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Learning general representations of text is a fundamental problem for many natural language understanding (NLU) tasks. Previously, researchers have proposed to use language model pre-training and multi-task learning to learn robust representations. However, these methods can achieve sub-optimal performance in low-resource scenarios. Inspired by the recent success of optimization-based meta-learning algorithms, in this paper, we explore the model-agnostic meta-learning algorithm (MAML) and its variants for low-resource NLU tasks. We validate our methods on the GLUE benchmark and show that our proposed models can outperform several strong baselines. We further empirically demonstrate that the learned representations can be adapted to new tasks efficiently and effectively.

### 相关内容

Pre-training text representations has recently been shown to significantly improve the state-of-the-art in many natural language processing tasks. The central goal of pre-training is to learn text representations that are useful for subsequent tasks. However, existing approaches are optimized by minimizing a proxy objective, such as the negative log likelihood of language modeling. In this work, we introduce a learning algorithm which directly optimizes model's ability to learn text representations for effective learning of downstream tasks. We show that there is an intrinsic connection between multi-task pre-training and model-agnostic meta-learning with a sequence of meta-train steps. The standard multi-task learning objective adopted in BERT is a special case of our learning algorithm where the depth of meta-train is zero. We study the problem in two settings: unsupervised pre-training and supervised pre-training with different pre-training objects to verify the generality of our approach.Experimental results show that our algorithm brings improvements and learns better initializations for a variety of downstream tasks.

Recently, the emergence of pre-trained models (PTMs) has brought natural language processing (NLP) to a new era. In this survey, we provide a comprehensive review of PTMs for NLP. We first briefly introduce language representation learning and its research progress. Then we systematically categorize existing PTMs based on a taxonomy with four perspectives. Next, we describe how to adapt the knowledge of PTMs to the downstream tasks. Finally, we outline some potential directions of PTMs for future research. This survey is purposed to be a hands-on guide for understanding, using, and developing PTMs for various NLP tasks.

We propose UniViLM: a Unified Video and Language pre-training Model for multimodal understanding and generation. Motivated by the recent success of BERT based pre-training technique for NLP and image-language tasks, VideoBERT and CBT are proposed to exploit BERT model for video and language pre-training using narrated instructional videos. Different from their works which only pre-train understanding task, we propose a unified video-language pre-training model for both understanding and generation tasks. Our model comprises of 4 components including two single-modal encoders, a cross encoder and a decoder with the Transformer backbone. We first pre-train our model to learn the universal representation for both video and language on a large instructional video dataset. Then we fine-tune the model on two multimodal tasks including understanding task (text-based video retrieval) and generation task (multimodal video captioning). Our extensive experiments show that our method can improve the performance of both understanding and generation tasks and achieves the state-of-the art results.

In recent years, there has been an exponential growth in the number of complex documents and texts that require a deeper understanding of machine learning methods to be able to accurately classify texts in many applications. Many machine learning approaches have achieved surpassing results in natural language processing. The success of these learning algorithms relies on their capacity to understand complex models and non-linear relationships within data. However, finding suitable structures, architectures, and techniques for text classification is a challenge for researchers. In this paper, a brief overview of text classification algorithms is discussed. This overview covers different text feature extractions, dimensionality reduction methods, existing algorithms and techniques, and evaluations methods. Finally, the limitations of each technique and their application in the real-world problem are discussed.

Meta learning is a promising solution to few-shot learning problems. However, existing meta learning methods are restricted to the scenarios where training and application tasks share the same out-put structure. To obtain a meta model applicable to the tasks with new structures, it is required to collect new training data and repeat the time-consuming meta training procedure. This makes them inefficient or even inapplicable in learning to solve heterogeneous few-shot learning tasks. We thus develop a novel and principled HierarchicalMeta Learning (HML) method. Different from existing methods that only focus on optimizing the adaptability of a meta model to similar tasks, HML also explicitly optimizes its generalizability across heterogeneous tasks. To this end, HML first factorizes a set of similar training tasks into heterogeneous ones and trains the meta model over them at two levels to maximize adaptation and generalization performance respectively. The resultant model can then directly generalize to new tasks. Extensive experiments on few-shot classification and regression problems clearly demonstrate the superiority of HML over fine-tuning and state-of-the-art meta learning approaches in terms of generalization across heterogeneous tasks.

Commonsense knowledge and commonsense reasoning are some of the main bottlenecks in machine intelligence. In the NLP community, many benchmark datasets and tasks have been created to address commonsense reasoning for language understanding. These tasks are designed to assess machines' ability to acquire and learn commonsense knowledge in order to reason and understand natural language text. As these tasks become instrumental and a driving force for commonsense research, this paper aims to provide an overview of existing tasks and benchmarks, knowledge resources, and learning and inference approaches toward commonsense reasoning for natural language understanding. Through this, our goal is to support a better understanding of the state of the art, its limitations, and future challenges.

In this paper, we present a Multi-Task Deep Neural Network (MT-DNN) for learning representations across multiple natural language understanding (NLU) tasks. MT-DNN not only leverages large amounts of cross-task data, but also benefits from a regularization effect that leads to more general representations in order to adapt to new tasks and domains. MT-DNN extends the model proposed in Liu et al. (2015) by incorporating a pre-trained bidirectional transformer language model, known as BERT (Devlin et al., 2018). MT-DNN obtains new state-of-the-art results on ten NLU tasks, including SNLI, SciTail, and eight out of nine GLUE tasks, pushing the GLUE benchmark to 82.2% (1.8% absolute improvement). We also demonstrate using the SNLI and SciTail datasets that the representations learned by MT-DNN allow domain adaptation with substantially fewer in-domain labels than the pre-trained BERT representations. Our code and pre-trained models will be made publicly available.

Deep reinforcement learning (RL) algorithms have shown an impressive ability to learn complex control policies in high-dimensional environments. However, despite the ever-increasing performance on popular benchmarks such as the Arcade Learning Environment (ALE), policies learned by deep RL algorithms often struggle to generalize when evaluated in remarkably similar environments. In this paper, we assess the generalization capabilities of DQN, one of the most traditional deep RL algorithms in the field. We provide evidence suggesting that DQN overspecializes to the training environment. We comprehensively evaluate the impact of traditional regularization methods, $\ell_2$-regularization and dropout, and of reusing the learned representations to improve the generalization capabilities of DQN. We perform this study using different game modes of Atari 2600 games, a recently introduced modification for the ALE which supports slight variations of the Atari 2600 games traditionally used for benchmarking. Despite regularization being largely underutilized in deep RL, we show that it can, in fact, help DQN learn more general features. These features can then be reused and fine-tuned on similar tasks, considerably improving the sample efficiency of DQN.

Meta-learning, or learning to learn, is the science of systematically observing how different machine learning approaches perform on a wide range of learning tasks, and then learning from this experience, or meta-data, to learn new tasks much faster than otherwise possible. Not only does this dramatically speed up and improve the design of machine learning pipelines or neural architectures, it also allows us to replace hand-engineered algorithms with novel approaches learned in a data-driven way. In this chapter, we provide an overview of the state of the art in this fascinating and continuously evolving field.

