1 worth of equity, the company was able to earn A\$0.30 in profit.
## What Is The Relationship Between ROE And Earnings Growth?
So far, we've learned that ROE is a measure of a company's profitability. Based on how much of its profits the company chooses to reinvest or "retain", we are then able to evaluate a company's future ability to generate profits. Assuming everything else remains unchanged, the higher the ROE and profit retention, the higher the growth rate of a company compared to companies that don't necessarily bear these characteristics.
## Wesfarmers' Earnings Growth And 30% ROE
First thing first, we like that Wesfarmers has an impressive ROE. Secondly, even when compared to the industry average of 8.4% the company's ROE is quite impressive. This likely paved the way for the modest 7.6% net income growth seen by Wesfarmers over the past five years.
As a next step, we compared Wesfarmers' net income growth with the industry and found that the company has a similar growth figure when compared with the industry average growth rate of 7.9% in the same period.
Earnings growth is a huge factor in stock valuation. The investor should try to establish if the expected growth or decline in earnings, whichever the case may be, is priced in. This then helps them determine if the stock is placed for a bright or bleak future. Is Wesfarmers fairly valued compared to other companies? These 3 valuation measures might help you decide.
## Is Wesfarmers Efficiently Re-investing Its Profits?
Wesfarmers has a significant three-year median payout ratio of 87%, meaning that it is left with only 13% to reinvest into its business. This implies that the company has been able to achieve decent earnings growth despite returning most of its profits to shareholders.
Besides, Wesfarmers has been paying dividends for at least ten years or more. This shows that the company is committed to sharing profits with its shareholders. Our latest analyst data shows that the future payout ratio of the company over the next three years is expected to be approximately 87%. Accordingly, forecasts suggest that Wesfarmers' future ROE will be 34% which is again, similar to the current ROE.
## Summary
On the whole, we feel that Wesfarmers' performance has been quite good. Especially the high ROE, Which has contributed to the impressive growth seen in earnings. Despite the company reinvesting only a small portion of its profits, it still has managed to grow its earnings so that is appreciable. We also studied the latest analyst forecasts and found that the company's earnings growth is expected be similar to its current growth rate. To know more about the latest analysts predictions for the company, check out this visualization of analyst forecasts for the company. | 873 | 3,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.948816 |
https://www.takshilalearning.com/matrices-matrix-for-ncert-maths-solutions-class-12/ | 1,680,232,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00224.warc.gz | 1,113,281,779 | 83,098 | # NCERT Maths Solutions Class 12 for Matrices – (Matrix)
## NCERT Maths Solutions Class 12 for Matrices – (Matrix)
12th Mathematics: The arrangement of real numbers in a rectangular array enclosed in brackets as [] or () is known as a Matrix (Matrices is plural of the matrix). Matrix operations are used in electronic physics, computers, budgeting, cost estimation, analysis, and experiments. They are also used in cryptography, modern psychology, genetics, industrial management etc. In general, an m x n matrix is matrix having m rows and n columns. it can be written as follows:
Order of a Matrix
There may be any number of rows and any number of columns in a matrix. If there are m rows and n columns in matrix A, its order is m x n and it is read as an m x n matrix.
Transpose of a Matrix
The transpose of a given matrix A is formed by interchanging its rows and columns and is denoted by A’.
Symmetric Matrix
A square matrix A is said to be a symmetric matrix if A’ = A.
Skew-Symmetric Matrix
A square matrix A is said to be a skew-symmetric if A’ = – A. all elements in the principal diagonal of a skew-symmetric matrix are zeroes.
If A and B are any two given matrices of the same order, then their sum is defined to be a matrix C whose respective elements are the sum of the corresponding elements of the matrices A and B and we write this as C = A + B.
Types:
Row matrix: A row matrix has only one row but any number of columns.
Column matrix: A column matrix has only one column but any number of rows.
Square matrix: A square matrix has the number of columns equal to the number of rows.
Rectangular Matrix: A matrix is said to be a rectangular matrix if the number of rows is not equal to the number of columns.
Diagonal matrix: a matrix having non-zero elements only in the diagonal running from the upper left to the lower right.
Scalar Matrix: A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant.
Zero or Null matrix: If all elements of a matrix are zero, then the matrix is known as zero matrices and denoted by O.
Unit or Identity matrix: If in a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I.
Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal.
Properties of Matrix
• When a matrix is multiplied by a scalar, then each of its elements is multiplied by the same scalar.
• If A and B are any two given matrices of the same order, then their sum is defined to be a matrix C whose respective elements are the sum of the corresponding elements of the matrices A and B and we write this as C = A + B.
• For any two matrices A and B of the same order, A + B = B + A. i.e. matrix addition is commutative.
• For any three matrices A, B and C of the same order, A + (B + C) = (A + B) + C i.e., matrix addition is associative.
• Additive identity is a zero matrix, which when added to a given matrix, gives the same given matrix, i.e., A + O = A = O + A.
• If A + B = O, then the matrix B is called the additive inverse of the matrix of A.
• If A and B are two matrices of order m x p and p x n respectively, then their product will be a matrix C of order m x n.
Invertible Matrix
A square matrix of order n is invertible if there exists a square matrix B of the same order such that AB = I = BA, Where I is identity matrix of order n.
Theorems of invertible matrices
• Theorem 1:Every invertible matrix possesses a unique inverse.
• Theorem 2:A square matrix is invertible if it is non-singular.
Example. Construct a 3 × 2 matrix whose elements are given by aij= 1/2|i-3j|
Solution. In general, a 3 × 2 matrix is given by A=
Now aij= ½ |i-3j| ,i = 1, 2, 3 and j = 1, 2.
Therefore a11= ½ |1-3*1|a12= ½|1-3*2| = 5/2
a21= ½ |2-3*1| = 1/2 a22 = ½ |2-3*2| = 2
a31 = ½ |3-3*1| = 0 a32 = ½ |3-3*2| = 3/2
Hence the required matrix is given by A=
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August 26, 2019 | 1,172 | 4,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-14 | latest | en | 0.903604 |
https://www.physicsforums.com/threads/series-of-even-and-odd-subsequences-converge.956951/ | 1,653,115,910,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00777.warc.gz | 1,093,712,561 | 19,348 | # Series of even and odd subsequences converge
I know the result that if ##\lim a_{2n} = L = \lim a_{2n-1}##, then ##\lim a_n = L##. I'm wondering, can this be generalized to series? i.e., if ##\displaystyle \sum_{n=1}^{\infty}a_{2n-1}## and ##\displaystyle\sum_{n=1}^{\infty}a_{2n}## converge, then ##\displaystyle \sum_{n=1}^{\infty}a_{n}## converges also? I think I can see how I would prove it using the definition of convergence, but is there a slicker way to prove it using just the algebraic properties of series?
FactChecker
Gold Member
If the individual series are absolutely convergent, then they can be rearranged at will and the combined series will converge. Otherwise, the convergence depends on the order of the terms and anything can happen.
If the individual series are absolutely convergent, then they can be rearranged at will and the combined series will converge. Otherwise, the convergence depends on the order of the terms and anything can happen.
So this hold only if the the two series are absolutely convergent?
fresh_42
Mentor
2021 Award
So this hold only if the the two series are absolutely convergent?
If they do, you can use the algebraic property you've asked for, because the order is irrelevant and we can add the infinite sums.
Otherwise you need the limit argument:
##(L_{2n}+L_{2n-1})-\varepsilon =L_{2n}-\varepsilon/2 + L_{2n-1}-\varepsilon/2 \leq S_n = S_{2n}+S_{2n-1} \leq L_{2n}+\varepsilon/2 + L_{2n-1}+\varepsilon/2 = (L_{2n}+L_{2n-1})+\varepsilon##
with the partial sums ##S_{*}## and its limits ##L_{*}##. Then ##\lim_{n \to \infty}S_n= L_{2n}+L_{2n-1}##.
WWGD
Gold Member
A series can be seen as a sequence of partial sums.
FactChecker
Gold Member
So this hold only if the the two series are absolutely convergent?
If any one of them is NOT absolutely convergent, then either the sum of the negative tems is ##-\infty## or the sum of the positive terms is ##+\infty##, or both. That gives enough terms of one sign to swing the partial sums as far as you want in that direction, over and over again. So that situation will also be true for the combined series. The result depends on how the terms are ordered and you can never count on the partial sums settling down. In general, that makes it impossible to apply algebraic properties.
If both series are absolutely convergent, then there is a limited ability of the terms of one sign to swing the sum in that direction. That also remains true of the combined series. The partial sums are well behaved and the final limits are the same regardless of the order of the terms. Algebraic properties can be applied.
fresh_42
Mentor
2021 Award
If any one of them is NOT absolutely convergent, then either the sum of the negative terms is ##\infty## or the sum of the positive terms is ##\infty##, or both.
Can you prove this, please? The minus signs don't have to be distributed in an alternating way.
FactChecker
Gold Member
Can you prove this, please? The minus signs don't have to be distributed in an alternating way.
I assume that we are talking about real numbers, not complex. Then, sure, I can prove it. If ##\sum \mid a_n \mid = \infty##, then either the sum of the absolute values of the negative terms is infinite or the sum of the positive terms is infinite or both.
fresh_42
Mentor
2021 Award
How? If both partial series were finite, then their sum is. But how can we conclude, that their sum also has to be absolute finite?
FactChecker
Gold Member
How? If both partial series were finite, then their sum is. But how can we conclude, that their sum also has to be absolute finite?
That is the opposite of what I said.
fresh_42
Mentor
2021 Award
That is the opposite of what I said.
No, it's the negation, not the opposite. ##\sum a_{2n} < \infty \wedge \sum a_{2n-1}<\infty \Longrightarrow \sum a_n<\infty##. You stated the stronger statement that ##\sum |a_n|<\infty## which I do not see. This is equivalent to what you said, namely ##\sum |a_n| = \infty \Longrightarrow \sum a_{2n} = \infty \vee \sum a_{2n-1}=\infty##.
I have no counterexample at hand, so I currently cannot rule out, that it is correct, but I doubt it. In any case, it is not obvious. We cannot conclude from convergence to absolute convergence, so your stronger statement needs to be proven. The point is, that convergence + not absolute convergence does not imply, that the distribution of signs goes along ##2n / 2n-1##, which is why I cannot see how you can conclude especially on these two subsequences.
For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.
FactChecker
Gold Member
For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.
Oh. I see your point. I meant to say that absolute convergence is the only way it is safe to apply simple rules of algebra as the OP asked. Otherwise, one must consider the order and it gets messy.
fresh_42
Mentor
2021 Award
That was what I said in post #4.
mathwonk
Homework Helper
just off the top of my head,
how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......
Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence. (OOPs! claim retracted below.)
Last edited:
fresh_42
Mentor
2021 Award
just off the top of my head,
how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......
Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence.
But if both subseries converge, then by transition to partial sums we can add those and the sum again converges. Was my argument wrong in post #4?
FactChecker | 1,648 | 5,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-21 | latest | en | 0.859201 |
https://forum.allaboutcircuits.com/threads/max-power-transfer.20390/ | 1,566,708,901,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323067.50/warc/CC-MAIN-20190825042326-20190825064326-00198.warc.gz | 461,608,070 | 18,282 | # Max power transfer
Discussion in 'Homework Help' started by minduka, Mar 8, 2009.
1. ### minduka Thread Starter New Member
Mar 5, 2009
2
0
Hi
I need some help on two max power transfer questions.
This is my approach to the first one.
$R_{L}= 8 \Omega$ - because max power occurs when $R_{TH}$ = $R_{L}$
Then I found the max power transfered in (b) by using
$P_{L,Max} = \frac{v_{L}^{2}}{4R_{TH}}$
$V_{L} = IR = 2A*8\Omega = 16v$
$R_{TH} = 16//8//16 = 4\Omega$
$P_{L,Max} = \frac{16^{2}}{16} = 16$
I'm confused when they give you $R_{L}$ and need to find the max power. I'm assuming you have to use $R_{L}$ to find the max power.
-----------------------------------------------------------------------
For the second problem:
$R_{TH} = 10//10+10 = 15\Omega$
$V = 20v$
$P = \frac{20^{2}}{4*15} = 6.67 W$
• ###### 1.jpg
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Last edited: Mar 8, 2009
2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
790
Hello minduka,
In the first question you recognized that RL = 8Ω for the left hand circuit.
The questioner is not asking for the maximum power transfer case for the right hand circuit - they are asking you to place the load RL = 8Ω into the right hand circuit and then solve for the power dissipated in the load, which with RL = 8Ω would give 32W - as indicated by the correct answer.
In the second question, you are nearly there, but just need to realize that the voltage you should use is not 20V. Suppose you reduce the circuit to its Thevenin equivalent. It would be a 10V source in series with a 15Ω resistance. So the maximum power case is ((10/2)λ2)/15 = 1.67W.
Hope this helps.
3. ### minduka Thread Starter New Member
Mar 5, 2009
2
0
I think the first problem is a little decieving
4. ### nirosha New Member
Nov 6, 2008
1
0
In the first problem,by using voltage division rule we get voltage across RL.use this voltage and given 2watts to find RL then place this RL value in the next ckt and solve the rest.
5. ### Ratch New Member
Mar 20, 2007
1,068
4
minduka,
Not necessily so. If the source impedance and voltage cannot be changed, then yes, max power occurs when the load impedance equals the source impedance. But if you can make the source impedance smaller while keeping the source voltage the same, then you will transfer more power with the smaller source impedance.
Ratch | 698 | 2,342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 12, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-35 | latest | en | 0.873782 |
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## Help with SUMX & IF
Hi All
I tried searching the forums for previous posts but all existing examples are so specific that none of the solutions were able to help me.
I am sure this is going to be a simple fix but I am having some beginers issues unfortunately.
---
Basically, I work out our SLA for our phone team but when the team gets 0 calls the SLA is at 100%. This confuses some readers of my report and so I want to build a DAX measure that would say:
```IF(Phones[Offered]=>0, "N/A"
ELSE
(perform equation to work out SLA%)
SUMX(Phones,Phones[SL]*Phones[Offered])/sum(Phones[Offered])```
Any help on this would be greatly appreciated.
Kind regards
Paul
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Super User II
## Re: Help with SUMX & IF
I assume the SLA should be calculated for Phones Offered >0, if it is as in your example it's easy to correct
```SLA% =
IF (
SUM ( Phones[Offered] ) > 0,
DIVIDE (
SUMX ( Phones, Phones[SL] * Phones[Offered] ),
SUM ( Phones[Offered] )
),
"N/A"
)```
if this doesn't work can you paste here few rows of sample data from Phones table? should be anonymised
Thank you for the kudos 🙂
Proud to be a Super User!
2 REPLIES 2
Highlighted
Super User II
## Re: Help with SUMX & IF
I assume the SLA should be calculated for Phones Offered >0, if it is as in your example it's easy to correct
```SLA% =
IF (
SUM ( Phones[Offered] ) > 0,
DIVIDE (
SUMX ( Phones, Phones[SL] * Phones[Offered] ),
SUM ( Phones[Offered] )
),
"N/A"
)```
if this doesn't work can you paste here few rows of sample data from Phones table? should be anonymised
Thank you for the kudos 🙂
Proud to be a Super User!
Highlighted
## Re: Help with SUMX & IF
Many thanks @Stachu that worked perfectly!!
Kind regards
Paul | 498 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-40 | latest | en | 0.885576 |
https://math.stackexchange.com/questions/2759028/the-relationship-between-positive-operators-and-positive-matrices-and-forms | 1,717,000,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00401.warc.gz | 310,726,323 | 35,712 | # The relationship between positive operators and positive matrices and forms
I'm a little confused about the definitions of positivity. I'm following Linear Algebra by Hoffmann and Kunze, and a positive operator is defined to be any operator such that $$T$$ is self adjoint and $$$$ is positive for a non-zero vector $$v$$. The positive matrix on the other hand is a Hermitian matrix which has the property $$X^{*}AX>0$$ for any complex $$nx1$$ matrices.
