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 |
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https://community.smartsheet.com/discussion/comment/258481/ | 1,638,167,332,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00465.warc.gz | 241,959,580 | 25,899 | or Explore Discussions
# Count IF for fixed dates
✭✭✭✭✭
07/31/20
I am trying to do a simple CountIF, where the 'if' is whether an Expiration Date field is between a certain range, for example for if Expiration Date in in = January 2020, I would want to count any expiries between 1/1/2020 and 1/31/2020...it doesn't have to be relative, the expiration dates are 'fixed' in the sheet so they won't change or be updated/don't need to be relative to today etc.
Tried something as simple as =COUNTIF([Expiration Date]:[Expiration Date],>12/31/19) but can't get that to even work, is there a trick to this one?
• mod
Hello @brian25736
When counting between a Range, you want to input the Start and End of the Range within the COUNTIFS.
Here is an Example of what the Formula may look like: =COUNTIFS([Expiration Date]:[Expiration Date], >=DATE(2020, 5, 1), [Expiration Date]:[Expiration Date], <=DATE(2020, 10, 1))
Furthermore, here is an Example of this working within my Sheet:
Let me know if you have any questions!
Regards
Sean | 289 | 1,039 | {"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-2021-49 | latest | en | 0.931362 |
https://au.mathworks.com/matlabcentral/cody/problems/41-cell-joiner/solutions/148420 | 1,606,209,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00271.warc.gz | 214,016,446 | 17,094 | Cody
# Problem 41. Cell joiner
Solution 148420
Submitted on 15 Oct 2012 by Wenjian Yu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = {'hello', 'basic', 'test', 'case'}; y_correct = 'hello basic test case'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
2 Pass
%% x = {'this', 'one', '', 'has', ' ', 'some tricky', 'stuff'}; y_correct = 'this one has some tricky stuff'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
3 Pass
%% x = {'delimiters', 'are', 'not', 'always', 'spaces'}; y_correct = 'delimiters?are?not?always?spaces'; assert(isequal(cellstr_joiner(x, '?'),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 239 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.489733 |
http://mathhelpforum.com/math-software/148867-derive-6-ti89-sum-product-terms-print.html | 1,526,847,180,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00157.warc.gz | 186,061,520 | 2,571 | # Derive 6 or Ti89 Sum or Product terms
• Jun 19th 2010, 05:38 AM
Gustavis
Derive 6 or Ti89 Sum or Product terms
Hi People (Hi)
I search a method for expand all terms of Sum or Product in Derive 6
example :
$\displaystyle \sum_{i=1}^{10} i^2 = 385$ not this (Shake)
$\displaystyle \sum_{i=1}^{10} i^2 = 1+4+9+...$ <-- I need this ! (Nod)
what I must try ? Thanks. (Bow)
• Jun 19th 2010, 06:33 AM
CaptainBlack
Firing up the old T2110 which has Derive 5 on it, it seems you should just simplify the $\displaystyle \sum_{i=1}^{10}i^2$ expression rather than expand it which appears to be what you have done.
CB | 215 | 614 | {"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} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.856064 |
http://www.math.sci.kobe-u.ac.jp/OpenXM/1.1.2/html/OpenXM-eg/OpenXM/node30.html | 1,547,958,240,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583700012.70/warc/CC-MAIN-20190120042010-20190120064010-00559.warc.gz | 335,104,786 | 2,307 | Next: CPU dependent double Up: CMOexpressions for numbers and Previous: Distributed polynomial Dpolynomial
## Recursive polynomials
#define CMO_RECURSIVE_POLYNOMIAL 27
#define CMO_POLYNOMIAL_IN_ONE_VARIABLE 33
Group CMObject/RecursivePolynomial requires CMObject/Primitive, CMObject/Basic.
Polynomial in 1 variable, Coefficient, Name of the main variable, Recursive Polynomial, Ring definition for recursive polynomials CMObject/RecursivePolynomial
Example:
(CMO_RECURSIEVE_POLYNOMIAL, ("x","y"),
(CMO_POLYNOMIAL_IN_ONE_VARIABLE, 2, 0, <--- "x"
3, (CMO_POLYNOMIAL_IN_ONE_VARIABLE, 2, 1, <--- "y"
5, 1234,
0, 17),
1, (CMO_POLYNOMIAL_IN_ONE_VARIABLE, 2, 1, <--- "y"
10, 1,
5, 31)))
This represents
x3 (1234 y5 + 17 ) + x1 (y10 + 31 y5)
We intend to represent non-commutative polynomials with the same form. In such a case, the order of products are defined as above, that is a power of the main variable a coeffcient.
sm1
sm1>(x^2-h). [(class) (recursivePolynomial)] dc /ff set ;
sm1>ff ::
Class.recursivePolynomial h * ((-1)) + (x^2 * (1))
Nobuki Takayama 平成12年4月13日 | 351 | 1,094 | {"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} | 2.609375 | 3 | CC-MAIN-2019-04 | latest | en | 0.681132 |
https://www.physicsforums.com/threads/e-field-of-a-long-charged-rod.918202/ | 1,521,751,018,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00157.warc.gz | 889,053,939 | 16,979 | # E field of a long charged rod
Tags:
1. Jun 21, 2017
### J6204
1. The problem statement, all variables and given/known data
A line of charge with a uniform density of 34.2 nC/m lies along the line y = -14.9 cm, between the points with coordinates x = 0 and x = 42.8 cm. Calculate the electric field it creates at the origin, entering first the x component then the y component
2. Relevant equations
3. The attempt at a solution
E_x = 1/(4πε₀) ∫ λ dx * 1/ (Y² + x²) * x/((Y² + x²)^½
where Y is the vertical distance 14.9 cm.
λ dx is the charge element dq, 1/ (Y² + x²) is the " 1/r^2 " and x/((Y² + x²)^½ is the geometric factor for the x-component ("sin(α)" ).
Then the integral gives
E_x = λ/(8πε₀) ∫ du * 1/ u^(3/2)
= λ/(8πε₀) [-2/√u]
= λ/(4πε₀) (1/Y - 1/√(Y² + X²)) [where X = 42.8 cm]
Along the same line of reasoning we have for the y-component
E_y = 1/(4πε₀) ∫ λ dx * 1/ (Y² + x²) * Y/((Y² + x²)^½
= Yλ/(4πε₀) ∫ dx /(Y² + x²)^(3/2)
= λ/(4πε₀) X/(Y√(Y² + X²))
When i substituted λ (34.2*10^-9 C/m), X ( 0.428 m ) and Y ( 0.149 m) and ε₀ ( 8.854 10^-12 F/m) to calculate Ex and Ey I got the following two numbers which were incorrect and I am not sure what I am doing wrong
Ex = 1385N/C
Ey = 1948N/C
2. Jun 21, 2017
### TSny
Your approach looks good. I have not checked the numerical evaluation. But do you expect both components of E to be positive?
3. Jun 21, 2017
### J6204
I thought so, do the answers look correct besides one or two of them being negative? Did I need to factor in the negative sign on the y coordinate?
4. Jun 21, 2017
### TSny
Yes. To determine the signs of the components, choose an arbitrary point along the line of charge and treat the point as a positive point charge. Consider the direction of E at the origin produced by the point charge. | 615 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-13 | latest | en | 0.8717 |
xugr.me | 1,653,126,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00793.warc.gz | 1,228,371,678 | 12,262 | # Formal Proofs in Combinatorics
This text is based on related reading about formal proofs in combinatroics. Mainly two papers, one is about formalization of two central theorems in combinatorics of finite posets [@singh2017formalization], one is the paper published in 2008 about the Four Color Theorem fully in formal proof [@gonthier2008formal]. 1
The goal of this note is to give a short introduction to the formal proofs and its applications in proving some theorems. The first half of the text is devoted to a brief review of how computer programs can do the proving job and some formalization examples in combinatorics. Then we introduce how it applied to prove the Four-Color Theorem and other related lemmas.
This note includs some of my own understandings, so it may be intuitive but not rigorously correct.
## Formal Methods and Proofs
In this section, we will explain why we need formal proofs and how a proof assistant (like Coq) works (in a shallow level).
### Motivation
A proof of a proposition is an argument, that can convince anyone that the proposition is correct. In mathematics, a handwritten proof often must be checked by many mathematicians before the corresponding proposition can be accepted as a theorem or lemma. Normally, these proofs are given in natural languages. But nature languages can be full of ambiguities. Also, a proof can have many details that are omitted. These ambiguities and the omitted details may lead to a wrong proof to be accepted.
When a proof has a very large size, or the correctness of a proposition is vitally important (normally in engineering, like the correctness of a chip design), it may be unacceptable written in natural languages. If we introduce a formal language to write the proof, and validate each step in the “formal proof” using the proof theory [@prawitz2006natural], then we will have a more convincing proof.
But another issue may rise: the proof size may be become very large when we use a limited formal language, and the validating work can be also very cumbersome and meaningless for humans. Thus, we can program the rules in proof theory and let computers to do this work. Since the computers are doing exactly the same as the limited proof rules tell it to do, it can not go wrong or beyond the rules if we ensure the rules are correct. What’s more, the computer will show the proof state in each step, and it can be served as an interactive assistant to help people to write the formal proof. This is as known as a proof assistant.
### Coq Formal Proof Assistant
Coq [@barras1997coq] is a well-known proof assistant developed by INRIA. It allows us to define terms, types, predicates and state mathematical theorems. It can also be interactive and help us to develop formal proofs and prove the theorems.
The core idea of how a proof and a program are equivalent is the Curry-Howard isomorphism/correspondence. It says that two families, the proof systems on one hand, and the models of computation on the other – are in fact the same kind of mathematical objects. For example, the implication from P to Q denoted as $P \rightarrow Q$ in the logic side, can be mapped into the field of programming side as a function that convert type $P$ into type $Q$. Other limited examples are shown in the table.
Logic side Programming side
universal quantification ($\forall$) generalised product type ($\Pi$ type)
existential quantification ($\exists$) generalised sum type ($\Sigma$ type)
implication ($P \rightarrow Q$) function type (type $P \rightarrow Q$)
conjunction ($P \wedge Q$) product type (type $(P, Q)$)
disjunction ($P \vee Q$) sum type (type $P | Q$)
true formula unit type (type $()$)
false formula bottom type
In this way, if we give a proof of a first-order logic proposition, then it can be easily converted into a program using a programming language that implements all of these types. And vice versa this program can be served as a proof. To enhance this idea and make it more expressive (not limited in the first-order logic or other trivial things), we can implement more sophisticated type systems in the proof system and achieve more advanced goal.
When the formal language in Coq is expressive enough, we can then define the concepts, state the theorems, and try to develop the proofs in this proof assistant. Once a statement is proved in Coq, the proof is certified without having to go through the proof-script.
### A Simple Example
Here we provide a simple example to show how formal proofs look like in Coq. The theorem that we want to prove is the addition is commutative between the natural numbers. We can see that the proof assistant is only a modeling tool that can describe the mathematical objects, and the lemmas or theorems can be proved by checking the derivation process. Thus, we can divide the whole process into two parts, the modeling stage and the proving stage.
#### Modeling Stage
The target of the modeling stage is to describe the mathematical objects and concept precisely so that it can be understood by the proof assistant to do the further checking.
Thus, we need to define the natural numbers and the plus operation in the formal system. We choose to formalize the Peano axioms. In Coq, it can be implemented as an inductive type definition. And the axiom 3, 4, 5 in Peano axioms will be implemented automatically. Note that the texts between “(*” and “*)” are comments that show current proof states.
1 2 3 Inductive nat : Type := | O (* O is a natural number. *) | S (n : nat). (* Natural number n has a successor (denoted as S n'') which is also a natural number.*)
Also, we need to formalize the plus operator in the form which seems like a recursive function. It considers the first operand in two cases, O or some other natural numbers which have the predecessor. As for O, simply return the other operand. As for another case, the plus operator will be wrapped into the successor structure to build up a new natural number.
1 2 3 4 5 Fixpoint plus (n m : nat) : nat := match n with | O => m | S n' => S (plus n' m) end.
#### Proving Stage
Then we need some auxiliary lemmas to complete the proof. We can also introduce the infix “$+$” of plus where $n + m$ denotes plus n m . And by default, “$+$” is left-associative.
Example Lemma 1. $\forall n \in \mathbb{N}, n = n + O$.
Its formal proof:
1 2 3 4 5 Lemma plus_n_O : forall n : nat, n = n + O. Proof. intros n. induction n as [| n' IHn']. - (* n = O *) reflexivity. - (* n = S n' *) simpl. rewrite <- IHn'. reflexivity. Qed.
We can see that the induction axiom in Peano axioms is used here. By induction on $n$, we can say that the lemma is correct for all natural numbers.
Example Lemma 2. $\forall n, m \in \mathbb{N}, S (n + m) = n + (S~m)$.
Its formal proof is as simple as the last one:
1 2 3 4 5 6 7 Lemma plus_n_Sm : forall n m : nat, S (n + m) = n + (S m). Proof. intros n m. induction n as [| n' IHn']. (* by induction on n *) - reflexivity. - simpl. rewrite -> IHn'. reflexivity. Qed.
Finally, we can give the proof of the commutativity.
Example Theorem 1. $\forall n, m \in \mathbb{N}, n + m = m + n$.
1 2 3 4 5 6 Theorem plus_comm : forall n m : nat, n + m = m + n. Proof. intros n m. induction n as [| n' IHn']. - simpl. rewrite <- plus_n_O. reflexivity. - simpl. rewrite <- plus_n_Sm. rewrite -> IHn'. reflexivity. Qed.
The “rewrite” tactic is to apply the lemma / theorem to current LHS or RHS. Then we completed a formal proof of the commutativity property for addition between natural numbers. This proof can be checked by machines and also can be translated easily into corresponding handwritten proofs.
Coq standard library and other formalized libraries can be reused in further proving since all of them have been checked correctly and no need to review the proof scripts.
## Formalized Theorems in Combinatorics
Though proof assistant is very powerful in describing the theorems and concepts, it needs much work to introduce the structure in the related theory. In this section, we want to formally prove two central theorems in combinatorics about chains, antichains and cover of finite posets. The proof scripts in this section can be found at GitHub repository.
### Definitions
Partial Order is defined as a record type (a type that holds several fields like a structure) in the Coq library. The four fields in the PO record represent the requirements that it must satisfy: it should have a carrier set, and there is a binary relation on this set and the relation is ordering.
Partially Ordered Set (Poset) can be then instantiated by PO U where U is a type (also the binary relation).
Finite Partially Ordered Set can be defined as a poset whose elements are finite. The “finite” property can be also found in the Coq standard library.
1 2 3 Record FPO (U : Type) : Type := Definition_of_FPO { PO_of :> PO U ; FPO_cond : Finite _ (Carrier_of _ PO_of ) }.
Chains are totally ordered subsets of a poset, so it can be formalized as:
1 2 3 4 Record Chain : Type := Definition_of_chain { PO_of_chain : PO U; Chain_cond : Totally_ordered U PO_of_chain (@Carrier_of _ PO_of_chain)}.
But this definition can not be well-used in the actual proof, since it can not carry the idea that “$A$ is a chain of the poset $B$”. Thus we can use a predicate to check if one is a chain in some poset: if the set $e$ is a subset of the poset, and each two elements ($x, y$) in $E$ is comparable (R $x$ $y$ $\vee$ R $y$ $x$, or equivalently, $x \le y \vee y \le x$), then we have:
1 2 Definition Is_a_chain_in (e: Ensemble U): Prop := (Included U e C /\ Inhabited U e) /\ (forall x y: U, (Included U (Couple U x y) e) -> R x y \/ R y x).
An antichain is a subset of the poset who has no two distinct elements are comparable.
A (anti)chain cover is a collection of (anti)chains whose union is the entire poset. The width of the poset is the size of the largest antichain. And the height of one poset is the size of the largest chain.
An element $b \in P$ is called a maximal element if there is no $a \in P$ such that $b \le a$. Similarly, an element $a \in P$ is called a minimal element if there is no $b \in P$ such that $b \le a$.
### Facts
Once we have these definitions, these facts can be proved trivially.
Fact 1. There exists a chain in every finite poset.
Fact 2. There exists a chain cover for every finite poset.
Fact 3. The set minimal($P$) and maximal($P$) are both non-empty for every finite poset $P$.
Proof. By induction on the size of $P$. ◻
Fact 4. Minimal($P$) and maximal($P$) are both antichains in $P$.
### Mirsky’s theorem
Mirsky’s theorem relates the size of an antichain cover and a chain in a poset. The definitions we have seen so far are sufficient to express the formal statement of Mirsky’s theorem in Coq.
Theorem 1. For any finite poset $P$, the maximum size of an chain (the height) of $P$ is equal to the size of the smallest antichain cover of P.
Proof. Firstly, split it to two proof goals:
1. Size of a chian is smaller than the size of an antichain cover in finite poset $P$.
2. There exists an antichain cover of size equal to the height of the poset $P$.
The first goal is easy to prove. Any chain has at most one common element with each antichain from an antichain cover. And each element of the chain must be covered by some antichain from the cover. Hence, the size of any chain is smaller than or equal to the size of any antichain cover.
To prove the second goal in Coq, we need to firstly prove several auxiliary lemmas. The induction tactic can be directly used, but to prove the proposition about the existence, we need to provide a concrete procedure of how we construct such an antichain cover. By induction on the height $h(P)$ of $P$. In the case $h(P) = 1$, then the poset itself is the antichain, since no two elements are comparable. In the case $h(P) = r + 1$, if we discard the largest elements in the largest chains (maybe several) of $P$ as $P’$, then $P’$ has the height $r$ and then can be covered by $r$ antichains by the hypothesis. The discarded elements are not comparable because they are from different chains. Thus, they form another antichain in $P$. By add this antichain into the former cover, we have an $r+1$ sized antichain cover of $P$. ◻
The related formal proof can be found at this link.
### Dilworth’s decomposition theorem
Theorem 2. For any finite poset $P$, the maximum size of an antichain (the width) is equal to the minimum number of chains in any chain cover.
Proof. Similar to the proof of Mirsky’s theorem, it can be proved by proving two goals:
1. Size of an antichain is smaller then size of a chain cover.
2. There exists a chain cover of size equal to the width of the poset $P$.
The first goal is easy by contradiction.
As for the second goal, we need to prove it by induction on the size of $P$. Let $m$ be the width of $P$. In the case the size $P$ is 1, the proof is trivial. In the case $P$ of size at $n+1$, we destruct the predicate: there exists an antichain $\mathcal{A}$ of size $m$ which is neither maximal($P$) or minimal($P$). The predicate can only be true or false. This destruct is supported intrinsically by Coq.
When it’s true, we define the sets $P^+$ and $P^-$ as follows:
\begin{aligned} P^+ & = \{x \in P | x \ge y \text{~for some~} y \in \mathcal{A}\} \\ P^+ & = \{x \in P | x \le y \text{~for some~} y \in \mathcal{A}\} \\ \end{aligned}
If $x \in \mathcal{A}$ then $x \in P^+ \cap P^-$, thus $x \in P^+ \cup P^-$. If $x \notin \mathcal{A}$, then there is must be an element $y \in \mathcal{A}$ is comparable with $x$, $x \le y$ or $x \ge y$. Thus $\forall x \in P, x \in P^+ \cup P^-$.
Therefore, $P^+ \cup P^- = P$. And $P^+ \neq P \wedge P^- \neq P$ since there at least one minimal or maximal element not in these two sets, respectively. We have the $\mathcal{A}$ as the largest antichain in $P^+$ and $P^+$ by contradiction. Using the hypothesis, there exists a chain cover sized $m$ to cover $P^+$ or $P^-$, denoted as $C_i$ or $D_i$, respectively. Elements in $\mathcal{A}$ are the minimal elements of $C_i$ ad the maximal elements of $D_i$. Therefore, we can join these two chains and form a larger chain cover of the origin poset $P$ since $P^+ \cup P^- = P$.
When the predicate is false (no antichain other than maximal($P$) or minimal($P$) has size $m$), we can remove the maximal element $a$ and minimal element $b$ in a chain in $P$ to get $P’$. Since $a$ must in the antichain maximal($P$) and $b$ must in the antichain minimal($P$), and other antichains have elements less than $m$ (the predicate is false), thus the width of $P’$ is $n-1$. By the hypothesis, there exists a cover of chains sized $m-1$. Add the chain just removed from the $P$, we can finally construct an $m$-sized chain cover of poset $P$. ◻
The related formal proof can be found at this link. In the formal proof, the author made some modification. Instead of using induction on the cardinality of posets, they use well-founded induction on the strict set-inclusion relation. When working with the Ensemble module of the Coq standard library it is easy to deal with the set-inclusion relation compared to the comparison based on set cardinalities.
## Four-Color Theorem
This section will talk about how programs help to do the proof, but not the proofs themselves. Only the core idea and proof steps are elaborated.
Theorem 3. Every planar map can be colored with at most four colors.
And it’s dual version:
Theorem 4. The vertices of every planar graph (without loops) can be colored with at most four colors in such way that no pair of vertices which lie on a common edge have the same color.
### First Attampt
The first computer-aided proof of 4CT is provided by K. Appel and W. Haken in 1976 [@appel1976every].
A drawing is a triangulation if every region is a triangle. A configuration is a subgraph of a planar triangulation consisting of a circuit and its interior. A configuration is called reducible if it can be shown by certain standard methods that it cannot be immersed in a minimal counterexample to the four color conjecture. A set of configurations is called unavoidable if every planar triangulation contains some member of the set.
From the definitions, 4CT can be proved by showing an unavoidable set of reducible configurations. By discharging algorithm, one can produce one of these unavoidable sets of configurations. The major effort in K. Appel’s work was involved in the development of the discharging procedure. The algorithm they developed produced a set $U$ of configurations less than 2000, each of ring size fourteen or smaller.
And the next step is to prove the reduciblility of the configurations in the unavoidable set. This step is also done by computer programs. Since in the producing step, the algorithm is avoiding configurations of ring size greater than fourteen, the reducibility of each configuration can be proved without exorbitant use of computer time. But it also cost more than 1000 hours to check all the configurations by computer.
The proof has not been fully accepted, basicly for two reasons:
1. part of the proof uses a computer and cannot be verified by hand.
2. even the part of the proof that is supposed to be checked by hand is extrordinarily compicated and tedious.
### An Elegant Revison
Using the same general approach in A&H proof, Neil Robertson et al. made their own proof of 4CT [@robertson1997four]. The basic idea of the proof is also exhibiting a set if configurations, but much less than before (633 cases). Where their method differs from A&H is in how they prove the unavoidability. Their proofs are easier to check: they not only rewrite the hand-checking of unavoidability, but also provide a “more” formally version which can be checked by a computer in a few minutes.
But there is still a problem, the proof still combines a textual argument, which could be checked by inspection, and a program that can not be verified. If there exists any “programmer error” in the program, we can not accept the output result. The computer aided proofs is still working in a verification way, but it can not be considered as a “proof”.
### Hybrid and Formal Proofs
In 2000, Georges Gonthier tried to produce such a formal proof as Section 1 for [@robertson1997four], and he succeeded. But soon another issue raised: there is a gap between the formal proof and the textual proof. One can not absolutely prove that the interface of the textual proof is fit for the formal proof. The specification used in the formal proof maybe wrong at first. To fix this issue, they formalized the entire proof of 4CT [@gonthier2008formal].
## Summary
As for large scale proofs, formal method may be very helpful. Proof assistants can clearly record the current proof state, and even find available lemmas or theorems from archives for users. A shortcoming is that formal proofs can be in large size and very cumbersome to develop. It may be fixed in the future by developing more advanced proof assistants, but textual proofs can not be replaced.
1. Recently I was working in the field of formal proofs and verification, so I choose this topic. ↩︎ | 4,815 | 19,210 | {"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": 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} | 3.921875 | 4 | CC-MAIN-2022-21 | longest | en | 0.924947 |
https://www.jiskha.com/display.cgi?id=1363671994 | 1,529,879,461,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00442.warc.gz | 858,057,669 | 4,097 | # Maths
posted by Stranger
ABC is a right angled triangle with ∠ABC=90∘ and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on AC and it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?
1. Steve
Hmm. Place point D such that ABCD is a rectangle.
Let the center of the semi-circle be at point O on AC.
Draw another semi-circle centered at O, to form a complete circle. Since it is tangent to AB, it is also tangent to DC.
So, the diameter of the circle is 7, and the radius is 7/2, so a+b=9.
2. student
its wrong steve
3. Kishan
199
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https://sports.answers.com/individual-sports/A_bowling_ball_with_a_negative_initial_velocity_slows_down_as_it_rolls_down_the_lane_toward_the_pins_Is_the_bowling_ball%27s_acceleration_positive_or_negative_as_it_rolls_toward_the_pins | 1,726,884,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00678.warc.gz | 486,770,713 | 50,066 | 0
# A bowling ball with a negative initial velocity slows down as it rolls down the lane toward the pins Is the bowling ball's acceleration positive or negative as it rolls toward the pins?
Updated: 12/12/2022
Wiki User
14y ago
In these situations, you usually define one direction as being positive and the other as negative. In this problem, the question does this for us: it clearly tells us that when the ball is moving down the lane, its velocity is negative. "Down the lane" is the negative direction.
Average acceleration is give by
a_ave = Δv/Δt
where Δv is a change in velocity and Δt is elapsed time.
Δv = v_final - v_initial
In this problem, the ball "slowed down," but did not change direction. This means that the NUMBER associated with "v" got smaller (5 ... 4 ... 3 ...) but the SIGN (-) did not change.
I hope it makes sense, then, that "v_final" was a smaller (slower) number with a negative sign (moving down the lane) while "v_initial" was a larger (faster) number with a negative sign (moving down the lane).
The subtraction of a large negative number from a small one (like, for instance, -4 - [-10]) is a POSITIVE number whose value is given by the difference (+6).
So Δv is positive, and Δt is ALWAYS positive (no matter what).
Therefore a_ave, the quotient of two positive numbers, will be positive.
Wiki User
14y ago
Earn +20 pts
Q: A bowling ball with a negative initial velocity slows down as it rolls down the lane toward the pins Is the bowling ball's acceleration positive or negative as it rolls toward the pins?
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Related questions
### What is the acceleration of a bowling ball when at rest at the end of the bowling lane?
The acceleration of a bowling ball at rest at the end of the bowling lane is 0 m/s^2. Since the ball is not changing its velocity, it is not experiencing any acceleration.
### When did High Velocity Bowling happen?
High Velocity Bowling happened in 2007.
### Are bowling scores and golf scores positive or negative correlations?
Bowling scores would be a positive correlation because the higher the score, the better the game. Golf scores would be negative correlations because the higher the score, the worse you are playing.
### When was High Velocity Bowling created?
High Velocity Bowling was created on 2007-12-07.
### What actors and actresses appeared in High Velocity Bowling - 2007?
The cast of High Velocity Bowling - 2007 includes: Chris Canning as Short Order Cook
### Is negative eight in bowling good or bad?
A negative eight in bowling is impossible. The lowest score one can have for a game is 0.
### Why does a bowling ball move without acceleration when it rolls along a alley?
Strictly speaking, it moves with negative acceleration. The forces of friction and air resistance both act to slow the ball down. If the lane were long enough, the ball would eventually come to a complete stop.
### How much force N is needed to knock over a bowling pin?
It typically takes about 110-130 Newtons of force to knock over a standard 15-inch-tall bowling pin. This force can vary depending on factors such as the weight of the bowling ball and the angle at which it strikes the pin.
### Examples where velocity is zero but not acceleration?
An object in free fall at its peak point has a velocity of 0, but there is still acceleration acting on it due to gravity. Similarly, a pendulum at the extreme points of its swing also has zero velocity but is undergoing acceleration towards the center of its swing.
### Where what a bowling ball and a napkin fall with the same acceleration?
In a vacuum, both a bowling ball and a napkin would fall with the same acceleration due to gravity, which is approximately 9.81 m/s^2. This is because in the absence of air resistance, all objects experience the same acceleration regardless of their mass.
### How much force is needed to accelerate at 25 kg bowling ball at?
The force required to accelerate a 25 kg bowling ball can be calculated using the equation F = ma, where F is the force, m is the mass of the bowling ball, and a is the acceleration. If the acceleration is given, you can plug in the numbers to find the force needed.
### Were would a bowling ball and a napkin fall with the same acceleration?
Both the bowling ball and the napkin would fall at the same rate of acceleration due to gravity, assuming no external forces are acting on them. This is because all objects experience the same acceleration due to gravity, regardless of their size, mass, or shape. | 999 | 4,546 | {"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-2024-38 | latest | en | 0.930012 |
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## SS1: MATHEMATICS - 3RD TERM
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# Bar Chart
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#### Topic Content:
• Bar Chart
Definition: These are rectangular bars with varying heights but the same width with equal spaces or gaps separating the bars. The height of each bar represents the frequency of each value. The bars may be drawn vertically or horizontally. A bar chart can be used to display qualitative and discrete quantitative data.
### Example 7.2.1:
The number of items produced by a company over a five-year period is given below:
(a) Plot a bar chart for this information.
(b) What is the average production for the five-year period? (WAEC)
Solution:
(a)
(b)
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Note: If you have Already Subscribed and you are seeing this message, it means you are logged out. Please Log In using the Login Button Below to Carry on Studying! | 259 | 1,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.841827 |
http://2ndinline.blogspot.com/2012/10/swaportunity-can-you-spell-it.html | 1,498,337,475,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320338.89/warc/CC-MAIN-20170624203022-20170624223022-00546.warc.gz | 2,403,826 | 53,593 | ## Saturday, October 6, 2012
### Swaportunity- Can you spell it?
SWAPORTUNITY. . . . you've all seen that commercial. . . where they child is in a spelling bee and the moderator gives the word. . . I love that word!
And today we are having a SWAPORTUNITY!
Jessica over at Mrs. Stanford's Class is hosting a swap between many many bloggers. You can go to her blog and find more people who are involved in this SWAPORTUNITY!
I had the wonderful opportunity to swap with Nicole Rios from Mrs. Rios Teaches Second Grade She is a 2nd grade teacher from California. If you haven't been to her blog, I would suggest you go find her as soon as you leave my blog!
She shared with me her Flip Book on 3D shapes. I did this with my students this week and I have to say I am truly amazed! This activity really hits home with my students. They loved it and I truly feel they learned a lot about shapes and their properties.
I teach my class in groups of about 8 kids. One group is with me, one group is on the computers and one group is doing either seat work or center activites.
We started with looking at these models of the shapes that I had in my class.
The first thing that amazed me was that she figured out HOW to make this so that it all folded together perfectly! She gave me very clear directions for copying it two sided so that it aligned right. And it did!
She said I could have the kids do the folding or do it myself. . . . I thought about having my aide do it, but then I thought I would jump right in and see how the students did with it.
I was very impressed. My students did well with this.
However here's an interesting twist. The first group that I did this with was my top math students. Very bright students, but they struggled with folding! They wanted to fold TO the dotted line, not on it! hmmmmm. . . . smart with intelligence, but no so on manipulatives! The 2nd and 3rd groups had no problem with the folding, but sometimes they struggle with math concepts!
There are 4 pages that they have to fold and staple together so they make the booklet.
Because I meet with these groups, I only have about 30 minutes with each group. The first day we put the books together. The seond day we started working on the pages and doing the work that was asked of us.
They learned how to count the faces, the vertices, and the edges. They tallied the number of shapes each face had on each shape, and they connected with real world images that had the same shapes.
They use highlighters to designate where the edges are.
They LOVE using highlighters!
We stopped and talked about each shape as we completed the booklet.
They love to touch the parts of the shapes.
Here we are all touching the VERTEX of the cone!
OOOOH a new word!
They glue real world pictures on for each of the shapes.
Our completed booklets about 3D Shapes!
I did this in 3 days with 30 minutes each day. I think it was a good idea to break it into at least a couple days. One day to put it all together and another day to complete the work on each page.
My students loved this activity! They kept telling me how fun it was.
I know that if my second graders can do this, your kids can too.
Don't forget! You can pick this up at Nicole's TPT shop here!
And . . . . if you like this flip book, she also has one about short vowels. . .