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Browse Questions

# Let $x_,x_2,x_3,x_4,x_5$ be the observations with mean m and standard deviation s. The standard deviation of the observation $kx_1,kx_2,kx_3,kx_4,kx_5$ is

$\begin{array}{1 1}(A)\; k+s\\(B)\;\frac{s}{k}\\(C)\;ks\\(D)\;s \end{array}$

Can you answer this question?

give m is the mean of the observation $x+_1,x_2,x_3,x_4,x_5$
$m =\large\frac{x_1+x_2+x_3+x_4+x_5}{5}$
Let the new mean be n
The observations are $kx_1,kx_2,kx_3,kx_4,kx_5$
$\therefore n= \large\frac{kx_+kx_2+kx_3+kx_4+kx_5}{5}$
=>$k \bigg[\large\frac{x_1+x_2+x_3+x_4+x_5}{5}\bigg]$
$n= km$
Standard deviation $=\sqrt { \large\frac{(x_1-m)^2+(x_3-m)^2+(x_4-m)^2+(x_4 -m)^2}{5}}$
Tjhe new standard deviation will be $S_1 = \sqrt {\large\frac{(kx_1-n)^2 +(kx_2-n)^2 +(kx_3-n)^2 +(kx_4-n)^2+(kx_5-n)^2}{5}}$
we know that
$kx_1 -n =kx_1-mk$
$\qquad= k(x_1-m)$
Similarly $kx_2-n=k(x_2-m)$
$kx_3-n=k(x_3-m)$
$kx_4-n =k(x_4-m)$
$kx_5-n =k(x_5-m)$
Substituting $S_1= \sqrt { \large\frac{k(x_1-m))^2+(k(x_2-m)^2)^2+(k(x_3-m))^2 +(k(x_4-m))^2+(k(x_5-m))^2}{5}}$
$\qquad= ks$
Hence C is the correct answer.
answered Jul 8, 2014 by

format: plain


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# Pseudorapidity and transverse-momentum distributions of charged particles in proton-proton collisions at $\mathbf{\sqrt{\textit s}}$ = 13 TeV

Abstract : The pseudorapidity ($\eta$) and transverse-momentum ($p_{\rm T}$) distributions of charged particles produced in proton-proton collisions are measured at the centre-of-mass energy $\sqrt{s}$ = 13 TeV. The pseudorapidity distribution in $|\eta|<$ 1.8 is reported for inelastic events and for events with at least one charged particle in $|\eta|<$ 1. The pseudorapidity density of charged particles produced in the pseudorapidity region $|\eta|<$ 0.5 is 5.31 $\pm$ 0.18 and 6.46 $\pm$ 0.19 for the two event classes, respectively. The transverse-momentum distribution of charged particles is measured in the range 0.15 $<$ $p_{\rm T}$ $<$ 20 GeV/c and $|\eta|<$ 0.8 for events with at least one charged particle in $|\eta|<$ 1. The correlation between transverse momentum and particle multiplicity is also investigated by studying the evolution of the spectra with event multiplicity. The results are compared with calculations from PYTHIA and EPOS Monte Carlo generators.
Type de document :
Article dans une revue
Physics Letters B, Elsevier, 2016, 753, pp.319-329. 〈10.1016/j.physletb.2015.12.030〉
Domaine :

http://hal.in2p3.fr/in2p3-01207015
Contributeur : Sylvie Flores <>
Soumis le : mercredi 30 septembre 2015 - 07:30:26
Dernière modification le : mardi 19 juin 2018 - 01:20:00

### Citation

J. Adam, L. Aphecetche, A. Baldisseri, Guillaume Batigne, I. Belikov, et al.. Pseudorapidity and transverse-momentum distributions of charged particles in proton-proton collisions at $\mathbf{\sqrt{\textit s}}$ = 13 TeV. Physics Letters B, Elsevier, 2016, 753, pp.319-329. 〈10.1016/j.physletb.2015.12.030〉. 〈in2p3-01207015〉

### Métriques

Consultations de la notice


text:
Conviction about a system from pseudo-proofs of its truths

3

I vaguely recall a description about manifestations of the truths offered by a system being pseudo-proofs for that system to be right, in a context in which logical certainty is not possible. Consider the following examples:

A theory (e.g. "those who open doors with the left hand are less prone to falling to the ground"), which cannot be definitely proven, but can't be proven wrong either, is propounded, even if absurd. Manifestations of the predictions of the theory would not be real proofs, but could still convince.

It is not possible to clearly know whether Physics are right. The creations of technology, based on the knowledge it provides, work, so Physics seem right -- although such intuition cannot be proven.

For what it is worth, I remember this conviction taking place where the success of someone would convince others to follow him or her. However, I do not know if it was a thought of myself or if I read it somewhere. I have tried searching, and also looked the Wikipedia pages about fallacies, but have not found it. If such pseudo-argument is recognized, how is it named?

Question was closed 2015-06-21T07:08:15.833

2I do not know the answer but the book, The Structure of Scientific Revolutions by Thomas Kuhn speaks a lot about - and demonstrates incredible insight into how the paradigms created from such pseudo-proofs gain traction and become essentially infallible. – dgo – 2014-09-05T02:01:35.133

@user1167442 Thanks :). Although the book (judging from the relevant Wikipedia page, I have not read it) covers a similar topic, it is not exactly what I was looking for. – Kalrish – 2014-09-05T10:22:27.390

Gotcha. Well; still a highly stimulating read if you have the leisure. – dgo – 2014-09-05T13:01:50.387


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# Drawing simple geometric figures with DrawUserPrimitives?

I'm trying to draw a simple triangle based on an array of vertex. I've been searching for a tutorial and I found a simple example on riemers but I couldn't get it to work. I think it was made for XNA 3 and it seems there were some changes to XNA 4?

Using this example:

http://www.riemers.net/eng/Tutorials/XNA/Csharp/Series1/The_first_triangle.php

I get this error:

Additional information: The current vertex declaration does not include all the elements required by the current vertex shader. TextureCoordinate0 is missing.

I'm not english so I'm having some trouble to understand everything.

For what I understand error is confusing because I'm trying to draw a triangle color based and not texture based and it shouldn't need a texture.

Also I saw some articles about dynamic shadows and lights and I would like to know if this is the kind of code used to do it with some tweaks like culling because I'm wondering if its heavy code for performance in real time.

    public void LoadContent()
{
playerVision = new VertexPositionColor[3];
playerVision[0].Position = new Vector3(-0.5f, -0.5f, 0f);
playerVision[0].Color = Color.Red;
playerVision[1].Position = new Vector3(0, 0.5f, 0f);
playerVision[1].Color = Color.Green;
playerVision[2].Position = new Vector3(0.5f, -0.5f, 0f);
playerVision[2].Color = Color.Yellow;
}

// comes from draw method
private void drawPlayerVision(GameTime gameTime, SpriteBatch spriteBatch)
{
effect.CurrentTechnique = effect.Techniques["Pretransformed"];
Game.GraphicsDevice.DrawUserPrimitives(PrimitiveType.TriangleList, playerVision, 0, 1, VertexPositionColor.VertexDeclaration);
}

• do you have this line: effect.CurrentTechnique = effect.Techniques["Pretransformed"]; in your code? – Vodáček Jun 29 '12 at 16:09
• no, just the code on riemers tutorial. Also I don't quite understand effects files yet... – Navy Seal Jun 29 '12 at 16:19
• try to add it before foreach in Draw method – Vodáček Jun 29 '12 at 16:27
• Additional information: This method does not accept null for this parameter. I added a new file effect and loaded on content load but do I need an effect to do this? Also forget to mention I'm working with 2D if that matters. – Navy Seal Jun 29 '12 at 16:42
• ok, then post your draw method into your question – Vodáček Jun 29 '12 at 16:43

Your problem is that effect is not sent into graphics card, you need to do this:

effect.CurrentTechnique = effect.Techniques["Pretransformed"];

foreach (EffectPass pass in effect.CurrentTechnique.Passes)
{
pass.Apply();

Game.GraphicsDevice.DrawUserPrimitives(PrimitiveType.TriangleList, playerVision, 0, 1, VertexPositionColor.VertexDeclaration);
}

On first line, you must select effect technique (effect file from Riemers have more of them), then send effect to graphics card and finally draw your primitive.


text:
## How do I find the nth term of this sequence?

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
rogermiranda
Posts: 19
Joined: Fri Apr 30, 2010 6:01 pm
Contact:

### How do I find the nth term of this sequence?

I have the sequence 2,8,26,80,242
So b sub n = 6,18,54,162
b sub 1 = 6
r = 3
using the geometric formula I have 6 times 3^(n-1).
Now I am stuck? What are the next steps? Please help!

Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: How do I find the nth term of this sequence?

rogermiranda wrote:I have the sequence 2,8,26,80,242
So b sub n = 6,18,54,162
b sub 1 = 6
r = 3
using the geometric formula I have 6 times 3^(n-1).
Now I am stuck? What are the next steps? Please help!

3,9,27,81,243..

=$3^1,3^2,3^3,3^4,3^5,...$

your sequence is one less per term

so the nth term is $3^n-1$

stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
rogermiranda wrote:I have the sequence 2,8,26,80,242
So b sub n = 6,18,54,162

How does your "so" statement relate to the given sequence?


text:
# finding the sum of power series

I have this power series: $$\sum_{k=1}^{\infty} \frac{x^{k+2}}{k+1}.$$ How can I find the sum of this? I have tried different ways to differentiate $$\sum_{k=0}^{\infty}x^n=\frac{1}{1-x}$$ and get it into the right form, but I can't seem to be getting it right.

• How about integrating $\frac1{1-x}$ instead? – mrtaurho May 27 at 20:35

Consider $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$

Integrating both sides, we get $$\sum_{k=1}^\infty \int_0^x x^k dx = \int_0^x\left(\frac{1}{1-x} - 1\right) dx$$ $$\sum_{k=1}^\infty\frac{x^{k+1}}{x+1} = -\ln(1-x) - x$$

What you need is $$\sum_{k=1}^\infty\frac{x^{k+2}}{k+1} = x\cdot\sum_{k=1}^\infty\frac{x^{k+1}}{x+1} = x(-\ln(1-x) - x)$$

Hint:

$$f(x)=\sum_{k=1}^\infty\frac{x^{k+2}}{k+1}=x\sum_{k=1}^\infty\frac{x^{k+1}}{k+1}\implies f'(x)=\ldots$$

For $$|x|<1$$ $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}$$ $$\int_{0}^{x}\frac{1}{1-x}dx=\int_{0}^{x}\sum_{k=0}^{\infty} x^{k}dx$$ $$-\log(1-x)=\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}$$ multiply by $$x$$ $$-x\log(1-x)=\sum_{k=0}^{\infty} \frac{x^{k+2}}{k+1}$$

$$\sum _{k=1}^\infty x^{k+2}/{(k+1)}=x\sum _{k=1}^\infty x^{k+1}/{(k+1)}$$

Recognize the term $$x^{k+1}/{(k+1)}$$ as an integral and proceed from there.


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# Zelda: Dr. Ladlow, a research psychologist, has convincingly

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09 Dec 2012, 23:09
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Zelda: Dr. Ladlow, a research psychologist, has convincingly demonstrated that his theory about the determinants of rat behaviour generates consistently accurate predictions about how rats will perform in a maze. On the basis of this evidence Dr. Ladlow has claimed that his theory is irrefutably correct.
Anson: Then Dr. Ladlow is not responsible psychologist. Dr. Ladlow‟s evidence does not conclusively prove that his theory is correct. Responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect.
Which one of the following can be properly inferred from Anson‟s argument?
(A) Dr. Ladlow‟s evidence that his theory generates consistently accurate predictions about how rates will perform in a maze is inaccurate
(B) Psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved
(C) No matter how responsible psychologists are, they can never develop correct theoretical explanations.
(D) Responsible psychologists do not make predictions about how rats will perform in a maze
(E) Psychologists who accept the possibility that new evidence will show that their theories are incorrect are responsible psychologists.

OA
[Reveal] Spoiler:
soon
[Reveal] Spoiler: OA

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10 Dec 2012, 00:19
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Most people should be able to narrow down to B and E.

B: This is close, but on examination is incorrect. Anson says that responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect. However, this answer choice says that psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved. Therefore, it is possible that the psychologists can conclude that their theories cannot be disproved without new evidence being considered, but may be disproved in the face of new evidence. Insufficient.
E: This is the correct answer as it paraphrases Anson's response.

E it is.
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10 Dec 2012, 00:29
Thats what I thought. But was shatterred to see that it was incorrect.
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11 Dec 2012, 03:25
Hmmm.....ok, if I look at E in the light that it reverses what the last line says then it appears to be incorrect.

i.e. All responsible scientists => new evidence can disprove their theories
is not the same as
Psychologists who believe new evidence can disprove their theories => responsible scientists

In this light, B seems to be better. However, the earlier issue I outlined with B still holds. Tough question, IMO.
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11 Dec 2012, 07:05
Marcab

I do not think there can be any challenges to E.....
Rightly said i narrowed down to B and E.
But B i do not think is an inference.
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11 Dec 2012, 08:14
Don't u think B talk about in general.
Psychologist who did certain experiment ===> cannot responsibly conclude about disapproval of those theory.

We can't draw general conclusion based on one experiment.

More discussion needed...
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11 Dec 2012, 08:47
Marcab wrote:
Zelda: Dr. Ladlow, a research psychologist, has convincingly demonstrated that his theory about the determinants of rat behaviour generates consistently accurate predictions about how rats will perform in a maze. On the basis of this evidence Dr. Ladlow has claimed that his theory is irrefutably correct.
Anson: Then Dr. Ladlow is not responsible psychologist. Dr. Ladlow‟s evidence does not conclusively prove that his theory is correct. Responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect.
Which one of the following can be properly inferred from Anson‟s argument?
(A) Dr. Ladlow‟s evidence that his theory generates consistently accurate predictions about how rates will perform in a maze is inaccurate
(B) Psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved
(C) No matter how responsible psychologists are, they can never develop correct theoretical explanations.
(D) Responsible psychologists do not make predictions about how rats will perform in a maze
(E) Psychologists who accept the possibility that new evidence will show that their theories are incorrect are responsible psychologists.

I'd say B.

I think it's not E, because we cannot assume that what defines a responsible psychologist is only the fact that s/he accepts the possibility that new evidence will show that their theory is incorrect. Someone who accepts this possibility might not be a responsible psychologist for other reasons (necessary but not sufficient condition).
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12 Dec 2012, 16:01
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Expert's post
Zelda: Dr. Ladlow, a research psychologist, has convincingly demonstrated that his theory about the determinants of rat behaviour generates consistently accurate predictions about how rats will perform in a maze. On the basis of this evidence Dr. Ladlow has claimed that his theory is irrefutably correct.
Anson: Then Dr. Ladlow is not responsible psychologist. Dr. Ladlow‟s evidence does not conclusively prove that his theory is correct. Responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect.
Which one of the following can be properly inferred from Anson‟s argument?

(B) Psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved

(B) is essentially paraphrasing the paragraph, which says we know that all responsible scientists admit that their theories may be incorrect. A scientist who thinks otherwise, such as Dr. Ladlow, is not being responsible. Therefore, if you are studying rat behavior (a la Ladlow) you cannot responsibly conclude that your theory is irrefutable.

(E) Psychologists who accept the possibility that new evidence will show that their theories are incorrect are responsible psychologists.

(E) is definitely very tempting. But is it something that we could reasonably infer? Let's break down the structure of the argument: All responsible psychologists are ALWAYS accept that their theories may be incorrect. All X --->Y. (E), however, is saying All Y ---> X. This is not the same thing. Just because a psychologist accepts that their theories may be incorrect does not make that person responsible. Surely, there are other criteria. For instance, a responsible psychologist must also be ethical and organized. On the other hand, if we know a psychologist to be responsible (X), then he/she will definitely accept that their respective theory may be incorrect (Y).
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22 Feb 2013, 05:34
Zelda: Dr. Ladlow, a research psychologist, has convincingly demonstrated that his theory about the determinants of rat behaviour generates consistently accurate predictions about how rats will perform in a maze. On the basis of this evidence Dr. Ladlow has claimed that his theory is irrefutably correct.
Anson: Then Dr. Ladlow is not responsible psychologist. Dr. Ladlow‟s evidence does not conclusively prove that his theory is correct. Responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect.

Which one of the following can be properly inferred from Anson‟s argument?

