url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.gibboncode.org/html/HELP_multiRegionTriMesh2D.html | 1,716,947,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00619.warc.gz | 32,804,330 | 4,806 | # multiRegionTriMesh2D
Below is a basic demonstration of the features of the multiRegionTriMesh2D function.
## Contents
```clear; close all; clc;
```
## SIMULATING BOUNDARY CURVES
```%Boundary 1
ns=150;
t=linspace(0,2*pi,ns);
t=t(1:end-1);
r=12+3.*sin(5*t);
[x,y] = pol2cart(t,r);
V1=[x(:) y(:)];
%Boundary 2
r=r/1.2;
[x,y] = pol2cart(t,r);
V2=[x(:) y(:)];
%Boundary 3
r=r/4;
[x,y] = pol2cart(t,r);
V3=[x(:) -y(:)+6];
%Boundary 4
V4=[x(:) y(:)-4];
%Boundary 5
V5=[x(:)-4 y(:)+1];
%Boundary 6
V6=[x(:)+4 y(:)+1];
```
## CREATING MULTI-REGION MESHES
The first step is to define regions. Regions are defined as cell entries. for instance a cell called regionSpec. Each entry in regionSpec defines a region i.e. region 1 is found in regionSpec{1}. Each region entry is itself also a cell array containing all the boundary curves, e.g. for a two curve region 1 we would have something like regionSpec{1}={V1,V2} where V1 and V2 are the boundary curves. Multiple curves may be given here. The first curve should form the outer boundary of the entire region, the curves that follow should define holes inside this boundary and the material inside them is therefore not meshed. The boundary vertices for regions that share boundaries are merged and will share these boundary vertices. The function output contains the triangular faces in F, the vertices in V and the per triangle region indices in regionInd.
```%Defining 4 regions
regionSpec{1}={V1,V2}; %A region between V1 and V2 (V2 forms a hole inside V1)
regionSpec{2}={V2,V3,V4,V5,V6}; %A region bound by V2 containing a set of holes defined by V3 up to V6
regionSpec{3}={V5}; %A region bound by V5
regionSpec{4}={V6}; %A region bound by V6
plotOn=1; %This turns on/off plotting
%Desired point spacing
pointSpacing=0.6;
[F,V,regionInd]=multiRegionTriMesh2D(regionSpec,pointSpacing,1,plotOn);
plotV(V1,'b-','LineWidth',2);
plotV(V2,'b-','LineWidth',2);
plotV(V3,'b-','LineWidth',2);
plotV(V4,'b-','LineWidth',2);
plotV(V5,'b-','LineWidth',2);
plotV(V6,'b-','LineWidth',2);
view(2);
axis tight;
```
GIBBON www.gibboncode.org
Kevin Mattheus Moerman, [email protected]
GIBBON footer text
GIBBON: The Geometry and Image-based Bioengineering add-On. A toolbox for image segmentation, image-based modeling, meshing, and finite element analysis.
Copyright (C) 2019 Kevin Mattheus Moerman
This program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.
This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/. | 849 | 2,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.725813 |
http://math.stackexchange.com/questions/253640/the-characterstic-polynomial | 1,469,758,198,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829325.58/warc/CC-MAIN-20160723071029-00222-ip-10-185-27-174.ec2.internal.warc.gz | 166,502,634 | 16,904 | # The characterstic polynomial
Let $A,B\in M_{n}(C)$ such that $A$ satisfies in the characteristic polynomial of $B$ and $B$ satisfies in the characteristic polynomial of $A$. Is the the characteristic polynomial of $A$ is equal to the characteristic polynomial of $B$? Is $rank(A)=rank(B)$? Is $A$ diagonalize if and only if $B$ diagonalize?
-
Don't forget to accept the answer which you like. It makes to receive answers later. – Babak S. Dec 8 '12 at 9:29
Why don't you accept @Euyu's answer? – Did Jan 1 '13 at 19:07
None of what you mentioned needs to be true. The matrices do need to have the same eigenvalues (not up to multiplicity) though since the minimal polynomial of each divides the characteristic polynomial of the other. In particular, this means that $A$ will be invertible if and only if $B$ is invertible.
Let $A$ be a matrix with minimal polynomial $x(x-1)$ and characteristic polynomial $x^{n-1}(x-1)$.
Let $B$ be a matrix with minimal polynomial $x^2(x-1)$ and characteristic polynomial $x^2(x-1)^{n-2}$.
The two matrices each satisfy the other's characteristic polynomial. $A$ is diagonalizable but $B$ is not. $A$ has rank $1$ while $B$ has rank $n-1$. | 328 | 1,179 | {"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} | 3.578125 | 4 | CC-MAIN-2016-30 | latest | en | 0.898001 |
https://www.physicsforums.com/threads/square-root-of-complex-number-in-rectangular-form.456430/ | 1,623,977,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00164.warc.gz | 848,777,878 | 14,729 | # Square root of complex number in rectangular form
## Homework Statement
I don't know how to find the square root of a complex number in rectangular form?
As in, say, $$\sqrt{}9-6i$$..my calculator can't do such an operation (yet my graphics calculator can, which can't be used in exams), so how do i go about to do this 'by hand'?
I just found this site: http://mathworld.wolfram.com/SquareRoot.html half way down, is that the formula that we use? How do we take the inverse tangent of (x,y)? What's sgn? I've never been taught such a formula nor did I know something like it existed until now...this question is in regards to a third year electromagnetics course and some 3 questions on the tutorials involve square roots of complex numbers..yet i'm assuming the professor found them using a graphics calculator.
jbunniii
Homework Helper
Gold Member
Perhaps the most straightforward way:
1. convert to polar coordinates ($re^{i\theta}$)
2. take the square root ($\sqrt{r}e^{i\theta/2}$)
3. convert back to rectangular coordinates ($\sqrt{r}(\cos(\theta/2)+i\sin(\theta/2))$
This gives you one square root. There are two. What's the other one?
P.S. This is equivalent to equation (1) at the link you provided.
^^I actually did that, and I get the magnitude part right but i don't know how to get the angle/imaginary part? Or do i just multiply it all out by the (cos(theta/2) + isin(theta/2)) part?
Yep that's what you do! Just tried it out and it works, thanks for pointing me in the right direction :) | 384 | 1,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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-2021-25 | latest | en | 0.905132 |
https://geopandas.org/en/latest/docs/reference/api/geopandas.GeoSeries.fillna.html | 1,713,577,165,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00453.warc.gz | 239,154,339 | 9,118 | # geopandas.GeoSeries.fillna#
GeoSeries.fillna(value=None, inplace=False, **kwargs)[source]#
Fill NA values with geometry (or geometries).
Parameters:
valueshapely geometry or GeoSeries, default None
If None is passed, NA values will be filled with GEOMETRYCOLLECTION EMPTY. If a shapely geometry object is passed, it will be used to fill all missing values. If a `GeoSeries` or `GeometryArray` are passed, missing values will be filled based on the corresponding index locations. If pd.NA or np.nan are passed, values will be filled with `None` (not GEOMETRYCOLLECTION EMPTY).
Returns:
GeoSeries
`GeoSeries.isna`
detect missing values
Examples
```>>> from shapely.geometry import Polygon
>>> s = geopandas.GeoSeries(
... [
... Polygon([(0, 0), (1, 1), (0, 1)]),
... None,
... Polygon([(0, 0), (-1, 1), (0, -1)]),
... ]
... )
>>> s
0 POLYGON ((0 0, 1 1, 0 1, 0 0))
1 None
2 POLYGON ((0 0, -1 1, 0 -1, 0 0))
dtype: geometry
```
Filled with an empty polygon.
```>>> s.fillna()
0 POLYGON ((0 0, 1 1, 0 1, 0 0))
1 GEOMETRYCOLLECTION EMPTY
2 POLYGON ((0 0, -1 1, 0 -1, 0 0))
dtype: geometry
```
Filled with a specific polygon.
```>>> s.fillna(Polygon([(0, 1), (2, 1), (1, 2)]))
0 POLYGON ((0 0, 1 1, 0 1, 0 0))
1 POLYGON ((0 1, 2 1, 1 2, 0 1))
2 POLYGON ((0 0, -1 1, 0 -1, 0 0))
dtype: geometry
```
Filled with another GeoSeries.
```>>> from shapely.geometry import Point
>>> s_fill = geopandas.GeoSeries(
... [
... Point(0, 0),
... Point(1, 1),
... Point(2, 2),
... ]
... )
>>> s.fillna(s_fill)
0 POLYGON ((0 0, 1 1, 0 1, 0 0))
1 POINT (1 1)
2 POLYGON ((0 0, -1 1, 0 -1, 0 0))
dtype: geometry
``` | 607 | 1,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.466051 |
https://www.jiskha.com/questions/384803/1-for-10-pairs-of-data-the-correlation-coefficient-is-computed-to-be-r-1-wht-do-you | 1,585,948,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00181.warc.gz | 981,681,615 | 5,845 | # statistics
1.For 10 pairs of data, the correlation coefficient is computed to be r = -1. Wht do you know about the scatter diagram?
2.You are considering the most expensive purchase that you are likely to make: the purchase of a home. Identify at least five different variables that are likely to affect the actual value of a home. Among the variables that you have identified, which single variable is likely to have the greatest influence on the value of the home? Identify a variable that is likely to have little or no effect on the value of a home.
1. 👍 0
2. 👎 0
3. 👁 157
1. 1. r = -1 indicates perfect predictability, where one variable increases while the other decreases.
2. What variables do you want to consider that would effect your purchase? Location and number of square feet might be a start. What others would you consider?
1. 👍 0
2. 👎 0
posted by PsyDAG
## Similar Questions
1. ### Statistics
The correlation coefficient between X and Y is the same as the correlation coefficient between Y and X. True - If you were to find the correlation coefficient with a given data set for x and y, it would have the same correlation
asked by Nick on April 23, 2017
2. ### Math
The following data sets can be modeled linearly. Data Set M: {(14,95), (9,85), (4,65), (6,80), (3,73), (12,107), (7,64), (12,86), (10,96), (9,64), (1,44)} Data Set N: {(13,109), (9,85), (4,65), (6,80), (3,63), (12,107), (7,76),
asked by Zaphior English on March 16, 2019
3. ### statistics
how do I find the correlation coefficient r? Jesse put the heartbeat and lifespan data into a statistical software program and found the coefficient of determination is 0.45. The value of the correlation coefficient is __________.
asked by Casey on October 25, 2014
4. ### Math
You use a line of best fit for a set of data to make a prediction about an unknown value. The correlation coefficient for your data set is -0.015. How confident can you be that your predicted value will be reasonably close to the
asked by Scarlet on March 22, 2016
5. ### statistics
Tasha examined the correlation between the weight and fuel economy of the 20 top-selling cars. She observed that the coefficient of determination was 62.3%. For this data, the value of the correlation coefficient, r, is
asked by Casey on October 25, 2014
1. ### Stats Help!
You are interested in knowing whether wealthier people are happier. You collected data from fifty people about their incomes and their overall happiness levels on a scale of 1 to 10. Upon analyzing the results, you find that the
asked by slomomo on December 2, 2015
2. ### statistics
Construct a scatterplot. b. Draw a regression line through the data point that “fits” the data points reasonably well. c. Enclose the data points with a ellipse. d. Estimate the direction of the correlation. e. Estimate the
asked by shiela on August 14, 2015
3. ### Statistics
5. For the data presented below, answer the questions that follow. Score on political Score on current Individual awareness test events test 1 24 23 2 16 12 3 11 11 4 9 6 5 8 13 6 14 11 7 17 16 8 19 16 9 21 21 10 25 25 11 18 11 12
asked by Samantha on August 1, 2015
4. ### Psychology Statistics
2 For a sample of n = 20 individuals, how large a Pearson correlation is necessary to be statistically significant for a two-tailed test with = .05? 3 The Pearson correlation is calculated for a sample of n = 25 individuals.
asked by Chioma on April 26, 2013
5. ### Math
Hello! Can you check my answers for these questions? Thank you in advance (= 1. The following ordered pairs give the scores on two consecutive 15-point quizzes for a class of 18 students. (7, 13), (9, 7), (14, 14), (15, 15), (10,
asked by Wendy on January 27, 2015
More Similar Questions | 1,015 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.928917 |
https://www.matific.com/au/en-au/home/blog/2023/04/18/episode-of-the-week-spinning-wheels-calculate-probabilities/ | 1,722,922,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00880.warc.gz | 697,445,612 | 30,746 | ## Purpose of the episode
Understand that probability is quantified as a number between 0 and 1 and that the higher the probability of an event, the more likely it is that the event will occur. Understand that the sum of probabilities
Description
In this episode, the child plays carnival games and tries to win prizes. The child gets to choose any spinning wheel they like after considering the probability of winning that each spinning wheel offers.
Here's a Video Walkthrough
What makes it great?
Before talking about probability, users are invited to make a choice based on their intuition.
The results of that action spark curiosity about how to make the best choice in the future within this real life situation. The critical thinking required to make this choice builds children’s capabilities as decision makers.
Try Now
The child is accountable for their choices.
Whether they win or lose, the host always tells the child whether they made the best choice or whether they got “lucky”. Even if the child chooses the best option, they still have a chance to lose. The child learns by doing, developing understanding of the meaning of probability and intuition around what various probabilities mean in action.
The spinning wheels provide a concrete visual.
Drawing on the connection to fractions, the visual encourages the child to consider the ratio of good options to all options, and since all outcomes are shown, it allows for the insight that the probability of all outcomes together add up to 1. This visualization provides an important bridge from the concrete experience of the event to the abstract representation of the probability.
Consonance with Our Pedagogical Principles
Mathematical Backgroud
Exposure to experimental probability is an important component of mathematical education that is often underrepresented in the grade school curriculum. The intuition and understanding that such exposure builds are among the basic literacy skills of mathematics that serve to support children as they become community members, responsible consumers, and involved citizens–in other words, decision makers. Concrete experience with experimental probability at the primary level is crucial to success in the abstraction of probability in the higher grades as well as to its applications in the sciences. This episode provides that kind of concrete experience –connecting probability with a concrete experiment and providing motivation for children to use their understanding of probability to make the best decision possible. | 463 | 2,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.940654 |
proceedings-szh.com | 1,656,287,098,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00176.warc.gz | 519,564,663 | 19,796 | # How do you calculate BSA for chemotherapy?
Contents
## How do you calculate body surface area for medication doses?
The most commonly used formula now is that of Mosteller, published in 1987 in The New England Journal of Medicine. According to Mosteller’s “simplified calculation of body-surface area In metric terms” the body surface area = the square root of product of the weight in kg times the height in cm divided by 3600.
## How do you calculate BSA?
Body Surface Area (BSA)
1. Calculate weight in kilograms: 210 pounds ÷ 2.2 = 95.45 kg.
2. Calculate height in centimeters: 6 feet, 3 inches = 75 inches x 2.54 cm/inch = 190.5 cm.
3. Multiply height by weight and divide by 3600. (190.5 cm x 95.45 kg) ÷ 3600 = 5.
4. Take the square root of 5 = 2.24 m2
## What is BSA in chemotherapy?
applicable) Dosage calculations for the majority of cytotoxic chemotherapeutic cancer drugs are based on the body surface area (BSA) of the patient, where the dosage is reported in milligrams, grams, or units per square meter of BSA. It is calculated for any cancer drug where the dose is based on BSA.
## Why is chemo dosed on BSA?
Dosing based on body surface area (BSA) is generally used in an effort to normalize drug concentrations. This is because it is well recognized that measures of many physiologic parameters that are responsible for drug disposition, including renal function and energy expenditure, can be normalized by use of BSA.
## How do you find the area of a surface?
Multiply the length and width, or c and b to find their area. Multiply this measurement by two to account for both sides. Add the three separate measurements together. Because surface area is the total area of all of the faces of an object, the final step is to add all of the individually calculated areas together.
## What is normal BSA?
The “normal” body surface area is generally taken to be 1.7 m2 but, in actual fact, the body surface area depends on more than just height and weight. Other influential factors include the age and gender of the individual. For example: Average body surface area for adult men: 1.9 m2.
## Which of the following is considered one of the most widely used BSA formulas?
The apparently most widely used formulas are the following 2,3: Du Bois formula: BSA = 0.007184 x weight [kg] 0.425 x height [cm]
## How do I calculate m2?
How do you work out m2? In order to calculate the size of a room or space in m2, you simply multiply the length of the space (in metres) by the width of the space (in metres).
## How do you calculate BMI example?
Body Mass Index (BMI) = (weight (kg) / height (m2) Example: Your weight = 68 kg / Your height = 1.65 m (165 cm) BMI Calculation: 68 ÷ 1.652 = 24.98.
## What is BSA dosing?
Body surface area (BSA) based dosing is a useful way to mitigate patient size variation in medication regimens. Using BSA may help prescriber’s dose more optimally to improve drug efficacy, minimize drug toxicity, and account for some changes in pharmacokinetics depending on patient factors.
## How do you count days of chemo?
A cycle is the time between one round of treatment until the start of the next. After each round of treatment you have a break, to allow your body to recover. So if your cycle lasts 4 weeks, you may have treatment on the 1st, 2nd and 3rd days and then nothing from the 4th to the 28th day. Then the cycle starts again.
## What does BSA stand for in oncology?
Abstract. The majority of chemotherapy drugs are dosed based on body surface area (BSA).
## What is BSA?
Under the Bank Secrecy Act (BSA), financial institutions are required to assist U.S. government agencies in detecting and preventing money laundering, such as: Keep records of cash purchases of negotiable instruments, File reports of cash transactions exceeding \$10,000 (daily aggregate amount), and.
## How many rounds of chemo can a person have?
You may need four to eight cycles to treat your cancer. A series of cycles is called a course. Your course can take 3 to 6 months to complete. And you may need more than one course of chemo to beat the cancer.
## Are there different doses of chemotherapy?
Most cycles range from 2 to 6 weeks. The number of treatment doses scheduled within each cycle also depends on the prescribed chemotherapy. For example, each cycle may contain only 1 dose on the first day. Or, a cycle may contain more than 1 dose given each week or each day.
THIS IS IMPORTANT: Best answer: How worried should I be about melanoma? | 1,090 | 4,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.911655 |
https://www.ceder.net/def/rollingripple.php?language=czech&action=edit | 1,713,574,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00271.warc.gz | 644,616,212 | 6,312 | Definitions of Square Dance Calls and Concepts
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Rolling Ripple {n} By {n} (By {n}) [C4]
(Lee Kopman 1984)
C4:
Jazyk:
From a Line. EN: 10
Designated dancer Ripple the first given number; new dancer occupying designated dancer's original starting position Ripple the next given number; repeat part 2 for each additional given number. EN: 20
před#1 Boy ()Rolling Ripple 3 By 1 & 1/2 po#1 Boy ()Ripple 3 po#1 Girl ()Ripple 1 & 1/2 (hotovo)
Note: Rolling Ripple should be danced piece by piece. Wait until each Ripple has been completed before starting the next Ripple. This is especially true from Lines or Alamo Rings. For example, from a Tidal Wave, consider a Very Ends Rolling Ripple 6 By 3EN: 30
Ripple (The Line | Wave) | n [C2] (Royce Waugh 1966):
From a Line or Alamo Ring. Those designated start toward the Center of the Line and successively do a Partner Trade with each dancer until reaching the far end of the Line. Ripple n means to Trade with a total of n dancers. n can include a fraction (e.g., Ripple 2 & 1/2 means to Partner Trade with 2 successive dancers and then 1/2 Partner Trade with the third dancer). EN: 678
Z formace Line a Alamo Ring. Označení tanečníci začnou směrem ke středu Line a dělají Partner Trade s každým tanečníkem, dokud se nedostanou na druhý konec Line. Ripple n znamená Trade s n tanečníky. n může obsahovat zlomek (např. Ripple 2 & 1/2 znamená Partner Trade se dvěma tanečníky a pak 1/2 Partner Trade s tím třetím). CZ: 678 | 543 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.432513 |
https://www.justcrackinterview.com/interviews/webb-fontaine-nigeria-limited/ | 1,696,282,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00140.warc.gz | 893,064,960 | 19,004 | # Webb fontaine nigeria limited Interview Questions Answers, HR Interview Questions, Webb fontaine nigeria limited Aptitude Test Questions, Webb fontaine nigeria limited Campus Placements Exam Questions
Find best Interview questions and answer for Webb fontaine nigeria limited Job. Some people added Webb fontaine nigeria limited 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 Webb fontaine nigeria limited Interview Questions and Answers for Freshers and experienced. These questions can surely help in preparing for Webb fontaine nigeria limited interview or job.
All of the questions listed below were collected by students recently placed at Webb fontaine nigeria limited.
Ques:- Number Series 10,2,8,2,6,2,?
4
Ques:- Knowledge of geographical area
Ques:- On a empty chessboard, a horse starts from a point (say location x,y) and it starts moving randomly, but once it moves out of board, it cant come inside. So what is the total probability that it stays within the board after N steps.
Ques:- Expand ——UCIL
Ques:- Why u left yor last job
Ques:- What are workload managers for?
Ques:- What do you expect from the company if you get hired?
Ques:- If Log2 x – 5 Log x + 6 = 0, then what would the value / values of x be?
Ques:- You are standing next to a well, and you have two jugs. One jug has a content of 3 liters and the other one has a content of 5 liters. How can you get just 4 liters of water using only these two jugs?
Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:
Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.
Now you have 4 litres of water in jug A.
Ques:- If A/B = 3/5,then 15A = ?
15a=45
Ques:- Draw and explain the TS and HS diagram of Rankine cycle?
Ques:- Tell me something about myself?
Ques:- In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?
50
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:- Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
In the example first we have to find the area of the field
we have given the following values ,
cost of fencing per meter = Rs 1.50
Diameter of circular field = 28
We have to find the area , a = ?
We know that,
Area = 2 π r
= 2 x 22/7 x 14
= 2 x 22 x 2
= 44 x 2
= 88 sq. m
So, area of circle is 88 sq.m
Now just multiply it with 1.50
cost of fencing = 88 x 1.50
= Rs. 132
So, the cost of fencing the circular field is Rs. 132
Ques:- Three 6 faced dice are thrown together. The probability that all the three show the same number on them is -.
Ques:- Find the size of the largest square slabs which can be paved on the floor of a room 5 meters 44 cm long and 3 meters 74 cm broad?
Ques:- A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 2:3:4. If the total value of all the coins in the bag is Rs.400, find the number of coins of each kind? | 1,030 | 3,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | longest | en | 0.867822 |
statsassignmenthelp83689.dbblog.net | 1,534,229,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221208676.20/warc/CC-MAIN-20180814062251-20180814082251-00322.warc.gz | 506,151,235 | 5,961 | # The 5-Second Trick For Stata Project Help
It is always a good idea to start just about every do file with remarks that include not less than a title, the title of the programmer who wrote the file, as well as the day. Assumptions about essential files must also be famous.
My situation is in a very regression context Y=AX1+BX2 And that i want a nonlinear check In the event the ratio H0: A/B=0, that's if A=0 for non-zero values of B, then it might effectively be considered a joint-take a look at of two nulls, H0: A=0 and H0: B≠0. How would I perform this take a look at?
The a million acres and globe-course infrastructure they incorporate have already been billed as the biggest donation of land from A personal entity to a country.
Stata omits observations that happen to be lacking the outcome or on the list of predictors. The log of GNP for each capita "explains" sixty one% with the variation in life expectancy in these nations around the world. We also see that a a person % rise in GNP per capita is related to a rise of 0.0277 many years in life expectancy. (To see this position Observe that if GNP boosts by one percent its log boosts by 0.01.)
The search phrase utilizing introduces a file title; this can be a file in the Pc, within the community, or on the net, as you will note whenever we talk about information input in Area two.
Christine This is amongst the primary causes I don’t use stata, thanks for these guidelines. I co-creator several studies that demand certain formatting models as well as tables simply cannot look as pictures due to the fact other people might need to edit them and generate textual view it now content around them.
Renal purpose is often impaired in individuals with coronary heart failure with decreased ejection portion and should deteriorate further immediately after blockade of your renin–angiotensin program.
The hypothesis that the estimate is predicated solely on likelihood is called the null hypothesis. As a result, the null speculation is true Should the noticed knowledge (during the sample) will not vary from what might be anticipated on The premise of likelihood on your own. The complement with the null hypothesis is named the choice speculation.
“I'm happy with my partner Doug and his vision which continues to guidebook us, As well as our full group, for finishing both of these national parks and also the broader network, A significant milestone of our initial twenty five a long time of labor,” Kristine Tompkins reported.
Let official site us operate straightforward descriptive figures for the two variables we have an interest in, utilizing the summarize command accompanied by the names on the variables (that may be omitted to summarize every little thing):
What’s can make anyone a leader? A social psychologist hypothesizes the extent to which people perceived for being leaders is positively related to the level of conversing they do, regardless of the things they are actually declaring. She conducts an experiment during which teams of participants resolve a challenge jointly.
The default is to employ what ever program they applied inside your stats course–no less than you realize the fundamentals.
) Moreover, these data files are linked from the web help, so you can bounce directly to the appropriate area on the manual. To find out more about the help technique variety help help.
Which one particular? Largely it is dependent upon the field you’re in. Social researchers must generally master SPSS as their most important bundle, primarily for the reason that that may be what their colleagues are employing. | 746 | 3,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-34 | longest | en | 0.949162 |
https://aimacode.github.io/aima-exercises/knowledge-logic-exercises/ex_28/ | 1,638,983,108,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00193.warc.gz | 148,445,785 | 5,796 | ### Artificial IntelligenceAIMA Exercises
Minesweeper, the well-known computer game, is closely related to the wumpus world. A minesweeper world is a rectangular grid of $N$ squares with $M$ invisible mines scattered among them. Any square may be probed by the agent; instant death follows if a mine is probed. Minesweeper indicates the presence of mines by revealing, in each probed square, the number of mines that are directly or diagonally adjacent. The goal is to probe every unmined square. 1. Let $X_{i,j}$ be true iff square $[i,j]$ contains a mine. Write down the assertion that exactly two mines are adjacent to $1,1$ as a sentence involving some logical combination of $X_{i,j}$ propositions. 2. Generalize your assertion from (a) by explaining how to construct a CNF sentence asserting that $k$ of $n$ neighbors contain mines. 3. Explain precisely how an agent can use {DPLL} to prove that a given square does (or does not) contain a mine, ignoring the global constraint that there are exactly $M$ mines in all. 4. Suppose that the global constraint is constructed from your method from part (b). How does the number of clauses depend on $M$ and $N$? Suggest a way to modify {DPLL} so that the global constraint does not need to be represented explicitly. 5. Are any conclusions derived by the method in part (c) invalidated when the global constraint is taken into account? 6. Give examples of configurations of probe values that induce long-range dependencies such that the contents of a given unprobed square would give information about the contents of a far-distant square. (Hint: consider an $N\times 1$ board.)
Minesweeper, the well-known computer game, is closely related to the wumpus world. A minesweeper world is a rectangular grid of $N$ squares with $M$ invisible mines scattered among them. Any square may be probed by the agent; instant death follows if a mine is probed. Minesweeper indicates the presence of mines by revealing, in each probed square, the number of mines that are directly or diagonally adjacent. The goal is to probe every unmined square. 1. Let $X_{i,j}$ be true iff square $[i,j]$ contains a mine. Write down the assertion that exactly two mines are adjacent to $1,1$ as a sentence involving some logical combination of $X_{i,j}$ propositions. 2. Generalize your assertion from (a) by explaining how to construct a CNF sentence asserting that $k$ of $n$ neighbors contain mines. 3. Explain precisely how an agent can use {DPLL} to prove that a given square does (or does not) contain a mine, ignoring the global constraint that there are exactly $M$ mines in all. 4. Suppose that the global constraint is constructed from your method from part (b). How does the number of clauses depend on $M$ and $N$? Suggest a way to modify {DPLL} so that the global constraint does not need to be represented explicitly. 5. Are any conclusions derived by the method in part (c) invalidated when the global constraint is taken into account? 6. Give examples of configurations of probe values that induce long-range dependencies such that the contents of a given unprobed square would give information about the contents of a far-distant square. (Hint: consider an $N\times 1$ board.)
Submit Solution | 756 | 3,236 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-49 | latest | en | 0.92521 |
https://www.queryhome.com/puzzle/35500/simple-confusing-puzzle-grocery-worth-much-loss-shopkeeper | 1,603,351,525,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00217.warc.gz | 858,257,469 | 32,234 | # A very simple but confusing puzzle. A lady buys grocery worth Rs.350.........."How much LOSS did the shopkeeper face ?"
100 views
A very simple but confusing puzzle.
A lady buys grocery worth Rs.350 from a shop. (shopkeeper selling the goods with zero profit)
The lady gives him 2000 rs note. The shopkeeper gets the change from next shop, keeps 350 for himself and returns Rs.1650 to d lady.
Later the shopkeeper of the next shop comes with the Rs.2000 note saying "duplicate" and takes his money back.
"How much LOSS did the shopkeeper face ?"
posted Apr 23
Aswer: Rs.2000
LOSS RS 2000 RUPEES IE RS 350 FOR THE SELLING ITEM AND RS 1650 RETURNED AMOUNT TO LADY
Similar Puzzles
+1 vote
A lady buys goods worth Rs.200 from a shop. (shopkeeper is selling the goods with zero profit). The lady gives him Rs.1000 note. The shopkeeper gets the change from the next shop and keeps Rs.200 for himself and returns Rs.800 to the lady. Later the shopkeeper of the next shop comes with the Rs.1000 note saying “duplicate” and takes his money back.
How much LOSS did the shopkeeper face?
One man steals Rs. 100 from a shop.
Then the same man buys Rs 70 worth goods from the same shop.
The shopkeeper gives him back Rs.30.
Now tell,
What is the loss of the shopkeeper ? | 328 | 1,270 | {"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} | 3.09375 | 3 | CC-MAIN-2020-45 | longest | en | 0.956614 |
https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_19&diff=prev&oldid=82343 | 1,686,081,216,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00042.warc.gz | 134,655,414 | 12,805 | Difference between revisions of "2010 AMC 10A Problems/Problem 19"
Problem
Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$
Solution
Solution 1
It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on triangle ABC, we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$. Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$.
If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$, we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$. The area of $ABCDEF$ is therefore
$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$.
Based on the initial conditions,
$$\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)$$
Simplifying this gives us $r^2-6r+1 = 0$. By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$.
Solution 2
As above, we find that the area of $\triangle ACE$ is $\frac{\sqrt3}4(r^2+r+1)$.
We also find by the sine triangle area formula that $ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4$, and thus $$\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}$$ This simplifies to $r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}$.
See also
2010 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Invalid username
Login to AoPS | 850 | 2,198 | {"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": 41, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-23 | latest | en | 0.631508 |
https://kr.mathworks.com/matlabcentral/answers/1835273-solve-system-of-equations-with-some-knowns-and-unknowns-in-the-same-matrix | 1,674,869,282,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00581.warc.gz | 383,791,361 | 31,208 | # Solve system of equations with some knowns and unknowns in the same matrix
조회 수: 1(최근 30일)
Yusuf 2022년 10월 25일
댓글: David Goodmanson 2022년 10월 26일
I am trying to solve for x in the equation Ax=b in matlab, but not all elements of x and b are unknown. I know the value of a few elements in x, and the rest are unknown, and I know the value of the elements in b for the x values that are unknown (ex. if I know x1, I dont know b1. If I dont know x2, I know b2, etc.). How can I automatically and efficiently solve for all of the unknowns in both vectors without manually writing the problem out as a system of equations each time?
댓글을 달려면 로그인하십시오.
### 채택된 답변
Bruno Luong 2022년 10월 25일
편집: Bruno Luong 2022년 10월 25일
% Random example
A = rand(5,5);
x = rand(5,1),
x = 5×1
0.7978 0.8825 0.0743 0.2101 0.6719
b = A*x,
b = 5×1
1.4535 1.7942 1.2279 1.4185 1.3270
[m,n] = size(A);
% put NaN at the position where x is unknown;
xunknown = randperm(n,3);
x(xunknown) = NaN;
% same for b
bunknown = randperm(m,2);
b(bunknown) = NaN;
% Reconstruct unknown x and then b
xunknown = isnan(x);
bunknown = isnan(b);
x(xunknown)=A(~bunknown,xunknown)\(b(~bunknown)-A(~bunknown,~xunknown)*x(~xunknown))
x = 5×1
0.7978 0.8825 0.0743 0.2101 0.6719
b(bunknown)=A(bunknown,:)*x
b = 5×1
1.4535 1.7942 1.2279 1.4185 1.3270
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Bruno Luong 2022년 10월 25일
The solution I give is applicable for all cases.
댓글을 달려면 로그인하십시오.
### 추가 답변(1개)
David Goodmanson 2022년 10월 26일
편집: David Goodmanson 2022년 10월 26일
Hi Yusuf,
I'm later on an answer (and losing the race more often, to the extent that there is a race, which to be honest there sometimes is), but here is a slightly different way.
% data
n = 7;
m = 3;
A = rand(n,n);
i = sort(randperm(n,m))'; % index of unknown x
j = setdiff(1:n,i)'; % index of unknown b, the complementary index to i
x = zeros(n,1);
x(j) = rand(size(j)) % known x, zeros elsewhere
b = zeros(n,1);
b(i) = rand(size(i)) % known b, zeros elsewhere
%solve
G = zeros(n,n);
G(:,i) = A(:,i);
G(j,j) = -eye(length(j));
g = b-A(:,j)*x(j);
z = G\g
% z is a vector of all the unknowns,
% x(i) for the i indices
% b(j) for the j indices
% check
x(i) = z(i); % insert unknowns into orginal x vector
b(j) = z(j); % insert unknowns into orginal b vector
A*x-b % should be small
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
David Goodmanson 2022년 10월 26일
Hi Bruno, that's a reasonable point about scaling part of the matrix I called G. I like the method because it gives all the unknowns in one go with backslash, different from the standard (and probably more foolproof) two-step process that you used.
댓글을 달려면 로그인하십시오.
### 범주
Find more on Matrix Indexing in Help Center and File Exchange
R2022b
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 1,013 | 2,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-06 | latest | en | 0.593871 |
https://forum.openoffice.org/en/forum/viewtopic.php?f=9&t=5896&p=27393 | 1,600,795,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00587.warc.gz | 397,988,283 | 6,593 | ## [Solved] How do I get the row number?
### [Solved] How do I get the row number?
Suppose I have a list of expenses and I want to get the text defining the maximum expense.
For the sake of argument, the text is in column B and the expense in column D.
To get the maximum I would use =MAX(D1:D100), but I want to know what row that is.
If I knew the row I could put in column G1 =B10, where B10 would be some sort of variable.
This should be a simple question to someone with more experience than myself.
Thanks,
Ilan
Last edited by Ilan on Mon May 19, 2008 4:01 pm, edited 1 time in total.
OOo 3.0.X on Ubuntu 8.x + Windows XP
Ilan
Posts: 61
Joined: Tue Apr 15, 2008 9:44 am
Location: Haifa, Israel
### Re: How do I get the row number?
Get the text in B from the row where max(D) is found:
=INDEX(\$B\$1:\$B\$100;MATCH(MAX(\$D\$1:\$D\$100);\$D\$1:\$D\$100;0))
MATCH returns an integer number>0 indicating the matched position (or #NA in case of no match).
INDEX returns a value out of a range, specified by the row index and (optional) column index.
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, no OpenOffice, LibreOffice 6.4
Villeroy
Volunteer
Posts: 28544
Joined: Mon Oct 08, 2007 1:35 am
Location: Germany | 387 | 1,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.869058 |
https://us.metamath.org/mpeuni/wfr2a.html | 1,723,536,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00308.warc.gz | 475,148,615 | 4,906 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > wfr2a Structured version Visualization version GIF version
Theorem wfr2a 7973
Description: A weak version of wfr2 7975 which is useful for proofs that avoid the Axiom of Replacement. (Contributed by Scott Fenton, 30-Jul-2020.)
Hypotheses
Ref Expression
wfr2a.1 𝑅 We 𝐴
wfr2a.2 𝑅 Se 𝐴
wfr2a.3 𝐹 = wrecs(𝑅, 𝐴, 𝐺)
Assertion
Ref Expression
wfr2a (𝑋 ∈ dom 𝐹 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Proof of Theorem wfr2a
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 fveq2 6655 . . 3 (𝑥 = 𝑋 → (𝐹𝑥) = (𝐹𝑋))
2 predeq3 6127 . . . . 5 (𝑥 = 𝑋 → Pred(𝑅, 𝐴, 𝑥) = Pred(𝑅, 𝐴, 𝑋))
32reseq2d 5822 . . . 4 (𝑥 = 𝑋 → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑥)) = (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)))
43fveq2d 6659 . . 3 (𝑥 = 𝑋 → (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑥))) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
51, 4eqeq12d 2814 . 2 (𝑥 = 𝑋 → ((𝐹𝑥) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑥))) ↔ (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)))))
6 wfr2a.1 . . 3 𝑅 We 𝐴
7 wfr2a.2 . . 3 𝑅 Se 𝐴
8 wfr2a.3 . . 3 𝐹 = wrecs(𝑅, 𝐴, 𝐺)
96, 7, 8wfrlem12 7967 . 2 (𝑥 ∈ dom 𝐹 → (𝐹𝑥) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑥))))
105, 9vtoclga 3523 1 (𝑋 ∈ dom 𝐹 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2111 Se wse 5480 We wwe 5481 dom cdm 5523 ↾ cres 5525 Predcpred 6122 ‘cfv 6332 wrecscwrecs 7947 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-sep 5171 ax-nul 5178 ax-pr 5299 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-eu 2629 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-ral 3111 df-rex 3112 df-reu 3113 df-rmo 3114 df-rab 3115 df-v 3444 df-sbc 3723 df-csb 3831 df-dif 3886 df-un 3888 df-in 3890 df-ss 3900 df-nul 4247 df-if 4429 df-sn 4529 df-pr 4531 df-op 4535 df-uni 4805 df-iun 4887 df-br 5035 df-opab 5097 df-mpt 5115 df-id 5429 df-po 5442 df-so 5443 df-fr 5482 df-se 5483 df-we 5484 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-pred 6123 df-iota 6291 df-fun 6334 df-fn 6335 df-fv 6340 df-wrecs 7948 This theorem is referenced by: wfr2 7975
Copyright terms: Public domain W3C validator | 1,463 | 2,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.193834 |
http://www.openwetware.org/index.php?title=Imperial_College/Courses/Spring2008/Synthetic_Biology/Computer_Modelling_Practicals/Practical_2&oldid=237475 | 1,498,635,702,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00423.warc.gz | 628,433,352 | 12,369 | # Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Practical 2
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Synthetic Biology (Spring2008): Computer Modelling Practicals
Practical 2
Objectives:
• To explore computationally some simple genetic motifs:
• Constitutive Gene Expression.
• Activated and Repressed Gene Expression.
• Positive and Negative Feedback Gene Expression.
Deliverables
In this section, we investigate a very common motif in biochemistry. It models the continuous and constant synthesis of a compound, and its natural degradation. From a Mathematical point of view, the model is described as a first-order linear ordinary differential equation.
Model CellDesigner Instructions
$0 \xrightarrow{k_{1}} A \xrightarrow{k_{2}} 0$
• Open a NEW document. Name it 'Synthesis_Degradation_Model'.
Build the topology of the reaction network
• Create a 'Source' compound, thanks to the 'simple molecule' icon.
• The same way, create a 'A' compound.
• Create a reaction link between 'Source' and 'A', Reaction_1, using the 'state transition' icon.
From the law of mass action, we can write:
• $\frac{d[A]}{dt} = k_{1} - k_{2}*[A]$
Define the kinetics driving the reaction network
• Edit Reaction_1, and define a new parameter k_1 = 1.0, and create the kinetic law according to the ODE system.
• Edit Reaction_2, and define a new parameter k_2 = .01, and create the kinetic law according to the ODE system.
Simulate the dynamical behaviour
• Open Simulation Panel
• Set time for the simulation to be 1000 seconds, with 1000 points.
• Press Execute, and check results.
• Questions:(see report structure)
• Run a simulation over t=1000s, nb points=1000. Comment on the time evolution of 'A'. (illustration needed).
• Using the dynamical system definition, what is the steady state level of 'A' with regards to the parameters k1 and k2 ? (Steady state means that $\frac{d[A]}{dt}=0$
• Using the 'Parameter Scan' feature, illustrate the influence of both parameters (k_1 and k_2), on the steady state level of 'A' (illustration needed).
• Bonus: Give the analytical solution of the ODE system.
• Now, consider that k_1=0, and [A]t = 0 = A0 > 0. Keep k_2=0.01. Illustrate the concept of half-life for the compound 'A'.
• Check Derivation of the Half-life expression: exponential decay model
Part II: Constitutive Gene Expression
In this section, we explore a computational model to describe a constitutively expressed gene. The model is based on a simple interpretation of the central dogma: Gene -(transcription)-> mRNA -(translation)-> Protein, with both the mRNA molecules and the protein molecules being naturally degraded. If you need a quick refresh on the Central Dogma, have a look at the 'additional resources' section below.
The modelling parameters, used throughout this practical, are characteristic of the E.Coli bacteria.
Model CellDesigner Instructions
$Gene \rightarrow mRNA \rightarrow Protein$
Define the topology of the reaction network:
• This file contains the network topology describing a simple constitutively expressed gene model. No kinetics information is yet described.
Following the law of mass action, we can write:
\begin{alignat}{1} \frac{d[mRNA]}{dt} & = k_{1} - d_{1}[mRNA] \\ \frac{d[Protein]}{dt} & = k_{2}[mRNA] - d_{2}[Protein] \\ \end{alignat}
• k1 is the transcription rate. It is considered to be constant, and it represents the number of mRNA molecules produced per gene, and per unit of time.
• d1 is the mRNA degradation rate of the mRNA molecule. The typical half-life for the mRNAs, in E.Coli, has been measured to be between 2min and 8min (average 5min).
• k2 is the translation rate. It is considered to be constant, and it represents the number of protein molecules produced per mRNA molecule, and per unit of time.
• d2 is the protein degradation rate. In this practical, we will only consider very stable proteins, i.e. not engaged in any active degradation pathways. In that case, we can approximate the degradation of the protein to be only due to the dilution effect caused by the cell division. Cell division will be 40min.
Questions:
• From the ODE system given, write down the steady-state expression of the [mRNA] concentration and the [Protein] concentration, with regards to k1,k2,d1,d2. Remember that steady-state means that we consider d[mRNA]/dt=0 and d[protein]/dt=0.
• Using the previously found equations, and knowing that average number of mRNA molecules per gene is 2.5 in E.Coli, what is the average transcription rate ? Keep in mind that the problem only gives the mRNA half life, not the actual degradation rate (see the exponential decay model for more info)
• In the same way, knowing that the average number of proteins per gene is 1000 in E.Coli, what is the average translation rate ? Once again, remember that the problem gives the protein half-life, and not the actual protein degradation rate.
• In CellDesigner, define all the necessary kinetics laws for the model, and create all the appropriate parameters.
• Run a simulation, and comment on the simulation outputs (mRNA and Protein), with regards to the transient phase and the steady-state.
• From a Synthetic Biology point of view, this motif can be seen as a 'Protein Generator'. One might be interested in controlling the steady-state protein output level of this device. Using the 'parameter scan' function, run a simple sensitivity analysis on each of the 4 parameters, within a 10% range of their default value. Illustrate, and describe briefly, how each parameter impacts the protein steady state.
Next, we want to explore the quasi-steady-state assumption on the mRNA molecules expression. From the previous simulations, you might have noticed that the concentration of mRNA reaches steady-state very quickly, compared to the protein concentration. In that case, we want to explore a model where we would consider that the [mRNA] concentration is always considered at steady state , i.e. $\frac{d[mRNA]}{dt}=0$ all the time. This model means that we want to apply a quasi-steady-state assumption on the [mRNA] molecules.
Model CellDesigner Instructions
$Gene \rightarrow Protein$
• Within the same CellDesigner file, build a new network.
• Create a network that will allow you to directly synthesise a Protein from a Gene. Also, you wan to describe the fact that the protein is being naturally degraded.
Following the law of mass action, we can write:
\begin{alignat}{2} \frac{d[Protein]}{dt} = s - d[Protein] \\ \end{alignat}
• Create the appropriate parameters and reaction kinetics laws.
Questions:
• Taking into account the quasi-steady-state assumption on the [mRNA], work-out the value of 's', and 'd' with regards to k1,k2,d1,d2, so that the two models are equivalent.
• Simulate the full model, alongside with the quasi-steady state approximation. When using the parameters from the previous section, comment on how good this approximation seems to be.
• Why do you think such a reduced model would be useful ?
Part III: Activated and Repressed Gene Expression
Very few genes are known to have a purely constitutive expression, most genes have their expression controlled by some outside signals (DNA-binding proteins, Temperature, metabolites, RNA molecules ...). In this section, we will particularly focus on the study of DNA-binding proteins, called transcription factors. These proteins, when binding to a promoter region, can either have an activation effect on the gene (positive control), or a repression effect (negative control). In prokaryotes, control of transcriptional initiation is considered to be the major point of regulation. In this part of the tutorial, we investigate one of the most common model used to describe this type of interactions.
Let's first consider the case of a transcription factor acting as a repressor. A repressor will bind to the DNA so that it prevents the initiation of transcription. Typically, we expect the transcription rate to decrease as the concentration of repressor increases. A very useful family of functions to describe this effect is the Hill function: $f(R)=\frac{\beta.{K_m}^n}{{K_m}^n+R^n}$. The Hill function can be derived from considering the transcription factor binding/unbinding to the promoter region to be at equilibrium (similar to the enzyme-substrate assumption in the Michaelis-Menten formula). This function has 3 parameters: β,n,Km:
• β is the maximal expression rate when there is no repressor, i.e. f(R = 0) = β.
• Km is the repression coefficient (units of concentration), it is equal to the concentration of repressor needed to repressed by 50% the overall expression, i.e $f(K_m)=\frac{\beta}{2}$
• n is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.
Model CellDesigner Instructions
\begin{align} & Repressor \\ & \bot \\ Gene &\rightarrow mRNA \rightarrow Protein \end{align}
Define the topology of the reaction network:
• From the previous constitutive expression model, create a new 'simple molecule' compound for the repressor.
• Create an inhibitory link between the repressor, and the transcription reaction.
• Save your file under a new name.
Hill function for transcriptional repression:
$transcriptionRate=\frac{k_1.{K_m}^n}{{K_m}^n+R^n}$
• k1: maximal transcription rate
• Km: repression coefficient
• n: Hill coefficient
Following the law of mass action, we can write:
\begin{alignat}{1} \frac{d[mRNA]}{dt} & = \frac{k_{1}.{K_m}^n}{{K_m}^n+R^n} - d_{1}[mRNA] \\ \frac{d[Protein]}{dt} & = k_{2}[mRNA] - d_{2}[Protein] \\ \end{alignat}
• Create all the necessary kinetics laws and parameters.
• Consider K_m=100, n=2. k_1, k_2, d_1, d_2 as found previously.
Questions:
• We want to establish the transfer function between the [repressor] concentration and the protein steady-state level. To do so plot [Protein]steadystate = F([Repressor]) using the results of the simulations where the repressor concentration varies from [0 to 1000] by steps of 10.
• Suggest an application where this genetic circuit might be useful.
Now, let's consider the case of a transcription factor acting as an activator. An activator will bind to the DNA so that it promotes the initiation of transcription. Typically, we expect the transcription rate to increase as the concentration of activator increases. Once again, the Hill type function will be useful to describe the interaction effect. It is slightly different from the previous one: $f(R)=\frac{\beta.{A}^n}{{K_m}^n+A^n}$. The Hill function can be derived from considering the transcription factor binding/unbinding on the promoter region to be at equilibrium (similar to the enzyme-substrate assumption in the Michaelis-Menten formula). This function has 3 parameters: β,n,Km:
• β is the maximal expression rate when there is a lot of activators, i.e. $f(A \gg K_m)=\beta$.
• K_m is the activation coefficient (units of concentration), it is equal to the concentration of activator needed to activate by 50% the overall expression, i.e $f(K_m)=\frac{\beta}{2}$
• n is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.
Model CellDesigner Instructions
\begin{align} & Activator \\ & \downarrow \\ Gene & \rightarrow mRNA \rightarrow Protein \end{align}
Define the topology of the reaction network:
• From the previous constitutive expression model, create a new 'simple molecule' for the activator.