I think I'm going to have to check that one out too!
She used 3D shapes graphics from Ginger Snaps Art
The fonts were from Kevin and Amanda.
The Cube Cover graphics were by Graphics from the Pond
If you want to see one of my activities head on over to Mrs. Rios Teaches 2nd Grade to see what she had to say about my stuff!
Patty
1. Patty, Thank you so much for using my 3D Flip Book with your students. I am so glad you and your students enjoyed it.
Nicole | 896 | 3,790 | {"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-2017-26 | latest | en | 0.985082 |
https://www.jiskha.com/display.cgi?id=1258646483 | 1,503,386,592,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110485.9/warc/CC-MAIN-20170822065702-20170822085702-00434.warc.gz | 897,058,642 | 4,076 | # maths
posted by .
im trying to rearrange the following equation for x but getting confused because of the brackets and squares. I understand the process of changing the subject but on more simple equations. Please can you help me. Here's the equation
y = 3(x+5)^2
• maths -
divide by three.
y/3=(x+5)^2
take the square root of each side.
+-sqrt(y/3)=x+5
solve for x. check both roots.
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https://www.physicsforums.com/threads/how-far-can-step-up-transformer-inflate-voltage.370457/ | 1,529,832,786,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00330.warc.gz | 884,077,254 | 14,114 | # How far can step up transformer inflate voltage
1. Jan 18, 2010
### aspardeshi
How far can a step up transformer increase voltage ? Why voltage is stepped up. Is the end result output power suitable for home use after using step up transformer ?
2. Jan 18, 2010
### samieee
voltage is stepped up to transmit in distant areas.When it is stepped up then value of voltage is increased but current value is decreased so (i^2*R) loss through wire is minimized .Generation company develops power & transmit through transmission line into sub station then sub-stations distribute it to the home users through distribution line.There are some catagories(in sub-station) ie:1KV,0.4 KV,though home users consume little power, they are connected with .4 KV.
3. Jan 18, 2010
### aspardeshi
Thanks. If the generator is designed from scratch and it delivers 1200 volts then how to make it suitable for home use. How to calculate the wattage of home and adjust it to the generator output.
4. Jan 18, 2010
### vk6kro
To convert 1200 volts to 120 or 240 volts you need a step down transformer.
To calculate the wattage your house uses just add up the wattages of all the appliances in the house.
5. Jan 18, 2010
### aspardeshi
Thank you for your overwhelming replies ! My problem is that If I make a generator 240 volts and not 1200 volts, will it be suitable for home use ? How to determine whether it is suitable for home use or not ? How to calculate the power load. Like to my knowledge varied devices use varied amps. So you mean to say that if I produce generator 240 volts or 1200 volts (with a step down transformer) it is suitable for home use ! is it so ? I think the power load requirement calculation is the biggest deal here.
6. Jan 19, 2010
### vk6kro
Assuming you supply the right voltage to the house, other necessary things are that the frequency is right (usually 60 Hz for 120 volts and 50 hz for 240 volts) and also that the generator can supply enough power at that voltage.
On each appliance, you will normally see a plate that says how much power it uses.
Maybe it is a coffee maker and it says on the label that it uses 840 watts at 120 volts. This is all you need to know, but you could work out from this that the coffee maker uses (840 watts / 120 volts) = 7 amps.
So, if you work out all the power requirements of the things that could be used at once, you can just add up the powers in watts.
In most houses there are a lot of things that just use a small amount of power like 20 or 30 watts. Things like computer modems or radios are like this. You can guess a bit with those, but big items like room heaters or washing machines you would need to read the label and get it right.
Having done all that, you would probably double it so that the generator is not being heavily loaded all the time.
Another thing you need to know is how much the generator voltage changes when you draw more or less current from it. It should be a fairly constant voltage out (and the right voltage) or the lights in the house will be changing in brightness when you turn on the coffee maker for example. | 736 | 3,121 | {"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-2018-26 | latest | en | 0.86722 |
https://branemrys.blogspot.com/2012_02_05_archive.html | 1,712,922,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00723.warc.gz | 127,616,005 | 45,479 | ## Saturday, February 11, 2012
### Buckley's
I don't know if it's a bug I've been made susceptible to by the oscillations of hot and cold, or if it's allergies, but I've come down with a nasty cough -- so nasty I had to break out the Buckley's. Buckley's has as its slogan, "Tastes Awful. And It Works" and had the very famous and quite funny advertising campaign in which Frank Buckley, the son of the founder, would just walk up to the microphone, say, "I wake up with nightmares that someone gives me a taste of my own medicine," or "Our largest bottle is 200 ml; anything more would be cruel," and then shuffle off. And they run ads saying things like, "People swear by us. And at us." Or "Feared by more people than ever before." Pure genius. I find it works pretty well. But with that advertising campaign -- even if it didn't work, I'd want it to.
The funny thing is, I've always thought it tasted pretty decent, and am apparently the only one on the planet who thinks this. When I was growing up, my mom always used Robitussin and I hated the stuff. It even took me ages to get used to red wine because at some point I would take a mouthful and it would suddenly start reminding me of Robitussin.
### Music on My Mind
Yakko's Universe Song! And he's right, you know. Suppose that this period is the size of the earth:
.
Then this is the size of the sun, more or less, in earth-size-units:
....................
....................
....................
....................
....................
If we were to take a period to indicate an earth-diameter, then the distance from the earth to the sun would be somewhat more than eleven thousand six hundred periods. Suppose that this is the entire distance between the earth and the sun (an AU):
.
Then this is the radius of the solar system in AUs, more or less, if we take it to include everything within the orbit of Pluto:
.......................................
That's the radius, mind you; the diameter would be twice as long. But it's bigger than that, really; if we include all the comets orbiting the sun, then it would end up being somewhat more than a thousand times longer. I will spare you that. Likewise, the distance to the Proxima Centauri, the nearest start to our sun, would be about six or seven thousand times longer than the above sequence.
Suppose that this is the diameter of the whole solar system, including the Kuiper Belt and the Oort Cloud (which is the measure that's more-or-less a thousand times the above sequence of periods):
.
Then the distance to Proxima Centauri, the star nearest to our sun, doesn't look so bad, being a bit shorter than this in whole-solar-system-units:
...
But if we wanted to indicate the radius of the Milky Way, we would need a fair number more than ten thousand periods.
Suppose that this is the diameter of the entire Milky Way galaxy:
.
Maybe you want to get to the nearest galaxy, not counting the Canis Major Dwarf galaxy, which is technically within the Milky Way itself. That would probably be the Sagittarius Dwarf Elliptical galaxy, which is (if you find the diameter of the Milky Way to be a conveniently traversable distance) quite close in diameter-of-Milky-Way units:
.
A little closer than that, actually.
But what if the real action is in the nearest spiral galaxy? That's Andromeda. Taking the same measure, diameter-of-Milky-Way units, the distance is, more or less:
.........................
The local neighborhood is (conveniently, if a bit self-centeredly) called the Local Group, and in diameter it is about a hundred times our diameter-of-Milky-Way unit. Suppose this is one galaxy, like the Milky Way:
.
There are at least this many galaxies in the Local Group:
..........
..........
..........
They are not all the same size. The Milky Way is the second largest; Andromeda is larger. (Although if we measure by mass, the Milky Way might be the largest.) Most of the others are small as galaxies go, which is still very, very big.
The Local Group is one of the two major groups in the Virgo Supercluster, the other being the Virgo Cluster. If this is diameter of the Local Group:
.
Then this is more or less the diameter of the Virgo Supercluster:
...........
The Virgo Supercluster is a small supercluster among a large population of superclusters. There are about 10 million superclusters in the observable universe. Truly, truly, it's a great big universe.
## Friday, February 10, 2012
### Bergamotto
I like Earl Grey and Lady Grey teas; they are probably my favorite teas after the standard Darjeeling. Earl Grey and Lady Grey are flavored with bergamot, and while I had a vague notion that bergamot was a citrus fruit, I knew nothing about it. So I've been looking things up. There are a number of plants called bergamot; the one used here is indeed a citrus fruit, Citrus bergamia. It is generally thought to have originated as a hybrid of lemon and sour orange, which gives a sense of what it tastes like on its own, and thus is found in fruit or pulp in almost nothing, although I have seen recipes for marmalades, preserves, and candies. The oils, however, are highly prized. Besides Earl Grey, the oil is a standard component of eau de cologne. The fruit is relatively difficult to find -- the overwhelming bulk of true bergamot is grown in a few thousand acres of southern Italy. It will grow elsewhere, but the quality of the oil is universally regarded as inferior -- it's the magic of the soil of Calabria, some mix of limestone and other minerals, that seems to bring out the flavor and aroma that has made the fruit's oil famous. Such local subtleties are not to be underestimated.
In any case, that explains the refreshing character of the scent of Earl Grey, and perhaps also the fact that Lady Grey, despite adding lemon and orange peel, always seems slightly milder.
## Thursday, February 09, 2012
### Links of Note
* Christopher Tollefsen argues that there are alternatives to thinking of moral principles in terms of commands at "Public Discourse".
* Jason Zarri has a critique of Alison Jaggar on abortion
* The February 2012 Biblioblog Carnival is Babylon 5 themed!
* A recent encyclopedia on the Tlingit language is facing a curious problem: no Tlingit speakers can make sense of the Tlingit presented in the encyclopedia.
* Some PDFs on Christine de Pizan:
Christine de Pizan and the Querelle de la Rose (Margaret E. Loebe)
Christine de Pizan and Jean Gerson on the Body Politic (Cary J. Nederman)
Christine de Pizan and the Moral Defence of Women (Rosalind Brown-Grant)
* Some interesting SEP articles on the philosophy of physics:
The Equivalence of Mass and Energy (Francisco Fernflores)
Space and Time: Inertial Frames (Robert diSalle)
Space and Time: The Hole Argument (John D. Norton)
Conventionality of Simultaneity (Allen Janis)
Early Philosophical Interpretations of General Relativity (Thomas A. Ryckman)
The first one, on E=mc2, is especially recommended: both readable and fascinating; this is what an article in a philosophical encyclopedia should be.
* Bill Vallicella suggests that the fallacy of composition is not an informal fallacy.
* John Wilkins has an interesting post on the history of the label 'Aristotelian essentialism' in contemporary philosophy of science.
* DarwinCatholic on the notion of material cooperation
* Dr. Seuss for the Symbolism-Challenged
### Seal of All the Fathers
Today is the feast of one of the most important theologians in Christian history, St. Cyril of Alexandria, confessor and Doctor of the Church, defender of the truth, consistent teacher of orthodoxy, glory of Alexandria, and many more titles. It's difficult to pin down a lot of details about Cyril's life, because he was born, was raised, and lived in Alexandria during a time of tumult that was unusual even for that famously tumultuous city. The Alexandrians were famous for rioting at the least notice, the politics was mean and ruthless, and the whole city constantly turned this way and that by political factions. Cyril seems to have become Patriarch of Alexandria in 412, and thrown himself into navigating the jungle of Alexandrian life and maintaining the eminence of Alexandria among other sees. There was not a single moment of his thirty-two year career as Pope of Alexandria that was not spent in some heated dispute or other. What is remarkable is how many of these disputes ended well, and in great measure this was due to Cyril's skill, indefatigability, and focus on argument. No one involved in Alexandrian politics, whether ecclesiastical or secular, could avoid making controversial decisions; Alexandrian politics was famously contentious, and it was notoriously difficult to make any decisions in that city without someone somewhere trying to incite people to a violent uprising over it. Even Cyril did not manage to keep things under control all the time. What is notable is how often he did. And his influence has been extraordinary, being found in the third, fourth, and fifth ecumenical councils (he was the major player in the third council, the Council of Ephesus).
From the Scholia on the Incarnation of Christ, section 36:
Saint Paul sets forth to us the Saving Passion, for he saith at one time, By the Grace of God for all tasted He death and also, For I delivered to you in the first place that which I too received, that Christ died for our sins according to the Scriptures and that He was buried and that He rose again the third day: moreover the most wise Peter also saith, Forasmuch as Christ suffered for us in the Flesh. Seeing therefore we believe that One is our Lord Jesus Christ, i. e. God the Word beheld in human form or made man as we, in what manner can we attribute Passion to Him and still hold Him impassible, as God?
The Passion therefore will belong to the Economy, God the Word esteeming as His own the things which pertain to His own Flesh, by reason of the Ineffable Union, and remaining external to suffering as far as pertains to His own Nature, for God is Impassible. And no wonder, since we see that the soul itself of a man, if its body suffer somewhat, remains external to the suffering as far as belongs to its own nature, yet is it not conceived of as external to suffering, in that the body which suffers is its very own: and albeit it be impalpable and simple, yet is that which suffers not foreign to it. Thus will you understand of Christ too the Saviour of all.
## Wednesday, February 08, 2012
### The Beauteous Pair
From Ariosto's Orlando Furioso, Canto XXII, as translated by William Stewart Rose:
XXXI
But it behoves that, ere the rest I say,
I Bradamant and good Rogero find.
After the horn had ceased, and, far away,
The beauteous pair had left the dome behind,
Rogero looked, and knew what till that day
He had seen not, by Atlantes rendered blind.
Atlantes had effected by his power,
They should not know each other till that hour.
XXXII
Rogero looks on Bradamant, and she
Looks on Rogero in profound surprise
That for so many days that witchery
Had so obscurred her altered mind and eyes.
Rejoiced, Rogero clasps his lady free,
Crimsoning with deeper than the rose's dyes,
And his fair love's first blossoms, while he clips
The gentle damsel, gathers from her lips.
XXXIII
A thousand times they their embrace renew,
And closely each is by the other prest;
While so delighted are those lovers two,
Their joys are ill contained within their breast.
Deluded by enchantments, much they rue
That while they were within the wizard's rest,
They should not e'er have one another known,
And have so many happy days foregone.
XXXIV
The gentle Bradamant, who was i' the vein
To grant whatever prudent virgin might,
To solace her desiring lover's pain,
So that her honour should receive no slight;
— If the last fruits he of her love would gain,
Nor find her ever stubborn, bade the knight,
Her of Duke Aymon through fair mean demand;
But be baptized before he claimed her hand.
The story of Bradamant/Bradamante and Ruggiero/Rogero/Roger is my favorite part of the Carolovingian tales; when I was in high school I did a re-write of their story (based on the version in Bulfinch's Mythology) which I called The Unicorns. Bradamante is interesting, because she is a knight, and, indeed, the stories are all quite clear that she is one of the best knights in France (and Ariosto makes quite clear that she is the equal of her brother, Rinaldo/Renault/Renaud, who is among the creme-de-la-creme of Carlovingian knighthood). She has a magic lance that never fails to unhorse its target and a magic ring that nullifies all enchantments, but even when she doesn't have these things she is virtually always successful. She rides around like any other knight, usually rescuing maidens in distress; the ruffian-knights she unhorses are often quite put out when later they discover that they were outmatched in arms by a woman. However, she happens to fall in love with a Moorish knight, who is Roger. She ends up having to rescue him from the clutches of an evil enchanter (getting captured by enchanters is almost as common for these Carlovingian knights as losing their horses), and in the course of this quest she meets Merlin, or, rather, his spirit, who is in the care of a good enchantress named Melissa. Merlin prophesies her great destiny. She manages to free Roger, but unfortunately a hippogriff absconds with him and Roger ends up flying away to many adventures, leaving Bradamante once more to look for him. Roger ends up finally getting the hippogriff to stop, and when he does, he finds Astolphe/Astolfo, her English cousin, who having been kidnapped by a whale had become turned into a myrtle tree. It's a long story. Anyway, they get into a terrible situation with the enchantress Alcina; but Melissa comes along to the save the day. Both Roger and Bradamante are eventually captured -- by an enchanter, as it happens -- and Astolphe, who has lost his horse Rabican but found the hippogriff, comes flying in to save them. This leads to the scene above. Now, finally, they are able to express their love -- at least, as Ariosto carefully notes, "whatever prudent virgin might" -- and they definitely want to marry. But there is an impediment: Bradamante is a good Catholic girl, of course, but Roger is Muslim. However, Roger for a number of reasons is somewhat cool at this point toward Moorish chivalry, and Bradamante is, after all, not only an extraordinary knight but a beautiful woman, so he'll be baptized and they'll marry. Since the end of the book hasn't come yet, it turns out not to be that easy.
## Tuesday, February 07, 2012
### Diamond Jubilee
This year is the Diamond Jubilee of Queen Elizabeth II; her reign started on February 6, 1952.. She is the second longest-reigning monarch in British history, after Queen Victoria, and all she has to do is last four more years to become the longest-reigning. Her precise style has two official versions, one Latin:
Elizabeth II, Dei Gratia Britanniarum Regnorumque Suorum Ceterorum Regina, Consortionis Populorum Princeps, Fidei Defensor
And one English:
Elizabeth II, by the Grace of God of the United Kingdom of Great Britain and Northern Ireland and of Her Other Realms and Territories Queen, Head of the Commonwealth, Defender of the Faith
"God Save the Queen" would seem fitting. The British usually only sing the first and sometimes the third verse, but as is often the case with anthems, the best verse is the one usually left out:
O Lord, our God, arise,
Scatter her enemies,
And make them fall.
Confound their politics,
Frustrate their knavish tricks,
On Thee our hopes we fix,
God save us all.
That's the anthem of a sensible people.
## Monday, February 06, 2012
### Admin Note, Especially for Australians
I've noticed in the statistics over the past few months a pretty heavy uptick in people ending up at http://branemrys.blogspot.com.au, which is the Australian version. It usually happens, of course, with Australians (sometimes New Zealanders -- given the Kiwis I've known, they probably find being lumped in with the Australians irritating, so I hope it's not a constant thing) who get here by Google, but also at least sometimes from elsewhere.
This can affect the reading of the blog, especially if you comment here -- comments show up on the .com edition, not the .com.au address.
If it keeps redirecting you to http://branemrys.blogspot.com.au no matter what you do, you should be able to get around this by adding a No Country Redirect. It's easy enough to do; just type in
http://branemrys.blogspot.com/ncr
This will always get you to the American version, which is the One True Version, at least of this blog.
Google is doing this with Blogspot because of increasingly specific laws for various countries regarding internet content. It will become more common in the future, and will certainly spread out from Australia, so it's worth knowing even if you are not Australian.
UPDATE: Also, additional testing shows that at present any comments made on .au posts disappear into oblivion. I'll look into seeing if I can get around this without going crazy.
UPDATE 2: OK, even further experimentation shows that comments put on the .au page do show up in the comments thread RSS -- but as far as I can discover, nowhere else. This is a problem; I'll still be looking into this.
UPDATE 3: Although I only occasionally get readers from India, it looks like it's an issue for India, as well (and checking my logs I find that this is certainly happening, although I hadn't noticed it before). I wonder if Google keeps a list of which country redirects they are currently using? The article linked to by the link mentions Brazil, Honduras, and Germany; I don't get many Brazilians or Hondurans, but I do get people from Germany, and haven't noticed this happening with them yet. Vikram Johri at the link wonders if NCR will last; if it doesn't, it will be a pain. But it probably is not going to last forever.
### The Strangest Whim
Relishing life with Chesterton today....
A Ballade of Suicide
by G.K. Chesterton
The gallows in my garden, people say,
Is new and neat and adequately tall;
I tie the noose on in a knowing way
As one that knots his necktie for a ball;
But just as all the neighbours on the wall
Are drawing a long breath to shout "Hurray!"
The strangest whim has seized me. . . After all
I think I will not hang myself to-day.
To-morrow is the time I get my pay
My uncle's sword is hanging in the hall
I see a little cloud all pink and grey
Perhaps the rector's mother will NOT call
I fancy that I heard from Mr. Gall
That mushrooms could be cooked another way
I never read the works of Juvenal
I think I will not hang myself to-day.
The world will have another washing-day;
The decadents decay; the pedants pall;
And H.G. Wells has found that children play,
And Bernard Shaw discovered that they squall;
Rationalists are growing rational
And through thick woods one finds a stream astray,
So secret that the very sky seems small
I think I will not hang myself to-day.
ENVOI
Prince, I can hear the trumpet of Germinal,
The tumbrils toiling up the terrible way;
Even to-day your royal head may fall
I think I will not hang myself to-day.
### The Bearer of Burdens
There were three men walking up a steep hill, each very tired, with a heavy burden on his back. Then there stepped up to them a strong kind man, who said to them, "Let me take your burdens; I will carry them for you." But the first man said, "I have no burden," for he had carried his burden so long that it seemed like his clothes, or like part of his body, so that he did not feel it, and did not know how much better he could walk without it. So the first man would not have his burden touched.
The second man was very selfish and unkind himself, and he thought that all other people must be selfish and unkind, so he said, "You want to play me some trick; I do not believe you want to carry my burden; I will not let you touch it."
The third man was very tired indeed, and was saying to himself, "Oh, who can help me? for I feel that I cannot carry this terrible weight any further;" and when he felt the stranger touch him on the shoulder, and offer to take his burden, he said at once, "It is very kind of you; I am very thankful; please take it, for I see you can bear it and I cannot."
Edwin Abbott Abbott, Parables for Children. Abbott is most famous for being the author of Flatland.
## Sunday, February 05, 2012
### That Inward Eye: Mill and Wordsworth
I Wandered Lonely as a Cloud
(also known as "The Daffodils")
by William Wordsworth
I wandered lonely as a cloud
That floats on high o'er vales and hills,
When all at once I saw a crowd,
A host, of golden daffodils;
Beside the lake, beneath the trees,
Fluttering and dancing in the breeze.
Continuous as the stars that shine
And twinkle on the milky way,
They stretched in never-ending line
Along the margin of a bay:
Ten thousand saw I at a glance,
Tossing their heads in sprightly dance.
The waves beside them danced; but they
Out-did the sparkling waves in glee:
A poet could not but be gay,
In such a jocund company:
I gazed--and gazed--but little thought
What wealth the show to me had brought:
For oft, when on my couch I lie
In vacant or in pensive mood,
They flash upon that inward eye
Which is the bliss of solitude;
And then my heart with pleasure fills,
And dances with the daffodils.
I spent some time on this poem in my Ethics class on Thursday. What's the connection with ethics? John Stuart Mill. | 4,986 | 21,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | longest | en | 0.972998 |
http://euclideanspace.com/maths/geometry/rotations/euler/examples/index.htm | 1,579,544,406,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00501.warc.gz | 61,803,580 | 4,728 | # Maths - Euler Angles - Sample Orientations
In order to try to explain things and give some examples we can try I thought it might help to show the rotations for a finite subset of the rotation group. We will use the set of rotations of a cube onto itself, this is a permutation group which gives 24 possible rotations as explaned on this page.
heading applied first giving 4 possible orientations:
reference orientation heading = 0 attitude = 0 bank = 0 rotate by 90 degrees about y axis heading = 90 degrees attitude = 0 bank = 0 rotate by 180 degrees about y axis heading = 180 degrees attitude = 0 bank = 0 rotate by 270 degrees about y axis heading = -90 degrees attitude = 0 bank = 0
Then apply attitude +90 degrees for each of the above: (note: that if we went on to apply bank to these it would just rotate between these values, the straight up and streight down orientations are known as singularities because they can be fully defined without using the bank value)
heading = 0 attitude = 90 degrees bank = 0 heading = 90 degrees attitude = 90 degrees bank = 0 heading = 180 degrees attitude = 90 degrees bank = 0 heading = -90 degrees attitude = 90 degrees bank = 0
Or instead apply attitude -90 degrees (also a singularity):
heading = 0 attitude = -90 degrees bank = 0 heading = 90 degrees attitude = -90 degrees bank = 0 heading = 180 degrees attitude = -90 degrees bank = 0 heading = -90 degrees attitude = -90 degrees bank = 0
Normally we dont go beond attitude + or - 90 degrees because thes are singularities, instead apply bank +90 degrees:
heading = 0 attitude = 0 bank = 90 degrees heading = 90 degrees attitude = 0 bank = 90 degrees heading = 180 degrees attitude = 0 bank = 90 degrees heading = -90 degrees attitude = 0 bank = 90 degrees
Apply bank +180 degrees:
heading = 0 attitude = 0 bank = 180 degrees heading = 90 degrees attitude = 0 bank = 180 degrees heading = 180 degrees attitude = 0 bank = 180 degrees heading = -90 degrees attitude = 0 bank = 180 degrees
Apply bank -90 degrees:
heading = 0 attitude = 0 bank = 90 degrees heading = 90 degrees attitude = 0 bank = 90 degrees heading = 180 degrees attitude = 0 bank = 90 degrees heading = -90 degrees attitude = 0 bank = 90 degrees
encoding of these rotations in quaternions is shown here.
encoding of these rotations in matricies is shown here.
encoding of these rotations in axis-angle is shown here.
The working to convert each of these to matrix is shown here. | 600 | 2,468 | {"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} | 4.34375 | 4 | CC-MAIN-2020-05 | latest | en | 0.829313 |
https://brilliant.org/problems/all-roots/ | 1,524,593,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00179.warc.gz | 576,507,213 | 10,752 | # All Roots
Algebra Level pending
$\large \sinh(z)=i$
The roots of the equation can be expressed as $$\displaystyle z = i \left(an + \frac{b}{c} \right)d$$ , where $$n$$ belongs to set of integers and $$i = \sqrt{-1}$$ with $$a,b,c,d$$ are real constants with $$b , c$$ as co-primes.
Evaluate $$\large \frac{abcd}\pi$$.
× | 104 | 326 | {"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} | 2.609375 | 3 | CC-MAIN-2018-17 | latest | en | 0.807421 |
https://physics.stackexchange.com/questions/771798/conditions-for-the-virial-theorem | 1,716,096,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00337.warc.gz | 401,695,755 | 41,670 | # Conditions for the Virial theorem
I've been investigating the Virial Theorem, and I've found different conditions, some more restrictive than others, for it to be applicable to a given system. According to my professor, said theorem can be applied in systems which fulfil all of these simultaneously:
1. There is a large number of point-like masses
2. These masses interact with one another exclusively through gravitational interactions
3. The system is in thermal equilibrium
Nonetheless, in other texts I've found the Virial Theorem can only be applied to systems which are in hydrostatic (not thermal) equilibrium. Furthermore, condition nº $$1$$ doesn't hold either because I find the theorem ($$2K+U=0$$) also applies to a single body orbiting another one, since $$E_{total}=\frac{1}{2}U$$
What are the exact conditions a system must fulfil in order for the Virial Theorem to be applicable?
I prefer to think of the Virial Theorem as an umbrella name for a set of related concepts, related in the sense that they all have one particular element in common.
The discussion of the Virial theorem by John Baez is a good reference. (My presentation here proceeds in a different order than the presentation by John Baez.)
Historically the name 'virial' was introduced by Clausius, in the context of statistical mechanics. I surmise that over time physicists noticed the same element in other contexts, so the scope of what got to be referred as instance of virial theorem widened.
The common element, as I see it, is interconversion rate of potential and kinetic energy.
I will work bottom-up; first simplest cases, and from there in ascending complexity.
I start with the two force laws that have the property that they give rise to closed orbits:
-Hooke's law
-inverse square law (such as gravity)
Let an object be circumnavigating, with the centripetal force according to Hooke's law.
In circular circumnavitating motion, with the centripetal acceleration according to Hooke's law the ratio of potential energy to kinetic energy is 1:1
That ratio is independent of radial distance; when you give the object additional tangential velocity the potential energy and kinetic energy of the object increase in that 1:1 ratio.
When you make that circumnavigating motion elongated then along that trajectory the values of the energies oscillate. Theorem: the averaged energies (over the full period of circumnavigation) are still in that 1:1 ratio that typifies Hooke's law orbiting motion.
Let an object be circumnavigating, with the centripetal force according to inverse square law.
With an inverse square law: when you shift an object from one circular orbit to another circular orbit the kinetic energy and potential energy do not co-change, instead they counterchange.
When you boost an orbiting spacecraft to a higher orbit the potential energy increases in two ways: the energy from the boost, and decrease of kinetic energy, because at higher altitude the kinetic energy of orbital motion is lower.
Conversely, when you apply reverse thrust to slow down an orbiting spacecraft the total energy of the spacecraft's orbital motion decreases, but the orbital kinetic energy goes up.
On Kevin Brown's website, in the article the virial theorem
[...] for any bound system of particles interacting by means of an inverse-square force, the average (negative) potential energy is twice the average kinetic energy.
In the case of a highly eccentric orbit: average over a period of revolution. As I understand it: with eccentric orbit: the average kinetic energy and average potential energy are in the same ratio to each other as in the case of circular orbit with the same total orbital energy.
Generalization to any force law
In general a force law will be a function of the radial distance $$r$$ in accordance with the following pattern:
$$... , \ r^{-3}, \ r^{-2}, \ r^{-1}, \ \text{constant force}, \ r^1, \ r^2, \ ...$$
[LATER EDIT]
In the previous version of this answer the mathematical statement with the label '(1)', was incorrect, and the correct statements (2) and (3) did not follow from it.
We have: for $$n$$ an integer, with the force proportional to $$r^n$$:
$$2 T = (n + 1) V \tag{1}$$
Subject to the condition that the applied force must accomodate the inertial mass:
In the case of Hooke's law (1) gives the relation:
$$T = V \tag{2}$$
And in the case of an inverse square force law that gives:
$$2T = - V \tag{3}$$
An issue that I should have covered when I first posted this answer: how does it come about that in (1) neither the left hand side nor the right hand side has a factor $$r$$ for radial distance?
Given a particular force law: we can use the expression for required centripetal force in reverse: given a force law, what is the required velocity (hence required kinetic energy) for sustained circumnavigating motion?
In the case of Hooke's law: magnitude of the centripetal force is given by the product of a coefficient $$k$$ and radial distance $$r$$: $$F=kr$$
Also, to accomodate the inertial mass of a particular object the centripetal force must be proportional to inertial mass $$m$$.
We set the centripetal force equal to the quantity $$mkr$$, for the purpose of arriving at an expression for kinetic energy as a function of radial distance.
$$\frac{mv^2}{r} = mkr \tag{4}$$
To transform the left hand side to the expression for kinetic energy: move the factor $$r$$ to the right hand side, and divide both sides by 2:
$$\tfrac{1}{2}mv^2 = \tfrac{1}{2}mkr^2 \tag{5}$$
In the case of Hooke's law we have that the potential energy is proportional to the square of the radial distance. That is, the kinetic energy and the potential energy are both proportional to $$r^2$$
In the case of the inverse square law of gravity: the magnitude of the centripetal force is given by:
$$F = \frac{GMm}{r^2} \tag{6}$$
We set the required force equal to the force given by (6):
$$\frac{mv^2}{r} = \frac{GMm}{r^2} \tag{7}$$
Transform to make the left hand side the expression for kinetic energy:
$$\tfrac{1}{2}mv^2 = \tfrac{1}{2} \frac{GMm}{r} \tag{8}$$
The examples of Hooke's law and inverse square law of gravity indicate why in (1) there is no explicit dependence on the radial distance $$r$$.
The process of transforming the expression for required centripetal force to the expression for the corrresponding kinetic energy raises the power of the factor $$r$$ by one. That is the same as what happens in the process of integrating the force to obtain the expression for the potential energy: the integration with respect to the radial distance $$r$$ raises the power of $$r$$ by 1. It follows: (1) is valid for any force law.
The Virial Theorem holds for any system of masses where the potential energy $$U$$ of the interaction is a homogeneous function of the coordinates (i.e. $$U(\alpha \cdot \vec{r})=\alpha^k\cdot U(\vec{r})$$, with $$k$$ the degree of homogeneity of the function) and where the motion takes place in a finite region of space. This is shown in detail in §10 of the book 'Mechanics' by Landau and Lifshitz. A slightly more popular approach (which however also is based on the derivation in the book by Landau and Lifshitz) is given on this web page.
The Virial Theorem states then that
$$2\bar{T}=k\cdot \bar{U}$$
where $$\bar{T}$$ and $$\bar{U}$$ are the time averaged total kinetic and total potential energy repectively.