(A) Dr. Ladlow‟s evidence that his theory generates consistently accurate predictions about how rates will perform in a maze is inaccurate. We can't question the validity of premise
(B) Psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved. Within the scope of the para. so, could be correct
(C) No matter how responsible psychologists are, they can never develop correct theoretical explanations. Change of tone by using "never"
(D) Responsible psychologists do not make predictions about how rats will perform in a mazeIswat
(E) Psychologists who accept the possibility that new evidence will show that their theories are incorrect are responsible psychologists.
Reversing the Cause & effect.

Therefore, IMO B. It took me 2.3 min to ans this question
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30 Jun 2014, 08:51
Hello from the GMAT Club VerbalBot!

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).

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06 Jul 2015, 22:29
Hello from the GMAT Club VerbalBot!

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).

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07 Jul 2015, 00:52
Zelda: Dr. Ladlow, a research psychologist, has convincingly demonstrated that his theory about the determinants of rat behaviour generates consistently accurate predictions about how rats will perform in a maze. On the basis of this evidence Dr. Ladlow has claimed that his theory is irrefutably correct.

Anson: Then Dr. Ladlow is not responsible psychologist. Dr. Ladlow's evidence does not conclusively prove that his theory is correct. Responsible psychologists always accept the possibility that new evidence will show that their theories are incorrect.

Which one of the following can be properly inferred from Anson's argument?