Can we say any operator is positive if and only if its matrix wrt an orthonormal basis is positive?
Similarly we have defined a positive sesqui linear form $$f$$ which has property $$f(v,v)>0$$ for any non- zero $$v$$, and proved $$f$$ is positive if and only if its matrix is positive, but we also have showed that $$f$$ is one to one correspondence with the set of operators on $$V$$, so again can we link the positivity on these two things?
• What inner product(s) are you interested in? Apr 29, 2018 at 16:59
• I think the definitions should work for any inner product? Apr 29, 2018 at 17:29
• $X^*AX$ is (a $1\times 1$-matrix, which, when identified with a number, is) $\langle AX,X\rangle$. This answers the question for the canonical basis. When working over $\Bbb C$, "orthogonal base change" should be replaced by "unitary base change". Such a base change via $U$ changes then $A$ in $U^*AU$, so for instance $X^*U^*AUX=(UX)^*A(UX)$, and $Y=UX$ covers all $n\times 1$ matrices, when $X$ does it. Apr 29, 2018 at 17:50
• $X^{*}AX > 0$ is associated with the inner product $\langle x,y \rangle=y^{H}x$ making $\langle AX,X \rangle > 0$. You can define some other inner product and perhaps change the situation. Apr 29, 2018 at 19:51
Yes, you can link the positivity of operator and sesqui linear form in the way that w.r.t. orthonormal basis matrix of sesqui linear form with respect to that basis is same as matrix of the operator $$T$$ which determines that form. If $$T$$ is positive then, form will be positive and correspondingly, matrices are positive. | 546 | 2,026 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.917088 |
http://www.jiskha.com/display.cgi?id=1290388141 | 1,495,997,693,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610374.3/warc/CC-MAIN-20170528181608-20170528201608-00075.warc.gz | 677,533,825 | 3,691 | # Math
posted by on .
How long will it take for an investment to double in value if it earns 9.5% compounded continuously?
• Math - ,
For r=0.095, with normal compound interest, compounded yearly, the number of years to double at a rate of r% is ln(2)/ln(1+r)=ln(2)/ln(1.095)=7.638 years.
from A=P(1+r)^n
take ln both sides,
ln(A/P)=n ln(1+r)
n=ln(2)/ln(1+r)
With continuous compounding,
A=Pern
Take logs
ln(A/P)=ern
ln(2)=rn
n=ln(2)/r
=ln(2)/0.095
=7.296 years | 168 | 467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-22 | latest | en | 0.854043 |
http://gmatclub.com/forum/investors-75350.html?fl=similar | 1,485,044,330,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00176-ip-10-171-10-70.ec2.internal.warc.gz | 118,443,110 | 48,730 | investors : GMAT Verbal Section
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# Events & Promotions
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# investors
Author Message
VP
Joined: 18 May 2008
Posts: 1286
Followers: 16
Kudos [?]: 411 [0], given: 0
### Show Tags
03 Feb 2009, 03:15
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Senior Manager
Joined: 26 May 2008
Posts: 431
Schools: Kellogg Class of 2012
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### Show Tags
03 Feb 2009, 03:28
I'll go with A
'Not one' is singular and 'if' has to be accompanied by 'were' not 'was'. Take out B, D and E. Between A and C, A is good because 'them' is ambiguous is C
Cheers,
unplugged
Director
Joined: 23 May 2008
Posts: 838
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### Show Tags
03 Feb 2009, 10:04
A for me too
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
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### Show Tags
03 Feb 2009, 17:17
good question.
D,E -- out.. one of investors .. singular
if the deal were not be concluded ..(talking about unreal ) subjunctive.. mood..
something .. like If I were rich..
B "was" incorrect-- out
between A and C
C -- them is ambigous.
Also agreement "be" --> be is incorrect it should be "is"
A is the best
_________________
Smiling wins more friends than frowning
Senior Manager
Joined: 02 Nov 2008
Posts: 282
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### Show Tags
03 Feb 2009, 18:33
x2suresh wrote:
good question.
D,E -- out.. one of investors .. singular
if the deal were not be concluded ..(talking about unreal ) subjunctive.. mood..
something .. like If I were rich..
B "was" incorrect-- out
between A and C
C -- them is ambigous.
Also agreement "be" --> be is incorrect it should be "is"
A is the best
C is also a bit wordy, so A for me to.
VP
Joined: 18 May 2008
Posts: 1286
Followers: 16
Kudos [?]: 411 [0], given: 0
### Show Tags
04 Feb 2009, 01:11
I also wanted 2 go for A. but 'were' in the end for singular deal confused me. I think wht u said abt unreal things is the answer 2 my question. Can u pls elaborate a bit on that?
x2suresh wrote:
good question.
D,E -- out.. one of investors .. singular
if the deal were not be concluded ..(talking about unreal ) subjunctive.. mood..
something .. like If I were rich..
B "was" incorrect-- out
between A and C
C -- them is ambigous.
Also agreement "be" --> be is incorrect it should be "is"
A is the best
Re: investors [#permalink] 04 Feb 2009, 01:11
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Display posts from previous: Sort by | 983 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-04 | latest | en | 0.891006 |
http://vedhavidhi.com/index.php/2022/05/27/the-combat-fun-math-games-for-kids/ | 1,721,144,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00225.warc.gz | 35,453,484 | 9,693 | # The Combat Fun Math Games For Kids
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Webslavery [pvc_stats postid="" increase="1" show_views_today="1"] | 869 | 4,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.929747 |
http://www.numdam.org/item/AIF_1991__41_3_651_0/ | 1,652,925,309,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00548.warc.gz | 99,913,814 | 8,136 | Unique continuation for the solutions of the laplacian plus a drift
Annales de l'Institut Fourier, Tome 41 (1991) no. 3, pp. 651-663.
Nous prouvons l’unicité du prolongement pour les solutions de l’inégalité $|\Delta u\left(x\right)|\le V\left(x\right)|\nabla u\left(x\right)|$, $x\in \Omega$$\Omega$ est une partie connexe de ${\mathbf{R}}^{n}$ et $V$ appartient aux espaces de Morrey ${F}^{\alpha ,p}$, avec $p\ge \left(n-2\right)/2\left(1-\alpha \right)$ et $\alpha <1$. Ces espaces contiennent ${L}^{q}$ pour $q\ge \left(3n-2\right)/2$ (voir L. Hörmander, Comm. PDE, 8 (1983, 21-64 et Barceló, Kenig, Ruiz, Sogge, Ill. J. of Math., 32-2 (1988), 230-245).
We prove unique continuation for solutions of the inequality $|\Delta u\left(x\right)|\le V\left(x\right)|\nabla u\left(x\right)|$, $x\in \Omega$ a connected set contained in ${\mathbf{R}}^{n}$ and $V$ is in the Morrey spaces ${F}^{\alpha ,p}$, with $p\ge \left(n-2\right)/2\left(1-\alpha \right)$ and $\alpha <1$. These spaces include ${L}^{q}$ for $q\ge \left(3n-2\right)/2$ (see [H], [BKRS]). If $p=\left(n-2\right)/2\left(1-\alpha \right)$, the extra assumption of $V$ being small enough is needed.
@article{AIF_1991__41_3_651_0,
author = {Ruiz, Alberto and Vega, Luis},
title = {Unique continuation for the solutions of the laplacian plus a drift},
journal = {Annales de l'Institut Fourier},
pages = {651--663},
publisher = {Institut Fourier},
volume = {41},
number = {3},
year = {1991},
doi = {10.5802/aif.1268},
zbl = {0772.35008},
mrnumber = {92k:35043},
language = {en},
url = {http://www.numdam.org/articles/10.5802/aif.1268/}
}
TY - JOUR
AU - Ruiz, Alberto
AU - Vega, Luis
TI - Unique continuation for the solutions of the laplacian plus a drift
JO - Annales de l'Institut Fourier
PY - 1991
DA - 1991///
SP - 651
EP - 663
VL - 41
IS - 3
PB - Institut Fourier
PP - Grenoble
UR - http://www.numdam.org/articles/10.5802/aif.1268/
UR - https://zbmath.org/?q=an%3A0772.35008
UR - https://www.ams.org/mathscinet-getitem?mr=92k:35043
UR - https://doi.org/10.5802/aif.1268
DO - 10.5802/aif.1268
LA - en
ID - AIF_1991__41_3_651_0
ER -
Ruiz, Alberto; Vega, Luis. Unique continuation for the solutions of the laplacian plus a drift. Annales de l'Institut Fourier, Tome 41 (1991) no. 3, pp. 651-663. doi : 10.5802/aif.1268. http://www.numdam.org/articles/10.5802/aif.1268/
[BKRS] Barceló, Kenig, Ruiz, Sogge, Weighted Sobolev inequalities and unique continuation for the Laplaciaan plus lower order terms, III. J. of Math., 32, n.2 (1988), 230-245. | MR 89h:35048 | Zbl 0689.35015
[C] S. Campanato, Proprietá di inclusione per spazi di Morrey, Ricerche Mat., 12 (1963), 67-896. | MR 27 #6157 | Zbl 0192.22703
[CS] S. Chanillo, E. Sawyer, Unique continuation for ∆ + V and the C. Fefferman-Phong class, preprint. | Zbl 0702.35034
[ChR] F. Chiarenza, A. Ruiz, Uniform L2 weighted inequalities, Proc. A.M.S., to appear. | Zbl 0745.35007
[ChF] F. Chiarenza, M. Frasca, A remark on a paper by C. Fefferman, Proc. A.M.S., (Feb. 1990), 407-409. | MR 91a:46030 | Zbl 0694.46029
[FeP] C. Fefferman, D.H. Phong, Lower bounds for Schrödinger equations, Journées Eqs. aux deriv. partielles, St. Jean de Monts, 1982. | Numdam | Zbl 0492.35057
[GL] N. Garofalo, F.H. Lin, Unique continuation for elliptic operators; a geometric variational approach, Comm. Pure Appl. Math., 40 (1987), 347-366. | MR 88j:35046 | Zbl 0674.35007
[H] L. Hörmander, Uniqueness theorem for second order differential operators, Comm. PDE, 8 (1983), 21-64. | Zbl 0546.35023
[Je] D. Jerison, Carleman inequalities for the Dirac and Laplace operators and unique continuation, Adv. In Math., 63 (1986), 118-134. | MR 88b:35218 | Zbl 0627.35008
[Jo] J.L. Journé, Calderón-Zygmund operators, pseudo-differential operators, and the Cauchyintegral of Calderón, Lecture Notes in Math., Springer Verlag, 1983. | Zbl 0508.42021
[K] C. Kenig, Restriction theorems, Carleman estimates, uniform Sobolev inequalities and unique continuation. Harmonic Analysis and PDE'S, Proceedings El Escorial 1987, Springer Verlag, 1384, (1989), 69-90. | Zbl 0685.35003
[P] J. Peetre, On the theory of Lp,λ spaces, J. Funct. Anal., 4 (1969), 71-87. | MR 39 #3300 | Zbl 0175.42602
[RV] A. Ruiz, L Vega, Unique continuation for Schrödinger operators in Morrey spaces, preprint. | Zbl 0809.47046
[St] G. Stampacchia, L(p,λ)-spaces and interpolation, Comm. on Pure and Appl. Math., XVII (1964), 293-306. | MR 31 #2608 | Zbl 0149.09201
[Se] E. Stein, Oscillatory integrals in Fourier Analysis. In: Beijing Lectures in Harmonic Analysis, Princeton Univ. Press, 112 (1986), 307-355. | MR 88g:42022 | Zbl 0618.42006
[T] P. Tomas, A restriction theorem for the Fourier transform, Bull. AMS, (1975), 477-478. | MR 50 #10681 | Zbl 0298.42011
Cité par Sources : | 1,781 | 4,782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.380798 |
http://wmbriggs.com/post/11231/ | 1,521,662,992,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647692.51/warc/CC-MAIN-20180321195830-20180321215830-00147.warc.gz | 336,228,516 | 15,687 | # Will Someone Please Laugh At These Jokes?
Well, I think these are funny.
Update And then for something not as funny.
Update This is for my number-two son.
1. MattS says:
What jokes?
2. JH says:
Yes, let’s have a Goofball day before the Super Bowl.
What did one statistician say to the other? “You are mean.”
3. Briggs says:
JH,
4. Mike Ozanne says:
Was it significant?
5. JH says:
Oh.. greater omentum is missing in the gut pic.
6. JH says:
Briggs, thank you! I am indeed a kind person. ^_*
7. Ye Olde Statisician says:
Drinking too much beer before the Super Bowl increases your p-value.
8. PaulH says:
Ha! A plate tectonics joke! 🙂
9. JH says:
YOS, p-value = potbelly value. Ha.
10. MattS says:
How many statisticians does it take to screw in a light bulb?
Two, but don’t ask me how they got in the light bulb.
None, they take the mean of the light and the darkness and think they can still see.
11. Bruce Foutch says:
“STUDY: People slow to react more likely to die prematurely…”
12. Ray says:
Statistics show that the celebration of birthdays is healthy. The people who celebrate the most birthdays live longest.
13. Scotian says:
This is the best all time statistician joke:
A statistics major was completely hung over the day of his final exam. It was a True/False test, so he decided to flip a coin for the answers. The stats professor watched the student the entire two hours as he was flipping the coin…writing the answer…flipping the coin…writing the answer. At the end of the two hours, everyone else had left the final except for the one student. The professor walks up to his desk and interrupts the student, saying:
“Listen, I have seen that you did not study for this statistics test, you didn’t even open the exam. If you are just flipping a coin for your answer, what is taking you so long?”
The student replies bitterly, as he is still flipping the coin: “Shhh! I am checking my answers!”
After taking a quick poll of my geologist colleagues I found that 40% think your lead joke is funny, 40% did not, and 20% did not get it.
14. Scotian says:
Of course the best economist joke goes:
An economist and a statistician are walking down the street together. The statistician says “Hey, look, there’s a \$20 bill on the sidewalk!†The economist replies by saying “That’s impossible- if it were really a \$20 bill, it would have been picked up by now.â€
15. Bruce Foutch says:
Anytime you think you’ve gotten in the last word in an argument with your wife, you need to understand that what you thought was the last word is just the beginning of the next argument. 🙁
16. Kip Hansen says:
I tried a mean of the three, still not funny.
What the trick? Anyone?
I’m gonna have to get out my Stats 101 book and se what I missed, has to be something in there somewhere….
17. Rich says:
An astronomer, a physicist and a mathematician are traveling through Scotland by train. They pass a field with a black sheep in it.
“Ooh, look!”, says the astronomer, “Scottish sheep are black!”
“No, no!”, says the physicist, “some Scottish sheep are black.”
The mathematician sighed. “In Scotland”, he said, “there is at least one field in which there is at least one sheep, at least one side of which is black.”
The other two throw him out of the window. | 828 | 3,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-13 | longest | en | 0.958987 |
https://community.powerbi.com/t5/DAX-Commands-and-Tips/DAX-Need-help-with-Weeks-of-supply-calculation/m-p/1218227 | 1,597,267,098,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00235.warc.gz | 257,675,913 | 92,153 | cancel
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Regular Visitor
## DAX: Need help with Weeks of supply calculation
Hi,
I am working on a DAX solution, where I would need to see the result in the below format.
I would like to see how many weeks it takes for the forecast to run out based on the on-hand value. I have rounded the weeks in the result.
Thanks for the help
Latha
5 REPLIES 5
Highlighted
Super User II
## Re: DAX: Need help with Weeks of supply calculation
what's a logic for your Result?
do not hesitate to give a kudo to useful posts and mark solutions as solution
LinkedIn
Highlighted
Regular Visitor
## Re: DAX: Need help with Weeks of supply calculation
I am checking how many weeks it takes the sum of the forecast to reach on-hand value.
Example: In the first row, on-hand is negative so I would need to get 0.
for the second row, 6 weeks of forecast (733+733+857+857+857+3099) is required to get the on-hand 6399.
Please let me know if you further have any question.