For the gravitational and Coulomb interaction, we have here $$k=-1$$ whereas for the harmonic oscillator $$k=2$$ (only in these two cases are actually closed orbits possible). | 1,800 | 7,600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-22 | latest | en | 0.946348 |
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Butane (C4H10) has actually a warm of vaporization of 22.44 kJ/mol and a regular boiling suggest of -0.4 degrees Celsius. A 250 mL flask contains 0.75g of butane at -22 levels Celsius. How much butane is present as a liquid
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why execute we usage 60:40 dichloromethane/hexane solvent mixture to separate 2-nitrophenol and 50:50 dichloromethane /ethylacetate solvent mixture to different 4-nitrophenol? What | 621 | 2,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.895484 |
http://vortextube.scienceontheweb.net/Propulsion/Advanced/advanced.html | 1,713,699,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00209.warc.gz | 33,153,202 | 8,789 | To break the vortex into bits, we create a very simple flow system - the building block of a vortex, or its discrete element.
The building block of a vortex
Consider a duct with compressed gas tank mounted to the inlet:
We shall move from simple to more complex concepts. For this reason, let's first study the behavior of the above duct system when the system is stationary. Then we can move on to the rectilinear motion of this duct system. Last, we can study its rotational motion.
The duct has adiabatic walls - no heat comes in or goes out through them. The same applies to the tank. The gas in the tank is stored at some high pressure and has room (static) temperature.
How does the duct system behave when not moving?
Let the system be stationary in the stationary frame of reference F. At some moment, gas is allowed to flow through the duct and leave through the outlet. This type of gas flow in adiabatic duct is known as "Fanno flow". The following takes place:
1.at inlet, gas velocity is lowest; at the outlet, gas velocity is highest
2.at inlet, gas pressure is highest; at the outlet, gas pressure is lowest (pressure difference drives the flow)
3.at inlet, gas density is highest; at the outlet, gas density is lowest
4.at inlet, the static temperature of the gas is highest; at the outlet, the static temperature of the gas is lowest
5.at any position in the duct, the mass flux is the same (continuity condition, mass conservation)
6.at any position in the duct, the total temperature is the same (adiabatic condition, enthalpy conservation)
We are specifically interested in the temperatures. At inlet, gas velocity is close to zero; at outlet, let the gas velocity be c. Then at inlet, the total temperature is equal to the static temperature of the tank. Denote this temperature with
Important!!!
The static temperature of the tank gas is denoted with "T-infinity" to show that it is the same as the ambient temperature (the temperature at infinity). This is a substantial part of the explanation of the vortex tube effect. Vortex tubes work with compressed, ambient-temperature air. But this air has first been compressed to some high pressure with a compressor. In the process, not just the pressure, but also the static temperature of the compressed air rises. Then the tank is left to cool to ambient temperature, which also leads to some decrease of its pressure. In this way, some invested energy is lost and this air is in essence, "pre-cooled" - its mere expansion to atmospheric pressure will already result in cold air. The vortex tube adds resistance to this expansion, thus cooling it even more, as we shall see below.
The total temperature at any cross-section of the duct is thus equal to this same value "T-infinity", including at outlet. At inlet,
At outlet,
Then
Then the observer in the stationary frame of reference F will see total temperature difference
and static temperature difference
This means that if the gas at the exit is stopped adiabatically, it will show the same thermodynamic temperature as the tank (all friction is neglected). Therefore, no cooling is observed yet.
(We neglect any cooling due to expansion before the gas reaches the inlet of the duct. At inlet, the gas is assumed to have exactly tank temperature, which, if the gas is injected leads to certain expansion and thus some cooling in real, non-ideal systems; in these, the static temperature difference for this reason can be slightly bigger)
What is the rectilinear motion of this duct system like?
We set the system in uniform rectilinear motion; the tank is at the leading end. Let the velocity of the system be equal to c. The inertial laboratory frame of the moving duct we denote with F'. The stationary frame we denote by F. Since the moving system F' is inertial, there is no acceleration. For the observer in F' the flow through the duct takes place just as before when the system was not moving. That is, the observer in F' sees Fanno flow in the duct. See all details above for the non-moving duct.
Does the moving observer see cooling?
No.
The observer in the moving frame of reference F' will see total temperature difference
and static temperature difference
This means that if the gas at the exit is stopped adiabatically by the observer in F', it will show the same thermodynamic temperature as the tank (all friction is neglected). Therefore, no cooling is observed F'.
Does the stationary observer see cooling?
Yes!!! Here is why.
First, let's look at the ejected gas. In the stationary frame, the duct system moves with velocity c, while in the moving frame, the gas leaves the duct with velocity c (in direction, opposite to the motion). This means, that the ejected gas has velocity of zero in the stationary frame.
Second, let's look at the moving gas tank. It moves with velocity c and the gas in it has static temperature
Then the stationary observer concludes, that the gas in the tank and at the inlet has total temperature
As we saw above, the static temperature of the gas at the outlet of the duct is
because
But this static temperature is the same in all frames of reference. This means that the stationary observer sees the ejected gas to have total temperature
Therefore, the observer in the stationary frame of reference F will see total temperature difference
(let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 108.25K separation in total temperature) and static temperature difference
(let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 54.1K maximum cooling of air, not counting any expansion effects when the gas may be injected from the tank into the duct inlet, similarly to the injection done in vortex tubes, where the maximum cooling may be somewhat larger due to nozzle expansion)
This means since the gas at the exit has been stopped adiabatically, it exhibits a drop in thermodynamic temperature compared to the tank (all friction is neglected). Therefore cooling is observed F. Not just cooling, but "temperature separation" is also observed. The leading end of the duct system appears hot, while its trailing end appears cold. This conclusion comes from the values of the total/static temperatures at the two ends of the system.
Where did the energy go?
The observer in the stationary frame F sees a gas parcel with high initial kinetic energy and room (static) temperature. Upon exiting the moving system, not only has the parcel stopped, it has also been cooled! This goes against intuition. Usually, when a moving parcel is forced to stop adiabatically, its static temperature goes up. The conclusion is:
The energy of the gas parcel was delivered as propulsion to the moving system!
Discretization of Vortex Tube Flow : Rotational Motion
The discrete vortex element (the duct) rotates about an axis with uniform angular velocity:
The rotation axis is perpendicular to the duct; the rotation axis is at the duct outlet. Gas from the tank flows towards the center of rotation. As before, the static temperature of the compressed gas in the tank is
Important!!!
The static temperature of the tank gas is denoted with "T-infinity" to show that it is the same as the ambient temperature (the temperature at infinity). This is a substantial part of the explanation of the vortex tube effect. Vortex tubes work with compressed, ambient-temperature air. But this air has first been compressed to some high pressure with a compressor. In the process, not just the pressure, but also the static temperature of the compressed air rises. Then the tank is left to cool to ambient temperature, which also leads to some decrease of its pressure. In this way, some invested energy is lost and this air is in essence, "pre-cooled" - its mere expansion to atmospheric pressure will already result in cold air. The vortex tube adds resistance to this expansion, thus cooling it even more, as we shall see below.
The rotating reference frame is F', the stationary frame is F.
What happens in the rotating frame?
The rotating observer notices that the compressed gas in the tank has ambient static temperature. The gas in the duct moves against the centrifugal force, as if "climbing up" a gravitational well. In doing so, it does work. This work can come only from the pressure energy contained in the gas. For this reason, as the gas moves towards the center, it expands, loses internal energy and becomes cold.
What happens to the total temperature T of the gas? In the "primed" frame F', this temperature is denoted by T' and is known as "relative total temperature", meaning "the total temperature as perceived by the rotating observer". Because the rotating frame is not inertial, the laws of physics here have different mathematical form. Remember the definition of "relative total temperature" in moving inertial frame? It was
where v' is the "relative velocity", or the gas velocity relative to the moving observer. In a moving, inertial frame T' is conserved throughout the duct. This meant, that the relative total temperature is the same at any cross-section of uniformly moving adiabatic duct. But in the non-inertial, rotating frame this is no longer true! In rotating frame, there is another temperature that is conserved. It is called "rotary total temperature" or "rotary stagnation temperature". Total temperature is related to the total enthalpy of the gas; Rotary total temperature is related to the rotational total enthalpy of the gas, also known as rothalpy. Therefore, the temperature that is conserved in rotating frames is
the rotary total temperature; v' is the relative velocity of the gas, omega is the angular velocity of the frame and r is the radial position. We omit the derivation of Trot; it can be derived from energy conservation considerations. At this time, it will suffice to say that in a rotating reference frame, at any cross-section of the adiabatic duct, the rothalpy of the gas is conserved, e.g.
What happens in the stationary frame?
The stationary observer monitors the total and static temperatures of the rotating gas. To derive an expression for the total temperature in F, use the velocity addition formula
where V is the velocity of the gas in the stationary frame. Now express v' in the velocity addition formula,
substitute it into the conservation of rothalpy statement
cancel the squared cross product and get
in other words
Now notice the sum of the first 2 terms is the definition of total temperature in the stationary frame; also cancel the factors of "2" in the 3rd fraction and thus
This is the vectorial form of Euler’s turbine equation, as shown in the previous section. This equation shows a quantity seen in F that is conserved at any cross-section of the rotating duct! Now, all ejected gas begins its motion at the same radial position R: the duct inlet. Denote the linear speed of the inlet (the linear speed of the rotating inlet, as seen in the stationary frame F) with c,
Look again at the conservation condition above. At the outlet, r=0; at inlet, v=c; then
which gives
The observer in the stationary frame sees this total temperature difference between inlet and outlet! (let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 108.25K separation in total temperature). What is the static temperature difference? At inlet,
At outlet,
Express the static temperatures at inlet/outlet through the total temperatures at inlet/outlet and take the difference:
precisely as in the case of rectilinear motion!
In the rotating frame of reference, the observer can see a room temperature reservoir; and all gas along the duct gradually becoming cold in both static temperature and relative total temperature (the total temperature observed in the moving frame). It is important to understand, that in the rotating frame, the entire system is COLD. The warmest spot in it is the tank at room temperature
This means radial heat transfer is not the reason for the radial temperature separation.
The stationary observer sees
in both static and total temperature! (let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 54.1K maximum cooling of air, not counting any expansion effects when the gas may be injected from the tank into the duct inlet, similarly to the injection done in vortex tubes, where the maximum cooling may be somewhat larger due to nozzle expansion)
In total temperatures, there is HEATING at periphery and COOLING at center! Air trajectories are SPIRALS! No doubt, the physics of this system is highly relevant to the vortex tube effect.
Where did the energy go?
The observer in the stationary frame F sees a gas parcel with high initial kinetic energy and room temperature (at periphery). Upon exiting the rotating system, not only has the parcel stopped, it has also been cooled! This goes against intuition. Usually, when a moving parcel is forced to stop adiabatically, its static temperature goes up. The conclusion is:
The energy of the gas parcel was delivered as propulsion to the rotating system! This is ANGULAR PROPULSION.
How To Model the Temperature Separation With CFD
How to prepare the geometry
The geometry is simple - a straight duct. The effect does not depend on size, thus choose the duct to be as long as you wish. For example, it can be chosen to be 15 m long, having rectangular cross-section of 0.3 m x 0.4 m (width x height). One end of the duct is the inlet; the other - the outlet. In this case, the geometries were prepared with Gambit (currently available only with legacy license, soon to be phased out). Alternatively, ICEM or other meshers may be used to build the grid for the duct. Here is a segment of meshed duct:
How to position the duct
Make sure to position the duct outlet face so that it goes through the origin (0,0,0) and is parallel to one of the coordinate axes, e.g. z. This will be the rotation axis. The direction of rotation (clockwise vs. counterclockwise) is unimportant. The face at (0,0,0) is outlet, the face at the other duct end is inlet.
What are the CFD simulation parameters in Ansys FLUENT?
The FLUENT solver parameters are:
•3d, double precision
•energy equation is ON
•viscous model: k-epsilon standard or realizable; k-omega SST is best; standard wall functions
•fluid: air, ideal gas
•operating conditions: all set to zero, no gravity
•fluid zone: set as "frame motion", choose rotation axis and constant rotation rate that will give c < 330 m/s
•Solution: "second order upwind"
•start with low Courant number and URFs: below 1.
•set all monitors to 1E-19
The FLUENT boundary conditions on the duct are:
•central outlet (exit) : pressure outlet, choose your pressure value (this value does not influence the cooling)
•inlet: either mass-flow-inlet or pressure-inlet.
•inlet reference frame: relative to adjacent cell zone. This ensures zero incidence angle of the inlet flow and proper inlet velocity magnitude!
•inlet flow direction: normal to boundary. Choose inlet temperature of 300 K.
•duct walls: no-slip; moving wall, rotational motion with 0 rad/s relative to adjacent fluid cell zone.
Ansys FLUENT simulation results:
FLUENT should predict cold temperatures very close (within 1.5 %) to their theoretical value given by the above formula.
•In total temperatures, heating and cooling should be symmetric with respect to the inlet temperature
•the temperature separation in thermodynamic (static) temperatures is 1/2 of the total temperature separation.
The information contained in this site is based on the following research articles written by Jeliazko G Polihronov and collaborators:
• “Thermodynamics of Angular Propulsion”
• “Vortex Tube Effect Without Walls”
• “Angular Propulsion - The Rotational Analog of Rocket Motion”
• “On the Thermodynamics of Angular Propulsion",
• Proceedings of the 10th International Conference on HEFAT, 14-16 July 2014, Orlando (2014). | 3,439 | 15,911 | {"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-2024-18 | latest | en | 0.933656 |
https://encyclopediaofmath.org/wiki/Continuity,_modulus_of | 1,723,411,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00411.warc.gz | 181,635,805 | 8,468 | # Continuity, modulus of
2020 Mathematics Subject Classification: Primary: 54C05 [MSN][ZBL]
One of the basic characteristics of (uniformly) continuous functions. The modulus of a continuity of a continuous function $f:\mathbb R^n \supset E \to \mathbb R^k$ is given by $\omega (\delta, f) := \sup_{|x-y|\leq \delta} |f(x)-f(y)|\, ,$ where we are implicitly assuming that $\lim_{\delta\downarrow 0} \omega (\delta, f) = 0$, which is indeed true if and only if $f$ is uniformly continuous. The definition of the modulus of continuity was introduced by H. Lebesgue in 1910 for functions of one real variable, although in essence the concept was known earlier. The definition can be readily extended to uniformly continuous maps $f$ between metric spaces $(X,d_X)$ and $(Y, d_Y)$, simply setting $\omega (\delta, f) := \sup_{d_X (x,y)\leq \delta} d_Y (f(x), f(y))\, .$
#### Examples
Very special classes of moduli of continuity give notable classes of functions. For instance
• If $\omega (\delta, f) \leq M \delta$ for some constant $M>0$, then $f$ satisfies the Lipschitz condition; the least constant $M$ for which such inequality holds is the Lipschitz constant of $f$; cf. also Lipschitz function.
• If $\omega (\delta, f) \leq M \delta^\alpha$ for some constants $M>0$, $\alpha\in ]0,1]$, then $f$ satisfies the Hölder condition of exponent $\alpha$.
• If
$\int_0^1 \frac{\omega (\delta, f)}{\delta}\, d\delta < \infty\, ,$ then $f$ is called, by some authors, Dini continuous (such condition plays a special role in the convergence of Fourier series, cf. Dini criterion).
#### Basic properties
It is elementary to derive bounds on the modulus of continuity of linear combinations, compositions and infima of uniformly continuous functions in term of their respective moduli of continuity. In particular
• $\omega (\delta, \lambda f + \mu g) \leq |\lambda| \omega (\delta, f) + |\mu| \omega (\delta, g)$;
• $\omega (\delta, g\circ f) \leq \omega (\omega (\delta, f), g)$;
• If $\{f_\lambda\}$ is a family of real-valued functions with $\omega (\delta, f_\lambda) \leq h (\delta)$ for some common function $h$, then $\inf_\lambda f_\lambda$ and $\sup_\lambda f_\lambda$ are also continuous with modulus of continuity bounded by $h$, provided the respective infima and suprema are finite (for which it is indeed necessary and sufficient that they are finite at some point).
For a non-negative function $\omega: [0,\infty[\to \mathbb R$ there is a continuous function $f: [0, \infty[ \to \mathbb R$ with $\omega (\delta, f) = \omega (\delta)$ for every $\delta$ if and only if the following properties hold:
• $\omega (0) = 0$
• $\omega$ is continuous
• $\omega$ is a subadditive function, namely $\omega (\alpha + \beta) \leq \omega (\alpha) + \omega (\beta)$ for every $\alpha, \beta \geq 0$.
#### Generalizations
As already mentioned, the notion can be easily generalized to maps between arbitrary metric spaces.
One can also consider moduli of continuity of higher orders. For instance, if $f$ is a function of one variable, the modulus of continuity can be rewritten as $\omega (\delta, f) = \sup_{|h| \leq \delta} \max_x \Delta_h f (x)\, ,$ where $\Delta_h f (x) = |f(x+h)-f(x)|$ is the usual finite difference of first order. Therefore, if we introduce the higher oder finite differences $\Delta^k_h (x) = \sum_{i =0}^k (-1)^{k-i} {{k}\choose{i}} f (x+ih)\, ,$ the higher oder moduli of continuity can be defined as $\omega_k (\delta, f) = \sup_{|h|\leq \delta} \max_x \Delta^k_h f (x)\, .$ See also Smoothness, modulus of. Moduli of continuity and smoothness are extensively used in approximation theory and Fourier analysis (cf. Harmonic analysis).
A further common generalization replaces pointwise maxima with integrals. For instance, if $f: \mathbb R^n \to \mathbb R$ is Lebesgue measurable, the $L^p$-modulus of continuity of $f$ is defined as $\omega^{(p)} (\delta, f) := \sup_{|\xi|\leq \delta} \int |f(x+\xi) - f(x)|^p\, dx\, .$ An obvious variant can be defined for maps on open domains $\Omega$ by simply restricting the domain of integration to $\{x\in \Omega : {\rm dist}\, (x, \partial \Omega) < \delta\}$. A classical characterization of the Sobolev space $W^{1,p} (\mathbb R^n)$ is then the following
Theorem Let $p \in ]1, \infty[$. $f\in L^p (\mathbb R^n)$ belongs to $W^{1,p}$ if and only if there is a constant $M$ such that $\omega^{(p)} (\delta, f) \leq M \delta^p$ for every $\delta$.
Cf. Theorem 3 in Section 5.8 of [Ev]. The limiting case $p=\infty$ of the above theorem is also valid and gives then the identity between $W^{1,\infty} (\mathbb R^n)$ and the space ${\rm Lip}_b (\mathbb R^n)$ of bounded Lipschitz functions. For $p=1$ the property $\omega^{(1)} (\delta, f) \leq M \delta$ characterizes instead the space of functions of bounded variation.
#### References
[AGS] N.I. [N.I. Akhiezer] Achiezer, "Theory of approximation" , F. Ungar (1956) (Translated from Russian) [AFP] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Dz] V.K. Dzyadyk, "Introduction to the theory of uniform approximation of functions by polynomials" , Moscow (1977) (In Russian) [Ev] L.C. Evans, "Partial differential equations", Graduate studies in mathematics. American Mathematical Society (1998). [St] K. G. Steffens, "The History of Approximation Theory", Birkhäuser (2006). [Zy] A. Zygmund, "Trigonometric series" , 1 , Cambridge Univ. Press (1988).
How to Cite This Entry:
Continuity, modulus of. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Continuity,_modulus_of&oldid=51338
This article was adapted from an original article by A.V. Efimov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 1,689 | 5,889 | {"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} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.812227 |
https://electronics.stackexchange.com/questions/337230/bandwidth-and-carrier-frequency | 1,601,437,360,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00408.warc.gz | 336,299,794 | 33,089 | # Bandwidth and Carrier frequency
What is exactly the difference between the Bandwidth and the carrier frequency ? And if we have two channels with the same bandwidth but with different carrier frequency will we have different speeds ? For example :
• System A : Bandwidth = 1 Khz , Carrier frequency = 1 Ghz
• System B : Bandwidth = 1 Khz , Carrier frequency = 1 Mhz
If we used 16-QAM in both systems will baudrate be the same ?
• Do you understand the answer to this question: Why is channel capacity a factor of bandwidth instead of frequency? – The Photon Oct 30 '17 at 22:57
• Actually I couldn't understand this part " A single frequency signal would be a continuous tone. It's amplitude would never change. It would just continue on repetitively forever. As such, it would not convey any information." – Ali Safaya Oct 30 '17 at 23:10
• I mean why would it continue repetitively ? If so wouldn't we be able to transmit data over it changing phase like in PSK ? – Ali Safaya Oct 30 '17 at 23:13
• What is the fourier transform of $\sin(\omega t)$? If you change the phase then you have $\sin(\omega t + m(t))$. If you take the F.T. of this, you won't get the same as for the simple sine, you'll get something spread over a band of frequencies. Put another way: once you modulate the phase, you no longer have a single-frequency signal. – The Photon Oct 30 '17 at 23:19
• So the answer to the question is yes and speed stay the same ? – Ali Safaya Oct 30 '17 at 23:25 | 379 | 1,477 | {"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} | 3.125 | 3 | CC-MAIN-2020-40 | latest | en | 0.913496 |
https://www.daeplatform.com/arduino-based-solar-tracker/ | 1,685,910,611,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00362.warc.gz | 778,705,204 | 53,030 | # Arduino Based Solar Tracker
Master Electronics, Programming, and Renewable Energy by Building Your Own Solar Tracking System with the Arduino Platfo
### If YOU are an Electrical Engineer then following quiz is for YOU
0%
2
Electrical Engineering
Smart Series Quiz-1
These questions are taken from smart series book which covers all the subjects of Electrical Engineering and Technology Field
1 / 30
Category: Power Plant
1) The average fossil fuel plant convers about is ______ % of the power going in to the power going out.
2 / 30
Category: Basic Electrical Engineering
2) Fourier series are infinite series of elementary trigonometric functions i.e. Sine and
3 / 30
Category: Basic Electrical Engineering
3) If the input capacitor of a power supply is shorted, it will result in
4 / 30
Category: Telecommunications
4) Modulation index in amplitude modulation
5 / 30
Category: Basic Electrical Engineering
5) Conjunction x ^ y behaves on digits 0 and 1 exactly as ____ does for ordinary algebra.
We have to maintain flatness of the surface.
6 / 30
Category: DC Machines
6) The Pole Shoes of DC Machines are fastened to the pole core by:
We have to maintain flatness of the surface.
7 / 30
Category: Power System Analysis
7) Which of the following are considered as disadvantages(s) of Gauss-Seidel method over Newton Raphson method in load flow analysis?
8 / 30
Category: Power Transmission and Distributions
8) Which of the following conditions relate line resistance ‘R’ and line reactance ‘X’ for maximum steady state power transmission on a transmission line?
9 / 30
Category: Singal and Systems
9) Main purpose of modulation process is to
10 / 30
Category: Electrical Machines
10) Fundamental property used in single node pair circuit analyzer is that ______ across all elements is same.
11 / 30
Category: Power Electroncis
11) If the firing angle in an SCR rectifier is decreased, output will be
12 / 30
Category: Electroncis
12) Discrete device field effect transistor is classified on the basis of their
13 / 30
Category: Electroncis
13) SCR (Silicon Controlled Rectifier) goes into saturation, when gate-cathode junction is
14 / 30
Category: Power Plant
14) Landfill gas is actually _______ used in thermal power plants.
15 / 30
Category: Electrical Machines
15) A transformer transforms
16 / 30
Category: Electroncis
16) A semiconductor device is connected in a series circuit with a battery and a resistance. If the polarity of battery is reversed, the current drops almost to zero. The device may be
17 / 30
Category: Network Analysis
17) Admittance is the reciprocal of
18 / 30
Category: Power Transmission and Distributions
18) For which purpose bundled conductors are employed to a power system
19 / 30
Category: Singal and Systems
19) The _____ time signal is described for all values of time.
20 / 30
Category: Power Plant
20) ______ is a by-product of paper industry, which is used by many industries as a primary source of electricity.
21 / 30
Category: Power System Analysis
21) There are ____ types of dependent sources, depending on the controlling variable and output of the source.
22 / 30
Category: Power System Analysis
22) Which of the following ideas means deciding at runtime what strategy to summon?
23 / 30
Category: Electrical Machines
23) Radio frequency chokes are air cored to
24 / 30
Category: Basic Electrical Engineering
24) Which line is obtained by the method of least square?
25 / 30
Category: Power System Analysis
25) The quantity “Rm” which relates dependent voltage to controlling current is called
26 / 30
Category: Power System Analysis
26) Power in AC circuit is found by
27 / 30
Category: Probability
27) What is the probability of a number “2” when a dice is thrown?
28 / 30
Category: Power Transmission and Distributions
28) Drop out to cut off ratio for most relays is of the order of
29 / 30
Category: Power Plant
29) If field resistance of DC shunt generator is increased beyond its critical value, the generator
30 / 30
Category: Basic Electrical Engineering
30) At very sunny places this source(s) of energy can be found
The average score is 28% | 966 | 4,204 | {"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-2023-23 | longest | en | 0.842948 |
einartysen.se | 1,531,744,583,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589270.3/warc/CC-MAIN-20180716115452-20180716135452-00243.warc.gz | 109,129,952 | 13,697 | ## Microsoft Excel Tip of the day: Using IF and cell ranges
Today I had one of thoose lovely “Ah! Never considered that before!”. I got a question from a colleague that had a formula that looked like this:
=””;”Missing”;”Present”)
Note that this is no hardcore crazy hackaway thing. Just a common “simple when you think of it”. I’m not claiming to be giving away awesome wisdom here. :)
Apart from not beeing very easy understandable it required a lot of handy work to stay up to date since columns where added periodically.
What it was supposed to do was check if all the cells between G3 and AK3 were empty, write “Missing” if they were and “Present” if they weren’t. If you have the energy to check you will notice that not all cells between G3 and AK3 was included. Some were lost when new columns were added.
Easy! I thougt, and entered: =IF(G3:AK3=””;”Missing”;”Present”)
Unexpected trouble. As far as I can remember I never tried using a range with the IF function that way before, lucky for me since it doesn’t work (neither do named areas).
I figured that I have to calculate a value from these cells so I asked what would be in the cells if present? Unfortunately the answer wasn’t digits. If it were numbers a simple: =IF(SUM(G3:AK3)=0;”Missing”;”Present”) would work.
Still, this should not be too tricky…and it wasn’t. As soon as I stoped thinking on how to identify the cell content at least (yes, that took me some beard scratching minutes).
Instead I count them. This is what I came up with:
IF(COUNTA(G3:AK3=0;”Missing”;”Present”)
=COUNTA simply counts the number of cells with values. Since I specify a range Excel will also update the formula to includedcells created within that range.
As with most questions about Excel, the solution was easy, you just have to think about it first :-)
## News or rumours or both?
Day4 is a Swedish company specialized in motion graphic and videos for TV, Internet and
so on. A week ago they got an idea (that right now makes any marketing company a bit jealous). They decided to create a rumour and see what happened with it.
That rumour was about Apple making there own weird screws and you can read all about what and how they did it directly from them here. They managed to get the rumour mill spinning either way.
The interesting thing is not that they managed to get a whole community (almost) talking about something made up but the perceived level of truth. The blogg Cult of Mac writes about it after 12 hours and then the ball was in motion. Yahoo, Wired and Mac World are among the news agents picking it up. Still, it’s a rumour and writers are sceptic. The longer from the (made up) source however…They note that when readers share and comment on these articles in there own media streams like Facebook, Twitter and Google+ – almost all questioning is gone.
Very interesting, if not so much of a chocker. Who values the news before it reaches you? I know a couple of journalists, both in old-school-paper-media and web publications. They might not always admit it but the faster our information wheel spins, the less source control. My humble guess is that that’s one of the main reasons we see so many articles stating another competitor as their source. Maybe we didn’t check it, but it’s not our fault, they wrote it first – sort of. Not that the source is a guarantee for accuracy, the Swedish wire service TT quickly gets a nation wide spread on their news. Both correct and incorrect ones.
Gets you thinking doesn’t it?
Disclaimer: the part about who was sceptic and the perceived level of truth is the opinion of Day4 and has not been thoroughly checked by me Ironic huh? :-)
## My new companion: Nexus7
If you’ve read my posts a month or so back, you’ll notice that I’ve been on the lookout for a tablet for a while. I decided to wait until Google I/O before deciding and suddenly I have 3 Nexus devices at home. Unfortunately it’s not released in Sweden yet but thanks to an awesome friend in the US that was solved with just some additional shipping charges.
Nexus 7 ships with the brand new Android 4.1 also known as Jelly Bean. All in all I’m very satisfied and will try to give you a quick write-up on why. This is my first tablet so this will be both, why do you (well, I) need a tablet and why I think the N7 is a good option.
When the iPad was introduced a lot of market analysts was, should we say sceptical? “Who will ever use that thing”, “Tried already by other companies, didn’t work” and so on was the verdict. Personally I was a bit more optimistic. I started using “handheld computers” at my first job. A glorious Palm m500 was replaced by a Tungsten that later became a TX. I used the calendar (not with a lot of meetings though, I was a technician after all), started reading books in PDF and played som games. Some e-mails was written but not many. After moving on with smartphones I had no trouble answering the question “Who will ever use a tablet/pad (or one of the many names)”. I will! I was however not that sure about the price Apple asked for it, and I wasn’t sure if enough “normal” users would like to use it, normal as in less geeky than I.
Fast forward a couple of years. I never got an iPad, I didn’t buy into the Apple ecosystem and considered it slightly expensive while at the same time the screen on my phone kept groving. Now that I got a tablet I regret waiting so long so here is why you should get a tablet if you have the possibility:
1. Reading e-books is cheaper, easier and by far cooler than regular books :)
2. Playing games are fun, and much more so on a bigger screen. Just played through classic RTS “Z Origins” and loved every minute (except when my grunts were mocking me for loosing).
3. Reading news and articles on a tablet is awesome. Some papers and magazines has their own apps, others are adjusted and available via one of the many apps for magazine reading like “Zinio Magazine Reader“. If your favourite news site haven’t made any mobile version several apps does a great adjustment job for you. Right now I’m trying to decide which one to use and run three different ones on and of: “Flipboard“, “Pulse” and “Google Currents“. Favourite right now is Pulse but they all have their pros and cons.
4. Blogging, mail and just about any writing at all is much much easier on the pad than on a smartphone. It’s not as smooth as on a laptop, but I seldom pick up my computer on the metro. With a nice sync-app everything is of course available on all your devices (Google Drive, Evernote and so on and so on, trying them out as well).
5. Control other devices, in my case I use it as a remote for my Mac Mini connected to my TV. Get one with infrared port and you can control basically every piece of your TV/home theater setup.
There you go. Now we move on to why Nexus7 is a good choice. This is of course much harder for me to answer since I’ve only owned this tablet but..I’ll give it a try anyway and you just remember my limited experience with other tablets :-)
1. It’s fast, really fast. I’ve been testing out both a Xoom and a Samsung tab before and this is way smoother (granted, it’s released much later).
2. You get a whole lot of bang for your buck because the Nexus 7 is cheap, no really, it’s cheap. Starting at \$199, compared to the new iPad \$499 (yes, the iPad have features the N7 lacks, more on that later) Samsung Tab 2 7″ starts somewhere around \$250-299. Finding USD prices from a Swedish IP is always a hassle :)
3. I absolutely love the form factor. 7″ is perfect for me. As long as I carry a bag (90-95% of the times I leave home) I got it with me. A larger more expensive tablet might be left at home more often. Holding it with one hand is comfortable, not just for a couple of minutes. That’s great if you’re using it for reading. I need to be able to stand up on the metro, hold on to the rail with one hand and still read, like I can with a paperback without trouble. Big enough to be a huge leap in usability compared to my Galaxy Nexus but small enough to be carried and used all the time, to sum it up. The material on the back deserves a mention as well, smooth and rubbery without being sticky, very well chosen by the designers!
4. Customization, this goes for all Android tabs. I will probably use my tab a bit different than you so why would we like them to look the same? Of course, you could just let it be and it will work just fine but the ability to change keyboard and features is a core feature for me. That includes widgets which are great on Android phones but is really awesome on a tablet.