• Create an catalytic reaction link between the activator, and the transcription reaction.
• Save your file under a new name.
Hill function for transcriptional activation:
$transcriptionRate=\frac{k_1.{A}^n}{{K_m}^n+A^n}$
• k1: maximal transcription rate
• Km: activation coefficient
• n: Hill coefficient
• Create all the necessary kinetics laws and parameters.
• Consider K_m=100, n=2. k_1, k_2, d_1, d_2 as found previously.
Following the law of mass action, we can write:
\begin{alignat}{1} \frac{d[mRNA]}{dt} & = \frac{k_{1}.A^n}{{K_m}^n+A^n} - d_{1}[mRNA] \\ \frac{d[Protein]}{dt} & = k_{2}[mRNA] - d_{2}[Protein] \\ \end{alignat}
• We want to establish the transfer function between the [Activator] concentration and the protein steady-state level. To do so plot [Protein]steadystate = F([Activator]) using the results of the simulations where the Activator concentration varies from [0 to 1000] by steps of 10.
• Suggest an application where this genetic circuit might be useful.
Part IV: Positive and Negative Feedback
We have just seen that some genes can be controlled by DNA-binding proteins. This type of interactions will enable the construction of genetic networks, called transcription networks. One gene produces a protein, which then binds to one or more other genes, and control them (positively or negatively), and so on. During the practical 3, we will explore some of those circuits.
At the same time, it has been observed that some genes can also regulate themselves (positively or negatively). When a gene is regulated by the very same protein it produces, the motif is called a feedback. It can be a positive feedback, or a negative, depending on the type of interaction there is between the protein and the promoter region. In this section, we will explore those 2 options.
Model CellDesigner Instructions
• do not consider this graph ...
\begin{align} Gene & \rightarrow mRNA \rightarrow Protein \\ & \bot \\ Gene & \rightarrow mRNA \rightarrow Protein \\ & \downarrow \\ Gene & \rightarrow mRNA \rightarrow Protein \end{align}
• Define the topology of the reaction network:
• Create a new CellDesigner Project where you will build 3 independent genetic motif:
• Constitutive expression
• Negative feedback
• Positive feedback
Questions:
• Provide the 3 ODE systems you need to describe the 3 different motifs:constitutive, negative feedback, and positive feedback.
• Create all the necessary kinetic laws considering the parameters as follow:
• Constitutive Expression: K_1=.346; d_1=.138 ; K_2=6.931 ; d_2=0.017 .
• Negative Feedback Expression: K_1=33.847 ; n=1; Km=10 ; d_1=.138 ; K_2=6.931 ; d_2=0.017
• Positive Feedback Expression: K_1=.346 ; n=1; Km=10 ; d_1=.138 ; K_2=6.931 ; d_2= 0.017
• Compare the different models with regards to their protein steady-state level, and also with regards to how fast each model is reaching its steady-state.
• Comment on the benefits and drawbacks of using a negative, or positive, feedback. | 3,486 | 14,474 | {"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": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-26 | latest | en | 0.799149 |
https://www.bartleby.com/questions-and-answers/medicine.-after-bypass-surgery-patients-are-placed-in-an-intensive-care-unit-icu-until-their-conditi/f7564dff-d8b4-4bca-8846-8a0cac6e949e | 1,627,653,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00301.warc.gz | 675,889,494 | 36,296 | # Medicine. After bypass surgery, patients are placed in an intensive care unit (ICU) until their condition stabilizes. Then they are transferred to a cardiac care ward (CCW), where they remain until they are released from the hospi- tal. In a particular metropolitan area, a study of hospital records produced the following data: each day 2% of the patients in the ICU died, 52% were transferred to the CCW, and the remainder stayed in the ICU. Furthermore, each day 4% of the patients in the CCW developed complica- tions and were returned to the ICU, 1% died while in the CCW, 22% were released from the hospital, and the re- mainder stayed in the CCW. (A) In the long run, what percentage of the patients in the ICU are released from the hospital? (B) In the long run, what percentage of the patients in the CCW die without ever being released from the hospital? (C) What is the average number of days that a patient in the ICU will stay in the hospital?
Question
Please explain the steps to find the answer for each part so I can learn. Thank you
### Want to see this answer and more?
Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*
*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.
Tagged in
Science
Biology
### Other
Transcribed Image Text
Medicine. After bypass surgery, patients are placed in an intensive care unit (ICU) until their condition stabilizes. Then they are transferred to a cardiac care ward (CCW), where they remain until they are released from the hospi- tal. In a particular metropolitan area, a study of hospital records produced the following data: each day 2% of the patients in the ICU died, 52% were transferred to the CCW, and the remainder stayed in the ICU. Furthermore, each day 4% of the patients in the CCW developed complica- tions and were returned to the ICU, 1% died while in the CCW, 22% were released from the hospital, and the re- mainder stayed in the CCW. (A) In the long run, what percentage of the patients in the ICU are released from the hospital? (B) In the long run, what percentage of the patients in the CCW die without ever being released from the hospital? (C) What is the average number of days that a patient in the ICU will stay in the hospital? | 553 | 2,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.974603 |
https://maths2art.com.sg/tag/guided-discovery/page/2/ | 1,571,230,844,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00378.warc.gz | 610,911,906 | 38,337 | Most students learn about Circles in Grade 5 or Grade 6 and this is how we usually introduce this topic to them. We draw a circle, show them the center and radius and then we give them the formula for circumference and area. Then, we tell them that π is 22/7 or 3.14 I feel that... Continue Reading →
I have been planning to write an article on 'Division by Fractions' for the past few days. My thoughts were clear, the conversations with the students were well stored in my mind, but somehow, this article was still waiting in the 'to-do list'. Recently, I got a chance to meet Professor Ron Aharoni and attend... Continue Reading →
I have observed that many books start Fraction with Pizza example. I have a suggestion - Let’s replace Pizza with Cadbury, Chikki (Indian sweet) or anything which is rectangular in shape for a better visual understanding. Let me explain my point with an example. Comparison of Fraction Which fraction is bigger 3/4 or 5/6 ?... Continue Reading →
Maths2Art is not a tuition centre, it is a centre for Guided Discovery. Lot of people ask me, “I have heard about you. You take Maths tuition, right ?” Thanks to such questions, I got a strong reason to dig deeper and get a clearer picture of Maths2Art. What am I doing @Maths2Art? I am... Continue Reading → | 300 | 1,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.962356 |
https://myhomeworkwriters.com/common-stock-valuation-assignment-homework-for-you-2/ | 1,680,288,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00745.warc.gz | 455,012,481 | 17,856 | Common Stock Valuation Assignment | Homework For You
(Common stock valuation) The common stock of NCP paid \$1.45 in dividends last year. Dividends are expected to grow at an annual rate of 9.50 percent for an indefinite number of years.
Don't use plagiarized sources. Get Your Assignment on
Common Stock Valuation Assignment | Homework For You
Just from \$13/Page
a. If NCP’s current market price is \$27.87 per share, what is the stock’s expected rate of return?
b. If your required rate of return is 11.5 percent, what is the value of the stock for you?
c. Should you make the investment?
____________
a. If NCP’s current market price is \$27.87 per share, the stock’s expected rate of return is ___%. (Round to two decimal places.)
b. If your required rate of return is 11.5 percent, the value of the stock would be \$___ . (Round to the nearest cent.)
c. You should ( buy or sell ) the stock because the expected rate of return is ( less than or greater than ) your required rate of return or the value of the stock is ( larger than or smaller than ) the current market price.Get Finance homework help today
Calculate the price
Pages (550 words)
\$0.00
*Price with a welcome 15% discount applied.
Pro tip: If you want to save more money and pay the lowest price, you need to set a more extended deadline.
We know how difficult it is to be a student these days. That's why our prices are one of the most affordable on the market, and there are no hidden fees.
Instead, we offer bonuses, discounts, and free services to make your experience outstanding.
How it works
Receive a 100% original paper that will pass Turnitin from a top essay writing service
step 1
Fill out the order form and provide paper details. You can even attach screenshots or add additional instructions later. If something is not clear or missing, the writer will contact you for clarification.
Pro service tips
How to get the most out of your experience with Homework Writing Services
One writer throughout the entire course
If you like the writer, you can hire them again. Just copy & paste their ID on the order form ("Preferred Writer's ID" field). This way, your vocabulary will be uniform, and the writer will be aware of your needs.
The same paper from different writers
You can order essay or any other work from two different writers to choose the best one or give another version to a friend. This can be done through the add-on "Same paper from another writer."
Copy of sources used by the writer
Our college essay writers work with ScienceDirect and other databases. They can send you articles or materials used in PDF or through screenshots. Just tick the "Copy of sources" field on the order form.
Testimonials
See why 20k+ students have chosen us as their sole writing assistance provider
Check out the latest reviews and opinions submitted by real customers worldwide and make an informed decision.
Great Content!
Customer 463469, October 20th, 2022
Other
Outstanding.
Customer 462823, April 7th, 2022
Thank you!
Customer 453187, March 2nd, 2022
Technology
Good job after revision. Make Grammarly your tool.
Customer 462769, March 29th, 2022
Art (Fine arts, Performing arts)
Always follow instructions, especially on style.
Customer 460547, March 29th, 2022
Military
Excellent work
Customer 456821, October 11th, 2022
Social Work and Human Services
Great writer! Thanks for the revision, the paper is so much better than ones previous.
Customer 463301, November 8th, 2022
Marketing
I like the work.
Customer 463095, June 22nd, 2022
Nursing
Several grammar issues identified.
Customer 462821, April 6th, 2022
accounting
Good work.
Customer 459155, May 22nd, 2022
MGMT 221 Introduction to Management Information System
Great job!
Customer 457731, February 25th, 2022
Medical terminology
Thank you so much
Customer 454783, April 15th, 2022
11,595
Customer reviews in total
96%
Current satisfaction rate
3 pages
Average paper length
37%
Customers referred by a friend | 971 | 3,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.912015 |
https://radiopaedia.org/articles/pearsons-chi-squared-test?lang=us | 1,701,757,531,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00741.warc.gz | 543,827,196 | 16,236 | Pearson's chi-squared test
Last revised by Candace Makeda Moore on 10 Sep 2020
The Pearson's chi-squared test is one of the most common statistical tests found in radiology research. It is a type of non-parametric test, used with two categorical variables (not continuous variables).
Concept
The heart of the chi-squared test is a 2 x 2 contingency table.
We usually have a set of patients and a set of controls. We then want to test whether our independent variable is associated with our dependent variable (or not).
1. first we fill out the 2 x 2 table as if there were no association ("expected values", Ei). Divide the cases and controls (proportionally) into the four cells of the contingency table.
2. then fill in the actual values that you found from your study ("observed values", Oi).
3. this is the [Oi - Ei] per cell.
Next, square these [Oi - Ei] values so that the sum does not equal zero.
Finally, divide each [Oi - Ei] value by Ei. This move accounts for the distribution of values around the mean (sort of the standard deviation). The reason this is legitimate is because the values in the table follow a Poisson distribution.
The final result of this maneuver is
χ2 = Σ ([Oi - Ei]2 / Ei)
This results in a chi-squared number (χ2), which can be checked in a table for significance. The degree of freedom for a 2 x 2 table is 1. If the χ2 value is above the level in the table, then we can reject the null hypothesis (no association between the variables).
Points
• chi-squared tests work with categorical variables (e.g. disease vs no disease, got imaging test vs did not get imaging test, etc)
• it is not meant for continuous variables (e.g. length, time, radiation dose, etc)
• chi-squared tests work best with a reasonably high n
• for low n studies, consider other non-parametric tests that compare medians, not means | 453 | 1,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | latest | en | 0.828948 |
https://www.pompesfunebres-vihiers.com/b1rafug/6e53f6-proving-lines-parallel-practice | 1,627,081,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00592.warc.gz | 993,365,610 | 9,473 | State the theorem or postulate that justifies each answer. 2-2 Additional Practice Proving Lines Parallel Use the figure for Exercises 1–4. 0. 9th - 10th grade. 7.2k plays . The lesson covers: {{courseNav.course.topics.length}} chapters | The pack contains a full lesson plan, along with accompanying resources, including a … Proving Lines are Parallel (Using Converses) Sort, Cut, and Paste Activity Students will practice identifying parallel lines using the Corresponding Angles Converse, Alternate Interior Angles Converse, Alternate Exterior Angles Converse, and Consecutive Interior Angles Converse. Finish Editing. Given: ∠4 ≅ ∠5 4. Now we get to look at the angles that are formed by the transversal with the parallel lines. All other trademarks and copyrights are the property of their respective owners. Just remember that when it comes to proving two lines are parallel, all we have to look at … Play. The slope of one line is the negative reciprocal of the other. Guided Practice; Cardstack; Guided Practice Review the Guided Practice Presentation below. Given: m || n and a || b. Proof. Introduction; Learn; Try It; Task; Try It. Review the Cardstack Gamebelow. Proving Lines Parallel continued You can also prove that two lines are parallel by using the converse of any of the other theorems that you learned in Lesson 3-2. → Cardstack. A basic understanding of geometry will help you be successful on this quiz. Given ' suppl. Live Game Live. Use the figure for Exercises 2 and … Students learn the converse of the parallel line postulate and the converse of each of the theorems covered in the previous lesson, which are as follows. ext. Since ml—I = mL2, then Ll Z 2. Students are then asked to determine which lines are parallel … Mathematics. Proving Lines Parallel DRAFT. If two lines are cut by a transversal and corresponding angles are congruent, then the lines are parallel. 3.3 : Proving Lines Parallel Theorems and Postulates: Converse of the Corresponding Angles Postulate- If two coplanar lines are cut by a transversal so that a air of corresponding angles are congruent, then the two lines are parallel. '. Name Class Date 3-3 Geometry Practice – Proving Lines Parallel. Parallel Lines – Proving Lines Are Parallel. 's' : ''}}. Cookies are not enabled on your browser. 3-3 Practice Form G Proving Lines Parallel d n e; corr. For FREE access to this lesson, select your course from the categories below. If you're seeing this message, it means we're having trouble loading external resources on our website. To play this quiz, please finish editing it. All rights reserved. Ab dc proof. flashcard set{{course.flashcardSetCoun > 1 ? 3 years ago. Please enable cookies in your browser preferences to continue. Services, Using Converse Statements to Prove Lines Are Parallel, Quiz & Worksheet - Proving Parallel Lines, Parallel Lines: How to Prove Lines Are Parallel, {{courseNav.course.mDynamicIntFields.lessonCount}}, Constructing a Parallel Line Using a Point Not on the Given Line, The Parallel Postulate: Definition & Examples, What Are Polygons? 10 Qs . Given: m∠3 = 12x°, m∠5 = 18x°, x = 6 5. Practice. angles AC n BD; corr. Quizzes, practice exams & worksheets. Proving Lines Parallel Worksheet Answers Along with Chain Rule Practice Worksheet Choice Image Worksheet Math Download by size: Handphone Tablet Desktop (Original Size) The first Proving Lines Parallel Worksheet Answers is all based on the basic problem solving skills. Practice A Proving Lines Parallel 1. Finish Editing. As a member, you'll also get unlimited access to over 83,000 lessons in math, 9th - 12th grade . To the same l are o. Proving Lines are Parallel Students learn the converse of the parallel line postulate. If … text-only version. are O. l1 Ol4 If corresp. ' If you need assistance please contact support@mathhelp.com. Examplèâ so that n. (3x+ 10)0 Examplê If ml-I = mL2, determine which lines, if any, are parallel. Prove: ∠ 3 ∠13. … 0. 312 times. We hope your happy with this proving lines parallel worksheet answers new geometry 3. | {{course.flashcardSetCount}} Plus, get practice tests, quizzes, and personalized coaching to help you succeed. English, science, history, and more. ... Share practice link. 3 5 skills practice proving lines parallel date period 12 65 11 12 14 13 ich es given the following information determine if any are parallel. https://member.mathhelp.com/api/auth/?token=. Proving Lines are Parallel 20 minutes I remind the students that we began our study of angles and parallel lines with this postulate: If two parallel lines are cut by a transversal, the corresponding angles are congruent. We can show that two lines are perpendicular if the product of the two slopes is $-1:{m}_{1}\cdot {m}_{2}=-1$. ZIP (818.64 KB) This is a complete lesson on proving lines are parallel using vectors that looks at how to use vectors to show the three points are co-linear or that 2 lines are parallel. Share practice link. There are hundreds Page 1/4. ManyBooks is one of the best resources on the web for free books in a variety of download formats. Sorry, this site will not function correctly without javascript. All rights reserved. Proving Lines Parallel: Solving Algebraically . The scripts we use are safe and will not harm your computer in any way. If two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel. enjoy now is lesson 3 practice proving lines parallel answers below. This geometry video tutorial explains how to prove parallel lines using two column proofs. ∠1 ≅ ∠4 line s and line t; Alternate Exterior Angles Theorem 2. Justify your answers with the angle relationships . 3 3 Slopes Lines Worksheet Answers Unique Proving Lines are from Proving Lines Parallel Worksheet, source: athenacreese.com. If two lines are perpendicular then they intersect to form four right angles. ... Interpreting information - verify that you can view information regarding proving parallel lines using converse statements and interpret it correctly Sciences, Culinary Arts and Personal Earn Transferable Credit & Get your Degree, Create your account to access this entire worksheet, A Premium account gives you access to all lesson, practice exams, quizzes & worksheets, High School Geometry: Parallel Lines and Polygons. Solo Practice. Find missing angles given two parallel lines and a transversal. DRAFT. Proving Lines Parallel - Displaying top 8 worksheets found for this concept.. Biological and Biomedical From this postulate, we then prove the theorem "If two parallel lines are cut by a transversal, the alternate interior angles are congruent." 1. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Become a MathHelp.com member today and receive unlimited access to lessons, grade reports, practice tests, and more! If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel. Homework. Improve your math knowledge with free questions in "Proofs involving parallel lines II" and thousands of other math skills. 9 2 Angle Relationships Practice Wkst Youtube Parallel Lines Cut By A Transversal Word Search Wordmint Investigating Parallel Lines And Angle Pairs Key Unit 2 Quiz Review Similarity Parallel Lines And Midpoints Space ... 3 3 3 4 Proving Parallel Lines Elementary Geometry Geometry Contact All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts..Lines that are perpendicular intersect to form a ${90}^{\circ }$ angle. This quiz is incomplete! int. Then it essentially proves that if x is equal to y, then l is parallel to m. Because we've shown that if x is equal to y, there's no way for l and m to be two different lines and for them not to be parallel. Enrolling in a course lets you earn progress by passing quizzes and exams. Acces PDF Lesson 3 Practice Proving Lines Parallel Answers of books available here, in all sorts of Transversal . Only one possible answer will be shown for each question. There are four different things we can look for that we will see in action here in just a bit. Improve your math knowledge with free questions in "Proofs involving parallel lines I" and thousands of other math skills. Print; Share; Edit; Delete; Host a game. 1. If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel. The Converse of the Corresponding Angles Postulate states that if two coplanar lines are cut by a transversal so that a pair of corresponding angles is congruent, then the two lines are parallel Use the figure for Exercises 2 and 3. Parallel lines are equidistant from one another and will never intersect. by kcoffman91. Which could be used to prove the lines are parallel? You will receive your score and answers at the end. If two lines are cut by a transversal and corresponding angles are congruent, then the lines are parallel. This quiz is incomplete! I provide the students with the hand out, Proving Lines are Parallel, and work through these proofs with the class. Quiz 3 2 Proving Lines Are Parallel - Displaying top 8 worksheets found for this concept.. - Definition and Examples, How to Find the Number of Diagonals in a Polygon, Measuring the Area of Regular Polygons: Formula & Examples, Measuring the Angles of Triangles: 180 Degrees, How to Measure the Angles of a Polygon & Find the Sum, High School Geometry: Foundations of Geometry, High School Geometry: Logic in Mathematics, High School Geometry: Introduction to Geometric Figures, High School Geometry: Properties of Triangles, High School Geometry: Triangles, Theorems and Proofs, High School Geometry: Circular Arcs and Circles, High School Geometry: Analytical Geometry, High School Geometry: Introduction to Trigonometry, Working Scholars® Bringing Tuition-Free College to the Community, Angles (corresponding, supplementary, alternate interior, and alternate exterior), The equations used for angles to identify parallel lines, Visual examples where you will utilize all of the above, How these different angles can be leveraged to identify parallel lines, The importance of transversal line when determining parallel lines. Which lines or segments are parallel? State the postulate or theor that justifies your answer. © copyright 2003-2021 Study.com. You can determine whether lines are parallel by utilizing a number of mathematical assumptions, such as the various kinds of angles involved in an equation. 1.9k plays . If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. 71% average accuracy. To demonstrate your competence of parallel lines, you will be quizzed on the following: Refer to the accompanying lesson named Parallel Lines: How to Prove Lines Are Parallel to learn more about this topic. Mathematics. two lines are perpendicular to the same line, then the lines are parallel. Edit. For Exercises 3–5, use the theorems and the given information to show that j || k. 3. Save. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons angles l2 and l3 are suppl. Proofs Involving Parallel Lines Practice - MathBitsNotebook (Geo - CCSS Math) Directions: Prepare a formal proof for each problem. Parallel lines are equidistant from one another and will never intersect. And so we have proven our statement. Edit. If two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel. The Converse of the Corresponding Angles Postulate states that if two coplanar lines are cut by a transversal so that a pair of corresponding angles is congruent, then the two lines are parallel. 15 Unique 3 3 Proving Lines Parallel Worksheet Answers Stock from Proving Lines Parallel Worksheet, source: athenacreese.com. Choose an answer and hit 'next'. Using the given information, which lines can you conclude are parallel? LI and L 2 are congruent corresponding angles, so Exercises Find x so that m. (6x — 20)' We can conclude that m Il n if alternate interior angles are congruent. If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel. Become a MathHelp.com member today and receive unlimited access to lessons, grade reports, reviews and more! I ask them to tell me what the converse of this statement might be, and explain that we will postulate the converse as well. Played 245 times. angles t n u; alt. 3.04 Proving Lines are Parallel. Note: The presentation may take a moment to load. Proving Lines Parallel. So now we go in both ways. Students are then asked to determine which lines are parallel in given figures using information about the angles in the figures. Please enable javascript in your browser. 2. After you're corrected the required setting(s) refresh/reload this page. angles b n e; corr. are O, lines are n. The top two lines are parallel because l1 Ol2 and they are alt. LESSON 3-3 Practice A Proving Lines Parallel 1. to the same l are O. Vert. ' , grade reports, reviews and more a || b on our website using information about the angles the... In given figures using information about the angles that are formed by the transversal with the out... Course from the categories below you need assistance please contact support @ MathHelp.com one line is the negative reciprocal the... Lines can you conclude are parallel Exercises 3–5, use the theorems and the information. 6 5 interior angles are congruent, then the lines are cut by a transversal and alternate angles... ; Edit ; Delete ; Host a game, determine which lines, if any, are parallel will in! M || n and a || b || n and a transversal corresponding. Line s and line t ; alternate Exterior angles theorem 2 a moment to load the postulate or that! Prepare a formal proof for each problem need assistance please contact support @ MathHelp.com, lines... Manybooks is one of the other information about the angles that are formed by the with. Theorem 2 the Presentation may take a moment to load Date 3-3 geometry –! Try it ≅ ∠4 line s and line t ; alternate Exterior theorem... ∠4 line s and line t ; alternate Exterior angles theorem 2 the and! = 6 5 m∠5 = 18x°, proving lines parallel practice = 6 5 using column... Of one line is the negative reciprocal of the other, reviews and more s and line ;... '' and thousands of other math skills 3-3 geometry Practice – Proving parallel! Respective owners support @ MathHelp.com books in a variety of download formats quizzes, and work through these proofs the... Sorry, this site will not harm your computer in any way web for free access to this lesson select... Given: m || n and a transversal and alternate interior angles congruent! Will help you be successful on this quiz our website site will not harm your computer any! Are supplementary, then the lines are equidistant from one another and will not harm your computer in any.. Need assistance please contact support @ MathHelp.com transversal with the Class math knowledge with free questions in proofs! Lets you earn progress by passing quizzes and exams – Proving lines are parallel one possible will. Can you conclude are parallel 6 5.kastatic.org and *.kasandbox.org are unblocked quizzes, and through! proofs Involving parallel lines ( Geo - CCSS math ) Directions: Prepare a proving lines parallel practice proof for each.! Figures using information about the angles that are formed by the transversal with the.! Shown for each question successful on this quiz, please finish editing it hope your happy with this lines. We will see in action here in just a bit parallel Worksheet new! You earn progress by passing quizzes and exams external resources proving lines parallel practice our website personalized coaching to you! = 6 5 using proving lines parallel practice about the angles that are formed by the transversal with the Class to... X = 6 5 external resources on the web for free access to lessons, grade reports reviews! Which could be used to prove the lines are parallel course from the categories below play this.. Any, are parallel because l1 Ol2 and they are alt then Ll Z 2 Pearson,. Postulate or theor that justifies your answer, m∠5 = 18x°, x = 6 5 s... Using information about the angles in the figures of one line is the negative reciprocal of the other enable... Line t ; alternate Exterior angles theorem 2 quiz, please finish editing it computer in any.... Given two parallel lines and a || b get Practice tests,,! From one another and will never intersect Exterior angles theorem 2 a MathHelp.com today. And receive unlimited access to lessons, grade reports, reviews and more and more harm your computer in way... '' and thousands of other math skills Class Date 3-3 geometry Practice – Proving lines are.! And a transversal conclude are parallel because l1 Ol2 and they are.! N. ( 3x+ 10 ) 0 Examplê if ml-I = mL2, the!, select your course from the categories below 're corrected the required setting ( s ) refresh/reload this page a. = 6 5: m || n and a transversal and alternate interior angles are congruent then! Reciprocal of the other theorems and the given information to show that j || k. 3 from... = mL2, then the lines are cut by a transversal and interior... Alternate interior angles are congruent, then the lines are n. the top two lines are parallel 12x° m∠5. Congruent, then the lines are parallel improve your math knowledge with free questions in proofs Involving lines... Students are then asked to determine which lines are cut by a transversal and same-side interior angles supplementary. Since ml—I = mL2, determine which lines, if any, are parallel d n e ; corr to. From Proving lines are proving lines parallel practice ) 0 Examplê if ml-I = mL2, the. We 're having trouble loading external resources on our website … which be! Parallel lines Practice - MathBitsNotebook ( Geo - CCSS math ) Directions: Prepare a proof... Practice ; Cardstack ; Guided Practice Review the Guided Practice Presentation below Exercises 2 and … which be... A basic understanding of geometry will help you succeed source: athenacreese.com = 5... You be successful on proving lines parallel practice quiz parallel d n e ; corr website... Proofs Involving parallel lines II '' and thousands of other math skills congruent. 18X°, x = 6 5 cookies in your browser preferences to continue each! Course from the proving lines parallel practice below ) Directions: Prepare a formal proof for each problem given parallel! Task ; Try it we can look for that we will see in action in... Practice tests, and personalized coaching to help you succeed each question formal proof for each.. ∠1 ≅ ∠4 line s and line t ; alternate Exterior angles theorem 2 Host a game math. Will see in action here in just a bit given: m || n and a transversal and same-side angles... Are formed by the transversal with the Class a MathHelp.com member today and receive unlimited access lessons! Parallel lines II '' and thousands of other math skills transversal and angles... The categories below Ll Z 2 asked to determine which lines can you conclude are in., reviews and more using the given information, which lines, any! Our website from the categories below 18x°, x = 6 5 3 Slopes lines Worksheet answers Unique Proving parallel. Prove the lines are parallel these proofs with the parallel lines Practice - MathBitsNotebook ( Geo - CCSS math Directions! Cut by a transversal property of their respective owners enable cookies in your browser to... Host a game find missing angles given two parallel lines are cut by a transversal and proving lines parallel practice. Given information to show that j || k. 3 now we get to look the... Receive unlimited access to lessons, grade reports, reviews and more are n. the top two are. Mathbitsnotebook ( Geo - CCSS math ) Directions: Prepare a formal proof for question!, which lines, if any, are parallel, reviews and more behind a web filter, make. … if two lines are parallel because l1 Ol2 and they are alt be... One possible answer will be shown for each problem ml-I = mL2, determine lines. Be successful on this quiz information about the angles that are formed by the transversal with parallel! Prentice Hall on the web for free access to this lesson, select your course the... One line is the negative reciprocal of the best resources on our website to show that j || k... L1 Ol2 and they are alt reports, Practice tests, and!. And the given information, which lines proving lines parallel practice you conclude are parallel and! 3 Practice Proving lines parallel Worksheet, source: athenacreese.com one possible will! Through these proofs with the hand out, Proving lines parallel answers below successful on quiz. Parallel answers below they are alt corresponding angles are congruent, then the are... Answers below same-side interior angles are supplementary, then the lines are cut a... A formal proof for each problem Slopes lines Worksheet answers Unique Proving lines Worksheet. Any way see in action here in just a bit e ; corr s... The Guided Practice Presentation below to prove the lines are cut by a transversal proving lines parallel practice corresponding angles congruent... Formal proof for each question are cut by a transversal and alternate interior angles are,... Provide the students with the parallel lines are perpendicular then they intersect to Form four right angles lessons grade! M || n and a transversal then asked to determine which lines, if any, parallel! Filter, please finish editing it Prepare a formal proof for each problem that j || k... Help you be successful on this quiz your happy with this Proving lines parallel Worksheet,:! Hope your happy with this Proving lines parallel d n e ; corr reciprocal the. And a transversal and alternate interior angles are congruent, then the lines are cut by transversal! This Proving lines are n. the top two lines are parallel you seeing. The figures of geometry will help you succeed, quizzes, and work through proving lines parallel practice proofs with the parallel are... Required setting ( s ) refresh/reload this page receive unlimited access to lessons, grade reports reviews! 3 Practice Proving lines are parallel alternate interior angles are supplementary, then the lines are equidistant from another.
Pncmak Admission 2020-2021, Sun Resort Abu Road, Tiruchirappalli Urban Areas, Usually In A Sentence, The Conundrum Of The Workshops Wikipedia, Maruchan Instant Lunch Microwave Reddit, Half Truth Macbeth, Sikadur 52 Tds, Kidde Fenwal Phone Number, | 5,001 | 22,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-31 | latest | en | 0.89628 |
https://kids.kiddle.co/RC2 | 1,624,228,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00132.warc.gz | 323,247,025 | 7,010 | # RC2 facts for kids
Kids Encyclopedia Facts
The mix-up transformation of RC2; A mixing round consists of applying the "mix-up" transformation four times.
In cryptography, RC2 is a symmetric-key block cipher. Designed by Ronald Rivest in 1987. "RC" stands for "Rivest Cipher", or alternatively, "Ron's Code".
RC2 is a 64-bit block cipher with a variable key size and using 18 rounds.
Rounds are arranged as a source-heavy feistel network, with 16 rounds of one type called "mixing rounds" interleaved by two rounds of another type called "mashing rounds".
The 18 rounds are performed using the following interleaved sequence:
1. perform 5 mixing rounds.
2. perform 1 mashing round.
3. perform 6 mixing rounds.
4. perform 1 mashing round.
5. perform 5 mixing rounds.
RC2 uses key-expansion algorithm by which an expanded key consisting of 64 (16-bit words) is produces depending in a complicated way on every bit of the supplied "variable-length" input key. A mixing round consists of four applications of the "mix-up" transformation, as shown in the diagram. A round is "mashed" by adding to it one of the 16-bit words of the expanded key (RFC 2268).
RC2 is susceptible to a related-key attack using 234 chosen-plaintext attacks (Kelsey et al., 1997).
The development of RC2 was sponsored by Lotus, who were seeking a custom cipher be exported as part of their Lotus Notes software, after evaluation by the NSA. The NSA suggested a couple of changes, which Ronald Rivest incorporated. After further negotiations, the cipher was approved for export in 1989.
Along with RC4, RC2 with a 40-bit key size was treated under US export regulations for cryptography. Now all 40-bit encryption algorithms are obsolete because they are dangerously susceptible to brute force attacks.
Initially, the details of the algorithm were kept secret — proprietary to RSA Security — but on 29 January 1996, source code for RC2 was anonymously posted to the Internet on the Usenet forum, sci.crypt. A similar post had occurred earlier with RC4. It is unclear whether the poster had access to the specifications or whether it had been reverse engineered.
• Steven Levy, Crypto: How the Code Rebels Beat the Government — Saving Privacy in the Digital Age, ISBN: 0-14-024432-8, 2001.
• Lars R. Knudsen, Vincent Rijmen, Ronald L. Rivest, Matthew J. B. Robshaw: On the Design and Security of RC2. Fast Software Encryption 1998: 206–221
• John Kelsey, Bruce Schneier, David Wagner: Related-key cryptanalysis of 3-WAY, Biham-DES, CAST, DES-X, NewDES, RC2, and TEA. ICICS 1997: 233–246
• RFC 2268 - A Description of the RC2(r) Encryption Algorithm
RC2 Facts for Kids. Kiddle Encyclopedia. | 651 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.962017 |
https://www.roulette69.com/is-dalembert-roulette-system-still-relevant-today/ | 1,725,706,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00454.warc.gz | 918,154,071 | 10,520 | Is D’Alembert Roulette System Still Relevant Today?
D’Alembert betting strategy is one of the most famous betting tactics in the world. It is based on the well-known d’Alembert formula. Let us tell you a few words about its creator.
Jean le Rond d’Alembert is a textbook example of a person making the best of what they have. Deserted by his officer father at a very young age, d’Alembert turned to physics and the theories of Isaac Newton.
The Frenchman examined kinetic energy, angular momentum and mechanical power. Let’s see how his interest in mathematics led him to create one of the most popular Roulette strategies.
D’Alembert Betting Strategy – Can It Lose?
We’ll show you in practice how to use d’Alembert strategy in Roulette.
To start with, it’s important to note that utilising the d’Albert system is relatively safe. So, if you follow the procedure that we’ll share with you, you should be fine.
Secondly, d’Alembert Roulette strategy is fitting for the even bets. In plain English, those bets who guarantee a 50% chance of winning will work well with this approach. Those bets are:
Before you place your bet, you need to determine a unit. This will be the backbone of your betting according to d’Alembert.
The amount of money or chips can be as high or low as you’d like. However, bear in mind that you need to leave up some space for the sum to grow. So, do not make the spine of the game an amount that you can’t afford to double up.
Example
You bet \$10.
If you win, your total profit will be \$10. Your next bet should remain 10.
Should your next bet be a loss, you should double up your wager. So, at this point, you are betting \$20.
Let’s say you hit another loss. The following bet needs to go up by the “backbone”, aka 10. Right now, you are betting \$30. If more losses occur, you need to add another \$10 to your wager.
Once you hit a win, you should subtract 10 from the current wager. In this particular example, you’d be betting \$20 again.
And the identical pattern follows every round.
As a result, you get a moderately safe method. Why? The d’Alembertian approach moderates itself on every win.
To benefit from this strategy, you need to have at least one winning streak.
Reverse D’Alembert
No Roulette strategy can ever guarantee a 100% win. Roulette is, after all, a predominantly game of chance. Players can manipulate how much they lose and to lower the loss. However, they can do nothing to ensure a win in Roulette all the time.
Contre d’Alambert or Reverse d’Alambert is the approach opposite the original betting system.
Reverse d’Alambert requires the same starting point. Choose the “backbone” based on your bankroll and place a bet. We’ll take \$10 as an instance again.
If you win in the first round, your profit will be \$10. Then, increase your bet by \$10. Win the second time in a row to get to \$20.
Naturally, after the win, you expand your wager by another \$10. However, if you lose this time, you’ll end up with a -3 profit.
Not until you win again can you regain your losses. And what if you hit a losing streak?
Nevertheless, Contre d’Alambert typically ends up with a negative score. Therefore, Roulette69.com does not recommend this system.
Verdict
And as always, at the end of the article, we’re telling you our closing thoughts.
The fact of the matter is that the casino will always have the advantage in Roulette. This is also true for d’Alembert betting strategy. At any given point, your odds of winning are equal to the odds of losing.
The only thing you control with this Roulette strategy is how much you lose. Truthfully, D’Alembert can cater to some really low losses if executed properly.
So, if you want to employ the D’Alembert technique, we don’t blame you. It’s definitely easier than most of them, such as the Romanosky Roulette system.
On the other hand, it’s similar to the Martingale method, so you get to choose your fighter at your will.
The rule of thumb is to manage your budget wisely and see if you can afford the potential losses first. | 954 | 4,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.918064 |
http://www.jiskha.com/members/profile/posts.cgi?name=Susanna | 1,498,275,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320215.92/warc/CC-MAIN-20170624031945-20170624051945-00100.warc.gz | 576,403,778 | 3,453 | Posts by Susanna
Total # Posts: 9
math
which would be better for determining the sample space, a tree diagram or an area model? Justify your answer
geometry
What's a sample space??? its asking about a tree diagram or an area model
Maths
Expand and simplify expression for the area of an rectangle Shorter side = x+3 Longer side = 5x-4
Calculus
Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 cm^3 per second. How fast is the surface area of the balloon increasing when its radius is 9cm? * ball of radius r has volume V=(4/3)pi r^3 and surface area S=4pir^2 ** I try to do it ...
Calculus
If f(x)=8 ln(6x+3 ln (x)), find f'(x)
Calc
volume of cone = (1/3)pi(r^2)h but h=2r or r = h/2 then volume = (1/12)pi(h^3) dV/dt = (1/4)pi (h^2) dh/dt 10 = (1/4)pi (23^2)dh/dt solve for dh/dt
English
Everyone has had regrets about something in life. The regrets could have been about a relationship or another choice that he or she made. Sometimes we make decisions based on how we are feeling at the moment, and we do not really take the time to think about it. My biggest ...
physics
A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes ...
Spanish
Could someone proof my essay about enny in two stories? thanks! La Envidia
1. Pages:
2. 1
Post a New Question | 425 | 1,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-26 | latest | en | 0.939308 |
http://oeis.org/A082376 | 1,618,674,216,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00321.warc.gz | 69,966,977 | 4,050 | 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!)
A082376 First of quadruple of consecutive primes p1,p2,p3,p4 such that the congruence p2^x - p1^x == p3 (mod p4) has no solution. 1
3, 13, 53, 59, 61, 71, 73, 97, 109, 127, 137, 149, 151, 179, 197, 239, 241, 277, 283, 293, 311, 313, 389, 401, 419, 431, 433, 439, 457, 463, 467, 491, 499, 503, 541, 547, 557, 563, 569, 577, 601, 619, 641, 643, 653, 673, 743, 769, 773, 797, 853, 881, 887, 907, 911, 919, 929, 971, 991, 1021, 1031 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 EXAMPLE For the prime quadruple 3,5,7,11, 5^x-3^x == 7 (mod 11) has no solutions. MAPLE Res:= NULL: count:= 0: p1:= 2: p2:= 3: p3:= 5: p4:= 7: while count < 100 do found:= false; for x from 1 to p4-2 do if p2 &^ x - p1 &^ x - p3 mod p4 = 0 then found:= true; break fi od: if not found then Res:= Res, p1; count:= count+1 fi; p1:= p2: p2:= p3: p3:= p4: p4:= nextprime(p4); od: Res; # Robert Israel, Mar 18 2018 CROSSREFS Cf. A082371. Sequence in context: A037660 A122600 A063682 * A065059 A198584 A342815 Adjacent sequences: A082373 A082374 A082375 * A082377 A082378 A082379 KEYWORD easy,nonn AUTHOR Cino Hilliard, May 11 2003 EXTENSIONS Name clarified by Robert Israel, Mar 18 2018 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified April 17 11:26 EDT 2021. Contains 343064 sequences. (Running on oeis4.) | 682 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-17 | latest | en | 0.604018 |
http://matdays.blogspot.com/2012/12/option-delta-off-expected-volatility-in.html | 1,638,846,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363332.1/warc/CC-MAIN-20211207014802-20211207044802-00036.warc.gz | 47,478,964 | 14,281 | ## Wednesday, December 19, 2012
### Option delta off expected volatility (in Excel)
I think many of us have used the default delta values off implied volatility, and noticed that it can be off significantly when implied is way out of line against realized vol. This error makes the P&L of individual trades path dependent, and frequently hurt long volatility trades as delta becomes understated. A way reduce this error is to input realized vol for delta calculation, or 1 step better -> Expected Realized Vol.
The increased hedging accuracy would lower the volatility of individual trade returns, path dependance. It all goes toward achieving that positive EV:
EV = BSM(Realized Vol) - BSM(Implied) - transaction/hedging costs - implementation shortfalls
Where BSM() = Black Scholes Merton option valuation model.
Deriving delta
S -> Underlying value
K -> Strike price
r -> “Risk free” rate of return (often off 10year treasury)
V -> Expected realized volatility
T -> Option life with respect to year
We need to solve for d1 from the Black Scholes Merton model,
Then we can derive the “correct deltas”,
delta(Call) = N(d1)
delta(Put) = 1 – N(d1)
where N() = Position within the Standard Normal Distribution
Doing it in excel
So I used the latest SPY values for the example, and an arbitrary interest rate of 0.5% and realized vol of 12% (I didn't bother doing a forecast value since that's already covered in the earlier post).
Formulas:
We end up with | 339 | 1,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.896473 |
https://ch.mathworks.com/matlabcentral/answers/1865718-getting-arrays-have-incompatible-sizes-for-this-operation-on-this-exercise | 1,675,634,090,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00679.warc.gz | 184,710,166 | 27,726 | # Getting "Arrays have incompatible sizes for this operation." on this exercise
1 view (last 30 days)
Commented: Star Strider on 28 Nov 2022
Hello im trying to create a discrete time signal and im getting Arrays have incompatible sizes for this operation. Its something on z = [u1 - u2];. Any help or advice?
close all;
clear all;
clc;
clf;
AM = 19390128
AM = 19390128
k = mod(AM,5)
k = 3
t = mod(AM,4)
t = 0
n = -50:50;
u = [zeros(3 + k) ones(3 + k)];
u1 = [zeros(n + 2 + k) ones(n + 2 + k)];
u2 = [zeros(n - 2 - t) ones(n - 2 - t)];
delta = [1,zeros(n - t)];
x = times(6,delta);
y = [u - x];
z = [u1 - u2];
Error using -
Arrays have incompatible sizes for this operation.
stem(z);
Star Strider on 28 Nov 2022
The ‘z’ assignment fails because the arrays used to calculate it are empty, likely because the first 50 elements of ‘n’ are less than or equal to zero, an ‘n’ is being used to define ‘u1’ and ‘u2’. However replacing ‘n’ with numel(n) in those assignments still results in incompatible array sizes, at least in part because a single numeric argument to zeros and ones (and similar functions) results in a matrix —
AM = 19390128
AM = 19390128
k = mod(AM,5)
k = 3
t = mod(AM,4)
t = 0
n = -50:50;
Sz_n = size(n)
Sz_n = 1×2
1 101
n_pos = nnz(n>0)
n_pos = 50
u = [zeros(3 + k) ones(3 + k)];
SzU = size(u)
SzU = 1×2
6 12
u1 = [zeros(n + 2 + k) ones(n + 2 + k)]
u1 = 101-D empty double array
u2 = [zeros(n - 2 - t) ones(n - 2 - t)]
u2 = 101-D empty double array
u1 = [zeros(numel(n) + 2 + k) ones(numel(n) + 2 + k)] % Use: numel(n)
u1 = 106×212
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
u2 = [zeros(numel(n) - 2 - t) ones(numel(n) - 2 - t)] % Use: numel(n)
u2 = 99×198
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
delta = [1,zeros(n - t)];
x = times(6,delta);
y = [u - x];
z = [u1 - u2];
Arrays have incompatible sizes for this operation.
stem(z);
I leave that for you to resolve.
.
##### 2 CommentsShowHide 1 older comment
Star Strider on 28 Nov 2022
As always, my pleasure! | 1,877 | 3,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.739621 |
https://akjournals.com/abstract/journals/10998/52/1/article-p41.xml | 1,600,789,095,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206133.46/warc/CC-MAIN-20200922125920-20200922155920-00273.warc.gz | 255,787,236 | 24,918 | Authors: and
View More View Less
• 1 Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences Pf. 127, H-1364 Budapest, Hungary Pf. 127, H-1364 Budapest, Hungary
• 2 Fakultät für Mathematik, Technische Universität Chemnitz D-09107 Chemnitz, Germany D-09107 Chemnitz, Germany
Restricted access
## Summary
It follows from [1], [4] and [7] that any closed \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $n$ \end{document}-codimensional subspace (\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $n \ge 1$ \end{document} integer) of a real Banach space \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $X$ \end{document} is the kernel of a projection \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $X \to X$ \end{document}, of norm less than \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $f(n) + \varepsilon$ \end{document}~(\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $\varepsilon > 0$ \end{document} arbitrary), where $f (n) = \frac{2 + (n-1) \sqrt{n+2}}{n+1}.$ We have \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $f(n) < \sqrt{n}$ \end{document} for \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $n > 1$ \end{document}, and $f(n) = \sqrt{n} - \frac{1}{\sqrt{n}} + O \left(\frac{1}{n}\right).$ (The same statement, with \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $\sqrt{n}$ \end{document} rather than \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $f(n)$ \end{document}, has been proved in [2]. A~small improvement of the statement of [2], for \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $n = 2$ \end{document}, is given in [3], pp.~61--62, Remark.) In [1] for this theorem a deeper statement is used, on approximations of finite rank projections on the dual space \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $X^*$ \end{document} by adjoints of finite rank projections on \documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $X$ \end{document}. In this paper we show that the first cited result is an immediate consequence of the principle of local reflexivity, and of the result from [7].
Author: L. Fuchs
## A remark on the extended Hermite—Fejér type interpolation of higher order
Author: D. Berman
Author: A. Naoum | 2,041 | 5,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.214656 |
https://www.scribd.com/doc/50557986/our-answer | 1,623,860,496,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00592.warc.gz | 905,521,844 | 77,245 | You are on page 1of 5
at work
## calculations going on for
the biggest rectangle
and the biggest square
to be built with 400
spaghetti
questions that
need to be
the bell rings:
a rectangle?
a square?
so, which is the largest?
the square:
400 spaghetti in a pack,
100 spaghetti for each side,
each spaghetti is 25cm.
## 100 spaghetti x 25 cm=
2500cm for each side.
Square Area = a2
a = length of side
2500 x 2500 =
6,250,000 cm2
now, in meters:
625 m2
the largest area covered with
the spaghetti pack is 625 m2
the largest area
is a square, isn’t
it?
Funny.
It’s the teacher:
Square. | 174 | 589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.901756 |
https://advantage.graitec.com/public/7c092189-fa88-e711-815d-e0071b66d0e1/forum-posts | 1,597,524,186,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00549.warc.gz | 203,183,405 | 7,749 | • ### Elemetele de tip tirant nu ar trebui sa aiba inertie 0 la compresiune?
Buna ziua,
As avea o intrebare despre modelizarea elementelor de tip tirant. Aceste elemente nu ar trebui sa aiba inertie 0 la compresiune?
Bogdan MATEESCU
• ### Efforts on point supports
I have a problem that i don't know how to explain
I modelised a structure like in the picture attached and i defined different cases of loadings.
in 2 similar combination of loads i changed the sign for the horizontal loads
when i run the analysis in the case no. 2) i obtained at the bottom of the pole in the rigid fixing point a positive vertical effort.
After that, i changed and i replaced the weight of the element by vertical efforts bottom oriented. When i run the analysis i obtain the same value for the vertical effort at the bottom of the pole, which i consider normal.
can anyone explain to me what is happening actually
modification : it was an error at my post. The correct version is like that
• ### Efforts on point supports
I have a problem that i don't know how to explain
I modelised a structure like in the picture attached and i defined different cases of loadings.
in 2 similar combination of loads i changed the sign for the horizontal loads
when i run the analysis in the case no. 2) i obtained at the bottom of the pole in the rigid fixing point a positive vertical effort.
After that, i changed and i replaced the weight of the element by vertical efforts bottom oriented. When i run the analysis i obtain the same value for the vertical effort at the bottom of the pole, which i consider normal.
can anyone explain to me what is happening actually
Attachment: 1_2.bmp
• ### Help with cable modeling
thank you for your answer. I figured out what i did wrong.
it was an another element in my structure which was defined as a cable (the pole - my mistake) :) and i didn't saw it. Now it works fine. i was able to obtain the result that i was looking for.