Highlighted
Super User II
## Re: DAX: Need help with Weeks of supply calculation
it's unclear for me why exactly 6 weeks for second row.
and to be clear, you need DAX for new Result column (highlited as yellow) or for SUM 733+733+857+857+857+3099 ?
do not hesitate to give a kudo to useful posts and mark solutions as solution
LinkedIn
Highlighted
Regular Visitor
## Re: DAX: Need help with Weeks of supply calculation
I want to do the calculation for the results column highlighted. And to clarify, I am counting the number of forwarding forecast rows needed to sum to equal the ending on-hand value. 6 is a rounded value in the above example, but the actual value would be 5. 6 approx.
Highlighted
Regular Visitor
## Re: DAX: Need help with Weeks of supply calculation
I was able to solve this problem. thanks.
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Top Kudoed Authors | 570 | 2,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-34 | latest | en | 0.896476 |
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# SUB: GEOTECH
MODULE: 43-46
TEST-1
________________________________________________________________________________________
1. If number of flow channels are 10, number of drops are 6, the total hydraulic head difference is 4m, the length of dam is 50m and the coefficient of permeability is 2 , the total loss of water per day in is a) 576 b) 5760 c) 57600 d) 576000 A flow line makes angles and with the normal to the interface of the soils having permeabilities 3 and 1 before and after deflection. According to the law of deflection of flow lines at the interface of the dissimilar soils, the is a) b) c) d) A compacted soil sample using 10% moisture content has a weight of 200g and mass unit weight of 2.0g/cc. if the specific gravity of soil particles and water are 2.7 and 1.0, the degree of saturation of the soil is a) 11.1% b) 55.6% c) 69.6% d) 80% Sieve analysis was carried out on a soil sample of 400gms. Out of 400gms, of soil 160gm retained on 4.75mm sieve, 120gm retained on 2mm sieve and 80gm retained on 600 sieve. The value of will be a) 7.4, 1.7 b) 1.7, 7.4 c) 7.92, 1.4 d) 1.4, 7.92 Degree of saturation of a natural soil deposit having content 15% specific gravity 2.50 and void ratio 0.5 is a) 50% b) 60% c) 75% d) 80% The liquidity index is defined as a ratio expressed as percentage of a) Plastic limit minus natural water content, to its plasticity index b) Natural water content minus its plastic limit to its plasticity index c) Natural water content plus in elastic limit, to its plasticity index d) Liquid limit minus natural water content to the plasticity index. 7. The weight of a Pycnometer containing 400g sand and water full to the top is 2150gr. The weight of Pycnometer full of clean water is 1950g. if specific gravity of the soil is 2.5, the water content is a) 5% b) 10% c) 15% d) 20% A 10m thick layer of clay is under lained by a 4m thick layer of sand. The water table is at a depth of 4m from the ground level. The void ratio of clay is 0.65 and specific gravity is 2.67 water content above the saturated soil is 17% and void ratio of the sand is 0.567 and G= 2.65. Determine total and effective stress. A barrow pit soil has a dry density of 17KN/ . How many cubic meters of this soil will be required to construct an embankment of 100 volume with a dry density of 16KN/ a) 94 b) 106 c) 100 d) 90
8. 2.
9.
3.
4.
5.
6.
10. Data from a sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 . The and of the finer fraction was found to be 45% and 33% respectively. The group symbol is a) SC b) MI c) CH d) MH 11. In a shrinkage limit test a 9.6 cc container was filled with soil slurry. The weight of the saturated soil was 17.46 g. The slurry was then gradually dried first in atmosphere and then in an oven at a constant temperature of C. The weight and volume of the dried soil were 11.58 g and 5.22 cc respectively. Determine the shrinkage limit of the soil. 12. A saturated clay stratum draining both at the top and bottom undergoes 50% consolidation in 16 years under an applied load. If additional drainage layer were present at the middle of the clay stratum, 50% consolidation would occur in a) 2 years b) 4 years c) 8 years d) 16 years 13. A well of diameter 20 cm fully penetrates a confined aquifer. After a long period of
Page 1
SUB: GEOTECH
pumping at rate of 2720 lit/min. The observations of draw down taken at 10 m and 100 m distances from the centre of the well are found to be 3m and 0.5m respectively. The transmissibility of the aquifer is a) 676 day b) 576 day c) 526 day d) 249 day At a reclamation site for which the soil strata is shown in Fig. a 3m thick layer of a full material is to be laid instaneous on the top surface. If the coefficient of volume compressibility for clay is 2.2 /KN, the consolidation settlement of the clay layer due to placing of fill material will be a) 69.5mm b) 139mm c) 228mm d) 278mm In the laboratory test on a clay sample of thickness 25mm drained at top only, 50% consolidation occurred in 11 minutes. Find the time required for the corresponding clay layer in the field 3m thick and drained at top and bottom, to undergo 70% consolidation. Assume In a constant head permeameter with cross section area of , when the flow was taking place under a hydraulic gradient of 0.5, the amount of water collected in 60 section is 600 cc. the permeability of the soil is a) 0.002 cm/s b) 0.02 cm/s c) 0.2 cm/s d) 2.0 cm/s In filling head permeability test the initial head of 1.0m dropped to 0.35m in 3 hours, the diameter of stand pipe 5mm. the soil specimen was 200mm long and 100mm diameter. The coefficient of permeability of the soil is a) 4.86 cm/sec b) 4.86 cm/sec c) 4.86 cm/sec d) 4.86 cm/sec Data from a sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 . The and of the finer fraction was found to be 45% and 33% respectively. The group symbol is a) SC b) MI c) CH d) MH A sand deposit has a porosity of 1/3 and its specific gravity is 2.5. the critical hydraulic gradient to cause sand boiling in the stratum will be a) 1.5 b) 1.25 c) 1.0 d) 0.75
MODULE: 43-46
14.
20. In a 6m thick stratum of fine sand having submerged density of 11KN/ , quicksand condition occurred at a depth of 4.2m of excavation. What is the depth of lowering of groundwater table required for making an excavation 5m deep? Take density of water as 10KN/ . a) 3.85m b) 1.68m c) 1.1m d) 0.897m
21. The porosity of a certain soil sample was found to be 80% and its specific gravity was 2.7; the critical hydraulic gradient will be estimated as a) 0.34 b) 0.92 c) 1.0 d) 1.5
15.
22. The total, neutral and effective vertical stresses(in t/ ) at a depth of 5m below the surface of a fully saturated soil deposit with a saturated density of 2 t/ ) would, respectively, be a) 5,5 and 10 b) 5,10 and 5 c) 10,5 and 10 d) 10,5 and 5
16.
23. A uniform sand stratum 2.5m thick has specific gravity of 2.62 and a natural void ratio of 0.62. The hydraulic head required to cause quick sand condition in the sand stratum is a) 0.5m b) 1.5m c) 2.5m d) 3.5m
17.
24. The figure below shows two flow lines for seepage across an interface between two soil media of different co-efficient of permeability. If entrance angle a1=30 , the exit angle a2 will be
18.
## (a) 7.50 (c) 66.59
(b) (d)
14.03 75.96
19.
25. If during a permeability test on a soil sample with a falling head permeameter, equal time intervals are noted for drop of head from to and again from to . Then which one of the following relations would hold good? a) = b) = c) = d) ( )=( ) | 1,963 | 6,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-35 | latest | en | 0.888561 |
https://groupprops.subwiki.org/wiki/Supercharacter_theories_for_symmetric_group:S3 | 1,618,092,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00397.warc.gz | 389,278,216 | 8,700 | # Supercharacter theories for symmetric group:S3
Jump to: navigation, search
## Contents
This article gives specific information, namely, supercharacter theories, about a particular group, namely: symmetric group:S3.
View supercharacter theories for particular groups | View other specific information about symmetric group:S3
This page discusses the various possible supercharacter theories for symmetric group:S3. Thus, it builds on a thorough understanding of the element structure of symmetric group:S3, subgroup structure of symmetric group:S3, and linear representation theory of symmetric group:S3.
We describe the group $S_3$ as the symmetric group on $\{ 1,2,3\}$, and elements of the group are described by means of their cycle decompositions.
## Character table
Below, the character table of $S_4$ is given. This table is crucial for understanding the possible supercharacter theories.
Representation/Conjugacy class representative $()$ (identity element) -- size 1 $(1,2,3)$ (3-cycle) -- size 2 $(1,2)$ (2-transposition) -- size 3
Trivial representation 1 1 1
Sign representation 1 1 -1
Standard representation 2 -1 0
## Supercharacter theories
### Summary
There are only two supercharacter theories possible for symmetric group:S3, namely the two extreme cases:
Quick description of supercharacter theory Number of blocks of conjugacy classes = number of blocks of irreducible representations Block sizes for conjugacy classses (in number of conjugacy class terms) (should add up to 3, the total number of conjugacy classes) Block sizes for conjugacy classes (in number of elements terms) (should add up to 6, the order of the group) Block sizes for irreducible representations (in number of representations terms) (should add up to 3, the total number of conjugacy classes) Block sizes for irreducible representations (in sum of squares of degrees terms) (should add up to 6, the order of the group)
ordinary character theory 3 1,1,1 1,2,3 1,1,1 1,1,4
all non-identity elements form one block 2 1,2 1,5 1,2 1,5
There are no other possibilities to consider because the group has only three conjugacy classes. | 500 | 2,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.792691 |
https://oeis.org/A162816 | 1,623,793,017,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00456.warc.gz | 401,615,898 | 3,917 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A162816 a(n) = 12*a(n-1)-33*a(n-2) for n > 1; a(0) = 5, a(1) = 33. 0
5, 33, 231, 1683, 12573, 95337, 729135, 5603499, 43180533, 333250929, 2574053559, 19891362051, 153752577165, 1188615978297, 9189556693119, 71050353033627, 549348865530597, 4247524736257473, 32841784272579975, 253933094974463091 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Binomial transform of A143647. LINKS Index entries for linear recurrences with constant coefficients, signature (12,-33). FORMULA a(n) = ((5+sqrt(3))*(6+sqrt(3))^n+(5-sqrt(3))*(6-sqrt(3))^n)/2. G.f.: (5-27*x)/(1-12*x+33*x^2). MATHEMATICA LinearRecurrence[{12, -33}, {5, 33}, 40] (* or *) Simplify[With[ {c=Sqrt[3]}, Table[(11(5+c)(6+c)^n+(53+7c)(6-c)^n)/(22(6+c)), {n, 30}]]] (* Harvey P. Dale, Jun 30 2011 *) PROG (MAGMA) [ n le 2 select 28*n-23 else 12*Self(n-1)-33*Self(n-2): n in [1..20] ]; (PARI) Vec((5-27*x)/(1-12*x+33*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 30 2011 CROSSREFS Cf. A143647. Sequence in context: A197533 A221441 A083076 * A128418 A335119 A284733 Adjacent sequences: A162813 A162814 A162815 * A162817 A162818 A162819 KEYWORD nonn,easy AUTHOR Klaus Brockhaus, Jul 14 2009 STATUS approved
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Last modified June 15 17:36 EDT 2021. Contains 345049 sequences. (Running on oeis4.) | 628 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-25 | latest | en | 0.56056 |
https://www.physicsforums.com/threads/how-to-calculate-tension-in-a-rope.712378/ | 1,675,410,921,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00004.warc.gz | 942,573,582 | 14,825 | # How to calculate tension in a rope?
BEECHBOY707
## Homework Statement
The link to the problem is here, sorry I couldn't put the diagram in so thought it was better to link!
I need help with both questions, the attempt is for the first question.
https://www.dropbox.com/s/2stb476hp738w9l/Physics question.jpg
Many thanks for any help!
## Homework Equations
Moments clockwise=Moments anticlockwise
F=ma
## The Attempt at a Solution
Moments clockwise=Moments anticlockwise
5*15=TD
75=TD
...?
Not sure first bit is right tho!
Staff Emeritus
Homework Helper
In all of these problems, do the standard analysis:
1. Draw a free-body diagram and indicate all known forces and moments applied.
2. Write equations of equilibrium.
3. Solve the equations of equilibrium for the unknown forces and reactions. | 200 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.872204 |
https://askthetask.com/114034/expression-represent-planned-garden-square-factors-dimensions | 1,679,476,456,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00206.warc.gz | 144,177,735 | 7,240 | 0 like 0 dislike
Bailey writes the expression g2 14g 40 to represent the area of a planned school garden in square feet. what factors can be used to find the dimensions of her garden? (g−4)(g−10) (g 4)(g 10) (g 4)(g−10) (g−4)(g 10)
0 like 0 dislike
The area of the planned school garden have dimensions of g + 4 and g + 10 respectively.
What is an equation?
An equation is an expression that shows the relationship between two or more variables and numbers.
Given the area of the planned school garden as:
A = g² + 14g + 40
A = g² + 10g + 4g + 40 = g(g + 10) + 4(g + 10)
A = (g + 4)(g + 10)
The area of the planned school garden have dimensions of g + 4 and g + 10 respectively.
Find out more on equation at:
#SPJ4
by | 229 | 727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-14 | latest | en | 0.908444 |
http://studyslide.com/doc/781985/electric-current---augusta-independent-schools | 1,597,363,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739104.67/warc/CC-MAIN-20200813220643-20200814010643-00568.warc.gz | 83,975,752 | 18,781 | #### Transcript electric current - Augusta Independent Schools
```Table of Contents
Chapter: Electricity
Section 1: Electric Charge
Section 2: Electric Current
Section 3: Electric Circuits
Electric Charge
1
Electricity
• All solids, liquids, and gases are made of tiny
particles called atoms.
• Atoms are
smaller
particles called
protons,
neutrons, and
electrons.
Electric Charge
1
Electricity
• Protons and neutrons are held together tightly
in the nucleus at the center of an atom, but
electrons swarm around the nucleus in all
directions.
• Protons and
electrons have
electric charge,
but neutrons
have no electric
charge.
Electric Charge
1
Positive and Negative Charge
• There are two types of electric charge—
positive and negative.
• Protons have a positive charge, and electrons
have a negative charge. The amount of
negative charge on an electron is exactly
equal to the amount of positive charge on a
proton.
Electric Charge
1
Positive and Negative Charge
• Because atoms have equal numbers of
protons and electrons, the amount of positive
charge on all the protons in the nucleus of an
atom is balanced by the negative charge on
all the electrons moving around the nucleus.
• Therefore, atoms are electrically neutral.
Electric Charge
1
Positive and Negative Charge
• An atom becomes negatively charged when it
gains extra electrons.
• If an atom loses electrons it becomes
positively charged.
• A positively or negatively charged atom is
called an ion (I ahn).
Electric Charge
1
Electrons Move in Solids
• If you have ever taken
clinging clothes from a
clothes dryer, you have
seen what happens when
electrons are transferred
from one object to
another.
• This imbalance of electric
charge on an object is
called a static charge.
Electric Charge
1
Electrons Move in Solids
• In solids, static charge is
due to the transfer of
electrons between objects.
• Protons cannot be
removed easily from
the nucleus of an atom.
Electric Charge
1
Ions Move in Solutions
• Sometimes, the movement of charge can be
caused by the movement of ions instead of
the movement of electrons.
• When table salt (NaCl)
dissolves in water, the
sodium ions and chloride
ions break apart.
• These ions now are able
to carry electric energy.
Electric Charge
1
Electric Forces
• All charged objects exert an electric force on
each other.
• The electric force between two charges can
be attractive or repulsive.
Electric Charge
1
Electric Forces
• The electric force between two charged
objects depends on the distance between
them and the amount of charge on each
object.
• The electric force between two electric
charges gets stronger as the charges get
closer together.
• The electric force between two charged
objects increases if the amount of charge on
at least one of the objects increases.
Electric Charge
1
Electric Fields
• Charged objects don’t have to be touching to
exert an electric force on each other.
• Electric charges exert a force on each other at
a distance through an electric field that exists
around every electric charge.
• An electric field gets stronger as you get
closer to a charge, just as the electric force
between two charges becomes greater as the
charges get closer together.
Electric Charge
1
Electric Fields
• The lines with arrowheads represent the
electric field around charges.