5. As a Nexus device it comes loaded with the latest Android version, the already mentioned Jelly Bean 4.1. It will also be updated directly from Google hopefully quick and easy :)
But… The Nexus7 isn’t perfect of course. If these are deal breakers for you, keep on looking.
1. If you have an iPhone already, you have to consider the ecosystem factor. Syncing between devices, already bought apps etcetera.
2. Films likes big screens. If you plan to watch a lot of video you might want to consider a bigger device.
3. In order to keep the low price on the Nexus Asus and Google removed some common features. There is no 3G/4G/LTE version, you get Wifi and that’s it. There are no camera on the back, just the front facing one. Like all devices I’ve tried the front camera is aimed towards video chat and similar, not snapping good pictures. They also left out the ability to add memory through external card. An unfortunate trend that we see among phones as well.
4. I would’ve loved an infra red sensor to get rid of my three remotes (sound, TV and IPTV Set Top box), just like the Sony Tab S and the just announced Samsung tab 10.1 has.
A bit longer than I intended, even though most of the post was written on the Nexus itself. That’s how nice typing on it is :-)
## Make your Galaxy Nexus yakjuxw to a yakju
I might translate it later but so far this is only for Swedish readers who are still waiting for their Jelly Bean update (the forumpost linked in the bottom are in Swedish). Chances are that your phone has the build “yakjuxw” that will get the update later. Have no fear, transforming it to the international “yakju” is quite simple. | 2,382 | 10,249 | {"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-2018-30 | latest | en | 0.967662 |
https://www.doubtnut.com/question-answer/compare-the-following-rational-numbers--9-27-8453 | 1,623,579,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00021.warc.gz | 686,362,728 | 61,982 | Class 7
MATHS
Integers
Compare the following Rational numbers -: (-9)/(27)
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 21-12-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
157.0 K+
96.9 K+
Text Solution
Solution :
(-9)/27 = -9/27 =- (3*3)/(3*3*3) =-1/3<br> 6/(-18) = -6/18 = -(2*3)/(2*3*3) = -1/3<br> :. -9/27 = -6/18<br>
Image Solution
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4:03 | 419 | 930 | {"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} | 3.625 | 4 | CC-MAIN-2021-25 | latest | en | 0.441057 |
http://www.mathworks.com/matlabcentral/fileexchange/38829-matlab-solutions-introduction-to-digital-signal-processing-a-computer-laboratory-textbook/content/Chapter%205/ex544.m | 1,436,130,358,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097710.59/warc/CC-MAIN-20150627031817-00212-ip-10-179-60-89.ec2.internal.warc.gz | 622,750,400 | 11,311 | Code covered by the BSD License
# MatLab Solutions: "Introduction to Digital Signal Processing: A Computer Laboratory Textbook".
### Ilias Konsoulas (view profile)
29 Oct 2012 (Updated )
These files are the MatLab solutions of exercises contained in the above DSP lab textbook.
ex544.m
```% Exercise 5.4.4. Poles and Zeros at Infinity.
clc; clear; close all;
%% System function definition:
b = [2 -3/4]; % Numerator Coefficients.
a = [1 -3/4 1/8]; % Denominator Coefficients.
% [z p k] = tf2zpk(b,a); % Find the poles and zeros of the original function.
%% a. Produce and display the Poles/Zeros plot.
figure('Name','Exercise 5.4.4. Poles and Zeros at the Origin and at Infinity');
subplot(3,2,1);
zplane(b,a);
xlim([-1.5 1.5]);
ylim([-1 1]);
title(['H(z) has a single zero at the origin';
' or a single pole at Infinity ']);
% Find and plot the implulse response.
delta = [1 zeros(1,15)];
h = filter(b,a,delta); % Calculate the first 16 points of h[n].
subplot(3,2,2);
stem(0:15,h);
grid on;
title('h[n] = \{(1/2)^n+(1/4)^n\}*u[n]');
axis tight;
%% b.(i). Insert a zero at z = 0. This happens by multipying the system function H(z)
% by z (or by dividing with z^-1). However, by this operation, the system function H(z) becomes
% improper and MatLab symbolic toolbox cannot handle it. Therefore, I illustrate what happens
% to the impulse response manually. All in all the impulse response sequence advances a single
% step forward and therefore it starts from n=-1. This makes the system noncausal.
% Inserting a zero at the origin is equivalent of adding a pole at Infinity.
% As noted above, the following lines of code do not work:
% The new system function H1(z) = z*H(z) is:
% H1 = (2*z - 3/4)/(1 - 3/4*z^(-1) + 1/8*z^(-2));
% h11= simplify(iztrans(H1))
n1 = -1:14;
%h1 = double(subs(h11,n,n1)); % If MatLab could figure this out, it would
% correspond to an impulse response of a non-causal system as shown on the next plot:
subplot(3,2,4);
stem(n1,h,'*r');
title('h_1[n] = \{(1/2)^{n+1}+(1/4)^{n+1}\}*u[n+1]');
grid on;
axis tight;
subplot(3,2,3);
zplane(b,a);
title(['H_1(z) = zH(z) has a double zero at the origin';
' or a double pole at Infinity ']);
xlim([-1.5 1.5]);
ylim([-1 1]);
%% c. Introduce a pole at z = 0. This happens when we multiply the numerator of H(z) by z^(-1):
% This is equivalent to adding a pole at Infinity.
b2 = [0 2 -3/4]; % Numerator Coefficients.
a2 = [1 -3/4 1/8]; % Denominator Coefficients.
% Produce and display the Poles/Zeros plot.
subplot(3,2,5);
zplane(b2,a2);
title(['H_2(z) = z^{-1}H(z) has no zero at the origin';
' and no pole at Infinity ']);
xlim([-1.5 1.5]);
ylim([-1 1]);
% Find and plot the implulse response.
delta = [1 zeros(1,15)];
h2 = filter(b2,a2,delta); % Calculate the first 16 points of h[n].
subplot(3,2,6);
stem(0:15,h2,'g.');
grid on;
title('h_2[n] = \{(1/2)^{n-1}+(1/4)^{n-1}\}*u[n-1]');
axis tight;``` | 961 | 2,946 | {"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} | 3.90625 | 4 | CC-MAIN-2015-27 | longest | en | 0.762673 |
http://stackoverflow.com/questions/12103322/understanding-twos-complement | 1,441,354,912,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645339704.86/warc/CC-MAIN-20150827031539-00102-ip-10-171-96-226.ec2.internal.warc.gz | 220,264,916 | 19,889 | # Understanding two's complement
I'm trying to understand two's complement:
Does two's complement mean that this number is invalid:
1000
Does two's complement disallow the use of the most significant bit for positive numbers. Ie. Could
1000
Ever represent 2^3? Or would it represent -0?
I'm also confused about why you need to add 1 to a one's complement.
-
twos complement only works on set bit widths, like 32 bits, with the most significant bit is the complement bit – Keith Nicholas Aug 24 '12 at 4:56
so 1000 is actually 00000000000000000000000000001000 – Keith Nicholas Aug 24 '12 at 4:57
Work through a few examples by hand, you'll get a much better feel for what 2's complement is, and why computers represent integers that way, than you will by reading more about it. – High Performance Mark Aug 24 '12 at 10:11
## 4 Answers
in two's complement the MSB (most significant bit) is set to one for negatives. to multiply by -1 a two's complement number you well do the following:
add one to the number.
reverse all bits after the first one.
for example:
the number 10010 after adding one you well get: 10011 after reversing you get: 01101.
that means that 10010 is negative 13.
the number 1000 after adding one is: 1001 after reversing: 0111.
that means that 1000 is negative 7.
now, to your last question: no. if you work with two's complement you can't use the MSB for positive numbers. but, you could define you are not using two's complement and use higher positive numbers.
-
It depends on how many bits you use to represent numbers.
The leftmost (largest) bit has a value of -1*(2**N-1) or in this case, -8. (N being the number of bits.) Subsequent bits are their normal values.
So
1000
is -8
1111
is -1 0111 is 7.
However, if you have 8 bits these become different values! 0000 1000
is positive 8. Only the leftmost bit adds a negative value to the answer.
In either case, the range of numbers is from 1000....0
for -2**(N-1) with N bits to
0111....1 Which is 2**(N-1) -1. (This is just normal base 2 since the leftmost bit is 0.)
-
Twos-complement is based on two requirements:
• numbers are represented by a fixed number of bits;
• x + -x = 0.
Assuming a four bit representation, say, we have
0 + -0 = 0000 + -0000 (base 2) = 0000 => -0000 = 0000
1 + -1 = 0001 + -0001 (base 2) = 0000 => -0001 = 1111 (carry falls off the end)
Now we have our building blocks, a drop of induction will show you the "flip the bits and add 1" algorithm is exactly what you need to convert a positive number to its twos-complement negative representation.
-
2's complement is mostly a matter of how you interpret the value, most math* doesn't care whether you view a number as signed or not. If you're working with 4 bits, 1000 is 8 as well as -8. This "odd symmetry" arises here because adding it to a number is the same as xoring it with a number (since only the high bit is set, so there is no carry into any bits). It also arises from the definition of two's complement - negation maps this number to itself.
In general, any number k represents the set of numbers { a | a = xk mod n } where n is 2 to the power of how many bits you're working with. This perhaps somewhat odd effect is a direct result of using modular arithmetic and is true whether you view number as signed or unsigned. The only difference between the signed and unsigned interpretations is which number you take to be the representative of such a set. For unsigned, the representative is the only such a that lies between 0 and n. For signed numbers, the representative is the only such a that lies between -(n/2) and (n/2)-1.
As for why you need to add one, the goal of negation is to find an x' such that x' + x = 0. If you only complemented the bits in x but didn't add one, x' + x would not have carries at any position and just sum to "all ones". "All ones" plus 1 is zero, so adding one fixes x' so that the sum will go to zero. Alternatively (well it's not really an alternative), you can take ~(x - 1), which gives the same result as ~x + 1.
*Signedness affects the result of division, right shift, and the high half of multiplication (which is rarely used and, in many programming languages, unavailable anyway).
- | 1,118 | 4,242 | {"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} | 4.28125 | 4 | CC-MAIN-2015-35 | latest | en | 0.933963 |
https://math.stackexchange.com/questions/4360695/constructing-geodesic-polygons-in-uniformly-normal-balls | 1,709,533,079,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00765.warc.gz | 368,173,981 | 34,804 | # Constructing geodesic polygons in uniformly normal balls
I am studying the book "Riemannian Manifolds: An Introduction to Curvature" by John Lee. In the chapter on Gauss-Bonnet Theorem, there is an exercise problem that outlines the proof of the fact that every compact Riemannian 2-manifold has a smooth triangulation. In this outline, the author asks us to prove the following:
Let $$M$$ be a smooth, compact, Riemannian $$2$$-manifold, $$v_1, v_2, \cdots, v_k \in M$$ and $$\epsilon > 0$$ be such that the geodesic balls $$B_{3 \epsilon} \left( v_i \right)$$ are convex and uniformly normal and the geodesic balls $$B_{\epsilon} \left( v_i \right)$$ cover $$M$$. Then, for each $$i$$, there is a convex geodesic polygon in $$B_{3 \epsilon} \left( v_i \right)$$ whose interior contains $$B_{\epsilon} \left( v_i \right)$$.
The hint given to solve it says, "Let the vertices be sufficiently nearby points on the circle of radius $$2 \epsilon$$."
I understand how this might work. If we consider points on the circle of radius $$2 \epsilon$$ in the Euclidean plane, we can get a polygon whose sides do not intersect the ball of radius $$\epsilon$$. While this is geometrically clear to me, I am not able to bring in the rigour and write the proof of this.
Also, another question that I have is that if we move to arbitrary Riemannian manifolds where the metric is not flat, how can we guarantee that the geodesic between points does not enter the smaller ball of radius $$\epsilon$$?
Any insights into this are appreciable!
Call the points on the circle centered at, say, $$v_1$$ and of radius $$2\epsilon$$ $$p_1,...,p_n$$ where the cyclic order is the one coming from the circle.
Make sure that for each $$i$$, the distance from $$p_i$$ to $$p_{i+1}$$ ($$i$$ is taken mod $$n$$) is $$<\epsilon$$) and apply the triangle inequality.
This will guarantee that the polygon that you formed is disjoint from $$B(v_1,\epsilon)$$ and is contained in $$B(v_1, 3\epsilon)$$. | 535 | 1,975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.871533 |
https://codedump.io/share/sAUgqMuWHfJD/1/multiple-if-conditions-optimisation | 1,527,144,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00635.warc.gz | 543,807,616 | 8,871 | John Gkikas - 1 year ago 71
C Question
# multiple if conditions optimisation
I am building a simple C project (for arduino) and I have come across this question.It's not actually that language specific, it's more of an algorithm optimisation thing.
So, I need to check a value of X against a sensor reading.
``````If X <5 ...
else if x<10...
else if x<15...
else if x<20...
``````
Then in each clause I have the same for loop,but the iterations change depending on the value of X.
In a general sense, how can these if conditions be replaced by something unified?I remember these "gradation" or "leveling" problems in highschool, but we still used if clauses.
Then in each clause I have the same for loop,but the iterations change depending on the value of X.
I'd set a variable to the number of iterations, then put the `for` loop after the `if`/`else` sequence.
``````int loops;
if (X < 5) {
loops = /*whatever*/;
} else if (X < 10) {
loops = /*whatever*/;
} else if (X < 15) {
loops = /*whatever*/;
// ...and so on...
} else {
loops = /*whatever for the catch-all case*/;
}
for (int i = 0; i < loops; ++i) {
// ...
}
``````
If you're trying to avoid the `if`/`else`, if there are only a small number of possible sensor values, you could use a `switch` instead, which in some languages is compiled to a jump table and so fairly efficient.
If you want to have the ranges held as data rather than in an `if`/`else` sequence, you could use an array of values:
``````int[][] values = {
{5, 500},
{10, 700},
{15, 800},
{20, 1200},
{0, 1500} // 0 is a flag value
};
``````
(There I'm using an array of `int[]`, but it could be a nice clean class instance instead.)
Then loop through the array looking for the first entry where `X < entry[0]` is true (or where `entry[0]` is `0`, to flag the last entry).
``````int loops = 0; // 0 will never be used, but the compiler doesn't know that
for (int[] entry : values) {
if (entry[0] == 0 || X < entry[0]) {
loops = entry[1];
break;
}
}
``````
...followed by the `for` loop using `loop`s.
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 607 | 2,146 | {"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} | 2.875 | 3 | CC-MAIN-2018-22 | latest | en | 0.816099 |
http://www.edaboard.com/showthread.php?t=277363 | 1,519,388,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814700.55/warc/CC-MAIN-20180223115053-20180223135053-00261.warc.gz | 450,242,572 | 24,418 | # Converting the battery voltage , help to a syrian xD
1. ## Converting the battery voltage , help to a syrian xD
( I have already posted this thread in another website )
hello my name is Manar 4th year (stopped studying) computer engineering bachelor , and I live still in Syria.
As it can be called the silliest thing happening here , but it is annoying , that we don't get electricity most of the day .
I am trying to build a battery for my internet router , I already have got the battery actually but I need to build the circuit so it gives what the router needs , the battery is 8.5 volts 2000 mA I got it from an old camera , and the router needs 12 volts and 1 A , the first thing is , how can I convert these 8,5 volts to 12 ? is there any way to do that ?
thanks you very much
•
2. ## Re: Converting the battery voltage , help to a syrian xD
Perhaps, first check if 12V 1A is really needed. Most routers have circuitry to step down 12V to 5 Volt,3.3Volt and so forth already on board and input can vary from the 12V. So depending on the exact design of router circuitry the input may have a range to get 12W power needs, i.e. so with a 12V power supply 12V*1A = 12W but 8.5V*1.5 is circa 12W.
3. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by cntube
Perhaps, first check if 12V 1A is really needed. Most routers have circuitry to step down 12V to 5 Volt,3.3Volt and so forth already on board and input can vary from the 12V. So depending on the exact design of router circuitry the input may have a range to get 12W power needs, i.e. so with a 12V power supply 12V*1A = 12W but 8.5V*1.5 is circa 12W.
you are totally right ,
how can I check if its needed ?
and if its like that , you mean that if i put this 8.5 battery with an lm317 and set the current to 1.5 A and plug it to the router it will be on and ok ?
thank u very much for the reply really
4. ## Re: Converting the battery voltage , help to a syrian xD
A simple check is to use a lab power supply to set the input voltage to 8.5V and check router function. Going to a lower voltage will not damage anything but depending on the router circuitry may not function at the lower voltage.
The LM317 is not needed because the router power supply will adjust automatically for input voltage variations, that is, increase current in take if the router has a switching converter, if a linear regulator is used, then current would remain about the same and actually the linear regulator will have to dissipate less power.
1 members found this post helpful.
5. ## Re: Converting the battery voltage , help to a syrian xD
Better open router and see inside whats on PCB. This can help you to understand what voltage router PCB need.
Its big chance that is only 5V.
Also You can post photos here.
6. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by cntube
A simple check is to use a lab power supply to set the input voltage to 8.5V and check router function. Going to a lower voltage will not damage anything but depending on the router circuitry may not function at the lower voltage.
The LM317 is not needed because the router power supply will adjust automatically for input voltage variations, that is, increase current in take if the router has a switching converter, if a linear regulator is used, then current would remain about the same and actually the linear regulator will have to dissipate less power.
ummm these r useful information , I should have known that , but what would be the difference if i dont use a lab power supply and I hook the battery directly ?
and u mean by that is that I am safe if I just hooked the V+ and the gnd of the battery to the router just like that without any driving circuit , it would works ?
danke schon yet again
•
7. ## Re: Converting the battery voltage , help to a syrian xD
Battery dont have constant voltage level, voltage will drop on load and during discharging.
This battery of 2Ah is not big. This is smaller in capacity from NiMh AA batteries (of course NiMh single cell is 1,2V bat in serie of 7-8 can make higher voltage). AA can have over 3Ah.
•
8. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by tpetar
Battery dont have constant voltage level, voltage will drop on load and during discharging.
This battery of 2Ah is not big. This is smaller in capacity from NiMh AA batteries (of course NiMh single cell is 1,2V bat in serie of 7-8 can make higher voltage). AA can have over 3Ah.
actully my battery is an NiMh , I didn't understand what do you mean , but actully today i have hooked the battery and it didnt work well, but it was giving only 6 volts , but now iam going to fully recharge it and try again , but shouldn't I make a circuit that would keep the voltage constant ?
thank uuuuu
9. ## Re: Converting the battery voltage , help to a syrian xD
Better first check router PCB design in input power section, to se what voltages that device use. If you give 6V to router, there is a possibility to make some damage to IC parts. Usually used voltage regulators needs 2,5V-3V minimal Vi-Vo difference. You need to take into account that the battery full and empty have different voltages, that voltage will change during disharge process.
1 members found this post helpful.
10. ## Re: Converting the battery voltage , help to a syrian xD
The cell voltage does not vary a great deal, for example fully charged 4 cell NiMH measured around 6V. Used this output through a diode to drop 0.7V for 5V+-10% until battery flat at circa 5.2V. The power requirements for your router are small enough that a boost circuit, if needed could power the device, possibly letting one get every maH out of the cell before overdischarging. At least, it is not a requirement like a have 6V 5AH battery and want to make 1kW inverter:D.
11. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by newbie noob
actully my battery is an NiMh , I didn't understand what do you mean , but actully today i have hooked the battery and it didnt work well, but it was giving only 6 volts , but now iam going to fully recharge it and try again , but shouldn't I make a circuit that would keep the voltage constant ?
thank uuuuu
If the battery voltage drops from 8.5V to 6V when loaded, then it is probably low on capacity and needs a recharge.
The router certainly has internal voltage regulators, so your source voltage doesn't need to be well regulated. But it does need to be above some minimum voltage at all times. That minimum voltage is not certain, but it might be low enough that your battery can work. If not, you will need to make a boost converter of some type. What resources do you have for making circuitry?
Good luck, please stay safe.
1 members found this post helpful.
12. ## Re: Converting the battery voltage , help to a syrian xD
If you assume the 'router' requirements are correct and that it does require 12V at 1 A, I have no idea what kind of a router would use 12W so im a bit suspicious but assuming it does. then you need your circuit to be at least 70.5% efficient. Now assuming this is a rechargeable battery and that its only holding 80% of the capacity as it did when new out of the box, then you are looking at requiring a step up converter efficiency of over 88% (this is getting tougher now). The fact that when you plugged in the battery and it dropped to 6 volts means its underpowered(not able to sink the current demanded from the load), however if you are plugging it in directly the 'router' really requires 12V then it might be trying to sink more current then normal do to operating voltages being lower then required and circuitry not functioning correctly. As mtwieg mentioned it will have internal regulators requiring a min range, but 8/12 is only 66% of the value, i double any regulator works with that much undervoltage and maintains any type of efficiency ( remember you loose some power with regulators as well as the step up converter).
13. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by mtwieg
If the battery voltage drops from 8.5V to 6V when loaded, then it is probably low on capacity and needs a recharge.
The router certainly has internal voltage regulators, so your source voltage doesn't need to be well regulated. But it does need to be above some minimum voltage at all times. That minimum voltage is not certain, but it might be low enough that your battery can work. If not, you will need to make a boost converter of some type. What resources do you have for making circuitry?
Good luck, please stay safe.
ok this was helpful ! , that is what I was thinking of but couldn't get the name " boost circuit" , and for resources , most of them will be available , but I guess some would be not available of a company and it would be available by another one , I mean for example if an ic by National Instrument was not found , you can find the same ic but from another company , anyhow I will be able to get all of the parts if I was lucky ,
I will try to look over the internet for a boost circuit and then come back here ,
thank you very much
14. ## Re: Converting the battery voltage , help to a syrian xD
Originally Posted by mtwieg
If the battery voltage drops from 8.5V to 6V when loaded, then it is probably low on capacity and needs a recharge.
The router certainly has internal voltage regulators, so your source voltage doesn't need to be well regulated. But it does need to be above some minimum voltage at all times. That minimum voltage is not certain, but it might be low enough that your battery can work. If not, you will need to make a boost converter of some type. What resources do you have for making circuitry?
Good luck, please stay safe.
mr mtwieg sorry if I was bothering you , but I found this circuit which steps up the voltage http://www.eleccircuit.com/dc-to-dc-...tor-by-lm2577/ but someone here has mentioned that using the lm317 with current regulating would disturb the circuit of the router and wont work probebly , does that so with lm2577 ? can you tell me if this circuit is suitable or not ? and I promise that this would be the last question
•
15. ## Re: Converting the battery voltage , help to a syrian xD
You will have a loss of energy, and your batteries will not last long.
16. ## Re: Converting the battery voltage , help to a syrian xD
This LM2577 circuit looks like it will meet requirements. However, the condition of the battery that you have on hand is the big question. Is the battery degraded or even rated for 2 Amp current draw? The conversion from 6V to 12V would require over 2A at 6V. Can the battery maintain 5-6V at over 2A draw? A load test would help determine this. That is just the POWER from battery requirements. The energy is stated to be 2000mAh so if the battery could give over 2A at 6V then the run time would be about 1 hour. There are a lot of details to consider. The efficiency of the converter would hinge on the inductor selection with due consideration for diode selection and output capacitor selection.
--[[ ]]-- | 2,681 | 11,152 | {"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-2018-09 | latest | en | 0.92914 |
https://www.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=3993 | 1,618,243,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00342.warc.gz | 778,148,843 | 17,245 | ### PC.FBF: Building Functions
#### PC.FBF.1: Write a function that describes a relationship between two quantities.
PC.FBF.1.b: Combine functions using the operations addition, subtraction, multiplication, and division to build new functions that describe the relationship between two quantities in mathematical and real-world situations.
#### PC.FBF.4: Understand that an inverse function can be obtained by expressing the dependent variable of one function as the independent variable of another, as f and g are inverse functions if and only if f(x)=y and g(y)=x, for all values of x in the domain of f and all values of y in the domain of g, and find inverse functions for one-to-one function or by restricting the domain.
PC.FBF.4.a: Use composition to verify one function is an inverse of another.
PC.FBF.4.b: If a function has an inverse, find values of the inverse function from a graph or table.
### PC.FIF: Interpreting Functions
#### PC.FIF.7: Graph functions from their symbolic representations. Indicate key features including intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior and periodicity. Graph simple cases by hand and use technology for complicated cases.
PC.FIF.7.a: Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
PC.FIF.7.b: Graph radical functions over their domain show end behavior.
PC.FIF.7.c: Graph exponential and logarithmic functions, showing intercepts and end behavior.
PC.FIF.7.d: Graph trigonometric functions, showing period, midline, and amplitude.
### PC.FT: Trigonometry
#### PC.FT.2: Define sine and cosine as functions of the radian measure of an angle in terms of the x - and y - coordinates of the point on the unit circle corresponding to that angle and explain how these definitions are extensions of the right triangle definitions.
PC.FT.2.a: Define the tangent, cotangent, secant, and cosecant functions as ratios involving sine and cosine.
PC.FT.2.b: Write cotangent, secant, and cosecant functions as the reciprocals of tangent, cosine, and sine, respectively.
### PC.NVMQ: Vector and Matrix Quantities
#### PC.NVMQ.4: Perform operations on vectors.
PC.NVMQ.4.a: Add and subtract vectors using components of the vectors and graphically.
PC.NVMQ.4.b: Given the magnitude and direction of two vectors, determine the magnitude of their sum and of their difference.
#### PC.NVMQ.11: Apply 2 × 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area.
Correlation last revised: 9/16/2020
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. | 630 | 2,823 | {"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-2021-17 | longest | en | 0.809198 |
https://metric-calculator.com/convert-g-to-gal.htm | 1,718,231,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00844.warc.gz | 363,093,278 | 5,783 | # Grams to Gallons [water] Converter
Select conversion type:
Rounding options:
Convert Gallons to Grams (gal to g) [water]▶
## Conversion Table
grams to gallons g gal 100 g 0.0264 gal 200 g 0.0528 gal 300 g 0.0793 gal 400 g 0.1057 gal 500 g 0.1321 gal 600 g 0.1585 gal 700 g 0.1849 gal 800 g 0.2113 gal 900 g 0.2378 gal 1000 g 0.2642 gal 1100 g 0.2906 gal 1200 g 0.317 gal 1300 g 0.3434 gal 1400 g 0.3698 gal 1500 g 0.3963 gal 1600 g 0.4227 gal 1700 g 0.4491 gal 1800 g 0.4755 gal 1900 g 0.5019 gal 2000 g 0.5283 gal
## How to convert
1 gram (g) = 0.000264172 gallon (gal). Gram (g) is a unit of Weight used in Metric system. Gallon (gal) is a unit of Volume used in Standard system. Please note this is weight to volume conversion, this conversion is valid only for pure water at temperature 4 °C.
US oz = 28.349523125 g
US fl oz = 29.5735295625 ml (milliliters) = 29.5735295625 g (grams) of pure water at temperature 4 °C.
US cup = 8 fl oz = 236.5882365 ml (milliliters) = 236.5882365 g (grams) of pure water at temperature 4 °C.
## Grams: A Unit of Weight
Grams are a unit of weight that are used in the International System of Units (SI), also known as the metric system. Grams are derived from the French word gramme, which was taken from the Late Latin term gramma, meaning a small weight. The symbol for gram is g.
## How to Convert Grams
Grams can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert grams to other units of weight in the US customary system and the SI system:
• To convert grams to ounces, divide by 28.349523125. For example, 100 g = 100 / 28.349523125 = 3.527 oz.
• To convert grams to pounds, divide by 453.59237. For example, 500 g = 500 / 453.59237 = 1.102 lb.
• To convert grams to tons (short), divide by 907184.74. For example, 1000 g = 1000 / 907184.74 = 0.0011 ton.
• To convert grams to kilograms, divide by 1000. For example, 200 g = 200 / 1000 = 0.2 kg.
• To convert grams to milligrams, multiply by 1000. For example, 50 g = 50 x 1000 = 50000 mg.
• To convert grams to micrograms, multiply by 1000000. For example, 10 g = 10 x 1000000 = 10000000 µg.
Grams also can be marked as grammes (alternative British English spelling in UK).
## Gallons: A Unit of Volume
Gallons are a unit of volume that are used to measure liquids, such as water, milk, oil, wine, etc. They are also used to measure some dry goods, such as grains, fruits, nuts, etc. They are different from cups, which are a smaller unit of volume. They are also different from liters, which are a larger unit of volume. They are also different from barrel of oil equivalent (BOE), which is a unit of energy based on the approximate energy released by burning one barrel of crude oil.
## How to Convert Gallons
To convert gallons to other units of volume, one can use the following formulas:
• To convert US liquid gallons to liters: multiply by 3.785
• To convert US liquid gallons to cubic inches: multiply by 231
• To convert US liquid gallons to fluid ounces: multiply by 128
• To convert US liquid gallons to UK gallons: multiply by 0.833
• To convert US liquid gallons to BOE: divide by 5
• To convert US dry gallons to liters: multiply by 4.405
• To convert US dry gallons to cubic inches: multiply by 268.8
• To convert US dry gallons to fluid ounces: multiply by 148.9
• To convert US dry gallons to UK gallons: multiply by 0.969
• To convert US dry gallons to BOE: divide by 4.6
• To convert UK gallons to liters: multiply by 4.546
• To convert UK gallons to cubic inches: multiply by 277.4
• To convert UK gallons to fluid ounces: multiply by 160
• To convert UK gallons to US liquid gallons: multiply by 1.2
• To convert UK gallons to BOE: divide by 6.1
The US gallon is equal to 3.785411784 liters and defined as 231 cubic inches.
Español Russian Français | 1,167 | 3,876 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.727391 |
http://www.calculatorx.com/convert/electric/kw-to-kva.htm | 1,527,428,309,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00464.warc.gz | 337,225,276 | 6,702 | # How to convert kW to kVA
## How to convert kW to kVA
How to convert real power in kilowatts (kW) to apparent power in kilovolt-amps (kVA).
### kW to kVA calculation formula
The apparent power S in kilovolt-amps (kVA) is equal to the real power P in kilowatts (kW), divided by the power factor PF:
S(kVA) = P(kW) / PF
So kilovolt-amps are equal to kilowatts divided by the power factor.
kilovolt-amps = kilowatts / PF
or
kVA = kW / PF
#### Example
What is the apparent power in kilovolt-amps when the real power is 3 kW and the power factor is 0.8?
Solution:
S = 3kW / 0.8 = 3.75kVA
How to convert kVA to kW ► | 202 | 625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-22 | latest | en | 0.792362 |
http://math.stackexchange.com/questions/383454/az-z-lead-to-xt-c-1e-lambda-1-tz-1c-2e-lambda2-t-z-2-c-pe/383474 | 1,398,013,749,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00538-ip-10-147-4-33.ec2.internal.warc.gz | 145,605,744 | 16,609 | # $Az=λz$ lead to $x(t) = c_1*e^{\lambda_1 t}z_1+c_2*e^{\lambda2 t} z_2+…+c_p*e^{\lambda_p t}z_p$ is a solution to $dx/dt=Ax$. Why?
I'm studying a course in dynamical systems. It's a pretty much linear algebra intensive course, and it's been a while since I did that sort of things.
In it, they say that if vector $z$ satisfies $Az=\lambda z$, then $x(t) = c e^{\lambda t} z$ is a solution to $dx/dt = Ax$.
So far so good.
Further, they claim that if there are several solutions to the equation $Az = \lambda z$, that is $A z_1 = \lambda_1 z_1, A z_2 = \lambda_2 z_2, ... , A z_p = \lambda_p z_p$, then according to the superposition principle it follows that for all $c_1, c_2, ... , c_p$ the vector function $x(t) = c_1*e^{\lambda_1 t}z_1+c_2*e^{\lambda_2 t} z_2+...+c_p*e^{\lambda_p t}z_p$ is a solution to $dx/dt=Ax$. I don't understand this reasoning. Could anybody explain why it follows from the superposition principle. Tried all day get my head around it.