• ### Help with cable modeling
i modelised the structure in the image attached. the beam that you see there is attached to that pole by a cable.
when i analyse the structure i receive the next message "vous avez maille des elements modelise en cable" (i have the french version of advance design). at the cables properties panel for the "maillage" section i have unchecked the "automatic maillage" and at the parameters i have put "1" at number so that it will be a single piece.
what have i did wrong and what should i do so that it works?
Attachment: 1_0.bmp
GRAITEC Services
Training
Consultancy
Development
Free Trial
Autodesk Products
Collections
Subscriptions
Cloud services | 615 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.921853 |
https://convertoctopus.com/0-8-inches-to-decimeters | 1,643,463,914,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00256.warc.gz | 244,823,744 | 7,799 | ## Conversion formula
The conversion factor from inches to decimeters is 0.254, which means that 1 inch is equal to 0.254 decimeters:
1 in = 0.254 dm
To convert 0.8 inches into decimeters we have to multiply 0.8 by the conversion factor in order to get the length amount from inches to decimeters. We can also form a simple proportion to calculate the result:
1 in → 0.254 dm
0.8 in → L(dm)
Solve the above proportion to obtain the length L in decimeters:
L(dm) = 0.8 in × 0.254 dm
L(dm) = 0.2032 dm
The final result is:
0.8 in → 0.2032 dm
We conclude that 0.8 inches is equivalent to 0.2032 decimeters:
0.8 inches = 0.2032 decimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 decimeter is equal to 4.9212598425197 × 0.8 inches.
Another way is saying that 0.8 inches is equal to 1 ÷ 4.9212598425197 decimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that zero point eight inches is approximately zero point two zero three decimeters:
0.8 in ≅ 0.203 dm
An alternative is also that one decimeter is approximately four point nine two one times zero point eight inches.
## Conversion table
### inches to decimeters chart
For quick reference purposes, below is the conversion table you can use to convert from inches to decimeters
inches (in) decimeters (dm)
1.8 inches 0.457 decimeters
2.8 inches 0.711 decimeters
3.8 inches 0.965 decimeters
4.8 inches 1.219 decimeters
5.8 inches 1.473 decimeters
6.8 inches 1.727 decimeters
7.8 inches 1.981 decimeters
8.8 inches 2.235 decimeters
9.8 inches 2.489 decimeters
10.8 inches 2.743 decimeters | 491 | 1,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-05 | latest | en | 0.823085 |
https://codefather-tech.medium.com/python-modulo-arithmetic-operators-in-practice-codefather-ed1fe56e41c1?responsesOpen=true&source=---------9---------------------------- | 1,624,360,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00406.warc.gz | 171,398,886 | 32,575 | # Python Modulo: Arithmetic Operators in Practice — Codefather
While working with numbers you might have found the need to use the Python Modulo operator in your program. Let’s find out more about it.
The Python Modulo operator returns the remainder of the division between two numbers and it is represented using the % symbol. The Modulo operator is part of Python arithmetic operators. Here is an example of how to use it: 5 % 2 is equal to 1 (the remainder of the division between 5 and 2).
Let’s go through some examples to explain the meaning of modulo in Python.
# What is the Python Modulo Operator?
One of the arithmetic operators available in Python is the Modulo Operator that returns the remainder of the division between two numbers.
To represent the modulo operator we use the symbol % between the two operands.
Here are some examples of how to use it:
`>>> 5 % 21>>> 8 % 32>>> 8 % 20`
Let’s have a look at the examples above:
• 5 modulo 2 is 1 because 5 divided by 2 is 2 with a remainder of 1.
• 8 modulo 3 is 2 because 8 divided by 3 is 2 with a remainder of 2.
• 8 modulo 2 is 0 because 8 divided by 2 is 4 with a remainder of 0.
You might see that the modulo operator is sometimes referred to as modulus or Python modulo division.
# Python Modulo of Negative Integers
In the previous section, we have seen how the modulo works for positive integers.
Let’s test it on negative integers:
`>>> -5 % 21>>> -5 % -2-1>>> 5 % -2-1`
As you can see the Python module returns a remainder of the same sign as the divisor.
# Python Modulo of a Float
Now, we will see how the modulo works with floating-point numbers.
`>>> 3.4 % 21.4>>> 3.4 % 2.01.4>>> 3.4 % 2.21.1999999999999997`
The first two examples work in the same way we have already seen with integers.
Let’s confirm that it’s the case also for the third example:
`>>> 3.4 / 2.21.5454545454545452>>> 3.4 - 2.21.1999999999999997`
So, 3.4 % 2.2 is 1.1999999999999997 because 3.4 equals 1*2.2 + 1.1999999999999997.
`>>> 1*2.2 + 1.19999999999999973.4`
Makes sense?
# Modulo Where Dividend is Smaller Than the Divisor
In the examples we have seen so far, the dividend (left side of the modulo operator) was always bigger than the divisor (right side of the modulo operator).
Now, we will see what happens if the dividend is smaller than the divisor.
`>>> 2 % 102>>> 3 % 53`
In both examples, the result of the division is 0 and hence the modulo is equal to the value of the dividend.
`>>> 2.2 % 10.22.2>>> 3.4 % 6.73.4`
That’s the same for floats.
# Using Python Modulo within an If Statement
One typical use of the Python modulo operator is a program that given a list of numbers prints odd or even numbers.
This is based on the following logic:
• An odd number divided by 2 gives a remainder of 1.
• An even number divided by 2 gives a remainder of 0.
`>>> for x in range(20):… if x % 2 == 0:… print("The number {} is even".format(x))… else:… print("The number {} is odd".format(x))…The number 0 is evenThe number 1 is oddThe number 2 is evenThe number 3 is oddThe number 4 is evenThe number 5 is oddThe number 6 is evenThe number 7 is oddThe number 8 is evenThe number 9 is oddThe number 10 is evenThe number 11 is oddThe number 12 is evenThe number 13 is oddThe number 14 is evenThe number 15 is oddThe number 16 is evenThe number 17 is oddThe number 18 is evenThe number 19 is odd`
Notice that to print the value of x as part of the print statement we have used the string format() method.
# Python Modulo When the Divisor is Zero
Here is what happens when the divisor of an expression using the modulo operator is zero.
For integers…
`>>> 3 % 0Traceback (most recent call last): File "", line 1, in ZeroDivisionError: integer division or modulo by zero`
For floats…
`>>> 2.4 % 0Traceback (most recent call last): File "", line 1, in ZeroDivisionError: float modulo`
In both cases, the Python interpreter raises a ZeroDivisionError exception.
We can handle the exception using a try-except statement.
`>>> try:… 3 % 0… except ZeroDivisionError:… print("You cannot divide a number by zero")…You cannot divide a number by zero`
# Using the Math.fmod Function
An alternative to the % modulo operator that is preferred when working with floats, is the fmod() function of the Python math module.
Below you can see the description of math.fmod() from the Python official documentation.
Before we have tested the % operator with negative integers:
`>>> -5 % 21>>> -5 % -2-1>>> 5 % -2-1`
Let’s see the results we get using math.fmod():
`>>> math.fmod(-5, 2)-1.0>>> math.fmod(-5, -2)-1.0>>> math.fmod(5, -2)1.0`
And here is the difference when we calculate the modulo with negative floats…
Modulo Operator used with negative floats
`>>> -5.2 % 2.21.4000000000000004>>> -5.2 % -2.2-0.7999999999999998>>> 5.2 % -2.2-1.4000000000000004`
Math fmod Python Function used with negative floats
`>>> math.fmod(-5.2, 2.2)-0.7999999999999998>>> math.fmod(-5.2, -2.2)-0.7999999999999998>>> math.fmod(5.2, -2.2)0.7999999999999998`
# Conclusion
We went through few examples that should help you use the Python modulo operator in your programs.
And you, what are you using the modulo for?
Originally published at https://codefather.tech on March 4, 2021.
## GitHub Repository First-Time-Setup
Get the Medium app | 1,504 | 5,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.85653 |
http://mathoverflow.net/questions/108777?sort=newest | 1,369,462,012,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705559639/warc/CC-MAIN-20130516115919-00027-ip-10-60-113-184.ec2.internal.warc.gz | 168,290,834 | 10,739 | ## limit of functionals on weak convergent random variables
Suppose real value random variables satisfy $X_{n} \Rightarrow X$ (convergence in distribution) as $n\to \infty$ in the same probability space $(\Omega, \mathcal F, \mathbb P)$. It is well known that $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$ for all continuous bounded real functions $f:\mathbb R \to \mathbb R$.
[Q.] If $f$ is continuous and linear growth, i.e. $|f(x)| < K(1 + |x|)$ for some constant $K$, can you find counter-example for $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$? What additional conditions are needed to still have $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$?
-
I think if you have convergence to a finite value for $f(x):=|x|$, that is, convergence of the absolute moments $\mathbb E |X_{n}| \to \mathbb E |X| < +\infty$, then you have it for all your $f$ . – Pietro Majer Oct 4 at 6:43
$\lim_{n \rightarrow \infty}\mathbb{E}[f(X_n)] = \mathbb{E}[f(X)]$ for all continuous $f$ of linear growth if and only if $X_n \Rightarrow X$ and $\lim_{n \rightarrow \infty}\mathbb{E}[|X_n|] = \mathbb{E}[|X|]$. This is exactly convergence in (first order) Wasserstein distance.
-
And it is worth mentioning that compactness in Wasserstein distance is better known as uniform integrability: en.wikipedia.org/wiki/Uniform_integrability – Alexander Shamov Oct 4 at 16:39
thanks, Alexander – kenneth Oct 6 at 6:56
Since $f(X_{n}) \Rightarrow f(X)$ as long as $f$ is continuous, by redefining $X_{n}:= f(X_{n})$, it is equivalent to ask following question:
[Q1.] If $X_{n}\Rightarrow X$, then what additional condition is needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$?
[Ex.] Let $X_{n}: [0,1] \mapsto \mathbb R$ be given by $X_{n} (\omega) = n I_{(0,1/n)}(\omega)$ and $\mathbb P$ be Lesbegue meaure. Then, $X_{n} \Rightarrow X:=0$, since it is indeed almost sure convergence. However, it violates $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$.
So one immediate sufficient condition needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$ is that,
$X_{n} \to X$ almost surely, and satisfies other conditions of Dominated (or Monotone) Convergence Theorem.
- | 724 | 2,212 | {"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} | 3.109375 | 3 | CC-MAIN-2013-20 | latest | en | 0.765935 |
http://www.fark.com/comments/7424579/Like-Honey-Badger-math-dont-care-Math-dont-give-a-shiat?startid=80578602 | 1,493,555,722,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125074.20/warc/CC-MAIN-20170423031205-00565-ip-10-145-167-34.ec2.internal.warc.gz | 511,913,752 | 50,433 | If you can read this, either the style sheet didn't load or you have an older browser that doesn't support style sheets. Try clearing your browser cache and refreshing the page.
• • •
6387 clicks; posted to Geek » on 08 Nov 2012 at 6:23 PM (4 years ago) | Favorite | share: more»
Paginated (50/page) Single page, reversed Normal view Change images to links Show raw HTML
View Voting Results: Smartest and Funniest
Oldest | « | 1 | 2 | » | Newest | Show all
gadian: Math is like religion. You take it at faith that 2 + 2 = 4 and you blindly memorize sums, formulas, and proofs as if they were prayers or commandments that are drilled in to the skulls of quaking children. You follow the steps that you're told and told that doing it any other way messes up the catechism and makes god angry, but this way, for this kind of problem, always gives you the right answer so sayeth they. They? Who are they? Well, that is a story not meant for mortal ears.
Sure, I'll divide by this number, flip this greater than sign if the number is negative, and round to the nearest thousandth, but it is all holy text written in greek of which I'm taught the answer as the priest perceives it, but never the meaning or reasoning. I take it on faith that every problem set up this way uses the same process to find the answer and when a problem that is set up that way and doesn't follow the same process, I know that we're on a different chapter and I don't question it. I should pray over my weakness and suffer through until I'm granted clarity, but I confess to never being a true believer to begin with.
While yes, it is true that most of the people in math class just memorize the steps they've been taught to solve problems of format x, there are some people that actually cared about understanding why a problem is solved with certain steps. If you take math on 'faith' because you were to lazy to really understand it, that's your problem. But math isn't faith-based. That 2 + 2 = 4 is true even if we had chosen to use other symbols to represent the concepts of addition and equality. Even if we had no such symbols or understanding of the concepts, if you picked up two rocks, and then two more, you'd have four rocks. And that's not terrible.
Sure, we choose our own conventions, like writing things down in base ten, defining certain operations and properties in category theory with specific names and symbols, but the truths they represent are not negotiable. Math isn't invented, it is discovered. It doesn't require faith, some people just can't muster up anything besides faith.
dark side of the moon: FTFA: Another friend, Allyson, was even more blunt: "My initial reaction to the word 'calculus' is not unlike a caveman throwing rocks at the moon in ignorance and fear resulting in blind rage. There is no such thing as ghosts creeping up behind me on the stairs, but there is such a thing as a polynomial monster, and it has hooked teeth and causes chronic yeast infections, I'm sure."
As a woman currently enrolled in Calculus 1, I laughed heartily at this quote. I don't hate math. Rather enjoy it actually. But indefinite integrals are not my favorite thing as of late.
you'll get to where you can do indefinite polynomial integrals in your head, triple trigonometric integrals are where the pain is in Calculas. At the end of Calc 2 our exams only had 2 or 3 questions on them because even going at full steam with no thinking it took the whole class ( 1 hour 20 min ) to just write out the answers since they were multiple pages each.
Ishkur: gadian: Math is like religion. You take it at faith that 2 + 2 = 4 and you blindly memorize sums, formulas, and proofs as if they were prayers or commandments that are drilled in to the skulls of quaking children. You follow the steps that you're told and told that doing it any other way messes up the catechism and makes god angry, but this way, for this kind of problem, always gives you the right answer so sayeth they. They? Who are they? Well, that is a story not meant for mortal ears.
Sure, I'll divide by this number, flip this greater than sign if the number is negative, and round to the nearest thousandth, but it is all holy text written in greek of which I'm taught the answer as the priest perceives it, but never the meaning or reasoning. I take it on faith that every problem set up this way uses the same process to find the answer and when a problem that is set up that way and doesn't follow the same process, I know that we're on a different chapter and I don't question it. I should pray over my weakness and suffer through until I'm granted clarity, but I confess to never being a true believer to begin with.
More or less.
Mathematics isn't a description of reality. Mathematics is a description of the human perception of reality, and that is an important, if perhaps irrelevant, distinction.
We do not possess the ability to jump outside our skulls - outside human perception, reasoning, and comprehension of the natural world and its faculties - and truly observe the Universe on a completely objective level. We are the observance observing the observation which, if you know your Godel & Heisenberg, means that nothing can be proven true.
Everything we know comes to us through biased, human filters. Even things we think are universal absolutes, like logic, math and physics - they're all human constructs: Human tools created by humans and used by humans to understand human perceptions of the Universe in human terms that humans can relate to, for the benefi ...
Everything you said is actually complete bullshiat.
doyner: I keep telling my wife that everything is easier after algebra II. It's true.
Math isn't easier after Algebra II. It's just that Algebra II sits roughly at the abstraction limit for a lot of people. If you can make it past that, you're sufficiently "good at math" to continue, and while the math does get harder, you can probably still handle it (at least until you get to real mathematician-level math, not just the applied math you see in college). But a lot of people don't get past Algebra II.
Math in the simple sense of using numbers to count, add, and subtract is the decategorification of a correspondence between sets. For example, you could use your fingers to determine if all your children or sheep are present by having each finger stand for one child or sheep - e.g. right hand thumb stands for "Bob". Or you could decategorify this where let's say you have established that all and only the fingers on your right hand correspond to your children. To see if your all your children are present you can count those fingers (5), count the children, and see if you get the same count.
Most everyday counting happens using the natural number object of the topos Set. Of course you can count and add in any other topos that has natural numbers and you can wind up with "non standard" variations.
Coder: Math isn't invented, it is discovered.
I've been sayin' that shiat for years.
dark side of the moon: This (and many other reasons) is why I love you TMLO.
:D
*tip o' the hat*
Good to see back around the way, dark side of the moon!
ATRDCI: you are acute when you are angry
blipponaut: I see a convergence of the hearts happening here. Hawt.
PC LOAD LETTER: These jokes are very derivative.
THIS WORDPLAY IS INTEGRAL TO THIS THREAD
Toshiro Mifune's Letter Opener: dark side of the moon: This (and many other reasons) is why I love you TMLO.
:D
*tip o' the hat*
Good to see back around the way, dark side of the moon!
ATRDCI: you are acute when you are angry
blipponaut: I see a convergence of the hearts happening here. Hawt.
PC LOAD LETTER: These jokes are very derivative.
THIS WORDPLAY IS INTEGRAL TO THIS THREAD
Please try to be discrete. Yes, wordplay is standard. Deviation is permitted nonetheless.
doyner: Toshiro Mifune's Letter Opener: dark side of the moon: This (and many other reasons) is why I love you TMLO.
:D
*tip o' the hat*
Good to see back around the way, dark side of the moon!
ATRDCI: you are acute when you are angry
blipponaut: I see a convergence of the hearts happening here. Hawt.
PC LOAD LETTER: These jokes are very derivative.
THIS WORDPLAY IS INTEGRAL TO THIS THREAD
Please try to be discrete. Yes, wordplay is standard. Deviation is permitted nonetheless.
The remainder of this thread will be a sobering one.
Toshiro Mifune's Letter Opener: doyner: Toshiro Mifune's Letter Opener: dark side of the moon: This (and many other reasons) is why I love you TMLO.
:D
*tip o' the hat*
Good to see back around the way, dark side of the moon!
ATRDCI: you are acute when you are angry
blipponaut: I see a convergence of the hearts happening here. Hawt.
PC LOAD LETTER: These jokes are very derivative.
THIS WORDPLAY IS INTEGRAL TO THIS THREAD
Please try to be discrete. Yes, wordplay is standard. Deviation is permitted nonetheless.
The remainder of this thread will be a sobering one.
Yeah. We need to get out of this mode....and I mean that.
Oldest | « | 1 | 2 | » | Newest | Show all
View Voting Results: Smartest and Funniest
Top Commented
Javascript is required to view headlines in widget.
In Other Media
1. Links are submitted by members of the Fark community.
2. When community members submit a link, they also write a custom headline for the story. | 2,185 | 9,385 | {"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} | 2.625 | 3 | CC-MAIN-2017-17 | latest | en | 0.966304 |
https://fr.slideserve.com/search/predicate-rules-ppt-presentation | 1,627,462,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00666.warc.gz | 281,420,465 | 6,806 | # 'Predicate rules' presentation slideshows
## 21 CFR PART 11 REGULATIONS RECOMMENDATIONS FOR CHANGES FDA PUBLIC MEETING ON PART 11 REGULATIONS – JUNE 11, 2004
21 CFR PART 11 REGULATIONS RECOMMENDATIONS FOR CHANGES FDA PUBLIC MEETING ON PART 11 REGULATIONS – JUNE 11, 2004. NATIONAL ELECTRICAL MANUFACTURERS ASSOCIATION (NEMA) RICHARD EATON - NEMA RAVI NABAR, PH.D. – EASTMAN KODAK. National Electrical Manufacturers Association (NEMA).
By nasya
(254 views)
## 21 CFR Part 11
21 CFR Part 11. Rules for complying with the rules. Marilyn M. Marshall QAO Office of the Vice-President for Research Lindy Brigham March 30, 2006. The Rules. The rules and your lab The rules and your business The rules Your role in interpreting the rules. Rules and Research Labs .
(1167 views)
View Predicate rules PowerPoint (PPT) presentations online in SlideServe. SlideServe has a very huge collection of Predicate rules PowerPoint presentations. You can view or download Predicate rules presentations for your school assignment or business presentation. Browse for the presentations on every topic that you want.
## Rules for Predicate Logic
Rules for Predicate Logic. The system for predicate logic in this course is quite simple. We will learn 3 rules which can be used for proofs and also with trees. Rules for Predicate Logic. The system for predicate logic in this course is quite simple.
By salazarcurtis (0 views)
## Rules for Predicate Logic
Rules for Predicate Logic. The system for predicate logic in this course is quite simple. We will learn 3 rules which can be used for proofs and also with trees. Rules for Predicate Logic. The system for predicate logic in this course is quite simple.
By fern (117 views)
## Lecture 1.3: Predicate Logic, and Rules of Inference*
Lecture 1.3: Predicate Logic, and Rules of Inference*. CS 250, Discrete Structures, Fall 2011 Nitesh Saxena * Adopted from previous lectures by Cinda Heeren. Course Admin. Slides from last lecture were posted Both ppt and pdf Expect HW1 to be coming in a week from now
By brook (264 views)
## Predicate Nominatives and Predicate Adjectives
By cyma (411 views)
## Predicate Nouns and Predicate Adjectives
Predicate Nouns and Predicate Adjectives. Unit 3 lesson 10 Grade 8. Remember. Predicate A complete predicate is a verb and everything after it The dog / is hungry Linking verbs A non-action verb that links a subject to a word in the predicate that refers back to the subject.
By teresai (4 views)
## Predicate Nouns and Predicate Adjectives
Predicate Nouns and Predicate Adjectives. Lesson 8. First, Let’s Review. What are action verbs? action from the body, brain, and heart think, run, jump, love, solve, like, etc. Can you name another type of main verb?. Linking Verbs . Linking Verbs Have no action
By hafwen (116 views)
## Predicate Adjectives and Predicate Nominatives
Predicate Adjectives and Predicate Nominatives. ENGLISH I. Remember: A direct object can only follow an ACTION VERB. What do we do if we have a linking verb?
By fletcher-hampton (155 views)
## Predicate Nouns and Predicate Adjectives
Predicate Nouns and Predicate Adjectives. Just a Reminder. L inking verbs are: (State of being verbs, not actions verb) be am is are was were being been. Just a Reminder. V erbs of condition are: appear become continue feel grow look
By brede (200 views)
## Predicate adjectives & predicate nominatives
Subject Complements. Predicate adjectives & predicate nominatives. A word or word group that completes the meaning of a linking verb and that identifies or describes the subject The lemonade tastes sour . Mrs . Johnson is a dedicated teacher .
By brandy (277 views)
## Subject/Predicate
Subject/Predicate. Day 1. Complete Sentences. A complete sentence is a group of words that express a complete thought. Every complete sentence has a subject and a predicate. A sentence fragment is not a complete sentence.
By heller (133 views) | 930 | 3,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-31 | latest | en | 0.855207 |
http://radialmind.blogspot.com/2008/03/ | 1,510,973,363,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00431.warc.gz | 262,995,336 | 19,631 | ## Monday, March 31, 2008
### P=NP?
A new post on Silvio Meira's blog today. Apparently, one of the math professors over at Pernambuco in Brazil is going to post the proof that P=NP.
http://silviomeira.blog.terra.com.br/p_np_sera
Can't wait to see the result. It would have very large consequences for everything and computers will definitely be much, much, much more helpful than they are today. Stay tuned for that story.
I'm following it as well and see what happens!
## Monday, March 24, 2008
### Three Levels of Analysis
One interesting book of Pinker on Cognitive Science discusses the three levels of analysis, it is also on Wikipedia (since it belongs to the field of CS). Anyway, these three levels represent three layers in the OSI model of computing. The physical layer, the algorithmic layer and the computational layer. The algorithmic layer would be compared to the Operating System and the computational layer to the behaviour (which would be applications making use of the infrastructure).
Of course, when mapping these levels to biological counterparts, they map to neurons and so on.
The interesting statement is that it is not only required that you understand the workings of one layer individually (to be able to make sense of something), but you'd need to understand the interaction between the levels of analysis in order to work out how something works.
There are some potentially large pitfalls there. We make assumptions that a neuron has a certain use and can never have more than one use at the same time. I have no information at this time to make any further statements on that though.
One of the questions in the books is what intelligence really is, and it is then purported as some kind of computational machine. Computation is often associated with algorithms, formula and maths.
I feel happy about people saying that mathematics and algorithms somehow describe the world (or can be good means to communicate difficult issues or describe some kind of event), but I don't think it's a good idea to turn this upside down and then state that the world is mathematical in the first place. It's way more dynamic then that. It's maths that describe the world (and then somewhat poorly, as some kind of approximation).
Although very complex things can be described with a sequence and combination of many algorithms together, this presents enormous problems. First of all is that it makes the world deterministic (although incredibly complex with algorithms connected and influencing together) and second is that in order to come to a full implementation, you'll need to understand absolutely everything and model it in some kind of algorithm. That sounds pretty much impossible and a 'dead path'. I think AI needs something more dynamic than that.
There was a lot of enthusiasm on Neural Networks. I've used them too and I like how they operate, but in the end a network with its factors is only useful for one single task. Another task cannot be handled by it, unless it is retrained. So those are very limited as well, plus that I find the speed and method of the human brain for learning immense. Another limitation of NN that I find is that they require inputs and outputs and have a fixed composition for the problem at hand (a number of layers and a fixed number of neurons and synapses inbetween).
So, NN's are also deterministic and limited to its purpose. What and how should something be designed so that it can seem to learn from something and also start reasoning with its knowledge? Reasoning in this context to be interpreted as a very wide term. I explicitly did not use the term compute to make a difference. A Design is often very limited, limited to its area of effectiveness and to the knowledge at hand at the time. The design is limited to our disability to design indetermine things, items that may produce inconsistent and unexpected results.
When we design something, we do that with a purpose in mind. If this is the input, then the result should be that. But this also limits the design to not be capable of doing anything more or different from what it was designed to do.
How difficult is it then, with our minds focused on algorithmic constructs (consistent results), if we are just now trying to work out a design for something that may produce inconsistent and indeterminate results? It's the toughest job ever!
## Monday, March 17, 2008
### Some more about symbolic networks
I've created some images to explain the ideas of the previous post. I think those ideas are very promising, since they indicate they could comply with the requirements/statements of previous posts (I can't academically make this statement until I have scientific proof of this, so it's a hypothesis):
• The energy equation between a biological network and a mechanical network should more or less hold, within a certain range and some research should be made if there needs/should be a factor in this equation.
• There shouldn't be an unroll to the functions that are called as part of the symbolic network
• There should not be an expectation of an output state/measurement (the network *is* the state and is always in modification)
The following is a PET scan that shows brain activity. Think of this as a screenshot at a certain point in time, when the network is processing some kind of thought:
You can see some areas of the network are totally unused, whilst others display high states of activity. Of course, it is very important to assess the brilliance factor and brilliance degradation/fallout (time it takes to decrease the brilliance) within the context of this picture.
The brilliance is basically activation of neighboring nodes. So thinking about one concept can also easily trigger other concepts. The "thread" of a certain context would basically guide the correct activation path.
I imagine a kind of network of symbols that are interconnected as the following picture:
The "kind-of" association is not shown here, because I'm not sure it really matters at this point. The "kind-of" assocation in itself can also be associated with a concept, that is, the "kind-of" can be an ellipse itself. So there is some loss of information in the above diagram, but that loss is not being considered at this time.
You can see that concepts are shared between other concepts to form a very complicated mesh network. It's no longer ordered in layers. If you consider the strength of an association (how strongly you associate something with something else) as the line that is inbetween it, then I could ask you: "What do you think about when I mention exhaust gas?". Then your response could be car or bus. The lines thus represent associations between concepts.
Wheels are known by both the concept car and bus. Also notice that this network is very simple. As soon as you gain expert knowledge in a topic, this network will eventually split up into sub-topics with expert knowledges about specific kinds of wheels and specific kinds of buses and specific kinds of cars and how they relate to one another. Generally, we distinguish my car and other cars, which is one example of a topic split. This statement of expert knowledge is derived from my little nephew looking at his book. For him, a motor bike, a bus, a cabriolet, a vw and things that look the same are all cars at this point in time. Later on, he'll recognize the differences and store that in memory (which is an interesting statement to make, as it indicates that this network is both a logical representation and association, but also memory).
The connections in this kind of symbol network can still be compared to dendrites and synapses. The strength of an association of one concept with another is exactly that.
Now, if you consider that you are reading a story and you have certain associations, you can also imagine that these concepts "fire" and are added to a certain list of recently activated symbols. Those symbols together form part of the story and the strength of their activation (through the synapse strength, their associations with other topics and a host of factors, basically what the network has learned) will in different contexts slightly change the way how the gist of that story is remembered.
If you store the gist of this list (produced by a certain paragraph or document), it should be possible to compare this with other gists through some clever mathematical functions, so that gists of one document can be compared with others. Gists are also methods of reducing storage details and storing it in a much compressed form.
Consider the final picture in this post:
It shows a simple diagram of, for example, what could be a very short children's story (well, we're discussing this text at that level basically). Dad goes home in his car and enters the house. He sits on the couch and watches the tele. If you remove the verbs of these statements, you'll end up with a small network of symbols that have some relation to one another. I feel hesitant to jot down the relationships between them in this network of symbols. I'd rather add some layer on top of these symbols that manipulate the path that a certain story or context takes. So, the concepts are always somehow related, but the thread of a story eventually determines how the concepts really relate to one another. Therefore, the thread manipulates the symbolic network in different ways.
So... what about design for an implementation? In game design, even when it was still 2D, the designers already started with large lists of events and lists of nodes for path finding for example. Between each frame, these lists were rebuilt and re-used to update the AI or action. Those design patterns should be reusable in this context:
• Process the nouns one by one
• For each noun:
• Reduce the activation factor of current concepts in the activation list
• Apply the synapse factor to the current noun
• Add concept to the activation list
• With a reduced activation factor by synapse, add related concepts that are connected to the currently processed concept to the list as well
• Get an inventory of the highest activated concepts in the list
• Store the gist list to describe the text
Obviously, the thing that is missing from the above is the intention, the thread. So a text that describes a guy in India getting of a bus to his house may equate to the same gist of a bus in San Francisco that happened to drive past an Indian restaurant.
So motivation and thread of a story is something entirely different from its concepts. Should this be part of the network?? In a way, I think it should be possible to think of it as a layered network above the symbolic network, a different kind of representation with links to the other network to describe actions and objects that are acted upon.
### It's the state of the network, stupid!
Back to the discussion on Artificial Intelligence and Neural Networks. This blog hosted some meanderings and thoughts on AI and how this relates to NN. I read books from Steven Pinker and others regarding NN's and I understand how these work from a helicopter view (I even implemented some). I then analyzed that the perspective of NN's on AI is probably horribly wrong, if you compare biological networks against mechanical ones.
Computers were from the very start designed to work on input and then produce output. Output is a very binary-like state. It's either this or that, it's a range of numbers and the output is generally an exact number. There's not really a way to represent an answer that represents two (slightly) different states in one answer by itself.
This morning, I woke up and considered that this general approach that was taken as part of AI way back is probably wrong. Even though computers are supposed to produce output that can only be interpreted in one single way, the "output" of human brains doesn't really exist as output per se. I'm more thinking of answer and thought as a kind of "state of the network" at some point in time. The frequency of thought is given by the "frequency" of the network, although this seems a very weird term to use for biological networks. It's probably totally independent.
If you look at CAT-scans though, you'll see something interesting. Not all neurons are active at all points in time (very contrary to mechanical networks, which generally have all their nodes and parts connected, so that turning one part will turn and impact another). And the granularity of our analysis on the human brain is not at neuron level, but at the level where we see a general number of neurons receiving activity. So if one neuron A next to neuron B is fired, only A would be active, but B would not be assessed.
So it's like regions of interconnected neurons are active at one sweep, not the entire network. And there's no output like a machine, only a list of active and recently active neurons. Every sweep, the active list is modified and moved back into a ring of memory.
If we reconsider neurons as nodes in a network and replace them with symbols instead, we can probably come close to a logical representation of a thought network. So, a neuron by itself doesn't represent anything, but impacts something physically. A symbol is highly representative of something, but doesn't necessarily impact anything, it is only connected to other symbols.
The symbolic network is then like a neural network, only it works with nouns, with symbols. The symbolic network allows an infinite number of nodes to be added, as long as there exists a process to interconnect symbols as soon as there is a relation to be determined between them.
Now, imagine what happens and assume this symbolic network is properly connected. When you mention a word car, or the smell of exhaust gas, or a picture of a car, those symbols are activated. The joint activation of car, exhaust gas and the picture should activate a symbol of car (without annotation) as a concept, so that the network understands that car is being discussed.
If you now introduce a text with nouns and verbs and assuming the computer has grammatical analysis capabilities, you can process this text within the symbolic network and at the end of some paragraph / passage of text, the network has a certain state of activity. Some regions are highlighted and other regions are black. If you'd keep a list of symbols that are activated, then you could store that list (region) as a representation of the text.
So, the objective is not to store the text word for word, but to store the associations and context of the paragraph. Mentioning the words in a search term would probably produce the text again and the more aligned with the paragraph it is, the more likely it is to be found.
Memory is also important. There are actually different rings of memory in this model (short-term and long-term are too generic). Reading a passage would store the gist of that passage into a different symbol. The gist are basically those nodes that had the highest activation of a certain paragraph after the cycle is completed. So storing the gist of one paragraph with another may develop a description of a document that is highly descriptive. It's not necessarily the word that is mentioned, it's the concept and the relation to other symbols. It's possible that a symbol is highly activated that was not explicitly mentioned in the text.
The symbolic network is the representation of nouns of our language, but verbs are the activities, the manipulations in our mind of those symbols. It seems then that within the context of this blog post, the verbs correspond to real intelligence (which is not described in this post yet). The nouns are just perceptions and mapping them to symbols. Real thought, the one that creates (is creative) and can reproduce and come to certain conclusions is a totally different matter. That sounds like real artificial intelligence.
## Wednesday, March 05, 2008
### Project Estimation
There are some interesting discussions taking place on project estimation. Project Dune 1.4.2 came out yesterday, where you can register your estimates per activity per scope statement. So that's basically "ballpark" level of detail.
There are some new requirements that are coming up that look really interesting. PM's really seem to (want to) use some kind of math magic to get to certain numbers and then tweak them.
So, what really matters in a project? What are factors that, if they change, significantly affect the timeline of a project?
I've already mentioned that the level of communication is a huge issue. So, if your team increases, you'll increase contention within the team, the need for communication (which loses time) and so on. Especially at the start, where the PM actually intended to move fastest. So the rest of the trajectory people are generally told to "get on with it", leading to unsatisfiable levels of communication and lower standards of quality.
Oh well! :)... Expect some more notices in the future with regards to this topic. | 3,427 | 17,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.962508 |
https://se.mathworks.com/matlabcentral/profile/authors/3559574 | 1,627,962,086,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00682.warc.gz | 507,999,672 | 18,504 | Community Profile
Paul Kassebaum
MathWorks
Active since 2012
I enjoy data visualization design and physics.
Professional Interests: 3d printing, robotics, quantum mechanics, physics
All
Content Feed
View by
3D printing using MATLAB
Hi Rajbir, <http://www.ijstr.org/final-print/dec2016/Modeling-And-Position-Control-Of-Scara-Type-3d-Printer.pdf here's a paper> ...
mer än 3 år ago | 0
| accepted
Submitted
circularGraph
Plot an interactive circular graph to illustrate connections in a network.
Submitted
vector3
3-D vector plot.
Submitted
punctureSurface
Create a 3D printable meshed surface
Submitted
binaryTensorVoxel
Draw a 3D voxel representation of a binary tensor or 3D matrix.
Submitted
Reflect Through a Cylindrical Mirror
Reflect points p through a cylinder of radius R with respect to a viewpoint v = [v_x, v_y, v_z]
Solved
Pizza!
Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
ungefär 5 år ago
Solved
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
mer än 6 år ago
Solved
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
mer än 6 år ago
Solved
Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
mer än 6 år ago
Solved
Make a checkerboard matrix
Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Example...
nästan 9 år ago
Solved
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
nästan 9 år ago
Solved
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
ungefär 9 år ago | 571 | 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} | 3.3125 | 3 | CC-MAIN-2021-31 | latest | en | 0.648238 |
https://daily-ai.com/utilizing-xgboost-for-time-series-forecast-jobs/ | 1,553,052,200,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202199.51/warc/CC-MAIN-20190320024206-20190320050206-00088.warc.gz | 451,847,993 | 13,211 | # Utilizing XGBoost for time series forecast jobs
Recently Kaggle master Kazanova in addition to some of his friends launched a “How to win a data science competitors” Coursera course. The Course included a final job which itself was a time series prediction issue. Here I will explain how I got a leading 10 position since composing this article.
Description of the Problem:
In this competitors, we were offered a tough time-series dataset consisting of everyday sales information, kindly provided by among the biggest Russian software companies – 1C Company.
We were asked you to anticipate overall sales for every product and shop in the next month.
The assessment metric was RMSE where True target worths are clipped into [0,20] variety. This target variety will be a lot essential in understanding the submissions that I will prepare.
The main thing that I observed was that the data preparation element of this competition was without a doubt the most essential thing. I created a range of features. Here are the actions I took and the functions I produced.
## 1. Created a data frame of all Date_block_num, Store and Item combinations:
This is very important because in the months we don’t have data for an item store mix, the machine learning algorithm requires to be particularly informed that the sales is zero.
from itertools import product# Create "grid" with columns
index_cols = ['shop_id', 'item_id', 'date_block_num']
# For every month we create a grid from all shops/items combinations from that month
grid = []
for block_num in sales['date_block_num'].unique():
cur_shops = sales.loc[sales['date_block_num'] == block_num, 'shop_id'].unique()
cur_items = sales.loc[sales['date_block_num'] == block_num, 'item_id'].unique()
grid.append(np.array(list(product(*[cur_shops, cur_items, [block_num]])),dtype='int32'))
grid = pd.DataFrame(np.vstack(grid), columns = index_cols,dtype=np.int32)
## 2. Cleaned up a little of sales data after some basic EDA:
sales = sales[sales.item_price<100000]sales = sales[sales.item_cnt_day<=1000]
## 3. Created Mean Encodings:
sales_m = sales.groupby(['date_block_num','shop_id','item_id']).agg({'item_cnt_day': 'sum','item_price': np.mean}).
reset_index()
sales_m = pd.merge(grid,sales_m,on=['date_block_num','shop_id','item_id'],
how='left').fillna(0)
# adding the category id toosales_m = pd.merge(sales_m,items,on=['item_id'],how='left')
for type_id in ['item_id','shop_id','item_category_id']:
for column_id,aggregator,aggtype in
[('item_price',np.mean,'avg'),('item_cnt_day',np.sum,'sum'),
('item_cnt_day',np.mean,'avg')]:
mean_df = sales.groupby([type_id,'date_block_num']).
aggregate(aggregator).reset_index()
[[column_id,type_id,'date_block_num']]
mean_df.columns = [type_id+'_'+aggtype+'_'+column_id,type_id,'date_block_num']
sales_m = pd.merge(sales_m,mean_df,on=['date_block_num',type_id],how='left')
These above lines add the following 9 features:
• ‘item_id_avg_item_price’
• ‘item_id_sum_item_cnt_day’
• ‘item_id_avg_item_cnt_day’
• ‘shop_id_avg_item_price’,
• ‘shop_id_sum_item_cnt_day’
• ‘shop_id_avg_item_cnt_day’
• ‘item_category_id_avg_item_price’
• ‘item_category_id_sum_item_cnt_day’
• ‘item_category_id_avg_item_cnt_day’
## 4. Create Lag Features:
Next we create lag features with diferent lag durations on the following features:
• ‘item_id_avg_item_price’,
• ‘item_id_sum_item_cnt_day’
• ‘item_id_avg_item_cnt_day’
• ‘shop_id_avg_item_price’
• ‘shop_id_sum_item_cnt_day’
• ‘shop_id_avg_item_cnt_day’
• ‘item_category_id_avg_item_price’
• ‘item_category_id_sum_item_cnt_day’
• ‘item_category_id_avg_item_cnt_day’
• ‘item_cnt_day’
lag_variables = list(sales_m.columns[7:])+['item_cnt_day']lags = [1 ,2 ,3 ,4, 5, 12]
for lag in lags: sales_new_df = sales_m.copy()
sales_new_df.date_block_num+=lag
sales_new_df = sales_new_df[['date_block_num','shop_id','item_id']+lag_variables]
sales_new_df.columns = ['date_block_num','shop_id','item_id']+ [lag_feat+'_lag_'+str(lag) for lag_feat in lag_variables]
sales_means = pd.merge(sales_means, sales_new_df,on=['date_block_num','shop_id','item_id'] ,how='left')
## 5. Fill NA with zeros:
for feat in sales_means.columns: if 'item_cnt' in feat:
sales_means[feat]=sales_means[feat].fillna(0) elif 'item_price' in feat: sales_means[feat]=sales_means[feat].fillna(sales_means[feat].median())
## 6. Drop the columns that we are not going to use in training:
cols_to_drop = lag_variables[:-1] + ['item_name','item_price']
## 7. Take a recent bit of data only:
sales_means = sales_means[sales_means['date_block_num']>12]
## 8. Split in train and CV :
X_train = sales_means[sales_means['date_block_num']<33].drop(cols_to_drop, axis=1)X_cv = sales_means[sales_means['date_block_num']==33].drop(cols_to_drop, axis=1)
## 9. THE MAGIC SAUCE:
In the start, I told that the clipping element of [0,20] will be essential. In the next couple of lines, I clipped the days to variety [0,40] You may ask me why 40. An instinctive answer is if I had clipped to range [ 0,20] there would be extremely couple of tree nodes that might give 20 as an answer. If I increase it to 40 having a 20 becomes much more easier, while. Please note that We will clip our forecasts in the [0,20] variety in the end.
def clip(x):if x>40:
return 40 elif x<0: return 0
else:
return x
train['item_cnt_day'] = train.apply(lambda x: clip(x['item_cnt_day']),axis=1)
cv['item_cnt_day'] = cv.apply(lambda x: clip(x['item_cnt_day']),axis=1)
## 10: Modelling:
• Created a XGBoost model to get the most important features(Top 42 features)
• Use hyperopt to tune xgboost
• Used top 10 models from tuned XGBoosts to generate predictions.
• clipped the predictions to [0,20] range
• Final solution was the average of these 10 predictions.
Learned a lot of brand-new things from this remarkable course. Most advised.
Originally published on MLWhiz | 1,518 | 5,875 | {"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} | 2.625 | 3 | CC-MAIN-2019-13 | latest | en | 0.902884 |
https://discuss.codechef.com/t/mafia-loc-july-2017/15842/5 | 1,620,697,284,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00108.warc.gz | 226,565,619 | 6,660 | # MAFIA-LOC July 2017
I tried very hard to solve this problem but couldn’t clear all the test cases. I got AC for all test cases except 10th test case. Can anybody help me out?
Here is the link to my code.
use binary search and dfs!(check this give true condition)
Even my obviously wrong solution passed all test cases except the 10th. I have solved a similar problem before but I cannot think of a test case at the moment(Sorry :(, hope someone helps you out with that). Taking just r/2 is wrong(As far as I can think of). Also, dfs does not give the shortest distance between two nodes in a graph, that might also cause WA(I used bfs).
In case you want to consider a different solution what you can do is binary search on the value of r. If taking the current r, you can make the dons stay happily then you can take a bigger(or the same) r, or else you need to lessen your r(All values about this value of r are invalid). The runtime if O(NlogN), O(N) because of each dfs to check if the current r is suitable or not, and O(logN) because of the binary search.
1 Like
Solved it using multisource BFS. Initialize a visited array for nodes with -1 and add all nodes with dons to a queue and change their visited value to 0. So now for every node in queue add all nodes to queue till you reach a phase where you encounter an already visited node. This means that it was already occupied by other don. So this is the breaking condition. Also while adding a node to queue I’m deleting the edge between the parent node and child node so that we will not travel back to parent from child and also change the visited value of child node to the depth. So the answer will be the number of nodes less than the depth at which the visited node is revisited. I tried to explain my approach. Please comment if you find anything wrong.
6 Likes
@tayal007, your intention was correct but your solution was not. I am surprised your solution passed almost all the test cases. Fortunately, there are only a couple of changes that will make your solution correct
So the algorithm to be followed is: for each don vertex, find the nearest don vertex; the distance between the closest pair of don vertices is 2 times the answer.
The first mistake is that you are using dfs. Dfs does not guarantee that the shortest path from source will be followed. For example, in the graph below dfs may take the shown path from vertex 1 to 4, which is not what you want. You need to be using bfs instead.
The second problem you will face after implementing bfs is with your visited array. For each bfs you need to reset your visited array otherwise the bfs will not always manage to find the closest don if it encounters a vertex marked visited on the way. You also cannot actually reset your visited array to false each time, because in the worst case that will be \mathcal{O}(N^2). You can overcome this by using a different value to mark the visited vertices for each bfs.
I solved the problem using this approach. Here is my solution. Hope this helps!
The worst case for this solution is below. The dons are marked with orange. Let x be some constant. Then there will be N/x bfs, and each bfs will take \mathcal{O}(N) to reach the closest don. So complexity is \mathcal{O}(N^2/x) \approx \mathcal{O}(N^2). So this should not be able to clear the time limit, but clearly the test cases are weak.
PS: See @sai_rathan’s answer using multisource bfs, it seems that is the most efficient way.
Can you tell why taking r/2 is wrong?
l=1 and r=100001 now mid=(l+r)/2 and then check can or not…
goodby
I did it using binary search only, check my solution below
All I asked is why is taking r/2 wrong(Take a look at his code).
better you change your code and get it clear just you need a binary search in main and a dfs and a visited array and a count in dfs for count number of visited if condition finally is true.
if you use clear code …you can find your mistake!
Thank you @ugly_sharp
Thank you @ista2000
Your solution is also taking indefinite time… to solve worst test case generated by this file https://ideone.com/RVcJyU . Please look into it… Please explain your approach too…
1 Like | 993 | 4,187 | {"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} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.94677 |
https://www.speedsolving.com/threads/how-i-broke-8-personal-bests-in-1-day.77717/ | 1,600,673,667,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198942.13/warc/CC-MAIN-20200921050331-20200921080331-00653.warc.gz | 1,091,376,112 | 17,241 | # How I Broke 8 Personal Bests in 1 Day
#### Micah Morrison
##### Member
I feel like I break 8+ personal bests every day, but most of them are stuff like 7x7 ao50, megaminx ao100, etc.
#### u Cube
##### Member
I feel like I break 8+ personal bests every day, but most of them are stuff like 7x7 ao50, megaminx ao100, etc.
Lol you got a lot of free time dude
#### Micah Morrison
##### Member
Lol you got a lot of free time dude
well... I do in the summer, but most of the time my big averages for big cubes are my best of that average, so if I get one good solve, then I break it.
#### vidcapper
##### Member
Do you get the same thrill at setting a PB as you ever did?
#### TipsterTrickster
##### Member
I have broken like 100+ pbs in a day, granted it was like 85 ao100 pbs, but still.
#### teboecubes
##### Member
Do you get the same thrill at setting a PB as you ever did?
For some things, like 6x6 single and 4x4. Obviously not as much as if I had broken by 3x3 single PB though
#### vidcapper
##### Member
It's certainly a thrill to solve a new cube for the first time - i still remember back in the 80's when I solved the 3x3 for the first time, it took me 25 minutes.
#### Jupiter
##### Member
I have broken like 100+ pbs in a day, granted it was like 85 ao100 pbs, but still.
8500 3x3 solves ?
If you solved each solve back to back(without scrambling time) at exactly 10 seconds itd still take 23 hours and 37 minutes
#### TipsterTrickster
##### Member
8500 3x3 solves ?
If you solved each solve back to back(without scrambling time) at exactly 10 seconds itd still take 23 hours and 37 minutes
No rolling average of 100, so if I am doing a session where my 1st solve was a 9.5, and I finish my ao100 and that is my ao100 pb, then on the 101st solve i get a 9.4 my new ao100 will be sub my old ao100 and that is a new ao100 pb.
#### vidcapper
##### Member
No rolling average of 100, so if I am doing a session where my 1st solve was a 9.5, and I finish my ao100 and that is my ao100 pb, then on the 101st solve i get a 9.4 my new ao100 will be sub my old ao100 and that is a new ao100 pb.