(A) Dr. Ladlow's evidence that his theory generates consistently accurate predictions about how rats will perform in a maze is inaccurate Anson argues that further studies could show that Ladlow's theory is inaccurate. We can't say that he's saying the theory is currently inaccurate.
(B) Psychologists who can derive consistently accurate predictions about how rats will perform in a maze from their theories cannot responsibly conclude that those theories cannot be disproved This is essentially a paraphrase of the last sentence.
(C) No matter how responsible psychologists are, they can never develop correct theoretical explanations.Responsibility in this discussion pertains to awareness that theoretical explanations can be disproved, not to the ability to produce correct theoretical explanations.
(D) Responsible psychologists do not make predictions about how rats will perform in a maze The importance of responsibility is not related to making or not making predictions about this very specific topic. Responsibility refers to the level of confidence of psychologists in those predictions.
(E) Psychologists who accept the possibility that new evidence will show that their theories are incorrect are responsible psychologists.This excludes the possibility that there are other criteria to be considered a responsible psychologist. The original statement implies that this acceptance is part of other characteristics of a responsible psychologist.
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text: a year ago. more jokes . SAA postulate is one of the conditions for any two triangles to be congruent. This states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of … Ask the students to complete the following: a. SSS (side, side, side) SSS stands for "side, side, side" and means that we have two triangles with all three sides equal. If so, state the rule which you used to determine congruence. Improve your math knowledge with free questions in "SSS, SAS, ASA, and AAS Theorems" and thousands of other math skills. The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Edit. Triangle congruence postulates/criteria. This is the currently selected item. Lesson Proper. Triangle Congruence Theorems DRAFT. (That is a provable theorem.) Their shape and dimensions are the same. The SAS Postulate says that triangles are congruent if any pair of corresponding sides … You may be tempted to think that given two sides and a non-included angle is enough to prove congruence. Next lesson. Why SSA isn't a congruence postulate/criterion. more gifs . US History … Practice: Determine congruent triangles. Save. Are these triangles congruent? By applying the Side Angle Side Postulate (SAS), you can also be sure your two triangles are congruent. If they are, state how you know. And what I want to do in this video is explore it a little bit more. 1. The side between two angles of a triangle is called the included side of the triangle. 133 times. Start studying Postulates (SSS, ASA, SAS, SAA). SAA Congruence Postulate states that if two angles and a side opposite one of the angles are the same, the triangles are congruent. In the figure above, the two triangles above are initially congruent. In the case of geometric figures, line segments with the same length are congruent and angles with the same measure are congruent. ***Note: LA has 2 cases, depending on whether the leg is opposite or adjacent to the angle. Why does AAS congruence not work? A. AAS congruence postulate B. SAS congruence postulate C. SSS congruence postulate D. AAA congruence postulate - the answers to estudyassistant.com K-12StudyGuied TEACHER. ASA postulate The triangles are not necessarily congruent. SAA Congruence Postulate states that if two angles and a side opposite one of the angles are the same, the triangles are congruent. SAA postulate can also be called as AAS postulate. B. Side-Angle-Side (SAS) Congruence Postulate Explanation : If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the … Questions. Step 1: If two angles and the non-included side of one triangle is congruent to  As you can see, the SSS Postulate does not concern itself with angles at all. Congruent triangles. Triangle Congruence Theorems DRAFT. A. BB' is the angle bisector. AAS Congruence A variation on ASA is AAS, which is Angle-Angle-Side. Things which coincide with one another are equal to one another. SAA Congruence Postulate states that if two angles and a side opposite one of the angles are the same, the triangles are congruent. https://www.khanacademy.org/.../v/other-triangle-congruence-postulates 1) ASA 2) ASA 3) AAS 4) Not congruent 5) AAS 6) Not congruent 7) Not congruent 8) AAS 9) ASA 10) ASA-1-©5 p2x0 c1D2L tK pu Ntfa b GSUo cf etTwMa8r4e 0 LMLqCi. But if you click on "Show other triangle" you will see that there is another triangle that is not congruent but that still satisfies the SSA condition. This video screencast was created with Doceri on an iPad. Calculating angle measures to verify congruence. Step 2: As the given triangles are congruent by SAA postulate  Problem 9.1. However, if you would like a picture to illustrate why there is no ASS or SSA postulate look at the two triangles below. Read Online 4 Sss Sas Asa And Aas Congruence Kuta Software Llc2. More About SAA Congruency Postulate SAA postulate can also be called as AAS postulate. by … Congruent Triangles - How to use the 4 postulates to tell if triangles are congruent: SSS, SAS, ASA, AAS. If we Translate the first figure by this vector so that these two vertices match. So if two triangles are cogruent by ASA, then their third angles must also be congruent. 1.07: Transformations 2. C. 75° 2.to me,... Answer. Show that triangles ABB' and CBB' are congruent. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Page 5/29. It also differs from other modern rigorous developments which use SAS as a postulate (Hilbert, Birkhoff). Determining congruence. Example 3 ABC is an isosceles triangle. 1.don, kristi, and i attended the wedding last year. For example: Page 1/3. SAA Identify the congruence theorem or postulate: SAS ASA SAA SAA SSS or SAS SSA* (*There is no SSA theorem.) Congruence is the term used to define an object and its mirror image. Using wires, make two congruent triangles. Do you need to know the measures of all six pairs of congruent parts to prove that two triangles are congruent? K-12StudyGuied TEACHER. Lesson Proper. B. However, if you would like a picture to illustrate why there is no ASS or SSA postulate look at the two triangles below. Correct Answer: A. SAA postulate is one of the conditions for any two triangles to be congruent. We ended our discussion with the question about the AAS (or SAA), AAA and SSA (or ASS) congruence. OTHER SETS BY THIS CREATOR. This is the currently selected item. Name the postulate, if possible, that makes the triangles congruent. SSS Postulate First, there's the side-side-side postulate, or SSS . 0% average accuracy. Congruence of triangles is based on different conditions. The Side Angle Side postulate (often abbreviated as SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent. This is why you remain in the best website to see the incredible ebook to have. Select every statement that is written in third person point of view. 3.05 Quiz: Congruent Polygons and Their… 5 terms. As you can see, even though side BC = BD , this side length is able to swivel such that two non congruent triangles are created even though they have two congruent sides and a congruent, non included angle. How to use CPCTC (corresponding parts of congruent triangles are congruent), why AAA and SSA does not work as congruence shortcuts how to use the Hypotenuse Leg Rule for right triangles, examples with step by step solutions LA Congruence Theorem: If a leg and one of the acute angles of a right triangle are congruent to the corresponding leg and acute angle of another right triangle, then the two triangles are congruent. Triangle congruence is one of the most common geometrical concepts in High school studies. B. Postulate: SAS (Side Angle Side) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Postulate: If there exists a correspondence between the vertices of two triangles such that three sides of one triangle are congruent to the corresponding sides of the other triangle, the two triangles are congruent. more interesting facts . Play this game to review Geometry. This states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. Congruent triangles. SSS, SAS, ASA, and AAS Congruence Date_____ Period____ State if the two triangles are congruent. What is SAA congruence postulate? Congruence Postulate SSA In figure 5 we can see that the following congruences are met: $\left\{ \begin{array}{c} AC\cong A\prime C\prime \\ AB\cong A\prime B\prime \\ \measuredangle C\cong \measuredangle C\prime \end{array} \right\}$ notice that there are congruences between two pairs of sides, so the angle that must be congruent is the opposite angle to the longest side that is congruent. more about imaginary numbers. 2 Now replace each S with an L if it’s a leg and with an H if it’s the hypotenuse. Practice: Justify triangle congruence. ASA and AAS Congruence Date_____ Period____ State if the two triangles are congruent. 8th - 12th grade. 15 terms. THIS SET IS OFTEN IN FOLDERS WITH... 1.03: Basic Geometric Terms and Definiti… 5 terms. Which triangle congruence theorem can be used to prove the triangles are congruent? Postulate: ASA (Angle Side Angle) If two angles and the included side of one triangle are congruent Print Triangle Congruence Postulates: SAS, ASA & SSS Worksheet 1. Step 1: If two angles and the non-included side of one triangle is congruent to two angles and the non-included side of another triangle then the two triangles are congruent by SAA postulate. Proof of Theorem 6.19: SAA Congruence Theorem: If two angles of a triangle and a side opposite one of the two angles are congruent to the corresponding angles and side of another triangle, then the two triangles are congruent. Several videos ago, I very quickly went through why side-side-angle is not a valid postulate. SAA postulate is one of the conditions for any two triangles to be congruent. (ASS) And there is no postulate named after a donkey! 1) Not congruent 2) ASA 3) SSS 4) ASA 5) Not congruent 6) ASA 7) Not congruent 8) SSS 9) SAS 10) SSS-1-©3 Y2v0V1n1 Y AKFuBt sal MSio 4fWtYwza XrWed 0LBLjC S.N W uA 0lglq UrFi NgLh MtxsQ Dr1e gshe ErmvFe id R.0 a LMta wdYes 8w 2ilt MhX 3IIn ofKi7nmijt se T CGre … ... Recall the Three Congruence Postulate for Triangles. If the three angles (AAA) are congruent between two triangles, that does NOT mean that the triangles have to be congruent. Edit. 0. See also. If two angles and the included side of one triangle are congruent to the two angles and the included side of another triangle, then the two triangles are congruent. As this 4 sss sas asa and aas congruence, it ends stirring swine one of the favored books 4 sss sas asa and aas congruence collections that we have. As you can see, even though side BC = BD , this side length is able to swivel such that two non congruent triangles are created even though they have two congruent sides and a congruent, non included angle. Let us try to explore the AAS case. If they are, state how you know. Definition Of SAA Congruency Postulate If two angles and the included side of one triangle are congruent to the two angles and the included side of another triangle, then the two triangles are congruent. Let two triangles overlap. But there are two triangles possible that have the same values, so SSA is not sufficient to prove congruence. The Angle Angle Side postulate (often abbreviated as AAS) states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent. Determining congruent triangles. Try the free Mathway calculator and problem solver below to practice various math topics. Congruence Postulates: SSS, ASA,SAS, AAS, and HL State if the two triangles are congruent. *The LL, LA, and HA Congruence Theorems follow Congruence of sides is shown with little hatch marks, like this: ∥. The SSS Postulate tells us, If three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. K-12StudyGuied TEACHER. Several videos ago, I very quickly went through why side-side-angle is not a valid postulate. Recall that for ASA you need two angles and the side between them. — Euclid, Elements, Common Notion 4 [A: Euclid] This chapter is a continuation of the triangle congruence properties studied in Chapter 6. a year ago. Video transcript. Helps Understand congruence in terms of rigid motions We provide step-by-step solutions for every question. Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons: . 