-
Also, I must say I'm a little bit unsure what x(t) is supposed to mean here. Is it a vector or a scalar? I'm almost 100% sure at least t is a scalar (because I guess it is time). If it is a vector, what is the meaning of $dx/dt$? Do you just take the derivative of x elementwise? – user1661303 May 6 '13 at 16:26
Yes, x(t) is a vector function: if $A$ is an $n\times n$ matrix, then $x(t)\in\mathbb R^n$. The rest is a consequence of linearity. – Ted Shifrin May 6 '13 at 16:27
It's because the differential equation is linear in $x$. – Milind May 6 '13 at 16:27
You are correct about $dx/dt$ - it is element-wise. If $x(t)$ were the location of a particle at time $t$, then $dx/dt$ would be the velocity (speed and direction) of the particle at time $t$. – Thomas Andrews May 6 '13 at 16:29
That's the thing, I guess I'm not 100% sure about the definition of the derivative of a vector. $c_1*e^{\lambda1*t} z_1$ should be a vector since $z_1$ is a vector. Is it just elementwise? – user1661303 May 6 '13 at 16:30
Given your final expression for $x(t)$,
$\displaystyle\frac{dx}{dt}=c_1\lambda_1 e^{\lambda_1 t} z_1+ c_2\lambda_2 e^{\lambda_2 t} z_2+ ... + c_p\lambda_p e^{\lambda_p t} z_p$.
Given your immediately preceding expressions relating $A$, $z_1, \lambda_1$, $z_2, \lambda_2$ etc.,
$\displaystyle\frac{dx}{dt}=c_1 e^{\lambda_1 t} Az_1 + c_2 e^{\lambda_2 t} Az_2 + ... + c_p e^{\lambda_p t} Az_p = A(c_1 e^{\lambda_1 t} z_1 + c_2 e^{\lambda_2 t} z_2 + ... + c_p e^{\lambda_p t} z_p)=$
$=Ax$
So your superposed expression is a solution of $\displaystyle\frac{dx}{dt}=Ax$.
-
Still not getting it. The prerequisites is that $c_1*e^{\lambda_1 t}z_1 = dx/dt, c_2*e^{\lambda_2 t}z_2 = dx/dt, c_3*e^{\lambda_3 t}z_3 = dx/dt$. Why does that mean that the sum of them equals $dx/dt$? – user1661303 May 6 '13 at 16:58
Assume $x(t)=c_1 e^{\lambda_1 t} z_1 + ....$ Now take the time derivative of this expression. Now you should be able to see how to replace all the $\lambda_n$'s in the time derivative with $A$. Now factor out the $A$ and you should be able to see that you have $Ax$. So $dx/dt = Ax$. – bob.sacamento May 6 '13 at 17:11
Actually, that makes sence. Thanks. – user1661303 May 6 '13 at 17:17
Glad to be of help. An upvote or an accept would be nice. Sorry to self-promote, but it's not like I'm the only one who does it. :) – bob.sacamento May 6 '13 at 18:48
@user1661303 Please consider upvote and accept the answer if you think it helps: meta.math.stackexchange.com/questions/3286/… – Shuhao Cao Aug 3 '13 at 23:41
show 1 more comment | 1,233 | 3,537 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2014-15 | latest | en | 0.914828 |
https://techcommunity.microsoft.com/t5/excel/help-i-forgot-how-to-write-a-quot-if-quot-statement-formula/td-p/3986259 | 1,726,402,849,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00637.warc.gz | 517,385,384 | 57,192 | SOLVED
# Help, I forgot how to write a "if" statement formula
Copper Contributor
# Help, I forgot how to write a "if" statement formula
Hello and thank you in advance.
I used to know how to do this but my brain is mush lately.
So, I am trying to create a :If" formula that looks at a cell with 2 choices in a drop down menu Bill or Pending (G6)
if the (G6) answer is Bill, then it takes the value in F6 and copies it to H6.
if the (G6) answer is Pending, then it writes Pending in H6
Thank you again,
GB
3 Replies
best response confirmed by PeterBartholomew1 (Silver Contributor)
Solution
# Re: Help, I forgot how to write a "if" statement formula
In H6
``=IF(G6="Bill", F6, IF(G6="Pending", G6, "something else") )``
or, if only two options exist
``=IF(G6="Bill", F6, G6 )``
# Re: Help, I forgot how to write a "if" statement formula
Thank you so much. Works great.
# Re: Help, I forgot how to write a "if" statement formula
@Glennbart , you are welcome
1 best response
Accepted Solutions
best response confirmed by PeterBartholomew1 (Silver Contributor)
Solution
# Re: Help, I forgot how to write a "if" statement formula
In H6
``=IF(G6="Bill", F6, IF(G6="Pending", G6, "something else") )``
or, if only two options exist
``=IF(G6="Bill", F6, G6 )`` | 385 | 1,280 | {"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-2024-38 | latest | en | 0.84358 |
https://web2.0calc.com/questions/how-many-months-will-money-become-three-times-itself | 1,580,105,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694908.82/warc/CC-MAIN-20200127051112-20200127081112-00349.warc.gz | 736,181,082 | 6,097 | +0
# How many months will money become three times itself at 6% compounded semiannually
+1
125
2
+9
How many months will money become three times itself at 6% compounded semiannually
Nov 4, 2019
#1
0
y = (1.06*0.5)/3
Nov 4, 2019
#2
+106539
+1
3A = A ( 1 + .06 / 2)^(N * 2) where N is in years
3 = (1.03)^(2N) take the log of both sides
log 3 = log (1.03)^(2N)
log 3 = (2N) log (1.03)
N = log (3) [ 2 log (1.03)] ≈ 18.58 years ≈ 18 years and 7 months = 223 months
Nov 4, 2019 | 230 | 508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-05 | longest | en | 0.802217 |
https://chat.stackexchange.com/transcript/13775/2019/11/17 | 1,576,454,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00170.warc.gz | 304,462,012 | 4,469 | 12:57 AM
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Bubble Sorting
Author Message
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Bubble Sorting
Hey There,
I need a bubble sorting algorithm for RobotC for the following numbers: Rand0, Rand1, Rand2, Rand3, Rand4, Rand5, Rand6, Rand7.
If someone could do that that would be amazing.
Thanks!
Mon Jul 23, 2012 2:36 pm
Joined: Wed Mar 05, 2008 8:14 am
Posts: 3654
Location: Rotterdam, The Netherlands
Re: Bubble Sorting
You're in luck, check out my tutorial
http://botbench.com/blog/2012/07/21/tut ... your-data/
It was published 2 days ago.
- Xander
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Mon Jul 23, 2012 2:43 pm
Joined: Thu May 24, 2012 12:15 pm
Posts: 722
Re: Bubble Sorting
Now, that's just good luck...
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Mon Jul 23, 2012 3:49 pm
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
Slight optimization for the bubble sort:
Code:while (true){ int swapCount = 0; for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (arr[j] > arr[j+1]) { temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; swapCount++; } } if (swapCount == 0) { break; }}
Mon Jul 23, 2012 4:37 pm
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Re: Bubble Sorting
Thanks Xander!
Could someone by any chance explain the code a little more detailed.
Code:for(int i = 0; i < MAX_NUMBERS; i++){ for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (arr[j] > arr[j+1]) { temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; } }}
How would I implement my own numbers? I'm not too familiar yet with Arrays.
Tue Jul 24, 2012 11:08 am
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
According to your code, you could easily make it into an array.
Code:#define MAX_NUMBERS 8task main(){ int randNum[MAX_NUMBERS]; for (int i = 0; i < MAX_NUMBERS; i++) { randNum[i] = random[1000]; } ....}
As the name suggested, air is lighter than water. So a bubble at the bottom of a water column will rise slowly to the top. The algorithm is basically doing an N-pass to the number column. In each pass, it compares each element with its adjacent neighbor below. If its neighbor below is smaller (lighter weight), it switches position with the neighbor. So in effect, the neighbor below rises. So as each pass progresses, each lighter element that is out of place will rise one position up. So for an N-element column, the worst case is having the lightest element at the bottom of the column, so it takes N passes for the bottom element to rise to the top of the column which is its proper place.
Last edited by MHTS on Tue Jul 24, 2012 11:39 am, edited 2 times in total.
Tue Jul 24, 2012 11:19 am
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Re: Bubble Sorting
Okay, what about if I have a number1 and number2 and my max number is 8.
How would I implement that into this code?
Code:for(int i = 0; i < MAX_NUMBERS; i++){ for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (arr[j] > arr[j+1]) { temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; } }}
Tue Jul 24, 2012 11:23 am
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
Code:#define MAX_NUMBERS 8task main(){ int randNum[MAX_NUMBERS]; // // Initialize the array with random numbers. // for (int i = 0; i < MAX_NUMBERS; i++) { randNum[i] = random[1000]; } // // Now, sort them. // for (int i = 0; i < MAX_NUMBERS; i++) { for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (randNum[j] > randNum[j+1]) { int temp; // // The random number at j is larger than its next neighbor j + 1. // So swap them. // temp = randNum[j]; randNum[j] = randNum[j+1]; randNum[j+1] = temp; } } }}
Tue Jul 24, 2012 11:36 am
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Re: Bubble Sorting
The problem is, I don't know how to implement my data. Can someone explain to me how to implement it? I'm so confused.. haha.
Tue Jul 24, 2012 11:44 am
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
The above algorithm will do MAX_NUMBERS passes no matter what. Even in the best case when the array is already sorted (i.e. all the numbers are in ascending order already), it will still do MAX_NUMBERS passes without any swapping. Therefore, that's where the slight optimization comes in. It basically keeps track of how many swaps we did in each pass. If we did not swap anything in a pass, it is safe to assume that all numbers are in order so we quit.
Code:#define MAX_NUMBERS 8task main(){ int randNum[MAX_NUMBERS]; // // Initialize the array with random numbers. // for (int i = 0; i < MAX_NUMBERS; i++) { randNum[i] = random[1000]; } // // Now, sort them. // while (true) { int swapCount = 0; for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (randNum[j] > randNum[j+1]) { int temp; // // The random number at j is larger than its next neighbor j + 1. // So swap them. // temp = randNum[j]; randNum[j] = randNum[j+1]; randNum[j+1] = temp; swapCount++; } } if (swapCount == 0) { break; } }}
Tue Jul 24, 2012 11:49 am
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Re: Bubble Sorting
Alright, since I'm still pretty confused I will show you what I have already. (Trying to make a program play some tones and then after they have played them they play them again but this time in the correct [smallest to largest] order.)
Code:task main(){ int rand0 = 1046; int rand1 = 659; int rand2 = 523; int rand3 = 784; int rand4 = 698; int rand5 = 988; int rand6 = 587; int rand7 = 880; PlayTone(rand0,10); nxtDisplayTextLine(0,"Tone: %d", rand0); wait1Msec(1000); PlayTone(rand1,10); nxtDisplayTextLine(1,"Tone: %d", rand1); wait1Msec(1000); PlayTone(rand2,10); nxtDisplayTextLine(2,"Tone: %d", rand2); wait1Msec(1000); PlayTone(rand3,10); nxtDisplayTextLine(3,"Tone: %d", rand3); wait1Msec(1000); PlayTone(rand4,10); nxtDisplayTextLine(4,"Tone: %d", rand4); wait1Msec(1000); PlayTone(rand5,10); nxtDisplayTextLine(5,"Tone: %d", rand5); wait1Msec(1000); PlayTone(rand6,10); nxtDisplayTextLine(6,"Tone: %d", rand6); wait1Msec(1000); PlayTone(rand7,10); nxtDisplayTextLine(7,"Tone: %d", rand7); wait1Msec(5000); #define MAX_NUMBERS 8 int randNum[MAX_NUMBERS]; for (int i = 0; i < MAX_NUMBERS; i++) { randNum[i] = random[1000]; } for (int i = 0; i < MAX_NUMBERS; i++) { for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (randNum[j] > randNum[j+1]) { int temp; temp = randNum[j]; randNum[j] = randNum[j+1]; randNum[j+1] = temp; } } }}
How would I get rand0 - rand7 implemented into this bubble algorithm?
Thanks so much!!!
Tue Jul 24, 2012 11:54 am
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
Simon1 wrote:The problem is, I don't know how to implement my data. Can someone explain to me how to implement it? I'm so confused.. haha.
Now I am confused. Doesn't the above code implement what you had on the other post? On the other thread, you had the code to generate 8 random numbers to 8 variables. I changed the 8 variables into one array variable with 8 elements. Then I applied the bubble sorting algorithm to it. At the end, you will have an array with all the numbers in sorting order. The only thing left to do is to add the code to play the tone.
Code:for (int i = 0; i < MAX_NUMBERS; i++){ PlayTone(randNum[i],10); nxtDisplayTextLine(i,"Tone: %d", randNum[i]); wait1Msec(1000);}
Tue Jul 24, 2012 11:56 am
Guru
Joined: Sun Nov 15, 2009 5:46 am
Posts: 1523
Re: Bubble Sorting
Code:#define MAX_NUMBERS 8task main(){ int randNum[MAX_NUMBERS]; // // Initialize the array with random numbers. // randNum[0] = 1046; randNum[1] = 659; randNum[2] = 523; randNum[3] = 784; randNum[4] = 698; randNum[5] = 988; randNum[6] = 587; randNum[7] = 880; // // Play the tone unsorted. // for (int i = 0; i < MAX_NUMBERS; i++) { PlayTone(randNum[i],10); nxtDisplayTextLine(i,"Tone: %d", randNum[i]); wait1Msec(1000); } // // Now, sort them. // while (true) { int swapCount = 0; for(int j = 0; j < (MAX_NUMBERS - 1); j++) { if (randNum[j] > randNum[j+1]) { int temp; // // The random number at j is larger than its next neighbor j + 1. // So swap them. // temp = randNum[j]; randNum[j] = randNum[j+1]; randNum[j+1] = temp; swapCount++; } } if (swapCount == 0) { break; } } // // Play the tones sorted. // for (int i = 0; i < MAX_NUMBERS; i++) { PlayTone(randNum[i],10); nxtDisplayTextLine(i,"Tone: %d", randNum[i]); wait1Msec(1000); }}
Tue Jul 24, 2012 12:03 pm
Rookie
Joined: Thu Jul 19, 2012 11:59 am
Posts: 10
Re: Bubble Sorting
YES! That's how you do it! Thanks so much! You have been so helpful. The program works wonderfully, and I think I finally understand how it works.
Thanks again.
Tue Jul 24, 2012 1:28 pm
Professor
Joined: Sat Aug 31, 2013 9:15 am
Posts: 256
Re: Bubble Sorting
Do you mean a bubble sort with 100 different numbers? If so, that's easy. You just change MAX_NUMBERS to 100, make an array with 100 elements, and put whatever data you want in them.
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Jump to: Select a forum ------------------ ROBOTC Applications ROBOTC for LEGO MINDSTORMS Third-party sensors ROBOTC for CORTEX & PIC ROBOTC for VEX IQ ROBOTC for Arduino Robot Virtual Worlds Multi-Robot Communications Issues and Bugs Competitions & Partners Mini Urban Challenge CS2N Robot Virtual Worlds Competitions VEX Skyrise Competition 2014-2015 VEX Toss Up 2013-2014 FTC Block Party! 2013-2014 Competitions using VEX - BEST, TSA, VEX, and RoboFest! FTC Programming RoboCup Junior and Other ROBOT Competitions Virtual Brick Robotics Discussions General Discussions Project Discussions Off-Topic ROBOTC Forum & ROBOTC.net Suggestions/Feedback ROBOTC Forums Suggestions/Comments ROBOTC.net Suggestions/Comments NXT Programming: Tips for Beginning with ROBOTC VEX Programming: Tips for Beginning with ROBOTC 2013 Robotics Summer Of Learning VEX Toss Up Programming Challenge FTC Ring It Up! Programming Challenge International Forums Spanish Forums ROBOTC for MINDSTORMS ROBOTC for VEX French Forums ROBOTC pour Mindstorms ROBOTC pour IFI VEX Japanese Forums （日本語のフォーラム） German Forums 2015 Spring Carnival Event PLTW (Project Lead The Way) Robotics Merit Badge 2014 Robotics Academy Summer of Learning | 3,938 | 12,528 | {"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-2016-40 | longest | en | 0.762266 |
http://betonvalue.org/sport-betting/maximize-your-wins-by-utilizing-the-right-arbitrage-formula-on-each-bet/ | 1,631,943,612,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00299.warc.gz | 7,480,804 | 9,993 | # Maximize your wins by utilizing the right arbitrage formula on each bet
If you have planned to engage in arbitrage betting to win money irrespective of the outcome of the match then you can certainly increase your wins by applying the best arbitrage formula on each bet. It is important to make correct calculations before placing your bet if you do not wish to end up with paltry wins or suffer a loss of revenue following the match.
Arbitrage betting involves betting on opposing players or teams in a match so that you win enough extra cash for the winning bet whenever a player or team wins while offsetting the loss caused by the losing bet. In this type of betting, you win money regardless of the end result of the bet as you would’ve placed different bets on opposing players and teams with various bookmakers. While one bookmaker might offer excellent odds on the favorite player or team, another bookie might favor the underdog and provide higher odds. The key is to quickly identify such bets and place your bets so that the total amount in winnings is more than both your bets placed together.
However, you will need a formula to decipher the offered odds and calculate the total amount which you might win or lose following the match. You’ll thus have to apply an arbitrage formula that may help you to calculate the precise amount that you will have to invest in the bet and the amount to be won when some of the players or teams wins, or even when the match heads towards a draw. This formula will have to factor in several components such as the possible outcomes of the match, the chances put on each player or team by the particular bookmaker, and the amount that you’re planning to invest in each bet. The answer to your arbitrage calculations should show you the amount you have collectively invested in the bets as well as the different amounts that you’ll win when either of the players or teams wins. In case there are more than two players or teams then this formula must also incorporate additional calculations.
If you are not really good at math and therefore are afraid of making mistakes while using the arbitrage formula, then you need not worry. It is possible to enlist the help of reputed betting experts such as Gert Gambell by going to his websites, gertgambell.net as well as win-every-time.com that will explain the entire concept of arbitrage betting in easy-to-understand terms. You can also lay your hands on helpful suggestions and advice offered by these websites as well as utilize the arbitrage calculator provided free on the site to calculate the exact amount easily that you can win after placing bets on opposing players or teams. Since it is crucial to bet only at trustworthy websites that delivers your winnings, you can surely rely on the bookmakers mentioned by Gert Gambell on his websites.
If you want to win money on each betting session irrespective of which player or team wins the match, then you can go in for arbitrage betting. However, rather than placing each bet excitedly without looking into all available facts or without undertaking the mandatory calculations, you should use the right arbitrage formula to increase your winnings with minimum investment in each bet. | 635 | 3,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.961894 |
https://www.hackmath.net/en/math-problem/56371 | 1,709,074,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00045.warc.gz | 782,700,813 | 10,461 | # Rajendra
Rajendra, a farmer, had two sons and two daughters. He decided to divide his property among his sons and daughters. So he wrote a "WILL" about the distribution of his property. According to his "WILL," he desired to give 3/5 of the property to his sons in equal proportion, 1/3 rd to his daughters in equal proportion, and the rest to a charitable trust. After his death, his "WILL" was opened and read out by the Advocate in the presence of all villagers. Now answer the following questions.
1. How much of the property each son gets? a) 3/5 b) 3/10 c) 1/5 d) 1/2
2. How much of the property each daughter gets? a) 1/3 b) 2/3 c) 1/6 d)
3. How much of the property Charitable Trust gets? b) 2/5 d) 1/15 a) c) 1/9 1/6 b) 1/2
4. How much total money was given to daughters and sons? c) 14/15 d) 3/10
s = 3/10
d = 1/6
c = 8/15
x = 7/15
### Step-by-step explanation:
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Showing 1 comment:
Dr Math
The choices seem fine but I think the correct answer for 3 is 1/15 and 4 would be 14/15. The explanation is not very thorough and uses decimals instead of fractions so I am struggling to follow i
1 year ago 4 Likes
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Need help calculating sum, simplifying, or multiplying fractions? Try our fraction calculator. | 425 | 1,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-10 | latest | en | 0.954067 |
https://pages.stern.nyu.edu/~adamodar/pc/archives/MReg99.html | 1,656,484,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00207.warc.gz | 489,323,495 | 1,933 | June 30, 1999
Regressions of Multiples on Fundamentals: Market Wide
I have run the latest versions of these regressions without an intercept term. Intuitively, my argument would be that firms with zero values for all of the fundamentals (independent variables) should trade at zero multiples. It does mean, however, that the R-squared cannot be directly compared with standard regressions run with an intercept. All of the percentages are entered in decimal format. Thus, a firm with an expected growth rate of 30%, a payout ratio of 10% and a beta of 1.25 can be expected to have a PE of:
PE = 291.27 (.30) + 37.74 (.10) - 21.62 (1.25) = 64.13
Equity Multiples
PE = 117.94 g + 48.79 Payout + 8.40 Beta (R2 = 0.7315) [Details]
PEG = 0.04 Beta + 2.77 Payout - 0.9688 ln(g) (R2 = 0.7327) [Details]
PBV= 17.25 ROE - 1.42 Payout -0.92 Beta + 17.62 g (R2=.7338) [Details]
PS= 4.36 g - 0.18 Payout + 0.09 Beta + 18.64 Margin (R2=0.7138) [Details]
Firm Value Multiples
Value/Sales = 5.13 g + 0.04 (Net Cap Ex/Total Assets) + 14.02 (Operating Margin) - 1.15 Std dev (R2 = 0.7460) [Details]
V/EBITDA= 3.97 + 7.51 (Return on Capital) -0.01 (Tax Rate) + 0.24 g (R2=0.1686) [Details]
To see the more detailed output from the regression, click on 'Details'.
g = Expected growth in earnings over the next 5 years (enter as decimals, i.e., 15% is .15)
PE = Price/ Current EPS: Companies with negative earnings were eliminated from the sample
PBV = Price/ Book Value per share: Companies with negative BV were eliminated
PS = Price/ Sales per share
Payout = DPS/EPS: from most recent year; if negative, it is set to 100%. (enter as decimals)
Std Dev = Standard Deviation in the Stock Price
Beta = Betas based upon 5 years of monthly data
MGN = Net Income / Sales (enter as decimals)
ROE = Net Income / BV of Equity (enter as decimals) | 583 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-27 | latest | en | 0.861483 |
http://academictorrents.com/details/0342ad0bd7ef06eb1500b5e7c8ef398060827c4e | 1,580,051,152,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00457.warc.gz | 5,286,748 | 10,487 | ## [Coursera] Introduction to Logic (Stanford University) (intrologic) Stanford University
intrologic-006 (347 files)
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08_Lesson_10_-_Induction/05_10.5_Structural_Induction_6_min.txt 5.47kB 09_Equality/01_Equality_3_min.html 30.32kB 09_Equality/01_Equality_3_min.mp4 2.00MB 09_Equality/01_Equality_3_min.pdf 1.33MB 09_Equality/01_Equality_3_min.srt 3.10kB 09_Equality/01_Equality_3_min.txt 2.11kB 09_Equality/02_Properties_of_Equality_3_min.mp4 2.48MB 09_Equality/02_Properties_of_Equality_3_min.srt 3.24kB 09_Equality/02_Properties_of_Equality_3_min.txt 2.21kB 09_Equality/03_Substitution_3_min.mp4 2.67MB 09_Equality/03_Substitution_3_min.srt 3.41kB 09_Equality/03_Substitution_3_min.txt 2.35kB 09_Equality/04_Fitch_with_Equality_3_min.mp4 2.48MB 09_Equality/04_Fitch_with_Equality_3_min.srt 3.87kB 09_Equality/04_Fitch_with_Equality_3_min.txt 2.66kB 09_Equality/05_More_examples_2_min.mp4 1.81MB 09_Equality/05_More_examples_2_min.srt 2.34kB 09_Equality/05_More_examples_2_min.txt 1.61kB 10_First_Order_Logic/01_First_Order_Logic_a_very_brief_introduction.html 44.94kB 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Type: Course
Tags:Coursera, intrologic
Bibtex:
```@article{,
title = {[Coursera] Introduction to Logic (Stanford University) (intrologic)},
author = {Stanford University}
}```
Hosted by users: | 5,494 | 13,772 | {"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-2020-05 | longest | en | 0.160069 |
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# Multiple choice questions on options
1. Answer questions 1-4 about insuring a portfolio identical to the S&P 500 worth \$12,500,000 with a 3-month horizon. The risk-free rate is 7%. Three-month t-bills are available at a price of \$98.64 per \$100 face value. The S&P 500 is at 385. Puts with an exercise price of 390 are available at a price of 13. Calls with an exercise price of 390 are available at a price of 13.125. Round off your answers to the nearest integer.
What is the minimum value of the insured portfolio?
1. 12,091,709
2. 16,672,344
3. 12,244,898
4. 12,500,000
5. 13,375,000
2. If the S&P 500 ends up at 401, determine the upside capture.
1. 96%
2. 96.7%
3. 100%
4. 99.3%
5. 94%
3. If the insured portfolio consisted entirely of calls and t-bills, how many would be used?
1. 19,143 calls and 124,176 t-bills
2. 31,397 calls and 122,449 t-bills
3. 933,238 calls and 2,547 t-bills
4. 31,407 calls and 119,977 t-bills
5. 32,468 calls 32,468 t-bills
4. If the insured portfolio were dynamically hedged with stock index futures, how many futures would be used? The call delta is .52 and the continuous risk-free rate is 5.48%. Each futures has a multiplier of 500 and a price of 388.65.
1. 60
2. 64
3. 32
4. none are correct
5. 30
5. If the stock price is currently 36, the exercise price is 35 and the stock ends up at 44, the value of an asset-or-nothing option at expiration is:
1. 35
2. 8
3. 9
4. none are correct
5. 44
6. Which of the following statements is correct about cash-or-nothing options?
1. they must be priced by the binomial model
2. none are correct
3. they have lower upside gains and lower downside losses than ordinary options
4. they are equivalent to short positions in asset-or-nothing options
5. they are subject to no credit risk
7. A contingent-pay option is replicated by which of the following combinations?
1. long an ordinary call and long an equity forward
2. long an ordinary call and long an ordinary put
3. long an ordinary call and short an asset-or-nothing call
4. long an ordinary call and long a risk-free bond
5. long an ordinary call and short a cash-or-nothing call
8. A lookback call option provides the right
1. to insure a stock against loss.
3. none are correct.
4. to buy the stock at its lowest price over the option's life.
5. to change the stock on which the option is written
9. Which of the following is a path-independent option
1. an up-and-out call option
2. a standard European call option
3. a fixed strike Asian call option
4. an American put option
5. none are correct
10. Weather derivative payoffs can be based on each of the following variables except
1. inches of snowfall
2. sunshine
3. total value of insurance claims
4. temperature above a given level
5. temperature below a given level
#### Solution Preview
1. Answer questions 1-4 about insuring a portfolio identical to the S&P 500 worth \$12,500,000 with a 3-month horizon. The risk-free rate is 7%. Three-month t-bills are available at a price of \$98.64 per \$100 face value. The S&P 500 is at 385. Puts with an exercise price of 390 are available at a price of 13. Calls with an exercise price of 390 are available at a price of 13.125. Round off your answers to the nearest integer.
What is the minimum value of the insured portfolio?
1. 12,091,709
2. 16,672,344
3. 12,244,898
4. 12,500,000
5. 13,375,000
N= Number of shares = Number of put = Value of portfolio / (Stock price + Put price)
12,500, 000 / (385+ 13) = 31,407.03
Minimum Value = N x Exercise Price of the Put option= 31,407.03 x \$390
= \$ 12,248,741
This is close to the answer choice 3. 12,244,898 but not exactly equal (there is a 0.03% difference)
This may be due to the fact that the values of call and put options have been interchanged in the problem. If we interchange the values, we get an exact answer.
N= Number of shares = Number of put = Value of portfolio / (Stock price + Put price)
12,500, 000 / (385+ 13.125) = 31,397.17
Minimum Value = N x Exercise Price of the Put option= 31,397.17 x \$390
= \$ 12,244,898
2. If the S&P 500 ends up at 401, determine the upside capture.
1. 96%
2. 96.7%
3. 100%
4. 99.3%
5. 94% | 1,238 | 4,178 | {"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-2016-50 | longest | en | 0.882651 |
https://www.cnblogs.com/water-mi/p/10198928.html | 1,638,893,049,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00044.warc.gz | 780,551,194 | 7,580 | # 【Bzoj4555】【Luogu P4091】求和（NTT）
Bzoj
Luogu
## 题解
$$\sum_{i=0}^n\sum_{j=0}^iS(i,j)2^jj! \\ =\sum_{j=0}^n2^jj!\sum_{i=0}^nS(i,j) \\ \because S(n, m)=\frac1{m!}\sum_{i=0}^m(-1)^i\binom{m}{i}(m-i)^n=\sum_{i=0}^m\frac{(-1)^i}{i!}\frac{(m-i)^n}{(m-i)!} \\ \therefore=\sum_{j=0}^n2^jj!\sum_{i=0}^n\sum_{k=0}^{j}\frac{(-1)^k}{k!}\frac{(j-k)^i}{(j-k)!} \\ =\sum_{j=0}^n2^jj!\sum_{k=0}^j\frac{(-1)^k}{k!}\frac{\sum_{i=0}^n(j-k)^i}{(j-k)!} \\ =\sum_{j=0}^n2^jj!\sum_{k=0}^j\frac{(-1)^k}{k!}\frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!}$$
#include <cstdio>
#include <algorithm>
using std::swap;
const int N = 2.7e5 + 10, Mod = 998244353, g = 3;
int n, m, P, jc[N], pow2[N], invjc[N];
int a[N], b[N], r[N], ret;
int qpow(int a, int b) {
int ret = 1;
while(b) {
if(b & 1) ret = 1ll * ret * a % Mod;
a = 1ll * a * a % Mod, b >>= 1;
} return ret;
}
void NTT (int f[], int opt) {
for(int i = 0; i < n; ++i) if(i < r[i]) swap(f[i], f[r[i]]);
for(int len = 1, nl = 2; len < n; len = nl, nl <<= 1) {
int rot = qpow(g, (Mod - 1) / nl);
if(opt == -1) rot = qpow(rot, Mod - 2);
for(int l = 0; l < n; l += nl) {
int w = 1, r = l + len;
for(int k = l; k < r; ++k, w = 1ll * w * rot % Mod) {
int x = f[k], y = 1ll * f[k + len] * w % Mod;
f[k] = (x + y) % Mod, f[k + len] = (x + Mod - y) % Mod;
}
}
}
}
int main () {
scanf("%d", &n), jc[0] = pow2[0] = invjc[0] = b[0] = 1, b[1] = n + 1;
for(int i = 1; i <= n; ++i)
jc[i] = 1ll * jc[i - 1] * i % Mod, pow2[i] = (pow2[i - 1] << 1) % Mod;
invjc[n] = qpow(jc[n], Mod - 2);
for(int i = n - 1; i; --i) invjc[i] = 1ll * invjc[i + 1] * (i + 1) % Mod;
for(int i = 0; i <= n; ++i) a[i] = 1ll * invjc[i] * (i & 1 ? Mod - 1 : 1) % Mod;
for(int i = 2; i <= n; ++i)
b[i] = 1ll * (qpow(i, n + 1) + Mod - 1) % Mod * qpow(i - 1, Mod - 2) % Mod * invjc[i] % Mod;
for(m = n << 1, n = 1; n <= m; n <<= 1, ++P);
for(int i = 0; i < n; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (P - 1));
NTT(a, 1), NTT(b, 1);
for(int i = 0; i < n; ++i) a[i] = 1ll * a[i] * b[i] % Mod;
NTT(a, -1); int invn = qpow(n, Mod - 2);
for(int i = 0; i <= n; ++i)
ret = (ret + 1ll * pow2[i] * jc[i] % Mod * a[i] % Mod * invn % Mod) % Mod;
printf("%d\n", ret);
return 0;
}
posted @ 2018-12-30 11:47 water_mi 阅读(395) 评论(0编辑 收藏 举报 | 1,156 | 2,209 | {"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} | 3.390625 | 3 | CC-MAIN-2021-49 | latest | en | 0.225313 |
https://becalculator.com/1-6-cm-to-inch-converter.html | 1,685,435,643,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00467.warc.gz | 161,233,498 | 16,668 | # 1.6 cm to inch converter
## FAQs on 1.6 cm to inch
### How many inches are in a cm?
If you wish to convert 1.6 centimeters to an inch-length number, first, you must determine how many inches one centimeter equals.