• The direction of each
arrow is the direction a
positive charge would
move if it were placed
in the field.
Electric Charge
1
Insulators and Conductors
• A material in which electrons cannot move
easily from place to place is called an
insulator.
• Examples of insulators are plastic, wood, glass,
and rubber.
• Charges placed on an insulator repel
each other but cannot move easily
on the surface of the insulator.
• As a result, the charges remain in
one place.
Electric Charge
1
Insulators and Conductors
• Materials that are
conductors
contain electrons
that can move
more easily in the
material.
• The electric wire is made from a conductor
coated with an insulator such as plastic.
Electric Charge
1
Insulators and Conductors
• Electrons move
easily in the
conductor but do
not move easily
through the
plastic insulation.
Electric Charge
1
Metals and Conductors
• The best conductors are metals such as
copper, gold, and aluminum.
• When metal atoms form a solid, the metal
atoms can move only short distances.
• However, the electrons that are loosely bound
to the atoms can move easily in the solid
piece of metal.
• In an insulator, the electrons are bound
tightly in the atoms that make up the insulator
and therefore cannot move easily.
Electric Charge
1
Induced Charge
• Has this ever happened to you?
• You walk across a carpet and as you reach for
a metal doorknob, you feel an electric shock.
• Maybe you even see a spark jump between
tour fingertip and the doorknob.
Electric Charge
1
Induced Charge
• This rapid movement of excess charge from
one place to another is an electric
discharge.
Electric Charge
1
Induced Charge
• Lightning is another example of an electric
discharge.
• In a storm cloud, air currents sometimes
cause the bottom of the cloud to become
negatively charged.
• This negative charge induces a positive
charge in the ground below the cloud.
• A cloud-to-ground lightning stroke occurs
when electric charge moves between the
cloud and the ground.
Electric Charge
1
Grounding
• Lightning is an electric discharge that can
cause damage and injury.
• Even electric discharges that release small
amounts of energy can damage delicate
circuitry in devices such as computers.
• One way to avoid the damage caused by
electric discharges is to make the excess
charges flow harmlessly into Earth’s surface.
Electric Charge
1
Grounding
• The process of providing a pathway to drain
excess charge into Earth is called
grounding.
• The pathway is usually a conductor such as a
wire or a pipe. Lightning rods at the top of
buildings and towers are made of metal and
are connected to metal cables that conduct
electric charge into the ground if the rod is
struck by lightning.
Section Check
1
Question 1
The two types of electric charge are _______
and _______.
The answer is positive and negative.
Section Check
1
Question 2
An imbalanced of electric charge on an object
is called a(n) _______.
A. electric discharge
B. electric current
C. insulator
D. static charge
Section Check
1
toward a comb you are witnessing the effect of
static charge.
Section Check
1
Question 3
Like charges _______ while unlike charges
_______.
Section Check
1
Like charges repel and unlike charges attract.
Electric Current
2
Flow of Charge
• Electrical energy in our
homes comes from an
electric current, which
is the flow of electric
charge.
• In solids, the flowing
charges are electrons.
Electric Current
2
Flow of Charge
• In liquids, the flowing
charges are ions.
• Electric current is
measured in units of
amperes (A).
• A model for electric
current is flowing water.
Electric Current
2
A Model for a Simple Circuit
• The gravitational potential energy of water is
increased when a pump raises the water above
Earth.
Electric Current
2
A Model for a Simple Circuit
• As the water falls and does work on the
waterwheel, the water loses potential energy
and the waterwheel gains kinetic energy.
• Electric charges will flow continuously only
through a
closed
conducting
loop called
a circuit.
Electric Current
2
Electric Circuit
• The simplest electric circuit contains a source
of electrical energy, such as a battery, and an
electric conductor, such as a wire, connected
to the battery.
• As long as there is a closed path for electrons
to follow, electrons can flow in a circuit.
Electric Current
2
Electric Circuit
• They move away from the negative battery
terminal and toward the positive terminal.
Electric Current
2
Voltage
• In a water circuit, a pump increases the
gravitational potential energy of the water by
raising the water from a lower level to a
higher level.
• In an electric circuit, a battery increases the
electrical potential energy of electrons.
• This electrical potential energy can be
transformed into other forms of energy.
Electric Current
2
Voltage
• The voltage of a battery is a measure of how
much electrical potential energy each
electron can gain.
• As voltage increases, more electrical
potential energy is available to be
transformed into other forms of energy.
• Voltage is measured in volts (V).
Electric Current
2
How a Current Flows
• When the ends of a wire are connected to a
battery, the battery produces an electric field
in the wire.
• The electric field forces electrons to move
toward the positive battery terminal. As an
electron moves, it collides with other electric
charges in the wire. After each collision, the
electron again starts moving toward the
positive terminal.
Electric Current
2
Batteries
• A battery supplies
energy to an electric
circuit.
• When the positive and
negative terminals in a
battery are connected
in a circuit, the electric
potential energy of the
electrons in the circuit
is increased.
Electric Current
2
Batteries
• As these electrons
move toward the
positive battery
terminal, this electric
potential energy is
transformed into other
forms of energy.
Electric Current
2
Batteries
• For the alkaline battery, the two terminals are
separated by a moist paste.
• Chemical reactions in the
moist paste cause the
negative terminal to
become negatively
charged and the positive
terminal to become
positively charged.
Electric Current
2
Batteries
• This produces the electric field in the circuit
that causes electrons to move away from the
negative terminal and toward the positive
terminal.
Electric Current
2
Battery Life
• Batteries contain only a limited amount of the
chemicals that react to produce chemical
energy.
• These reactions go on as the battery is used
and the chemicals are changed into other
compounds.
• Once the original chemicals are used up, the
chemical reactions stop and the battery is
Electric Current
2
Resistance
• The measure of how difficult it is for
electrons to flow through a material is called
resistance.
• The unit of resistance is the ohm (Ω).
Insulators generally have much higher
resistance than conductors.
Electric Current
2
Resistance
• As electrons flow through a circuit, they
collide with the atoms and other electric
charges in the materials that make up the
circuit.
Electric Current
2
Resistance
• These collisions cause some of the electrons’
electrical energy to be converted into thermal
energy—heat—and sometimes into light.
• The amount of electrical energy that is
converted into heat and light depends on the
resistance of the materials in the circuit.
Electric Current
2
Buildings Use Copper Wires
• Copper has low resistance
and is one of the best electric
conductors.
• Less heat is produced as
electric current flows in
copper wires, compared to
• As a result, copper wire is used in
household wiring because the wires usually
don’t become hot enough to cause fires.
Electric Current
2
Resistance of Wires
• The electric resistance of a wire also depends
on the length and thickness of the wire, as
well as the material it is made from. The
electric resistance of a wire increases as the
wire becomes longer or as it becomes
narrower.
Electric Current
2
Lightbulb Filaments
• In a lightbulb, the filament is made of wire so
narrow that it has a high resistance.
Electric Current
2
Lightbulb Filaments
• When electric current flows in the filament, it
becomes hot enough to emit light.
• The filament is made of tungsten metal,
which has a much higher melting point than
most other metals.
• This keeps the filament from melting at the
high temperatures needed to produce light.
Section Check
2
Question 1
The flow of an electric charge is known as
_______.
The answer is electric current. In solids, this
is the flow of electrons.
Section Check
2
Question 2
Electric charges will flow continuously only
through a closed conducting loop called a
_______.
A. circuit
B. battery
C. conductor
D. filament
Section Check
2
The answer is A. However, if the path is not
closed the circuit is broken and the electrons
do not flow.
Section Check
2
Question 3
The amount of electrical energy converted
to thermal energy in a wire increases as
the _______ of the wire increases.
A. thickness
B. static charge
C. resistance
D. force
Section Check
2
The answer is C. Because the electric resistance
of copper is low compared to other conductors,
less electrical energy is converted into thermal
energy as current flows in copper wires.
Electric Circuits
3
Controlling the Current
• When you connect
a conductor, such as
a wire or a
lightbulb, between
the positive and
negative terminals
of a battery,
electrons flow in
the circuit.
Electric Circuits
3
Controlling the Current
• The amount of
current is
determined by the
voltage supplied
by the battery and
the resistance of
the conductor.
Electric Circuits
3
Controlling the Current
• To help understand this relationship, imagine
a bucket with a hose at the bottom.
• If the bucket is
raised, water will
flow out of the hose
faster than before.
• Increasing the
height will increase
the current.
Electric Circuits
3
Voltage and Resistance
• Just as the water current increases when the
height of the water increases, the electric
current in a circuit increases as voltage
increases.
• If the diameter of the tube is decreased,
resistance is greater and the flow of the water
decreases.
• In the same way, as the resistance in an
electric circuit increases, the current in the
circuit decreases.
Electric Circuits
3
Ohm’s Law
• A nineteenth-century German physicist,
Georg Simon Ohm, carried out experiments
that measured how changing the voltage in a
circuit affected the current.
• He found a simple relationship among
voltage, current, and resistance in a circuit
that is now known as Ohm’s law.
Electric Circuits
3
Ohm’s Law
• According to Ohm’s law, when the voltage in
a circuit increases the current increases.
• However, if the voltage in the circuit doesn’t
change, then the current in the circuit
decreases when the resistance is increased.
Electric Circuits
3
Series and Parallel Circuits
• For current to flow, the circuit must provide
an unbroken path for current to follow.
• There are two kinds of basic circuits—series
and parallel.
Electric Circuits
3
Wired in a Line
• A series circuit is a circuit that has only one
path for the electric current to follow.
• If this path is broken, then the current no
longer will flow and all the devices in the
circuit stop working.
• In a series circuit, electrical devices are
connected along the same current path.
Electric Circuits
3
Wired in a Line
• However, each new device that is added to
the circuit decreases the current because each
device has electrical resistance. The total
resistance to the flow of electrons increases
circuit. By Ohm’s law, if the voltage doesn’t
change, the current decreases as the
resistance increases.
Electric Circuits
3
Branched Wiring
• Houses, schools, and other buildings are
wired using parallel circuits.
• A parallel circuit is a circuit that has more
than one path for the electric current to
follow.
• If one path is broken, electrons continue to
flow through the other paths.
Electric Circuits
3
Branched Wiring
• In a parallel circuit, the resistance in each
branch can be different, depending on the
devices in the branch.
• The lower the resistance is in a branch, the
more current flows in the branch.
Electric Circuits
3
Protecting Electric Circuits
• In a parallel circuit, the current that flows out
of the battery or electric outlet increases as
• As the current through the circuit increases,
the wire heats up.
• To keep the wire from becoming hot enough
to cause a fire, the circuits in houses and
other buildings have fuses or circuit breakers.
Electric Circuits
3
Protecting Electric Circuits
• When the current becomes larger than 15 A
or 20 A, a piece of metal in the fuse melts or
a switch in the
circuit breaker
opens, stopping
the current.
Electric Circuits
3
Protecting Electric Circuits
• The cause of the overload can then be
removed, and the circuit can be used again by
replacing the
fuse or resetting
the circuit
breaker.
Electric Circuits
3
Electric Power
• The rate at which electrical energy is
converted into other forms of energy is
electric power.
• In an electric appliance or in any electric
circuit, the electric power that is used can be
calculated from the electric power equation:
Electric Circuits
3
Electric Power
• The electric power is
equal to the voltage
provided to the
appliance times the
current that flows into
the appliance.
• In the electric power
equation, the SI unit of
power is the watt.
Electric Circuits
3
Cost of Electric Energy
• Power is the rate at
which energy is
used, or the amount
of energy that is
used per second.
• Using electrical
energy costs money.
Electric Circuits
3
Cost of Electric Energy
• Electric companies
generate electrical
energy and sell it in
units of kilowatthours to homes,
schools, and
• One kilowatt-hour,
kWh, is an amount of electrical energy equal
to using 1 kW of power continuously for 1 h.
Electric Circuits
3
Electrical Safety
• Electricity can have dangerous effects.
• In 1997,
electric
shocks
killed an
estimated
490 people
in the
United
States.
Electric Circuits
3
Electric Shock
• You experience an electric shock when an
• In some ways your body is like a piece of
insulated wire.
• The fluids inside your body are good
conductors of current.
• The electrical resistance of dry skin is much
higher.
• Skin insulates the body like the plastic
insulation around a copper wire.
Electric Circuits
3
Electric Shock
• A current can enter your
body when you
accidentally become
part of an electric
circuit. Whether you
depends on the amount
of current that flows
Electric Circuits
3
Lightning Safety
• On average, more people are killed every
year by lightning in the United States than by
• If you are outside and can see lightning or
hear thunder, take shelter indoors
immediately.
Electric Circuits
3
Lightning Safety
• If you cannot go
indoors, you
should take these
precautions:
∙ Avoid high places and open fields.
∙ Stay away from tall objects such as
trees, flag poles, or light towers.
Electric Circuits
3
Lightning Safety
∙ Avoid object that
conduct current
such as bodies
of water, metal
fences, picnic
shelters, and
bleachers.
Section Check
3
Question 1
If the computer listed in
the table was plugged into
a 110-V outlet, how much
current would flow in the
computer?
Section Check
3
The current into the computer would be 3.2 A.
Section Check
3
Question 2
What is Ohm’s Law?
Ohm’s law can be stated as V = IR, or voltage
equals current (in amperes) times resistance
(in ohms).
Section Check
3
Question 3
A circuit with more than a single path for the
wiring to follow is called a _______.
It is a parallel circuit. The electrical outlets in
your house are on a parallel circuit. If they
weren’t, no appliance would work if any one
appliance was turned off.
Chapter 22 Review 1 of 2
• Two types of electric charges: positive and
negative.
• Like charges repel and unlike charges attract.
• Electrons do not move freely through an
insulator.
• The human body is a good conductor of
electricity.
• Electric discharge: rapid movement of excess
charge from one place to another.
Chapter 22 Review 2 of 2
• Materials have electric resistance because
electrons collide with atoms in the materials,
releasing thermal energy and light.
• Increase in resistance = increase in energy lost
as heat and light.
• Voltage increase = current increase.
• Series circuit: electricity only has one path to
follow.
• Parallel circuits: are used in houses, schools,
and other buildings.
``` | 4,626 | 19,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-34 | latest | en | 0.904729 |
https://tunxis.commnet.edu/view/9s-multiplication-worksheet.html | 1,721,670,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00134.warc.gz | 523,860,918 | 5,818 | # 9S Multiplication Worksheet
9S Multiplication Worksheet - On this printable, students begin by skip counting from 0 to 90 by 9s. Web multiplication facts with 9's. Web learn to multiply by 9s free. Then they fill in a scrambled multiplication table with 9s. Web grab this freebie here. After that, they compare numbers and complete a magic multiplication wheel.
There are several multiplication activities that the students enjoy including tic tac toe games, bump, number puzzles and sorting worksheets. On this printable, students begin by skip counting from 0 to 90 by 9s. Worksheets and games introduction to multiplication with groups Students multiply 9 times numbers between 1 and 12. After that, they compare numbers and complete a magic multiplication wheel.
Worksheets and games introduction to multiplication with groups Students multiply 9 times numbers between 1 and 12. Then they fill in a scrambled multiplication table with 9s. Web grab this freebie here. After that, they compare numbers and complete a magic multiplication wheel.
## 9 Times Table
9S Multiplication Worksheet - Then they fill in a scrambled multiplication table with 9s. Web grab this freebie here. Web multiplication facts with 9's. After that, they compare numbers and complete a magic multiplication wheel. Once we understand multiplying by 9 using arrays, pictures and manipulatives, we begin practicing facts for accuracy and fluency. Web learn to multiply by 9s free. On this printable, students begin by skip counting from 0 to 90 by 9s. There are several multiplication activities that the students enjoy including tic tac toe games, bump, number puzzles and sorting worksheets. Students multiply 9 times numbers between 1 and 12. Worksheets and games introduction to multiplication with groups
On this printable, students begin by skip counting from 0 to 90 by 9s. Worksheets and games introduction to multiplication with groups After that, they compare numbers and complete a magic multiplication wheel. Web grab this freebie here. Web multiplication facts with 9's.