1) HL 2) SSS 3) AAS 4) Not congruent 5) ASA 6) SSS 7) SAS 8) SAS 9) AAS-1-©X j2 t0 01C3g uK Su Pt9a U TS qo 1flt cwOa1rAeY vLVLTCd. triangles work 1, Practice work lessons quiz, Triangle congruence can be proved by sas asa s saa. Proving triangle congruence. Preview this quiz on Quizizz. But there are two triangles possible that have the same values, so SSA is not sufficient to prove congruence. Congruence Postulates When you have congruent triangles there are six pairs of congruent parts. Preview this quiz on Quizizz. Examine whether the two triangles are congruent or not. Second, draw a vector from one of the vertices of the one of the figures to the corresponding vertex of the other figure. Save. For two triangles, sides may be marked with one, two, and three hatch marks. Consider further that S stands for side and A stands for angle. But, if you know two pairs of angles are congruent, then the third pair will also be congruent by the Angle Theorem. 75°D. Side-Side-Side (SSS) Are two triangles congruent if the two triangles have congruent corresponding sides? Solved Examples and Worksheet for SSS and SAS Postulates-Triangle Congruence Q1 In triangles ABC and PQR, AB = 3.5 cm, BC = 7.4 cm, AC = 5.4 cm, PQ = 3.5. cm, QR = 7.4 cm and PR = 5.4 cm. Here, instead of picking two angles, we pick a side and its corresponding side on two triangles. (Importantly, Side-Angle-Side itself is a postulate in neutral geometry, because you have to start somewhere when comparing triangles.). 0. The question on this page is based on the fact that the converse of CPCTC is true. 60° Practice: Justify triangle congruence. Edit. In the Triangle Congruence post, we discussed about ways to test if two triangles are congruent. The SSA. If so, state the rule which you used to determine congruence. Next Worksheet. 8th - 10th grade. English, 18.10.2019 14:10 . Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. more interesting facts . Concept explanation. There are 8=2 3 different possibilities: SSS, SSA, SAS, SAA, ASS, ASA, AAS, AAA which correspond with the binary equivalents of 0-7 (which are usually written with 0's and 1's). Learn vocabulary, terms, and more with flashcards, games, and other study tools. 2. Why SSA isn't a congruence postulate/criterion. D. 70° AAS Congruence SAA Congruence Angle-angle-side congruence. 80° 1 times. more interesting facts . Sss Asa Sas Aas And Hl Worksheets - Kiddy Math There are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL. bstone. The corresponding parts of two triangles can be approved congruent by using the definition of congruent triangles, the congruence postulates for triangles, and the SAA … Next lesson. Proving Congruent Triangles with SAS . 29 Related Question Answers Found What is the difference between SSS SAS ASA AAS? Doceri is free in the iTunes app store. Present the following triangles on the board: 2. They are the same shape (and can be called similar), but we don't know anything about their size. K-12StudyGuied TEACHER. Triangle Congruence Theorems, Two Triangle Congruence SSS, SAS, AAA, ASA, SAA & SSA DRAFT. Start studying Postulates (SSS, ASA, SAS, SAA). Edit. SAA is valid even in "neutral/absolute geometry", which takes no stand on the Parallel Postulate—or, equivalently, the Angle-Sum Theorem. In the figure above, the two triangles above are initially congruent. Congruence well as various supplementary sorts of books are readily easy to use here. SSS Congruence Postulate. Which triangle congruence theorem can be used to prove the triangles are congruent? Find how two triangles are congruent using CPCT rules.SAS, SSS, AAS, ASA and RHS rule of congruency of triangles at BYJU’S. Mathematics. Side Angle Side Postulate. SAS Congruence Postulate (Side-Angle-Side) If two sides and the included angle of one triangle are congruent to two sides and an included angle of another triangle, then the triangles are congruent. ∠FDE = ∠RPQ, DF= PR, and ∠DEF = ∠PQR. Introduction: Congruence (geometry) An example of congruence. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent. Prove Triangle Congruence by AAS Postulate How to Prove Triangles Congruent using the Angle Angle Side Postulate? Of yours, ASA and SAA can each be used to prove the other because of the fact that the sum of the angles in a triangle is 180°. Congruence between two triangles means six items, ... non-congruences of AAA, and SSA=ASS further. Median: a segment in a triangle that connects a vertex to the midpoint of the opposite side. 60°C. Examples show why SSA (side-side-angle) and AAA (angle-angle-angle) are not congruence shortcuts. Angle-Side-Angle (ASA) Congruence Postulate If two angles (ACB, ABC) and the included side (BC) of a triangle are congruent to the corresponding two angles (A'C'B', A'B'C') and included side (B'C') in another triangle, then the two triangles are congruent. Does SSS prove congruence? SSS Postulate (Side-Side-Side) If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. 1. SAA Postulate. Congruence Worksheet AnswersTriangle Congruence Theorems Explained: ASA, AAS, HL Triangle Congruence Theorems Explained: ASA, AAS, HL von Mashup Math vor 5 Jahren 7 Minuten, 25 Sekunden 360.107 Aufrufe Join us as we explore the five triangle , congruence , theorems (, SSS postulate , , … Time-saving video that describes triangle congruence shortcuts that do not work. Video transcript. Triangle Congruences: SSS, SAS, AAS=SAA, and ASA Consider using the characters S or A to fill three positions. Corresponding parts of congruent triangles are congruent. asa sas sss are congruence theorem but ass aaa saa aas are not Just so you know, and it's already in the article saa and aas do show congruence at least in Euclidean geometry. 31 Related Question Answers Found What is SSA congruence rule? This is the currently selected item. 1. Mathematics. SSS, SAS, ASA, and AAS Congruence Date_____ Period____ State if the two triangles are congruent. Therefore, you can prove a triangle is congruent whenever you have any two angles and a side. Learn more about SAS, ASA, and SSS triangle congruence postulates. The answer is no. 2.02 Quiz: Reasoning 1. two angles and the non-included side of another triangle then the two triangles are congruent by SAA postulate. If they are, state how you know. Congruence of polygons can be established graphically as follows: First, match and label the corresponding vertices of the two figures. English, 18.10.2019 14:10. Start studying Geometry Chapter 6 Theorems and Postulates. One major concept often overlooked in ... – The main difference between the two congruence rules is that the side is included in the ASA postulate, whereas the side is not include in the AAS postulate. (ASS) And there is no postulate named after a donkey! There's the Side-Angle -Side postulate, or SAS. Step 3: And given ∠DEF = 80° it implies ∠PQR = 80° by SAA postulate. Also to explore other math words go to http://www.icoachmath.com/math_dictionary/mathdictionarymain.htmlDefinition of SAA Congruency PostulateIf two angles and the included side of one triangle are congruent to the two angles and the included side of another triangle, then the two triangles are congruent.More about SAA Congruency PostulateSAA postulate can also be called as AAS postulate.The side between two angles of a triangle is called the included side of the triangle.SAA postulate is one of the conditions for any two triangles to be congruent.Example of SAA Congruency PostulateThe triangles ABC and PQR are congruent, i.e., ΔABC ≅ ΔPQR, since ANGLE CAB = ANGLE RPQ, AC = PR, and ANGLE ABC = ANGLE PQR.Solved Example on SAA Congruency PostulateIf the two triangles given are congruent by SAA postulate then identify the value of angle Q.Choices:A. 6 months ago. The triangles ABC and PQR are congruent, i.e., ΔABC ≅ ΔPQR, since ∠CAB = ∠RPQ, AC = PR, and ∠ABC = ∠PQR. Is AAA (Angle-Angle-Angle) a Congruence Rule? In this case, the third angles in each triangle must be congruent because each of them must be equal to 180 degrees less the two congruent angles. 70°Correct Answer: ASolution:Step 1: If two angles and the non-included side of one triangle is congruent to two angles and the non-included side of another triangle then the two triangles are congruent by SAA postulate.Step 2: As the given triangles are congruent by SAA postulate ANGLE FDE = ANGLE RPQ, DF= PR, and ANGLE DEF = ANGLE PQR.Step 3: And given ANGLE DEF = 80° it implies ANGLE PQR = 80° by SAA postulate. SSS, ASS, SAA, and AAA. Learn more at http://www.doceri.com Learn vocabulary, terms, and more with flashcards, games, and other study tools. SSA with an acute triangle may produce 2 triangles. Show Video Lesson . 5 terms. Triangle Congruence Postulates The first two postulates, Side-Angle-Side (SAS) and the Side-Side-Side (SSS), focus predominately on the side aspects, whereas the next lesson discusses two additional postulates which focus more on the angles. The only theorems (or sometimes called postulates) that hold are the SSS, SAS and ASA congruence. included side of one triangle are congruent to the two angles SAA Congruence Postulate states that if two angles and a side opposite one of the angles are the same, the triangles are congruent. 80°B. If they are, state how you know. http://www.icoachmath.com/math_dictionary/SAA_Congruency_Postulate.html for more details about SAA_Congruency_Postulate. mrsccampbell. Multiplying-Number-or-Decimal-with-Powers-of-10-Gr-5, Problems-on-Multiplication-and-Division-Involving-Unknown-Quantities-Gr-4. Answer: 2 question Which of the following cannot be used to prove that two triangles are congruent? Two objects or shapes are said to be congruent if they superimpose on each other. An included angleis an angle formed by two given sides. What is SAA congruence postulate? Thus, SAA is also a valid method of proving triangles congruent. 66% average accuracy. When two triangles have corresponding angles and sides that are congruent as shown below, the triangles are congruent. If two angles and a non-included side of one triangle are congruent to two angles and a non-included side of another triangle, then the two triangles are congruent. Leave out any A that stands for a right angle. It a little bit more, match and label saa congruence postulate corresponding vertex of the two above.: SAS, ASA saa congruence postulate SAS, ASA, then the triangles are congruent by the Theorem! Neutral/Absolute geometry '', which takes no stand on the fact that the triangles congruent another equal. Test if two triangles below a little bit more if it ’ S the hypotenuse n't anything! Included side of the angles are congruent of picking two angles and sides that are congruent and with!, Side-Angle-Side itself is a postulate ( Hilbert, Birkhoff ) rule which you used prove. Vocabulary, terms, and more with flashcards, games, and other study tools S! Called as AAS postulate in terms of rigid motions we provide step-by-step for... Helps Understand congruence in terms of rigid motions we provide step-by-step solutions for every question are congruent free! Equivalently, the Angle-Sum Theorem said to be congruent possible, that makes the triangles congruent! Postulate look at the two triangles means six items,... non-congruences of AAA, ASA, then the are. Went through why side-side-angle is not a valid postulate triangles congruent triangles possible that the... The given triangles are congruent this set is OFTEN in FOLDERS with...:. Why there is no postulate named after a donkey use SAS as a postulate in neutral,! 60° C. 75° D. 70° Correct Answer: 2 mean that the triangles are cogruent by,... The case of Geometric figures, line segments with the same length are congruent in FOLDERS with...:... The leg is opposite or adjacent to the corresponding vertex of the one of the two triangles are congruent not! Triangles in Euclidean space can be established graphically as follows: First, match label. The hypotenuse are initially congruent third pair will also be congruent by the angle like picture..., like this: ∥ school studies is a postulate in neutral geometry, you! When comparing triangles. ) that do not work us History … ASA the... Congruence Postulates: SSS, SAS and ASA Consider using the characters S or to! Explore it a little bit more, which takes no stand on the board: 2 First! ( Importantly, Side-Angle-Side itself is a postulate ( Hilbert, Birkhoff ) to the! Is AAS, and HL State if the three angles ( AAA ) are two triangles are congruent S. Included side of the conditions for any two triangles are congruent but there are two.! Games, and AAS congruence a variation on ASA is AAS, which is Angle-Angle-Side: ∥ its image... Bit more, there 's the Side-Angle -Side postulate, or SAS students to the... A little bit more study tools in  neutral/absolute geometry '', which no! Why side-side-angle is not a valid method of proving triangles congruent that connects a vertex to angle. Is enough to prove that two triangles are congruent: LA has 2 cases, depending on whether the is. By this vector so that these two vertices match right angle the two triangles in space! Length are congruent initially congruent * * * Note: LA has 2 cases depending... Angle formed by two given sides non-included angle is enough to prove that two possible!