Here I can give you a direct indication that one centimeter equals 0.3937 inches.
### How do you convert 1 cm into inches?
To convert 1 cm into inches, simply multiply 1cm with the conversion rate of 0.3937.
This will help you to easily calculate 1.6 cm to inches.
Also, 1 cm into inches = 1 cm x 0.3937 = 0.3937 inches, precisely.
Based on this, you can answer the following question very lightly and simply.
• What is one centimeter into inches?
• What is cm to inches conversion formula?
• How many inches equals 1 cm?
• What does 1 cm mean in inches?
### Definition:Centimeter
Centimeter is the International Standard Unit of Length. It is equal to one hundredth of a millimeter. It’s approximately equivalent to 39.37 inches.
### Definition:Inch
Anglo-American units of length are in inches. 12 inches equals one foot, and 36 inches equals one yard. In modern times, one inch equals 2.54 cm.
### How do I convert 1.6 cm to inches?
You have fully understood cm to inches by the above.
The following is the specific algorithm:
Value in inches = value in cm × 0.3937
So, 1.6 cm to inches = 1.6 cm × 0.3937 = 0.062992 inches
You can use this formula to answer related questions:
• What is 1.6 cm in inches?
• How do I convert inches from cm?
• How do you change cm to inches?
• How to turn cm to inches?
• What is 1.6 cm equal to in inches?
cm inch 1.4 cm 0.055118 inch 1.425 cm 0.05610225 inch 1.45 cm 0.0570865 inch 1.475 cm 0.05807075 inch 1.5 cm 0.059055 inch 1.525 cm 0.06003925 inch 1.55 cm 0.0610235 inch 1.575 cm 0.06200775 inch 1.6 cm 0.062992 inch 1.625 cm 0.06397625 inch 1.65 cm 0.0649605 inch 1.675 cm 0.06594475 inch 1.7 cm 0.066929 inch 1.725 cm 0.06791325 inch 1.75 cm 0.0688975 inch 1.775 cm 0.06988175 inch 1.8 cm 0.070866 inch
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 | 673 | 2,168 | {"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-2023-23 | longest | en | 0.80722 |
https://community.glideapps.com/t/math-on-things-with-more-than-one-variable/27587 | 1,623,976,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00308.warc.gz | 175,142,754 | 6,300 | # Math on things with more than one variable
I’m working on transactional math. A consumer gets loyalty points from many merchants, a consumer can spend these points at many merchants. The consumer can choose 5% of their purchase at the merchant as loyalty points to spend anywhere, or 10% of the purchase to come back and spend at the same place. A point represents a dollar of buying power. I need to display the total points available for a consumer to spend at a particular merchant, and separately, the amount of points to spend anywhere.
I currently have the transactions related between the consumer and merchant, this allows me to show a history of transactions for each party. I have a math column that adds up all the transactions for each consumer to show their total wallet, but I get stuck when I want to do this addition and split by merchant which would allow me to show total points a consumer can spend back at a particular merchant.
Can you show some screenshots of what you have so far to get a better visual?
1 Like
yeah, what Jeff said.
You’ve done a decent job of explaining your challenge, but it’s still quite difficult to visualise.
A picture tells a thousand words
2 Likes
Hey guys, I’ve made a Loom, probably the clearest way is to talk and show.
Loom | Free Screen & Video Recording Software
Tricky! I imagine you’ll first need to create an if then else column in the transactions sheet to determine if it’s a 10% in store for each line (else 0). Then, fabricate a user&merchant key value using a template column in the transactions sheet, and create an identical value in your users+merchants sheet. Then, do a multiple relation between the two. afterwards do a Rollup of that if then else column to see total bucks per store.
This will give you a total amount of Goose bucks earned per store but you’ll need to subtract any Goose bucks spent at that store as well.
I imagine if they’re spending goose bucks, then it’s a value within the transaction line. So, you’ll need to do a roll up of that column as well and subtract it from the dollars earned per store.
Of course this is all theory, you’ll have to play with it.
2 Likes
Oh man I’ve been going round and round in circles. you’ve outlined some fantastic ideas but i think there’s a concept I’m not getting, how to do a rollup of multiple entries in a single row, and split them by type…
doing a rollup on the transaction key (made up of the merchant and consumer email addresses) only offers total count of transactions, not per merchant?
A rollup is done on a column, not on a row.
If you’re really trying to do a rollup on a single row, it could be that you’re trying to insert a square peg in a round hole. It might be that you need to re-think how your data is organised.
One thing I would add: I did have a look at your video the other day, but didn’t respond as I didn’t really have time to study it and get my head around it. But one thing that did jump out at me is that having your buyers and merchants mixed in the same table probably makes things a bit more complicated than they otherwise could be. If it was me, I think I’d be splitting those into separate tables. I realise that some users could be in both, so there would be a little bit of duplication, but that shouldn’t create any problems.
I had originally started with them split, but because so much of the info was identical it felt smoother to put them in the same sheet - but now that i’ve got these complex rollups i think you might be right… | 773 | 3,521 | {"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-2021-25 | latest | en | 0.918006 |
https://www.physicsforums.com/threads/vapor-pressure-curve-on-the-pvt-diagram.721071/ | 1,519,299,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00232.warc.gz | 898,900,550 | 24,948 | # Vapor Pressure Curve on the PVT Diagram
1. Nov 5, 2013
### LunaFly
Hi all,
I am wondering about where the vapor pressure curve would be located on the PVT surface of a substance.
I know that as the temperature of a liquid increases, its vapor pressure increases. This continues until the vapor pressure is equal to the atmospheric pressure, at which point the substance begins to boil and goes through a phase transition.
What I am wondering is what kind of path does the vapor curve make on the PVT surface? I am using a closed flask being heated as an example. I would think that it would begin somewhere in the liquid region of the surface. Then as the temperature increases, the volume would decrease slightly as the water evaporates and the vapor pressure increases. Once the liquid reaches boiling point and the liquid-vapor region of the diagram, the volume would immediately increase (as vapor is now included as a part of the volume) then remain constant (as the size of the flask is constant). During this time the pressure would continue to increase until all of the liquid had turned into vapor.
I am very unsure if this is really how the vapor pressure curve would appear on the PVT surface or not. I am also curious as to how it would change if the flask was a different size but with the same amount of liquid initially. Any input would be much appreciated! Thanks.
-Luna
2. Nov 6, 2013
### Staff: Mentor
The vapor pressure curve doesn't stop at the boiling point. It goes all the way up to the critical point. It tells you the temperature and pressure of pure water within a closed container in which there is both water vapor and liquid present at equilibrium with one another. In your experiment, even if there is air in your closed vessel, the water won't boil. The pressure will just keep getting higher as you increase the temperature; from one temperature to the next, some liquid evaporates and forms more vapor at a higher pressure, and the pressure of the air also increases as it gets hotter.
Chet
3. Nov 7, 2013
### LunaFly
If the water in the flask could never boil, then what would be happening on the PVT surface? It would remain in the liquid phase region until the critical point is reached? Also, what would the volume be doing during this process? If the liquid never reaches boiling point it would seem that the volume would continue to decrease as the liquid evaporates, and if we are still in the liquid region of the PVT diagram then the overall volume would continue to decrease as well.
I am also not clear on why the water would not boil in the first place. Is it because it is not exposed to the atmosphere (i.e. vapor pressure could never overcome atmospheric pressure because atmospheric pressure isn't even present?), or is there a different reason?
Last question: if the vapor pressure curve continues past the boiling point, then past that point there is no liquid left and the data is just relating the temperature of the gas with the pressure of the gas, correct? Thanks for the input.
4. Nov 7, 2013
### Staff: Mentor
The easiest way to answer all these questions is to solve a specific problem. Are you willing to participate? Here is the problem: You have a 2 liter container containing 1 liter of liquid water and 1 liter of water vapor in equilibrium with the liquid water. We are going to assume that the density of the liquid water remains constant at 1 gm/cc, and that the vapor above the water behaves like an ideal gas. The container is rigid, and its volume can't change. Hydrostatic variations is pressure are to be assumed negligible. The initial condition of the container contents is 20 C. We are going to do calculations in which we gradually increase the temperature, and determine the pressure, the amount of liquid, and the amount of vapor as a function of the temperature (with the contents at equilibrium at each temperature). Temperatures above 100 C will be included in latter parts of the problem. Part 1 of the problem is to determine at 20 C the pressure in the container (mm Hg), the mass of liquid (gm) water in the container, the mass of water vapor (gm) in the container, and the total mass of water in the container. Also, how does the mass of water within the container change when we increase the temperature?
Chet
5. Nov 8, 2013
### LunaFly
Ok, I think that's a good idea.
So far, I have the initial pressure of the vapor at 17.5 mmHg (found from a vapor pressure table).
The mass of the liquid water is V*ρ= 1000g.
Using the ideal gas law, I found n, the number of moles of vapor, to be 9.58*10-4 moles, from which I found the mass by multiplying n*molar mass of H2O (18 g/mol) to get 0.017 g of water vapor.
From here the total mass is a simple addition resulting in 1000.017g of water present. Because we have a closed system, the mass is conserved throughout the processes.
So far, so good?
6. Nov 8, 2013
### Staff: Mentor
Excellent job!!!! Now let's heat it up to 60 C. When the system equilibrates at 60C, there will no longer be 1000 g of liquid present. Some of the liquid will have evaporated and entered the gas phase. There will also no longer be 1000 cc of liquid present. At 60 C, what is the new pressure of the system? What will be the new (a) grams of liquid water, (b) grams of water vapor, (c) volume of liquid water, and (d) volume of water vapor? To get you started with this, let G be the new number of grams of liquid water. Solve for G.
Chet
7. Nov 10, 2013
### LunaFly
Now the system is at 60C = 333K. The mass of the liquid water can be found by solving:
1.000017 kg = G + n*M, where M is the molar mass of water, and n is the number of moles of vapor.
Using ideal gas law, n can be substituted with PV/(RT), in which V can be substituted with
V= (0.002m3 - G/ρ).
The expression becomes 1.000017 = G + [MP(0.002 - G/ρ)]/(RT).
Solving for G yields:
G= ρ(1.000017RT - 0.002MP)/(ρRT - MP)
Substituting in values of:
ρ= 1000 kg/m3
R= 8.31 J/(mol*K)
M= 0.018 kg/mol
P= 149.4 mmHg= 19,918 Pa
T= 60C= 333K
results in the mass of liquid water G= 0.9989 kg= 999.89 g.
The mass of the vapor (due to conservation of mass) is 1000.017 g - 999.89 g = 0.127 g of vapor.
The volume of the liquid is G/ρ= 0.99989 L.
The volume of the vapor is 2L - 0.999898L =1.00011 L
These answers seem reasonable. The temperature of the system increased, and so did the vapor pressure. Higher vapor pressure means more molecules in the vapor phase. The volume and mass of the liquid has slightly decreased, and the volume and mass of the vapor has increased.
8. Nov 10, 2013
### Staff: Mentor
OK. Nice work.
Next let's move up to 100 C, water's atmospheric boiling point. Does anything unusual happen there? We'll see that the answer is no.
We're going to do things a little differently this time. Instead of G representing the number of grams of liquid water, we are going to let G represent the number of grams of water vapor. Why are we making this change? Not for any particular physical reason. However, we will find that it will reduce the numerical roundoff error in the calculation.
We are also going to do the calculation two different ways, and compare the results:
Method 1: The same way we have been doing it up to now.
Method 2: Not assuming that the density of the liquid water stays constant, and not assuming that the water vapor behaves as an ideal gas. Look up the density of saturated liquid water and saturated water vapor at 100 C. This should be in the steam tables, or in some other set of tables that you can lay your hands on. Use these values in your calculations.
I think that the comparison of results between methods 1 and 2 will be pretty interesting, and method 2 will give you more practice in doing calculations of this type.
Chet
9. Aug 5, 2015
### chutonio
Dear Lunafly and Chestermiller,
first of all thank you for working on this topic. Phase diagrams teaching is always too far from real condition to be efficiently and safely applied to real world situations.
I hope you are already there and could help me with a development of this topic.
I reached to this post because I was trying to figure out how (and why) pressure builds up while heating a liquid in a sealed tank, with or without a second inert component (air). I am a chemist, and this affect very much the way one should properly use an autoclave to perform hydrothermal syntheses.
In such a process one typically fills with water (and reactants) a teflon lined reactor, seals everything up and starts heating up to a desired temperature.
Let´s forget about reactants and focus on the physics though
The teflon tube is a fixed volume reactor, in which are present a) a portion of water and b) a portion of gas (air + water vapour fraction).
It turns out experimentally that the pressure, that builds up upon heating, is linear with temperature. The steepness of this line is a function of the degree of filling of the reactor (ratio liquid/gas portions in the reactor). The higher the filling the higher the steepness and the pressure build-up. In practical cases one should avoid filling the reactor with more than 70%vol of liquid, otherwise a tremendous pressure will quickly build up with consequent rupture of the reactor.
I was wondering why this happens.
Is it because the liquid also expands considerably with temperature and constrains the expanding headspace gas containing air?
[At low filling the volume eaten up by the expanding gas is relatively small and expanding air is only building up pressure as it was alone (PV=nRT). On the contrary, when the initial volume of liquid is big, by expanding it eats up a relatively high portion of the initial gas volume, boosting up the compression of the headspace and multiplying the normal pressure increase produced by temperature (V' is forced to be much smaller)]
If this is true, I think this effect would not happen in presence of a single component system (air and its wapour, without air). Water vapour in equilibrium with liquid water would just be compensated by condensation. I suppose a critical point would be only reached if the final volume of water expanded at volumes higher than the tank.
Do you think this is a reasonable explanation or am I missing details?
Thank you
10. Aug 5, 2015
### Staff: Mentor
The volume of liquid water in the container is going to decrease with increasing temperature, not increase. So the explanation is not reasonable.
But all of this can be precisely calculated. In the case of only water being present in the container, the pressure is going to be equal to the equilibrium vapor pressure of water at the imposed temperature, as long as the liquid water lasts.
In the case of water and air in the container, as long as the total pressure is less than a few of atmospheres, the partial pressure of the water in the gas phase is going to be equal to the equilibrium vapor pressure of water at the imposed temperature, and the partial pressure of the air is going to be determined by the ideal gas law. The calculations to analyze this are pretty simple. The constraint is that the mass of water and air in the container are constant, and the total volume is constant. Part of the liquid water can evaporate to enter the gas phase as the temperature rises, so the volume of the gas phase is going to increase. It is reasonable to assume that none of the air dissolves in the liquid water and that the density of the liquid water is constant at 1000 kg/m^3. The game plan would be to set the initial conditions as: initial temperature (say 20 C) and initial fraction of liquid water. Then calculate what happens at equilibrium when the temperature is raised: new fraction liquid water, new partial pressure of water vapor, new partial pressure of air, new total pressure. Do this at a sequence of temperatures. Then repeat the calculations with a different initial fraction of liquid water.
Chet
11. Aug 5, 2015
### chutonio
I think your assumptions are not correct.
Under the boundaries condition you state the pressure variability vs. degree of filling would be relatively little.
What I´m talking about is the hundreds of bars variability well described by the picture 4.37 of this section
Extract from Schubert´s book Synthesis of Inorganic Materials
I think the author is right in saying that the liquid expands, it does not shrink.
Its density is decreasing upon temperature increase. At the same time the gas density is increasing (more pressure and molecules in the gas phase). Further increase in T follows the same trend up to the critical point, where the density of gas and liquid meet i.e. a single phase is formed.
It seems that the fact that higher T allows for more molecules to escape the liquid phase into the gas is completely overcome by the density change in the liquid. Still according to the author, and to give a practical example, density of water at 300 °C is only 0.75 g/ml.
Therefore if you fill your tank up to the 75% you are already going through the roof at that temperature (tank full, pressure exploding).
Do you feel my statements are correct and that you could reformulate your reply according to this?
Or I´m still missing something in the picture?
Thanks again
12. Aug 5, 2015
### Staff: Mentor
Why speculate? Let's do an actual calculation and see how it plays out. You pick the initial conditions. Then we'll do a sequence of calculations for increasing temperatures using the steam tables, and see what we come up with.
Chet
13. Aug 5, 2015
### Staff: Mentor
OK. So please specify the initial conditions for our calculations. I suggest using a tank volume of 1 m^3.
Chet
14. Sep 7, 2015
### chutonio
Hi Chet,
sorry for taking so long to reply.
I was thinking over it during my free time and I could not get a nice grip on it. The thing is density of liquid water varies substantially with temperature as I mentioned. Not only that. Density is somehow a function of what we want to calculate, that is pressure, since the latter affects the bulk modulus of elasticity (eventhough to a smaller extent apparently).
So basically P= f(T,ρ) but ρ is a function of both temperature and pressure.
That's why I guess practical people prefer to work with tables. Deriving a generally valid equation seems non-trivial to me.
Where do I start to clean up this self-biting snake mess? Do we have to work in a limited temperature range and start considering the bulk modulus invariant in that range?
Thanks
15. Sep 7, 2015
### Staff: Mentor
I still haven't seen a single indication from you as to a starting condition. Please define an initial condition for a constant volume container with a certain fraction of liquid water and fraction water vapor in the head space, and we can work the problem together. By doing modelling like this, all your uncertainties can be removed and we can stop with all the hand waving and speculation.
Using the steam tables will give the most accurate answer. Are you comfortable with using the steam tables? If not, we can model the liquid water using bulk compressibility and coefficient of thermal expansion. This is less accurate than the steam tables, particularly as the temperature increases. A still easier (but less accurate method) is to take the liquid density as constant and use the ideal gas law for the vapor. This method should be pretty good up to temperatures of about 200 C. So, what approach do you want to use? Whatever approach you choose, I will lead you through the calculations. Or, we can do all three approaches and compare the results.
Chet
16. Sep 8, 2015
### chutonio
Alright. I am not confident with steam tables, but it sounds like the perfect occasion to learn.
As to conditions I would do as follows. 1 m3 container is fine.
The container is filled at T1 = 293 K, sealed, and heated up to T2 = 586K (to make it nasty)
The question is, what is the final pressure inside the tank at 586 K varying the following conditions:
- Initial degree of filling of water in the tank a) 10% b) 50% c) 80%
- Composition: i) only water in equilibrium with its vapour (ideal condition); ii) Air in the headspace (realistic condition of sealing a tank, headspace is a mix of air and water vapour in equilibrium )
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Do you agree on the selected points?
I read from steam table the saturated water pressure at 586 K ===> 103 bar , finished
It feels to me that points i,b and i,c are exactly the same , as the equilibrium liquid-vapour is an intensive property (as long as there's enough initial liquid water to inject into the headspace, but this is not the point of the discussion at the moment).
I suppose the tank will blast only if the expansion of the liquid fills completely the tank. At 586 K water density is 0.68
So the final volume of
i,a ) 0.1 / 0.68 = 0.15 m3 SAFE
i,b) 0.5 / 0.68 = 0.73 m3 SAFE
i,c) 0.8 / 0.68 = 1.17 m3 BLAAST (or containment of water expansion if the container is super resistant)
[I approximated density of water to 1 at 293 K ]
let´s consider now cases ii), in presence of air headspace.
How can I use correctly the steam tables here? It is possibly my ignorance here stopping me from going forward. I don´t like to wave speculating, I just don´t know how to proceed.
Can I treat independently the variation of physical properties of air and water upon heating?
Here air pressurizes upon heating, and even more because of water expansion. But water expansion and air pressurization it´s an equilibrium (I think): are there models describing it or can we look up a separate set of steam tables?
Do we approximate density of water as only dependent on temperature?
In this case pressure depens dramatically on degree of filling I think, but I don´t know where to start from do it correctly.
Thank
Cheers
17. Sep 8, 2015
### Staff: Mentor
yes.
I thought you were also interested in getting the volumes and masses of liquid water and water vapor at 586. Were you?
Let's wait on the air in head space cases until we are sure we are really through with the pure water cases.
Chet
18. Sep 8, 2015
### chutonio
Not really.
But should we? Don´t steam tables already account for it?
Was my pressure derivation directly form the steam tables correct?
In case we were interested...
Initial conditions
pressure : 0.023392147667769 [ bar ]
density water : 998.16080927879 [ kg / m3 ]
density steam : 0.017312574945658 [ kg / m3 ]
starting with 0.1 m3 of liquid water in equilibrium with its vapour.
23.4 mbar in V0 = 0.9 m3 of headspace,
n = RT/PV0=8.314E-5*293 / (0.0234*0.9) = 1.156 mol
m = 18 * 1.156 = 20.8 g of water vapour (0.2% of liquid phase mass)
Now I heat up.
At 586 K I would do the same starting from the 100 bar pressure given from the table, but how do I know now the residual headspace volume after some water evaporated/expanded? I should anyway staple on the new density value given by the table shouldn't I?
Chut
19. Sep 8, 2015
### Staff: Mentor
Yes.
If you are interested in the actual pressure at 586K of compressed liquid water at a density of 1.25 gm/cc, you can find it in steam tables that include compressed (subcooled) liquid water. The pressure under these conditions is about 800 bars.
This result does not look correct. (0.9)(0.0173126)=0.0156 kg = 15.6 g. Applying the ideal gas law, I get 15.8 g. It looks like you got your gas law relation inverted.
So the total mass of water is 99.82+.02=99.84 kg
Yes. The total mass is constant and the total volume is constant. What is the density of saturated liquid and saturated vapor at 586? Let m = mass of liquid remaining at 586 and let (99.82-m) = mass of vapor at 586. In terms of m, what is the volume of liquid? In terms of m, what is the volume of vapor? These have to add up to the total volume. What is the value of m?
Chet
20. Sep 8, 2015
### insightful
I start with a 1m3 rigid vessel with 0.8m3 liquid water and 0.2m3 vapor in equilibrium at 20oC giving 798.4kg liquid and 0.00346kg vapor. Then heat to 100oC, I get 0.0961kg liquid water transfers to the vapor space giving 0.167m3 vapor and 0.833m3 liquid. | 4,775 | 20,317 | {"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-2018-09 | longest | en | 0.968796 |
https://wordpandit.com/catx-cat-quantitative-aptitude/?amp=1 | 1,723,422,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00314.warc.gz | 472,537,515 | 10,051 | # How to approach CAT Quantitative Aptitude in the next few months?
How should you approach CAT Maths, or as the purists would call it, CAT Quantitative Ability?
What should be done over the course of next few months and how should you plan your preparation? Before I dig into the advice bit, let’s go through the list of topics that make an appearance in the CAT exam.
List of CAT Quantitative Aptitude Topics:
• Number System: Important keeping in mind that this is where your preparation begins. Ensure that you are clear with the basic concepts of this topic.
• Arithmetic: Though Arithmetic has a vast number of sub-topics, in the last 3 years, only a couple of questions have featured from this topic. Arithmetic remains very important for non-CAT papers.
• Algebra: Considering this has been one of the most dominant topics in CAT in the last three years, it makes a lot of sense to place extra emphasis on this topic.
• Geometry: Another topic which has featured quite a lot in the last 3 years.
• Permutation & Combination/Probability: This is another topic that contributes around 1 to 2 questions to the exam.
• Miscellaneous Topics: This includes Set Theory, Trigonometry (one question appeared from this topic in CAT-2013) and Logarithms
Break-up of the CAT Quantitative Ability and Data Interpretation Section in the last 2 Years:
The following tables provide you with a comprehensive break-up of the exams held in 2015 and 2016.
### CAT 2016 Slot-2 Quantitative Aptitude Break-up
The above break-up points to some very important facts:
1. Algebra and Geometry have been the most important topics in CAT in the last 3 years.
2. Arithmetic has made a comeback in the exam and surprisingly, Number System has not been that visible in the exam., and generally, not a lot of questions are featuring from this area.
Key points keeping the above in mind:
The above chart presents an interesting conundrum: in the last few years, the two topics that give the maximum amount of trouble (algebra and geometry) are the ones that are the most frequent ones on the exam. Keeping this in mind, the following are some important suggestions for you:
1. Cover Concepts for Most topics
The best approach for CAT is one where you cover the basic concepts for most topics. Let’s assume you are not that good with geometry. But you get an absolute sitter question for the topic. Forced to skip such a question in the exam, because you did not study the topic in totality? Since you do not want this happen, ensure you study the basic concepts for every topic.
2. Solve previous year CAT questions
A lot of students are generally averse to the idea of solving previous year CAT questions as they feel they have done most of these questions somewhere or the other. Irrespective of that argument, I highly recommend that you solve all previous year questions, and that too topic-wise. This is the best pool of questions for any topic, and a number of CAT questions are based on logic similar to the ones already asked.
Whenever and wherever you can use answer options to solve questions, kindly do that. This would ensure you save time in the exam and with this saved time, you can place extra focus on your weak areas.
4. Work on Arithmetic
Arithmetic is one of those topics that you essentially covered in school. The basic nature of the topic can be misleading at times and you should be fully prepared for it. For basic levels concepts and problems, you can use books by Abhijit Guha and RS Agarwal for this topic. Remember, this is one topic where you can solve almost any problem with the help of basic logic and reasoning.
5. Use a planned and phase-wise approach
In general, try to adopt a planned approach. What this means is that make a schedule, wherein you list the number of days you would be spending on a particular topic. Ensure you break down the schedule into two stages: concept stage and practice stage. Give sufficient time to each area in a phase-wise manner.
6. Solve timed topic/area tests for a particular topic
It is super important that you solve timed topic tests for a particular topic. This is the only way that you can learn whether you are a master of a topic or not.
7. Don’t get lost in shortcuts and super-fundas
Learning shortcuts and super-fundas sounds very good but do not allow this to distract you. Hardly any CAT question is based on obscure formulas. In fact, most concepts check conceptual clarity. Always remember this.
8. Build a small pocket diary of your learning
Whatever you learn, jot it down in a small diary. These might be simple tricks or concept explanations, just keep noting these down and keep revising them. Essentially, you are building your own flashcards for quantitative aptitude.
Personally speaking, I always used to run two topics side by side: one was my strong area (used to solve advanced problems for this area) and one was my weak/mediocre area (used to study basic concepts and practice exercises for this area). This way, I used to ensure that every week, I am working on my strengths as well as my weaknesses, and at the end of the week, I do not feel that I have missed out in any way.
Why am I not providing a day-wise breakup for individual topics?
Each one of us has his strengths and weaknesses. A common prescriptive time schedule cannot be outlined for all as each individual would have different needs. Keeping that in mind, I have left out the time recommendation.
But I can add something on a personal basis. If I was preparing for CAT for the first time (assuming that my conceptual knowledge is fairly decent), I would make the following schedule for myself:
• Number System: 20 days
• Arithmetic: 30 days (even though this topic does not appear in CAT, I would give it 10 days as it has massive application in DI)
• Algebra: 30 Days
• Geometry: 20 Days
• Permutation & Combination/Probability: 10 Days
• Miscellaneous Topics: 10 Days
This makes a total of 120 days. Remember, as I said above, I would study two topics at the same time, so essentially Arithmetic and Algebra might run for 30 days side by side, but I would ensure that I gain confidence in two topics.
These are some of the tips that you can follow for CAT maths. In case you have any doubts, just leave a comment here and I will try to post a reply for the same. | 1,363 | 6,331 | {"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-2024-33 | latest | en | 0.948541 |
https://socratic.org/questions/how-do-you-simplify-8-sqrt-50c-6 | 1,621,323,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00093.warc.gz | 554,276,086 | 5,736 | # How do you simplify 8\sqrt { 50c ^ { 6} }?
Oct 15, 2017
The answer is $40 \sqrt{2} {c}^{3}$.
#### Explanation:
$\sqrt{n} = {n}^{\frac{1}{2}}$
Using this, rewrite the variable as $8 \cdot {\left(50 {c}^{6}\right)}^{\frac{1}{2}}$
Since ${\left({a}^{n}\right)}^{\frac{1}{m}} = {a}^{\frac{n}{m}}$,
$\therefore 8 \cdot \sqrt{50} {c}^{\frac{6}{2}}$
$= 8 \cdot 5 \sqrt{2} {c}^{3}$
$= 40 \sqrt{2} {c}^{3}$. (Answer). | 194 | 419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 4.40625 | 4 | CC-MAIN-2021-21 | latest | en | 0.576615 |
https://matsci.org/t/is-it-possible-to-control-the-pressure-of-certain-group/36780 | 1,721,813,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00815.warc.gz | 338,807,149 | 5,531 | # Is it possible to control the pressure of certain group?
I am simulating some liquid between two fixed solid surface. I want to control the pressure of the liquid in x and y direction to be 1atm. But the fix npt command control the pressure of whole system. Is it possible to do npt on one group?
No, that is not possible.
To elaborate on my answer. You can change box dimensions only for the whole system. But it is possible to adjust only `x` and/or `y` and have them changed individually or isotropically (= coupled). Details are - as always - in the LAMMPS manual.
Axel, thanks for your reply. I noticed someone mentioned that developing a new compute pressure/group for the fix_modify press command with npt could achieve what I said. I am not sure if it is proper to only control the pressure of one group and change the whole box size while the other groups remain steady. For example, I have two solid surfaces with certain structure and set them as rigid groups. And the liquid between them should reach equilibrium under NPT ensemble. It seems to be a rational system.
First off. You may be using fix npt, but that doesn’t mean that you will be simulating a real NPT ensemble. That would only be true for a homogeneous bulk system and no additional modifications of forces.
That said, measuring pressure in a confined volume is a tricky issue for MD simulations, since the volume itself is often not a very well defined property (after all you are simulating point particles that have no real volume).
For a system confined by walls, things can be even more complex, if the walls are represented by real particles. If you change the box, you need to scale all positions, but then you are changing the walls, too. If you keep them unchanged (check out the dilate keyword of fix npt) then you have problems at the boundaries due to generating unphysical close contacts.
A more physical approach to relaxing liquids between solids would be to move the solid (as a whole) in z-direction rather than changing the box in x- and y-direction. This could be done with fix controller. It is particularly straightforward with wall potentials, but not confined to that.
Please note that a lot of issues like you are describing have their origin in trying to apply an understanding of macroscopic processes and methods to the atomic scale, and that is not always possible. The connection of those two “worlds” is what Statistical Mechanics and Thermodynamics is all about.
Does the controller approach you mentioned means that move the solid in z direction and control the pressure in x and y direction to be expected values? I am trying similar method now. Thanks again! | 557 | 2,680 | {"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-2024-30 | latest | en | 0.943653 |
https://www.storyboardthat.com/storyboards/agarcia35/math-project | 1,561,421,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999779.19/warc/CC-MAIN-20190624231501-20190625013501-00311.warc.gz | 930,487,864 | 16,779 | # MATH project
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• Hey Andrew do you wanna play football?
• Uhh yea sure, lets go after school?
• Hey do you wanna go to the fair I have a shortcut so we can get their quick?
• yea sure.
• Hey instead of going through the school lets go through the football field to the fair.
• Yea that sounds good. But how far is it?
• from were my house is its 6 miles away. Walking down a different neighborhood to the fair is 4 miles. So if we use the Pythagorean theorem we'll find out whats the difference.
• Fair
• My house
• 62+42=C2 36+16=52 if you get the square root of 52 it is approximately 7.2
• Pythagorean Theorem is A*2+B*2=C*2
• A=6 (length of house to fair. B=4 (length from fair going through different neighborhood) C?=(distance using the shortcut)
• 6
• 4
• 7.2
• Were here finally!
• Lets go!
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Video Marketing | 305 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-26 | latest | en | 0.794447 |
https://nztrailrun.com/distance-5k-miles-training-guide/ | 1,696,268,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00285.warc.gz | 470,681,167 | 23,970 | # What is a 5k race?