There are several multiplication activities that the students enjoy including tic tac toe games, bump, number puzzles and sorting worksheets. After that, they compare numbers and complete a magic multiplication wheel. Web multiplication facts with 9's. Then they fill in a scrambled multiplication table with 9s.
There are several multiplication activities that the students enjoy including tic tac toe games, bump, number puzzles and sorting worksheets. Worksheets and games introduction to multiplication with groups Web grab this freebie here.
Web learn to multiply by 9s free. Web grab this freebie here. On this printable, students begin by skip counting from 0 to 90 by 9s.
## After That, They Compare Numbers And Complete A Magic Multiplication Wheel.
Web learn to multiply by 9s free. Once we understand multiplying by 9 using arrays, pictures and manipulatives, we begin practicing facts for accuracy and fluency. On this printable, students begin by skip counting from 0 to 90 by 9s. Students multiply 9 times numbers between 1 and 12.
## There Are Several Multiplication Activities That The Students Enjoy Including Tic Tac Toe Games, Bump, Number Puzzles And Sorting Worksheets.
Worksheets and games introduction to multiplication with groups Web multiplication facts with 9's. Then they fill in a scrambled multiplication table with 9s. Web grab this freebie here. | 709 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.911543 |
http://www.exceltip.com/summing/summing-values-based-on-text-criteria.html | 1,481,374,181,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543170.25/warc/CC-MAIN-20161202170903-00335-ip-10-31-129-80.ec2.internal.warc.gz | 440,822,841 | 16,151 | # How to use excel sumif function?
http://www.exceltip.com/summing/summing-values-based-on-text-criteria.html
In this article we will learn how to use Sumif function for count the number text contains cells on the bases of criteria in Excel.
While working in MS Excel when we prepare reports and dashboards, we come across situations wherein we need to sum multiple criteria based. Here we can use excel SUMIF function.
Let’s understand how to use the SUMIF formula in different ways. Let’s take an excel file where each row of columns A & B contains text and any value respectively.
We want to sum all the numbers in column B corresponding with the text values in column A that meet certain criteria. These criteria have been listed in column F of each screenshot below. We will review one criteria a t time.
1. Text is “Excel”
The formula will be =SUMIF(A2:A6,”Excel”,B2:B6)
This will sum all the values in column B where the corresponding cell in column A is equal to “Excel”.
2. Text starts with “Excel”
=SUMIF(A2:A6,”Excel*”,B2:B6)
This will sum all the values in column B where the corresponding cell in column A has text starting with “Excel”.
3. Text ends with “Excel”
=SUMIF(A2:A6,”*Excel”,B2:B6)
This will sum all the values in column B where the corresponding cell in column A has text ending with “Excel”.
4. Text contains “Excel”
=SUMIF(A2:A6,”*Excel*”,B2:B6)
This will sum all the values in column B where the corresponding cell in column A has text containing “Excel”.
5. Text length is 3 characters
=SUMIF(A2:A6,”???”,B2:B6)
This will sum all the values in column B where the corresponding cell in column A contains text of length 3 characters.
Please follow and like us:
0
## 6 thoughts on “How to use excel sumif function?”
1. Hi!
I’ve tried to use this formula but it doesn’t work on my Excel file, I use Excel 2003…
Thank you for your help!
• Hey,
This formula will work in MS Excel-2003 as well.
2. Is there a way to do this with an “OR” function? So it searches for multiple terms?
Thanks!
• Hi Alex,
You can use SUMIFS function. | 519 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-50 | longest | en | 0.812153 |
scientificchick.blogspot.com | 1,569,117,975,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00283.warc.gz | 169,705,458 | 15,633 | ## Sunday, November 7, 2010
I have fond memories of high school math classes. Numbers came easy for me, and I derived a lot of satisfaction from solving problems (and even more so when I solved them fast!). I was lucky to have excellent teachers, especially in grade 11 and in CEGEP, who turned math into something of a game, a code I needed to crack. However, I’m well aware that math class was not a party for everyone. About 15 to 20% of the population struggle with some form of difficulty in learning or understanding mathematics. Obviously, this can be an obstacle to success in school, and in employment. To address this issue, a team of researchers from the UK set out to test whether brain stimulation could improve someone’s math abilities.
The researchers used a technique called transcranial direct current stimulation (TDCS, similar to the method used in this post on morality). TDCS consists of applying a weak current to a brain region (in this case, the parietal lobe, a region important for learning and understanding of math) over a given time period (in this case, 20 minutes). The technique is non-invasive, meaning they don’t open up your scalp to get at your brain: electrodes are simply place on your head (volunteers are much easier to recruit when the electrodes are on the outside, not the inside). Depending on the type of current that the researchers apply, TDCS can increase or reduce the excitation of the brain cells in the targeted region.
To test the impact of this kind of brain stimulation on math capabilities, the researchers delivered the stimulation while the participants (15 healthy adults) were learning the relative values between nine arbitrary symbols (for example, square is bigger than triangle). The learning session lasted 90 to 120 minutes. The participants received either the brain stimulation during the first 20 minutes of the session (the experimental group), or only during the first 30 seconds of the session (the control group, as 30 seconds of the stimulation is not long enough to see any effects, but still gives you the “tingles” associated with the protocol). After this learning phase, the researchers assessed the participants’ newly created sense of numerical value for the symbols with two different math tasks using the symbols. This whole process was then repeated over six days.
The results show that brain stimulation leads to better and more consistent performance on both math tasks. Mathematical ineptitude is cured! To make the matters even more interesting, the researchers called the participants back six months later and re-tested them (no brain stimulation this time). And six months later, the brain stimulation group still performed better at the math tasks involving the fake digits.
While this may sound great, don’t start tazing your brain just yet. It’s worthwhile to note that on the last day of the initial six-day study, the researchers had the participants perform the same two math tasks, but with normal numbers. In this case, there were no difference between the experimental group and the control group. This means that the brain stimulation paradigm only worked for the specific task that was learned during the stimulation, and didn’t extend to math in general. Nonetheless, the researchers suggest that brain stimulation may be a tool for intervention for those who have “developmental and acquired disorders in numerical cognition” (read: for people who are bad at math).
Must we all be good at math? There’s a French saying that goes: "ça prend toute sorte de monde pour faire un monde" (roughly translates into: "it takes all sorts of people to make a world"). What’s your take on this? Do you think this is a great advance? Do you have any concerns? Let’s hear it!
Reference: Modulating neuronal activity produces specific and long-lasting changes in numerical competence. (2010) Cohen Kadosh R et al. Current Biology 20:1-5.
#### 3 Responses to “Math made easy”
JLoh said...
Very interesting read! I wonder what would happen if they did the same experiment with the same amount of brain stimulation but with the participants reading Shakespeare. My feeling is that the results would be similar because any form of brain stimulation would make one focus more on the task/subject at hand.
I believe that it's important to have a nice, well-rounded understanding of all things...but we do definitely need specialists in every field.
Anonymous said...
I enjoy reading your analysis, SC. But I would virtually explode if a media outlet got a-hold of this study and inflated it for public view. Since you did not hyperbole the results, I enjoy our exchange here and might even re-tweet it.
To even consider this little study has value one has to accept that the groups were evenly homogeneous.
Since you read the study - but I cannot help but ask - you had no questions about the physiological effects in the brain of "weak current" on the surface of the head? Honestly, I can't imagine/believe a single neural cell molecule was changed during stimulation.
Physical therapists use electricity as a treatment modality for muscle activation and pain relief (read on the 'gate theory'). Depth of the current through skin is variable and I'm not sure about current through bone.
Perhaps the method should not be called "brain stimulation" but scalp stimulation - and if it is effective - a different hypothesis for how it works needs to be formulated. Thanks so much for an interesting read. Barbara
@ JLoh: You ask a good question. It's well established that the parietal lobe is important for math, but I wonder if any learning task would work.
@ Barbara: You're absolutely right: things can quickly get out of hand when studies like this one are misinterpreted or described out of context.
It's always difficult to get perfectly homogeneous groups, but in this case, they should start with a larger sample. I think it was mostly a proof of concept type of article.
In terms of the physiological effects of the stimulation, they seem legit based on previous studies. Yes, the scalp is quite the barrier, but neuronal transmission is tiny in magnitude, and so a change of a few millivolts in the membrane potential could impact firing rates. That said, the precise location and magnitude of this effect is hard to evaluate, and so I absolutely agree that this type of experiment is a little bit vague and the cause-and-effect could be assessed more convincingly.
Thanks for the comment! | 1,318 | 6,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-39 | latest | en | 0.948841 |
https://www.justcrackinterview.com/interviews/tremco-incorporated/ | 1,696,278,351,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00423.warc.gz | 902,138,615 | 19,266 | # Tremco Incorporated Interview Questions Answers, HR Interview Questions, Tremco Incorporated Aptitude Test Questions, Tremco Incorporated Campus Placements Exam Questions
Find best Interview questions and answer for Tremco Incorporated Job. Some people added Tremco Incorporated interview Questions in our Website. Check now and Prepare for your job interview. Interview questions are useful to attend job interviews and get shortlisted for job position. Find best Tremco Incorporated Interview Questions and Answers for Freshers and experienced. These questions can surely help in preparing for Tremco Incorporated interview or job.
All of the questions listed below were collected by students recently placed at Tremco Incorporated.
Ques:- Which of the following is always odd?
A. Sum of two odd numbers
B. Product of two odd numbers
C. Difference of two odd numbers
D. None of these
B
Ques:- if a ball is dropped from 8 meter and it bounce back half the distance then what is the distance it travel before going to rest?
to reach land
first time it travels 8 mtrs
next time (half=4) 4+4 mtrs (up&down)
next time (half=4) 2+2 mtrs (up&down)
next time (half=4) 1+1 mtrs (up&down)
and so on.
Altogether,close to 24 mtrs.
Ques:- 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?
The definition of left and right depends on the observer and
is reversed when facing the opposite direction. The
definition of up and down does not depend on the orientation
of the observer.
Ques:- 1/8 is devided by ‘s’ , if ‘s’ is incresed by 2 times, what is the result?
decrease two times
Ques:- Were you not satisfied with the old company
Ques:- A Roman was born the first day of the 35th year before Christ and died the first day of the 35th year after Christ. How many years did he live?
A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69
Ques:- My father’s name A.Kasi, he is car driver. My mother’s name K.Lakshmi, she is housewife. I have one elder brother.
Ques:- TELL US SOMETHING ABOUT OUR COLLEGE?
Ques:- Are you willing to put the interests of the organization ahead of your own?
Ques:- A light bulb is hanging in a room. Outside of the room there are three switches, of which only one is connected to the lamp. In the starting situation, all switches are 'off' and the bulb is not lit. If it is allowed to check in the room only once to see if the bulb is lit or not (this is not visible from the outside), how can you determine with which of the three switches the light bulb can be switched on?
Ques:- Tell me something negative you?ve heard about our company?
Ques:- What is the punchline of Corporation bank?
Prosperity for all
Ques:- Where are you working at present?
Ques:- Why you looking change
Ques:- Wat will u do if not selected?????
Ques:- A number when, 28 subtracted from it reduces to its one third. What is the value of 50% of that number?
Recent Answer : Added by K R ESWAR REDDY On 2021-09-29 16:49:53:
x-28=(1/3)x
x= 42
50 %of 42 = 21
Ques:- 10 books are placed at random in a shelf. The probability that a pair of books will always be together is -. | 907 | 3,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-40 | longest | en | 0.929563 |
https://cstheory.stackexchange.com/questions/11087/layman-interpretation-quantum-factoring-algorithm | 1,656,551,224,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103646990.40/warc/CC-MAIN-20220630001553-20220630031553-00025.warc.gz | 225,736,460 | 66,241 | Layman Interpretation: Quantum Factoring Algorithm
I must firstly express that I know only a little about quantum computing and my knowledge comes largely from popular science texts and the media.
So, I'm hoping that somebody will be able to help me to correct my understanding of quantum computing.
My understanding is as follows:
• a qubit acts as though it is in both states at once (1 and 0)
• a register of n qubits can act as though it is in any of $2^n$ states.
• this has obvious benefits in terms of a factoring algorithm: we can identify whether any of these states is a 'correct' solution to a problem.
• however, I understand that although we can identify whether or not there is a correct state, we can not necessarily observe the state which is correct
So, my assumption up to now has been that we can 'pin' one or more of the qubits by replacing with a classical bit, and observe whether or not the remaining set of states still contain a correct solution. (Simple!)
My problem is that this would seem to lead to a solution to the factoring problem in $O(log n)$ time, by passing down and pinning each of the bits. I haven't proved that out but it feels right, based on my assumptions.
However, Shor's algorithm takes $O((logn)^3)$ and doesn't seem that simple. I'd like to know which of my assumptions are wrong, but Wikipedia's description of Shor's algorithm seems intractable to me.
Can you help identify my misconception? Which of my points of understanding are correct/incorrect? Thanks!
• There's a great blog post by Scott Aaronson explaining Shor's algorithm in non-specialist terms: scottaaronson.com/blog/?p=208 Apr 14, 2012 at 18:07
The most general representation of the quantum state of $n$ qubits is the $2^n \times 2^n$ density matrix $\rho$, where the diagonal entries correspond to the probability of finding the system in a particular classical state, and the off diagonal entries indicate the phase of the superposition (going to 0 for classical probability distributions).
Any quantum algorithm can be seen as a unitary operation followed by some measurement in the computational basis. Given an initial input state $\rho$, it will evolve to $\rho' = U\rho U^\dagger$. The most general set of measurements that can be performed are called positive operator valued measurements(POVMs), where each measurement outcome $i$ is associated with a positive sei-definite Hermitian operator $P_i$ and the probability of outcome $i$ is given by $p_i = \mbox{Tr}(P_i \rho)$ (and hence $\sum_i P_i = \mathbb{I}$). | 583 | 2,549 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-27 | longest | en | 0.950997 |
http://ww2010.atmos.uiuc.edu/(Gh)/wwhlpr/airmasses_activity.rxml?hret=/indexlist.rxml | 1,506,332,880,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690591.29/warc/CC-MAIN-20170925092813-20170925112813-00091.warc.gz | 371,333,320 | 3,253 | .
Air Masses
scaffolding activity
Introduction:
An air mass is a large body of air with similar temperature and moisture properties throughout. The best source regions for air masses are large flat areas where air can be stagnant long enough to take on the characteristics of the surface below. As an air mass moves away from its source region, it is modified as it encounters conditions different than those found in the source region. Air masses typically clash in the middle latitudes, producing some very interesting weather.
The purpose of this activity is to introduce air masses that commonly influence the weather in the United States, characteristics of these air masses, and how to identify air masses on weather maps. Key words throughout this activity link directly to helper resources that provide useful information for answering the questions.
Characteristics of Air Masses:
1) The diagram below depicts two types of air masses that commonly influence weather in the United States. For each air mass, identify the following characteristics.
Air Mass #1 Air Mass #2
Type of Air Mass:
Source Region:
Relative Temperature:
Wind Direction:
Moisture Content:
Find the Air Masses:
2) One way of identifying a tropical air mass on the weather map below is to look for a region of higher temperatures. To find a polar air mass, look for a region of colder temperatures. The image below is a map of surface observations and for this part of the activity, use the temperature field to draw two lines; a red line to outline the edge of a tropical air mass and a blue line to identify a polar air mass. Here is an example. You may label the diagram in one of two ways; 1) by printing out a copy of this activity and marking your answers directly onto the printout or 2) by saving the image into your favorite graphics software and modifying the image using that graphics package.
3) Now examine the regions you have outlined. Look particularly close at the wind barbs for wind direction and also examine the reports of dew point temperature. In question #1, you determined typical wind direction and dew point temperatures associated with a tropical air mass and a polar air mass. Use this additional information to again identify the tropical and the polar air masses in the diagram below. Label the edge of a tropical air mass with a red line and use a blue line to indicate the outer edge of a polar air mass.
4) Are there any differences between your first analysis and the second analysis? Explain why they are different (if they indeed are).