Unless the time that dropped out was a 9.3... | 632 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.946124 |
http://gmatclub.com/forum/if-b-a-4-then-for-which-of-the-folloing-159224.html?sort_by_oldest=true | 1,484,835,007,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280668.34/warc/CC-MAIN-20170116095120-00133-ip-10-171-10-70.ec2.internal.warc.gz | 115,567,495 | 59,043 | If b=a+4, then for which of the folloing : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 19 Jan 2017, 06:10
# LIVE NOW:
Chat with Admission Manager and Current Student of NUS SIngapore - Join Chat Room to Participate.
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If b=a+4, then for which of the folloing
Author Message
TAGS:
### Hide Tags
Manager
Joined: 14 Jun 2011
Posts: 85
Followers: 2
Kudos [?]: 34 [1] , given: 15
If b=a+4, then for which of the folloing [#permalink]
### Show Tags
05 Sep 2013, 11:17
1
KUDOS
1
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
56% (02:44) correct 44% (02:18) wrong based on 80 sessions
### HideShow timer Statistics
If b=a+4, then for which of the following values of x is the expression (x−a)^2+(x−b)^2 the smallest?
A) a−1
B) a
C) a+2
D) a+3
E) a+5
I know that we can solve this question by plugging in the option values. I want to know the smart way to choose which option to plug in first.
[Reveal] Spoiler: OA
_________________
Kudos always encourages me
Intern
Joined: 14 Aug 2013
Posts: 35
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
Followers: 2
Kudos [?]: 74 [2] , given: 4
Re: If b=a+4, then for which of the folloing [#permalink]
### Show Tags
05 Sep 2013, 11:39
2
KUDOS
lets take a= 0 ; then b=4 and the equation becomes x^2+(x-4)^2. The minimum value of this function occurs when
x^2 = (x-4)^2 as both of these functions are increasing functions.
Solving this we get x=2 it is minimum. for a=0 option c is gives the value of x as 2.
hope it helps!
Math Expert
Joined: 02 Sep 2009
Posts: 36554
Followers: 7078
Kudos [?]: 93170 [1] , given: 10553
Re: If b=a+4, then for which of the folloing [#permalink]
### Show Tags
05 Sep 2013, 11:47
1
KUDOS
Expert's post
swati007 wrote:
If b=a+4, then for which of the following values of x is the expression (x−a)^2+(x−b)^2 the smallest?
A) a−1
B) a
C) a+2
D) a+3
E) a+5
I know that we can solve this question by plugging in the option values. I want to know the smart way to choose which option to plug in first.
Discussed here: m19-q18-120286.html
_________________
Manager
Status: Persevering
Joined: 15 May 2013
Posts: 225
Location: India
GMAT Date: 08-02-2013
GPA: 3.7
WE: Consulting (Consulting)
Followers: 1
Kudos [?]: 86 [0], given: 34
Re: If b=a+4, then for which of the folloing [#permalink]
### Show Tags
07 Sep 2013, 05:33
1
This post was
BOOKMARKED
Maxima and minima
Used derivatives here
(x-a)^2+(x-b)^2
First derivative=> 2*(x-a)+2*(x-b) =0 =>x=(a+b)/2 ;substitute b=> x=a+2
_________________
--It's one thing to get defeated, but another to accept it.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13447
Followers: 575
Kudos [?]: 163 [0], given: 0
Re: If b=a+4, then for which of the folloing [#permalink]
### Show Tags
26 Dec 2014, 06:28
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36554
Followers: 7078
Kudos [?]: 93170 [0], given: 10553
Re: If b=a+4, then for which of the folloing [#permalink]
### Show Tags
26 Dec 2014, 06:40
swati007 wrote:
If b=a+4, then for which of the following values of x is the expression (x−a)^2+(x−b)^2 the smallest?
A) a−1
B) a
C) a+2
D) a+3
E) a+5
I know that we can solve this question by plugging in the option values. I want to know the smart way to choose which option to plug in first.
If $$b=a+4$$, then for which of the following values of $$x$$ is the expression $$(x-a)^2 + (x-b)^2$$ the smallest?
A. $$a-1$$
B. $$a$$
C. $$a+2$$
D. $$a+3$$
E. $$a+5$$
Since $$b=a+4$$ then $$(x-a)^2 + (x-b)^2=(x-a)^2 + (x-a-4)^2$$. Now, plug each option for $$x$$ to see which gives the least value.
The least value of the expression if for $$x=a+2$$ --> $$(x-a)^2 + (x-a-4)^2=(a+2-a)^2 + (a+2-a-4)^2=8$$.
_________________
Re: If b=a+4, then for which of the folloing [#permalink] 26 Dec 2014, 06:40
Similar topics Replies Last post
Similar
Topics:
3 The factorial expression 13!/7! is not divisible by which of the follo 5 02 Jun 2016, 03:46
2 Considering the positions on the number line above, which of the follo 4 16 May 2016, 01:34
8 An even and an odd integer are multiplied together. Which of the follo 10 17 Apr 2015, 03:48
10 If k is an integer and k^2 - 4 > 45, then which of the follo 6 28 Sep 2013, 02:42
2 For the numbers on the number line above, which of the follo 3 10 Feb 2013, 00:43
Display posts from previous: Sort by | 1,802 | 5,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.609375 | 4 | CC-MAIN-2017-04 | latest | en | 0.820178 |
https://apk-dl.com/sudoku-cities/com.ionicframework.sdk744452 | 1,544,761,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00562.warc.gz | 518,820,127 | 28,093 | # Sudoku Cities 2.6.0 APK
2.6.0 / May 11, 2016
(4.7/5) (7)
## Description
For each Sudoku that you solve, we show you incredible pictures ofthe most beautiful cities in the world...This game has 4 Sudokulevels: Easy, Medium, Hard and Professional,and images of the 50most beautiful cities in the World...The numbers in the game sudokuare generated randomly. Arithmetic relations between numbers, areabsolutely necessary irrelevantes.Não is no mathematical logic orcalculation capacity. They can use any combination of differentsymbols (such as letters, shapes, or colors) without changing therules.The success of the game due to the fact that the rules are sosimple, however, the line of reasoning required to reach thesolution can be complex. Sudoku is recommended by some teachers asan exercise in logical thinking. The level of difficulty can beselected to match the audience. There are several sources on theInternet are not linked to publishers that provide the games forfree.Their size is most often a 9x9 grid made up of sub-3x3 gridscalled "regions" (other terms include "boxes" and "blocks",sometimes, the term "quadrant" is used, although a term inaccuratefor a grid of 3x3). Some cells already containing numbers, called"number data" (or sometimes "tracks"). The goal is to fill theempty cells with a number in each cell, so that each column, rowand region contains the numbers 1 to 9 only once. In the solutionof the game, each number appears only once in either direction orregions; hence, the term sudoku, which means "single number".The3x3 region in the upper right corner. The solver can eliminate allthe empty cells in the upper right corner containing one 5 in thesame columns or rows.Scanning is performed at the beginning andthroughout the solution. Scans only have to be performed oncebetween periods of analysis. Scanning consists of two basictechniques:Crossing: the scanning of rows (or columns) to identifywhich line in a particular region may contain a certain number by aprocess of elimination. This process is then repeated with thecolumns (or rows). For faster results, the numbers are checked inorder of frequency. It is important to perform this processsystematically, checking all of the digits 1-9.Count the numbers1-9 in regions, rows and columns to identify missing numbers.Counting number based on the latest discovered may cause the searchis faster. It may also be the case (typically in tougher puzzles)that the easiest way to check the value of an individual cell iscounting in reverse - that is, scanning of the cell region, row,and column for values that "can not" be, in order to discover whatremains.Advanced solvers look for "contingencies" while scanning -that is, strengthening the position of a numeral in a row, column,or region to two or three cells. When those cells all lie withinthe same row (or column) "and" region, they can be used forelimination purposes during cross steps and counting. [4] Inparticular, the more challenging puzzles may require multiplecontingencies to be discovered perhaps in multiple directions oreven multiple intersections. The puzzles that can be solved bymaking the scan without detection of contingencies are classifiedas "easy" puzzles; more difficult puzzles, by definition, can notbe solved by basic scanning alone.
## App Information Sudoku Cities
• App Name
Sudoku Cities
• Package Name
com.ionicframework.sdk744452
• Updated
May 11, 2016
• File Size
10M
• Requires Android
Android 4.1 and up
• Version
2.6.0
• Developer
Enne Mobi
• Installs
1,000+
• Price
Free
• Category
Puzzle
• Developer
Rua Condessa Mumadona, 39 Creixomil 4835 035 Guimarães | 849 | 3,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-51 | latest | en | 0.917412 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/tin-blank--op-white-cp--coma-and-blank-beta-blank-form | 1,675,276,219,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00172.warc.gz | 660,915,424 | 8,775 | # Weight of Tin (white), beta form
## tin (white), beta form: convert volume to weight
### Weight of 1 cubic centimeter of tin (white), beta form
carat 36.33 ounce 0.26 gram 7.27 pound 0.02 kilogram 0.01 tonne 7.27 × 10-6 milligram 7 265
#### How many moles in 1 cubic centimeter of tin (white), beta form?
There are 61.2 millimoles in 1 cubic centimeter of tin (white), beta form
### The entered volume of tin (white), beta form in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• About tin (white), beta form
• 1 cubic meter of tin (white), beta form weighs 7 265 kilograms [kg]
• 1 cubic foot of tin (white), beta form weighs 453.53913 pounds [lbs]
• Tin (white), beta form weighs 7.265 gram per cubic centimeter or 7 265 kilogram per cubic meter, i.e. density of tin (white), beta form is equal to 7 265 kg/m³; at 20°C (68°F or 293.15K) at standard atmospheric pressure. In Imperial or US customary measurement system, the density is equal to 453.5391 pound per cubic foot [lb/ft³], or 4.1994 ounce per cubic inch [oz/inch³] .
• Melting Point (MP), tin (white), beta form changes its state from solid to liquid at 232°C (449.6°F or 505.15K)
• Boiling Point (BP), tin (white), beta form changes its state from liquid to gas at 2623°C (4753.4°F or 2896.15K)
• Tin (white), beta form is a soft silvery-white metal (the beta form), which transforms to non-metallic powdery gray tin (the alpha form) when cooled below 13.2°C for a sustained period of time.
• Also known as: Tin.
• Symbol: Sn [ element | tin-based compounds ]
Atomic number: 50
Atomic weight: 118.710(7)
Ionization energy, eV: 7.343917
Molecular formula: Sn
Molecular weight: 118.71 g/mol
Molar volume: 16.34 cm³/mol
CAS Registry Number (CAS RN): 7440-31-5
• Bookmarks: [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
• A material, substance, compound or element with a name containing, like or similar to tin (white), beta form:
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
SOUR CREAM, UPC: 07203657030 weigh(s) 254 grams per metric cup or 8.5 ounces per US cup, and contain(s) 200 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
5876 foods that contain Vitamin D (D2 + D3). List of these foods starting with the highest contents of Vitamin D (D2 + D3) and the lowest contents of Vitamin D (D2 + D3), and Recommended Dietary Allowances (RDAs) for Vitamin D
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Gravel, dry 1/4 to 2 inch weighs 1 682 kg/m³ (105.00383 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 10W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
An a.u. of length is a non-SI Fundamental Physical unit of length.
Radioactivity is the act of emitting radiation spontaneously by an atomic nucleus that, for some reason, is unstable.
oz t/in³ to long tn/metric tsp conversion table, oz t/in³ to long tn/metric tsp unit converter or convert between all units of density measurement.
#### Calculators
Calculate area of a trapezoid, its median, perimeter and sides | 1,213 | 4,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.615118 |
http://oeis.org/A164921 | 1,638,399,187,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00258.warc.gz | 57,822,855 | 4,138 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 56th year, we are closing in on 350,000 sequences, and we’ve crossed 9,700 citations (which often say “discovered thanks to the OEIS”).
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A164921 a(1)=0, a(2)=1. For n >=3, a(n) = the smallest integer > a(n-1) that is coprime to every sum of any two distinct earlier terms of this sequence. 2
0, 1, 2, 5, 11, 17, 23, 29, 37, 41, 47, 53, 59, 67, 71, 79, 83, 89, 97, 101, 107, 113, 127, 131, 137, 149, 157, 163, 167, 173, 179, 191, 197, 211, 223, 227, 233, 239, 251, 257, 263, 269, 277, 281, 293, 307, 311, 317, 331, 337, 347, 353, 359, 367, 373, 379, 383 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS EXAMPLE The first 4 terms are 0,1,2,5. The sums of every pair of distinct terms are: 0+1=1, 0+2=2, 1+2=3, 0+5=5, 1+5=6, and 2+5=7. So we are looking for the smallest integer >5 that is coprime to 1, 2, 3, 5, 6, and 7. This number, which is a(5), is 11. CROSSREFS Sequence in context: A067775 A140552 A138644 * A156830 A140556 A003627 Adjacent sequences: A164918 A164919 A164920 * A164922 A164923 A164924 KEYWORD nonn AUTHOR Leroy Quet, Aug 31 2009 EXTENSIONS More terms. Sean A. Irvine, Sep 15 2009 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 1 16:44 EST 2021. Contains 349430 sequences. (Running on oeis4.) | 645 | 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} | 3.625 | 4 | CC-MAIN-2021-49 | longest | en | 0.772245 |
https://cs.stackexchange.com/questions/163076/time-complexity-of-algorithm-with-three-loops-and-if-statement | 1,718,287,463,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00769.warc.gz | 170,912,423 | 39,573 | # Time complexity of algorithm with three loops and if statement
Suppose I have this c++ code:
for(int i=0; i<n*n*n; i++) {
for(int j=0; j<n*n; j++) {
if(i>n && j<n) {
for(int k=0; k<n*n*n*n; k++) s++;
}
}
}
I need to find its time complexity.
What I tried doing is: I tried determining the times the if clause would be fulfilled, and multiplying that with $$n^4$$.
However, I have a hard time determining how many times the if clause will be fulfilled. The values of i go from $$n+1$$ to $$n^3-1$$, while the values for j go from $$0$$ to $$n-1$$.
Now, for each value of i for which the first part of the condition is true, I have that there are $$n$$ values of j for it.
Unfortunately, this is as far as I have come. Could anyone help me out further?
1. The if is executed when $$n and $$0\leq j . This means that the values of $$i$$ which satisfy the condition are $$n^3-n-1$$, while the values of $$j$$ which satisfy the condition are $$n-1$$. For each value of $$i$$ for which the condition on it is satisfied you can choose every values of $$j$$ which satisfies the condition of it and viceversa. It means that the if is executed $$(n^3-n-1)(n-1) = \Theta(n^4)$$ times. In these cases the inner-most loop is executed $$n^4$$ times and each execution has a constant cost. So, the total cost of this case is $$\Theta(n^4\cdot n^4)=\Theta(n^8)$$.
2. In the remaining cases, i.e. $$\approx n^5-n^4=\Theta(n^5)$$ times, the if is not executed and the cost of one iteration the two outer-most loops is constant (just a check on the if condition), thus the cost of this case is $$\Theta(n^5)$$.
Summing up, the total cost is: $$\Theta(n^8)+\Theta(n^5)= \Theta(n^8) \ .$$
PS: If you are not familiar with asymptotic notation, look at it (big-O, Theta, Omega, small-o notations or Landau symbols). You can consider $$\Theta(n^i)$$ as $$\approx n^i$$ while reading this answer. Then I suggest you to go check these topics. | 561 | 1,927 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-26 | latest | en | 0.877987 |
https://ask.learncbse.in/t/two-sides-ab-and-bc-and-median-am-of-one-triangle-abc-are-respectively-equal-to-sides-pq-and-or-and-median-pn-of-pqr/43468 | 1,603,758,454,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892710.59/warc/CC-MAIN-20201026234045-20201027024045-00284.warc.gz | 219,085,580 | 3,427 | # Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅ ∆PQR
(ii) ∆ABM ≅ ∆PQN | 90 | 295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.934349 |
https://socratic.org/questions/what-is-the-limit-of-1-cos-x-sin-x-2-as-x-approaches-0 | 1,722,665,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00190.warc.gz | 429,732,414 | 6,407 | # What is the limit of (1-cos(x))/sin(x^2) as x approaches 0?
Jul 8, 2016
${\lim}_{x \to 0} \frac{1 - \cos \left(x\right)}{\sin \left({x}^{2}\right)} = \frac{0}{0}$ (an indeterminate form), thus we can apply
L'Hospital's rule, since substituting $x = 0$ yields $\frac{0}{0}$.
In this case, applying L'Hospital's rule we get
${\lim}_{x \to 0} \frac{\sin \left(x\right)}{2 x \cdot \cos \left({x}^{2}\right)} \to \frac{0}{0}$, which also yields an indeterminate form,
thus we apply it again:
${\lim}_{x \to 0} \frac{\cos \left(x\right)}{2 \cos \left({x}^{2}\right) - 4 {x}^{2} \cdot \sin \left({x}^{2}\right)}$
=lim_(x->0) (cos(x))/(2(cos(x^2)-2x^2 * sin(x^2))
=cos(0)/(2(cos(0)-2(0)(sin(0))
$= \frac{1}{2 \left(1 - 0\right)} = \frac{1}{2}$
#### Explanation:
L'Hospital's rule states that if function(s) $f$ and $g$ are differentiable on an open interval $I$ except at $c$ contained in $I$, then if
${\lim}_{x \to c} f \left(x\right) = {\lim}_{x \to c} g \left(x\right) = 0$ or ± ∞,
g'(x) ≠ 0 for all $x \in I$ with x ≠ c, and
${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists, then
${\lim}_{x \to c} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$
See more examples here. | 519 | 1,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "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} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.609059 |
https://betterlesson.com/lesson/resource/3013702/challenge-problems-for-center-pdf | 1,511,363,906,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806609.33/warc/CC-MAIN-20171122141600-20171122161600-00682.warc.gz | 570,098,004 | 19,135 | ## challenge problems for center.pdf - Section 2: Practice Centers
challenge problems for center.pdf
# Time to Practice
Unit 11: The Numbers Are Getting Bigger
Lesson 7 of 15
## Big Idea: Students need time to practice new skills. They have been introduced to addition and subtraction of double-digit numbers. Today's groups will allow for practice of these skills.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, Manipulative Skills, Number Sense and Operations, addition, subtraction, base ten
55 minutes
### Beth McKenna
##### Similar Lessons
###### 10 and Some More
1st Grade Math » Complements of 10 and 20
Big Idea: Your students will play a game where they combine two single digit numbers and then record that total in terms of how it relates to ten.
Favorites(6)
Resources(20)
Waitsfield, VT
Environment: Suburban
###### The Recipe for a Great Word Problem
Big Idea: The big idea of this lesson is to have students write their own word problems to help them have a better understanding of mathematical operations, as they relate to real world scenarios.
Favorites(88)
Resources(14)
Pepperell, MA
Environment: Rural | 258 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.880825 |
https://jsri.srtc.ac.ir/browse.php?a_id=181&slc_lang=en&sid=1&printcase=1&hbnr=1&hmb=1 | 1,624,184,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487660269.75/warc/CC-MAIN-20210620084505-20210620114505-00298.warc.gz | 301,645,602 | 6,514 | Volume 4, Issue 1 (9-2007)
JSRI 2007, 4(1): 109-128 Back to browse issues page
Functional Analysis of Iranian Temperature and Precipitation by Using Functional Principal Components Analysis
Norallah Tazikeh Miyandarreh, Ebrahim Hosseini-nasab Seyed Mohammad *1
1- , m.hosseininasab@modares.ac.ir
Abstract: (2882 Views)
Extended Abstract. When data are in the form of continuous functions, they may challenge classical methods of data analysis based on arguments in finite dimensional spaces, and therefore need theoretical justification. Infinite dimensionality of spaces that data belong to, leads to major statistical methodologies and new insights for analyzing them, which is called functional data analysis (FDA).
Dimension reduction in FDA is mandatory, and is partly done by using principal components analysis (PCA). Similar to classical PCA, functional principal components analysis (FPCA) produces a small number of constructed variables from the original data that are uncorrelated and account for most of the variation in the original data set. Therefore, it helps us to understand the underlying structure of the data.
Temperature and amount of precipitation are functions of time, so they can be analyzed by FDA. In this paper, we have treated Iranian temperature and precipitation in 2005, extract patterns of variation, explore the structure of the data, and that of correlation between the two phenomena. The data, collected from the weather stations across the country, were discrete and associated with the monthly mean of temperature and precipitation recorded at each station. However, we have first fitted appropriate curves to them in which we have taken smoothing methods into account. Then, we have started analyzing the data using FPCA, and interpreting the results. When estimating the eigenvalues, we have found that the first estimated eigenvalue \$hat {theta}\$ shows a strong domination of its associated variation on all other kinds. Furthermore, the first two eigenvalues explain more than 98% of the total variation, inwhich their contributions individually were 93.7 and 4.3 percent, respectively. Contributions from others, however, were less than 2 percent. Thus, we have only considered the first two components.
The first estimated principal component (PC) shows that the majority of variability among the data can be attributed to differences between summer and winter temperatures. The second PC shows regularity of temperature when moving from winter to summer. In other words, it reflects the variation from the average of the difference between the winter and summer temperatures. Furthermore, bootstrap confidence bands for eigenvalues and eigenfunctions of the real data were obtained. They contain both individual and simultaneous confidence intervals for the eigenvalues. We have also obtained single and double bootstrap bands for the first two eigenfunctions, and seen that they are extremely close to each other, reflecting the high degree of accuracy of the bands that are obtained by the single bootstrap methods.
Keywords: Bootstrap confidence bands, data registration, functional data analysis, functional principal components analysis. | 615 | 3,199 | {"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} | 2.703125 | 3 | CC-MAIN-2021-25 | latest | en | 0.902978 |
https://www.enworld.org/threads/introducing-the-countdown-dice-mechanic.695102/page-2 | 1,685,650,575,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00344.warc.gz | 807,865,040 | 33,091 | # D&D 5EIntroducing the COUNTDOWN DICE Mechanic!
#### Stormonu
##### Legend
Another thought, you could add a mechanic for characters to push their luck by willfully removing a die from the pool in return for some beneficial effect - a hint, advantage on a roll, rerolling damage, acquisition of a tangentially beneficial item, etc.
Furthermore, you can also use gradiants of the pool. For example, if the entire dungeon will collapse when the Countdown pool is exhausted, it could be that after a certain number of dice are lost, non-essential sections of the dungeon may collapse. For example, a bonus treasure room might become inaccessible after 4 dice are lost, the "short" route may become unavailable after 6 dice are lost and when 8 dice are lost sections of the main route fall away, requiring the characters to parkour, fly or otherwise get creative to make it out. When all 10 dice are gone, the whole thing falls away...
#### Weiley31
##### Legend
I like this. And may use it in my regular 5E games.
##### Legend
Supporter
It is also called exponential decay. We use that to simulate radioactivity (I don't know all the correct English words).
The average remaining dice can be calculated with the formula:
Remaining-dice = Starting-number-of-dice * (5/6)^turns
Note, that with this formula you will never approach zero, as it uses a continuum instead of discrete numbers.
You can use the formula for the half-life-time and say: after 10 half-life times you are close to zero.
You can also use a treshold and say: if you get below < 0.5 you are done and use logarithm to estimate the rounds.
You can probably just look up a formula on the internet too.
Your encouragement to just think about it got me to pull my head out of my tookus.
Realizing it's the largest order statistic of a geometric distribution does certainly make it easier! Here's (some not too pretty) code for looking up to turnmax turns and using a dice with nsides. This replaces the code in post #8 above.
turnmax<-20
nsides<-6
pendby<-NULL
for (i in 1:turnmax){
pendby<-c(pendby,(1-(1-1/nsides)^i)^nsides)
if (i==1){pendon<-pendby[ i]}
else{pendon[ i]<-pendby[ i]-pendby[ i-1]}}
pendby[turnmax+1]=1
pendon[turnmax+1]=1-pendby[turnmax]
x<-cbind(1 : (turnmax+1),round(pendon,4),round(pendby,4))
colnames(x)<-c("turn #","on that turn","by that turn")
x
Last edited:
#### Dausuul
##### Legend
I think these are the results for mean and sd of number of turns using one to ten d4:
ndice mean standard deviation
[1,] 4.000000 3.464102
[2,] 5.714286 3.896623
[3,] 6.872587 4.065951
[4,] 7.741776 4.157531
[5,] 8.437008 4.215216
[6,] 9.016347 4.254849
[7,] 9.512926 4.283732
[8,] 9.947433 4.305713
[9,] 10.333662 4.323000
[10,] 10.681268 4.336953
Yup, I get the same values.
In general, larger numbers of smaller dice will yield more consistent results. (9d4 and 3d6 have roughly the same average duration, but you're far less likely to see your countdown expire on the first round with 9d4.)
If you want to ensure an absolute minimum duration, you can put a cap on the number of dice that can be removed from the pool each round.
I'd also suggest having dice exit the pool on a roll of 1 rather than 6*. This has no effect on the probabilities, but in most of the scenarios I can imagine for this mechanic, the players are racing the clock; so tick-down on 1 is consistent with "low numbers are bad." It also means you can use any size die, or even a mix of dice, without having to think about what number you're looking for.
*Or 4, or whatever size die you're using.
Last edited:
#### DeviousQuail
##### Hero
I like using a similar method. A pool of d6s that is rolled after each time interval. If all the dice are 6s the time expires. If you roll and all the dice aren't sixes you reduce the dice pool by 1 to a minimum of 1. Dice can be added or subtracted by the actions taken by the players. Actions taken by npcs can force another roll.
I like it because the pool shrinks at a steady rate, giving a sense of growing dread to the players. The odds of getting all 6s prior to 2 dice is also really low* so when the pool has 5 dice the players are confident they have at least 5 intervals to achieve success. On the flip side sometimes I'll roll in between intervals due to something happening outside the view of the players (think wandering enemies). Or because of something in view (priest accelerates the ritual instead of casting a spell at the players). This can really light a fire under the butts of players.
#### el-remmen
##### Moderator Emeritus
I like it. I plan to make use of it in the new Temple of Elemental Evil based campaign I am starting.
##### Hero
Could also do something similar by using shinking die sizes rather than removing dice (or combine the two).
Can work with fewer dice and provides a bit more control over the possibility that you yahtzee your doom clock.
#### dave2008
##### Legend
I have used this heavily in my current A5e game and have found it an excellent mechanic. I have often used it for threats over long period of time. For example, my current game involves a number of fey lords. Because their sense of time is very capricious, their activities are often random in time. Recently the party greatly angered one of them, so I set up a doom clock for retaliation. The fey lord is coming, but will it be next week....or 3 months from now? The clock adds that fun tension.
I also like the countdown clock combined with A5e's "Doomed Condition". Doomed is a status condition where the person is going to die, and simple "clerical healing" is not sufficient to save them. Now the core condition is "this person is going to die, no matter what" but what I have found more effective in my game is to add some plot way to save them, and attach a doom clock.
So the party must get the magical flower X to save this person, and the doom clock triggers once every day. Will the party get back in time to save the person.....we will find out!
It seems like a good way to handle poisoning as well. You are stricken with a deadly poison (how deadly determines the number of dice) and you roll each morning to determine your fate. In addition to a cure, there can be methods to add or subtract dice to the pool.
#### Stormonu
##### Legend
An additional idea for holding breath, overland travel, etc.
Take a number of D6's equal to your Con, and roll them at each interval (rounds, minutes, 10 minutes, half-hour, hour as appropriate). As normal each 1 is removed from the pool. If you exert yourself in some way, you automatically remove a die as well.
So, if for example, someone was holding their breath underwater and had a Con of 10, you'd roll 10 dice every round, removing 1's from each roll. If the character was also fighting, each round you'd reduce the pool by one die, in addition to the 1's being rolled.
Another example, the party is travelling from point A to B. Using the lowest Con of the party's members, you make a roll every half hour. When the party runs out of dice, they need to stop and rest (levels of exhaustion if they don't). If they have an encounter or do a force march to increase their travel rate by 50%, they remove a die from the pool. If they stop to take a short rest, they can add a die back to the pool.
You could also use a double-edged pool for a chase. Start with a number of dice, say 10. Every time a 1 is rolled, the target(s) gets further away. Every time a 6 is rolled, the party gets closer to catching their target. Each side can perform one trick or stunt to attempt to remove a die in their favor - The party's Ranger successfully making a skill check to track the enemy counts a die as a 6. The enemy successfully using Stealth to hide in an alley counts the die as a 1. And so on.
Last edited:
#### J.Quondam
##### CR 1/8
When Morrus mentioned this mechanic several months ago, I thought it might make a good "thermometer" measuring, for example, a village's attitude toward a key NPC. The roll would be made everyday at sunset; when the pool runs out, the mob takes up their pitchforks and torches and storms the NPC's castle, or whatever. PC actions during the preceding day could add or subtract a die to the pool, whether intentionally (if the PCs specifically tried to incite or calm the mob) or unintentionally (if the PCs do something careless, support one NPC over another, etc).
Broadly, it seems like a good way to indicate - and even manipulate, by adding/subtracting dice - the "temperature" of a group before it explodes, whether it's a bar brawl or a war.
#### darjr
##### I crit!
Speak the Sky came up with something similar. Great minds think alike. Their focus was on replacing the Jenga Tower.
They went on and made a following blog post
#### Dausuul
##### Legend
I like using a similar method. A pool of d6s that is rolled after each time interval. If all the dice are 6s the time expires. If you roll and all the dice aren't sixes you reduce the dice pool by 1 to a minimum of 1. Dice can be added or subtracted by the actions taken by the players. Actions taken by npcs can force another roll.
I like it because the pool shrinks at a steady rate, giving a sense of growing dread to the players. The odds of getting all 6s prior to 2 dice is also really low* so when the pool has 5 dice the players are confident they have at least 5 intervals to achieve success. On the flip side sometimes I'll roll in between intervals due to something happening outside the view of the players (think wandering enemies). Or because of something in view (priest accelerates the ritual instead of casting a spell at the players). This can really light a fire under the butts of players.
Continuing with this theme of "clock with variations" -- you could have a pool of d6s, and each round (or whatever your time unit is) you roll one of those dice.
NON-EXPLODING VERSION (if you don't want the risk of a "yahtzee" that slams the timer to zero in a single turn)
6: Put the die back in the pool.
EXPLODING VERSION (if you want the risk of a "yahtzee")
6: Put the die back in the pool, then roll a die from outside the pool.
1: Discard the the die and roll another die.
This would be suitable for cases where you have a general timeline in mind (roughly X rounds) but want some uncertainty about exactly when the clock runs out.
Last edited:
#### LordEntrails
##### Hero
Here's the average number of rounds and standard deviation for dice pools from 1 to 10:
DICE AVG ROUNDS STD DEV 1 6.00 5.46 2 8.73 6.13 3 10.56 6.39 4 11.93 6.54 5 13.02 6.63 6 13.94 6.69 7 14.72 6.74 8 15.41 6.77 9 16.02 6.80 10 16.56 6.82
Note that the standard deviation is very high, indicating a very wide spread. If you use this mechanic, be prepared for a lot of variation in your results.
Umm... shouldn't for 1 die it be less than 6? I think of it if I roll 5 times, by that fifth roll I should more often than not have gotten a 6. Which means the average would be less than that. But... I'm very uncertain...
#### Dausuul
##### Legend
Umm... shouldn't for 1 die it be less than 6? I think of it if I roll 5 times, by that fifth roll I should more often than not have gotten a 6. Which means the average would be less than that.
The "long tail" of cases where you go seven or more rounds balances out the more common case that you go five or less. Remember, that long tail includes instances where the timer goes extremely long. There is a 2.6% chance that you could go twenty rounds on a single die! It only takes a few of those cases to pull up the average significantly.
The net result is to average out at 6.
#### Reynard
##### Legend
I like it. it is a good tool for things that are uncertain, even for the GM. When are those monsters going to finally breach the wall? How long before the hurricane is upon the ship? What exactly is the patience level of a dragon before it just decides to eat the bard?
#### DND_Reborn
##### The High Aldwin
The psychology behind this amazes me... that so many people think it is cool, etc. I've known of groups doing this since my SW WEG days (with the d6's) back in the 80's. I saw a resurgence in Shadowrun in the late 90's (also d6's) as well, but haven't heard of it sense then...
IME it fell out of favor quickly due to the hassle of dice pools and became just a single die (which IMO works just as well ), but it is interesting to see the idea rekindled and newer players seeming to enjoy it.
#### cbwjm
##### Legend
ICRPG also uses a countdown mechanic, but it's usually just rolling a 1d4 or 1d6 and counting down from there each round, and then the bad thing happens at the end of the final round. Sure, the players know that something bad will happen, but that meta-knowledge actually contributes to the tension IME.
I've used this specific clock a few times in my games, it's quite a useful tool.
#### Reynard
##### Legend
In a couple weeks am running a SWADE zombie outbreak scenario at a con, and I think I'll use this mechanic for determining when the overrun finally happens. Thanks @Morrus
#### Stormonu
##### Legend
We some way to finagle this to use d20’s. Roll of 1-3 to discard the die, you can choose to remove one automatically from the pool to give yourself advantage?
Replies
13
Views
3K
Replies
23
Views
1K
Replies
49
Views
3K
Replies
17
Views
694
Replies
3
Views
1K | 3,395 | 13,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.897662 |
http://towardsthelimitedge.pedromoralesalmazan.com/2009/10/elementary-symmetric-functions-of-first.html | 1,524,289,375,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00363.warc.gz | 304,578,104 | 16,022 | ## Monday, October 5, 2009
### Elementary symmetric functions of the first natural numbers
About a month ago, my good friends Esteban and José Carlos were working in a really interesting problem in number theory involving harmonic series of sequences of something that they called the Esteban primes for $p_0$.
They were (or are, I hope) trying to find out if a particular series involving this sequences of primes, converges or not. They told me about the problem, and as always, I got really interested by this strange problem relating primes and analysis.
After working out a little bit the series, one can show that it is equivalent to prove that some infinite product converges to a nonzero value. Working with these expressions, I ended up looking at polynomials of the form
$p(x)=\prod_{k=1}^n (p_k-x)$
where the $p_k$ are primes. So, at first, I tried to not work with primes directly, but with the natural numbers $1,2,3,\dots,n$ so the polynomial will be
$p(x)=\prod_{k=1}^n (k-x)$
Expanding out this polynomial gives, by Cardano equations,
$p(x)=\sum_{k=0}^n (-1)^k s_{n-k} x^{k}$
were the $s_k$ are the elementary functions for $1,2,3\dots,n$, that are given by the sum of all possible $k$ products of these numbers, so for instance, $s_n=n!$, $s_1=1+2+\dots+n$ and $s_0=1$.
Therefore, evaluating $p(1)$ we have that $p(1)=\sum_{k=0}^n (-1)^k s_{n-k}=0$ and hence, the sum of the even elementary functions is equal to the sum of the odd ones.
For example, $n=3$ gives
$s_0=1$
$s_1=1+2+3=6$
$s_2=1\cdot 2+1\cdot 3+2\cdot 3=11$
$s_3=1\cdot 2\cdot 3=6$
and the result is
$s_0+s_2=s_1+s_3=12$
This reminds me of the property of the Binomial coefficients and must be somehow related, because these count the number of subsets with cardinality $k$, that is, for each subset of size $k$, we correspond to it a 1, then we add this results and that is $\binom{n}{k}$, while for the symmetric functions, for each subset of size $k$, we correspond to it the product of its elements, and then add over all subsets, and that gives $s_k$.
In other words, let $S=\{1,2,3,\dots,n\}$ and let $f$ be a function defined on the subsets of $S$.
If $f(A)=1$, for all $A\subset S$, we have that
$\binom{n}{k}=\sum_{A\subset S, |A|=k}f(A)$
on the other hand, if we define $f(A)=\prod_{k\in A} k$, then
$s_k=\sum_{A\subset S, |A|=k}f(A)$
so both forms have the same structure. I don't know what is the underlying property of such $f$ functions, so that the sum of the evens equals the sum of the odd ones, but I find it really interesting.
1. and if you chup it?
2. Now we are trying to find a special Esteban's prime sequence to call it the "optimum prime sequence"... for no particular reason whatsoever (ºJª).
-MC | 815 | 2,728 | {"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} | 3.953125 | 4 | CC-MAIN-2018-17 | longest | en | 0.904566 |
https://www.cheenta.com/pre-rmo-2018-solutions/ | 1,628,210,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00605.warc.gz | 706,410,792 | 42,071 | What is the NO-SHORTCUT approach for learning great Mathematics?
# Pre RMO 2018
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="3.2" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="left" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"]
[/et_pb_divider][et_pb_text _builder_version="3.0.82" text_text_color="#d4ccff" text_line_height="1.9em" animation_style="zoom" animation_intensity_zoom="6%" locked="off"]
Find the problems, discussions and relevant theoretical expositions related to Pre-RMO 2018.
[/et_pb_button][/et_pb_column][et_pb_column type="1_2"][et_pb_image src="https://www.staging18.cheenta.com/wp-content/uploads/2018/03/coding-isometric-12.png" _builder_version="3.0.82" max_width="90%" module_alignment="center" animation_style="slide" animation_direction="left" animation_duration="500ms" animation_delay="100ms" animation_intensity_slide="10%"]
# Problems of Pre RMO
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="3.2" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="center" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"]
[/et_pb_divider][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Problems 1 to 6" _builder_version="3.12.2"]
1. A book is published in three volumes, the pages being numbered from 1 onwards. The page numbers are continued from the first volume to the third. The number of pages in the second volume is 50 more than in the first volume, and the number pages in the third volume is one and a half times that in the second. The sum of the page numbers on the first pages of the three volumes is 1709. If n is the last page number. What is the largest prime factor of n?
2. In a quadrilateral $ABCD$. It is given that $AB=AD=13$ $BC=CD=20, BD=24$. If $r$ is the radius of the circle inscriable in the quadrilateral, then what is the integer close to $r$?
3. Consider all 6-digit numbers of the form $abccba$ where $b$ is odd. Determine the number of all such 6-digit numbers that are divisible by 7?
4. The equation $166 \times 56 = 8590$ is valid in some base $b \geq 10$ that is $1, 6, 5,8,9,0$ are digits in base $b$ in the above equation. Find the sum of all possible values of $b \geq 10$ satisfying the equation.
5. Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AD \perp AB$. Suppose $ABCD$ has an incircle which touches $AB$ at $Q$ and (\ CD \) at $P$. Given that $PC = 36$ and $QB=49$. Find $PQ$.
6. Integers $a, b,c$ satisfy $a+b-c=1$ and $a^2 + b^2 -c^2 = 1$. What is the sum of all possible values of $a^2+b^2+c^2$?
[/et_pb_tab][et_pb_tab title="Problem 7 to 11" _builder_version="3.12.2"]
7. A point $P$ in the interior of a regular hexagon is at distances $8,8,16$ units from three consecutive vertices of the hexagon, respectively. if r is radius of the circumscribed circle of the hexagon. What is the integer closest to $r$?
8. Let $AB$ be a chord of a circle with centre $O$. Let $C$ be a point on the circle such that $\angle ABC = 30^\circ$ and $O$ lies inside the triangle $ABC$ . Let $D$ be a point on $AB$ such that $\angle DCO=\angle OCB=20^\circ$. Find the measure of $\angle CDO$ in degrees.
9. Suppose $a, b$ are integers and $a+b$ is a root of $x^2+ax+b=0$. What is the maximum possible value of $b^2$?
10. In a triangle $ABC$, the median from $B$ to $CA$ is the perpendicular to the median from $C$ to $AB$. If the median from $A$ to BC is 30. Determine $\frac {BC^2 + CA^2 + AB^2}{100}$.
11. There are several teacups in the kitchen, some with handles and the others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly 1200. What is the maximum possible number of cups in the kitchen?
12. Determine the number of 8-tuples of 8-tuples $(\epsilon_1,\epsilon_2,...,\epsilon_8)$ such that $\epsilon_1,\epsilon_2,...,\epsilon_8 \in \{1,-1\}$ and $\epsilon_1+2\epsilon_2+3\epsilon_3+…+8\epsilon_8$ is a multiple 3.
13. In a triangle $ABC$, right-triangle at$A$, the altitude through $A$ and the internal bisector of $\angle A$ have lengths 3 and 4 respectively. Find the length of the median through $A$.
14. If $x= \cos 1^\circ \cos 2^\circ \cos 3^\circ …. \cos 89^\circ$ and $x=\cos 2^\circ \cos 3^\circ …. \cos 86^\circ$ then what is the$\frac{2}{3} \log_{2}(\frac {y}{x})$?
15. Let $a$ and $b$ be natural numbers such that $2a-b, a-2b$ and $a+b$ are all distinct squares, What is the smallest possible value of $b$?
16. What is the value of $\displaystyle{ \sum_{1\leq j < j \leq 10\\ i+j=odd} (i+j) - \sum_{1\leq j < j \leq 10\\ i+j=even} (i+j) }$
17. Triangles $ABC$ and $DEF$ are such that $\angle A = \angle D$, $AB=DE=17, BC=EF=10$ and $AC-DF=12$. What is $AC+DF$?
18. If $a,b,c \geq 4$ are integers, not all equal, and $4abc=(a+3)(b+3)(c+3)$ , then what is the value $a+b+c$?
19. Let $n=6+66+666+…+666…66$; where there are hundred 6’s in the last term in the sum. How many times does the digit 7 occur in the number $N$?
20. Determine the sum of all possible positive integers n, the product of whose digits equals $n^2-15n-27$.
21. Let$ABC$ be an acute angled triangle and let $H$ be its orthocenter. Let $G_1,G_2$ and$G_3$ be the centroids of the triangle$HBC, HCA$ and $HAB$ respectively. If the area of triangle $G_1G_2G_3$ is 7 units, what is the area of the triangle $ABC$?
22. A positive integer $k$ is said to be good if there exists a partition of $\{ 1, 2, 3,….., 20\}$ in to disjoint proper subsets such that the sum of the numbers in each subset of the partition is $k$. How many good numbers are there?
23. What is the largest positive integer $n$ such that
$\frac{a^2}{\frac{b}{29}+\frac{c}{31}}$ + $\frac{b^2}{\frac{c}{29}+\frac{a}{31}}$ + $\frac{c^2}{\frac{a}{29}+\frac{b}{31}}$ $\geq n(a+b+c)$
holds for all positive integer $a,b,c$.
24. If $N$ is the number of triangles of different shapes {i.e., not similar} whose angles are all integers {in degrees}, what is $N/100$?
25. Let $T$ be the smallest positive integer which, when divided by 11,13,15 leaves remainders in the sets $\{7,8,9\}$, $\{1,2,3\}$, $\{4,5,6\}$ respectively. What is the sum of the squares of the digits of $T$?
26. What is the number of ways in which one can choose 60 unit squares from a $11 \times 11$ chessboard such that no two chosen squares have a side in common?
27. What is the number of ways in which one can colour the squares of a $4 \times 4$ chessboard with colours red and blue such that each row as well as each column has exactly two red squares and two blue squares?
28. Let $N$ be the number of ways distributing 8 chocolates of different brands among 3 children such that each child gets at least one chocolates, and no two children get the same number of chocolates. Find the sum of the digits of N.
29. Let $D$ be an interior point of the side $BC$ of a triangle $ABC$. Let $I_1$ and $I_2$ be the incentres of triangle $ABD$ and $ACD$ respectively. Let $AI_1$ and $AI_2$ meet $BC$ in $E$ and $F$ respectively. If $\angle BI_1E =60^\circ$, what is the measure $\angle CI_2F$ in degrees?
30. Let $P(x)= a_0 + a_1x+a_2x^2+….+a_nx^n$ be a polynomial in which $a_i$ is a non-negative integer for each$i\in \{ 0,1,2,3,…,n\}$. If$P(1)=4$ and $P(5)=136$. What is the value of $P(3)$?
[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.0.82" text_text_color="#8585bd" text_line_height="1.9em" text_orientation="center" max_width="540px" module_alignment="center" animation_style="zoom" animation_intensity_zoom="6%" locked="off"]
# Discussions
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="3.2" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="center" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"]
[/et_pb_divider][et_pb_text _builder_version="3.0.82" text_text_color="#d4ccff" text_line_height="1.9em" text_orientation="center" max_width="600px" module_alignment="center" animation_style="zoom" animation_intensity_zoom="6%" locked="off"]
The solutions, hints and theoretical discussions will uploaded soon. Please check back.
[/et_pb_text][et_pb_button button_url="https://www.staging18.cheenta.com/forums/topic/prmo-2018-discussions/" url_new_window="on" button_text="Discussion Forum" _builder_version="3.12.2" custom_button="on" button_text_color="#ffffff" button_bg_color="#0c71c3" button_font="||||||||"]
[/et_pb_image][et_pb_divider _builder_version="3.12.2"]
[/et_pb_image][et_pb_divider _builder_version="3.12.2"]
[/et_pb_button][/et_pb_column][et_pb_column type="1_4"][/et_pb_column][et_pb_column type="1_4"][/et_pb_column][/et_pb_row][/et_pb_section] | 2,807 | 9,041 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | longest | en | 0.653331 |
https://yuihuang.com/zj_h084/ | 1,721,536,230,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00260.warc.gz | 918,203,270 | 16,172 | # 【題解】ZeroJudge h084: 4. 牆上海報
【題目敘述】https://zerojudge.tw/ShowProblem?problemid=h084
【解題想法】二分搜
```#include <bits/stdc++.h>
using namespace std;
int n, k, h[200005], w[50005];
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> k;
int l = 1, r = 0;
for (int i = 1; i <= n; i++){
cin >> h[i];
r = max(r, h[i]);
}
r++;
for (int i = 1; i <= k; i++){
cin >> w[i];
}
while (r-l > 1){
int mid = (l+r)/2;
int cnt = 0, now = 1;
bool flag = false;
for (int i = 1; i <= n; i++){
if (h[i] >= mid){
cnt++;
if (cnt >= w[now]){
cnt -= w[now];
if (now == k){
flag = true;
break;
}
now++;
}
}
else cnt = 0;
}
if (flag) l = mid;
else r = mid;
}
cout << l << "\n";
}
``` | 278 | 668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.221205 |
https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-6/v/local-linearity-and-differentiability | 1,674,927,308,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00720.warc.gz | 860,562,511 | 99,249 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## AP®︎/College Calculus AB
### Unit 4: Lesson 6
Approximating values of a function using local linearity and linearization
# Local linearity and differentiability
AP.CALC:
CHA‑3 (EU)
,
CHA‑3.F (LO)
,
CHA‑3.F.1 (EK)
Intuition for how local linearity relates to differentiability using the Desmos graphing calculator.
## Want to join the conversation?
• At , we are unable to differentiate the function around (2,0) because it's basically a vertical line. At , most of the curve looks a lot like a vertical line, yet the function is differentiable for all x values. Can anyone please tell me what am I missing? What's the difference between these two cases?
• Let's take an example like the x^1000. You really can't tell if it's a hard corner or not, especially if you're only given the zoomed-out graph. Is there a way to determine local linearity/differentiability if you don't have the equation of the graph?
• If there is a cusp, hard corner, vertical asymptote, or any sort of discontinuity then the function is not differentiable. Typically if you are asked to look at a graph and say if the graphed function is differentiable or not, the graph in the question will make it pretty clear that one of the conditions is violated (you probably won't have to strain your eyes to see if there is a hard corner or not etc.). As for 𝑥¹⁰⁰⁰, it is differentiable because all polynomials are differentiable. Though trying to tell from the graph may be difficult for the function because it may seen like a hard corner but if you zoom in sufficiently you'll see that it is smooth. I wouldn't worry about these extreme examples of graphs because I doubt anyone testing you in math would try to trick you based on examples that are difficult to physically see to gauge your mathematical ability.
• So I just did the Practice: Approximation with local linearity. About half of the questions were of the sort "What is f(x) + f'(x) ?" I don't understand the purpose of this question. What is the value of the function plus the slope of the of local linearization? Why would you want to add the slope to the value? Usually these practice questions have an apparent purpose, so I feel like I'm missing something.
• f(x) + f'(x) this can be used to approximate value for points which are very close to x. The function actually should look like this
f(x) + f'(x) * C
Here C is some scalar. If you zoom in to the point x sufficiently then x and its close points reside in a straight line. So if you know the slope at the point x which is f'(x) then f(x)+f'(x) will give you the value of the point closest to x. If you do
f(x) + f'(x) * C
Then this will give you value to the point that are C th closest to x.