A 5k race is a popular distance in the running community, typically held as a charity event, fun run, or competitive race. The “5k” in the name refers to the distance of the race, which is approximately 5 kilometers. In miles, a 5k race is equivalent to 3.1 miles. It is a great distance for beginners who are looking to challenge themselves or for seasoned runners who want to improve their speed.
## Converting kilometers to miles
Converting kilometers to miles is essential for understanding the distance of a 5k race. To convert kilometers to miles, you can use the conversion factor of 0.621371. Simply multiply the number of kilometers by 0.621371 to get the equivalent distance in miles. For example, if a race is 5 kilometers long, you would multiply 5 by 0.621371 to get 3.106855, which can be rounded to 3.1 miles.
Here’s a helpful table to convert common 5k distances from kilometers to miles:
KilometersMiles
10.6
31.9
53.1
106.2
2113.1
4226.2
## Understanding the 5k distance
The 5k distance is a manageable challenge for runners of all levels. It provides a great opportunity to set goals, measure progress, and participate in the vibrant running community. Whether you are a beginner or an experienced runner, understanding the 5k distance can help you plan your training effectively.
A 5k race is typically run on a combination of roads, trails, or tracks. It is a great starting point for those who are new to running or looking to improve their fitness. The distance allows runners to push their limits without overwhelming themselves. It also serves as a benchmark for measuring one’s running abilities and improvements over time.
## Essential tips for 5k training
When preparing for a 5k race, it is important to have a well-structured training plan. Here are some essential tips to keep in mind:
1. Start gradually: If you are new to running, begin with shorter distances and gradually increase your mileage over time. This will help prevent injuries and allow your body to adapt to the demands of running.
2. Incorporate cross-training: Include cross-training activities such as swimming, cycling, or strength training to improve your overall fitness and prevent overuse injuries.
3. Consistency is key: Aim to run at least three to four times a week to build endurance and improve your running technique.
4. Mix up your workouts: Incorporate different types of runs into your training plan, including easy runs, tempo runs, interval training, and long runs. This will help you build speed, endurance, and mental toughness.
5. Listen to your body: Pay attention to any signs of fatigue or pain. Rest and recover when needed to avoid burnout and injuries.
6. Stay hydrated and fuel properly: Drink plenty of water and consume a balanced diet with adequate carbohydrates, proteins, and healthy fats to support your training and recovery.
7. Get proper gear: Invest in a good pair of running shoes that fit well and provide adequate support. Comfortable clothing and accessories, such as moisture-wicking socks and a GPS watch, can also enhance your running experience.
## Creating a 5k training plan
To create an effective 5k training plan, consider the following factors:
1. Goal setting: Determine your specific goals for the race, whether it’s completing the distance, improving your time, or achieving a personal best. This will help you tailor your training plan accordingly.
2. Time frame: Decide on the duration of your training plan. It is generally recommended to allow 8-12 weeks to prepare for a 5k race, depending on your current fitness level and experience.
3. Weekly mileage: Gradually increase your weekly mileage, aiming for a total distance that challenges you but remains manageable. Be sure to include rest days to allow your body to recover.
4. Variety of workouts: Include a mix of short runs, long runs, speed or interval training, and cross-training activities to improve your speed, endurance, and overall fitness.
5. Tapering: In the final weeks leading up to the race, reduce your mileage and intensity to allow your body to recover and prepare for peak performance on race day.
By following a well-designed training plan, you can maximize your potential and increase your chances of achieving your 5k race goals.
## Overcoming challenges in a 5k race
Running a 5k race can present various challenges, but with proper preparation and mindset, you can overcome them. Here are some common challenges and strategies to overcome them:
1. Mental toughness: Running a 5k requires mental strength and perseverance. Stay focused, set small achievable goals throughout the race, and maintain a positive mindset to push through any discomfort or fatigue.
2. Pacing: Start the race at a manageable pace, gradually increasing your speed as you build confidence. Avoid starting too fast, as it may lead to exhaustion later on.
3. Hills and terrain: Familiarize yourself with the race course and train on similar terrain to prepare for any hills or uneven surfaces. Incorporate hill workouts into your training to build strength and endurance.
4. Weather conditions: Be prepared for various weather conditions on race day. Train in different weather conditions beforehand to adapt and make necessary adjustments to your race strategy.
5. Hydration and nutrition: Plan your hydration and nutrition strategy during the race. Practice fueling during your training runs to determine what works best for you. Be sure to hydrate adequately before, during, and after the race to maintain optimal performance.
By anticipating and preparing for these challenges, you can stay focused, confident, and resilient throughout your 5k race.
## Reaching your goals: 5k race day tips
Race day is the culmination of your training efforts, and with proper planning, you can maximize your performance and enjoyment. Here are some tips to make the most out of your 5k race day:
1. Arrive early: Give yourself plenty of time to check-in, warm up, and familiarize yourself with the race course.
2. Warm-up: Perform a dynamic warm-up routine to activate your muscles and prepare your body for the race. Include dynamic stretches, light jogging, and strides.
3. Stick to your race plan: Trust the training you have done and stick to your predetermined race strategy. Avoid getting caught up in the excitement of the race and starting too fast.
4. Find your rhythm: Settle into a comfortable pace that allows you to maintain a steady effort throughout the race. Focus on your breathing and running form.
5. Stay positive and motivated: Encourage yourself throughout the race by focusing on positive thoughts and visualizing your success. Use the energy of the crowd and fellow runners to stay motivated.
6. Enjoy the experience: Remember to enjoy the race and celebrate your accomplishments. Take in the atmosphere, appreciate the support from spectators, and savor the moment as you cross the finish line.
7. Reflect and learn: After the race, take time to reflect on your performance. Identify what went well and areas for improvement. Use this feedback to adjust your training for future races.
By following these tips, you can make your 5k race day a memorable and successful experience.
## FAQs
### Q: How long is a 5k in miles?
A: A 5k race is approximately 3.1 miles in length.
### Q: What is the average time to complete a 5k race?
A: The average time to complete a 5k race varies depending on individual fitness levels and experience. However, a typical range for beginners is between 30 to 40 minutes.
### Q: How should I train for a 5k race as a beginner?
A: As a beginner, it is important to start gradually and build your endurance over time. Incorporate a mix of running and walking intervals, gradually increasing your running time until you can run the full 5k distance.
### Q: What gear do I need for a 5k race?
A: The essential gear for a 5k race includes a pair of well-fitting running shoes, comfortable and moisture-wicking clothing, and accessories such as a GPS watch or smartphone app to track your distance and pace.
### Q: How often should I run when training for a 5k race?
A: It is generally recommended to run at least three to four times a week when training for a 5k race. Include a mix of easy runs, speed workouts, and long runs to build endurance and improve your running performance.
### Q: Can I walk during a 5k race?
A: Yes, walking during a 5k race is perfectly acceptable. Many beginners choose to incorporate walk breaks into their race strategy to conserve energy and maintain a steady pace.
### Q: How can I improve my 5k race time?
A: To improve your 5k race time, incorporate speed workouts such as interval training and tempo runs into your training plan. Focus on building your endurance, strength, and speed gradually over time. | 1,885 | 8,842 | {"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-2023-40 | latest | en | 0.945736 |
https://van.physics.illinois.edu/qa/listing.php?id=2368&t=field-multiplications | 1,631,917,649,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00474.warc.gz | 624,330,791 | 7,204 | # Q & A: field multiplications
Q:
Re: Positive/Negative Think the intended question was how positive x negative relates to EM fields since positive + negative seems to correlate to the sum of EM fields. Does multiplication correlate to some interaction between EM fields in a similar way? Multiplication most naturally describes volume and EM fields occupy volume. Maybe additon describes the composit EM field strength at a given point whereas multiplication describes the EM field over an area or volume, but how would that relate to two negatives fields describing a positive field?
- jason (age 28)
ca
A:
It’s important to realize that there is just one value of the E field at a particular point in space and time. The same is true for the magnetic (B) field. So when you want to figure out the field somewhere, you simply add contributions from all the sources.
If you were to multiply fields, the result would not be a field of the same type. It wouldn’t even have the right dimensions. Furthermore, since the fields are all vectors, not numbers, you have to figure out what type of product is relevant for a particular calculation.
Some important field products are E dot E and B dot B. These are NOT other E and B field vectors. They are not ordinary numbers, and their dimensions are those of energy per volume, in the simple cgs unit system. Multiplied by constants, these represent the energy present in the EM fields. Both terms are SQUARES so each is always positive.
Another important field product is E cross B. It also has units of energy density (in cgs units). However it is a vector, not a number. Multiplied by the speed of light (and another numerical constant) it represents the flow of energy in the EM fields. It’s called the Poynting vector.
Some of your remarks (’multiplication most naturally describes volume’) seem to assume that the things being multiplied represent ordinary lengths. That assumption doesn’t apply here.
Mike W.
(published on 10/22/2007) | 419 | 1,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.930006 |
https://www.solutioninn.com/the-rolling-resistance-of-car-depends-on-its-weight | 1,708,694,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00019.warc.gz | 1,023,131,441 | 16,261 | # The rolling resistance of a car depends on its weight as: F = 0.006 mg. How far
## Question:
The rolling resistance of a car depends on its weight as: F = 0.006 mg. How far will a car of 1200 kg roll if the gear is put in neutral when it drives at 90 km/h on a level road without air resistance?
Fantastic news! We've located the answer you've been seeking! | 95 | 361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-10 | latest | en | 0.939199 |
https://www.gams.com/latest/testlib_ml/nlcode5.532 | 1,566,211,162,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-35/segments/1566027314721.74/warc/CC-MAIN-20190819093231-20190819115231-00559.warc.gz | 822,528,162 | 1,503 | \$title Test for NL code bug from Dist 23.6 (NLCODE5,SEQ=532) \$ontext This example turfs up a bug in the NL code generation. The 23.5.2 system (last in the 23.5 series) was OK. The problem exists in 23.6.0 alpha, in 23.6.5, and probably all of 23.6 Contributor: Kent Zhao, June 2011 \$offtext SETS t Time periods / 1*5 / tfirst(t) tlast(t) t2(t); tfirst(t) = ord(t)=1; tlast (t) = ord(t)=card(t); t2 (t) = ord(t) <= card(t)-2; display tfirst,tlast,t2; PARAMETERS alpha /0.33/ theta /0.7/ gamma_n(t) Nbar /0.7921/ Am(t) An(t) beta /0.95/ delta /0.04/ gamma_m /0.04/ K0 /1/ gamma_n(t); gamma_n(t) = -0.000321469; * use integer power instead of real power, ord(t)-1 > t.off Am(t) = 1*power(1+gamma_m ,t.off); An(t) = 1*power(1+gamma_n(t),t.off); Variables h(t) total available time K(t) capital stock ; Equation EQ_h(t) h(t) EQ_K(t) K(t) EQ_KT ; EQ_h(t).. h(t) =E= 1 - (Nbar/An(t))**(1/theta); EQ_K(t2(t)).. K(t+2) =E= (Am(t+1)*(K(t+1)/h(t+1))**alpha)*h(t+1) + (1-delta)*K(t+1) - beta*(1-delta + alpha*Am(t+1)*((K(t+1)/h(t+1))**(alpha-1))) *(Am(t)*((K(t)/h(t))**alpha)*h(t) + (1-delta)*K(t) -K(t+1)); EQ_KT(tlast(t)).. (K(t )-(1-delta)*K(t-1)) /(K(t-1)-(1-delta)*K(t-2)) - (Am(t-1)*((K(t-1)/h(t-1))**alpha)*h(t-1)) /(Am(t-2)*((K(t-2)/h(t-2))**alpha)*h(t-2)) =E= 0; model m /EQ_h, EQ_K, EQ_KT/; K.lo(t) = 0.1; h.lo(t) = 0.01; h.up(t) = 1; K.fx(tfirst) = K0; \$ontext h.l(t) = 0.8; K.l(t) = K0; \$offtext execute_loadpoint 'nlcode5'; option cns = conopt; solve m using CNS; abort\$(m.solvestat <> %solvestat.NormalCompletion%) 'Expected solvestat Normal Completion'; abort\$(m.modelstat <> %modelstat.Solved%) 'Expected modelstat Solved'; abort\$(m.iterusd > 3) 'Expected to start at a solution', m.iterusd; \$ontext save a solution for comparison & restarting execute_unload 'badnl.gdx', h, k; \$offtext | 741 | 1,800 | {"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-2019-35 | latest | en | 0.460863 |
http://www.mfooz.com/bblog/?p=879 | 1,657,214,278,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00156.warc.gz | 89,975,187 | 12,899 | # Player Value
This is a description of the player values that I compute for the ABL. I have bits and pieces of the explanation in various places, but thought it would be good to have everything in one place.
The basis of my calculations is linear weights, which is a method for estimating the number of runs produced by a player using the number of each play outcome for the batter. The particular variety of linear weights I use is called Extrapolated Runs. (See note below.) Each outcome is associated with a run value. A home run is 1.44 runs, a single is 0.5 runs, a strikeout is -0.098 runs. Note that the calculation can be done for both batters and pitchers. Of course, good batters will produce more runs, and good pitchers will allow fewer runs.
Now let’s consider a particular batter’s Triple Play Baseball card. If I can estimate the outcomes of each possible roll (000-999), then I can add up the run values (Extrapolated Runs) for each of those outcomes. If I divide that by 1000, then I have an average run estimate for one plate appearance by that batter. Note that I can do the same thing for a particular pitcher’s card.
To get all those outcomes requires a lot of data and a lot of estimates. The data part involves all the numbers in the main area of the card: this much of a home-run range, this much of an easy-fly range, etc. Then we need to create an average pitcher to face each batter, and vice versa. Then we need to estimate the number of times a batter will face righty and lefty arms, then weight those two values appropriately. We need to calculate the average outcomes of range plays and Deeps! But in the end we can get an estimated runs per plate appearance for every player.
Run values do not take into account the following ratings: injury, jump, steal, speed, hold, catcher throw, outfield throw, and double-play turn.
What’s missing at this point is defense. The Range and Error charts can be used to determine the runs saved by a defender using the same linear weights concept. These adjustments can be applied to a particular player, but if that player is carded at multiple positions, then the combined offensive-defensive run estimate is different for each position.
The goal is to calculate a player “value” that is something like WAR (Wins Above Replacement). Replacement players at different positions have different run-producing capacities. That holds true for both MLB and the ABL. For the ABL I set replacement levels close to the estimated run levels of the best available free agents at each position during the regular season. That level of runs at each position becomes the zero point of my calculated player value. The zero-adjusted run values are then scaled such that only the best players have a player value above 100. Players can have negative player values when free agents with higher run estimates are available at a position.
Defensive ability and position value can lead to very different player values for the same player. For example, an average-hitting catcher may have a significant value behind the plate, but a very low value playing first base, especially if his defense at first is FR/8.
Values are adjusted according to the average number of appearances as a full-time pitcher or position player. For example, on average closers will face fewer batters than a starter, so a closer’s value is adjusted down relative to a starter.
When I total the value of all players on a team, I do not count players with negative player value, because such players are unlikely to get lots of playing time. If a player plays multiple positions, I use the position with the highest value.
Since all free agents are rated, I can use historical ABL draft data to estimate the player value for various points in the draft.
TL;DR: The numbers and ratings on the cards are used to estimate the frequency of outcomes (single, home run, walk, strikeout, etc.). The outcomes are converted into runs using linear weights. The run estimate is adjusted for defense, then adjusted to a scale with zero indicating that an equivalent free-agent player is available, and 100 indicating an arbitrary superstar level.
A note on Extrapolated Runs
Extrapolated Runs (XR) appealed to me, because it is an estimate of absolute runs, unlike Palmer’s Batting Runs, which is measured relative to an average player. XR also includes double plays, which can be estimated from TPB cards.
The big weakness of XR is that it’s formulated to apply over a large span of seasons, specifically 1955-1997. I don’t find any XR coefficients for single, recent seasons.
Jim Furtado wrote an article about the development of XR in 1999.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,001 | 4,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-27 | longest | en | 0.925575 |
https://solvedlib.com/wallis-company-manufactures-only-one-product-and,57015 | 1,696,090,223,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00373.warc.gz | 576,630,398 | 18,904 | # Wallis Company manufactures only one product and uses a standard cost system. The company uses a...
##### How is natural selection related to the concept of niche?
Niches are restricted because neighboring species with similar ecological characteristics prevent expansion into other niches or even narrow down niches. This continual struggle for existence is an important assumption of natural selection....
##### A small cannon is placed on top of a fortification The cannonball leaves the muzzle of the cannon with a speed of 85 m/s at an angle of 259 above the horizontal. Just before the cannonball hits the ground, the vertical component of the velocity is 48 m/s downward; What is the speed of the cannonball just before it hits the ground? Ignore air resistance.91 m/s133 m/s85 m/s68 m/s53 m/s
A small cannon is placed on top of a fortification The cannonball leaves the muzzle of the cannon with a speed of 85 m/s at an angle of 259 above the horizontal. Just before the cannonball hits the ground, the vertical component of the velocity is 48 m/s downward; What is the speed of the cannonball...
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##### C01 pt 0100Question 11A variable is normally distributed with mean 19 and standard deviation 3. Use your graphing calculator to find each of the following areas Write your answers in decimal form. Round to the nearest thousandth as needed:a) Find the area to the left of 18. b) Find the area to the left of 13. c) Find the area to the right of 16. d) Find the area to the right of 27. e) Find the area between 13 and 28,
C01 pt 0100 Question 11 A variable is normally distributed with mean 19 and standard deviation 3. Use your graphing calculator to find each of the following areas Write your answers in decimal form. Round to the nearest thousandth as needed: a) Find the area to the left of 18. b) Find the area to th...
##### Qujod Bullloq Ja48yy e 3eY IIIM pipaud nok op anjaiouIoueyiaIoue1joHOHO_:Mojaq sjoyo3e OM
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# What is the load capacity of a steel bar 0.5 inches x 6 inches x 0.5 inches?
Updated: 12/12/2022
Wiki User
9y ago
capacity of rectangular bar = l X b X h .
here,capacity = 0.5 X 6 X 0.5 .
=1.5 inches ^ 3
Wiki User
9y ago
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Q: What is the load capacity of a steel bar 0.5 inches x 6 inches x 0.5 inches?
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### Could you help me with this equationA piece of bar is 12mm in diameter calculate the stress in the bar?
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### What can steel bar make?
A steel bar can make a: Steel dagger, Steel Medium helm, and a Steel sword if you're a non member. There are various other things you can make if you're a member. | 619 | 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} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.855866 |
http://www.wikihow.com/Work-out-Salary-Increase-Percentage | 1,506,405,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695066.99/warc/CC-MAIN-20170926051558-20170926071558-00121.warc.gz | 609,321,868 | 51,412 | Expert Reviewed
# wikiHow to Work out Salary Increase Percentage
Salary increases can take on many forms. You may have gotten a raise or a promotion, or you may have accepted a new, higher-paying job entirely. Regardless of circumstance, you probably want to know how to calculate your pay raise as a specific percentage of your old rate. Since inflation rates and cost of living statistics are often expressed as percentages as well, calculating an increase as a percentage can help you compare the increase to other forces like inflation. Learning how to work out a salary increase percentage will also help you to compare your compensation against others in your field.
### Part 1 Calculating Your Salary Increase Percentage
1. 1
Subtract your old salary from your new salary. Say you made \$45,000 per year at your old job and that you accepted a new position making \$50,000 per year. This means you would take \$50,000 and subtract \$45,000. \$50,000 - \$45,000 = \$5,000.
• If you receive hourly pay and don’t know your total yearly earnings, you can simply use the old and new hourly rate in place of the salary. For instance, if the raise was from \$14/hour to \$16/hour, then you would use \$16 - \$14 = \$2.
2. 2
Divide the salary difference by your old salary. In order to turn the increase amount into a percentage, you must first calculate it as a decimal. To get the decimal you want, take the difference calculated in Step 1, and divide it by the amount of your old salary.
• Based on the example from Step 1, this would mean taking \$5,000 and dividing it by \$45,000. \$5,000 / \$45,000 = 0.111.
• If you’re calculating the hourly percentage increase, this would still work the same way. From the previous hourly example, take \$2 / \$14 = 0.143
3. 3
Multiply the decimal figure by 100. To turn a number expressed in a decimal format into a percentage, simply multiple it by 100. Using the previous example, you would multiply 0.111 by 100. 0.111 x 100 = 11.1% This means that the new salary of \$50,000 is roughly 111.1% of the previous \$45,000 salary or that you received an 11.1% increase.
• For the hourly rate example, you would still multiply the decimal number by 100. This would make the previous hourly example 0.143 x 100 = 14.3%.
• To check your work, multiply your original salary or hourly rate by the increase percentage. For example, if you multiply \$45,000 x 1.111, the answer is is \$49,995, which rounds up to \$50,000. Likewise, \$14 x 1.143 = \$16.002.
4. 4
Factor in additional benefits if applicable. If you’re comparing a new job at a new company rather than just a salary increase or promotion at your current company, then the salary might be just one part of the overall benefits package to consider. You’ll have a wide range of other items to factor into your increased bottom line. Some of these items include:
• Insurance benefits/premiums - If both jobs offer employer-based insurance coverage, then you’ll have to compare the coverage of the insurance plans. You’ll also need to factor the premium (if applicable) taken out of your paycheck into your decision. Going from paying \$100/month to \$200/month in insurance premiums for the same coverage would negate some of your salary increase, for instance. Also consider the depth of the coverage (do they include dental or vision?), the overall yearly deductibles you may have to pay, etc.
• Bonuses or commissions - Though not part of your standard salary, don’t forget to include bonuses and/or commissions in each calculation. The new salary might offer more each paycheck, but if your current job has the potential for quarterly bonuses, for instance, does the increase still pan out? Keep in mind that this amount may not be consistent because it will probably depend on your performance and/or the company’s performance.
• Retirement plans - Most companies offer a 401k retirement plan that allows you to take pre-taxed wages and put them aside for your retirement. Many companies match up to a certain percentage of an employee’s contribution to his or her personal 401k. If your current company does not match and your new company matches up to 6%, then that’s essentially free additional money toward your retirement to consider.
• Pensions - Jobs that offer pensions for certain numbers of years of continuous service also require consideration. If your current position offers a great pension after twenty-five years, but the new position doesn’t offer any type of pension, then you should consider that as well. A higher yearly salary might be more money immediately, but it’s also worth considering the lifetime earning potential of each. However, keep in mind that pensions are not the norm today. They still exist, but they do not always pay out as expected. In some cases, the funds have been mismanaged and there is little or nothing left for people to retire on.
### Part 2 Determining How Your Increase Relates to Inflation
1. 1
Understand inflation. Inflation is an increase in the prices of goods and services, so it has an effect on your cost of living. High inflation, for instance, often means an increase in food, utility, and gas prices. People tend to buy less during periods of high inflation because these periods mean higher prices.
2. 2
Look up inflation. A wide range of factors determine the inflation of currency. In the U.S., the Department of Labor’s Bureau of Labor Statistics releases a monthly report following and calculating inflation.[1] You can find a month-by-month breakdown of U.S. inflation rates for the past fifteen years here.
3. 3
Subtract the inflation rate from your increase percentage. To determine the effect inflation has on your increased salary, simply subtract the rate of inflation from the increase percentage you calculated in Part 1. For example, the average inflation rate in 2014 was 1.6%.[2] Using the 11.1% increase rate calculated in Part 1, you would determine the effect of inflation on the raise like so: 11.1% - 1.6% = 9.5%. This means that once you take into account the inflated prices of standard goods and services, the increase is only worth an extra 9.5% because the money is worth 1.6% less than the previous year.
• In other words, it took on average 1.6% more money in 2014 to buy the same items as in 2013.
4. 4
Relate the effect of inflation to purchasing power. Purchasing power refers to the comparative costs of goods and services over time. For instance, say you have the salary of \$50,000 per year from Part 1. Now say that inflation stays at a flat 0% the year you get the raise, but rises 1.6% the following year without you receiving another raise. This means it will take you an additional 1.6% to purchase the same basic goods and services. 1.6% of \$50,000 is equal to 0.016 x 50,000 = \$800. Your overall purchasing power based on inflation actually decreased by \$800 over the previous year.
## Community Q&A
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Add New Question
• A new employee is requesting a starting salary of \$40,000 per year. The base salary is \$38,696.82. How do you figure the percentage over base salary?
Michael R. Lewis
Entrepreneur & Retired Financial Advisor
Michael R. Lewis is a retired corporate executive, entrepreneur, and investment advisor in Texas. He has over 40 years of experience in business and finance.
Subtract the base salary (\$38,696.82) from the starting salary (\$40,000) and divide the difference (\$1,303.18) by the starting salary to get the difference in a percentage (3.36%).
• A 5% monthly salary increase resulted in a salary increase of tk6000 per year for an employee. What was his monthly salary before the increase?
Jill Newman, CPA
Certified Public Accountant
Jill Newman is a Certified Public Accountant in Ohio. She received her CPA from the Accountancy Board of Ohio in 1994.
To calculate the salary before the increase, take the ending annual salary of tk6000 and divide it by 1.05, which will give you the annual amount before the 5% increase. Divide the annual amount by 12 to get the original monthly amount.
• How do I calculate my new salary when given the percentage of increase?
wikiHow Contributor
Translate the percentage increase into decimal form. For example, a 5% increase in decimal form would be 0.05. Multiply your current salary by the decimal and the answer is the amount of your salary increase. Add the increase to your current salary to find your new salary.
• I'd like to know how to calculate the 5% of 10,000. What can I do?
wikiHow Contributor
To work out what 5% of 10,000, all you do is, 0.05 x 10 000, or to increase it by 5%, you would do 1.05 x 10,000.
• How do I figure out a 7% increase?
wikiHow Contributor
Multiply your income right now, example given, \$1000 income, by 1.07 and that will be your new income. The example given, 1000*1.07=1070.
• What is the purpose of a pension?
It is a savings account in your name, funded in part by your employer, and designed to pay you money periodically after you retire.
• If a person making \$33,000 a year receives a cost-of-living increase of 2.3 percent, what will the new salary be?
\$33,759.
Unanswered Questions
• My salary has increased 10%, how to find my previous salary?
• How do I increase my salary of \$3000?
• How do I calculate the maximum amount as a percentage of base salary (%) of entire company?
• How do I prepare a salary increment percentage summary? If I prepare salary increment percentage, how is this summarized between the old salary and the new salary?
• How do I calculate my salary?
Show more unanswered questions
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## Tips
• There are several online calculators you can use to quickly determine salary increases in terms of percentage.
• The examples above will work equally well when expressed in other currencies.
• Calculator
## Article Info
Categories: Personal Income
In other languages:
Español: averiguar el porcentaje de un aumento salarial, Português: Verificar a Percentagem do Aumento Salarial, Русский: вычислить процент повышения заработной платы, Italiano: Calcolare la Percentuale di Incremento dello Stipendio, Français: calculer le pourcentage d’augmentation de son salaire, Bahasa Indonesia: Menghitung Persentase Kenaikan Gaji, Deutsch: Den Prozentsatz einer Gehaltserhöhung herausfinden, 日本語: 昇給の割合（％）を計算する
Thanks to all authors for creating a page that has been read 447,711 times.
Did this article help you? | 2,482 | 10,515 | {"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} | 3.984375 | 4 | CC-MAIN-2017-39 | latest | en | 0.947528 |
https://list.indology.info/pipermail/indology/1997-May/008162.html | 1,716,433,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058588.75/warc/CC-MAIN-20240523015422-20240523045422-00065.warc.gz | 328,718,752 | 2,368 | # Nines (Was Re: yuga, VarNa and colour)
Gregory {Greg} Downing downingg at is2.nyu.edu
Mon May 19 13:42:51 UTC 1997
```At 02:18 PM 5/19/97 BST, you (jacob.baltuch at euronet.be (Jacob Baltuch)) wrote:
>>Actually the digits of any number that is a multiple of 9 will have a
>>sum of 9 as well.
>
>Not true: 9 * 11 = 99
>
I think what was meant was that if you start with 99, then 9 + 9 = 18 and 1
+ 8 = 9. I.e., supposedly all numbers that are multiples of nine will "add
down" to nine if you make sums of their component digits till you get to a
single-digit number. But I do not know anything remotely like enough math
theory to say that this is true in all cases, or if so why.
As long as I'm on for a moment, I suspect the important thing
(culture-historically) about nine might be its status as a cube of three and
maybe its status as the proportion between cycles of moon/menstruation and
the average human gestation period. Three is connected to this aspect of
nine in that sexual reproduction involves three (two parents and a child;
"and baby makes three" as the old song goes). That is the case anyway in
Chap. 14 of Joyce's Ulysses, set in a maternity hospital, where threes and
nines figure prominently.
The neoplatonist Plotinus's Enneads (from the Greek for nine) is a treatise
of philosophical and metaphysical analysis set up in nine books. Etc....
Greg Downing/NYU
greg.downing at nyu.edu
or
downingg at is2.nyu.edu
``` | 414 | 1,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.948015 |
https://forum.creative.gimkit.com/t/text-operations-collab-w-getrithekd-wip-dont-click-or-edit-please/8024 | 1,725,977,455,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00497.warc.gz | 239,664,302 | 9,763 | # Text Operations (collab w/ getrithekd, wip, dont click or edit please)
It can be really annoying when you want to perform standard changes to text, like removing characters, splitting strings, and replacing letters. However, gimkit doesn’t offer those capabilities by default, only the ability to get the length of text, combine text, or convert numbers to text. This can be really annoying, and this guide will explain a way to do more with text, using some math.
# The Idea
Think about the device you’re using to read this post. How does it store data? How does it know what words to show you? The answer is numbers. Text and images are stored as numbers on computers, and if a computer can do it then so can we. Let’s make a way to store text as a number. Because computers store text as numbers, and computers can do anything with text, if we store text as a number, then WE can do anything with text.
To do: Boil down to base 72. Add functions. Use decimal places. Rule the world. Autocorrect .
# The System
We need a protocol for storing the text. First, we can encode the alphabet with the numbers 10 to 36. This is done so that there isn’t ever a situation where a number starts with zero. Here is an example of helloworld in this text system.
``````18152222253325282214
``````
If we want to store more characters, like capital letters, punctuation, and spaces, the numbers 37-99 can be used.
And that is how we’ll encode the numbers!
# Manipulating Data
A number isn’t what we want, text is! Below are some formulas I came up with for reading/changing this text.
### Modulo
This won't directly help you with custom text, but it is used in most of the equations here, and while gimkit does not support it by default, it can be recreated like so:
### Backspace
A is the number you want to modify.
mod(a, 100) will extract the last two digits of the number. They are subtracted from the number, then the remaining number is divided by 100 to remove the two zeros created by the subtraction.
### Appending a letter
A is the number you want to modify, L is the letter (in number code) you want to append.
This equation just multiplies the number by 100, then replaces the zeros at the end with whatever number from 1 to 98 you are appending.
### Length of string
A is the number you are getting the length of.
The ceiling ceiling log base 10 returns the number of digits in the base 10 representation of the number. Since each encoded character is two digits long, we divide by two to get the number of encoded characters.
### Nth character of string
Where a is the number you are working with, l1 is the length of the string, and n is the index of the letter you want to get.
This equation basically just clips out the front and end of the number, to leave the original remaining. a divided by 100 to the power of l1 minus n will result in all the digits we don’t want being moved to decimal places in the number. Then, taking the mod of that number and 100 will get the last two numbers in the remaining number, which is the letter code we want.
### A slice of a string
Where a is the string you want to slice, s_1 is the index of the letter you want to start the slice at and s_2 is the index of the letter you want to end the slice at.
This equation can be broken down into two sections. First, we have a/100^l_1-s_2. This clips off everything after the end of the slice by moving it into the decimal place values and rounding down.
Then, by taking the modulo of that and 100^s2-s1+1, we can clip off the front of the number. Now, we have the slice we want!
### Combining strings
Where a and b are both strings, and l_2 is the length of string b.
This equation “pushes” the a string out of the way by multiplying it by the 100^l_2, leaving a bunch of zeros in the lower place values. Then, by adding the b-string, the zeros are filled, and a new string is created.