Find the Current Air Masses:
5) Using what you learned from this activity, go to the Weather Visualizer ( CoVis version | public version) and create a map of the latest surface observations. From this map, identify the boundaries of warm and cold air masses (questions #2 and #3) currently affecting the United States.
You may label your image in one of two ways; 1) by printing out a copy of the image and marking your answers directly onto the printout or 2) by saving the image into your favorite graphics software and modifying the image using that graphics package If the Weather Visualizer is too busy, here are additional web sites for accessing current weather data.
Terms for using data resources. CD-ROM available. Credits and Acknowledgments for WW2010. Department of Atmospheric Sciences (DAS) atthe University of Illinois at Urbana-Champaign. | 683 | 3,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | longest | en | 0.902802 |
https://www.jefftk.com/p/r0-is-not-counterfactual | 1,723,605,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00440.warc.gz | 636,430,620 | 8,471 | ### R0 Is Not Counterfactual
April 13th, 2023
covid-19
Someone who helps organize an N95-required dance recently wrote to a group of organizers:
R0 is the number of people that an infected individual is likely to infect. R0 was 5.4 in Dec 2022. ... People who assume increased risk for themselves are also assuming increased risk for 5.4 other people.
This is wrong in two main ways, and I responded on the list, but I wanted to share my response here as well because the claim illustrates two common misconceptions.
The first issue is that R0 is for an entirely "susceptible" population, one in which no one has any gained any immunity from exposure or vaccination. What an R0 of 5.4 would mean is that if it had suddenly appeared in 2019, each infected person would on average infect 5.4 others. This is very different from the current situation where most people have had several shots, plus most people have had covid at least once. The term for the expected number of people an infected person will directly infect given current conditions is Rt, currently about 1. Which is really just another way of saying that covid levels have been changing relatively slowly: if Rt were 5.4 we'd have rapid growth tearing through the population.
The other issue is that even Rt doesn't tell you how many infections you getting sick would cause. You may know various things about your behavior that make the expected number of people you'd directly infect higher or lower, but that's not the main issue. Instead it's that (a) people you infect can go on to infect other people and (b) people you infect might otherwise have been infected by other people. These two factors push in opposite directions, but both can be quite large. Here are a pair of toy situations showing how, holding Rt fixed, one infection can lead to either very many or almost no counterfactual infections:
• A new epidemic is starting, and Rt (which is R0 in this case) is 5.4. There are very clear symptoms, and people are just starting to catch on. Very soon there will be massive behavior changes to suppress Rt, and at this stage those might or might not be enough. One more person getting infected could have a 5% impact on the chance that this becomes a pandemic infecting ~half the world. In which case the expected number of additional infections is ~200M people. Here (a) is the main factor.
• An new pandemic is well under way, and it has easily missed symptoms. Even with lots of precautions in place, Rt is still a very high 5.4. There are so many paths by which a person can get infected that one additional infection has almost no effect on how many people eventually get infected. Here (b) is the main factor.
Now, "how many counterfactual infections would I cause if I got sick", and "how would my getting sick shift the distribution of when other people get sick" are really valuable questions to know the answers to if you're trying to understand the social impacts of more risky behavior, and it would be great if we did know these. But the progress epidemiologists have put into estimating R0 is not most of what you'd draw on in trying to get better answers.
### Recent posts on blogs I like:
#### How good can you be at Codenames without knowing any words?
About eight years ago, I was playing a game of Codenames where the game state was such that our team would almost certainly lose if we didn't correctly guess all of our remaining words on our turn. From the given clue, we were unable to do this. Altho…
via Posts on August 11, 2024 | 778 | 3,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.985695 |
http://steamcommunity.com/app/4920/discussions/0/540732888758071632/?l=spanish | 1,508,565,609,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00434.warc.gz | 324,785,367 | 12,054 | Natural Selection 2
### Natural Selection 2
twitch.tv/AtlantisThief 5 FEB 2014 a las 3:36
Hi,
i was just playing a round NS2 and i tried to do a bit more gorging this time. I fail'd ...
Question is: Can i actually replace the second places tunnel exit instead of the first one? So for example, let the base entrance stay and replace the outcome point?
Mostrando 1-10 de 10 comentarios
Kaeru 5 FEB 2014 a las 4:49
Yes if you hold E at the second tunnel you'll destroy it and the first tunnel is still there.
twitch.tv/AtlantisThief 5 FEB 2014 a las 5:04
Can i do it also remotly?
R1B\$ 5 FEB 2014 a las 6:21
I don't really gorge but from what I understand:
You can only build two holes maximum, when you have two there's a connection. There really is no concept of entrance or exit. If you destroy a hole, tunnel stops working until both holes are built up.
twitch.tv/AtlantisThief 5 FEB 2014 a las 6:34
I know, but for the purpose of explaining, i said entrance (the first i placed) and exit (the second i place). ^^
Mr. Guggleywubbins 5 FEB 2014 a las 9:55
Destroying Gorge tunnels may not be done remotely by the Gorge.
100% Recycled Awesome 5 FEB 2014 a las 10:56
Tunnels are first built first replaced to my knowledge. Thus the first tunnel part you placed will be replaced first. So if you build A then B, then place C, A will be replaced. Then if you drop D, B will be replaced. Note this is only for completed tunnels.
Última edición por 100% Recycled Awesome; 5 FEB 2014 a las 11:09
100% Recycled Awesome 5 FEB 2014 a las 10:58
Destroying Gorge tunnels may not be done remotely by the Gorge.
You can "destroy" them remotely by placing a new set of tunnel entrances.
Hinsonator 5 FEB 2014 a las 12:39
I don't really gorge but from what I understand:
You can only build two holes maximum, when you have two there's a connection. There really is no concept of entrance or exit. If you destroy a hole, tunnel stops working until both holes are built up.
Technecally the tunnel still "works" as in you can still enter it but there isn't a point B to go to so it just ends up being a place to hide people or ambush marines silly enough to enter.
ShadowSoul 5 FEB 2014 a las 13:24
Technecally the tunnel still "works" as in you can still enter it but there isn't a point B to go to so it just ends up being a place to hide people or ambush marines silly enough to enter.
Doesn't the tunnel close after one end has been destroyed? If it is empty that is, had it happen more than once that the other end got destroyed while I was in there, but of course I could still get out.
100% Recycled Awesome 5 FEB 2014 a las 17:54 | 735 | 2,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-43 | latest | en | 0.914038 |
https://se.mathworks.com/matlabcentral/cody/problems/76-de-dupe/solutions/2164133 | 1,590,386,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387219.0/warc/CC-MAIN-20200525032636-20200525062636-00510.warc.gz | 543,849,234 | 15,510 | Cody
# Problem 76. De-dupe
Solution 2164133
Submitted on 18 Mar 2020 by Yin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [5 3 6 4 7 7 3 5 9]; b_correct = [5 3 6 4 7 9]; assert(isequal(dedupe(a),b_correct));
2 Pass
a = [1 0 0 0 1 1 1 1 0 0 1 2 2 0 1] b_correct = [1 0 2]; assert(isequal(dedupe(a),b_correct));
a = 1 0 0 0 1 1 1 1 0 0 1 2 2 0 1
3 Pass
a = [-1 -1 -1 -1 -1 -1]; b_correct = [-1]; assert(isequal(dedupe(a),b_correct)); | 246 | 567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-24 | latest | en | 0.661018 |
https://www.yumpu.com/en/document/view/24644536/quantum-scattering-chaos-classical-and-quantum | 1,568,723,982,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573071.65/warc/CC-MAIN-20190917121048-20190917143048-00077.warc.gz | 1,095,205,630 | 31,448 | # Quantum scattering - Chaos: Classical and Quantum
Quantum scattering - Chaos: Classical and Quantum
Chapter 35
Quantum scattering
Scattering is easier than gathering.
—Irish proverb
(A. Wirzba, P. Cvitanović and N. Whelan)
SO FAR the trace formulas have been derived assuming that the system under
consideration is bound. As we shall now see, we are in luck - the semiclassics
of bound systems is all we need to understand the semiclassics for open,
scattering systems as well. We start by a brief review of the quantum theory of
elastic scattering of a point particle from a (repulsive) potential, and then develop
the connection to the standard Gutzwiller theory for bound systems. We do this
in two steps - first, a heuristic derivation which helps us understand in what sense
density of states is “density,” and then we sketch a general derivation of the central
result of the spectral theory of quantum scattering, the Krein-Friedel-Lloyd
formula. The end result is that we establish a connection between the scattering
resonances (both positions and widths) of an open quantum system and the poles
of the trace of the Green function, which we learned to analyze in earlier chapters.
35.1 Density of states
For a scattering problem the density of states (31.18) appear ill defined since formulas
such as (34.6) involve integration over infinite spatial extent. What we will
now show is that a quantity that makes sense physically is the difference of two
densities - the first with the scatterer present and the second with the scatterer
absent.
In non-relativistic dynamics the relative motion can be separated from the
center-of-mass motion. Therefore the elastic scattering of two particles can be
treated as the scattering of one particle from a static potential V(q). We will study
the scattering of a point-particle of (reduced) mass m by a short-range potential
638
V(q), excluding inter alia the Coulomb potential. (The Coulomb potential decays
slowly as a function of q so that various asymptotic approximations which apply
to general potentials fail for it.) Although we can choose the spatial coordinate
frame freely, it is advisable to place its origin somewhere near the geometrical
center of the potential. The scattering problem is solved, if a scattering solution
to the time-independent Schrödinger equation (31.5)
(− 2
∂ 2 )
2m ∂q 2 + V(q) φ ⃗ k (q) = Eφ (q) (35.1)
⃗k
can be constructed. Here E is the energy, ⃗p = ⃗k the initial momentum of the
particle, and ⃗k the corresponding wave vector.
When the argument r = |q| of the wave function is large compared to the typical
size a of the scattering region, the Schrödinger equation effectively becomes
a free particle equation because of the short-range nature of the potential. In the
asymptotic domain r ≫ a, the solution φ ⃗ (q) of (35.1) can be written as superposition
of ingoing and outgoing solutions of the free particle Schrödinger equation
k
for fixed angular momentum:
φ(q) = Aφ (−) (q) + Bφ (+) (q) , (+ boundary conditions) ,
where in 1-dimensional problems φ (−) (q), φ (+) (q) are the “left,” “right” moving
plane waves, and in higher-dimensional scattering problems the “incoming,” “outgoing”
radial waves, with the constant matrices A, B fixed by the boundary conditions.
What are the boundary conditions? The scatterer can modify only the
outgoing waves (see figure 35.1), since the incoming ones, by definition, have yet
to encounter the scattering region. This defines the quantum mechanical scattering
matrix, or the S matrix
φ m (r) = φ (−)
m (r) + S mm ′φ (+)
m ′ (r) . (35.2)
All scattering effects are incorporated in the deviation of S from the unit matrix,
the transition matrix T
S = 1 − iT . (35.3)
For concreteness, we have specialized to two dimensions, although the final formula
is true for arbitrary dimensions. The indices m and m ′ are the angular momenta
quantum numbers for the incoming and outgoing state of the scattering
wave function, labeling the S -matrix elements S mm ′. More generally, given a set
of quantum numbers β, γ, the S matrix is a collection S βγ of transition amplitudes
β → γ normalized such that |S βγ | 2 is the probability of the β → γ transition. The
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
Figure 35.1: (a) Incoming spherical waves running
into an obstacle. (b) Superposition of outgoing
spherical waves scattered from an obstacle. (a) (b)
total probability that the ingoing state β ends up in some outgoing state must add
up to unity
|S βγ | 2 = 1 , (35.4)
γ
so the S matrix is unitary: S † S = SS † = 1.
We have already encountered a solution to the 2-dimensional problem; free
particle propagation Green’s function (33.48) is a radial solution, given in terms
of the Hankel function
G 0 (r, 0, E) = − im
2 2 H(+) 0 (kr) ,
where we have used S 0 (r, 0, E)/ = kr for the action. The mth angular momentum
eigenfunction is proportional to φ m (±) (q) ∝ H m (±) (kr), and given a potential
V(q) we can in principle compute the infinity of matrix elements S mm ′. We will
not need much information about H m (t) (kr), other than that for large r its asymptotic
form is
H ± ∝ e ±ikr
In general, the potential V(q) is not radially symmetric and (35.1) has to be
solved numerically, by explicit integration, or by diagonalizing a large matrix in
a specific basis. To simplify things a bit, we assume for the time being that a radially
symmetric scatterer is centered at the origin; the final formula will be true
for arbitrary asymmetric potentials. Then the solutions of the Schrödinger equation
(31.5) are separable, φ m (q) = φ(r)e imθ , r = |q|, the scattering matrix cannot
mix different angular momentum eigenstates, and S is diagonal in the radial basis
(35.2) with matrix elements given by
S m (k) = e 2iδ m(k) . (35.5)
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
The matrix is unitary so in a diagonal basis all entries are pure phases. This means
that an incoming state of the form H m (−) (kr)e imθ gets scattered into an outgoing state
of the form S m (k)H m (+) (kr)e imθ , where H m (∓) (z) are incoming and outgoing Hankel
functions respectively. We now embed the scatterer in a infinite cylindrical well
of radius R, and will later take R → ∞. Angular momentum is still conserved so
that each eigenstate of this (now bound) problem corresponds to some value of m.
For large r ≫ a each eigenstate is of the asymptotically free form
φ m (r) ≈ e imθ ( S m (k)H (+)
m (kr) + H (−)
m (kr) )
≈ · · · cos(kr + δ m (k) − χ m ) , (35.6)
where · · · is a common prefactor, and χ m = mπ/2+π/4 is an annoying phase factor
from the asymptotic expansion of the Hankel functions that will play no role in
what follows.
The state (35.6) must satisfy the external boundary condition that it vanish at
r = R. This implies the quantization condition
k n R + δ m (k n ) − χ m = π (n + 12) .
We now ask for the difference in the eigenvalues of two consecutive states of
fixed m. Since R is large, the density of states is high, and the phase δ m (k) does
not change much over such a small interval. Therefore, to leading order we can
include the effect of the change of the phase on state n + 1 by Taylor expanding. is
k n+1 R + δ m (k n ) + (k n+1 − k n )δ ′ m(k n ) − χ m ≈ π + π(n + 12) .
Taking the difference of the two equations we obtain ∆k ≈ π(R + δ ′ m(k)) −1 . This
is the eigenvalue spacing which we now interpret as the inverse of the density of
states within m angular momentum sbuspace
d m (k) ≈ 1 ( R + δ
π m (k) ) .
The R term is essentially the 1 − d Weyl term (34.8), appropriate to 1 − d radial
quantization. For large R, the dominant behavior is given by the size of the circular
enclosure with a correction in terms of the derivative of the scattering phase shift,
approximation accurate to order 1/R. However, not all is well: the area under
consideration tends to infinity. We regularize this by subtracting from the result
from the free particle density of states d 0 (k), for the same size container, but this
time without any scatterer, figure 35.2. We also sum over all m values so that
d(k) − d 0 (k) = 1 π
m
δ ′ m (k) = 1 ∑
2πi
m
=
d
dk log S m
(
1
2πi Tr S † dS
dk
)
. (35.7)
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
Figure 35.2: The “difference” of two bounded reference
systems, one with and one without the scattering
system.
b
-
b
The first line follows from the definition of the phase shifts (35.5) while the second
line follows from the unitarity of S so that S −1 = S † . We can now take the limit
R → ∞ since the R dependence has been cancelled away.
This is essentially what we want to prove since for the left hand side we already
have the semiclassical theory for the trace of the difference of Green’s functions,
d(k) − d 0 (k) = − 1
2πk Im (tr (G(k) − G 0(k)) . (35.8)
There are a number of generalizations. This can be done in any number of
dimensions. It is also more common to do this as a function of energy and not
wave number k. However, as the asymptotic dynamics is free wave dynamics
labeled by the wavenumber k, we have adapted k as the natural variable in the
above discussion.