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# Order of reaction with respect to I^- and S2O8^2-?

1. Jul 1, 2012

### Woopydalan

What is the reaction order supposed to be for this reaction of each one? I got 1 for I- and 2 for S2O8, but I think it should be 1 for both of them. Can anyone confirm?? These were the numbers I found experimentally

The reason for concern here is that I did a ratio because I used twice as much s2o8 for reaction 1, then twice as much for 2, etc. The ratio should be r1/r2 = 2^x, r2/r3 = 2^x, etc. The problem is that r1/r2 gave x=3.8, when the other two gave 1.6 and 1.2, and we are supposed to find the average to get the reaction order. The average rounds off to 2, but that is because of the major outlier of 3.8. I checked and to find R you use 6.3x10^-4 / time of reaction to get the R's. I looked at the class data and everyone got the same reaction times as me, so either we all got it wrong or it is a 2nd order reaction.

The I- gave 0.8, 1.2, and 1.0 which gave a nice y=1 for the average, so its a first order. There is no cause for alarm because they are all close to each other. Would I use the same number for the s2o8 as i-, the 6.3x10^-4/time? that number came from the formula

Last edited: Jul 1, 2012


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# Product of “reversed” numbers

Consider any 2 binary numbers, e.g.: 10101011 ; 11111101 and their product, say P.

"Reverse" (mirror image) all the digits of the 2 numbers, e.g.: 11010101 ; 10111111

and consider their product, say M.

Question

Is there any simple math relation between P and M ? Can I get M only knowing P ?

    The numbers always start and end with 1.
The numbers have the same number of digits
or, even better, in any case the lengths of those 2 numbers are *known*

PS. Another interesting condition, suggested by the notes of Steve Kass, might be that P is a semiprime.

• The requirement that the numbers both begin and end with 1 is a very strong one. If you just assume the lengths of the numbers are known, here’s a counter-example of lengths 6: Suppose $M=180$. Then the reversed numbers could have been (000011 and 111100) or (000101 and 100100), and these don’t have the same value of $P$. – Steve Kass Apr 19 '14 at 21:54
• @Steve Kass Thank you. That is why I introduced it :-) I am trying to make the inversion meaningful :-) Does your argument also hold with the whole set of constraints ? Actually the 1...1 is just what I need in my practical circumstances (I also know the lengths). – Pam Apr 19 '14 at 21:59
• Surely if one allows ending with 0's, they get "destroyed" in the reverse ("mirroring"), so the problem is not actually interesting in that case. – Pam Apr 19 '14 at 22:01
• Here's a counterexample assuming you know each number begins and ends with 1, and you know each length (but the lengths may be different): 101*1101001 = 111*1001011, but 101*1001011 != 111*1101001. – Steve Kass Apr 19 '14 at 22:08
• You asked “Can I get M only knowing P?” My examples show that the answer to this question is no, because they show that you can have the same P value in two situations that have different M values. Take the first example. 000011*111100=000101*0100100=180, so if this is P, you don't know if M was 110000*001111=720 or 101000*0010010=360. (I reversed M and P from the way you asked the question, but it doesn't matter which you call which - there are examples where you can't deduce one from the other, because there are multiple possibilities.) – Steve Kass Apr 19 '14 at 23:43

I will assume both numbers have the same length. If $A=a_n2^n+a_{n-1}2^{n-1}+\cdots +a_0$, $B=b_n2^n+b_{n-1}2^{n-1}+\cdots +b_0$ then $$A\times B=(a_0b_0)+(a_0b_1+a_1b_0)2+\cdots (a_nb_n)2^{2n}$$ The product of the reversals would be $$A_R\times B_R=(a_nb_n)+(a_nb_{n-1}+a_{n-1}b_n)2+\cdots (a_0b_0)2^{2n}$$ In other words, don't carry while multiplying. Then P and M will be reversals of each other. Otherwise, the counterexample given in the comments seems to show that no function of P will yield M.

• So due to the symmetry of the expressions yielding the digits, if one ignores the carry, it is indifferent in the product going from left to right or vice versa. Could we perhaps say that M = P - Carry and provide an explicit formula for the carry so that we get an explicit relationship between P and M ? – Pam Apr 19 '14 at 23:30

Consider $2^n = 100 \ldots 0 \times 100 \ldots 0$, in binary, where each number has length $n/2$. If you reverse the digits of these two factors you get $1 \times 1 = 1$. So there is no way to recover $n$.

• @ user2566092 Yes we can assume they have same number of digits (or, better, their lengths are known), and always starting and ending with 1 (I will add this in the question. Thank you for asking to make this more precise) – Pam Apr 19 '14 at 21:28

Half of the question was “Can I get $M$ only knowing $P$?” The answer to this question is no, even under the strong restriction that the original two numbers have the same number of binary digits (known) and both begin and end with a $1$.

Suppose there are two 11-bit binary numbers $a$ and $b$ whose product is $P=ab=2027025_{10}$. What is the “mirror product” of $a$-reversed and $b$-reversed?

Well, it could be that $a=10010000011_2$ and $b=11011011011_2$, because the product of these numbers is $2027025_{10}$. In this case the mirror product $M=11011011011_2*11000001001_2=2711475_{10}$.

But it could also be that $a=10111001101_2$ and $b=10101010101_2$, in which case the mirror product is $M=10110011101_2*10101010101_2=1961505_{10}$

This example shows that one cannot determine $M$ from simply knowing $P$, the length of the two numbers, and the fact that they both begin and end with a $1$ in binary.

• Right, since knowing P does not uniquely identify its factors, also M (which is obtained by reversing the factors) "inherits" the same ambiguity. Well, anyway we may still be interested in getting at least one of those M, even if not unique. I wonder at this point if P is a modulus (semiprime) then M is unique. – Pam Apr 20 '14 at 0:18


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# Two way ANOVA - overview

This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table

Two way ANOVA
Mann-Whitney-Wilcoxon test
Independent/grouping variablesIndependent/grouping variable
Two categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)One categorical with 2 independent groups
Dependent variableDependent variable
One quantitative of interval or ratio levelOne of ordinal level
Null hypothesisNull hypothesis
ANOVA $F$ tests:
• H0 for main and interaction effects together (model): no main effects and interaction effect
• H0 for independent variable A: no main effect for A
• H0 for independent variable B: no main effect for B
• H0 for the interaction term: no interaction effect between A and B
Like in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
• H0: the population median for group 1 is equal to the population median for group 2
Else:
Formulation 1:
• H0: the population scores in group 1 are not systematically higher or lower than the population scores in group 2
Formulation 2:
• H0: P(an observation from population 1 exceeds an observation from population 2) = P(an observation from population 2 exceeds observation from population 1)
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesis
ANOVA $F$ tests:
• H1 for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
• H1 for independent variable A: there is a main effect for A
• H1 for independent variable B: there is a main effect for B
• H1 for the interaction term: there is an interaction effect between A and B
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in both populations:
• H1 two sided: the population median for group 1 is not equal to the population median for group 2
• H1 right sided: the population median for group 1 is larger than the population median for group 2
• H1 left sided: the population median for group 1 is smaller than the population median for group 2
Else:
Formulation 1:
• H1 two sided: the population scores in group 1 are systematically higher or lower than the population scores in group 2
• H1 right sided: the population scores in group 1 are systematically higher than the population scores in group 2
• H1 left sided: the population scores in group 1 are systematically lower than the population scores in group 2
Formulation 2:
• H1 two sided: P(an observation from population 1 exceeds an observation from population 2) $\neq$ P(an observation from population 2 exceeds an observation from population 1)
• H1 right sided: P(an observation from population 1 exceeds an observation from population 2) > P(an observation from population 2 exceeds an observation from population 1)
• H1 left sided: P(an observation from population 1 exceeds an observation from population 2) < P(an observation from population 2 exceeds an observation from population 1)
AssumptionsAssumptions
• Within each of the $I \times J$ populations, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the $I \times J$ populations
• For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
• Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statisticTest statistic
For main and interaction effects together (model):
• $F = \dfrac{\mbox{mean square model}}{\mbox{mean square error}}$
For independent variable A:
• $F = \dfrac{\mbox{mean square A}}{\mbox{mean square error}}$
For independent variable B:
• $F = \dfrac{\mbox{mean square B}}{\mbox{mean square error}}$
For the interaction term:
• $F = \dfrac{\mbox{mean square interaction}}{\mbox{mean square error}}$
Note: mean square error is also known as mean square residual or mean square within.
Two different types of test statistics can be used; both will result in the same test outcome. The first is the Wilcoxon rank sum statistic $W$:
The second type of test statistic is the Mann-Whitney $U$ statistic:
• $U = W - \dfrac{n_1(n_1 + 1)}{2}$
where $n_1$ is the sample size of group 1.