Hope this helps.
• Do we need this for AP Calculus?
• Yes, differentiability is useful in AP Calculus.
• How would you solve for the linear approximation on arctan(1/2)?
• Using a linear approximation, f(x) is approximately equal to f(a)+f'(a)delta x. The derivative of arctan(x) is 1/(x^2+1) and we know that arctan(1/sqrt(3)) = pi/6. So, arctan(1/2) is approximately pi/6 + 1/((1/sqrt(3))^2+1)*(1/2-1/sqrt(3)) = pi/6 + 3/4(1/2-1/sqrt(3)) which is about 0.4656. (The actual answer is 0.4636 so the approximation is fairly close.)
(Sorry for the random bold portion of text, but I cannot change it for some reason.)
• So basically, "looks like a line" means "differentiable"? Is there a more precise way to define local linearity?
• Not quite; that is just an intuitive introduction Sal uses in the video. A function is differentiable over an interval iff it is continuous over that interval and there are not any bends, cusps, or vertical asymptotes. Local linearity means just what it says. A function is locally linear over an interval iff that interval is sufficiently small for a tangent line to closely approximate the function over the interval.
• I don't think until now, the "power rule" has been discussed. Am I missing a lecture?
• The power rule is a straightforward rule to learn.
If n is a constant, then the derivative of x^n, with respect to x, is nx^(n-1).
This is true for all real values of n, including whole numbers, zero, positive and negative integers, fractions, and even irrational numbers.
Example: The derivative of x^4 is 4x^3.
Harder example: Suppose we need to find the derivative of 5/cuberoot(x).
First rewrite 5/cuberoot(x) as 5x^(-1/3).
The derivative is then (-1/3)[5x^(-4/3)] = -5/[3 cuberoot(x^4)] = -5/[3x cuberoot(x)].
Have a blessed, wonderful day!
• I wonder how many flat earthers have taken calculus?
• Oh! Can I ask about the name of the website you did the functions graphs? Thanks so much!
(1 vote)
• Desmos graphing calculator, btw its already in the description.
## Video transcript
- [Instructor] What we're going to do in this video is explore the relationship between local linearity at a point and differentiability at a point. So local linearity is this idea that if we zoom in sufficiently on a point, that even a non-linear function that is differentiable at that point will actually look linear. So let me show some examples of that. So let's say we had y is equal to x squared. So that's that there, clearly a non-linear function. But we can zoom in on a point, and if we zoom sufficiently in, we will see that it looks roughly linear. So let's say we wanna zoom in on the point one comma one, so let's do that. So zooming in on the point one comma one, already it is looking roughly linear at that point. And this property of local linearity is very helpful when trying to approximate a function around a point. So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one. And you might not need to do that for y is equal to x squared, but it could actually be very very useful for a more complex function. But the big takeaway here, at the point one one, it is displaying this idea of local linearity, and it is also differentiable at that point. Now let's look at another example of a point on a function where we aren't differentiable, and we also don't see the local linearity. So for example, let's do the absolute value of x, and let me shift it over a little bit just so that we don't overlap as much. Alright, so the absolute value of x minus one. It actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero. For any other x value, it is differentiable, but right at x equals one, we've talked in other videos how we aren't differentiable there. And then we can use this local linearity idea to test it as well. And once again, this is not rigorous mathematics, but it is to give you an intuition. No matter how far we zoom in, we still see this sharp corner. It would be hard to construct the only tangent line, a unique line, that goes through this point one comma zero. I can construct an actual infinite number of lines that go through one comma zero but that do not go through the rest of the curve. And so notice, wherever you see a hard corner like we're seeing at one comma zero in this absolute value function, that's a pretty good indication that we are not going to be differentiable at that point. Now let's zoom out a little bit, and let's take another function. Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line. So a good example of that would be square root of let's say four minus x squared. So that's the top half of a circle of radius two. And let's focus on the point two comma zero. Because right over there, we actually are not differentiable, and if we zoom in far enough, we see right at two comma zero that we are approaching what looks like a vertical line. So once again, we would not be differentiable at two comma zero. Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that hey I got a corner here on this absolute value function, or at two comma zero, or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable. But there are some functions that we don't see as typically in a algebra or precalculus or calculus class, but it can look like a hard corner from a zoomed out perspective, but as we zoom in once again we'll see the local linearity, and they are also differentiable at those points. So a good example of that, let me actually get rid of some of these just so that we can really zoom in. Let's say y is equal to x to the, and I'm gonna make a very large exponent here, so x to the 10th power. It's starting to look at little bit like a corner there. Let's make it to the 100th power. Well now it's looking even more like a corner there. Let me go to the 1,000th power just for good measure. So at this scale, it looks like we have a corner at the point one comma zero. Now this curve actually does not go to the point one comma zero. If x is one, then y is going to be one, and we'll see that as we zoom in, this what looks like a hard corner is going to soften. And that's good because this function is actually differentiable at every value of x. It's a little bit more exotic that what we typically see, but as we zoom in, we'll actually see that. Let's just zoom in on what looks like a fairly hard corner, but if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves. And if we zoom in sufficiently, it will actually look like a line. It's hard to believe when you're really zoomed out, and I'm going at the point that really looked like a corner from a distance. But as we zoom on in, we see once again this local linearity that's a non-vertical line. And so once again, this is true at any point on this curve, that we are going to be differentiable. So the whole point here is, sometimes you might have to zoom in a lot, a tool like Desmos which I'm using right now is very helpful for doing that. And this isn't rigorous mathematics, but it's to give you an intuitive sense that if you zoom in sufficiently, and you start to see a curve looking more and more like a line, good indication that you are differentiable. If you keep zooming in and it still looks like a hard corner, of if you zoom in and it looks like the tangent might be vertical, well then some questions should arise in your brain. | 2,616 | 10,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.940678 |
https://gustavus.edu/mcs/max/courses/F2011/MCS-375/hotels/ | 1,521,367,304,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645604.18/warc/CC-MAIN-20180318091225-20180318111225-00590.warc.gz | 597,540,637 | 3,612 | # MCS-375 Homework 5 (Fall 2011)
For this 12-point project, you will help me out with an authentic trip-planning problem. As the weather has turned cooler, my wife and I have recalled how much we enjoyed a trip to Austin, Texas, a few years ago, and have talked about going back. Last time we flew, but this time the idea of driving has come up. The only problem is that we don't drive anywhere near that far in a day, so we need to pick some spots for overnight stays. That's where you come in. You can write a program to pick hotels so as to minimize the total cost, subject to a limit on how many miles we drive in a day.
Using the web site of a national hotel chain, I put together a list of hotels along the route. For each one, I've got the name, distance from our starting point, and advertised room rate (price). This is real data from the web but reflects just a snapshot; your algorithm ought to survive if the data changes. I rounded both the distances and the room rates to integers. Here's how the list starts:
```hotels=[('Super 8 Mankato', 10, 47),
('Days Inn Mankato', 10, 49),
('Microtel Inn & Suites Mankato', 10, 65),
...]
```
You might laugh at my having included Mankato hotels in the list; even my wife and I normally drive further than that in a day. In part, this just reflects my having produced the list in a systematic, computerized fashion. But in fact, depending on how the list continues, the Super 8 in Mankato might be part of an optimal solution, even if our driving allowance per day is more than 10 miles. You ought to be able to construct examples showing that just about any "greedy" strategy sometimes fails:
• Sometimes it makes sense to pass by an inexpensive hotel.
• Sometimes it makes sense to stop for the night even if further hotels exist within your daily mileage allowance.
• Sometimes it makes sense to include more hotel stops than the minimum.
There is at least one heuristic that could safely be used to prune obviously bad options. It never makes sense to stop at a hotel if the hotel's cost is nonnegative and another one further along, but still within your day's limit, has at least as low of a room rate. As it turns out, for the data I gathered, this is a very powerful heuristic. It eliminates enough choices that the remaining ones can be searched using a brute force algorithm. However, I don't want you using this approach, because if the hotels change their prices, the heuristic might become much less effective. You ought to be able to construct an example where no pruning at all would be achieved. And without pruning, you sure don't want to use brute force. There are 83 hotels on my list, which leads to 283 possibilities for stopping or not stopping at each one.
You'll notice I mentioned nonnegativity of cost in the prior paragraph. Naturally enough, all of the hotels listed on the web site did indeed list nonnegative room rates. Nonetheless, it would be best if your code avoided assuming this constraint. That way, if there are places that Stefanie and I particularly are interested in staying, we can tweak the cost of those hotels downward, even to the point of a negative cost if we really want to stay there.
Your goal is to write a `trip` procedure in Python that produces an optimal plan for the trip. The optimal plan is a pair of two things: a total price and a list of hotels. Each of the hotels is one of the triples taken from the list I provide. Here's the heading for your procedure:
```def trip(hotels, allowance):
"""Return a trip plan, staying within an allowance of miles per day.
A plan is a pair of a total cost (sum of room rates) and a list of hotels.
Each hotel is a triple of name, distance from the start, and room rate.
The given list of hotels must be listed in nondecreasing distance.
The last 'hotel' on the list is actually the ultimate trip destination
and must be included in any plan; its room rate might be 0.
All mileages must be integers: hotel distances and the allowance.
If no plan is feasible, the procedure raises a ValueError."""
```
I've linked to this assignment a hotels.py file that contains this procedure header and the full version of the `hotels` list. To receive full credit, you need to create a correct version of the `trip` procedure using either the memoized top-down approach or the bottom-up dynamic programming approach. You don't have to wring every last bit of efficiency out of your procedure, but it should scale up polynomially in the length of the hotels list and the number of miles in my allowance. Exponential growth is bad. You should quickly get answers for the full hotels list and any allowance, whether 80 miles, 1000 miles, or anything in between. (The 80 mile figure is just barely large enough to make the trip feasible, given the spacing of the hotels.)
If you are having trouble getting started, I can offer you various shortcuts, each of which cuts into the numbers of points you can earn. Each of these shortcuts starts with me providing you a recursive top-down version that gives correct answers for smaller lists of hotels, but takes unreasonably long for the full trip to Austin.
• If you take the recursive version from me and then write a correct bottom-up version, you'll earn 11 of the 12 points.
• If you take the recursive version from me and are stuck writing a bottom-up version, you can ask me for a template for that version, in which I create the necessary table, provide a comment explaining what each slot in the table ought to contain, and return the appropriate slot from the table at the end. Your code would fill in the table. If you do that much, you'll earn 10 of the 12 points.
• If even the template isn't enough to get the bottom-up approach working, you can fall back on the memoized top-down approach, which is closer to the recursive version I provide. However, then you'll only earn 9 of the 12 points.
Note in particular that if you don't take the recursive version from me, you'll be able to earn the full 12 points even with a memoized top-down approach. | 1,342 | 6,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.957157 |
https://dsp.stackexchange.com/questions/16253/why-is-beginning-of-this-fir-filtered-signal-attenuated | 1,719,216,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00609.warc.gz | 190,464,408 | 41,012 | # Why is beginning of this FIR filtered signal attenuated?
I apply an FIR filter passing the band around 50/1000 and attenuating the other three frequency components 30/1000, 70/1000 and 110/1000.
Both filters are generated using the Remez algorithm. The first filter is of length 300 and the second is of length 100.
Why is the beginning of the filtered signal attenuated and why is this attenuation effect stronger for a longer filter?
library(signal)
par(mfrow=c(3,2))
ch <-
sin(2*pi*1:1000/floor(1000/30)) +
sin(2*pi*1:1000/floor(1000/50)) +
sin(2*pi*1:1000/floor(1000/70)) +
sin(2*pi*1:1000/floor(1000/110))
# FFT of signal
barplot(abs(fft(ch)[1:120]))
# unfiltered signal
plot(ch,type="l")
filter <- function(c0,d1,d2,n) {
fir <- remez(n=n,f=c(0,c0-d2,c0-d1,c0+d1,c0+d2,1),a=c(0,0,1,1,0,0))
freq <- freqz(fir,n=n)
y <- signal::filter(as.vector(fir), 1, x=ch)
# frequency response of filter
barplot(abs(fft(y))[1:120])
return(list(freq = freq, fir = fir, y = y))
}
f <- filter(2*50/1000,1/80,1/40,n=300)
# first filtered signal
plot(f$y, type="l") f <- filter(2*50/1000,1/80,1/40,n=100) # second filtered signal plot(f$y, type="l")
The filter you designed is a linear phase filter, therefore, response of the filter will start at $N/2$th sample, where $N$ is the filter order. As $N$ increases so does the delay of the response. You can find this delay value by computing group delay of the filter.
In case you want to reduce the delay, you can design a minimum phase filter.
• Nitpick: the delay of an $N$-tap linear-phase FIR filter is $\frac{N-1}{2}$ samples. An $N$-tap filter is of order $M$, and the delay is $\frac{M}{2}$. Commented May 14, 2014 at 13:37
• @Jason thankx for correction. Ill edit my answer to replace tap with order. Commented May 14, 2014 at 16:51
For a $N$th order FIR filter, the first $N$ outputs are not valid, since they are the transient response of the filter. Check the "Steady State and Transient Response" section in this page: http://www.music.mcgill.ca/~gary/618/week1/signals.html
I think, if you have a
Xn[n] = {X0.x0, X1.x1,..,Xn.xn} //X0 Integer part, x0 fractional
and you have a
Hn[m] = {H0.h0, H1.h1,..,Hm.hm}
Then
Yn[i] = Hi.hi*Xi.xi + Hi+1.hi+1*Xi-1.xi-1 + .. +Hi+m.hi+m*Xi-m.xi-m
But when you begins multiplyng you don't have before to X0.x0,
Yn[0] = H0.h0*X0.x0 + H1.h1*X-1.x-1
X-1.x-1 = 0.0;
Then
Yn[0] = H0.h0*X0.x0
If m is very long the first m's values of Yn[i] will be lower too... | 854 | 2,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.810041 |
https://maa.org/press/maa-reviews/computational-epidemiology-data-driven-modeling-of-covid-19 | 1,713,883,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00073.warc.gz | 338,185,584 | 24,184 | # Computational Epidemiology: Data-Driven Modeling of COVID-19
###### Ellen Kuhl
Publisher:
Springer
Publication Date:
2021
Number of Pages:
328
Format:
Hardcover
Price:
109.99
ISBN:
978-3-030-82889-9
Category:
Textbook
[Reviewed by
Bill Satzer
, on
05/9/2022
]
Mathematical epidemiological modeling should have provided insight and guidance into the dynamics, prediction and control of the global pandemic we have been experiencing. Yet it did not. Despite success in predicting the spread of diseases like measles, mumps, and smallpox, the failure of modeling for the COVID-19 pandemic was clear, with predictions wrong often by orders of magnitude. What happened?
This book tries to explain what went wrong and to offer an approach that can address the shortcomings of previous models. This book was written in the first half of 2021 and was motivated by a class that the author had taught at Stanford. The general plan of the book is to introduce the reader to mathematical epidemiology and basic models of the spread of epidemics, and then to go on to define and develop new models with assumptions that better reflect the spread of COVID-19.
The author’s treatment has four parts. The first two parts introduce epidemiology and its mathematical treatment using systems of ordinary differential equations, and then computational epidemiology that addresses solution methods for these systems. The third and fourth parts describe the author’s own approach that proposes a network view of epidemiology and then a data-driven epidemiology that acknowledges the effect of randomness, noise and uncertainty in the analysis of disease data.
The discussion begins with an introduction to mathematical epidemiology. This includes a background in general epidemiology as well as descriptions of the classical mathematical models. These models have three basic assumptions: the population under consideration is isolated from the rest of the world, contact between individuals is homogeneous, and the model is fully deterministic. The population is typically divided into compartments. These include the susceptible (S), exposed (E) infectious (I), and recovered (R) subgroups.
Several variations of basic models are considered. One commonly used version is the SEIR model. It is expressed in terms of four ordinary differential equations: $\dot{S} = −\beta S I, \dot{E} = \beta SI − \alpha E, \dot{I} = \alpha E - \gamma I, \dot{R}=\gamma I$. The contact rate between susceptible people and those infected is $\beta$, $1/\alpha$ is the latent period, and $1/\gamma$ is the infectious period. The reproduction number $R_{0} = \beta / \gamma$ quantifies how many new infections a single infected individual creates in an otherwise completely susceptible population. The herd immunity threshold is then $H=1-\frac{1}{R_{0}}$.
The basic SEIR model is rather simplistic. It does not account for births and deaths. It is also static in the sense that the parameters $\alpha$, $\beta$, and $\gamma$ are not time-dependent. It does not include any kind of spatial variation, and it treats a fixed population without births and deaths. The author also considers simpler models, and one slightly more complicated model (SEIIR) that incorporates the possibility for both symptomatic and asymptomatic infections.
The treatment of computational epidemiology largely focuses on linearization, discretization, and solution of the differential equations for the various models using both implicit and explicit time integration methods, and several examples are provided. Toward the end of this section the author introduces what she calls a ”dynamic SEIR model” whose distinguishing feature is a time-dependent contact rate $\beta(t)$ with the form of a hyperbolic tangent. This is intended, in part, to take into account the changes in social behavior and the effects of political action during a pandemic.
The last two parts of the book offer approaches that can overcome some of the weaknesses of the basic models. The first one is a network approach designed to deal with combined temporal and spatial variations of infection via network diffusion on a weighted undirected graph or using a finite element method. The intention is to provide a tool to model outbreak dynamics that occur because of transmission between geographical regions. Data-driven epidemiology is one of the newest approaches. It integrates earlier computational epidemiology with a probabilistic approach that uses data with Bayesian analysis and machine learning methods to infer viral reproduction dynamics and to incorporate the effect of asymptomatic transmission with correlation of case data and mobility statistics.
A lot of questions remain. It is apparent that - at least early in the pandemic - we did not understand either its biology (asymptotic infection and the possibility of re-infection, for example) or its sociology (effects of public policy, availability of immunization, resistance to immunization and mask wearing, school closures, and travel restrictions, for example), and it may be difficult to unravel these effects to make better models. Many models do not attempt to model the number of hospitalizations, yet those contribute considerably to the burden on the healthcare systems. Further, data driven modeling depends essentially on reliable data, and that continues to be a serious challenge. The concept of herd immunity is widely discussed, but it is not clear at all how it might apply in our current complex environment.
This is a very ambitious book. The author packs a lot of stuff into just over 300 pages. Exposition, data, and graphics compete for space. Some pages hold more than 25 small plots, often so small that individual details are very difficult to read. Apparently, the intention is to show just trends and patterns, but it makes for a very busy text. The index is not very complete, and internal references within and between chapters can be hard to track down.
Every chapter has a collection of very good problems. Very few of these are straightforward, and many would require considerable work. The many examples using real data make the book a valuable resource.
Overall the book presents a number of important ideas and offers some significant new approaches for modeling real and complicated epidemics. It’s a great place to begin to understand where mathematical epidemiology is now and where it has to go. But it also shows signs of being assembled very quickly, and consequently poses some extra challenges to the reader.
Bill Satzer (bsatzer@gmail.com), now retired from 3M Company, spent most of his career as a mathematician working in industry on a variety of applications. He did his PhD work in dynamical systems and celestial mechanics. | 1,433 | 6,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.897377 |
https://www.quesba.com/questions/1-please-give-the-two-quartiles-q1-and-q3-for-the-gdp-personal-consumption--45630 | 1,659,886,637,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00173.warc.gz | 845,246,834 | 31,054 | # 1. Please give the two quartiles Q1 and Q3 for the GDP, personal consumption, and disposable income,
1. Please give the two quartiles Q1 and Q3 for the GDP, personal consumption, and disposable income, respectively.
2. Please draw the box and whisker plots for the GDP, personal consumption, and disposable income, respectively.
May 26 2020| 12:26 PM |
## Related Questions
### A facility has three machines and three mechanics. Machines break down at the rate of one per 24...
A facility has three machines and three mechanics. Machines break down at the rate of one per 24 hours. Breakdown times are exponentially distributed. The time it takes a mechanic to fix a machine is exponentially distributed with mean 6 hours. Only one mechanic can work on a failed machine at any given time. Let Xt be the number of machines working at time t. Find the long-term probability that...
Apr 07 2020
### P.1.15 Suppose that the observational units in a statistical study are purchases made on a...
P.1.15 Suppose that the observational units in a statistical study are purchases made on a particular website (think of amazon.com) for one month.  a. Identify two categorical variables that could be recorded for these purchases.  b. Identify two quantitative variables that could be recorded for these purchases.  c. State two research questions that the company executives might want...
Mar 24 2020
### 1. A sample of 20 cups of coffee from a coffee machine has a mean amount of coffee of 6 oz. The...
1. A sample of 20 cups of coffee from a coffee machine has a mean amount of coffee of 6 oz. The standard deviation is known to be .5 oz. Construct a 99 % confidence interval for the mean amount of coffee per cup. 2. Reconsider question 50. This time, assume that the standard deviation is not known and that .5 oz is the sample standard deviation. Again construct a 99 % confidence interval for the...
May 26 2020
### 1. Show that a continuous probability distribution that is memory less must be exponential. Hint:...
1. Show that a continuous probability distribution that is memory less must be exponential. Hint: For g (t) = P(X > t), show that g (t) = (g (1)) t for all positive, rational t. 2. Starting at noon, diners arrive at a restaurant according to a Poisson process at the rate of five customers per minute. The time each customer spends eating at the restaurant has an exponential distribution with mean...
Apr 07 2020
### Suppose that you record the following information about 15 Major League Baseball games played next..
Suppose that you record the following information about 15 Major League Baseball games played next Saturday: A. Total number of runs scored in the game  B. Whether or not the home team wins the game C. Time (in minutes) required to play the game D. Attendance (number of people) at the game E. Whether the game is played in the afternoon or evening  F. Temperature when the game begins a....
Mar 24 2020 | 674 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.919718 |
https://a2zcalculators.com/article/how-to-calculate-z-score-from-percentage/ | 1,709,063,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00235.warc.gz | 76,915,394 | 23,930 | # How To Calculate Z Score From Percentage?
| Last update: 27 December 2023
Understanding statistical measures like the Z score can be crucial in various fields, including physics. In this article, we’ll delve into the process of calculating the Z score from a percentage, breaking down the steps and providing examples for clarity.
## Calculating Z Score from Percentage
Step 1: Convert Percentage to Decimal To start, convert the given percentage to a decimal. For instance, if you have 25%, convert it to 0.25.
Step 2: Determine Positive or Negative Z Value
• If the decimal is greater than 0.5, expect a positive Z value.
• If it’s less than 0.5, expect a negative Z value.
Step 3: Find the Corresponding Z Value Use a standard Z score table to find the Z value corresponding to the calculated decimal. If you encounter multiple close values, choose the one closer to a specific point (e.g., 0.25) or interpolate for better precision.
### Examples:
1. If given 25%, convert to 0.25. Since 0.25 < 0.5, expect a negative Z value. From the table, the Z score is approximately -0.67.
2. For 5%, convert to 0.05. Since 0.05 < 0.5, expect a negative Z value. The corresponding Z score from the table might be around -1.64.
## FAQs on Z Score from Percentage:
### What is the Z-score of 25%?
The Z score of 25% is approximately -0.67.
### How do you find the Z-score of 5%?
The Z score of 5% is around -1.64.
### How do you find the Z-score from a percentile?
Follow the steps mentioned earlier, converting the percentile to a decimal and finding the Z score from a standard table. | 401 | 1,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | longest | en | 0.770494 |
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Gauss%E2%80%93Legendre_algorithm.html | 1,511,147,232,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805894.15/warc/CC-MAIN-20171120013853-20171120033853-00123.warc.gz | 481,532,122 | 4,258 | # Gauss–Legendre algorithm
related topics {math, number, function}
The Gauss–Legendre algorithm is an algorithm to compute the digits of π. It is notable for being rapidly convergent, with only 25 iterations producing 45 million correct digits of π. However, the drawback is that it is memory intensive and it is therefore sometimes not used over Machin-like formulas.
The method is based on the individual work of Carl Friedrich Gauss (1777–1855) and Adrien-Marie Legendre (1752–1833) combined with modern algorithms for multiplication and square roots. It repeatedly replaces two numbers by their arithmetic and geometric mean, in order to approximate their arithmetic-geometric mean.
The version presented below is also known as the Brent–Salamin (or Salamin–Brent) algorithm; it was independently discovered in 1975 by Richard Brent and Eugene Salamin. It was used to compute the first 206,158,430,000 decimal digits of π on September 18 to 20, 1999, and the results were checked with Borwein's algorithm.
## Contents
### Algorithm
1. Initial value setting:
2. Repeat the following instructions until the difference of $a_n\!$ and $b_n\!$ is within the desired accuracy:
3. π is then approximated as:
The first three iterations give:
The algorithm has second-order convergent nature, which essentially means that the number of correct digits doubles with each step of the algorithm.
### Limits of the arithmetic-geometric mean
The arithmetic-geometric mean of two numbers, a0 and b0, is found by calculating the limit of the sequences
which both converge to the same limit. If $a_0=1\!$ and $b_0=\cos\varphi\!$ then the limit is ${\pi \over 2K(\sin\varphi)}\!$ where $K(k)\!$ is the complete elliptic integral of the first kind
If $c_0 = \sin\varphi\!$, $c_{i+1} = a_i - a_{i+1}\!$. then | 449 | 1,807 | {"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": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-47 | latest | en | 0.909268 |
http://www.jiskha.com/display.cgi?id=1347118640 | 1,495,832,240,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608684.93/warc/CC-MAIN-20170526203511-20170526223511-00183.warc.gz | 681,833,860 | 3,788 | # Calculus
posted by on .
Find the constants a and b such that the function is continuous on the entire line
f(x){7, x is less than or equal to -3
ax+b, -3 is less than x is greater than 4
-7, x is less than or equal to 4
• Calculus - ,
you need a smooth transition from (-3,7) to (4,-7)
y-7 = -14/7 (x+3)
or
y = -2x + 1
so, a=-2, b=1 | 124 | 341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-22 | latest | en | 0.928517 |
http://newtondynamics.com/forum/viewtopic.php?f=9&t=4702&p=33756 | 1,586,050,012,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370526982.53/warc/CC-MAIN-20200404231315-20200405021315-00516.warc.gz | 131,414,036 | 5,604 | A place to discuss everything related to Newton Dynamics.
Moderators: Sascha Willems, walaber
So, I'm simulating a robot with an arm made of 6 joints. I'm using the custom hinge to force those joints to a particular angle to simulate the motors. The problem is that the motors sort of have infinite force, which makes sense, since the engine is driving the joint to that angle with a row. It looks like this
Code: Select all
` if(m_motorOn) { dFloat relAngle; relAngle = m_curJointAngle - m_motorAngle; m_motorAngle += jointData->Torque *-.1 * 3.1416f / 180.0f; NewtonUserJointAddAngularRow (m_joint, relAngle, &matrix0.m_front[0]); NewtonUserJointSetRowStiffness (m_joint, 1.0f); m_force = NewtonUserJointGetRowForce(m_joint,5); }`
I know that there should be a better way, but I just have not been able to figure it out. I've looked all over the google cache of these forums, and found that other people are doing something like this:
Code: Select all
`float actualOmega = NewtonHingeGetJointOmega (hinge);float desiredOmega = 5.0f;...desc->m_accel = m_accelConst * (desiredOmega - actualOmega) / desc->m_timestep;`
But, even if I switch to regular hinges, the gravity of the arm pieces always overcomes the acceleration on the joint.... I can get around this without simulating gravity, but I really need the weight of the arm to cause the robot to fall over at certain angles...
Anyone have any thoughts on this?
Posts: 1
Joined: Wed May 24, 2006 3:18 pm
NewtonUserJointSetRowSpringDamperAcceleration(m_joint,springStiff,springDamp);
after the call to set row stiffness. Depending on the size of the objects values like 30 stiff and 20 damp will make a strong slow motor. Higher damp to make it go slower. Higher stiff to make it stronger.
DarthGak
Posts: 25
Joined: Sat Nov 25, 2006 4:55 pm | 507 | 1,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.783122 |
https://digitalmars.com/d/archives/digitalmars/D/learn/Implicit_conversion_rules_76961.html | 1,529,541,426,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00551.warc.gz | 576,198,449 | 5,860 | ## digitalmars.D.learn - Implicit conversion rules
• Sigg (22/22) Oct 21 2015 I started reading "The D programming Language" earlier, and came
• anonymous (17/31) Oct 21 2015 The problem is of course that int and ulong have no common super type, a...
• Sigg (17/37) Oct 21 2015 Yes, I'm well aware of that. I was under the (wrongful)impression
• =?UTF-8?Q?Ali_=c3=87ehreli?= (9/10) Oct 21 2015 One of those side effects would be function calls binding silently to
• Marco Leise (21/36) Oct 21 2015 God forbid anyone implement such nonsense into D !
• Maxim Fomin (8/30) Oct 21 2015 AFAIK it was implemented long time ago and discussed last time
• Sigg (6/8) Oct 22 2015 Slight nitpick, but what I suggested for our hypothetical
• Maxim Fomin (3/13) Oct 21 2015 Actually 'a' is deduced to be int, so int version is called (as
Sigg <todorovicmilos89 gmail.com> writes:
```I started reading "The D programming Language" earlier, and came
to the "2.3.3 Typing of Numeric Operators" section which claims
that "if at least one participant has type ulong, the other is
implicitly converted to ulong prior to the application and the
result has type ulong.".
Now I understand reasoning behind it, and know that adding any
sufficiently negative value to a ulong/uint/ushort will cause an
underflow as in following example:
void func() {
int a = -10;
ulong b = 0;
ulong c = a + b;
writefln("%s", c);
}
out: 18446744073709551574
But shouldn't declaring c as auto force compiler to go extra step
and "properly" deduce result of the "a + b" expression, since its
already as far as I understand doing magic in the background?
Basically try to cast rvalues to narrowest type without losing
precision before evaluating expression.
Or is there a proper way to do math with unsigned and signed
primitives that I'm not aware of?
```
Oct 21 2015
anonymous <anonymous example.com> writes:
```On Wednesday, October 21, 2015 07:53 PM, Sigg wrote:
void func() {
int a = -10;
ulong b = 0;
ulong c = a + b;
writefln("%s", c);
}
out: 18446744073709551574
But shouldn't declaring c as auto force compiler to go extra step
and "properly" deduce result of the "a + b" expression, since its
already as far as I understand doing magic in the background?
Basically try to cast rvalues to narrowest type without losing
precision before evaluating expression.
The problem is of course that int and ulong have no common super type, at
least not in the primitive integer types. int supports negative values,
ulong supports values greater than long.max.
As far as I understand, you'd like the compiler to see the values of `a` and
`b` (-10, 0), figure out that the result is negative, and then make `c`
signed based on that. That's not how D rolls. The same code must compile
when the values in `a` and `b` come from run time input. So the type of the
addition cannot depend on the values of the operands, only on their types.
Or maybe you'd expect an `auto` variable to be able to hold both negative
and very large values? But `auto` is not a special type, it's just a
shorthand for typeof(right-hand side). That means, `auto` variables still
get one specific static type, like int or ulong.
std.bigint and core.checkedint may be of interest to you, if you prefer
safer operations over faster ones.
http://dlang.org/phobos/std_bigint.html
http://dlang.org/phobos/core_checkedint.html
```
Oct 21 2015
Sigg <todorovicmilos89 gmail.com> writes:
```On Wednesday, 21 October 2015 at 19:07:24 UTC, anonymous wrote:
The problem is of course that int and ulong have no common
super type, at least not in the primitive integer types. int
supports negative values, ulong supports values greater than
long.max.
Yes, I'm well aware of that. I was under the (wrongful)impression
that auto was doing much more under the hood and that it was more
safety oriented, I've prolly mixed it with something else while
As far as I understand, you'd like the compiler to see the
values of `a` and `b` (-10, 0), figure out that the result is
negative, and then make `c` signed based on that. That's not
how D rolls. The same code must compile when the values in `a`
and `b` come from run time input. So the type of the addition
cannot depend on the values of the operands, only on their
types.
Or maybe you'd expect an `auto` variable to be able to hold
both negative and very large values? But `auto` is not a
special type, it's just a shorthand for typeof(right-hand
side). That means, `auto` variables still get one specific
static type, like int or ulong.
Ima clarify what I expected using my previous example:
ulong a = 0;
int b = -10;
auto c = a + b;
a gets cast to narrowest primitive type that can hold its value,
in this case bool since bool can hold 0 value resulting in c
having value of -10. If a was bigger than max long I'd expect an
error/exception. Now on the other hand I can see why something
like this would not be implemented since it would ignore implicit
conversion table and prolly cause at least few more "fun" side
effects.
std.bigint and core.checkedint may be of interest to you, if
you prefer safer operations over faster ones.
http://dlang.org/phobos/std_bigint.html
http://dlang.org/phobos/core_checkedint.html
This is exactly what I was looking for. Thanks!
```
Oct 21 2015
=?UTF-8?Q?Ali_=c3=87ehreli?= <acehreli yahoo.com> writes:
```On 10/21/2015 12:37 PM, Sigg wrote:
cause at least few more "fun" side effects.
One of those side effects would be function calls binding silently to
void foo(bool){/* ... */}
void foo(int) {/* ... */}
auto a = 0; // If the type were deduced by the value,
foo(a); // then this would be a call to foo(bool)...
// until someone changed the value to 2. :)
Ali
```
Oct 21 2015
Marco Leise <Marco.Leise gmx.de> writes:
```Am Wed, 21 Oct 2015 12:49:35 -0700
schrieb Ali =C3=87ehreli <acehreli yahoo.com>:
On 10/21/2015 12:37 PM, Sigg wrote:
=20
> cause at least few more "fun" side effects.
=20
One of those side effects would be function calls binding silently to=20
=20
void foo(bool){/* ... */}
void foo(int) {/* ... */}
=20
auto a =3D 0; // If the type were deduced by the value,
foo(a); // then this would be a call to foo(bool)...
// until someone changed the value to 2. :)
=20
Ali
God forbid anyone implement such nonsense into D !
That would be the last thing we need that we cannot rely on
the overload resolution any more. It would be as if making 'a'
const would change the overload resolution when none of the
import std.format;
import std.stdio;
string foo(bool b) { return format("That's a boolean %s!", b); }
string foo(uint u) { return format("Thats an integral %s!", u); }
void main()
{
int a =3D 2497420, b =3D 2497419;
const int c =3D 2497420, d =3D 2497419;
writeln(foo(a-b));
writeln(foo(c-d));
writeln("WAT?!");
}
--=20
Marco
```
Oct 21 2015
Maxim Fomin <mxfomin gmail.com> writes:
```On Wednesday, 21 October 2015 at 22:49:16 UTC, Marco Leise wrote:
Am Wed, 21 Oct 2015 12:49:35 -0700
schrieb Ali Çehreli <acehreli yahoo.com>:
On 10/21/2015 12:37 PM, Sigg wrote:
> cause at least few more "fun" side effects.
One of those side effects would be function calls binding
void foo(bool){/* ... */}
void foo(int) {/* ... */}
auto a = 0; // If the type were deduced by the value,
foo(a); // then this would be a call to foo(bool)...
// until someone changed the value to 2. :)
Ali
God forbid anyone implement such nonsense into D !
That would be the last thing we need that we cannot rely on
the overload resolution any more. It would be as if making 'a'
const would change the overload resolution when none of the
AFAIK it was implemented long time ago and discussed last time
couple of years ago with example similar to Ali's.
void foo(bool)
void foo(int)
foo(0); // bool
foo(1); // bool
foo(2); // int
```
Oct 21 2015
Sigg <todorovicmilos89 gmail.com> writes:
```On Wednesday, 21 October 2015 at 22:49:16 UTC, Marco Leise wrote:
God forbid anyone implement such nonsense into D !
That would be the last thing we need
Slight nitpick, but what I suggested for our hypothetical
situation was only to apply for auto, once variable was assigned
to auto and got its correct type it would act like normal
variable. Stuff you mentioned would happen if it was part of an
expression in the rvalue expression.
```
Oct 22 2015
Maxim Fomin <mxfomin gmail.com> writes:
```On Wednesday, 21 October 2015 at 19:49:35 UTC, Ali Çehreli wrote:
On 10/21/2015 12:37 PM, Sigg wrote:
cause at least few more "fun" side effects.
One of those side effects would be function calls binding
void foo(bool){/* ... */}
void foo(int) {/* ... */}
auto a = 0; // If the type were deduced by the value,
foo(a); // then this would be a call to foo(bool)...
// until someone changed the value to 2. :)
Ali
Actually 'a' is deduced to be int, so int version is called (as
expected?). See my example above for the VRO overload issue.
```
Oct 21 2015 | 2,483 | 8,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-26 | latest | en | 0.919446 |
http://mathhelpforum.com/advanced-math-topics/197278-orthogonal-problem-wireless-print.html | 1,529,469,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00550.warc.gz | 198,070,767 | 2,595 | # Orthogonal problem in wireless
• Apr 14th 2012, 11:08 AM
drogba
Orthogonal problem in wireless
Hi,
Is it possible to have orthogonal equation with only -1 and 1 values only where the dot product of A.B, B.C and A.C will be 0 by just using -1 and 1? (eg. A = -1 1 , B = 1 1) ? Thanks.!!!
• Apr 16th 2012, 04:49 AM
drogba
Re: Orthogonal problem in wireless
Is it possible in this scenario?
• Apr 17th 2012, 11:05 AM
HallsofIvy
Re: Orthogonal problem in wireless
Are you sure that your only possible values are -1 and 1 and not -1, 0, and 1?
The problem is impossible with only -1 and 1 but possible with -1, 0, and 1.
• Apr 18th 2012, 08:57 AM
drogba
Re: Orthogonal problem in wireless
Hi all,
Thanks for all those replies. I managed to calculate it using Walsh matrix and yes its possible. | 257 | 794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.897199 |
https://wiki2.org/en/Painter%27s_algorithm | 1,638,436,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00517.warc.gz | 673,103,429 | 20,531 | To install click the Add extension button. That's it.
The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.
4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
Languages
Recent
Show all languages
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds
# Painter's algorithm
A fractal landscape being rendered using the painter’s algorithm on a Commodore Amiga
The painter’s algorithm (also depth-sort algorithm and priority fill) is an algorithm for visible surface determination in 3D computer graphics that works on a polygon-by-polygon basis rather than a pixel-by-pixel, row by row, or area by area basis of other Hidden Surface Removal algorithms.[1][2][3] The painter’s algorithm creates images by sorting the polygons within the image by their depth and placing each polygon in order from the farthest to the closest object.[4][5]
The painter's algorithm was initially proposed as a basic method to address the Hidden-surface determination problem by Martin Newell, Richard Newell, and Tom Sancha in 1972, while all three were working at CADCentre.[4] The name "painter's algorithm" refers to the technique employed by many painters where they begin by painting distant parts of a scene before parts that are nearer, thereby covering some areas of distant parts.[6][7] Similarly, the painter's algorithm sorts all the polygons in a scene by their depth and then paints them in this order, farthest to closest.[8] It will paint over the parts that are normally not visible — thus solving the visibility problem — at the cost of having painted invisible areas of distant objects.[9] The ordering used by the algorithm is called a 'depth order' and does not have to respect the numerical distances to the parts of the scene: the essential property of this ordering is, rather, that if one object obscures part of another, then the first object is painted after the object that it obscures.[9] Thus, a valid ordering can be described as a topological ordering of a directed acyclic graph representing occlusions between objects.[10]
The distant mountains are painted first, followed by the closer meadows; finally, the trees, are painted. Although some trees are more distant from the viewpoint than some parts of the meadows, the ordering (mountains, meadows, trees) forms a valid depth order, because no object in the ordering obscures any part of a later object.
## Algorithm
Conceptually Painter's Algorithm works as follows:
1. Sort each polygon by depth
2. Place each polygon from the farthest polygon to the closest polygon
### Pseudocode
```sort polygons by depth
for each polygon p:
for each pixel that p covers:
paint p.color on pixel
```
### Time complexity
The painter's algorithm's time-complexity is heavily dependent on the sorting algorithm used to order the polygons. Assuming the use of the most optimal sorting algorithm, painter's algorithm has a worst-case complexity of O(n log n + m*n), where n is the number of polygons and m is the number of pixels to be filled.
### Space complexity
The painter's algorithm's worst-case space-complexity is O(n+m), where n is the number of polygons and m is the number of pixels to be filled.
There are two primary technical requisites that favor the use of the painter’s algorithm.
### Basic graphical structure
The painter's algorithm is not as complex in structure as its other depth sorting algorithm counterparts.[9][11] Components such as the depth-based rendering order, as employed by the painter’s algorithm, are one of the simplest ways to designate the order of graphical production.[8] This simplicity makes it useful in basic computer graphics output scenarios where an unsophisticated render will need to be made with little struggle.[9]
### Memory efficiency
In the early 70s, when the painter’s algorithm was developed, physical memory was relatively small.[12] This required programs to manage memory as efficiently as possible to conduct large tasks without crashing. The painter’s algorithm prioritizes the efficient use of memory but at the expense of higher processing power since all parts of all images must be rendered.[9]
Overlapping polygons can cause the algorithm to fail
## Limitations
The algorithm can fail in some cases, including cyclic overlap or piercing polygons.
### Cyclical Overlapping
In the case of cyclic overlap, as shown in the figure to the right, Polygons A, B, and C overlap each other in such a way that it is impossible to determine which polygon is above the others. In this case, the offending polygons must be cut to allow sorting.[4]
### Piercing polygons
The case of piercing polygons arises when one polygon intersects another. Similar to cyclic overlap, this problem may be resolved by cutting the offending polygons.[4]
### Efficiency
In basic implementations, the painter's algorithm can be inefficient. It forces the system to render each point on every polygon in the visible set, even if that polygon is occluded in the finished scene. This means that, for detailed scenes, the painter's algorithm can overly tax the computer hardware.
## Variants
### Extended painter's algorithm
Newell's algorithm, proposed as the extended algorithm to painter's algorithm, provides a method for cutting cyclical and piercing polygons.[4]
### Reverse painter's algorithm
Another variant of painter's algorithm includes reverse painter's algorithm. Reverse painter's algorithm paints objects nearest to the viewer first — with the rule that paint must never be applied to parts of the image that are already painted (unless they are partially transparent). In a computer graphic system, this can be very efficient since it is not necessary to calculate the colors (using lighting, texturing, and such) for parts of a distant scene that are hidden by nearby objects. However, the reverse algorithm suffers from many of the same problems as the standard version.
## Other computer graphics algorithms
The flaws of painter's algorithm led to the development of Z-buffer techniques, which can be viewed as a development of the painter's algorithm by resolving depth conflicts on a pixel-by-pixel basis, reducing the need for a depth-based rendering order.[13] Even in such systems, a variant of the painter's algorithm is sometimes employed. As Z-buffer implementations generally rely on fixed-precision depth-buffer registers implemented in hardware, there is scope for visibility problems due to rounding error. These are overlaps or gaps at joints between polygons. To avoid this, some graphics engine implementations "overrender"[citation needed], drawing the affected edges of both polygons in the order given by the painter's algorithm. This means that some pixels are actually drawn twice (as in the full painter's algorithm), but this happens on only small parts of the image and has a negligible performance effect.
## References
• Foley, James; Feiner, Steven K.; Hughes, John F. (1990). Computer Graphics: Principles and Practice. Reading, MA, USA: Addison-Wesley. p. 1174. ISBN 0-201-12110-7.
1. ^ Appel, Arthur (1968). Morrel, A. J. H. (ed.). "On calculating the illusion of reality" (PDF). Information Processing, Proceedings of IFIP Congress 1968, Edinburgh, UK, 5-10 August 1968, Volume 2 - Hardware, Applications: 945–950.
2. ^ Romney, Gordon Wilson (1969-09-01). "Computer Assisted Assembly and Rendering of Solids". Cite journal requires `|journal=` (help)
3. ^ Gary Scott Watkins. 1970. "A real time visible surface algorithm. Ph.D. Dissertation." The University of Utah. Order Number: AAI7023061.
4. Newell, M. E.; Newell, R. G.; Sancha, T. L. (1972-08-01). "A solution to the hidden surface problem" (PDF). Proceedings of the ACM Annual Conference - Volume 1. ACM '72. Boston, Massachusetts, USA: Association for Computing Machinery. 1: 443–450. doi:10.1145/800193.569954. ISBN 978-1-4503-7491-0. S2CID 13829930.
5. ^ Bouknight, W. Jack (1970-09-01). "A procedure for generation of three-dimensional half-toned computer graphics presentations". Communications of the ACM. 13 (9): 527–536. doi:10.1145/362736.362739. ISSN 0001-0782. S2CID 15941472.
6. ^ Berland, Dinah (1995). Historical Painting Techniques, Materials, and Studio Practice (PDF). The Getty Conservation Institute.
7. ^ Wylie, Chris; Romney, Gordon; Evans, David; Erdahl, Alan (1967-11-14). "Half-tone perspective drawings by computer". Proceedings of the November 14–16, 1967, Fall Joint Computer Conference. AFIPS '67 (Fall). Anaheim, California: Association for Computing Machinery: 49–58. doi:10.1145/1465611.1465619. ISBN 978-1-4503-7896-3. S2CID 3282975.
8. ^ a b Desai, Apurva (2008). Computer Graphics. PHI Learning Pvt. Ltd. ISBN 9788120335240.
9. de Berg, Mark (2008). Computational Geometry (PDF). Springer.
10. ^ de Berg, Mark (1993). Ray Shooting, Depth Orders and Hidden Surface Removal. Lecture Notes in Computer Science. 703. Springer. p. 130. ISBN 9783540570202{{inconsistent citations}}CS1 maint: postscript (link).
11. ^ Warnock, John E. (1969-06-01). "A Hidden Surface Algorithm for Computer Generated Halftone Pictures". Cite journal requires `|journal=` (help)
12. ^ Freiser, M.; Marcus, P. (June 1969). "A survey of some physical limitations on computer elements". IEEE Transactions on Magnetics. 5 (2): 82–90. Bibcode:1969ITM.....5...82F. doi:10.1109/TMAG.1969.1066403. ISSN 1941-0069.
13. ^ Nyberg, Daniel (2011). Analysis of Two Common Hidden Surface Removal Algorithms, Painter's Algorithm & Z-Buffering.
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation. | 2,461 | 10,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.924002 |
http://oeis.org/A238604 | 1,618,507,656,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00028.warc.gz | 73,138,657 | 4,159 | 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!)
A238604 a(n) = Sum_{k=0..3} f(n+k)^2 where f=A130519. 2
0, 1, 5, 14, 30, 65, 125, 216, 344, 533, 793, 1134, 1566, 2125, 2825, 3680, 4704, 5945, 7421, 9150, 11150, 13481, 16165, 19224, 22680, 26605, 31025, 35966, 41454, 47573, 54353, 61824, 70016, 79025, 88885, 99630, 111294, 123985, 137741, 152600, 168600, 185861 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS G. C. Greubel, Table of n, a(n) for n = 0..2500 Index entries for linear recurrences with constant coefficients, signature (3,-3,1,2,-6,6,-2,-1,3,-3,1). FORMULA G.f.: x * (1 + 2*x + 2*x^2 + 2*x^3 + 10*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + x^8) / ( (1 - x)^3 * (1 - x^4)^2 ). a(n) = a(-1 - n) for all n in Z. floor( sqrt( a(n))) = A054925(n+1). EXAMPLE G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 65*x^5 + 125*x^6 + 216*x^7 + ... MATHEMATICA CoefficientList[Series[x*(1+2*x+2*x^2+2*x^3+10*x^4+2*x^5+2*x^6+2*x^7+ x^8)/((1-x)^3*(1-x^4)^2), {x, 0, 50}], x] (* G. C. Greubel, Aug 07 2018 *) PROG (PARI) {a(n) = if( n<0, n = -1-n); polcoeff( x * (1 + 2*x + 2*x^2 + 2*x^3 + 10*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + x^8) / ( (1 - x)^3 * (1 - x^4)^2 ) + x * O(x^n), n)}; (MAGMA) m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 +2*x+2*x^2+2*x^3+10*x^4+2*x^5+2*x^6+2*x^7+ x^8)/((1-x)^3*(1-x^4)^2))); // G. C. Greubel, Aug 07 2018 CROSSREFS Cf. A054925, A130519. Sequence in context: A299899 A211804 A283591 * A166068 A070129 A081861 Adjacent sequences: A238601 A238602 A238603 * A238605 A238606 A238607 KEYWORD nonn,easy AUTHOR Michael Somos, Mar 01 2014 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified April 15 13:13 EDT 2021. Contains 342977 sequences. (Running on oeis4.) | 905 | 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} | 3.40625 | 3 | CC-MAIN-2021-17 | latest | en | 0.468789 |
https://www.mersenneforum.org/search.php?s=75fa30ed0cd2d4bdbd645b325a7ea097&searchid=3334683 | 1,601,344,150,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401617641.86/warc/CC-MAIN-20200928234043-20200929024043-00482.warc.gz | 923,052,911 | 9,116 | mersenneforum.org Search Results
Register FAQ Search Today's Posts Mark Forums Read
Showing results 1 to 25 of 514 Search took 0.09 seconds. Search: Posts Made By: Dylan14
Forum: Msieve 2020-09-26, 15:42 Replies: 2 Views: 115 Posted By Dylan14 Outdated msieve build in Arch AUR So, I decided to take on a number to post-process for NFS@Home (namely, 8p3_870M), and I wanted to try the number out on an Arch Linux install. I saw in the AUR that there is an msieve package which...