Finally, we state without proof that the relation (35.7) applies even when there
is no circular symmetry. The proof is more difficult since one cannot appeal to the
phase shifts δ m but must work directly with a non-diagonal S matrix.
35.2 Quantum mechanical scattering matrix
The results of the previous section indicate that there is a connection between the
scattering matrix and the trace of the quantum Green’s function (more formally
between the difference of the Green’s function with and without the scattering
center.) We now show how this connection can be derived in a more rigorous
manner. We will also work in terms of the energy E rather than the wavenumber
k, since this is the more usual exposition. Suppose particles interact via forces of
sufficiently short range, so that in the remote past they were in a free particle state
labeled β, and in the distant future they will likewise be free, in a state labeled γ.
In the Heisenberg picture the S -matrix is defined as S = Ω − Ω † + in terms of the
Møller operators
Ω ± = lim
t→±∞ eiHt/ e −iH 0t/ , (35.9)
where H is the full Hamiltonian, whereas H 0 is the free Hamiltonian.
interaction picture the S -matrix is given by
In the
S = Ω † +Ω − = lim
t→∞
e iH 0t/ e −2iHt/ e iH 0t/
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
(
= T exp −i
∫ +∞
−∞
)
dtH ′ (t)
, (35.10)
where H ′ = V = H − H 0 is the interaction Hamiltonian and T is the time-ordering
operator. In stationary scattering theory the S matrix has the following spectral
representation
such that
S =
∫ ∞
0
dE S (E)δ(H 0 − E)
S (E) = Q + (E)Q −1
− (E), Q ± (E) = 1 + (H 0 − E ± iɛ) −1 V , (35.11)
Tr
[
S † (E) d ] [
]
dE S (E) 1
= Tr
H 0 − E − iɛ − 1
H − E − iɛ − (ɛ ↔ −ɛ)
. (35.12)
The manipulations leading to (35.12) are justified if the operators Q ± (E) can be
appendix J
We can now use this result to derive the Krein-Lloyd formula which is the
central result of this chapter. The Krein-Lloyd formula provides the connection
between the trace of the Green’s function and the poles of the scattering matrix,
implicit in all of the trace formulas for open quantum systems which will be presented
in the subsequent chapters.
35.3 Krein-Friedel-Lloyd formula
The link between quantum mechanics and semiclassics for scattering problems is
provided by the semiclassical limit of the Krein-Friedel-Lloyd sum for the spectral
density which we now derive. This derivation builds on the results of the last
section and extends the discussion of the opening section.
In chapter 33 we linked the spectral density (see (31.18)) of a bounded system
d(E) ≡ δ(E n − E) (35.13)
via the identity
n
1
δ(E n − E) = − lim
ɛ→0 π Im 1
1
= − lim
ɛ→0
=
1
2π i lim
ɛ→0
E − E n + iɛ
π Im〈E 1
n|
E − H + iɛ |E n〉
〈 ∣ ∣∣∣∣ 1
E n
E − H − iɛ − 1
E − H + iɛ ∣ E n
(35.14)
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
to the trace of the Green’s function (34.1.1). Furthermore, in the semiclassical
approximation, the trace of the Green’s function is given by the Gutzwiller trace
formula (34.11) in terms of a smooth Weyl term and an oscillating contribution of
periodic orbits.
Therefore, the task of constructing the semiclassics of a scattering system is
completed, if we can find a connection between the spectral density d(E) and the
scattering matrix S . We will see that (35.12) provides the clue. Note that the right
hand side of (35.12) has nearly the structure of (35.14) when the latter is inserted
into (35.13). The principal difference between these two types of equations is that
the S matrix refers to outgoing scattering wave functions which are not normalizable
and which have a continuous spectrum, whereas the spectral density d(E)
refers to a bound system with normalizable wave functions with a discrete spectrum.
Furthermore, the bound system is characterized by a hermitian operator,
the Hamiltonian H, whereas the scattering system is characterized by a unitary
operator, the S -matrix. How can we reconcile these completely different classes
of wave functions, operators and spectra? The trick is to put our scattering system
into a finite box as in the opening section. We choose a spherical conatiner with
radius R and with its center at the center of our finite scattering system. Our scattering
potential V(⃗r) will be unaltered within the box, whereas at the box walls we
will choose an infinitely high potential, with the Dirichlet boundary conditions at
the outside of the box:
φ(⃗r)| r=R = 0 . (35.15)
In this way, for any finite value of the radius R of the box, we have mapped our
scattering system into a bound system with a spectral density d(E; R) over discrete
eigenenergies E n (R). It is therefore important that our scattering potential
was chosen to be short-ranged to start with. (Which explains why the Coulomb
potential requires special care.) The hope is that in the limit R → ∞ we will
recover the scattering system. But some care is required in implementing this.
The smooth Weyl term ¯d(E; R) belonging to our box with the enclosed potential V
diverges for a spherical 2-dimensional box of radius R quadratically, as πR 2 /(4π)
or as R 3 in the 3-dimensional case. This problem can easily be cured if the spectral
density of an empty reference box of the same size (radius R) is subtracted
(see figure 35.2). Then all the divergences linked to the increasing radius R in
the limit R → ∞ drop out of the difference. Furthermore, in the limit R → ∞
the energy-eigenfunctions of the box are only normalizable as a delta distribution,
similarly to a plane wave. So we seem to recover a continous spectrum. Still the
problem remains that the wave functions do not discriminate between incoming
and outgoing waves, whereas this symmetry, namely the hermiticity, is broken in
the scattering problem. The last problem can be tackled if we replace the spectral
density over discrete delta distributions by a smoothed spectral density with a
small finite imaginary part η in the energy E:
d(E + iη; R) ≡ 1
i 2π
∑ {
n
1
E − E n (R) − iη − 1
E − E n (R) + iη
}
. (35.16)
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
Note that d(E + iη; R) d(E − iη; R) = −d(E + iη; R). By the introduction of the
positive finite imaginary part η the time-dependent behavior of the wave function
has effectively been altered from an oscillating one to a decaying one and the
hermiticity of the Hamiltonian is removed. Finally the limit η → 0 can be carried
out, respecting the order of the limiting procedures. First, the limit R → ∞ has
to be performed for a finite value of η, only then the limit η → 0 is allowed. In
practice, one can try to work with a finite value of R, but then it will turn out (see
below) that the scattering system is only recovered if R √ η ≫ 1.
Let us summarize the relation between the smoothed spectral densities d(E +
iη; R) of the boxed potential and d (0) (E + iη; R) of the empty reference system and
the S matrix of the corresponding scattering system:
lim lim
(
d(E+iη; R) − d (0) (E+iη; R) ) =
η→+0 R→∞
= 1
2πi Tr d
dE ln S (E) = 1
2πi
[
1
2πi Tr S † (E) d ]
dE S (E)
d
ln det S (E) . (35.17)
dE
This is the Krein-Friedel-Lloyd formula. It replaces the scattering problem by
the difference of two bounded reference billiards of the same radius R which finally
will be taken to infinity. The first billiard contains the scattering region or
potentials, whereas the other does not (see figure 35.2). Here d(E + iη; R) and
d (0) (E + iη; R) are the smoothed spectral densities in the presence or in the absence
of the scatterers, respectively. In the semiclassical approximation, they are
replaced by a Weyl term (34.10) and an oscillating sum over periodic orbits. As in
(34.2), the trace formula (35.17) can be integrated to give a relation between the
smoothed staircase functions and the determinant of the S -matrix:
lim lim
(
N(E+iη; R) − N (0) (E+iη; R) ) =
η→+0 R→∞
1
2πi
ln det S (E) . (35.18)
Furthermore, in both versions of the Krein-Friedel-Lloyd formulas the energy argument
E +iη can be replaced by the wavenumber argument k+iη ′ . These expressions
only make sense for wavenumbers on or above the real k-axis. In particular,
if k is chosen to be real, η ′ must be greater than zero. Otherwise, the exact left
hand sides (35.18) and (35.17) would give discontinuous staircase or even delta
function sums, respectively, whereas the right hand sides are continuous to start
with, since they can be expressed by continuous phase shifts. Thus the order of
the two limits in (35.18) and (35.17) is essential.
The necessity of the +iη prescription can also be understood by purely phenomenological
considerations in the semiclassical approximation: Without the iη
term there is no reason why one should be able to neglect spurious periodic orbits
which are there solely because of the introduction of the confining boundary.
The subtraction of the second (empty) reference system removes those spurious
periodic orbits which never encounter the scattering region – in addition to the removal
of the divergent Weyl term contributions in the limit R → ∞. The periodic
orbits that encounter both the scattering region and the external wall would still
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
survive the first limit R → ∞, if they were not exponentially suppressed by the
+iη term because of their
e iL(R)√ 2m(E+iη) = e iL(R)k e −L(R)η′
behavior. As the length L(R) of a spurious periodic orbit grows linearly with the
radius R. The bound Rη ′ ≫ 1 is an essential precondition on the suppression of
the unwanted spurious contributions of the container if the Krein-Friedel-Lloyd
formulas (35.17) and (35.18) are evaluated at a finite value of R. exercise 35.1
Finally, the semiclassical approximation can also help us in the interpretation
of the Weyl term contributions for scattering problems. In scattering problems the
Weyl term appears with a negative sign. The reason is the subtraction of the empty
container from the container with the potential. If the potential is a dispersing billiard
system (or a finite collection of dispersing billiards), we expect an excluded
volume (or the sum of excluded volumes) relative to the empty container. In other
words, the Weyl term contribution of the empty container is larger than of the
filled one and therefore a negative net contribution is left over. Second, if the
scattering potential is a collection of a finite number of non-overlapping scattering
regions, the Krein-Friedel-Lloyd formulas show that the corresponding Weyl
contributions are completely independent of the position of the single scatterers,
as long as these do not overlap.
35.4 Wigner time delay
The term d
dE
ln det S in the density formula (35.17) is dimensionally time. This
suggests another, physically important interpretation of such formulas for scattering
systems, the Wigner delay, defined as
d(k) = d Argdet (S(k))
dk
= −i d log det (S(k)
dk
(
= −i tr S † (k) dS )
dk (k)
(35.19)
and can be shown to equal the total delay of a wave packet in a scattering system.
We now review this fact.
A related quantity is the total scattering phase shift Θ(k) defined as
det S(k) = e +i Θ(k) ,
so that d(k) = d dk Θ(k).
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
The time delay may be both positive and negative, reflecting attractive respectively
repulsive features of the scattering system. To elucidate the connection
between the scattering determinant and the time delay we study a plane wave:
The phase of a wave packet will have the form:
φ = ⃗k · ⃗x − ω t + Θ .
Here the term in the parenthesis refers to the phase shift that will occur if scattering
is present. The center of the wave packet will be determined by the principle of
stationary phase:
0 = dφ = d⃗k · ⃗x − dω t + dΘ .
Hence the packet is located at
⃗x = ∂ω
∂⃗k t − ∂Θ
∂⃗k .
The first term is just the group velocity times the given time t. Thus the packet is
retarded by a length given by the derivative of the phase shift with respect to the
wave vector ⃗k. The arrival of the wave packet at the position ⃗x will therefore be
delayed. This time delay can similarly be found as
τ(ω) = ∂Θ(ω)
∂ω .
To show this we introduce the slowness of the phase ⃗s = ⃗k/ω for which ⃗s · ⃗v g = 1,
where ⃗v g is the group velocity to get
d⃗k · ⃗x = ⃗s · ⃗x dω = x
v g
dω ,
since we may assume ⃗x is parallel to the group velocity (consistent with the
above). Hence the arrival time becomes
t = x
v g
+ ∂Θ(ω)
∂ω .
If the scattering matrix is not diagonal, one interprets
(
∆t i j = Re
−i S −1
i j
)
∂S i j
= Re
∂ω
( ) ∂Θi j
∂ω
scattering - 29dec2004 ChaosBook.org version13, Dec 31 2009
as the delay in the jth scattering channel after an injection in the ith. The probability
for appearing in channel j goes as |S i j | 2 and therefore the average delay for
the incoming states in channel i is
〈∆t i 〉 =
|S i j | 2 ∆t i j = Re (−i
j
= −i
(
S † · ∂S )
,
∂ω
ii
j
S i ∗ ∂S i j
j
∂ω ) = Re (−i S† · ∂S
∂ω ) ii
where we have used the derivative, ∂/∂ω, of the unitarity relation S · S † = 1 valid
for real frequencies. This discussion can in particular be made for wave packets
related to partial waves and superpositions of these like an incoming plane wave
corresponding to free motion. The total Wigner delay therefore corresponds to the
sum over all channel delays (35.19).
Commentary
Remark 35.1 Krein-Friedel-Lloyd formula. The third volume of Thirring [35.1], sections
3.6.14 (Levison Theorem) and 3.6.15 (the proof), or P. Scherer’s thesis [35.15] (appendix)
discusses the Levison Theorem.
namely for the radially symmetric potential in a symmetric cavity. Have a look at
the book of K. Huang, chapter 10 (on the ”second virial coefficient”), or Beth and Uhlenbeck
[35.5], or Friedel [35.7]. These results for the correction to the density of states are
particular cases of the Krein formula [35.3]. The Krein-Friedel-Lloyd formula (35.17)
was derived in refs. [35.3, 35.7, 35.8, 35.9], see also refs. [35.11, 35.14, 35.15, 35.17,
35.18]. The original papers are by Krein and Birman [35.3, 35.4] but beware, they are
mathematicans.
Also, have a look at pages 15-18 of Wirzba’s talk on the Casimir effect [35.16]. Page
16 discusses the Beth-Uhlenbeck formula [35.5], the predecessor of the more general
Krein formula for spherical cases.
Remark 35.2 Weyl term for empty container. For a discussion of why the Weyl term
contribution of the empty container is larger than of the filled one and therefore a negative
net contribution is left over, see ref. [35.15].
Remark 35.3 Wigner time delay. Wigner time delay and the Wigner-Smith time delay
matrix, are powerful concepts for a statistical description of scattering. The diagonal
elements Q aa of the lifetime matrix Q = −iS −1 ∂S/∂ω, where S is the [2N×2N] scattering
matrix, are interpreted in terms of the time spent in the scattering region by a wave packet
incident in one channel. As shown by Smith [35.26], they are the sum over all ouput
channels (both in reflection and transmission) of ∆t ab = Re [(−i/S ab )(∂S ab /∂ω)] weighted
by the probability of emerging from that channel. The sum of the Q aa over all 2N channels
is the Wigner time delay τ W = ∑ a Q aa , which is the trace of the lifetime matrix and is
proportional to the density of states.
exerScatter - 11feb2002 ChaosBook.org version13, Dec 31 2009
Exercises
35.1. Spurious orbits under the Krein-Friedel-Lloyd contruction.
Draw examples for the three types of period
orbits under the Krein-Friedel-Lloyd construction: (a)
the genuine periodic orbits of the scattering region, (b)
spurious periodic orbits which can be removed by the
subtraction of the reference system, (c) spurious periodic
orbits which cannot be removed by this subtraction.
What is the role of the double limit η → 0, container size
b → ∞?
a
R
a
35.2. The one-disk scattering wave function. Derive the
one-disk scattering wave function.
(Andreas Wirzba)
35.3. Quantum two-disk scattering. Compute the quasiclassical
spectral determinant
Z(ε) = ⎜⎝ 1 −
p, j,l
t p
Λ j+2l
p
⎟⎠
j+1
for the two disk problem. Use the geometry
The full quantum mechanical version of this problem
can be solved by finding the zeros in k for the determinant
of the matrix
M m,n = δ m,n + (−1)n
2
J m (ka)
H n
(1) (ka)
(
H
(1)
m−n(kR) + (−1) n H (1)
m+n(k
where J n is the nth Bessel function and H n
(1) is the Hankel
function of the first kind. Find the zeros of the determinant
closest to the origin by solving det M(k) = 0.
(Hints: note the structure M = I + A to approximate the
determinant; or read Chaos 2, 79 (1992))
so that it computes the topological index for each
orbit it finds.