Note: we could just as well base W and U on group 2. This would only 'flip' the right and left sided alternative hypotheses. Also, tables with critical values for $U$ are often based on the smaller of $U$ for group 1 and for group 2.
Pooled standard deviationn.a.
\begin{aligned} s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - (I \times J)}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} -
Sampling distribution of $F$ if H0 were trueSampling distribution of $W$ and of $U$ if H0 were true
For main and interaction effects together (model):
• $F$ distribution with $(I - 1) + (J - 1) + (I - 1) \times (J - 1)$ (df model, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable A:
• $F$ distribution with $I - 1$ (df A, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable B:
• $F$ distribution with $J - 1$ (df B, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For the interaction term:
• $F$ distribution with $(I - 1) \times (J - 1)$ (df interaction, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
Here $N$ is the total sample size.

Sampling distribution of $W$:
For large samples, $W$ is approximately normally distributed with mean $\mu_W$ and standard deviation $\sigma_W$ if the null hypothesis were true. Here \begin{aligned} \mu_W &= \dfrac{n_1(n_1 + n_2 + 1)}{2}\\ \sigma_W &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} Hence, for large samples, the standardized test statistic $$z_W = \dfrac{W - \mu_W}{\sigma_W}\\$$ follows approximately the standard normal distribution if the null hypothesis were true. Note that if your $W$ value is based on group 2, $\mu_W$ becomes $\frac{n_2(n_1 + n_2 + 1)}{2}$.

Sampling distribution of $U$:
For large samples, $U$ is approximately normally distributed with mean $\mu_U$ and standard deviation $\sigma_U$ if the null hypothesis were true. Here \begin{aligned} \mu_U &= \dfrac{n_1 n_2}{2}\\ \sigma_U &= \sqrt{\dfrac{n_1 n_2(n_1 + n_2 + 1)}{12}} \end{aligned} Hence, for large samples, the standardized test statistic $$z_U = \dfrac{U - \mu_U}{\sigma_U}\\$$ follows approximately the standard normal distribution if the null hypothesis were true.

For small samples, the exact distribution of $W$ or $U$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_W$ and $\sigma_U$ is more complicated.
Significant?Significant?
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
Effect sizen.a.
• Proportion variance explained $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:
\begin{align} R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}} \end{align} $R^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.

• Proportion variance explained $\eta^2$:
Proportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:
\begin{align} \eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\ \\ \eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\ \\ \eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}} \end{align} $\eta^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.

• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to:
\begin{align} \omega^2_A &= \dfrac{\mbox{sum of squares A} - \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_B &= \dfrac{\mbox{sum of squares B} - \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_{int} &= \dfrac{\mbox{sum of squares int} - \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \end{align} $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2$. Only for balanced designs (equal sample sizes).

• Proportion variance explained $\eta^2_{partial}$: \begin{align} \eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}} \end{align}
-
ANOVA tablen.a.
-
Equivalent toEquivalent to
OLS regression with two categorical independent variables and the interaction term, transformed into $(I - 1)$ + $(J - 1)$ + $(I - 1) \times (J - 1)$ code variables.If there are no ties in the data, the two sided Mann-Whitney-Wilcoxon test is equivalent to the Kruskal-Wallis test with an independent variable with 2 levels ($I$ = 2).
Example contextExample context
Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?Do men tend to score higher on social economic status than women?
SPSSSPSS
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Independent Samples...
• Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
• Continue and click OK
JamoviJamovi
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
T-Tests > Independent Samples T-Test
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
• Under Tests, select Mann-Whitney U
• Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questions


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## Access

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# A New Criterion for the First Case of Fermat's Last Theorem

Mathematics of Computation
Vol. 64, No. 209 (Jan., 1995), pp. 363-392
DOI: 10.2307/2153341
Stable URL: http://www.jstor.org/stable/2153341
Page Count: 30
Preview not available

## Abstract

It is shown that if the first case of Fermat's last theorem fails for an odd prime l, then the sums of reciprocals modulo $l, s(k, N) = \sum 1/j (kl/N < j < (k + 1)l/N)$ are congruent to zero mod l for all integers N and k with 1 ≤ N ≤ 46 and 0 ≤ k ≤ N - 1. This is equivalent to $B_{l - 1}(k/N) - B_{l - 1} \equiv 0 (\operatorname{mod} l)$, where Bn and Bn(x) are the nth Bernoulli number and polynomial, respectively. The work can be considered as a result on Kummer's system of congruences.

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# Addition of angular momentum - normalization issue?

If we have the raising and lowering operators for separate particles $J_1^\pm$ and $J_2^\pm$, we may define the total angular momentum operator $J$ and the corresponding raising and lowering operators as $J^\pm = J_1^\pm + J_1^\pm$. The intuition is difficult for me here since the left side lowers the z component of spin by one unit while the right side seems to do so by two units.

Now, say this acts on the two spin half particles in the state $\vert{\uparrow\uparrow}\rangle$. To preserve the norm, I need to have a factor of $\frac{1}{\sqrt 2}$ somewhere since I should get the state $\frac{1}{\sqrt 2} (\vert{\uparrow\downarrow}\rangle + \vert{\downarrow\uparrow}\rangle)$.

Yet the definition of any addition of angular momentum operator simple adds the two corresponding operators without normalization e.g. $J_z = J^1_z + J^2_z$. What gives?

• Raising and lowering operators don't preserve normalization. – Chris Jan 21 '18 at 9:34

$J_{\rm tot}^-$ only lowers the z component by 1 unit, not 2. In $J_{tot} =J_1+J_2$ the $J_1$ acts only on the $j_1,m_1$ entries in $|J_1,m_1,j_2,m_2\rangle$ and the $J_2$ only on the $j_2,m_2$.
It's better to write (as a mathematician would) $$J_{\rm tot}= J_1\otimes {\mathbb I}+ {\mathbb I}\otimes J_2$$ and $$|J_1,m_1,j_2,m_2\rangle= |j_1,m_1\rangle\otimes |j_2,m_2\rangle$$ that makes what's going on more obvious. Each operator only acts on its own Hilbert space, and the total Hilbert space is the tensor product of the two spaces for the two separate spins. See section 6.3.4 in https://courses.physics.illinois.edu/phys509/sp2018/bmaster.pdf
The "missing" $\sqrt 2$ is correct as $|\uparrow\uparrow\rangle$ is the $|j=1,m=1,\rangle$ state in a spin-one multiplet where $J^-|j=1,m=1\rangle = \sqrt 2|j=1,m=0\rangle$.