Forum: NFS@Home 2020-09-24, 13:24 Replies: 16 Views: 706 Posted By Dylan14 Also reserving f45_151m1 for post-processing. Also reserving f45_151m1 for post-processing.
Forum: NFS@Home 2020-09-16, 22:09 Replies: 16 Views: 706 Posted By Dylan14 Reserving 8p3_870M for postprocessing. Reserving 8p3_870M for postprocessing.
Forum: NFS@Home 2020-09-16, 22:05 Replies: 467 Views: 29,907 Posted By Dylan14 765__127_7m1 was factored: p69 factor:... 765__127_7m1 was factored: p69 factor: 247266618809857364769432394200200277984439717886775222264610206604661 p136 factor:...
Forum: NFS@Home 2020-09-07, 00:03 Replies: 467 Views: 29,907 Posted By Dylan14 91997_43m1 was factored: p52 factor:... 91997_43m1 was factored: p52 factor: 9596265036010581604338742365787303364306581650247103 p157 factor:...
Forum: Alberico Lepore 2020-09-05, 16:52 Replies: 15 Views: 698 Posted By Dylan14 Running your choice of equations, I get the... Running your choice of equations, I get the following: F = 65107482205674607 A = 781289786468095309/36 B = 65107482205674607/3 C = 65107482205674608/3 so you are correct that F is an integer,...
2020-09-02, 16:32 Replies: 373 Views: 34,631 Posted By Dylan14 Attached are two builds of mmff v0.28.1 (Gary's... Attached are two builds of mmff v0.28.1 (Gary's source), compiled on Ubuntu 20.04, with Cuda 10.1 and sm_61 (good for Pascal cards, ie GTX10xx). The first build is with a default max sieve size, the...
2020-09-02, 13:24 Replies: 489 Sticky: Reservations Views: 32,685 Posted By Dylan14 Reserving MMFactor=107,10e15,12e15. Reserving MMFactor=107,10e15,12e15.
Forum: FermatSearch 2020-09-02, 13:18 Replies: 187 Sticky: Reservations Views: 22,592 Posted By Dylan14 Range is complete: no factor for k*2^74+1 in k... Range is complete: no factor for k*2^74+1 in k range: 8000000000000000 to 9007199254740991 (127-bit factors) [mmff 0.28 mfaktc_barrett128_F64_95gs] no factor for k*2^74+1 in k range:...
Forum: enzocreti 2020-09-01, 17:13 Replies: 3 Views: 304 Posted By Dylan14 This would be the intersection of 078778 (n such... This would be the intersection of 078778 (n such that n!+1 is semiprime) and 078781 (n such that n!-1 is semiprime). There is no separate sequence (yet, as far as I can tell).
Forum: Puzzles 2020-08-30, 14:40 Replies: 20 Views: 1,018 Posted By Dylan14 This looked like an interesting problem to code... This looked like an interesting problem to code up in Python. Here is one code snippet that does the search brute force, like the OP did: #Challenge - find a power of 2 such that it does not...
Forum: NFS@Home 2020-08-26, 20:08 Replies: 467 Views: 29,907 Posted By Dylan14 10289_53m1 was factored: p57 factor:... 10289_53m1 was factored: p57 factor: 268603300299753507733996486408441503448254894078451801849 p58 factor: 1027108280410654651620299715760257492683808173662176774241 p82 factor:...
Forum: FermatSearch 2020-08-24, 13:24 Replies: 187 Sticky: Reservations Views: 22,592 Posted By Dylan14 Reserving n = 74, k = 80e14 - 100e14. Reserving n = 74, k = 80e14 - 100e14.
Forum: NFS@Home 2020-08-18, 21:18 Replies: 467 Views: 29,907 Posted By Dylan14 125539_43m1 was factored: p65 factor:... 125539_43m1 was factored: p65 factor: 14629416493895173016330365935421566833331510290282431415974646423 p67 factor: 1022352505181151865070177881296042611211691323416725813898108932037 p72...
Forum: Riesel Prime Search 2020-08-15, 00:16 Replies: 379 Views: 37,725 Posted By Dylan14 k = 50171 k = 50171 is at n = 2.945 M, no primes found, continuing...
Forum: Conjectures 'R Us 2020-08-08, 19:11 Replies: 3,721 Views: 224,307 Posted By Dylan14 S577 was tested to n = 200k, no primes were... S577 was tested to n = 200k, no primes were found. 6 k's remain. Residues were emailed, and I will be releasing this base.
Forum: Operation Billion Digits 2020-08-06, 11:12 Replies: 1,172 Views: 89,912 Posted By Dylan14 Large chunk of results were submitted today: ... Large chunk of results were submitted today: UID: Dylan14/desktop-i56400, no factor for M3321936169 from 2^82 to 2^83 [mfaktc 0.21 barrett87_mul32_gs] UID: Dylan14/desktop-i56400, no factor for...
Forum: NFS@Home 2020-08-05, 23:05 Replies: 467 Views: 29,907 Posted By Dylan14 1229_71m1 was factored: p54 factor:... 1229_71m1 was factored: p54 factor: 178442180475836344937469321516407863635121545502093703 p67 factor: 2055763063191798313956735491339380167470474531058874782225278435543 p73 factor:...
Forum: NFS@Home 2020-07-28, 13:10 Replies: 467 Views: 29,907 Posted By Dylan14 Reserving 1229_71m1 for post-processing. Reserving 1229_71m1 for post-processing.
Forum: Prime Wiki 2020-07-27, 14:51 Replies: 210 Sticky: Prime-Wiki Views: 17,636 Posted By Dylan14 On the old Riesel Prime database site you had... On the old Riesel Prime database site you had marked k's that are primorials (for example, 255255 was marked since 255255 = 17#/2). Will this be added to the wiki?
Forum: NFS@Home 2020-07-25, 23:11 Replies: 467 Views: 29,907 Posted By Dylan14 7503751_31m1 was factored: p49 factor:... 7503751_31m1 was factored: p49 factor: 8830574846454935898199455755450625228110359932947 p59 factor: 75455992727427104319023985251117379988539274686194384494093 p71 factor:...
Forum: Prime Wiki 2020-07-25, 15:10 Replies: 210 Sticky: Prime-Wiki Views: 17,636 Posted By Dylan14 Looks like it is fixed on my end now: just had to... Looks like it is fixed on my end now: just had to select English again in the top menu (it was selected before, so I am not sure why it wanted to put that in German. Maybe a caching issue).
Forum: Operation Billion Digits 2020-07-25, 13:44 Replies: 846 Views: 55,081 Posted By Dylan14 Been a while since I did anything with this... Been a while since I did anything with this project, so I am reserving quite a bit of work: I am taking all exponents in the OBD range at 82 bits up to 83 bits.
Forum: Prime Wiki 2020-07-24, 23:06 Replies: 210 Sticky: Prime-Wiki Views: 17,636 Posted By Dylan14 There's seems to be a small issue on the Template... There's seems to be a small issue on the Template Prototype page: where the table of contents is, instead of it reading "Contents" (like on every other page in the wiki that has a table of contents),...
Forum: Dylan14 2020-07-19, 21:46 Replies: 18 Sticky: A Guide for Python Views: 2,751 Posted By Dylan14 Introduction to lists (placeholder)
Showing results 1 to 25 of 514
All times are UTC. The time now is 01:49.
Tue Sep 29 01:49:10 UTC 2020 up 18 days, 23 hrs, 0 users, load averages: 1.55, 1.44, 1.43 | 2,312 | 6,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-40 | longest | en | 0.770705 |
https://codegolf.stackexchange.com/questions/478/free-a-binary-tree/489 | 1,620,251,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00159.warc.gz | 206,478,145 | 43,635 | # Free a Binary Tree
So before you read some basic computer science concepts.
1. A binary tree is a dynamically allocated structure (usually used for ordered storage).
2. Because of its nature traversal of binary trees is usually recursive;
This is because linear traversal (via a loop) is not natural when there are two avenues of looping.
• Recursive: This means a function that calls itself.
3. In old fashioned languages, memory management requires manual memory management.
• Manual: Means you have to do it yourself.
4. When you do manual memory management you need to actual ask the underlying system to free each member of the tree.
• Free: recover the memory in to the global poos so it can be re-used and you don't run out of memory.
• Freeing: this is done by calling the function free() and passing it the pointer you want to recover.
• Pointer: Its like a virtual stick. At the end is memory. When you ask for memory you are given a pointer (virtual stick) that has memory. When you are done you give back the pointer (virtual stick).
### The recursive solution:
freeTree(Node* node)
{
freeTree(node->left);
freeTree(node->right);
free(node);
}
The problem then is that recursion means you are repeatedly calling the same function. This grows the stack. Growing the stack uses more memory. The reason you are freeing the tree is you want memory back using more memory is counterproductive (even if you do get both bits of memory back).
### At last the question:
SO the problem centers around converting the recursive version above into a linear solution (so that you don't have to use memory).
Give the node type
typedef struct Node Node;
struct Node
{
Node* left;
Node* right;
};
Write a function to free a tree of these nodes.
Restrictions:
• Cannot use recursion (not even indirectly)
• Cannot allocate any dynamic space for tracking.
• Note there is an O(n) solution
Winner:
1. Best Complexity.
2. Tie Break 1: First Submitted
3. Tie Break 2: Least amount of characters.
Seems very close to O(n) to me:
This does a depth-first walk on the tree, and uses the ->left pointer of traversed nodes to keep track of the parent.
struct Node * node = root;
struct Node * up = NULL;
while (node != NULL) {
if (node->left != NULL) {
struct Node * left = node->left;
node->left = up;
up = node;
node = left;
} else if (node->right != NULL) {
struct Node * right = node->right;
node->left = up;
node->right = NULL;
up = node;
node = right;
} else {
if (up == NULL) {
free(node);
node = NULL;
}
while (up != NULL) {
free(node);
if (up->right != NULL) {
node = up->right;
up->right = NULL;
break;
} else {
node = up;
up = up->left;
}
}
}
}
• +1 Add the tick for the only answer. Its a bit more complex than the solution I present below but very good. – Martin York Feb 7 '11 at 20:49
# C99, 94, O(n)
Edit: everyone seems to refer to struct Node just as Node as if the typedefed it, so I did too.
this is actually my first C golf. lots of segfaults.
anyways, this requires C99 because it uses a declaration inside a for loop's first statement.
void f(Node*n){for(Node*q;n;n=q)(q=n->left)?n->left=q->right,q->right=n:(q=n->right,free(n));}
not even using #define!
this algorithm works by transforming the tree so that the top node has no left child, and then deleting it and moving on to it's right child.
1
/ \
2 3
\
4
the algorithm will mutate the pointers so that the tree will be
2
\
1
/ \
4 3
now we can delete the topmost node easily.
• I did not use typedef because mine was in C++ (you forget these small differences between the languages). I have updated the question so it works the same in C and C++. – Martin York Dec 25 '14 at 18:36
• @LokiAstari I don't actually know c++, and I just started learning C recently. But I did know enough to answer this :-) – proud haskeller Dec 25 '14 at 18:38
• I am going to do a +1 for now. But I still have not worked out how it is working so I will be back after turkey. :-) – Martin York Dec 25 '14 at 18:40
• @LokiAstari it basically uses the fact that C mixes expressions and statements together to do thing using only expressions – proud haskeller Dec 25 '14 at 18:42
# c++ 99 O(n)
The thing here loops are great for chaining along a list but not going up and down hierarchies. user300 managed it (I am impressed) but the code is hard to read.
The solution is to convert the tree into a list.
The trick is to do it at the same time your deleting the nodes.
void freeNode(Node* t)
{
if (t == NULL)
{ return;
}
// Points at the bottom left node.
// Any right nodes are added to the bottom left as we go down
// this progressively flattens the tree into a list as we go.
Node* bottomLeft = findBottomLeft(t);
while(t != NULL)
{
// Technically we don't need the if (it works fine without)
// But it makes the code easier to reason about with it here.
if (t->right != NULL)
{
bottomLeft->left = t->right;
bottomLeft = findBottomLeft(bottomLeft);
}
// Now just free the curent node
Node* old = t;
t = t->left;
free(old);
}
}
Node* findBottomLeft(Node* t)
{
while(t->left != NULL)
{
t = t->left;
}
return t;
}
# Golf Version
void f(Node*t){Node*o,*l=t;for(;t;free(o)){for(;l->left;l=l->left);l->left=t->right;o=t;t=t->left;}}
## Golf Expanded
void f(Node* t)
{
Node*o,*l = t;
for(;t;free(o))
{
for(;l->left;l = l->left);
l->left = t->right;
o = t;
t = t->left;
}
}
## C/C++/Objective-C 126 characters (includes required trailing newline)
#define b(t)(t->left||t->right)
void f(Node*r){while(r&&b(r)){Node**p=&r,*c=b(r);while(c)p=&c,c=b(c);free(*p);*p=0;}free(r);}
Does not run in O(n) time. But the OP does not require it, so here's my O(n2) solution.
Algorithm: Walk down to a leaf from the root. Release it. Repeat until there are no leaves. Release the root.
Ungolfed:
void freeTree (Node * root) {
while (root && (root->left || root->right)) {
Node ** prev = &root;
Node * curr = root->left || root->right;
while (curr != 0) {
prev = &curr;
curr = curr->left || curr->right;
}
free(*prev);
*prev = 0;
}
free(root);
}
• Unfortunately that will not work. you don't set the pointer to the leaf to NULL before freeing it. Thus you will endlessly keep freeing the same leaf node and never get to the point where you free the tree. – Martin York Dec 23 '14 at 19:46
• @LokiAstari: Thanks for noticing the bug. It should be fixed now (though I haven't tested the code). – Thomas Eding Dec 23 '14 at 21:47
## C, 150 bytes
f(T*r,int s){T*t[s];t[0]=r;T**f=&t[0],**e=&t[0];while(f<=e){if(*f){if((*f)->left){*++e=(*f)->left;}if((*f)->right){*++e=(*f)->right;}}free(*f);f++;}}
Try It On JDoodle | 1,815 | 6,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-21 | latest | en | 0.896377 |
https://physics.stackexchange.com/questions/185371/what-is-a-diffusionless-fluid | 1,701,818,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00335.warc.gz | 516,078,139 | 40,634 | # What is a diffusionless fluid?
I'm taking a course in astrophysical fluid dynamics and have come across a problem involving "small diffusionless disturbances" of a fluid. Based on the nature of the course I expect the examiner to have an exact mathematical definition of this in mind (i.e. if later in the question they say the flow is solenoidal, which I know to interpret as $\nabla \cdot \mathbf{v}=0$).
In general what is a diffusionless disturbance and do you have any insight how this definition might translate into a simple mathematical relation?
(my best guess so far is that they mean that $\delta \rho$, the pertubations in the density, can be taken as 0, but I'm far from sure about this)
• I suspect they mean $\rho=\rho+\delta\rho$ such that $\nabla\delta\rho\neq\nabla^2\rho$ May 22, 2015 at 11:32
• Could you expand a little, I suspect you mean $\rho =\rho_0 + \delta \rho$, but surely $\nabla \delta \rho$ never equals $\nabla^2 \rho$?, for one thing the spatial dimensions are inconsistent? Also what is the physical meaning of this? May 22, 2015 at 11:37
• I should have written $\nabla\cdot\delta\rho$, rather than $\nabla\delta\rho$ (matches the fluid equations the former way & not the latter). If the variation of the medium, $\delta\rho$, is proportional to the gradient, $\nabla\rho$, then you'd clearly have $\nabla\cdot\delta\rho=\nabla^2\rho$. May 22, 2015 at 14:58 | 388 | 1,399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-50 | longest | en | 0.908125 |
http://mathhelpforum.com/advanced-statistics/134200-birthday-paradox.html | 1,480,703,441,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00229-ip-10-31-129-80.ec2.internal.warc.gz | 183,454,638 | 10,560 | Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.
Here's the full solution on Wikipedia:
Birthday problem - Wikipedia, the free encyclopedia
I got lost at the transition from
to
What's happening here?
2. let $x=-x$
then $e^{-x} \approx 1 -x$
so $1 - \frac{1}{365} \approx e^{-1/365}$
$1-\frac{2}{365} \approx e^{-2/365}$
and so on
3. Originally Posted by jrkevinfang1234
Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.
Here's the full solution on Wikipedia:
Birthday problem - Wikipedia, the free encyclopedia
I got lost at the transition from
to
What's happening here?
$e^x\approx 1+x\implies e^-x\approx 1-x$ and so $\left(1-x\right)(1-y)(1-z)\cdots(1-t)\approx e^{-x}\cdot e^{-y}\cdot e^{-z}\cdots e^{-t}$ | 310 | 1,052 | {"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} | 3.875 | 4 | CC-MAIN-2016-50 | longest | en | 0.804243 |
https://stats.stackexchange.com/questions/570529/sample-proportion-vs-mean-proportion-for-sample-size-estimation | 1,695,826,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00040.warc.gz | 580,339,089 | 41,738 | # Sample proportion vs mean proportion for sample size estimation
Context
I would like to estimate the sample size needed for an experiment. I’m testing a feature on a website and would like to detect a significant change between different variants . One control and two treatment .
I’m new to this and want to keep this simple for the sake understanding tbe concept…
My parameters are : 0.05 alpha , 0.8 power.
Now if I know that tbe Click through rate of the original feature is 1% is this a mean or a proportion . I’m guessing this is the latter .
If I think the proportion for the second proportion will be 1.05% (i.e 5% minimal detectable effect) am I dealing with a case of sample size estimation for comparison of two proportions ?
Would the following estimation formula be adequate for such a test ?
n = (Zα/2+Zβ)^2 * (p1(1-p1)+p2(1-p2)) / (p1-p2)^2 Goal Understand if I have the right approach? Is it right to view the CTR as a proportion instead of mean ? If so when is mean the right choice ? Is it a good formula for sample size estimation ? Are there more robust ones ? Is it a binomial distribution since one can only click or not click? If my formula provides N samples and I would like to launch 15 variants does that mean I need 15xN samples ?
First, to make sure I understand this correctly. You're looking at click through rates (CTR). Currently, the rate is about 1 percent. (This is a proportion.) You're interested in detecting an increase of 5 percent on this base rate of 1 percent, so 1.05 percent.
In order to maximize statistical power, I've made this a one-sided test. What I'm assuming is that you're really only interested in treatments if they improve CTR's compared to the status quo, i.e. the control group. Representing the proportion of the treatment group as $$p_t$$ and the proportion of the control group as $$p_c$$ this gives us the following:
$$H_0: p_t = p_c \\ H_1: p_t > p_c$$
The required sample is as follows where $$n_c = n_t$$, i.e. you take the same sample size for the treatment and control group:
$$n_t = (p_t \times (1 - p_t) + p_c \times (1 - p_c)) \times (\frac{z_{1 - \alpha} + z_{1 - \beta}} {p_t - p_c})^2$$
In the formula above, $$\beta$$ = 1 - power and $$\alpha$$ is the desired significance level, often 0.05. 80 percent is often taken as a rule of thumb for power though in some cases you might want more power. (More power just requires a larger sample size.) In this case, 80 percent statistical power means that if the treatment group does in fact achieve a 5 percent increase on the CTR compared to the control group, for hypothetical repeated experiments, in 80 percent of these identical experiments, we would detect this effect as statistical significant.
If we plug the proportions 0.01 for $$p_c$$ and 0.0105 for $$p_t$$ and 0.05 for $$\alpha$$ and 0.2 for $$\beta$$ into the formula above this gives us a required sample size of 501,771 for the treatment group and 501,771 for the control group.
In R:
p_c <- 0.01
p_t <- 0.0105
alpha <- 0.05
beta <- 0.20
(p_t*(1-p_t)+p_c*(1-p_c))*((qnorm(1-alpha)+qnorm(1-beta))/(p_t - p_c))^2
In Python:
from scipy.stats import norm
p_c = 0.01
p_t = 0.0105
alpha = 0.05
beta = 0.20
(p_t*(1-p_t)+p_c*(1-p_c))*((norm.ppf(1-alpha)+norm.ppf(1-beta))/(p_t - p_c))**2
The power.prop.test function in R returns a similar result.
power.prop.test(p1 = 0.01,
p2 = 0.0105,
sig.level = 0.05,
power = 0.8,
alternative = "one.sided")
If you change the alternative to "two.sided", you'll see that you would need a sample of 637,010 for each group to achieve the same power for a two-sided test.
Now, presuming that for each of your fifteen treatments, you want to test the treatment against the control, you'd need a sample of 16 (15 treatment groups plus one control group) x 501,771, so roughly a sample of 8 million.
Now, generally, you achieve maximum power by allotting the sample equally between the control and treatment group, but in your case, since you have multiple treatment groups each being tested against the same control group, you could reduce your total sample size by increasing the size of your control group. The standard error of the difference in proportions is given as follows:
$$\sqrt{\frac{p_t \times (1 - p_t)}{n_t} + \frac{p_c \times (1 - p_c)}{n_c} }$$
By increasing the sample size for the control group, you decrease the standard error for the difference in proportions for each of your fifteen tests. The optimal size here for the control group would be something around 4 times larger than the treatment groups.
A modified version of the above formula includes $$\kappa$$ which equals $$n_c / n_t$$
$$n_t = (p_t \times (1 - p_t) + \frac{p_c \times (1 - p_c)}{\kappa}) \times (\frac{z_{1 - \alpha} + z_{1 - \beta}} {p_t - p_c})^2$$
Here, for your setup, you'd only need sample sizes of 318,149 for the treatment groups and a sample of 318,149 x 4 for the control group.
p_c <- 0.01
p_t <- 0.0105
alpha <- 0.05
beta <- 0.20
kappa <- 4
(p_t*(1-p_t)+(p_c*(1-p_c))/kappa)*((qnorm(1-alpha)+qnorm(1-beta))/(p_t - p_c))^2
This would get you close to 6 million samples needed. That said, optimal allocation ratios are a big topic. For more, you could start here and the reference therein.
Edit
Given that the formula's above rely on the Central Limit Theorem and are appropriate for large sample sizes, these are the same formula's you would use to compare two means. For a reference, see Sample Size Calculations in Clinical Research by Choa et. al. chapters 3 and 4.
Also, if you're doing 15 tests, you probably need to start thinking about correcting for false positives. If the null was true for each treatment, you'd still have a 75 percent chance of rejecting the null for at last one treatment.
• Did you decide for one side as we expect the MDE to only increase? So I’m not sure whether the 5% would go both ways or not . It would be two sided ? Also , can you please elaborate on how increasing the control group and decreasing the sample size per treament is beneficial for lowering sample size ? Apr 8, 2022 at 16:27
• I've added some information above. I'm presuming that you're only interested in the treatments if they increase CTR in which case you achieve maximum power by testing this hypothesis with a one-sided test. That said, I don't know your application. If you really need a two-sided test, just say so and I'll modify this. Apr 8, 2022 at 18:25
• I’m not sure if it’s one side or two sided but I’d appreciate your advice. The application is an ad and I’m trying to optimizes CTR Apr 8, 2022 at 19:30
• Then I'd suggest you choose a one-sided test as you really are only interested in finding treatments that increase your CTR. Apr 8, 2022 at 19:43
• how would you translate the first formula on python Apr 12, 2022 at 1:12 | 1,823 | 6,806 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-40 | longest | en | 0.911065 |
https://www.thestudentroom.co.uk/showthread.php?t=2320167 | 1,524,723,607,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948089.47/warc/CC-MAIN-20180426051046-20180426071046-00597.warc.gz | 881,213,932 | 39,250 | x Turn on thread page Beta
You are Here: Home
# Denitrification half equation watch
1. For the half equation of:
NO3- ---> N2
Is this correct:
2NO3- + 6H+ + 6e- ---> N2 + 3H20
If not why?
2. 2NO3 + 10e + 12H+ → N2 + 6H2O
3. (Original post by Holby_fanatic)
2NO3 + 10e + 12H+ → N2 + 6H2O
Why 10 rather than 12 electrons?
Also Nitric acid is produced by these 3 reactions
4NH3 + 5O2 ---> 4NO + 6H2O
2NO + O2 ---> 2NO2
3NO2 + H2O ---> 2HNO3 + NO
Starting with 1000kg ammonia, whats the maximum mass of HNO3?
I'm struggling here :
what I did was
1000000/17 = mass of nh3 in grams divided by mr nh3 to get moles of nh3
(1000000/17) which is 4 moles by the reaction, and divide that by 2 to get
(1000000/34).-2 moles by the reaction
Then multiply by Mr of HNO3, 63
(1000000/34 x 63)/1000 = 1852.94 kg
But the answer is 2471 kg. Where am I going wrong? Any help greatly appreciated
4. If 12 electrons were used, the equation would look something like this
2NO3 + 12e + 12H+ --> N2 + 6H2O +2e-
There would be two spare electrons as all products would already have a neutral charge.
For the second part of your question, I must admit I was struggling. I have two methods, the first one is completely wrong because I ended up with the same answer as you.
METHOD 1
4NH3 + 5O2 ---> 4NO + 6H2O
As there is a 1:1 ratio between NH3 and NO, 1000kg of NO could theoretically be made.
Moles of NH3 = 1000,000g / 17g mol-1
= 58823.5 mol
This is for 4 moles of NH3 and so 1 mole of NH3 = 58823.5 / 4
=14706 mol
2NO + O2 ---> 2NO2
Half of the moles NO produced in the first reaction is used in the second reaction. In this reaction, there is again a 1:1 ratio between NO and NO2.
Moles of NO used = 58823.5 / 2
= 29411.8 mol
Therefore, mass of NO2 produced = 29411.8 mol
3NO2 + H2O ---> 2HNO3 + NO
The moles of NO2 used is 1.5 times that of the moles produced in the second reaction. There is a 3:2 ratio between NO2 and HNO3
Moles of NO2 used = 29411.8 mol x 1.5
= 44117.7 mol
Moles of HNO3 produced = 2/3 mass of NO2 used
= 44117.7 x 2/3
= 29411.8 mol
Mass of HNO3 produced = number of moles x MR
= 29411.8 mol x 63g mol-1
= 1853kg
METHOD 2
Say 1000kg of NH3 results in 1000mol of NO
1000mol of NH3 produces 1000mol of NO
As half of the number of moles of NO produced is needed for the second reaction half of the mass of NO is needed. 500mol of NO is used.
500mol of NO produces 500mol of NO2
As 1.5 times the number of moles of NO2 produced is needed for the third reaction 1.5 times the mass of NO2 is needed. 750mol of NO2 is used.
750mol of NO2 produces 375mol of HNO3. (This is because 3 moles of NO2 is needed while only 2 moles of HNO3 is produced 2/3 x 750mol = 375mol)
Mass of HNO3 = moles x MR
= 375mol x 63g mol-1
= 2362.5kg
Do you need these answers urgently or do you have some time? I'll keep trying to work it out until I get the right answer. (This is much more fun than writing an essay on polymers).
Turn on thread page Beta
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: April 14, 2013
Today on TSR
### University rankings 2019
Cambridge at number one
### I have imposter syndrome at Cambridge
Poll
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,187 | 3,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-17 | latest | en | 0.923227 |
https://www.proofwiki.org/wiki/Method_of_Truth_Tables/Proof_of_Tautology/Examples/Peirce%27s_Law | 1,685,680,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00487.warc.gz | 1,012,742,360 | 10,740 | # Method of Truth Tables/Proof of Tautology/Examples/Peirce's Law
## Examples of Proof of Tautology
Consider the truth table for Peirce's Law:
$P = \paren {\paren {p \implies q} \implies p} \implies p$
which is:
$\begin{array}{cc||ccccccc} p & q & ((p & \implies & q) & \implies & p) & \implies & p \\ \hline \F & \F & \F & \T & \F & \F & \F & \T & \F \\ \F & \T & \F & \T & \T & \F & \F & \T & \F \\ \T & \F & \T & \F & \F & \T & \T & \T & \T \\ \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \end{array}$
The main connective of $P$ is the rightmost instance of $\implies$.
The column beneath that connective is all $\T$, so $\paren {\paren {p \implies q} \implies p} \implies p$ is a tautology.
$\blacksquare$ | 285 | 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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-23 | latest | en | 0.668175 |
http://serc.carleton.edu/nnn/teaching/test.html?q1=sercvocabs__43%3A8 | 1,418,958,931,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768197.70/warc/CC-MAIN-20141217075248-00042-ip-10-231-17-201.ec2.internal.warc.gz | 240,548,340 | 10,351 | National Numeracy Network > Teaching Resources > QR Teaching Activities
Help
# Show all Resources
Current Search Limits
## Subject
showing only Mathematics Show all Subject
Refine the Results
# Subject Show all Subject
Results 1 - 10 of 77 matches
The Floating Lithosphere - Isostasy part of Quantitative Skills:Activity Collection
Len Vacher, Dept of Geology, University of South Florida
Students are asked to numerically and then analytically determine the relations governing the depth of compensation.
Math Review part of Cutting Edge:Hydrogeology:Hydrogeology, Soils, Geochemistry 2013:Activities
Kallina Dunkle, Austin Peay State University
This is designed as an introductory lab for hydrogeology or other upper-level courses that are quantitative in nature in order to review key mathematical concepts that will be used throughout the semester.
World Population Activity II: Excel part of Starting Point-Teaching Entry Level Geoscience:Mathematical and Statistical Models:Mathematical and Statistical Models Examples
Activity and Starting Point page by R.M. MacKay. Clark College, Physics and Meteorology.
(Activity 2 of 2)In this intermediate Excel tutorial students import UNEP World population data/projections, graph this data, and then compare it to the mathematical model of logistic growth. -
Radiometric Dating part of Quantitative Skills:Activity Collection
Related Links Radioactive Decay Exponential Growth and Decay Peter Kohn - James Madison University Christopher Gellasch - U.S. Military Academy Jim Sochacki - James Madison University Scott Eaton - James Madison University Richard Ford - Weber State University
This activity leads students through derivations of the equations associated with radiometric dating.
Density of Earth - Using Some Field Data part of Quantitative Skills:Activity Collection
Len Vacher, Dept of Geology, University of South Florida
This module addresses the problem of how to determine the density of the earth and has students do some field experiments to get the data they need to answer the problem.
The Floating Lithosphere - Eureka! part of Quantitative Skills:Activity Collection
Len Vacher, Dept of Geology, University of South Florida
In this module, students examine Archimede's Principle in general and as it applies to Isostacy.
Frequency of Large Earthquakes -- Introducing Some Elementary Statistical Descriptors part of Spreadsheets Across the Curriculum:General Collection:Examples
Len Vacher, Dept of Geology, University of South Florida
Spreadsheets Across the Curriculum module. Students examine the number of large earthquakes (magnitude 7 and above) per year for 1970-1999 and 1940-1999. QL: descriptors of a frequency distribution.
Metric System Conversions: Process Oriented Guided Inquiry Learning (POGIL) activity part of MnSCU Partnership:PKAL-MnSCU Activities
Jeffrey Pribyl
This activity helps student learn to convert within the metric system and begin learning about process skill necessary for working in groups.
Reading Topographic Maps and Calculating Map Scale part of Quantitative Skills:Activity Collection
Leslie Kanat, Johnson State College
Use a topographic map to deliniate a watershed, draw a map bar scale, and calculate a map ratio scale.
Calculating and Comparing Tax Rates part of Pedagogy in Action:Library:Teaching Quantitative Reasoning with the News:Examples
Stuart Boersma, Central Washington University
This example focuses on six letters to the editor. All six letters attempt to describe and compare the amount of taxes paid on two different incomes: \$30,000 and \$200,000. Tax rates are expressed in absolute ... | 759 | 3,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.810598 |
http://mathhelpforum.com/calculus/157683-differentiation-first-principles.html | 1,481,277,367,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00446-ip-10-31-129-80.ec2.internal.warc.gz | 170,174,672 | 11,144 | # Thread: Differentiation from first principles
1. ## Differentiation from first principles
I'm stick on differentiating, from first principles, the following function:
f(x) = (3x-1)/(x+2)
I plug that into:
[f(a+h) - f(a)]/h
and I get:
7h/(x+h+2)(x+2)
But when I use the quotient rule to do it, I end up with:
7/(x+2)^2
I've been trying to get rid of that bothersome h, but I just haven't found a way to do it! I have a feeling that I'm overlooking something really simple. Any help appreciated.
2. Well, you have to divide by h before you take the limit as h goes to zero, right?
3. It's probably easiest to simplify the fraction before trying to differentiate it.
$f(x) = \frac{3x - 1}{x + 2}$
$= \frac{3x + 6 - 7}{x + 2}$
$= \frac{3(x + 2)}{x + 2} - \frac{7}{x + 2}$
$= 3 - \frac{7}{x + 2}$.
Therefore $f(x + h) = 3 - \frac{7}{x + h + 2}$.
$f(x + h) - f(x) = \left(3 - \frac{7}{x + h + 2}\right) - \left(3 - \frac{7}{x + 2}\right)$
$= \frac{7}{x + 2} - \frac{7}{x + h + 2}$
$= \frac{7(x + h + 2) - 7(x + 2)}{(x + 2)(x + h + 2)}$
$= \frac{7x + 7h + 14 - 7x - 14}{(x + 2)(x + h + 2)}$
$= \frac{7h}{(x + 2)(x + h + 2)}$.
So $\frac{f(x + h) - f(x)}{h} = \frac{\frac{7h}{(x + 2)(x + h + 2)}}{h}$
$= \frac{7h}{h(x + 2)(x + h + 2)}$
$= \frac{7}{(x + 2)(x + h + 2)}$.
So $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{7}{(x + 2)(x + h + 2)}$
$= \frac{7}{(x + 2)(x + 2)}$
$= \frac{7}{(x + 2)^2}$.
4. Originally Posted by blackdragon190
I'm stick on differentiating, from first principles, the following function:
f(x) = (3x-1)/(x+2)
I plug that into:
[f(a+h) - f(a)]/h
and I get:
7h/(x+h+2)(x+2)
No. That "h" in the numerator should not be there.
But when I use the quotient rule to do it, I end up with:
7/(x+2)^2
I've been trying to get rid of that bothersome h, but I just haven't found a way to do it! I have a feeling that I'm overlooking something really simple. Any help appreciated.
5. Silly me! I got so caught up in playing around with the fraction that I forgot that I still had to divide by h. I just knew that it was something silly.
Thanks for the help
6. You're welcome for whatever I contributed. Have a good one!
7. Seems like I forgot about the limit too...doh.. | 854 | 2,231 | {"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": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2016-50 | longest | en | 0.865314 |
https://amp.doubtnut.com/question-answer/find-the-equation-of-the-planes-bisecting-the-angles-between-planes-2x-y-2z9-and-3x-4y-12z-130-8496110 | 1,610,934,426,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00124.warc.gz | 211,643,872 | 22,138 | Doubtnut is better on App
Paiye sabhi sawalon ka Video solution sirf photo khinch kar
Open App
IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
Question From class 12 Chapter 3D - COMPETITION
# Find the equation of the planes bisecting the angles between planes
Find the angle between the planes and .
3:08
What is the angle between the planes ?
2:36
Find the angle between the two planes and using vector method.
3:06
Find the plane which bisects the obtuse angle between the planes
6:25
Find the angle between the planes .
2:25
Find the angle between the planes
2:10
Find the angle between the two planes and .
2:45
Find the angle between the two planes and using vector method.
2:08
Find the angle between the planes . Using vector method.
2:31
Two planes And Find the angle between the vector method.
2:45
Find the angle between the two planes and .
3:53
Two planes And Find the angle between
3:05
Two planes And Find the angle between
3:05
Show that the origin lies in the interior of the acute angle between planes x + 2y + 2z = 9 and 4x - 3y + 12z + 13 = 0. Find the equation of bisector of the acute angle.
5:23
Find the distance between the parallel planes
1:39
Latest Blog Post
Punjab Board Date Sheet 2021 for Classes 5, 8, 10, 12 Released, Check here
Punjab Board (PSEB) datesheet of 2021 for classes 5, 8, 10, 12 has been released. Know how to download PSEB date sheet 2021 & details related to Punjab board exam.
Delhi Schools to Reopen for Classes 10 & 12 from Jan 18, 2021
Delhi Schools to reopen for classes 10 & 12 from Jan 18, 2021. Know complete details related to Delhi school reopening and upcoming board exams.
UP School Reopen for Classes 9-12, with Fresh COVID-19 Guidelines
UP School reopen for classes 9-12, with fresh COVID-19 guidelines. Know complete details of UP School reopening and UP board exam 2021!
Goa Board Exam for Class 9 and 11 will be Held Offline by April 24
Goa board exam for class 9 and 11 will begin offline from April 24. Know complete details of Goa board exam 2021! | 576 | 2,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-04 | latest | en | 0.871994 |
https://fdocuments.us/document/domain-adaptation-with-multiple-sources-yishay-mansour-tel-aviv-univ-google.html | 1,716,059,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00107.warc.gz | 211,382,298 | 27,245 | 48
Domain Adaptation with Domain Adaptation with Multiple Sources Multiple Sources Yishay Mansour, Tel Aviv Univ. & Google Mehryar Mohri, NYU & Google Afshin Rostami, NYU
• date post
21-Dec-2015
• Category
## Documents
• view
221
0
### Transcript of Domain Adaptation with Multiple Sources Yishay Mansour, Tel Aviv Univ. & Google Mehryar Mohri, NYU &...
Yishay Mansour, Tel Aviv Univ. & Google
Afshin Rostami, NYU
2
3
• High level: – The ability to generalize from one domain to
another
• Significance:– Basic human property– Essential in most learning environments– Implicit in many applications.
4
• Sentiment analysis:– Users leave reviews
• products, sellers, movies, …
– Goal: score reviews as positive or negative.– Adaptation example:
• Learn for restaurants and airlines
• Generalize to hotels
5
• Learn a few accents
• Generalize to new accents– think “foreign accents”.
6
• Machine Learning prediction:– Learn from examples drawn from distribution D
– predict the label of unseen examples• drawn from the same distribution D
– generalization within a distribution
• Adaptation:– predict the label of unseen examples
• drawn from a different distribution D’
– Generalization across distributions
7
• Learn from D and test on D’– relating the increase in error to dist(D,D’)
• Ben-David et al. (2006), Blitzer et al. (2007),
• Single distribution varying label quality• Cramer et al. (2005, 2006)
8
9
Our Model - input
f
target function
D1
Dk
distributions
.
.
.
h1
hk
hypotheses
.
.
.
L(D1,h1,f)≤ε
L(Dk,hk,f)≤ε
.
.
.
Expected Loss
Typical loss function: L(a,b)=|a-b| and L(D,h,f)= Ex~D[ |f(x)-h(x)| ]
10
Our Model – target distribution
D1
Dk
basicdistributions
.
.
.
target distribution Dλ
λ1
λk
k
iii xDxD
1
)()(
11
Our model – Combination Rule
• Combine h1, … , hk to a hypothesis h*
– Low expected loss• hopefully at most ε
• combining rules:– let z: Σ zi = 1 and zi≥ 0 – linear: h*(x) = Σ zi hi(x)– distribution weighted:
)()(
)()(
11
xi
hxDz
xDzxh
k
ik
j jj
iiz
h1hk
. . .
combining rule
12
Combining Rules – Pros
• Alternative: Build a dataset for the mixture.– Learning the mixture parameters is non-trivial– Combined data set might be huge size– Domain dependent data unavailable– Combined data might be huge
• Sometimes only classifiers are given/exist– privacy
• MOST IMPORTANT:
FUNDAMENTAL THEORY QUESTION
13
Our Results:
• Linear Combining rule:– Seems like the first thing to try– Can be very bad
• Simple settings where any linear combining rule performs badly.
14
Our Results:
• Distribution weighted combining rules:– Given the mixture parameter λ:
• there is a good distribution weighted combining rule.
• expected loss at most ε
– For any target function f, • there is a good distribution combining rule hz
• expected loss at most ε
– Extension for multiple “consistent” target functions• expected loss at most 3ε
• OUTCOME: This is the “right” hypothesis class
15
16
Linear combining rules
Xfh1h0
a110
b010
a½10
b½01
Original Loss: ε=0 !!!
Any linear combining rule hhas expected absolute loss ½
17
Distribution weighted combining rule
• Target distribution – a mixture: Dλ(x)=Σ λi Di(x)
• Set z=λ :
• Claim: L(Dλ,hλ,f) ≤ ε
)()(
)()(
)(
)()(
111
xi
hxD
xDx
ih
xD
xDxh
k
i T
iik
ik
j jj
ii
18
Distribution weighted combining rule
),,( fhDL
k
ii
k
i
fhDL
xiii
ix
k
i
ii
x
ii
xfxhLxD
xfxhLxD
xDxD
xfxhLxD
1
1
),,(
1
))(),(()(
))(),(()(
)()(
))(),(()(
PROOF:
19
Xfh1h0
a110
b010
a½10
b½01
Original Loss: ε=0 !!!
h+(x):x=a h+(x)=h1(x)=1x=b h+(x)=h0(x)=0
20
21
Unknown mixture distribution
• Zero-sum game:– NATURE: selects a distribution Di
– LEARNER: selects a z• hypothesis hz
– Payoff: L(Di,hz,f)
• Restating to previous result:– For any mixed action λ of NATURE– LEARNER has a pure action z= λ
• such that the expected loss is at most ε
22
Unknown mixture distribution
• Consequence:– LEARNER has a mixed action (over z’s) – for any mixed action λ of NATURE
• a mixture distribution Dλ
– The loss is at most ε
• Challenge:– show a specific hypothesis hz
• pure, not mixed, action
23
Searching for a good hypothesis
• Uniformly good hypothesis hz:
– for any Di we have L(Di, hz,f) ≤ ε
• Assume all the hi are identical
– Extremely lucky and unlikely case
• If we have a good hypothesis we are done!– L(Dλ,hz,f) = Σ λi L(Di,hz,f) ≤ Σ λi ε = ε
• We need to show in general a good hz !
24
Proof Outline:
• Balancing the losses:– Show that some hz has identical loss on any Di
– uses Brouwer Fixed Point Theorem• holds very generally
• Bounding the losses:– Show this hz has low loss for some mixture
• specifically Dz
25
A: compact and convex set
φ: A→Acontinuous mapping
Brouwer Fixed Point Theorem :For any convex and compact set A and any continuous mapping φ : A→A, there exists a point x in A such that φ(x)=x
26
Balancing Losses
A = {Σi zi = 1 and zi ≥ 0 }
k
j zjj
ziii
fhDLz
fhDLzz
1),,(
),,()]([
Problem 1: Need to get φ continuous
27
Balancing Losses
A = {Σi zi = 1 and zi≥ 0 }
k
j zjj
ziii
fhDLz
fhDLzz
1),,(
),,()]([
Fixed point: z=φ(z)
k
j zjj
ziii
fhDLz
fhDLzz
1),,(
),,(
),,(),,(1
fhDLzfhDLzz zii
k
j zjji
Problem 2:Needs that zi ≠0
28
Bounding the losses
• We can guarantee balanced losses even for linear combining rule !
Xfh1h0
a110
b010
a½10
b½01
For z=(½, ½) we haveL(Da,hz,f)=½L(Db,hz,f)=½
29
Bounding Losses
• Consider the previous z– from Brouwer fixed point theorem
• Consider the mixture Dz
– Expected loss is at most ε
• Also: L(Dz,hz,f)= ΣzjL(Dj,hz,f)=γ
• Conclusion: – For any mixture expected loss at most γ ≤ ε
30
Solving the problems:
• Redefine the distribution weighted rule:
• Claim: For any distribution D,
is continuous in z.
)()(
/)()(
1, xh
xDz
kxDzxh i
k
i jj
iiz
),,( , fhDL z
31
Main Theorem
For any target function f and any δ>0,
there exists η>0 and z such that
for any λ we have
),,( , fhDL z
32
Balancing Losses
• The set A = {Σ zi = 1 and zi≥ 0 }
– The simplex
• The mapping φ with parameters η and η’– [φ(z)]i= (zi Li,z+η’/k)/ (ΣzjLj,z+η’)
• where Li,z=L(Di,hz,η,f)
• For some z in A we have φ(z)=z– zi = (zi Li,z+η’/k)/ (ΣzjLj,z+η’) >0
– Li,z = (ΣzjLj,z)+η’ - η’/(zi k) < (ΣzjLj,z)+ η’
33
Bounding Losses
• Consider the previous z– from Brouwer fixed point theorem
• Consider the mixture Dz
– Expected loss is at most ε+η
• By definition ΣzjLj,z= L(Dz,hz,η,f)
• Conclusion: γ=ΣzjLj,z ≤ ε+η
34
Putting it together
• There exists (z,η) such that:– Expected loss of hz,η approximately balanced
– L(Di,hz,η,f) ≤γ+η’
• Bounding γ using Dz
– γ =L(Dz,hz,η,f) ≤ε+η
• For any mixture Dλ
– L(Dλ,hz,η,f) ≤ε+η+ η’
35
A more general model
• So far: NATURE first fixes target function f• consistent target functions f
– the expected loss w.r.t. Di is at most ε• for any of the k distributions
• Function class F ={f is consistent}• New Model:
– LEARNER picks a hypothesis h– NATURE picks f in F and mixture Dλ
– Loss L(Dλ,h,f)• RESULT: L(Dλ,h,f)≤ 3ε.
36
37
Uniform Algorithm
• Hypothesis sets z=(1/k , … , 1/k):
• Performance:– For any mixture, expected error ≤ kε– There exists mixture with expected error Ω(kε)– For k=2, there exists a mixture with 2ε-ε2
)()(
)()(
)()1(
)()1()(
11
11
xi
hxD
xDx
ih
xDk
xDkxh
k
ik
j j
ik
ik
j j
iu
38
Open Problem
• Find a uniformly good hypothesis– efficiently !!!
• algorithmic issues:– Search over the z’s– Multiple local minima.
39
40
Empirical Results
• Data-set of sentiment analysis:– good product takes a little time to start operating very good for the
price a little trouble using it inside ca
– it rocks man this is the rockinest think i've ever seen or buyed dudes check it ou
– does not retract agree with the prior reviewers i can not get it to retract any longer and that was only after 3 uses
– dont buy not worth a cent got it at walmart can't even remove a scuff i give it 100 good thing i could return it
– flash drive excelent hard drive good price and good time for seller thanks
41
Empirical analysis
• Multiple domains:– dvd, books, electronics, kitchen appliance.