References
[35.1] W. Thirring, Quantum mechanics of atoms and molecules, A course in
mathematical physics Vol. 3 (Springer, New York, 1979). (Springer, Wien
1979).
[35.2] A. Messiah, Quantum Mechanics, Vol. I (North-Holland, Amsterdam,
1961).
[35.3] M.G. Krein, On the Trace Formula in Perturbation Theory, Mat. Sborn.
(N.S.) 33, 597 (1953) ; Perturbation Determinants and Formula for Traces
of Unitary and Self-adjoint Operators, Sov. Math.-Dokl. 3, 707 (1962).
[35.4] M.Sh. Birman and M.G. Krein, On the Theory of Wave Operators and
Scattering Operators, Sov. Math.-Dokl. 3, 740 (1962); M.Sh. Birman and
D.R. Yafaev, St. Petersburg Math. J. 4, 833 (1993).
[35.5] E. Beth and G.E. Uhlenbeck, Physica 4, 915 (1937).
refsScatter - 11aug2005 ChaosBook.org version13, Dec 31 2009
[35.6] K. Huang, Statistical Mechanics (John Wiley & Sons, New York (1987).
[35.7] J. Friedel, Phil. Mag. 43, 153 (1952); Nuovo Cim. Ser. 10 Suppl. 7, 287
(1958).
[35.8] P. Lloyd, Wave propagation through an assembly of spheres. II. The density
of single-particle eigenstates, Proc. Phys. Soc. 90, 207 (1967).
[35.9] P. Lloyd and P.V. Smith, Multiple-scattering theory in condensed materials,
Adv. Phys. 21, 69 (1972), and references therein.
[35.10] R. Balian and C. Bloch, Ann. Phys. (N.Y.) 63, 592 (1971)
[35.11] R. Balian and C. Bloch, Solution of the Schrödinger Equation in Terms
of Classical Paths Ann. Phys. (NY) 85, 514 (1974).
[35.12] R. Balian and C. Bloch, Distribution of eigenfrequencies for the wave
equation in a finite domain: III. Eigenfrequency density oscillations, Ann.
Phys. (N.Y.) 69,76 (1972).
[35.13] J.S. Faulkner, “Scattering theory and cluster calculations,” J. Phys. C 10,
4661 (1977).
[35.14] P. Gaspard and S.A. Rice, Semiclassical quantization of the scattering
from a classically chaotic repeller, J. Chem. Phys. 90, 2242 (1989).
[35.15] P. Scherer, Quantenzustände eines klassisch chaotischen Billards, Ph.D.
thesis, Univ. Köln (Berichte des Forschungszentrums Jülich 2554, ISSN
0366-0885, Jülich, Nov. 1991).
[35.16] A. Wirzba, “A force from nothing into nothing: Casimir interactions,”
(2003).
[35.17] P. Gaspard, Scattering Resonances: Classical and Quantum Dynamics,
in: Proceedings of the Int. School of Physics “Enrico Fermi”, Course CXIX,
Varena, 23 July - 2 August 1991, eds G. Casati, I. Guarneri and U. Smilansky
(North-Holland, Amsterdam, 1993).
[35.18] A. Norcliffe and I. C. Percival, J. Phys. B 1, 774 (1968); L. Schulman,
Phys. Rev. 176, 1558 (1968).
[35.19] W. Franz, Theorie der Beugung Elektromagnetischer Wellen (Springer,
Berlin 1957); “Über die Greenschen Funktionen des Zylinders und der
Kugel,” Z. Naturforschung 9a, 705 (1954).
[35.20] G.N. Watson, Proc. Roy. Soc. London Ser. A 95, 83 (1918).
[35.21] M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions
with Formulas, Graphs and Mathematical Tables, (Dover, New York, 1964).
[35.22] W. Franz and R. Galle, “Semiasymptotische Reihen für die Beugung
einer ebenen Welle am Zylinder,” Z. Naturforschung 10a, 374 (1955).
refsScatter - 11aug2005 ChaosBook.org version13, Dec 31 2009
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• May 15th 2011, 01:42 AM
iva
Monotone sequence - how can this be a monotone sequence if it bounces between numbers
My understanding of showing a sequence is monotone is to show that it either increases or decreases by working out
\$\displaystyle (a_n+1) - (a_n)\$
and if it's > 0 its increasing, if less than, decreasing, right?
But now i have this problem that I must say if the sequence is monotone:
\$\displaystyle (a_n) = (-1)^n + 2n\$
which to me works out that it is bounded by 0 and 4, ie that it bounces between 0 and 4 ( if n is odd then the value is 0 if even then its 4)
Surely this goes against monotone sequence??? But my answer says that it IS monotone. How can it be monotone when it is neither increasing nor decreasing??
• May 15th 2011, 01:51 AM
Mondreus
Quote:
A monotonically increasing sequence is one for which each term is greater than or equal to the term before it; if each term is strictly greater than the one preceding it, the sequence is called strictly monotonically increasing.
You have a monotonically increasing sequence since \$\displaystyle a_{n+1}-a_n \geq 0\$, but it's not strictly monitonically increasing.
• May 15th 2011, 01:55 AM
iva
Thanks, I was just replying as I picked up my error (I was confusing the \$\displaystyle a_n + 1 - a_n\$ with the original sequence. Ok I see it is increasing.
But what do you mean by "it is not strictly monotonically increasing?"
Thank you
• May 15th 2011, 02:13 AM
Mondreus
In order for it to be strictly monitonically increasing, you need \$\displaystyle a_{n+1} - a_n > 0\$ | 465 | 1,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.945377 |
http://www.sparknotes.com/physics/rotationalmotion/rotationaldynamics/section2.rhtml | 1,371,650,444,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708789647/warc/CC-MAIN-20130516125309-00026-ip-10-60-113-184.ec2.internal.warc.gz | 684,430,223 | 19,488 | Rotational Dynamics
Work, Energy and Combined Motion
Having established the dynamics of rotational motion, we can now extend our study to work and energy. Given what we already know, the equations governing energetics are quite easy to derive. Finally, with the equations that we have derived, we will be able to describe the complicated situations involving combined rotational and translational motion.
Work
Given our definition of work as W = Fs , can we generate an expression for work done on a rotational system? To derive our expression we begin by taking the simplest case: when the force applied to a particle in rotational motion is perpendicular to the radius of the particle. In this orientation, the force applied is parallel to the displacement of the particle, and would exert the maximum work. Given this situation the work done is simply W = Fs , where s is the arc length that the force acts through in a given period of time. Recall, however, that arc length can also be expressed in terms of the angle swept out by the arc: s = . Our expression for work in this simple case becomes:
W = Frθ = τμ
Since Fr gives us our torque, we can simplify our expression in terms of only τ and μ .
What if the force is not perpendicular to the radius of the particle? Let the angle between the force vector and the radius vector be θ , as shown below.
Figure %: A force acting at angle θ to the radius of rotation of point P
To compute the work we calculate the component of the force acting in the direction of the particle's displacement. In this case, this quantity is simply F sinθ . Again, this force acts over an arc length given by . Thus the work is given by:
W = (F sinθ)() = (Fr sinθ)μ
Recall that
τ = Fr sinθ
Thus W = τμ Surprisingly enough, this equation is exactly the same as our special case when the force acted perpendicular to the radius! In any case, the work done by a given force is equal to the torque it exerts multiplied by the angular displacement.
For you calculus types, there is also an equation for work done by variable torques. Instead of deriving it, we can just state it, as it is quite similar to the equation in the linear case:
W = τdμ
Thus we have quickly gone through deriving our expression for work. The next thing after work we studied in linear motion was kinetic energy, and it is to this topic that we turn.
Rotational Kinetic Energy
Consider a wheel spinning in place. Clearly the wheel is moving, and has a kinetic energy attached to it. But the wheel is not engaged in translational motion. How do we calculate the kinetic energy of the wheel? Our answer is similar to how we calculated the result of a net torque on a body: by summing over each particle.
Given a rotating body, we state that the body is made up of n single rotating particles, each at a different radius from the axis of rotation. When each particle is considered individually, we can see that each one does in fact have a translational kinetic energy:
K = m 1 v 1 2 + m 2 v 2 2 + ... m n v n 2
However, we also know from our relation between linear and angular variables that v = . Substituting this expression in, we see that:
K = m 1 r 1 2 σ 2 + m 2 r 2 2 σ 2 + ... m n r n 2 σ 2
Since all particles are part of the same rigid body, we can factor our σ 2 :
K = ( mr 2)σ 2
This sum, however, is simply our expression for a moment of inertia. Thus:
K = Iσ 2
As we might expect, this equation is of the same form as our equation for linear kinetic energy, but with I substituted for m , and σ substituted for v . We now have rotational analogues for nearly all of our translational concepts. The last rotational equation that we need to define is power.
Power
The equation for rotational power can be easily derived from the linear equation for power. Recall that P = Fv is the equation that gives us instantaneous power. Similarly, in the rotational case:
P = τσ
With the equation for rotational power we have generated rotational analogues to every dynamic equation we derived in linear motion and completed our study of rotational dynamics. To provide a summary of our results, the two sets of equations, linear and rotational, are given below: Linear Motion:
F = ma W = Fx K = mv 2 P = Fv
Rotational Motion:
τ = Iα W = τμ K = Iσ 2 P = τσ
Equipped with these equations, we can now turn to the complicated case of combined rotational and translational motion. | 1,009 | 4,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2013-20 | longest | en | 0.928921 |
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# In recent years cattle breeders have increasingly used
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In recent years cattle breeders have increasingly used [#permalink]
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07 Apr 2005, 12:36
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In recent years cattle breeders have increasingly used crossbreeding,
in part that their steers should acquire certain characteristic and partly because crossbreeding is said to provide hybrid vigor.
(A)....
(B) in part for the acquisition of certain characteristic in their steers
(C)partly because of their steers acquiring certain characteristic
(D)partly because certain characteristic should be acquired by their steers
(E)partly to acquire certain characteristic in their steers
Last edited by tonglin on 08 Apr 2005, 08:31, edited 1 time in total.
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07 Apr 2005, 13:19
agree "D"....never knew what steers meant before this
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07 Apr 2005, 13:28
D.
Tonglin, could you edit the question stem so that the underlined part can be easily distinguished by others?
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07 Apr 2005, 13:36
E) "because" is not necessary to maintain parallelism "partly...party" is sufficient. moreover, active voice is preferred, cause i dont think the steers can choose whether they like to acquire the new characteristics
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07 Apr 2005, 14:12
christoph wrote:
E) "because" is not necessary to maintain parallelism "partly...party" is sufficient. moreover, active voice is preferred, cause i dont think the steers can choose whether they like to acquire the new characteristics
I was debating with E as well, reason I dropped E becose "partly to acquire certain characteristic in their steers" sounds awkward. "Acquire certain characteristic in their steers" doesn't sound right.
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07 Apr 2005, 14:56
I go for E too the awkwardness of E mentioned earlier is there is D too
how in gods name does "certain characteristic" fit in here
shouldnt we have
"certain characteristics" or "a certain characteristic"
if it is just one characteristic then there is all the more justification to choose E
someone plz xplain
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08 Apr 2005, 07:03
The QA is E. could anyone who choose E explain the reason..
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08 Apr 2005, 07:42
tonglin wrote:
In recent years cattle breeders have increasingly used crossbreeding,
in part that their steers should acquire certain characteristic and partly because crossbreeding is said to provide hybrid vigor.
(A)....
(B) in part for the acquisition of certain characteristic in their steers
(C)partly because of their steers acquiring certain characteristic
(D)partly because certain characteristic should be acquired by their steers
(E)partly to acquire certain characteristic in their steers
Here is my explanation:
A is wrong because "that" introduce a restrictive clause. It should be preceded by a noun.
For e.g.: The dog that ate my homework died.
B is wrong because it lacks parallelism.
<in part for the aquisition of certain characteristics> & <partly because crossbreeding is said to provide hybrid vigor> are NOT parallel.
D is wrong one of the "parallel" phrases is PASSIVE voice [should be acquired] and the other is active [crossbreeding is].
That leaves us with C and E.
C: The phrase "partly because of their steers acquring certain characteristics" has modifier issues. The cattle breeders want their steers to acquire certain characteristics. In C acquiring certain chracteritics modifies steers...Not correct.
Hence the OA is E.
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08 Apr 2005, 07:46
banerjeea_98 wrote:
christoph wrote:
E) "because" is not necessary to maintain parallelism "partly...party" is sufficient. moreover, active voice is preferred, cause i dont think the steers can choose whether they like to acquire the new characteristics
I was debating with E as well, reason I dropped E becose "partly to acquire certain characteristic in their steers" sounds awkward. "Acquire certain characteristic in their steers" doesn't sound right.
"...used...to acquire...in their steers" => you have to add the full context. that makes more sense, does it ?
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08 Apr 2005, 09:51
christoph wrote:
E) "because" is not necessary to maintain parallelism "partly...party" is sufficient. moreover, active voice is preferred, cause i dont think the steers can choose whether they like to acquire the new characteristics
I can understand passive vs active, but why "Because" is not necessary to maintain parallelism? Why "partly...partly" is sufficient? Please elaborate. Thanks.
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08 Apr 2005, 12:14
I try to write my understanding after reading those explanations.
why not D
1. D maintains parallelism but does not follow the rule that subject should be starting at the head of the sentence.
2. D does not clear the reason why cattle breeders have increasingly used corssbreeding.
If we break the sentence into two parts
"In recent years cattle breeders have increasingly used crossbreeding, partly because certain characteristic should be acquired by their steers"
does not sound very clear why breeders have increasingly used crossbreeding.
Compare to the second reason
"In recent years cattle breedings have increasingly used crossbreeding,
partly because crossbreeding is said to provide hybrid vigor"
Correct me if I am wrong.
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08 Apr 2005, 13:46
Is the OA "E" ? .....Can someone explain how "Acquire in their XYZ" is a correct use ? And what that means if it is correct. Generic examples wud be helpful.
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08 Apr 2005, 21:18
I too have problem picking between D and E. The only thing I could say in D is not good is that it is passive. But I might very well have picked D myself.
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09 Apr 2005, 00:00
tonglin wrote:
In recent years cattle breeders have increasingly used crossbreeding,
in part that their steers should acquire certain characteristic and partly because crossbreeding is said to provide hybrid vigor.
(A)....
(B) in part for the acquisition of certain characteristic in their steers
(C)partly because of their steers acquiring certain characteristic
(D)partly because certain characteristic should be acquired by their steers
(E)partly to acquire certain characteristic in their steers
This is a parallel problem. "...and partly because crossbreading" signals you to look for "partly because (noun)" in your choice. This logic leaves you with only (D).
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Re: In recent years cattle breeders have increasingly used [#permalink]
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20 Feb 2015, 02:44
Would like to go with 'D'. Parallelism seems correct in this.
Also I feel that certain characteristics should be acquired BY the steers.Partly to acquire in option E sounds awkward to me.
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Re: In recent years cattle breeders have increasingly used [#permalink]
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22 Feb 2015, 15:20
In recent years cattle breeders have increasingly used crossbreeding,
in part that their steers should acquire certain characteristic and partly because crossbreeding is said to provide hybrid vigor.
(A) in part that their steers should acquire certain characteristic
Incorrect. in part is not parallel with partly.
(B) in part for the acquisition of certain characteristic in their steers
Incorrect. in part is not parallel with partly.
(C)partly because of their steers acquiring certain characteristic
Incorrect. a prepositional phrase (because of ...) is not parallel to aclause.
(D)partly because certain characteristic should be acquired by their steers
Correct. [main clause], partly [subordinate clause] and partly [subordinate clause].
(E)partly to acquire certain characteristic in their steers
Incorrect. [main clause], partly [noun phrase] and partly [subordinate clause].
I am going with D. Is really the OA E?
Re: In recent years cattle breeders have increasingly used [#permalink] 22 Feb 2015, 15:20
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A great addition to your elementary math corner, these idea cards encourage kids to make geometric shapes with Magna-Tiles, Picasso Tiles, and other magnetic tiles etc. The pack includes 60 cards with various building challenges!
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