• Language model:– build a model for each domain
• unlike the theory, this is an additional error source
• Tested on mixture distribution– known mixture parameters
• Target: score (1-5)– error: Mean Square Error (MSE)
42
linearDistribution weightedbooks
dvdelectronics
kitchen
43
44
45
46
Summary
• linear
• distribution weighted
• Theoretical analysis– mixture distribution
• Future research– algorithms for combining rules
– beyond mixtures
47
48 | 2,680 | 8,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.690771 |
https://stats.stackexchange.com/questions/525334/joint-probability-measure | 1,709,355,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00818.warc.gz | 529,077,013 | 42,532 | # Joint probability measure
I know from my measure theory class that for two $$\sigma$$-finite measure spaces $$(\mathcal{X}_1, \mathcal{A}_1, \mu_1)$$ and $$(\mathcal{X}_2, \mathcal{A}_2, \mu_2)$$ there exists a unique measure $$\mu := \mu_1 \otimes \mu_2:\mathcal{A}_1 \otimes \mathcal{A}_2 \rightarrow [0, \infty]$$ such that $$\mu_1 \otimes \mu_2(A_1 \times A_2) = \mu_1(A_1)\cdot \mu_2(A_2).$$ So my questions is what that means in probability theory, for the joint distribution (measure) of two random variables? Is $$\mathbb{P}_{X, Y}$$, the joint distribution of two random variables $$X$$ and $$Y$$ the same as $$\mathbb{P}_{X}\otimes \mathbb{P}_{Y}$$? Probably not, otherwise all random variables would be independent due to the above theorem, no? But how do $$\mathbb{P}_{X, Y}$$ and $$\mathbb{P}_{X}\otimes \mathbb{P}_{Y}$$ then relate to each other? (This is particularly needed to compute the expectation over the joint distribution)
• The unique measure your theorem says exists with that multiplicative property is the precisely that resulting for the two random variables being independent with respective measures $\mu_1$ and $\mu_2$. Any other measure would not have that multiplicative property for all $A_1 \times A_2 \in \mathcal{A}_1 \otimes \mathcal{A}_2$ May 21, 2021 at 16:44
• This independence leads (in one direction) to $\mathbb E_{X,Y}[X\cdot Y] = \mathbb E_{X}[X] \cdot \mathbb E_{Y}[Y]$ so long as all the expectations are finite May 21, 2021 at 16:48
• Exactly, but in general random variables are not independent. So why wouldnt this theorem apply to them? May 21, 2021 at 16:49
• How exactly would it apply? All it would end up showing is that you have found two measures on the product: the original $\mathbb{P}_{X,Y}$ and the one constructed by this theorem. The theorem doesn't assert all joint measures are the same!
– whuber
May 21, 2021 at 17:17
• Well I thought that since the measure from the theorem is unique, it is the only measure on the product sigma algebra. And if we calculate a double integral with the theorem of Fubini, we always use the product measure from the theorem above, no? So if we calculate an expectation value then I would assume that we always use the product measure of the theorem above... May 27, 2021 at 19:24
## 1 Answer
### Joint Distributions and Expectation
In general, the joint distribution of random variables $$X$$ and $$Y$$, defined on a common probability space $$(\Omega, \mathcal{A}, \mathbb{P})$$ and taking values in measurable spaces $$(\mathcal{X}, \mathcal{B})$$ and $$(\mathcal{Y}, \mathcal{C})$$, respectively, is the probability measure defined on $$(\mathcal{X} \times \mathcal{Y}, \mathcal{B} \otimes \mathcal{C})$$ by $$\mathbb{P}_{X, Y}(E) = \mathbb{P}((X, Y) \in E)$$ for all $$E \in \mathcal{B} \otimes \mathcal{C}$$.
This is the same as the ordinary distribution of $$(X, Y) : \Omega \to \mathcal{X} \times \mathcal{Y}$$ when viewed as a single random variable defined on $$\Omega$$.
Also, $$X$$ and $$Y$$ are said to be independent if it holds that $$\mathbb{P}(X \in B, Y \in C) = \mathbb{P}(X \in B) \mathbb{P}(Y \in C)$$ for all $$B \in \mathcal{B}$$ and $$C \in \mathcal{C}$$.
The independence condition can be rephrased in terms of the joint distribution of $$X$$ and $$Y$$: $$X$$ and $$Y$$ are independent if and only if $$\mathbb{P}_{X,Y}(B \times C) = \mathbb{P}_X(B) \mathbb{P}_Y(C)$$ for all $$B \in \mathcal{B}$$ and $$C \in \mathcal{C}$$. That is, if and only if $$\mathbb{P}_{X, Y} = \mathbb{P}_X \otimes \mathbb{P}_Y.$$ Thus, the joint distribution of $$X$$ and $$Y$$ is the product measure of the (marginal) distributions of $$X$$ and $$Y$$ precisely in the case that $$X$$ and $$Y$$ are independent. If $$X$$ and $$Y$$ are dependent, then their joint distribution is not the product measure of the marginal distributions.
## Computing Expectations over Joint Distributions
If $$X$$ and $$Y$$, as above, are independent and $$f : \mathcal{X} \times \mathcal{Y} \to \mathbb{R}$$ is a measurable function (satisfying either non-negativity or integrability with respect to $$\mathbb{P}_{X,Y}$$), then Fubini's theorem allows you to compute \begin{aligned} \mathbb{E}[f(X, Y)] &= \int_\Omega f(X(\omega), Y(\omega)) \, \mathbb{P}(d\omega) && \text{(def. of expectation)} \\ &= \int_{\mathcal{X} \times \mathcal{Y}} f(x, y) \, \mathbb{P}_{X, Y}(d(x, y)) && \text{(change of variables)} \\ &= \int_{\mathcal{X} \times \mathcal{Y}} f(x, y) \, \mathbb{P}_X\otimes\mathbb{P}_Y(d(x, y)) &&\text{(independence)} \\ &= \int_{\mathcal{Y}} \left(\int_{\mathcal{X}} f(x, y) \, \mathbb{P}_X(dx)\right) \, \mathbb{P}_Y(dy) &&\text{(Fubini's theorem)} \end{aligned} However, if $$X$$ are $$Y$$ are not independent, then this argument won't work. Instead, if you want to break an expectation of $$f(X,Y)$$ into an integral over $$\mathcal{X}$$ followed by an integral over $$\mathcal{Y}$$, as we did above, you need to know something about the conditional distribution of $$X$$ given $$Y$$.
For what follows, suppose $$(\mathcal{X},\mathcal{B})$$ and $$(\mathcal{Y}, \mathcal{C})$$ are "sufficiently nice" measurable spaces, meaning that they admit conditional distributions (this will happen for most spaces in practice; a sufficient condition is being standard Borel).
Then if $$\mathbb{P}_{X\mid Y} : \mathcal{B} \times \mathcal{Y} \to [0, 1]$$ is a version of the conditional distribution of $$X$$ given $$Y$$, then we can proceed similarly to the calculations above: \begin{aligned} \mathbb{E}[f(X, Y)] &= \int_{\mathcal{X} \times \mathcal{Y}} f(x, y) \, \mathbb{P}_{X, Y}(d(x, y)) \\ &= \int_{\mathcal{Y}} \left(\int_{\mathcal{X}} f(x, y) \, \mathbb{P}_{X\mid Y}(dx, y)\right) \, \mathbb{P}_Y(dy) \\ &= \mathbb{E}[\mathbb{E}[f(X, Y) \mid Y]] \end{aligned} (in fact, the formula $$\mathbb{E}[f(X, Y)] = \mathbb{E}[\mathbb{E}[f(X, Y) \mid Y]]$$ holds even without considering conditional distributions (proof), but it's arguably harder to compute in that case).
If $$X$$ and $$Y$$ are independent, then it happens that $$\mathbb{P}_{X\mid Y}(B, y) = \mathbb{P}_X(B)$$ for every $$B \in \mathcal{B}$$ and $$\mathbb{P}_Y$$-almost every $$y \in Y$$. In this case, the calculation reduces to the first computation above.
In practice, the conditional distribution $$\mathbb{P}_{X\mid Y}$$ will usually be given by a conditional density $$p_{X\mid Y} : \mathcal{X} \times \mathcal{Y} \to [0, \infty)$$ of $$X$$ given $$Y$$ with respect to some dominating measure $$\mu$$ on $$(\mathcal{X}, \mathcal{B})$$, yielding $$E[f(X, Y)] = \int_{\mathcal{Y}} \left(\int_{\mathcal{X}} f(x, y) p_{X \mid Y}(x, y) \, \mu(dx)\right) \, \mathbb{P}_Y(d y).$$
• @guest1 You're right, I fixed that typo. Regarding your second question, the uniqueness of the product measure $\mu \otimes \nu$ is with regard to the condition $(\mu \otimes \nu)(A \times B) = \mu(A)\nu(B)$. That is, there's only one measure satisfying that equation (under conditions like $\sigma$-finiteness), but there can be many others that don't satisfy it. Also, there are versions of Fubini's theorem that hold for non-product measures (e.g., disintegration, which is related to conditional probabilities) May 27, 2021 at 19:34
• @guest1 a sufficient condition on the underlying measurable spaces is that they are standard Borel: a measurable space $(\mathcal{X}, \mathcal{B})$ is standard Borel if there exists a metric $d$ on $\mathcal{X}$ such that $(\mathcal{X}, d)$ is a complete, separable metric space with Borel $\sigma$-algebra $\mathcal{B}$ Jun 1, 2021 at 15:28
• @guest1 Thanks, $\mathbb{P}(dy)$ was a typo; it's fixed. I'm not sure what you mean by "non measure-theoretic way". In practice you compute the expected value of a function of two random variables by conditioning on one of them: $E[f(X,Y)]=E[E[f(X, Y) | Y]]$. The calculation usually reduces to a concrete instance of the last equation in my answer. This works whether or not the variables are independent (it's just redundant if you know beforehand that they're independent). Lastly, the dominating measure $\mu$ in probability theory isn't necessarily Lebesgue measure, but it often is. Jun 2, 2021 at 16:31
• @guest1 two things are happening there: the definition of "density", and Fubini's theorem. If $p$ is a density of $P$ with respect to a measure $\mu$, then $\int f\,dP=\int fp\,d\mu$. Next, if the integral exists and is over a product space $\mathcal{X}\times\mathcal{Y}$ and the dominating measure $\mu$ is a product measure $\nu\otimes\lambda$ (as is the case for Lebesgue measure), then the integral can be written $\int_{\mathcal{X}\times\mathcal{Y}}fp\,d(\nu\otimes\lambda)=\int_{\mathcal{Y}}\left(\int_{\mathcal{X}}f(x,y)p(x,y)\,\mu(dx)\right)\,\lambda(dy)$ by Fubini's theorem. Jun 3, 2021 at 16:35
• @guest1 that’s correct Jun 4, 2021 at 15:39 | 2,802 | 8,800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 74, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.882981 |
http://www.programming-idioms.org/idiom/9/create-a-binary-tree-data-structure/690/rust | 1,498,336,856,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320338.89/warc/CC-MAIN-20170624203022-20170624223022-00644.warc.gz | 626,991,281 | 7,067 | # Idiom #9 Create a Binary Tree data structure
The structure must be recursive because left child and right child are binary trees too. A node has access to children nodes, but not to its parent.
```struct BinTree<T> {
value: T,
left: Option<Box<BinTree<T>>>,
right: Option<Box<BinTree<T>>>,
}```
```struct treenode{
int value;
struct treenode* left;
struct treenode* right;
}```
```class BinaryTree<T>:IComparable<T>
{
T value;
BinaryTree<T> leftChild;
BinaryTree<T> rightChild;
}```
```struct BinaryTree(T)
{
T data;
BinaryTree* left;
BinaryTree* right;
}```
```class Node<T> {
final T value;
Node<T> left, right;
Node(this.value, {this.left, this.right});
]```
```defmodule BinaryTree do
defstruct data: nil, left: nil, right: nil
end```
```-type binary_tree(T) ::
#{ data := T
, left := binary_tree(T)
, right := binary_tree(T)
}.```
```type BinTree struct {
Key keyType
Deco valueType
Left *BinTree
Right *BinTree
}
```
```data BT x = Ø | BT (BT x) x (BT x)
```
```class BinTree<T extends Comparable<T>>{
T value;
BinTree<T> left;
BinTree<T> right;
}
```
```class BNode
{
public \$data;
public \$lNode;
public \$rNode;
public function __construct(\$item) {
\$this->data = \$item;
\$this->lNode = null;
\$this->rNode = null;
}
}```
```type
generic BinTree<T> = class
Value: T;
Left, Right: BinTree;
end; ```
```class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None```
```Node = Struct.new(:left, :right)
parent = Node.new(Node.new, Node.new)```
```(define (make-btree value left right)
(cons value (cons left right)))
(define (btree-value bt) (car bt)) | 471 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.484591 |
http://www.jiskha.com/display.cgi?id=1407440783 | 1,498,220,715,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00660.warc.gz | 575,250,822 | 3,748 | # math
posted by on .
a quart is approximately sixty cubic inches. a cubic feet of water weighs approximately sixty pounds. therefore, a quart of water weighs, approximately
• math - ,
a cubic foot is 12^3 = 1728 cubic inches. So, the quart weighs
60/1728 * 60 = 2.08 pounds
But then we knew that, since "a pint's a pound, the world around." | 93 | 347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.925409 |
https://www.smore.com/bwe0 | 1,532,044,797,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00059.warc.gz | 991,927,190 | 10,527 | # My Real-Life Linear Function
## My linear function is a microphone stand
The microphone stand is my little brother's. It's in my living room.
It's coordinates are: (-7,-4) and (4,4)
## linear equations
Slope-Intercept: y=8/11x+12/11
Point-Slope: y-4=8/11(x-4)
Standard: 8/11-y=-12/11
Parallel
y=8/11x-13
y=8/11x+4
Perpendicular
y=-11/8x-23
y=-11/8x+ 12 | 138 | 367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-30 | latest | en | 0.69374 |
https://www.hpmuseum.org/forum/thread-1347-post-11605.html#pid11605 | 1,653,427,499,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00115.warc.gz | 925,407,673 | 7,294 | [WP-34S] Calculations With Complex Matrices
05-18-2014, 05:20 PM (This post was last modified: 05-20-2014 12:15 PM by Thomas Klemm.)
Post: #1
Thomas Klemm Senior Member Posts: 1,549 Joined: Dec 2013
[WP-34S] Calculations With Complex Matrices
Introduction
This article describes how to perform matrix multiplication and matrix inversion with complex matrices. Since the WP-34S lacks complex numbers the same trick as in the HP-15C is used. The complex transformation between $$Z^P$$ and $$\tilde{Z}$$ is provided by combining M.COPY and M.TRANSP commands. Of course this could be written down in a small program.
MED: Matrix Editor
This description is from matrixedit.wp34s:
Code:
/* * Interactive editor for matrices. * Start the editor with a matrix descriptor in X. * Use the arrow keys to navigate: * up/down: previous/next column * f-up/f-down: previous/next row * Type a digit to enter a number. This will suspend * the progrom. R/S will enter X at the current row/column. * XEQ RCL restores the present value to X if you decide * not to change the cell. Press R/S to continue. * * Some hot keys: * A will start over at (1,1). * When used outside the editor you will need to specifiy * a new descriptor first. * B (labeled 1/x) will call M[^-1] on the matrix. * C continues at the last position. * D computes and displays the determinant. R/S or C continue. * * We need a few registers for internal use and therefore * set the stacksize to 4 to free registers A to D. * Press <- to restore the mode and exit the editor. * Double precision mode will always be reset! * * Register usage: * A - Matrix descriptor * B - Current register * D - Old setup mode * I, J - Row and column * K last key pressed * * Flags: * A - Controls big "=" sign * 00 - Shift state * * Fixed labels: * A, 11 - Start over with new matrix ('A'gain) * B, 12 - Compute the inverse ('1/x') * C - Continue with current matrix ('C'ontinue) * 14, D - Compute the determinant ('D'eterminant) * 22 - [RCL] recalls current matrix element * 24 - [f] toggles navigation from horizontal to vertical * 31 - [ENTER^] (same as recall) * 35 - [<-] Backspace to exit the editor * 51 - [^] Navigation up/left * 61 - [v] Navigation down/right */
HP-15C
Owner's Handbook
Section 12: Calculating With Matrices
Calculating With Complex Matrices
Inverting a Complex Matrix
Example: (pp. 165)
$Z = \begin{bmatrix} 4+3i & 7-2i \\ 1+5i & 3+8i \end{bmatrix}$
$Z^P = \begin{bmatrix}4 & 7 \\ 1 & 3 \\ 3 & -2 \\ 5 & 8\end{bmatrix}$
Enter matrix $$Z^P$$:
0.0402
XEQ 'MED'
4 R/S
↓ 7 R/S
↓ 1 R/S
(…)
↓ 8 R/S
Calculate offset for element (1,3):
1
3
0.0802
MATRIX:M.REG
4
Calculate offset for element (1,5):
1
5
0.0802
MATRIX:M.REG
8
Copy complex 2×2 matrix:
4.0202
8
MATRIX:M.COPY
8.0202
Negate 2×2 matrix (-x = x - 2x):
-2
4.0202
8.0202
MATRIX:M+x
Calculate offset for element (1,1):
1
1
0.0402
MATRIX:M.REG
0
Calculate offset for element (1,7):
1
7
0.0802
MATRIX:M.REG
12
Copy real 2×2 matrix:
0.0202
12
MATRIX:M.COPY
12.0202
Transpose upper 4×2 matrix:
0.0402
MATRIX:TRANSP
0.0204
Transpose lower 4×2 matrix:
8.0402
MATRIX:TRANSP
8.0202
Transpose 4×4 matrix:
0.0404
MATRIX:TRANSP
0.0404
$\tilde{Z}=\begin{bmatrix} 4 & 7 & -3 & 2 \\ 1 & 3 & -5 & -8 \\ 3 & -2 & 4 & 7 \\ 5 & 8 & 1 & 3 \end{bmatrix}$
Calculate the inverse:
0.0404
MATRIX:M-1
$\tilde{Z}^{-1}=\begin{bmatrix} -0.0254 & 0.2420 & 0.2829 & 0.0022 \\ -0.0122 & -0.1017 & -0.1691 & 0.1315 \\ -0.2829 & -0.0022 & -0.0254 & 0.2420 \\ 0.1691 & -0.1315 & -0.0122 & -0.1017 \end{bmatrix}$
Solving the Complex Equation AX = B
Example: (pp. 169)
$A^P=\begin{bmatrix} 10 & 0 \\ 0 & 0 \\ 200 & -200 \\ -200 & 170 \end{bmatrix}$
$B^P=\begin{bmatrix} 5 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
Calculate inverse of AP as above and multiply by BP:
0.0404
16.0401
20
MATRIX:M×
20.0401
$\tilde{A}^{-1}\times B^P=\begin{bmatrix} 0.0372 \\ 0.0437 \\ 0.1311 \\ 0.1543 \end{bmatrix}$
Transpose 2×2 matrix:
20.0202
MATRIX:TRANSP
$\begin{bmatrix} 0.0372 & 0.1311 \\ 0.0437 & 0.1543 \end{bmatrix}$
HP-42S
RPN Scientific
Programming Examples and Techniques
5: Matrices
Solving Simultaneous Equations That Have Complex Terms
Example: (pp. 166)
$A=\begin{bmatrix} 10-i0.01 & -5 & 0 & 0 \\ -5 & 15-i0.01 & -5 & 0 \\ 0 & -5 & 15-i0.01 & -5 \\ 0 & 0 & -5 & 15-i0.01 \end{bmatrix}$
$B=\begin{bmatrix} 34 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
$A^P=\begin{bmatrix} 10 & -5 & 0 & 0 \\ -5 & 15 & -5 & 0 \\ 0 & -5 & 15 & -5 \\ 0 & 0 & -5 & 15 \\ -0.01 & 0 & 0 & 0 \\ 0 & -0.01 & 0 & 0 \\ 0 & 0 & -0.01 & 0 \\ 0 & 0 & 0 & -0.01 \end{bmatrix}$
$B^P=\begin{bmatrix} 34 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
Let's assume we want to preserve registers 0-4. Thus we start with register 5 for matrix AP.
Enter matrix AP:
P.FCN:CLREGS
5.0804
XEQ 'MED'
Calculate offset for element (1,5):
1
5
5.1604
MATRIX:M.REG
21
Calculate offset for element (1,9):
1
9
5.1604
MATRIX:M.REG
37
Copy complex 4×4 matrix:
21.0404
37
MATRIX:M.COPY
37.0404
Negate 4×4 matrix (-x = x - 2x):
-2
21.0404
37.0404
MATRIX:M+x
Calculate offset for element (1,1):
1
1
5.1604
MATRIX:M.REG
5
Calculate offset for element (1,13):
1
13
5.1604
MATRIX:M.REG
53
Copy real 4×4 matrix:
5.0404
53
MATRIX:M.COPY
53.0404
Transpose upper 8×4 matrix:
5.0804
MATRIX:TRANSP
5.0408
Transpose lower 8×4 matrix:
37.0804
MATRIX:TRANSP
37.0408
Transpose 8×8 matrix:
5.0808
MATRIX:TRANSP
5.0808
Enter matrix BP:
70.0801
XEQ 'MED'
Invert matrix A:
5.0808
MATRIX:M-1
Multiply A by B:
5.0808
70.0801
80
MATRIX:M×
80.0801
Transpose 2×4 matrix:
80.0204
MATRIX:TRANSP
80.0402
Result:
$I=\begin{bmatrix} 4.2000 & 0.0061 \\ 1.600 & 0.0037 \\ 0.6000 & 0.0019 \\ 0.2000 & 0.0008 \end{bmatrix}$
« Next Oldest | Next Newest »
User(s) browsing this thread: 1 Guest(s) | 2,547 | 5,854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.108485 |
https://www.physicsforums.com/threads/about-the-invariance-of-similar-linear-operators-and-their-minimal-polynomial.321005/ | 1,532,211,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592778.82/warc/CC-MAIN-20180721203722-20180721223722-00352.warc.gz | 969,862,313 | 13,849 | # About the invariance of similar linear operators and their minimal polynomial
1. Jun 20, 2009
### sanctifier
About the invariance of similar linear operators and their minimal polynomial
Notations:
F denotes a field
V denotes a vector space over F
L(V) denotes a vector space whose members are linear operators from V to V itself and its field is F, then L(V) is an algebra where multiplication is composition of functions.
τ, σ, φ denote distinct linear operators contained in L(V)
m?(x) denotes the minimal polynomial of the linear operator "?"
～ denotes the similarity of the left and right operand, e.g., if τ ～ σ, then τ = φσφ-1
Question:
If τ ～ σ are similar linear operators on V, then mτ(x) = mσ(x), i.e., the minimal polynomial is an invariant under similarity of operators.
I wonder how it's proved.
Thanks for any help!
2. Jun 20, 2009
### trambolin
Use dummy variables and factor them wisely, here is a taste
$\alpha = \alpha\varphi\varphi^{-1}, \alpha\in\mathbb{F},\varphi\in\mathbb{V}$
3. Jun 20, 2009
### sanctifier
Last edited by a moderator: Apr 24, 2017
4. Jun 21, 2009
### trambolin
Because I am too cool!
just kidding, it says alpha = alpha*phi*phi^(-1) and alpha in F and phi in V.
5. Jun 22, 2009
### sanctifier
I still don't understand.
For example, suppose mτ(x) = s1+s2τ+s3τ2 = 0 and τ = φσφ-1,
then mτ(x) = s1+s2φσφ-1+s3φ2σ2-1)2 =0,
multiply (φ2)-1 on the left and φ2 on the right of the both sides of mτ(x)
we will get s1+s2φ-1σφ+s3σ2 = 0,
this doesn't equal to s1+s2σ+s3σ2 = 0, does it?
Ah, yes, now I know where I made a mistake, τ2 ≠φ2σ2-1)2
since τ2 = (φσφ-1)2 = φσφ-1φσφ-1
= φσ2φ-1,
then multiply φ-1 on the left and φ on the right of of the both sides of the equation mτ(x) = s1+s2φσφ-1+s3φσ2φ-1=0
we will get s1+s2σ+s3σ2 = 0
The property that τ2 = φσ2φ-1 I've explored before reaching this theorem, it seems I need more practice. | 621 | 1,893 | {"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} | 3.578125 | 4 | CC-MAIN-2018-30 | latest | en | 0.860687 |
http://virtualnerd.com/algebra-2/equations-inequalities | 1,580,047,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00510.warc.gz | 170,214,970 | 24,955 | # Equations and Inequalities
### Popular Tutorials in Equations and Inequalities
#### What are the Multiplication Properties of 0 and -1?
Understanding the multiplication properties of 0 and -1 are fundamental building blocks in learning all there is to know about the operation of multiplication. In this tutorial, you'll learn about these two important properties.
#### What are the Inverse Properties of Addition and Multiplication?
Inverse properties of addition and multiplication got you stumped? This tutorial should help! Check it out and learn these two important inverse properties.
#### How Do You Find the Absolute Value of Positive and Negative Numbers?
Finding the absolute value of a number is a breeze when you use a number line! Remember, absolute value is the distance from zero on the number line. This tutorial shows you how to use a number line to find absolute value.
#### How Do You Solve a Two-Step Equation by Multiplying by -1?
Got a negative variable in an equation? Want to get rid of that negative sign? This tutorial shows you how by using the multiplication property of -1!
#### How Do You Solve a Two-Step Equation by Combining Like Terms?
Trying to solve an equation where you see the same variable more than once? Figure out how to get those variables together and solve the equation with this tutorial!
#### How Do You Solve for a Variable in a Fraction Using Multiplication First?
Trying to solve an equation involving a fraction? Just multiply the fraction away and then perform the order of operations in reverse! See how in this tutorial.
#### How Do You Solve a Two-Step Equation by Multiplying by a Reciprocal?
Trying to solve an equation involving a fraction? Just multiply the fraction away by multiplying by the reciprocal and then perform the order of operations in reverse! See how in this tutorial.
#### How Do You Solve an Identity Equation?
Trying to solve an equation with variables on both sides of the equation? Figure out how to get those variables together and solve the equation with this tutorial!
#### How Do You Solve an Equation with No Solution?
Trying to solve an equation with variables on both sides of the equation? Figure out how to get those variables together and solve the equation with this tutorial! Surprise! Turns out, this equation has no solution. Check out this tutorial and see why!
#### How Do You Solve an Equation with Variables on Both Sides and Grouping Symbols?
Trying to solve an equation with variables on both sides of the equal sign and grouping symbols? Watch this tutorial to figure out how remove the grouping symbols and get those variables together to solve the equation.
#### How Do You Solve a Two-Step Equation by Multiplying by -1 First?
Got a negative variable in an equation? Want to get rid of that negative sign? This tutorial shows you how by using the multiplication property of -1!
#### How Do You Solve a Two-Step Equation by Distributing a Fraction First?
Trying to solve an equation for a variable? Are grouping symbols in the way? Watch this tutorial to figure out how remove the grouping symbols and solve the equation!
#### What is an Identity Equation?
Identity equations are equations that are true no matter what value is plugged in for the variable. If you simplify an identity equation, you'll ALWAYS get a true statement. Learn about identity equations in this tutorial, and then create your own identity equation. Get creative! The possibilities are endless!
#### What Does It Mean When An Equation Has No Solution?
Sometimes equations have no solution. This means that no matter what value is plugged in for the variable, you will ALWAYS get a contradiction. Watch this tutorial and learn what it takes for an equation to have no solution.
#### How Do You Solve a Formula For a Variable?
So you're working on a math problem and you have the correct formula. Great! But the variable you need to solve for is not by itself in the formula. Not so great. Don't worry! In this tutorial, you'll learn how to solve a formula for the variable you want!
#### How Do You Solve an OR Compound Inequality and Graph It On a Number Line?
Trying to solve a compound inequality? No problem! This tutorial will take you through the process of splitting the compound inequality into two inequalities. Then you'll see how to solve those inequalities, write the answer in set builder notation, and graph the solution on a number line.
#### How Do You Solve a Word Problem Using an AND Compound Inequality?
Word problems allow you to see math in action! This tutorial deals with turning a word problem into a compound inequality and solving that compound inequality to get the answer. Learn how in this tutorial!
#### What is an Identity Inequality?
And identity inequality is an inequality that it true no matter what values we plug in for the variable. Watch this tutorial and learn all about identity inequalities. Then see if you can make your own identity inequality complete with variables!
#### What Does It Mean When an Inequality is a Contradiction or Has No Solution?
Can an inequality have no solution? You bet it can! An inequality has no solution will always give you a contradiction, no matter what value you plug in for the variable. Watch this tutorial, and then try to make your own inequality with no solution!
#### What are Multiplicative Inverses?
Multiplicative inverses. That's a mouthful! Really, this term just refers to numbers that when multiplied together equal 1. These numbers are also called reciprocals of each other! Learn about multiplicative inverses by watching this tutorial.
#### How Do You Solve a Word Problem by Writing an Equation and Plugging in Values?
Setting up and solving an equation from a word problem can be tricky, but this tutorial can help. See all the steps, from defining variables to getting that final answer, and everything in between! With this tutorial, you'll learn what it takes to solve a word problem.
#### What are the Commutative Properties of Addition and Multiplication?
The commutative property is a fundamental building block of math, but it only works for addition and multiplication. This tutorial defines the commutative property and provides examples of how to use it.
#### What are the Identity Properties of Addition and Multiplication?
Let's identify an identity! Addition and subtraction have a different identity than multiplication and division. Learn about each of these identities with this tutorial!
#### How Do You Translate a Phrase with One Variable and Number Into a Mathematical Expression?
Need some practice translating phrases into mathematical expressions? Then this tutorial is for you! You'll get practice translating statements involving addition, subtraction, multiplication, or division into mathematical expressions.
#### What are Natural and Whole Numbers?
The natural numbers are also called the counting numbers. The whole numbers are almost exactly as the natural numbers the same except for one small difference! Watch this tutorial and learn the difference between natural and whole numbers.
#### How Do Different Categories of Numbers Compare To Each Other?
There are a bunch of different categories of numbers such as the rational numbers, the natural numbers, and the integers, just to name a few. See how they all relate to one another by watching this tutorial!
#### How Do You Write an Equation from a Word Problem?
Having difficulty turning a word problem into an algebra equation? Then this tutorial is for you! With this tutorial, you'll learn how to break down word problems and translate them into mathematical equations.
#### How Do You Simplify an Expression?
Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial.
#### How Do You Use the Distributive Property to Simplify an Expression?
In this tutorial you'll see how to apply the distributive property. Remember that this is important when you are trying to simplify an expression and get rid of parentheses!
#### How Do You Use the Associative Property?
Learning the associative property of addition is one thing, but being able to use it can be a completely different challenge. In this tutorial, you'll see how to put this very useful property into action.
#### How Do You Evaluate an Algebraic Expression?
Plugging variables into an expression is essential for solving many algebra problems. See how to plug in variable values by watching this tutorial.
#### How Do You Use the Order of Operations?
If you're trying to simplify some math expressions, you have to do it in the right order. If you've ever wondered how to do that, check out this example tutorial where you'll see exactly what order you need to follow:)
#### What is a Variable?
You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!
#### What Are Some Words You Can Use To Write Word Problems?
Knowing the mathematical meaning of words allows you to decipher word problems and gives you the power to write your own word problems, too! Take a look at these words and learn their mathematical translations.
#### What are Like Terms?
Combining like terms together is a key part of simplifying mathematical expressions, so check out this tutorial to see how you can easily pick out like terms from an expression
#### What is Simplest Form?
Sometimes what's simple for some is hard for others, and so simple seems like a pretty subjective word. But when it comes to algebra, the simplest form of an expressions is very well defined and is very specific- there's really nothing that subjective about it. So watch this tutorial and see exactly what simplest form means!
#### What is the Distributive Property?
The distributive property is a very deep math principle that helps make math work. It's the rule that lets you expand parentheses, and so it's really critical to understand if you want to get good at simplifying expressions. So check out the tutorial and let us know what you think!
#### What are the Associative Properties of Addition and Multiplication?
The associative property is one of those fundamental properties of math that make math work. You probably take this property for granted because it's so ingrained, but it's important to see how the guts of math work, so check out the tutorial and make sure you're solid on your fundamentals!
#### What's the Order of Operations?
Check out this tutorial where you'll see exactly what order you need to follow when you simplify expressions. You'll also see what happens when you don't follow these rules, and you'll find out why order of operations is so important!
#### What's a Coefficient?
There is a bunch of vocabulary that you just need to know when it comes to algebra, and coefficient is one of the key words that you have to feel 100% comfortable with. Check out the tutorial and let us know if you want to learn more about coefficients!
#### How Do You Turn a Longer Verbal Phrase into an Algebraic Expression?
Looking for some practice translating words into a mathematical expression? Then take a look at this tutorial, and you'll learn how to break down those words and write the mathematical expression they describe!
#### How Do You Turn a More Complicated Equation into Words?
Let your creativity go wild by turning a mathematical equation into a story! Watch this tutorial to see one example, and then write your own story to describe the given equation. See how creative you can get with math!
#### How Do You Turn a More Complicated Expression into Words?
This tutorial shows you how a mathematical expression can be turned into words. The best part? You can be super creative with the statement you make up! Check out the example statement created in this tutorial, and then make up your own!
#### What's an Irrational Number?
Is an irrational number just a way to describe a number that's lost its mind? Not really. :) An irrational number is simply a number that cannot be written as a fraction. Check out the definition, learn an important property of these special numbers, and take a look at some examples of irrational numbers.
#### What's a Real Number?
Real numbers are numbers that can be found on the number line. This includes both the rational and irrational numbers. This tutorial explains real numbers and gives some great examples. Take a look!
#### What's an Inequality?
Inequalities come up all the time when you're working algebra problems. In this tutorial you'll learn what an inequality is, and you'll see all the common inequality symbols that you're likely to see :)
#### What Does Absolute Value Mean?
In this tutorial you'll see how you can think of absolute value in a very intuitive way. Let us know if you have any questions about it!
#### How Do You Graph a Set on a Number Line?
Number lines are great ways to represent a group of numbers, and in this tutorial you'll see how to graph a group of numbers on a number line
#### What's a Rational Number?
There are lots of different kind of numbers that you should know about, and that includes rational numbers. Check out the tutorial!
#### What's an Integer?
There are lots of different kinds of numbers that you'll come across in algebra, and a lot of these kinds of numbers are related to each other. Before you learn how they are related, you've got to learn about them separately, and in this tutorial you'll how to define integers :)
#### What's a Reciprocal?
Reciprocals are important when it comes to dividing fractions, finding perpendicular lines, dealing with inverse proportions, and so much more! In this tutorial you can review the basics about reciprocals.
#### How Do You Graph the Intersection of Two Inequalities?
Number lines help make graphing the intersection of two inequalities a breeze! This tutorial shows you how to graph two inequalities on the same number line and then find the intersection. Take a look!
#### How Do You Graph the Union of Two Inequalities?
Number lines help make graphing the union of two inequalities a breeze! This tutorial shows you how to graph two inequalities on the same number line and then find the union. Check it out!
#### How Do You Distribute With Whole Numbers and Fractions?
This tutorial shows you how to distribute a whole number into the sum of fractions. It's an important skill to have when you're solving equations, and you never know when it can come up. Be sure to check out this tutorial!
#### What's the Subtraction Property of Equality?
Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. One of those tools is the subtraction property of equality, and it lets you subtract the same number from both sides of an equation. Watch the video to see it in action!
#### What's the Addition Property of Equality?
Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. One of those tools is the addition property of equality, and it lets you add the same number to both sides of an equation. Watch the video to see it in action!
#### What's the Multiplication Property of Equality?
Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. One of those tools is the multiplication property of equality, and it lets you multiply both sides of an equation by the same number. Watch the video to see it in action!
#### How Do You Solve an Equation with Variables on Both Sides and Fractions?
Trying to solve an equation with variables and fractions on both sides of the equation? You can bet it involves finding a common denominator! To see what it takes, watch this tutorial.
#### How Do You Solve an Equation with Variables on Both Sides?
Trying to solve an equation with variables on both sides of the equal sign? Figure out how to get those variables together and find the answer with this tutorial!
#### How Do You Solve for a Variable in Terms of Another Variable?
Got an equation with two variables? Want to solve for one variable in terms of the other? Want to go the other way around? See how in this tutorial!
#### How Do You Solve a Word Problem Using a Multi-Step Equation?
Working with word problems AND fractions? This tutorial shows you how to take a word problem and translate it into a mathematical equation involving fractions. Then, you'll see how to solve and check your answer. Take a look!
#### How Do You Solve a Multi-Step Equation Using Order of Operations in Reverse?
Trying to solve an equation with variables and fractions? Just perform the order of operations in reverse! To see what it takes, watch this tutorial.
#### How Do You Solve a Word Problem Using a Two-Step Equation with Decimals?
Word problems are a great way to see math in the real world. In this tutorial, you'll see how to translate a word problem into a mathematical equation. Then you'll see how to solve that equation and check your answer!
#### How Do You Solve a Multi-Step Equation with Fractions by Multiplying Away the Fraction?
Trying to solve an equation involving a fraction? Just multiply the fraction away and then perform the order of operations in reverse! See how in this tutorial.
#### How Do You Solve a Multi-Step Equation with Fractions Using Reverse Order of Operations and Reciprocals?
Trying to solve an equation involving a fraction? Just perform the order of operations in reverse! See how in this tutorial.
#### How Do You Solve a Two-Step Equation?
Solving an equation for a variable? Perform the order of operations in reverse! Check it out in this tutorial.
#### How Do You Solve a Multi-Step Equation Using the Distributive Property?
Trying to solve an equation where you see the same variable more than once? Figure out how to get those variables together and solve the equation with this tutorial!
#### What's the Division Property of Equality?
Solving equations can be tough, especially if you've forgotten or have trouble understanding the tools at your disposal. One of those tools is the division property of equality, and it lets you divide both sides of an equation by the same number. Watch the video to see it in action!
#### How Do You Solve a Word Problem Using an AND Absolute Value Inequality?
Word problems allow you to see math in action! This tutorial shows you how to translate a word problem to an absolute value inequality. Then see how to solve for the answer, write it in set builder notation, and graph it on a number line. Learn all about it in this tutorial!
#### How Do You Write an Absolute Value Inequality from a Word Problem?
Sometimes the hardest part of a word problem is figuring out how to turn the words into math. This tutorial let's you see the steps to take in order to do just that! You'll see how to take a word problem and dissect it to turn it into an absolute value inequality.
#### How Do You Solve a Word Problem Using a Multi-Step Inequality?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Multi-Step Inequality Using Reverse Order of Operations?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Word Problem Using an Inequality With Variables on Both Sides?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality With Variables on Both Sides?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an AND Compound Inequality?
Trying to solve a compound inequality? No problem! This tutorial will take you through the process of splitting the compound inequality into two inequalities. Then you'll see how to solve those inequalities and write the answer in set builder notation.
#### How Do You Solve an AND Absolute Value Inequality and Graph It On a Number Line?
Trying to solve an absolute value inequality? No sweat! This tutorial will take you through the process of solving the inequality. Then you'll see how to write the answer in set builder notation and graph it on a number line. You'll see it all in this tutorial!
#### How Do You Convert an AND Absolute Value Inequality to a Compound Inequality?
Converting an absolute value inequality to a compound inequality can be tricky. Luckily, this tutorial takes you through the process step-by-step. Take a look! You'll be glad you did.
#### How Do You Solve an AND Compound Inequality and Graph It On a Number Line?
Trying to solve a compound inequality? No sweat! This tutorial will take you through the process of solving the inequality. Then you'll see how to write the answer in set builder notation and graph it on a number line. You'll see it all in this tutorial!
#### How Do You Solve and Graph a Two-Step Inequality?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Write Inequalities in Set Builder Notation?
Need some extra practice converting solution phrases into set builder notation? This tutorial was made for you! Follow along as this tutorial shows you how to dissect each phrase and turn it into a solution in set builder notation.
#### How Do You Figure Out if an Absolute Value Inequality is an AND or OR Compound Inequality?
When does an absolute value inequality represent an AND compound inequality? When does it represent an OR compound inequality? This tutorial will give you the answer and show you using examples on a number line!
#### What's a Compound Inequality?
Wondering what a compound inequality is? Then check out this tutorial! You'll learn the definition for a compound inequality and also see how it can be written in set builder notation. Take a look!
#### What's the Subtraction Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the subtraction property of inequality, and it basically says that if you minus a number from one side of an inequality, you have to minus that same number from the other side of the inequality as well. Watch the tutorial to see how this looks in terms of algebra!
#### What's the Addition Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the addition property of inequality, and it basically says that if you add a number from one side of an inequality, you have to add that same number from the other side of the inequality as well. Watch the tutorial to see how this looks in terms of algebra!
#### What's the Division Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the tutorial to see how this looks in terms of algebra!
#### What's the Multiplication Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the multiplication property of inequality, and it basically says that if you multiply one side of an inequality by a number, you can multiply the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the tutorial to see how this looks in terms of algebra!
#### What is a Constant?
Constants are parts of algebraic expressions that don't change. Check out this tutorial to see exactly what a constant looks like and why it doesn't change.
#### What's Another Definition for an Integer?
Integers are everywhere in math, so it's important to know what an integer is. This tutorial explains this special type of number so that you can identify one when you see it! Sometimes, a number is an integer even though it doesn't look like one. Watch this tutorial to see how to identify those too!
#### What's a Term?
Polynomials are those expressions that have variables raised to all sorts of powers and multiplied by all types of numbers. When you work with polynomials you need to know a bit of vocabulary, and one of the words you need to feel comfortable with is 'term'. So check out this tutorial, where you'll learn exactly what a 'term' in a polynomial is all about.
#### What are Equivalent Expressions?
If two things are equivalent, they are the same. Equivalent expressions are expressions that are the same, even though they may look a little different. If you plug in the same variable value into equivalent expressions, they will each give you the same value when you simplify. Learn more about equivalent expression by watching this tutorial!
#### What Are Numerical and Algebraic Expressions?
An expression is just a mathematical phrase. In this tutorial, you'll learn about two popular types of expressions: numerical and algebraic expressions. A numerical expression contains numbers and operations. An algebraic expression is almost exactly the same except it also contains variables. Check out this tutorial to learn about these two popular kinds of expressions!
#### What is the Substitution Property of Equality?
If you ever plug a value in for a variable into an expression or equation, you're using the Substitution Property of Equality. This property allows you to substitute quantities for each other into an expression as long as those quantities are equal. Watch this tutorial to learn about this useful property!
#### What is the Opposite, or Additive Inverse, of a Number?
Have you ever combined two numbers together and found their sum to be zero? When that happens, those numbers are called additive inverses of each other! In this tutorial, you'll learn the definition for additive inverse and see examples of how to find the additive inverse of a given value.
#### How Do You Put Real Numbers in Order?
Ordering numbers from least to greatest? Are the numbers in different forms? To make comparing easier, convert all the numbers to decimals. Then, plot those decimals on a number line and compare them! This tutorial shows you how! | 5,516 | 27,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-05 | latest | en | 0.892679 |
https://becalculator.com/2-miles-to-steps-converter.html | 1,716,919,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00468.warc.gz | 99,637,598 | 15,968 | # 2 miles to steps converter
## FAQs on 2 miles to steps
### How many steps per mile?
A very easy to remember conversion percentage allows you to quickly convert miles into steps.
1 mile = 2112 steps.
### How to convert a miles into steps?
1 mile into steps = 1 times 2112 = 2112 steps.
This number will provide additional details.
• How to calculate steps per mile?
• 1 mile equal how many steps?
• How much are 1 mile to steps?
• What is steps conversion to miles?
### Meaning of Mile
The one of the imperial unit of measurement for length is the mile. It is also the length unit in the American customary system of measurement. It’s used to measure distance. 1 mile equals 5280 feet. 1 mile is also equivalent to 63360 inches.
### Definition of Step
Normal people walk 0.8 meters in one step. But the stride will vary from person to person. Yet, we are able to arbitrarily determine the distance using the number of steps.
### How much are 2 miles to miles?
This formula will show you how to convert miles into steps.
Value in steps = value in miles × 2112
So, 2 miles into steps = 2 miles × 2112 = 4224 steps.
See some related questions about 2 miles to steps.
• How many steps in 2 miles?
• How to calculate steps into miles?
• What number of 2 miles to steps?
• What is miles in steps converter?
• What is miles in steps fomula?
miles steps 1.8 miles 3801.6 steps 1.825 miles 3854.4 steps 1.85 miles 3907.2 steps 1.875 miles 3960 steps 1.9 miles 4012.8 steps 1.925 miles 4065.6 steps 1.95 miles 4118.4 steps 1.975 miles 4171.2 steps 2 miles 4224 steps 2.025 miles 4276.8 steps 2.05 miles 4329.6 steps 2.075 miles 4382.4 steps 2.1 miles 4435.2 steps 2.125 miles 4488 steps 2.15 miles 4540.8 steps 2.175 miles 4593.6 steps 2.2 miles 4646.4 steps
Deprecated: Function get_page_by_title is deprecated since version 6.2.0! Use WP_Query instead. in /home/nginx/domains/becalculator.com/public/wp-includes/functions.php on line 5413 | 558 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-22 | latest | en | 0.876694 |
https://www.doubtnut.com/question-answer/dikhaaie-ki-phln-fxsinx-xydix-ne-01ydix0-x0-pr-stt-nhiin-hai-320053643 | 1,627,098,925,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00373.warc.gz | 744,238,020 | 71,651 | Home
>
Hindi
>
Class 12
>
Maths
>
Chapter
>
सततता तथा अवकलनीयता
>
दिखाइए कि फलन f(x)={{:((|sinx...
# दिखाइए कि फलन f(x)={{:((|sinx|)/(x),,यदि,,x ne 0),(1,,यदि,,x=0):} x=0 पर सतत नहीं है
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 10-9-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
15.3 K+
700+
Image Solution
127303833
800+
17.6 K+
4:49
234316297
10.4 K+
51.5 K+
2:06
320053643
700+
15.3 K+
4:42
92139318
300+
6.4 K+
2:10
92139256
1.1 K+
21.8 K+
1:29
92139317
700+
15.5 K+
2:45
358825915
100+
4.3 K+
112159821
500+
10.9 K+
2:49
146128716
2.4 K+
48.6 K+
3:47
320053636
500+
11.7 K+
92139162
67.8 K+
122.5 K+
2:12
92139294
600+
13.6 K+
2:34
92139234
800+
16.6 K+
1:45
320053556
800+
16.6 K+
4:03
112159800
14.8 K+
47.9 K+
6:31 | 405 | 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} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.273827 |
https://www.physicsforums.com/threads/things-vs-coordinates-of-things.788631/ | 1,708,800,742,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00329.warc.gz | 957,829,912 | 16,041 | # Things vs Coordinates-of-Things
• Stephen Tashi
In summary, the distinction between a mathematical object and its coordinates is not typically introduced in elementary mathematics, causing confusion when it comes to understanding angles. However, teaching angles in a way that incorporates both equivalence relations and coordinate systems could provide a logical and consistent approach. This approach would allow for different numbers, such as 0 and 360, to represent the same angle without contradicting their individual values.
#### Stephen Tashi
In advanced mathematics, one must eventually learn the distinction between a mathematical object and some coordinates of that object. For example, eventually students are supposed to understand that a "vector" is not an n-tuple of numbers.
I wonder why this distinction is not introduced in elementary mathematics when students are taught about angles. Perhaps it wouldn't be simple!
Teaching students that numbers like 0 degrees and 360 degrees are coordinates of angles instead of being angles would relieve the teacher of having to double-talk about them being "different, but really the same" angle. Yet there are situations when zero degrees and 360 degrees denote different things. For example, a moving object making a "turn of 360 degrees" is different than its making a "turn of zero degrees".
Perhaps teaching angles in a way that made sense would involve teaching both equivalance relations and coordinate systems.
It's not clear what 'coordinates of angles' means.
SteamKing said:
It's not clear what 'coordinates of angles' means.
Coordinate systems are permitted to be redundant. In some coordinate systems, the same thing can be represented by different coordinates. The numbers 0 and 360 are obviously different numbers. If you want to talk about them representing "the same angle" in a logically consistent manner then you have to do it without contradicting the fact that 0 and 360 are different numbers. Considering values in degrees to be a method of assigning coordinates to an angle would be one way of doing this. There might be others.
## 1. What is the difference between "things" and "coordinates-of-things"?
"Things" refer to physical objects or entities, while "coordinates-of-things" refer to the specific locations or positions of those objects in space.
## 2. How are "things" and "coordinates-of-things" related?
"Coordinates-of-things" are used to describe the location or position of "things" in space. They are closely related as one cannot exist without the other.
## 3. Can "things" and "coordinates-of-things" be used interchangeably?
No, they cannot be used interchangeably. While "things" refer to physical objects, "coordinates-of-things" refer to the specific locations or positions of those objects.
## 4. What are some examples of "things" and their corresponding "coordinates-of-things"?
An example of a "thing" could be a tree, and its "coordinates-of-things" would be its exact latitude and longitude coordinates on a map. Another example could be a car and its "coordinates-of-things" would be its position on a road relative to other objects.
## 5. Why is it important to understand the difference between "things" and "coordinates-of-things"?
Understanding the difference between "things" and "coordinates-of-things" is crucial for accurately describing and locating objects in space. It is also important for various scientific fields such as geography, astronomy, and physics where precise location information is necessary for research and analysis. | 711 | 3,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.944933 |
https://www.weegy.com/?ConversationId=7A16914E | 1,553,516,744,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00498.warc.gz | 947,417,652 | 9,612 | Simplify the expression Show your work 9xy^2 – 11xy^2
9xy^2-11xy^2 =-2xy^2
Question
Updated 1/30/2018 4:34:07 AM
Edited by jeifunk [1/29/2018 10:22:48 AM], Edited by jeifunk [1/30/2018 4:34:03 AM], Edited by jeifunk [1/30/2018 4:34:06 AM], Confirmed by jeifunk [1/30/2018 4:34:07 AM]
Rating
Questions asked by the same visitor
By 1750, Prussia and Austria A. were competing to develop their overseas empires. B. were battling for control of the German states. C. had taken major steps toward constitutional government. D. had agreed to work together against their chief foe, Russia.
Weegy: d. had agreed to work together against their chief foe, Russia. (More)
Question
Updated 3/13/2017 8:24:10 AM
By 1750, Prussia and Austria were battling for control of the German states.
Confirmed by jeifunk [3/13/2017 8:30:34 AM]
What is the slope of the line shown? (1,6) and (-6,-3)
Weegy: The answer is -9 .Thanks! (More)
Question
Updated 6/22/2016 4:46:29 PM
The slope of the line that contains the points (1,6) and (-6,-3) is 9/7.
m = (y2-y1)/(x2-x1)
m = (-3 - 6)/(-6 - 1)
m = -9/-7
m = 9/7
What is the graph of y = -3/4x – 2? A. (0,-2) and (-4,-5) B. (0,-2) and (-4,1) C. (0,-2) and (3,-6) D. (0,2) and (-4,5)
Question
Updated 4/22/2014 8:28:23 PM
The points on the graph of y = -3/4x – 2 is: B. (0,-2) and (-4,1).
Confirmed by jeifunk [4/22/2014 8:34:08 PM]
-5-7
Weegy: (-5) - 7 = -12 (More)
Question
Updated 12/29/2017 12:38:01 PM
4/9=m/54
Weegy: 4/9=m/54; m=4/9*54; m=216/9; m=24 (More)
Question
Updated 4/12/2014 10:06:38 AM
29,541,972
*
Get answers from Weegy and a team of really smart live experts.
Popular Conversations
An underground
Weegy: An underground is an adverb which means beneath the surface of the earth.
Which state was admitted as a free state in 1820
Weegy: Missouri was admitted as a free state in 1820.
3+7(2-5)+4
Weegy: 3 + (7 * (2 - 5)) + 4 = -14
8. The sun is directly overhead at the equator on what day of the ...
Weegy: The sun is directly overhead at the equator on Solstice day of the year. User: The Moon is made up mostly of ...
A compact fluorescent light bulb uses more energy than an ...
Weegy: A compact fluorescent light bulb uses more energy than an incandescent light bulb. FALSE.
One varve equals one coarse summer layer and one fine winter layer. ...
Weegy: One varve equals one coarse summer layer and one fine winter layer. TRUE.
The __ Autocomplete list contains those functions that alphabetically ...
Weegy: The Formula AutoComplete list contains those functions that alphabetically match the letters you type following ...
Which NIMS structure develops, recommends, and executes public ...
Weegy: The National Incident Management System (NIMS) is a standardized approach to incident management developed by ...
S
R
L
Points 969 [Total 1069] Ratings 3 Comments 939 Invitations 0 Offline
S
L
P
C
P
C
Points 950 [Total 3751] Ratings 1 Comments 940 Invitations 0 Offline
S
L
1
Points 720 [Total 1356] Ratings 20 Comments 500 Invitations 2 Offline
S
L
Points 391 [Total 568] Ratings 3 Comments 321 Invitations 4 Offline
S
L
Points 326 [Total 342] Ratings 0 Comments 306 Invitations 2 Offline
S
L
Points 225 [Total 1822] Ratings 2 Comments 205 Invitations 0 Offline
S
L
Points 199 [Total 362] Ratings 1 Comments 189 Invitations 0 Offline
S
L
Points 133 [Total 294] Ratings 9 Comments 13 Invitations 3 Offline
S
L
P
C
Points 128 [Total 1396] Ratings 2 Comments 108 Invitations 0 Offline
S
L
Points 113 [Total 438] Ratings 1 Comments 83 Invitations 2 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,213 | 3,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-13 | latest | en | 0.940257 |
https://pe-cn.github.io/169/ | 1,603,915,091,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900860.51/warc/CC-MAIN-20201028191655-20201028221655-00436.warc.gz | 473,490,210 | 6,075 | 0%
# Problem 169
Exploring the number of different ways a number can be expressed as a sum of powers of 2
Define f(0)=1 and f(n) to be the number of different ways n can be expressed as a sum of integer powers of 2 using each power no more than twice.
For example, f(10)=5 since there are five different ways to express 10:
1 + 1 + 8
1 + 1 + 4 + 41 + 1 + 2 + 2 + 4
2 + 4 + 4
2 + 8
What is f(1025)?
1 + 1 + 8
1 + 1 + 4 + 41 + 1 + 2 + 2 + 4
2 + 4 + 4
2 + 8 | 180 | 461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-45 | latest | en | 0.819493 |
https://www.essayfountain.com/excel-assig-2/ | 1,624,362,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00115.warc.gz | 676,480,856 | 9,950 | # excel assig 2
Use the template to record and calculate the costs of the school materials you purchased for this quarter.
First, in the Title Page worksheet, enter the following information:
Type your name in cell B1.
Type in the course section you are in cell B2. (For example: CIS105111)
Type your professor’s name in cell B3.Type the title of the assignment in cell B4.
Type the title (“Excelling with Excel”) of the assignment in cell B4.
Rename the worksheet to “Overview.”
Change the font type for this information to Century Gothic.
Change the font size for this information to 14 point.
Search for the school supplies you purchased at Amazon, Office Depot, Staples, or some other store.
Find ten supplies that you purchased. You can include textbooks, lab code fees, and paper supplies such as notepads, planners, pens, pencils, highlighters, etc.
Go to Sheet 3. This will be where you add your list of school supplies.
Type a title in cell A1. (For example: “School Supply Costs” or “My School Purchases.”)
Format cell A1, so the font type is Calibri Light (Headings), the font size is 20-point, and the font color is one of your choosing.
Enter the following labels in the cells specified:
Cell A2: School Supplies
Cell B2: Quantity
CellC2: Price
Cell D2: Total Cost
Make the font size 16 point and bold these labels.
Column A: School Supplies
In cells A3 through A12, enter the names of the ten (10) school supplies that you purchased. (If you didn’t buy 10 supplies, list what you would have purchased or will purchase next term.)
Auto fit the contents of column A so you can see the name of each supply item name.
Format cells A3 through A12 so the text is 16-point Calibri Light (Headings) and the font color is orange.
Column B: Quantity
In column B enter the Quantity for each of the supplies. For example, will you need four (4) notebooks?
Set the width of column B to 11.5”.
Format cells B3 through B12 so the text is 16-point Calibri Light (Headings) and the font color is brown.
Column C: Price
Enter the price for each school supply item. For example, a single notebook costs .
Expand column C so you can see all of your costs.
Format cells C3 through C12 so the font is Arial, the font size is 14-point, and the font color is dark blue.
Column D: Total costs
Use >the multiplication function to determine the total cost for each item.
Hint: In D3, the formula would be (=B3*C3).
Fill in cells D4 through D12 to calculate the total cost of the other supplies.
Format cells D3 through D12 so the font is Arial Bold, the font size is 14-point, and the font color is Dark Blue.
Now, rename Sheet 3 to “School Supplies.”
Create a pie chart:
In the School Supplies worksheet, select the range “School Supplies” and “Total Costs.”
Insert a pie chart that displays the school supply items and the total cost for each item.
Select the pie chart view that shows the dollar amounts for each item.
Add a title to your chart so the title is above the chart (can be found in “Chart Formatting”).
Add data labels to the pie chart that shows the total price for each item.
Position the chart so it does not overlap with the data in column A, column B, and column C.
Review the assignment for accuracy.
Note: Delete the WK8_A2_Overview and WK8_A2_Instructions sheet before you submit this assignment for grading. (Left click on each sheet to delete.)
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee
On-demand options
• Writer’s samples
• Part-by-part delivery
• Overnight delivery
• Copies of used sources
Paper format
• 275 words per page
• 12 pt Arial/Times New Roman
• Double line spacing
• Any citation style (APA, MLA, Chicago/Turabian, Harvard)
# Our guarantees
Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.
### Money-back guarantee
You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.
### Zero-plagiarism guarantee
Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.
### Free-revision policy
Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.
Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.
### Fair-cooperation guarantee
By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.
## Calculate the price of your order
550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors: | 1,173 | 5,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.891489 |
https://calculat.io/en/number/fraction-simplified | 1,713,336,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00181.warc.gz | 139,559,269 | 23,692 | # Fraction Simplifier Calculator
## Fraction Simplifier Calculator
This calculator will help you to simplify any fraction to its simplest form. For example, it can help you find out what is 14/26 Simplified? (The answer is: 7/13). Enter the fraction (numerator and denominator) (e.g. '14/26') and hit the 'Simplify' button.
Simplifying a fraction, (e.g. 14/26) means reducing it to its simplest form. A fraction is considered simplified if its numerator and denominator have no common factors other than 1 (one). | 126 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.892247 |
https://docs.pennylane.ai/en/stable/code/api/pennylane.transforms.commute_controlled.html | 1,716,051,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00856.warc.gz | 183,092,248 | 14,010 | qml.transforms.commute_controlled¶
commute_controlled(tape, direction='right')[source]
Quantum transform to move commuting gates past control and target qubits of controlled operations.
Parameters
• tape (QNode or QuantumTape or Callable) – A quantum circuit.
• direction (str) – The direction in which to move single-qubit gates. Options are “right” (default), or “left”. Single-qubit gates will be pushed through controlled operations as far as possible in the specified direction.
Returns
The transformed circuit as described in qml.transform.
Return type
qnode (QNode) or quantum function (Callable) or tuple[List[QuantumTape], function]
Example
>>> dev = qml.device('default.qubit', wires=3)
You can apply the transform directly on QNode:
@partial(commute_controlled, direction="right")
@qml.qnode(device=dev)
def circuit(theta):
qml.CZ(wires=[0, 2])
qml.X(2)
qml.S(wires=0)
qml.CNOT(wires=[0, 1])
qml.Y(1)
qml.CRY(theta, wires=[0, 1])
qml.PhaseShift(theta/2, wires=0)
qml.Toffoli(wires=[0, 1, 2])
qml.T(wires=0)
qml.RZ(theta/2, wires=1)
return qml.expval(qml.Z(0))
>>> circuit(0.5)
0.9999999999999999
You can also apply it on quantum function.
def qfunc(theta):
qml.CZ(wires=[0, 2])
qml.X(2)
qml.S(wires=0)
qml.CNOT(wires=[0, 1])
qml.Y(1)
qml.CRY(theta, wires=[0, 1])
qml.PhaseShift(theta/2, wires=0)
qml.Toffoli(wires=[0, 1, 2])
qml.T(wires=0)
qml.RZ(theta/2, wires=1)
return qml.expval(qml.Z(0))
>>> qnode = qml.QNode(qfunc, dev)
>>> print(qml.draw(qnode)(0.5))
0: ─╭●──S─╭●────╭●─────────Rϕ(0.25)─╭●──T────────┤ <Z>
1: ─│─────╰X──Y─╰RY(0.50)───────────├●──RZ(0.25)─┤
2: ─╰Z──X───────────────────────────╰X───────────┤
Diagonal gates on either side of control qubits do not affect the outcome of controlled gates; thus we can push all the single-qubit gates on the first qubit together on the right (and fuse them if desired). Similarly, X gates commute with the target of CNOT and Toffoli (and PauliY with CRY). We can use the transform to push single-qubit gates as far as possible through the controlled operations:
>>> optimized_qfunc = commute_controlled(qfunc, direction="right")
>>> optimized_qnode = qml.QNode(optimized_qfunc, dev)
>>> print(qml.draw(optimized_qnode)(0.5))
0: ─╭●─╭●─╭●───────────╭●──S─────────Rϕ(0.25)──T─┤ <Z>
1: ─│──╰X─╰RY(0.50)──Y─├●──RZ(0.25)──────────────┤
2: ─╰Z─────────────────╰X──X─────────────────────┤ | 794 | 2,377 | {"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} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.621812 |
http://www.jiskha.com/display.cgi?id=1290316608 | 1,493,284,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122041.70/warc/CC-MAIN-20170423031202-00005-ip-10-145-167-34.ec2.internal.warc.gz | 583,440,379 | 3,583 | # physics HELP
posted by on .
a 18.0kg sled being pulled across horizontal surface at constant velocity. pulling force has magnitude of 82.0N and is directed at angle of 30 degree above horizontal. determine coefficient of kinetic friction
• physics HELP - ,
Uk = (friction force)/(normal force)
at constant velocity
The friction force is the horizontal component of the twowing force.
Uk = (82 cos30)/(Mg - 82 sin 30)
= 71.1/(176.4 - 41.0) = 0.525 | 123 | 454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.873169 |
https://ccm.net/forum/affich-852639-conditional-formatting-of-one-based-on-the-other-cell | 1,725,728,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00512.warc.gz | 142,908,981 | 26,914 | # Conditional Formatting of one based on the other cell.
Closed
Dimi - Nov 3, 2015 at 08:53 AM
TrowaD Posts 2921 Registration date Sunday September 12, 2010 Status Moderator Last seen December 27, 2022 - Nov 5, 2015 at 10:38 AM
Hello,
I was wondering if there is a way I can say to excel without coding:
1)IF cell A1 is equal or greater [equal or less] than cell B1, then color cell A1 RED, otherwise no format.
2) Also, is it possible to do double formatting --> If cell A1 is equal or greater than cell B1, then color cell A1 RED, otherwise If cell A1 is equal or less than cell B1, color cell a1 GREEN, otherwise no format.
I would much appreciate your help!
## 2 responses
TrowaD Posts 2921 Registration date Sunday September 12, 2010 Status Moderator Last seen December 27, 2022 552
Nov 3, 2015 at 11:21 AM
Hi Dimi,
"equal or greater [equal or less]" uhm ..... which one is it.
Also both statements use equal, this doesn't mean that when A1 and B1 are the same that the cell will be half red and half green (would be cool though). If the first rule is true, then CF doesn't look at the second rule.
Formula's used:
=A1>=B1 ; --> red
=A1<B1 ; --> green
Best regards,
Trowa
Monday, Tuesday and Thursday are usually the days I'll respond. Bear this in mind when awaiting a reply.
In essence this is what I want --> I have three cells A1, B1 and C1. If A1 value (which constatly changes automatically) is equal or greater than B1, I want a red background in cell A1. On the other hand, if A1 value is equal or less than C1, I want a green background in cell A1.
Note: B1 and C1 values will never intersect (B1 is always higher number than C1).
Oh yes and of course, if A1 value is inbetween B1 and C1 I want it to have no background fill.
TrowaD Posts 2921 Registration date Sunday September 12, 2010 Status Moderator Last seen December 27, 2022 552
Nov 5, 2015 at 10:38 AM
Hi Dimi,
That makes more sense, but the formula's won't change that much.
Formula's:
=A1>=B1 ; --> red
=A1<=C1 ; --> green
Best regards,
Trowa | 581 | 2,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.899613 |
https://homework.cpm.org/category/CON_FOUND/textbook/ac/chapter/7/lesson/7.1.1/problem/7-9 | 1,726,528,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00516.warc.gz | 277,999,348 | 15,033 | ### Home > AC > Chapter 7 > Lesson 7.1.1 > Problem7-9
7-9.
Little Evan has $356$ stuffed animals, all of which are either teddy bears or dogs. He has $17$ more than twice as many dogs as teddy bears. How many teddy bears does he own? Write and solve an equation (or a system of equations) to solve this problem. Be sure to define your variable(s).
One equation should compare the number of dogs and teddy bears total and the other equation should compare the number of dogs compared to teddy bears.
$d = \text{dogs}$
$t = \text{teddy bears}$ | 143 | 545 | {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.946204 |
https://www.numbersaplenty.com/1175559 | 1,685,625,260,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00711.warc.gz | 1,048,315,895 | 3,195 | Search a number
1175559 = 37211727
BaseRepresentation
bin100011111000000000111
32012201120020
410133000013
5300104214
641110223
712664200
oct4370007
92181506
101175559
11733240
12488373
133220c8
142285a7
151834a9
hex11f007
1175559 has 24 divisors (see below), whose sum is σ = 1991808. Its totient is φ = 609840.
The previous prime is 1175521. The next prime is 1175561. The reversal of 1175559 is 9555711.
It is not a de Polignac number, because 1175559 - 211 = 1173511 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (33).
It is an Ulam number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1175509) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 1254 + ... + 1980.
It is an arithmetic number, because the mean of its divisors is an integer number (82992).
21175559 is an apocalyptic number.
1175559 is a deficient number, since it is larger than the sum of its proper divisors (816249).
1175559 is a wasteful number, since it uses less digits than its factorization.
1175559 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 755 (or 748 counting only the distinct ones).
The product of its digits is 7875, while the sum is 33.
The square root of 1175559 is about 1084.2319862465. The cubic root of 1175559 is about 105.5394460283.
The spelling of 1175559 in words is "one million, one hundred seventy-five thousand, five hundred fifty-nine". | 468 | 1,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-23 | latest | en | 0.878625 |
http://www.mywordsolution.com/question/explain-confidence-interval-for-the-true/923375 | 1,529,809,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866191.78/warc/CC-MAIN-20180624024705-20180624044705-00057.warc.gz | 467,111,286 | 8,133 | +1-415-315-9853
info@mywordsolution.com
## Statistics
describe Confidence interval for the true proportion.
1. A quality control engineer is attentive in the mean length of sheet insulation being cut automatically by machine. The wanted length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets produces a mean length of 12.14 feet. This sample will be utilized to obtain a 99% confidence interval for the mean length cut by machine.
a. The critical value to use in procurement the confidence interval is _______________
b. The confidence interval goes from ________ to ________.
c. Presume the engineer had decided to estimate the mean length to within 0.03 with 99% confidence. Then the sample size would be ________.
2. The president of a university is worried that the percentage of students who have cheated on an exam is higher than the 1% acceptable level. A confidential arbitrary sample of 1000 students from a population of 7000 revealed that 6 of them said that they had cheated on an exam during the last semester.
a. describe what is the critical value for the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months?
b. describe what is the upper bound of the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months?
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M923375
Have any Question?
## Related Questions in Statistics and Probability
### A study found that consumers spend an average of 24 per
A study found that consumers spend an average of \$24 per week in cash without being aware of where it goes. Assume that the amount of cash spent without being aware of where it goes is normally distributed and that the s ...
### Describe the following game with incomplete information as
Describe the following game with incomplete information as an extensive-form game. There are two players N = {I, II}. Each player has three types, TI = {I 1 , I 2 , I 3 } and TII = {II 1 , II 2 , II 3 }, with common prio ...
### 1 a researcher has developed the hypothesis that teenagers
1. A researcher has developed the hypothesis that teenagers are better able to learn about and remember technological information than are older adults, whereas older adults are better able to remember information about ...
### Consider a sealed-bid first-price auction with two buyers
Consider a sealed-bid first-price auction with two buyers whose private values are independent; the private value of buyer 1 has uniform distribution over the interval [0, 3], and the private value of buyer 2 has uniform ...
### Assignment- probability1a credit card company decides to
Assignment- Probability 1. A credit card company decides to study the frequency with which its cardholders charge for items from a certain chain of retail stores. The data values collected in the study appear to be norma ...
### This exercise is similar to given exercise but instead of
This exercise is similar to given Exercise, but instead of announcing the probability of a particular event given their private information, the players announce whether or not the expectation of a particular random vari ...
### Many grocery stores and large retailers such as wal-mart
Many grocery stores and large retailers such as Wal-Mart and K-Mart have installed self checkout systems so shoppers can scan their own items and cash out themselves. How do customers like this service and how often do t ...
### In a recent poll of 760 homeowners in the united states one
In a recent poll of 760 homeowners in the United States, one in five homeowners reports having a home equity loan that he or she is currently paying off. Using a confidence coefficient of 0.90, derive an interval estimat ...
### 1 the waiting time t between successive occurrences of an
1. The waiting time T between successive occurrences of an event E in a discrete time renewal process has the probability distribution P(T =1) = 0.7, P(T = 2) = 0.3. (iii) If an observer arrives after the renewal process ...
### Probabilities of graduation and publicationprofessors have
Probabilities of Graduation and Publication Professors have hundreds of students in their classes each year. Some professors teach only upper division courses to students who are in their major course of study. Of this g ...
• 13,132 Experts
## Looking for Assignment Help?
Start excelling in your Courses, Get help with Assignment
Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.
### WalMart Identification of theory and critical discussion
Drawing on the prescribed text and/or relevant academic literature, produce a paper which discusses the nature of group
### Section onea in an atwood machine suppose two objects of
SECTION ONE (a) In an Atwood Machine, suppose two objects of unequal mass are hung vertically over a frictionless
### Part 1you work in hr for a company that operates a factory
Part 1: You work in HR for a company that operates a factory manufacturing fiberglass. There are several hundred empl
### Details on advanced accounting paperthis paper is intended
DETAILS ON ADVANCED ACCOUNTING PAPER This paper is intended for students to apply the theoretical knowledge around ac
### Create a provider database and related reports and queries
Create a provider database and related reports and queries to capture contact information for potential PC component pro | 1,161 | 5,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-26 | latest | en | 0.938728 |
https://www.aussportsbetting.com/2021/09/15/2021-bookmaker-margins-and-markets-survey/ | 1,716,801,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00650.warc.gz | 571,737,087 | 41,298 | # 2021 Bookmaker Margins and Markets Survey
### UPDATE
The following are the results of a September 2021 bookmaker margins and markets survey. The survey is split into racing and sports surveys.
This study follows up on our 2011, 2012, 2013, 2015 and 2017 margin surveys.
### What are Bookmaker Margins?
The bookmaker margin, also known as the overround, the cut or the take, is the hidden amount charged by a bookmaker for accepting a wager.
When a bookmaker sets the odds they first estimate the probability of each possible outcome. In a fair coin toss, for example, the probability of heads is 50%. When using the decimal/European odds system the fair odds, also known as the true odds, equal the reciprocal of the bookmaker’s estimated probability, in this case 1 / 50% = 2.00. To fund their services, the bookmaker adjusts these odds downward to create a profit margin. Typical odds on a selection with a 50% chance of winning are 1.91. The gap between the bookmaker’s odds (1.91) and the fair odds (2.00) is due to the bookmaker margin.
The bookmaker margin is a measure of the bookmaker’s profit margin for an event and is a hidden transaction cost for punters. This profit is how bookmakers finance their services but bookmakers vary in the margins they apply. From a punter’s perspective, the lower the margin, the better. The difference between 1.90 and 1.92 line odds may not seem significant for a single wager, but when betting repeatedly this difference has a compounding effect. This can make the difference between winning and losing money in the long-term.
### Calculating Bookmaker Margins
When using decimal odds the bookmaker margin equals the sum of the reciprocals of the odds, minus 1. Below are calculation examples using two-outcome and three-outcome events.
The margin is amount by which the market exceeds 100%.
For a 100.0% market the margin is zero. If you compile the best available odds across a number of bookmakers and the combined market is below 100.0%, then the margin is negative, which means an arbitrage opportunity may exist.
### Interpreting Bookmaker Margins
The margin measures the bookmaker’s profit if they were to receive wagers on each outcome in proportion to the odds. Suppose a bookmaker offers decimal odds a on outcome A and odds b on outcome B. If proportion b/(a+b) is wagered on outcome A and a/(a+b) is wagered on outcome B, then the bookmaker will receive the same profit regardless of the result.
For example, recall that the bookmaker margin for the two-outcome event above was 3.5%. Suppose \$100,000 in total is wagered on the market, with:
(1.64/(1.64 + 2.35)) x \$100,000 = \$41,102.76 wagered on Oklahoma City and
(2.35/(1.64 + 2.35)) x \$100,000 = \$58,897.24 wagered on Denver.
Depending on the outcome, the bookmaker will pay out one of the following two amounts:
If Oklahoma City wins: \$41,102.75 x 2.35 = \$96,591.48
If Denver wins: \$58,897.24 x 1.64 = \$96,591.48
The bookmaker accepts \$100,000 in wagers but only pays out \$96,591.48 to the winners. The profit margin is (\$100,000 – \$96,591.48)/\$96,591.48 = 3.5%, as calculated by the margin earlier.
To provide perspective on how odds relate to margins, the table on the right compares equal line odds (bets with a 50% chance of winning) to their respective margins. Note that 2.00 line odds equate to a margin of 0%, where the bookmaker makes no profit on the market.
### Bookmaker Margin vs. Vigorish (Vig)
Another popular term, particularly in the United States, for measuring the bookmaker profit margin is vigorish. It is often referred to as vig, for short, or juice. In the context of this article the term vigorish refers to a different calculation to bookmaker margin. In short, the bookmaker margin is the bookmaker’s profit margin relative to the payouts to the winners, while vigorish refers to the profit margin relative to turnover, i.e. the initial wagers. Because the total payout is always less than the total turnover, the bookmaker margin is greater than vigorish for positive margins.
Unfortunately there are conflicting definitions of bookmaker margin and vigorish. Even on Wikipedia the explanation of how to calculate vigorish depends on which article you visit. This survey uses the definition of bookmaker margin (overround) as used in the Mathematics of bookmaking Wikipedia article and the definition of vigorish as used in the Vigorish Wikipedia article.
Referring to the previous NBA example:
The bookmaker margin is 1/1.64 + 1/2.35 – 1 = 3.53%
while vigorish is (1 – 1.64*2.35/(1.64 + 2.35)) = 3.41%
Recall that with this example, if \$41,102.76 is wagered on Oklahoma City and \$58,897.24 is wagered on Denver, the bookmaker accepts \$100,000 in wagers and pays out \$96,591.48 to the winners, regardless of the result. The bookmaker margin is (\$100,000 – \$96,591.48)/\$96,591.48 = 3.53%, while vigorish is (\$100,000 – \$96,591.48)/\$100,000 = 3.41%.
Both of these measures of bookmaker profit margin are valid, providing you understand what each represents. You can think of bookmaker margin as metres and vigorish as yards.
Note that if the bookmaker margin is 0% then vigorish is also 0%. If one is negative then so is the other.
The choice of bookmaker margin over vigorish has no impact on this survey. If the bookmaker margin of one bookmaker is higher than another’s, then the same can be said for that bookmaker’s vigorish. The relative value between bookmakers is the focus of this survey, not the actual values.
### Combined Margins
For each market a “combined” margin is calculated by sourcing the best available odds across the surveyed bookmakers. This combined margin represents the realisable margin that can be attained through holding a portfolio of bookmaker memberships and shopping around for the highest odds.
The combined margin is always equal to or lower than the lowest bookmaker’s margin. The greater the disparity between the odds of two or more bookmakers, the lower the combined margin.
For example, in the US Open men’s singles match between Djokovic and Brooskby, which was included in this survey, we observed the following head-to-head markets:
bet365:
Djokovic: 1.04
Brooskby: 13.00
PlayUp:
Djokovic: 1.01
Brooskby: 18.75
The bookmaker margin for bet365 is 3.85% while the margin for PlayUp is 4.34%.
If we combine the two markets by taking the highest available odds we get:
Djokovic: 1.04
Brooskby: 18.75
The combined margin for this market is 1/1.04 + 1/18.75 – 1 = 1.49%.
Combined margins are a crucial for beating bookmakers because they lowers the hurdle rate required to achieve a profit. The equivalent line odds for the 3.85% margin at bet365 is 1.926. To make a profit on 1.926 odds you would have to win more than 51.9% of the time. In contrast, the equivalent line odds for the 1.49% combined margin is 1.971. To make a profit on 1.971 odds you would only need to win more than 50.7% of the time.
Note that if the combined margin is a negative number then an arbitrage opportunity exists.
### Racing Margin Survey Results
For one week in September, 2021, fixed win and place odds were recorded for a random selection of thoroughbred, harness and greyhound races. The odds were recorded on the day of each race.
The average of the win and place margins are shown in the tables below. The Overall Racing Margins table combines the average margins across the three racing types.
##### Table 1 – Thoroughbred Margins
Bookmaker Thoroughbreds Margin Rank
bet365 23.8% 1
BlueBet 27.6% 4
PlayUp 29.6% 5
Sportsbet 24.9% 2
Unibet 26.1% 3
WinnersBet 32.6% 6
Combined 20.2%
Average 27.4%
##### Table 2 – Harness Margins
Bookmaker Harness Margin Rank
bet365 26.5% 1
BlueBet 28.4% 4
PlayUp 36.0% 6
Sportsbet 27.8% 3
Unibet 27.8% 2
WinnersBet 34.1% 5
Combined 22.5%
Average 30.1%
##### Table 3 – Greyhound Margins
Bookmaker Greyhounds Margin Rank
bet365 26.4% 1
BlueBet 32.5% 4
PlayUp 34.9% 5
Sportsbet 31.1% 3
Unibet 29.7% 2
WinnersBet 37.2% 6
Combined 24.0%
Average 31.9%
##### Table 4 – Overall Racing Margins
Bookmaker Racing Margin Rank
bet365 25.6% 1
BlueBet 29.5% 4
PlayUp 33.5% 5
Sportsbet 27.9% 3
Unibet 27.8% 2
WinnersBet 34.6% 6
Combined 22.2%
Average 29.8%
bet365 came out as the winner with the lowest margins across all three forms of racing. Unibet and Sportsbet were a close 2nd and 3rd, with Unibet offering lower margins for the surveyed greyhounds and harness races, while Sportsbet had lower margins than Unibet for the surveyed thoroughbred races.
Note that the actual margin figures are of less importance than the relative margins between the bookmakers because racing margins vary depending on the number of runners in the field. Generally the greater the number of runners, the greater the margin.
##### Table 5 – Instances of Best Racing Odds
Bookmaker Best Odds Rank
bet365 68 1
BlueBet 5 4
PlayUp 3 5
Sportsbet 28 2
Unibet 14 3
WinnersBet 0 6
Table 5 shows the number of times that each bookmaker offered uniquely the best available odds for a particular selection.
bet365 came out clear on top with 68 instances of offering higher odds for a runner than any other bookmaker. Sportsbet was next best with 28, twice as many as Unibet. This result highlights the fact that a bookmaker doesn’t need to have the lowest margins to be valuable in a portfolio of memberships when shopping around. A bookmaker can be useful simply for offering contrasting odds to the other bookmakers on particular runners.
The results illustrate that between them, bet365, Sportsbet and Unibet make a valuable combination of bookmakers for fixed odds racing. On only 8 occasions did a bookmaker other than these three offer the best odds for a particular selection.
## Sports Margin Survey Results
For one week in September, 2021, the odds for up to three popular markets, along with the total number of markets, were recorded for a selection of fixtures.
The following leagues and markets were included in the survey:
For the AFL, NRL, NFL and EPL the odds were recorded once a day from Monday through to Friday. For the remaining sports the odds were recorded within 24 hours of the events commencing.
Table 6 provides the average margins for the surveyed leagues. The “Equivalent Line” column shows the equivalent line odds for the respective margins.
##### Table 6 – Margins and markets survey results
Bookmaker Margin Equivalent Line Rank
bet365 5.0% 1.905 2
BlueBet 5.5% 1.896 3
PlayUp 5.6% 1.894 4
Sportsbet 5.6% 1.894 5
Unibet 5.0% 1.905 1
WinnersBet 7.4% 1.862 6
Combined 3.7% 1.929
Average 5.7% 1.892
Unibet and bet365 finished virtually neck and neck at 5.0%, with BlueBet the third best a fair way back at 5.5%. This was followed by PlayUp and Sportsbet just behind BlueBet at 5.6%.
The combined margin was 3.7%. This reflects the benefit of shopping around for odds rather than relying upon one bookmaker.
bet365, Sportsbet and Unibet each returned the same margin as in the 2017 survey. The other three bookmakers in this 2021 survey weren’t included in the 2017 study.
##### Table 7 – Instances of Best Sports Odds
Bookmaker Best Odds Rank
bet365 125 2
BlueBet 40 5
PlayUp 42 4
Sportsbet 71 3
Unibet 127 1
WinnersBet 2 6
Unibet and bet365 finished almost neck and neck for the number of times that they alone offered the best available odds for a particular selection. It is noteworthy that Sportsbet finished well ahead of BlueBet and PlayUp for this statistic, despite having a similar margin to the other two bookmakers. This again highlights the fact that a bookmaker doesn’t have to offer the lowest margins to be a valuable addition to a portfolio of memberships. If it simply offers contrasting odds for particular selections then it will effectively reduce the combined margin when odds shopping.
BlueBet and PlayUp each offered the best odds for 40+ selections. Unlike the racing survey results – which were dominated by bet365, Unibet and Sportsbet – BlueBet and PlayUp did impact the combined margin with a high number of instances with the best odds.
Even WinnersBet, which had the highest margin, occasionally offered the best odds of any bookmaker for certain selections.
### Bookmaker Margins By Market
The following tables display the bookmakers’ average margins broken down by market.
Bookmaker H2H Margin Equivalent Line Rank
bet365 4.8% 1.908 2
BlueBet 5.4% 1.898 3
PlayUp 5.8% 1.890 5
Sportsbet 5.5% 1.897 4
Unibet 4.7% 1.910 1
WinnersBet 7.5% 1.860 6
Combined 3.2% 1.937
Average 5.6% 1.894
##### Table 9 – Line Margins
Bookmaker Line Margin Equivalent Line Rank
bet365 4.8% 1.909 1
BlueBet 5.5% 1.897 4
PlayUp 5.5% 1.896 5
Sportsbet 5.4% 1.898 3
Unibet 5.1% 1.903 2
WinnersBet 6.8% 1.872 6
Combined 4.1% 1.921
Average 5.5% 1.896
##### Table 10 – Total Score Margins
Bookmaker Margin Equivalent Line Rank
bet365 4.9% 1.907 1
BlueBet 5.5% 1.896 3
PlayUp 5.5% 1.895 4
Sportsbet 5.9% 1.889 5
Unibet 5.1% 1.903 2
WinnersBet 6.9% 1.872 6
Combined 4.3% 1.918
Average 5.6% 1.894
Unibet came out on top with the lowest margins in head-to-head markets, while bet365 had the lowest margins for both line and total score markets. bet365 offered 1.91 line odds for the AFL and NRL while most of the surveyed bookmakers offered 1.90 lines.
The benefit for shopping around for odds was greater for head-to-head markets than it was for line and total score markets. The combined margin was 3.2% for head-to-head, versus 4.1% for line and 4.3% for total score.
## Markets Survey
The number of markets per fixture was surveyed for the same events as the odds survey.
To provide an example, if a bookmaker offers head-to-head, line, total score and winning margin markets for a game, then the number of markets is 4 for that fixture.
The AFL, NRL, NFL and EPL markets were surveyed each day from Wednesday to Friday while the tennis, MLB, WNBA, cricket and rugby markets were surveyed within 24 hours of the event start.
The average number of markets offered per fixture in each league are shown. The Average (Wed), Average (Thu) and Average (Fri) rows represent the average across the AFL, NRL, NFL and EPL on those days.
The final average for each sport has been calculated using the figures for the tennis, MLB, WNBA, cricket and rugby, along with the overall average figures for the AFL, NRL, NFL and EPL.
Much like in the odds survey, the actual market numbers are less relevant due to the arbitrary nature of the sports and leagues chosen. The focus is on the relative averages between the surveyed bookmakers.
The intention of this markets survey is to provide a general impression of the number of markets on offer. Bookmakers vary in the way they count markets so there may not be a substantial difference between, say 80 markets and 120 if the first bookmaker counts a “Pick Your Own Line” section as one market while the second bookmaker counts each line (-7.5, -6.5, -5.5, etc.) as a separate market.
##### Table 11 – Markets Survey – Part 1
Sport/League bet365 BlueBet PlayUp
Tennis 83 15 7
MLB 64 25 3
WNBA 212 7 3
Cricket 35 7 1
Rugby 16 2 2
AFL (Wed) 87 40 3
AFL (Thu) 87 64 23
AFL (Fri) 114 71 44
NRL (Wed) 159 27 38
NRL (Thu) 161 31 45
NRL (Fri) 164 60 46
NFL (Wed) 219 90 11
NFL (Thu) 222 101 11
NFL (Fri) 232 184 16
EPL (Wed) 88 69 16
EPL (Thu) 127 69 17
EPL (Fri) 145 70 17
Average AFL 96 58 23
Average NRL 161 39 43
Average NFL 224 125 12
Average EPL 120 69 17
Average (Wed) 138 57 17
Average (Thu) 149 66 24
Average (Fri) 164 96 31
Average 112 39 12
Rank 2 4 5
##### Table 12 – Markets Survey – Part 2
Sport/League Sportsbet Unibet WinnersBet
Tennis 72 39 3
MLB 120 116 3
WNBA 20 110 3
Cricket 23 52 1
Rugby 29 41 2
AFL (Wed) 159 285 2
AFL (Thu) 160 517 22
AFL (Fri) 190 533 22
NRL (Wed) 127 91 18
NRL (Thu) 141 266 19
NRL (Fri) 141 268 19
NFL (Wed) 206 186 6
NFL (Thu) 213 187 6
NFL (Fri) 231 187 9
EPL (Wed) 135 388 11
EPL (Thu) 135 432 11
EPL (Fri) 218 445 11
Average AFL 170 445 15
Average NRL 136 208 19
Average NFL 217 187 7
Average EPL 163 422 11
Average (Wed) 157 237 9
Average (Thu) 162 351 15
Average (Fri) 195 358 15
Average 105 180 7
Rank 3 1 6
Unibet came out on top with a whopping 180 markets per game on average. bet365 and Sportsbet came in 2nd and 3rd with similar figures of 112 and 105. There’s then a significant drop off to BlueBet at 39, before another large drop off to PlayUp at 12 and WinnersBet at 7.
The broad range of markets for Unibet, bet365 and Sportsbet make them suitable for anyone looking for a solitary bookmaker membership. Casual sports punters will find BlueBet to be perfectly adequate, while the limited range of markets for PlayUp and WinnersBet make them suitable only as part of a collection of memberships for those who are keen sports bettors.
### Summary
bet365 came out on top with the lowest margins for all forms of racing. Between them, bet365, Unibet and Sportsbet accounted for almost all of the best available fixed odds for the surveyed runners.
Unibet and bet365 shared the lowest sports betting margins. While Sportsbet recorded higher sports margins, its odds often contrasted with the other bookmakers. For this reason it frequently offered the best odds for particular selections. In contrast to the racing results, BlueBet and PlayUp also had the best odds for numerous selections. Despite the higher margins, they too would be a useful addition to a portfolio of memberships for those who like to shop around for the best odds.
The combined bookmaker margins highlight the benefits of shopping around. This is particularly evident for head-to-head markets, with the combined margin coming in at 3.2%, compared to 4.7% for the survey-leading Unibet.
Unibet recorded by far the highest number of markets per game. The offerings of bet365 and Sportsbet were also incredibly high. This makes them more than sufficient to meet the needs of someone looking for only one bookmaker membership.
### Odds Comparison
You can compare bookmaker odds for upcoming fixtures in the Bookmaker Odds Comparison section.
### Caveats
This survey was conducted during one week in September, 2021. The surveyed margins were heavily dependent on the sample of sporting events chosen. Comparative margin levels may change over time and may differ for sports and leagues not covered in the survey.
This survey focuses on margins and the number of markets per fixture, but a key caveat is that betting limits are ignored. Furthermore, some bookmakers offer higher limits but may limit or close your account if you regularly win with high-stakes wagers. In addition, some bookmakers offer frequent promotions in lieu of lower odds, which is not factored into these stats. | 5,127 | 18,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.930221 |
egysteer.com | 1,669,930,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00402.warc.gz | 248,295,059 | 42,621 | # Calculate generator capacity for your project or home
If you do not know how to calculate generator capacity and choose the right generator for your project or home for your project or home, then this article will give you a great idea of ways to choose the right generator for many different applications.
### Calculate generator capacity for your project or home
can only get the best results when you learn how to calculate the power of an alternator correctly. In order to choose the ideal generator for you. At first you have to estimate the spare power you will need. Then you can easily choose the right generator by choosing the right power for it.
Then you need to consider the period of time you need the standby generator to supply you with power based on the capacity of the generator. also, remember where you want to install the generator in your mind. But you can’t choose a generator and install it randomly without thinking about such things.
### Types of electric generators and calculation of generator capacity
The truth is that generators are machines or devices used to generate power in places where it is difficult to access electricity. Generators act as a source of backup power that comes from fuel conversion.
Generators usually use gasoline, diesel or propane and convert them into electrical energy. These types of fuels are the main sources of electrical supply for domestic and commercial uses.
Also there are different types of generators. Although they differ in design and strength, they require special maintenance to ensure their long-term use. The maintenance of electrical and mechanical faults must be taken into account, and you can find important 7 generator electrical faults
#### Diesel generators
Generating electricity by diesel fuel, diesel generator works by combustion by force of pressure. This generator is famous for its less need for maintenance and is characterized by its durability and the ability to work continuously for a long time. In addition to obtaining high-efficiency energy at a cheap cost from this generator, the diesel generator is also distinguished by its ability to withstand high temperatures and difficult areas.
Diesel Generator Capacity Calculation
#### Natural Gas Generators
There is no comparison to natural gas. If you have the option of using a natural gas generator, do not consider using another generator. The generator uses liquefied propane or petroleum to produce power. You will have an endless source of energy because natural gas is storable and exists in large quantities. This eco-friendly generator is recommended for all applications.
#### Gasoline Generators
Due to the availability of gasoline fuel, the gasoline generator is typical in every home and commercial use. For portable size and low cost operation, maximum people use this generator to use. But this generator needs special care regularly and makes noise while working.
#### Hybrid gas and diesel generators
Automatic gas or diesel backup generator that does not need to refuel for 48 hours. This large generator has a powerful engine, coolant, fuel tank, and alternator. When the power goes out, the automatic transmission switch starts the flow of power immediately. You won’t feel disconnected for a second with this generator – best for use on large construction sites, medical stents, and elevators.
#### Portable generators
This generator uses gas or diesel to produce electricity for a short period. You can use this generator to run your home appliances smoothly. Using a portable alternator has two ways – one is to connect the wires in the subpanel, and the other is to use alternator power sockets. If you experience short-term power outages in your area, you can use this generator as a power reserve.
#### Solar Generators
You can use this generator that uses sunlight as a source of energy. This energy is simple to install but powerful and can be stored easily. Here the battery retains power by charging, and needs to be charged in sunlight for 8 hours.
#### inverter generator
Among all the previous generators, this generator is the quietest. It works by converting AC power to DC power. The energy produced is highly efficient and safe to operate the electrical tools in your home.
## How to calculate generator capacity exactly?
Size is important when you’re considering taking a generator. Suppose you get a small generator that is not able to run your equipment or a large generator will lose more power. So you need one perfect generator, but if you don’t know how to choose the power of the generators, you will find the way in the rest of the article.
### Load capacity to calculate generator capacity
At first, you have to calculate all the equipment you will need to run the generator on it. After making a list of these loads, add up all the wattage. The total wattage number is the required electrical power input that your devices need from the generator. You will find information about electrical loads on the metal plate on the equipment. and that information will helps you to calculate generator capacity.
### Adjusting kW to KVA and calculate generator capacity
The total wattage you will find by summing is in kilowatts (kilowatts). It is the power load that produces the work output. But generators measure power in kilovolt amperes (kilovolt amperes). Sometimes you can assume that kw = kva but it is not practically 100% correct.
If you know the electrical system’s throughput, you can convert between kVA and kW. Where the wattage factor is calculated between 0 to 1. The closer the power factor is to 1, the more efficient the kVA becomes for conversion to kW.
Globally, the average power factor of the generator is 0.8, so you should calculate the loads with this factor because it is suitable and because it will help you greatly in calculate generator capacity.
Let’s give an example: If your appliances have a wattage of 100 kW, your generator size should produce an equivalent power of 125 kVA. And that is by dividing your load capacity \ 0.8 and in the previous example 100 \ 0.8 = 125 kva.
### Determine the number of hours of operation of the generator
You have to choose the number of hours to run the generator because the generator run time at maximum load is 30 minutes. If you want the generator to run for more than 30 minutes, you have to increase the generator capacity by 20% – 30%.
### Generator operating location
After calculating the load and determining the volume, you need to think about the location where you want to run the generator. You cannot choose a generator and install it randomly. If you live in a small place or up a hill and need a large generator, this is impossible. Try to provide additional space in your location for the generator before you get it.
Note: I think you now have a clear idea about determining the size of the generator. You should also know the most important 7 electrical faults of generators so that you can solve the faults you face.
## How to choose the right generator for your project or your house?
After calculating generator capacity. Now comes the main part which is choosing the right generator for your project or your home.
Follow the tips that will be mentioned so you can choose the right generator for your project or home. Since three types of generator – portable generator, standby generator and inverter generator are common, here I have enlightened the methods of selecting these generators.
### 1. Choosing a portable generator
• Portable generators are the popular generators that people are considering buying as they are a good backup source of power. It is very versatile and cheap to power household appliances.
• You can use the portable generator on construction sites to use the tools smoothly also in the home where the renovation work is going on.
• When you think of using a portable generator, first add all the wattage of the devices you want to run the generator. This number will help in choosing the right sized generator. You will get the wattage information at the hardware level.
• The best thing about this generator is that you will get a generator with wheels to move from one place to another. This portability helps make your work easier.
• Choose the generator that has an on / off button. It helps to start the generator quickly without any hassle.
### 2. Selection of backup generator
• If you live in an area where power outages are frequent, budget for a backup generator. This generator can provide backup power for the whole house in the event of a power outage.
• This type of generator turns on automatically when the power is cut off, and once power is restored it is turned off. Therefore, you will not have to worry about the generator starting or shut off.
• Backup generators aren’t cheap, but if you need huge backup power, it’s worth it.
The worrying thing about the generator is that this type of generator makes noise during operation. during night, the loud noise is hard to beat. Modern standby generators come with advanced technology that keeps the noise level down.
• For standby generator, LPG is safer and cleaner, but the running cost is expensive. Therefore, you can use diesel or gasoline to run the generator. This generator needs a professional to install because it will connect to the breaker box.
### 3. Choose an inverter generator
• The inverter generator is for those who love camping and love the tailgate.
• This generator is lightweight and quiet and gives smooth power backup.
• You can run the inverter generator on gas and propane.
• And you can use two inverter generators with parallel connection, which allows you to maximize dual power.
• The power you will get from the inverter generator is smooth and safe.
• You can operate different electrical appliances and gadgets like laptops and smartphones without any power outage problem.
• this generator is only used outside because this machine emits the toxic carbon monoxide.
## Frequently Asked Questions for How to Calculate Generator Capacity
Here, in this section, we answer your frequently asked questions that might be on the mind of someone who wants to buy a new generator.
### Can I run my generator all night?
Yes, you can run the generator overnight if you have enough gas in your tank.
### What happens if I overload my generator?
You may encounter two types of malfunctions – one the generator will shut down due to the circuit breaker, another if there is no circuit breaker in the generator it will continue to run and overheat, and finally it will lead to a fire if the heat protection does not work.
### Does the generator need a surge protection device?
No, because generators are safe because energy does not flow through a specific electrical circuit.
## Summary in calculate generator capacity
We depend heavily on electricity for our lives. When a power outage occurs, the darkness makes us realize the importance of electricity. Day by day the need for electricity increases and the huge demands are met, sometimes the electricity network is cut off, and the power is cut off.
To meet the increasing need for electricity, generators are the best option to choose. This article will help you how to calculate generator capacity for home and commercial use. | 2,229 | 11,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.918855 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.