# Implementing This in GKC
So we have a bunch of equations. You know the theory. Now to get these in GKC.
### Functions
Use the guide below to make functions. I'll explain the theory once more. You'll want to use functions for some of these since the block code can get a bit chunky (Like the slicing a string). I'll post the block code, and then show how to make it into a function, if the block code is above 50 blocks.
Function Guide:
Another Explanation:
There is a property that counts the number of times a function called in the block. Based on that, the block will choose which code to run. Think about it like this: We are using a characteristic of functions in code. After 1 function has run, only the code where only 1 function will run. After this, we either hit the end of the code or a second function called. After the 2nd function, only the code where 2 functions have run will run. Don’t overthink it. If it sounds too simple to be true, then it is simple.
### Modulo
Here is the modulo block code:
In this case, since we are using 100, every time it says the variable base, put the number 100 instead. Also, the round down block is in the dropdown for the round block.
# Conclusion
And that’s that math! It should be noted that due to properties only being able to store numbers up to 99,999,999,999, each property can only store four characters of encoded text. That is the big tradeoff, less text per property for more control over the text. This system would receive a massive buff if the gimkit team set the max for number properties to the 64 bit limit, but for now it’s only four characters per property. If you’re interested in playing around with all the math here, here is a desmos graph.
29 Likes
@getrithekd here
3 Likes
Now this is what I call Advance!
3 Likes
It’s not done lol
2 Likes
We can also use the 5 decimal places, not in a base less than one, but the nth digit represents the n-1th character or something like that.
1 Like
still tho its pretty smart how you came up with this
1 Like
See this is why I don’t call myself the smartest.
3 Likes
It’s standard cs.
2 Likes
i’m not smart enough
also him
(not trying to be mean)
3 Likes
why doesn’t this have the “wiki” tag?
1 Like
I just made it a wiki so that me and getrithekid could both work on it, its not really a proper wiki
3 Likes
I’m only able to get 3 decimal places of accuracy.
So 6 characters
5 Likes
I gotta go to another class, will try to rework the math during lunch in an hour and a half
3 Likes
nice math this could be useful to coders
2 Likes
My name in the title is spelled wrong. Its getrithekd, without an i.
1 Like
Fixed, also am halfway through reworking the math
3 Likes
Wait are we doing 99 characters or 26 characters?
1 Like
Isn’t it just a translation, so like subtract 3 from n each time we see it?
1 Like
What do you mean 99 characters?
The current system supports 99 characcters, the new one is just the same but with numbers 10-99 (so no numbers that start with 0)
3 Likes
Ok I’ll add a section on how to get more characters per property by reducing the base.
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https://www.webassign.net/features/textbooks/scalcet6/details.html?l=publisher | 1,632,051,034,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00147.warc.gz | 1,095,891,988 | 19,929 | # Calculus: Early Transcendentals 6th edition
James Stewart
Publisher: Cengage Learning
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• Chapter 0: Diagnostic Tests
• Diagnostic Tests (4)
• Chapter 1: Functions and Models
• 1.1 Four Ways to Represent a Function (56)
• 1.2 Mathematical Models: A Catalog of Essential Functions (19)
• 1.3 New Functions from Old Functions (48)
• 1.4 Graphing Calculators and Computers (15)
• 1.5 Exponential Functions (21)
• 1.6 Inverse Functions and Logarithms (59)
• Chapter Review
• True -False
• True -False (13)
• Chapter 2: Limits and Derivatives
• 2.1 The Tangent and Velocity Problems (13)
• 2.2 The Limit of a Function (37)
• 2.3 Calculating Limits Using the Limit Laws (52)
• 2.4 The Precise Definition of a Limit (15)
• 2.5 Continuity (28)
• 2.6 Limits at Infinity: Horizontal Asymptotes (46)
• 2.7 Derivatives and Rates of Change (50)
• 2.8 The Derivative as a Function (46)
• Chapter Review
• True -False
• True -False (20)
• Chapter 3: Differentiation Rules
• 3.1 Derivatives of Polynomials and Exponential Functions (67)
• 3.2 The Product and Quotient Rules (55)
• 3.3 Derivatives of Trigonometric Functions (48)
• 3.4 The Chain Rule (79)
• 3.5 Implicit Differentiation (53)
• 3.6 Derivatives of Logarithmic Functions (58)
• 3.7 Rates of Change in the Natural and Social Sciences (29)
• 3.8 Exponential Growth and Decay (24)
• 3.9 Related Rates (51)
• 3.10 Linear Approximation and Differentials (41)
• 3.11 Hyperbolic Functions (39)
• Chapter Review
• True -False
• True -False (12)
• Chapter 4: Applications of Differentiation
• 4.1 Maximum and Minimum Values (66)
• 4.2 The Mean Value Theorem (15)
• 4.3 How Derivatives Affect the Shape of a Graph (66)
• 4.4 Indeterminate Forms and L'Hospital's Rule (75)
• 4.5 Summary of Curve Sketching (51)
• 4.6 Graphing with Calculus and Calculators (18)
• 4.7 Optimization Problems (69)
• 4.8 Newton's Method (40)
• 4.9 Antiderivatives (65)
• Chapter Review
• True -False
• True -False (20)
• Chapter 5: Integrals
• 5.1 Areas and Distances (23)
• 5.2 The Definite Integral (66)
• 5.3 The Fundamental Theorem of Calculus (71)
• 5.4 Indefinite Integrals and the Net Change Theorem (59)
• 5.5 The Substitution Rule (89)
• Chapter Review
• True -False
• True -False (15)
• Chapter 6: Applications of Integration
• 6.1 Areas between Curves (51)
• 6.2 Volumes (60)
• 6.3 Volumes by Cylindrical Shells (43)
• 6.4 Work (29)
• 6.5 Average Value of a Function (23)
• Chapter Review
• True -False
• True -False
• Chapter 7: Techniques of Integration
• 7.1 Integration by Parts (69)
• 7.2 Trigonometric Integrals (69)
• 7.3 Trigonometric Substitution (43)
• 7.4 Integration of Rational Functions by Partial Fractions (62)
• 7.5 Strategy for Integration (70)
• 7.6 Integration Using Tables and Computer Algebra Systems (46)
• 7.7 Approximate Integration (46)
• 7.8 Improper Integrals (76)
• Chapter Review
• True -False
• True -False (14)
• Chapter 8: Further Applications of Integration
• 8.1 Arc Length (33)
• 8.2 Area of a Surface of Revolution (28)
• 8.3 Applications to Physics and Engineering (40)
• 8.4 Applications to Economics and Biology (18)
• 8.5 Probability (18)
• Chapter Review
• True -False
• True -False
• Chapter 9: Differential Equations
• 9.1 Modeling with Differential Equations (14)
• 9.2 Direction Fields and Euler's Method (25)
• 9.3 Separable Equations (42)
• 9.4 Models for Population Growth (20)
• 9.5 Linear Equations (28)
• 9.6 Predator-Prey Systems (8)
• Chapter Review
• True -False
• True -False (7)
• Chapter 10: Parametric Equations and Polar Coordinates
• 10.1 Curves Defined by Parametric Equations (38)
• 10.2 Calculus with Parametric Curves (59)
• 10.3 Polar Coordinates (66)
• 10.4 Areas and Lengths in Polar Coordinates (43)
• 10.5 Conic Sections (48)
• 10.6 Conic Sections in Polar Coordinates (25)
• Chapter Review
• True -False
• True -False (10)
• Chapter 11: Infinite Sequences and Series
• 11.1 Sequences (68)
• 11.2 Series (66)
• 11.3 The Integral Test and Estimates of Sums (37)
• 11.4 The Comparison Tests (38)
• 11.5 Alternating Series (32)
• 11.6 Absolute Convergence and the Ratio and Root Tests (35)
• 11.7 Strategy for Testing Series (33)
• 11.8 Power Series (38)
• 11.9 Representations of Functions as Power Series (35)
• 11.10 Taylor and MacLaurin Series (64)
• 11.11 Applications of Taylor Polynomials (33)
• Chapter Review
• True -False
• True -False (20)
• Chapter 12: Vectors and the Geometry of Space
• 12.1 Three-Dimensional Coordinate Systems (32)
• 12.2 Vectors (38)
• 12.3 The Dot Product (48)
• 12.4 The Cross Product (41)
• 12.5 Equations of Lines and Planes (61)
• 12.6 Cylinders and Quadric Surfaces (40)
• Chapter Review
• True -False
• True -False (18)
• Chapter 13: Vector Functions
• 13.1 Vector Functions and Space Curves (27)
• 13.2 Derivatives and Integrals of Vector Functions (43)
• 13.3 Arc Length and Curvature (48)
• 13.4 Motion in Space: Velocity and Acceleration (36)
• Chapter Review
• True -False
• True -False (12)
• Chapter 14: Partial Derivatives
• 14.1 Functions of Several Variables (56)
• 14.2 Limits and Continuity (39)
• 14.3 Partial Derivatives (71)
• 14.4 Tangent Planes and Linear Approximations (38)
• 14.5 The Chain Rule (47)
• 14.6 Directional Derivatives and the Gradient Vector (47)
• 14.7 Maximum and Minimum Values (45)
• 14.8 Lagrange Multipliers (38)
• Chapter Review
• True -False
• True -False (12)
• Chapter 15: Multiple Integrals
• 15.1 Double Integrals over Rectangles (16)
• 15.2 Iterated Integrals (33)
• 15.3 Double Integrals over General Regions (49)
• 15.4 Double Integrals in Polar Coordinates (32)
• 15.5 Applications of Double Integrals (30)
• 15.6 Triple Integrals (42)
• 15.7 Triple Integrals in Cylindrical Coordinates (24)
• 15.8 Triple Integrals in Spherical Coordinates (40)
• 15.9 Change of Variables in Multiple Integrals (23)
• Chapter Review
• True -False
• True -False (8)
• Chapter 16: Vector Calculus
• 16.1 Vector Fields (27)
• 16.2 Line Integrals (40)
• 16.3 The Fndamental Theorem for Line Integrals (30)
• 16.4 Green's Theorem (26)
• 16.5 Curl and Divergence (31)
• 16.6 Parametric Surfaces and Their Areas (51)
• 16.7 Surface Integrals (39)
• 16.8 Stokes' Theorem (16)
• 16.9 The Divergence Theorem (27)
• 16.10 Summary
• Chapter Review
• True -False
• True -False (8)
• Chapter 17: Second-Order Differential Equations
• 17.1 Second-Order Linear Equations (27)
• 17.2 Nonhomogeneous Linear Equations (23)
• 17.3 Applications of Second-Order Differential Equations (15)
• 17.4 Series Solutions (10)
• Chapter Review
• True -False
• True -False (4)
• Chapter A: Appendix
• A.A: Numbers, Inequalities, and Absolute Values (69)
• A.B: Coordinates Geometry and Lines (61)
• A.C: Graphs of Second-Degree Equations (40)
• A.D: Trigonometry (82)
• A.E: Sigma Notation (49)
• A.F: Proofs of Theorems
• A.G: The Logarithm Defines as an Integral (10)
• A.H: Complex Numbers (50)
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##### Question Group Key
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TF - True / False Quiz
Tut - Tutorial Question
XP - Extra Problem
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GRAY questions are under development
Group Quantity Questions
Chapter 0: Diagnostic Tests
0.Diagnostic 4 Algebra AnalyticGeometry Functions Trigonometry
Chapter A: Appendix
A.A 69 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 062 063 065 066 067 068 069 070
A.B 61 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 055 056 057 058 059 060 061 062
A.C 40 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040
A.D 82 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 042 043 044 045 046 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 062 063 064 065 066 067 068 069 070 071 072 073 074 075 076 077 078 079 080 081 082 084 088 089
A.E 49 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 040 041 042 043 044 045 046 047 048 049 050
A.G 10 001 002 003 004 005 006 007 008 009 010
A.H 50 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050ab
Chapter 1: Functions and Models
1.TF 13 001 002 003 004 005 006 007 008 009 010 011 012 013
1.1 56 AE.05 AE.07 AE.11 001 002 005 006 007 008 010 021 022 023 024.MI 024.MI.SA 025 026 027 028 029 030 031 032 033 034 035 036 037.MI 037.MI.SA 038 039 040 041 042 043 044 045 046 047 048 051 052 053 054 055 056 057 061 062 063 065 066 067 068 069 070
1.2 19 AE.01 AE.02 AE.03 001 002 008 009.MI 009.MI.SA 010 011.MI 011.MI.SA 013 014 015 016 017 018 022 023
1.3 48 AE.01 AE.05 AE.07 002 003 006.MI 006.MI.SA 007 026 029 030 031 032 033 034 035 036 037 038 039.MI 039.MI.SA 040 042 043 044 045 046 047 048 049 050 051 052 053 054.MI 054.MI.SA 055 056 057 058 059 060 061 062 063 064 065 066
1.4 15 AE.04 AE.05 AE.08 001 002 017 018 019 020 021 022 023 024 025 026
1.5 21 AE.02 AE.03 001 002 013 014 015 016.MI 016.MI.SA 017 018 020.MI 020.MI.SA 022 023 024 025 026.MI 026.MI.SA 027 028
1.6 59 AE.03 AE.04 AE.09 003 004 005 006 007 008 009.MI 009.MI.SA 010 011 012 013 014 015 016 017 018 021 022.MI 022.MI.SA 023 024 025 026 033 034 035 036 037 038 039 040 043 044 047 048 049 050 051 052 053 054 057.MI 057.MI.SA 058 059.MI 059.MI.SA 060 061 062 063 064 066 067 068 071
Chapter 2: Limits and Derivatives
2.TF 20 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020
2.1 13 AE.01 AE.02 AE.03 001.MI 001.MI.SA 002 003 004 005 006.MI 006.MI.SA 007 008
2.2 37 AE.03 AE.04 AE.06 AE.07 004 005.MI 005.MI.SA 006 007 008 009 010.MI 010.MI.SA 011 017 018 019 020 021 022.MI 022.MI.SA 023 024 025 026.MI 026.MI.SA 027 028 029 030 031 032 033 034 035 038 040
2.3 52 AE.05 AE.08 AE.11 001 002 003 004 005 006 007 008 009 011 012 013 014.MI 014.MI.SA 015 016 017 018 019 020.MI 020.MI.SA 021 022 023 024 025 026 027 028 029 030 035 036 039 040.MI 040.MI.SA 041 042 043 044 045 046 047 048 049 055 056 060 061
2.4 15 AE.02 AE.03 AE.05 001 002 003 004 007 008 009 011 012 013 014 041
2.5 28 AE.02 AE.09 AE.10 002 003.MI 003.MI.SA 004 004.alt 008 009 029 030 031.MI 031.MI.SA 032 033 034 037 038 039 040 041.MI 041.MI.SA 042 043 059 060 061
2.6 46 AE.03 AE.06 AE.11 003 004 011 012 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030.MI 030.MI.SA 031 032 033 034 035 036 037 038 039 040.MI 040.MI.SA 041 042 043 044 047 048 055 057 058 059 066
2.7 50 AE.01 AE.03 AE.04 AE.05 AE.06 AE.07 003 004 005 006 007.MI 007.MI.SA 008 009 010 011 013.MI 013.MI.SA 014 015 016 018 021 022 023 024 025 026 027.MI 027.MI.SA 028 029 030 031 032 033 034 035 036 037 038 040 041 042 043 048 049 050 051 052
2.8 46 AE.01 AE.02 AE.05 001 002 003.MI 003.MI.SA 004 005 006 007 008 009 010 011 018 019.MI 019.MI.SA 020 021 022 023 024 025 026 027.MI 027.MI.SA 028 029 031 032 035 036 037 038 041 042 043 044 045 046 049.MI 049.MI.SA 050 052 057
Chapter 3: Differentiation Rules
3.TF 12 001 002 003 004 005 006 007 008 009 010 011 012
3.1 67 AE.03 AE.06 AE.08 001 01.XP 02.XP 003 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023.MI 023.MI.SA 024 025 026 027 028 029 030 031 032 034 035 036 037 038.MI 038.MI.SA 039 040 041 042 045 046 047 048 049 050 051 052.MI 052.MI.SA 054 056 057 058 059 060 061 062 063 065 066 067 071 073 074 076 078 079
3.2 55 AE.04 AE.05 001 01.XP 002 02.XP 004 005 006 007 008 009 010 011.MI 011.MI.SA 012 013 014 015 016 017 018 020 021 022 024 025 026 027 028 029 030 031 032.MI 032.MI.SA 033 034 035 036 037 038 039 040 041 043 044 045.MI 045.MI.SA 046 047 048 050 051 052 053 054
3.3 48 AE.01 AE.03 AE.06 001 002 003 004.MI 004.MI.SA 005 006 007 008 009.MI 009.MI.SA 010 011 012 013 014 015 016 021 022 023 024 025 026 027 028 030 032 034 035.MI 035.MI.SA 036 038 039 040 041 042 043 044 045 046 047 048 050 051
3.4 79 AE.02 AE.04 AE.08 001 01.XP 002 02.XP 003 004 005 006 009 010 011 012 013 014 015 016 017 018.MI 018.MI.SA 019 020 021 023 024 025 026 027 028 029 030 031 032 034 035 036 037 038 039 040 041 042 044 045 046 047 048 049 050 051 052 053 054 055 056 057 059 060 061 062 063 064 065 069 071 072 074 075 076 077 078 079 080 081.MI 081.MI.SA 085 092
3.5 53 AE.01 AE.02 AE.05 001 01.XP 002 003 03.XP 004 04.XP 005.MI 005.MI.SA 05.XP 006 007 008.MI 008.MI.SA 009 011 012 013 014 015 016 017 018 019 020.MI 020.MI.SA 021 023 024 025 026 027 028 030 031 032 039 045 046 047.MI 047.MI.SA 049 050 051 052 053 054 060 066 069
3.6 58 AE.01 AE.06 AE.08 01.XP 002 02.XP 003.MI 003.MI.SA 03.XP 004 04.XP 005 05.XP 006 06.XP 007 07.XP 008 08.XP 009 011 012 013 014 016 017 018 019 020 021 022 023 024 025 026 027 028 029 031 032 033 034.MI 034.MI.SA 035 036 037 038 039 041 042.MI 042.MI.SA 043 044 045 046 047 048 049
3.7 29 AE.01 AE.08 001 01.XP 002 02.XP 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018.MI 018.MI.SA 019 020 023 024 030 033
3.8 24 AE.01 AE.02 001.MI 001.MI.SA 002 003 004 005 006 008 009.MI 009.MI.SA 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019 020
3.9 51 AE.01 AE.04 AE.05 003 004.MI 004.MI.SA 005.MI 005.MI.SA 006 007 008.MI 008.MI.SA 010 011 012 013 014 015.MI 015.MI.SA 016.MI 016.MI.SA 017 018 020.MI 020.MI.SA 021 022 023 024.MI 024.MI.SA 025 026 027 028 029 030 031.MI 031.MI.SA 032 033 034 035 036 037 038 039 040 041 042 043 044
3.10 41 AE.01 AE.04 001.MI 001.MI.SA 01.XP 002 003 004 005 006.MI 006.MI.SA 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024.MI 024.MI.SA 025 026 028 033 034 035 036 038 040 043 044
3.11 39 AE.01 AE.04 AE.05 001 01.XP 002 02.XP 003 03.XP 004 005 006 008 018 020 021 023 024 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 051 054
Chapter 4: Applications of Differentiation
4.TF 20 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020
4.1 66 AE.04 AE.07 AE.08 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031.MI 031.MI.SA 032 033.MI 033.MI.SA 034 035.MI 035.MI.SA 036 037.MI 037.MI.SA 038 039 040 041 042 043 044 047 048 049.MI 049.MI.SA 050 051 052 053 054.MI 054.MI.SA 055.MI 055.MI.SA 056 057.MI 057.MI.SA 058 059 060 061.MI 061.MI.SA 062 063 064 065 066 067 068 071 073
4.2 15 AE.03 AE.04 AE.05 001 002 003 004 007 009 010 011 012 013 014 023 024
4.3 66 AE.01 AE.02 AE.05 AE.06 001 002 005 006 007.MI 007.MI.SA 008 009.MI 009.MI.SA 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019 020 021 022 023.MI 023.MI.SA 031 032 034 035 036.MI 036.MI.SA 037 038 039 040.MI 040.MI.SA 041 042 043 044 045.MI 045.MI.SA 046 047 048 049 050 051 052 053.MI 053.MI.SA 055 056.MI 056.MI.SA 061 062 065 066 067.MI 067.MI.SA 068 069 081
4.4 75 AE.01 AE.03 AE.06 001 002 003 004 005.MI 005.MI.SA 006 007 008 010.MI 010.MI.SA 011 012.MI 012.MI.SA 015.MI 015.MI.SA 016 018 019 020 021 022 023 024 025 026 027 028.MI 028.MI.SA 029 031 032 033 034 035 036 037.MI 037.MI.SA 039 040 041.MI 041.MI.SA 042 043 044 046 047 048 049 050 051.MI 051.MI.SA 052 053.MI 053.MI.SA 054 055.MI 055.MI.SA 056 057 058 059 060 061 063 064 065 066 072 078 079 083
4.5 51 AE.01 AE.03 AE.06 001 002 004 005 007 009 010 011 012 013 015 016 018 019 020 022 023 024 025 026 027 028 029 032 034 035 036 037 039 040 042 043 045 046 047 049 051 052 056 057 058 059 060 061 062 063 065 066
4.6 18 AE.02 AE.03 AE.05 009.MI 009.MI.SA 010.MI 010.MI.SA 011 012 013 014 017 018 019 020 021 023 024
4.7 69 AE.02 AE.03 AE.05 AE.06 001 002.MI 002.MI.SA 003 004 005.MI 005.MI.SA 006 007.MI 007.MI.SA 008 010 011 012 013.MI 013.MI.SA 014 017 018 019.MI 019.MI.SA 020 022 024 026 028 030.MI 030.MI.SA 031 032 033 034 036 038 040 042 043 044.MI 044.MI.SA 045 046 047.MI 047.MI.SA 048 049 050.MI 050.MI.SA 052 053 054 055 056 057 058.MI 058.MI.SA 060 061 062 065 066 068 069 070 073 074
4.8 40 AE.01 AE.02 AE.03 003 005.MI 005.MI.SA 006 007.MI 007.MI.SA 008 009 010 011.MI 011.MI.SA 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026.MI 026.MI.SA 027 028 032 035 036.MI 036.MI.SA 037 038 039.MI 039.MI.SA
4.9 65 AE.04 AE.05 AE.06 001 002 003 004.MI 004.MI.SA 005.MI 005.MI.SA 006 007 008 010.MI 010.MI.SA 011 012.MI 012.MI.SA 013 014.MI 014.MI.SA 015 018 020 023 024 025 026.MI 026.MI.SA 031.MI 031.MI.SA 033 036 037 038 039 040.MI 040.MI.SA 041 042 043 044 044.alt 045 046 048 050 058 059 060 062 063 066 067.MI 067.MI.SA 068 069.MI 069.MI.SA 070 071 072 073 074 075 076 077
Chapter 5: Integrals
5.R 066
5.TF 15 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015
5.1 23 AE.01 AE.02 AE.04 001 002 003.MI 003.MI.SA 004 005 011.MI 011.MI.SA 012 013 014.MI 014.MI.SA 015 016 017 018 019 020 021 022
5.2 66 AE.04 AE.05 AE.07 001.MI 001.MI.SA 002.MI 002.MI.SA 003.MI 003.MI.SA 004 005 006 007 008 009.MI 009.MI.SA 010 011 012 017.MI 017.MI.SA 018 019 020 021.MI 021.MI.SA 022 023 024 025 026 029 030 033 034.MI 034.MI.SA 035.MI 035.MI.SA 036 037 038.MI 038.MI.SA 039 040 041 042 043 044 045.MI 045.MI.SA 046 047.MI 047.MI.SA 048.MI 048.MI.SA 049 050 051 055 056 057 058 059 060 069 070
5.3 71 AE.01 AE.02 AE.05 002 002.alt 003 004 005 007.MI 007.MI.SA 008 009 010 011 012.MI 012.MI.SA 013.MI 013.MI.SA 014 015 016 017 018 019 020 021 022.MI 022.MI.SA 023.MI 023.MI.SA 024 025 026 027 028 029.MI 029.MI.SA 030 031 032.MI 032.MI.SA 033.MI 033.MI.SA 034 035 036 037.MI 037.MI.SA 038.MI 038.MI.SA 039 040 041 042 047 048 049 050 053 054 055 056 057 058 059 060 063 064 065 066 072 074
5.4 59 AE.02 AE.04 AE.06 005 006 007 008 009 010.MI 010.MI.SA 011 012 013 014 015 016 017 018 021 022 023 024 025 026 027 028 029 030 031.MI 031.MI.SA 032 033 034 035 036 037 038.MI 038.MI.SA 039 040 041.MI 041.MI.SA 042 043 044 047 048 050 056 057 058 059 060.MI 060.MI.SA 061 062 063 064 065 066
5.5 89 AE.03 AE.06 AE.09 001 002.MI 002.MI.SA 003.MI 003.MI.SA 004 005 006.MI 006.MI.SA 007.MI 007.MI.SA 008.MI 008.MI.SA 009 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019.MI 019.MI.SA 020 021 022 023 024 025.MI 025.MI.SA 026 027 028.MI 028.MI.SA 029 030 031.MI 031.MI.SA 032 033 034.MI 034.MI.SA 035 036 037 038 039 040 042.MI 042.MI.SA 043 044 045 046 048 050 051 052 053.MI 053.MI.SA 054.MI 054.MI.SA 055 056 058 059 060 061 062 063 064 065 066 067 068 069 076 077 078 079 080 081 082
Chapter 6: Applications of Integration
6.1 51 AE.02 AE.05 AE.06 001.MI 001.MI.SA 004 005 006 006.alt 007 008 009 009.alt 010 011 012 013 014 014.alt 015 015.alt 016 017 018 019 020 021.MI 021.MI.SA 022 022.alt 023 024 024.alt 025 026 027 027.alt 028 029 030 032 032.alt 040 042 043 044 045 048 049 051 052
6.2 60 AE.02 AE.04 AE.08 001 002.MI 002.MI.SA 003 004 005 006.MI 006.MI.SA 007.MI 007.MI.SA 008 009 010.MI 010.MI.SA 011 012 013 014 015.MI 015.MI.SA 016.MI 016.MI.SA 017 018 019 020 021 023 025 027 029 042 045 046 049.MI 049.MI.SA 050 051 052 053 054 055 056 057 058 059 060 061 062 063 064 065 068 069 070 071 072
6.3 43 AE.02 AE.03 AE.04 001 002 003.MI 003.MI.SA 004 005.MI 005.MI.SA 006.MI 006.MI.SA 007 008 009.MI 009.MI.SA 010 011 012 013 014 015.MI 015.MI.SA 016 017 018 019 020.MI 020.MI.SA 022 027 028 030 037 038 039 040 041 042 043 044 045 046
6.4 29 AE.01 AE.03 AE.04 001 002 003.MI 003.MI.SA 004 005 006 007.MI 007.MI.SA 008 009 010 012 013.MI 013.MI.SA 015 016 021.MI 021.MI.SA 022 023 023.alt 024 026 028.MI 028.MI.SA 030
6.5 23 AE.01 AE.02 AE.03 001.MI 001.MI.SA 002 003 004 005 006 008.MI 008.MI.SA 009 010 011 012 014 015 016 017.MI 017.MI.SA 019 022
Chapter 7: Techniques of Integration
7.TF 14 001 002 003 004 005 006 007 008 009 010 011 012 013 014
7.1 68 AE.02 AE.03 AE.04 001 01.XP 002 02.XP 003.MI 003.MI.SA 03.XP 004 04.XP 005.MI 005.MI.SA 05.XP 006 007.MI 007.MI.SA 008 009 010 011.MI 011.MI.SA 012 013 014 015 016 017.MI 017.MI.SA 018 019 020 021 022 023.MI 023.MI.SA 024 025 026 027 028 030 031 032 033.MI 033.MI.SA 034.MI 034.MI.SA 035 036 037 038 039 040 041 042 045 047 048 049 050 053 054 058 060 061 062 065
7.2 69 AE.02 AE.03 AE.05 001.MI 001.MI.SA 01.XP 002 02.XP 003 03.XP 004.MI 004.MI.SA 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 054 055 061 062 063 064 065 066
7.3 43 AE.01 AE.02 AE.03 001 01.XP 002 02.XP 003 03.XP 004 005.MI 005.MI.SA 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 029 030 032 033 034 036 038 041 042
7.4 62 AE.01 AE.02 AE.05 001 01.XP 002 02.XP 003 03.XP 004 005 006 007 008.MI 008.MI.SA 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025.MI 025.MI.SA 026 027 028 029 030 031 032 033 034 035 036 037 038 041 043 044 045 047 048 049 050 051 055 056 059 062 063 064 070
7.5 70 AE.02 AE.04 AE.05 001 01.XP 002 02.XP 003 03.XP 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 040 041 042 043 044 045 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 066 068 069 070 072 073 076
7.6 46 AE.02 AE.04 001 01.XP 002 02.XP 003 004.MI 004.MI.SA 005 006 007.MI 007.MI.SA 008 009 010.MI 010.MI.SA 011 012 013 014 015 016.MI 016.MI.SA 017 018 019 020 021 022 023 024 025 026 027 028 029.MI 029.MI.SA 030 031 032 034 040 041 045 046
7.7 46 AE.02 AE.03 AE.05 001 01.XP 002 003.MI 003.MI.SA 004 005 006 007.MI 007.MI.SA 008 009 010 011 012 013 014 015 015.alt 016 017 018 019.MI 019.MI.SA 020 020.alt 021 022.MI 022.MI.SA 025 026 027 028 030 031 032.MI 032.MI.SA 035 036 038 039 040 041 042
7.8 76 AE.01 AE.06 AE.09 01.XP 002 02.XP 003.MI 003.MI.SA 03.XP 004 04.XP 005 05.XP 006 007 008.MI 008.MI.SA 009.MI 009.MI.SA 010.MI 010.MI.SA 011 012 013 014.MI 014.MI.SA 015.MI 015.MI.SA 016 017 018 019 020 021 022.MI 022.MI.SA 023.MI 023.MI.SA 024 025 026 027.MI 027.MI.SA 028 029 030 031 032 033 034 035 036 037 038 039 040 042.MI 042.MI.SA 043 044 049 050 051 052 053 054 057 058.MI 058.MI.SA 059 060 068 069 071 077 078
Chapter 8: Further Applications of Integration
8.1 33 AE.01 AE.03 AE.04 001 002 003 004 005 007 008.MI 008.MI.SA 009 010 011 014 015 016 017 018 019 020 023 024 025 026 032 033 035 036 037 039 040 041
8.2 28 AE.01 AE.02 AE.03 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 025 026 027 028 030
8.3 40 AE.01 AE.03 AE.07 001 002 003 004 005 006.MI 006.MI.SA 007 008 008 009 010 011 012 013 014 015 016 017 019 021 022 023.MI 023.MI.SA 024 025.MI 025.MI.SA 026 027 028 029 030 031 032 033 034 035 046
8.4 18 AE.01 AE.02 001.MI 001.MI.SA.SA 002 003 004.MI 004.MI.SA.SA 005 006 007 008 011 012 014.MI 014.MI.SA 015 017
8.5 18 AE.02 AE.04 AE.05 002 003 004 005.MI 005.MI.SA 006 008 010 011.MI 011.MI.SA 012 013 014 015 017
Chapter 9: Differential Equations
9.TF 7 001 002 003 004 005 006 007
9.1 14 AE.01 AE.02 001 002 003.MI 003.MI.SA 004 005 006 007 008 009.MI 009.MI.SA 010
9.2 25 AE.01 AE.02 AE.04 001 002 003 004 005 006 007 008 009 017 019.MI 019.MI.SA 020 021 022 023.MI 023.MI.SA 024 025 026 027 028
9.3 42 AE.02 AE.04 AE.05 001 002 003.MI 003.MI.SA 004 005 006 007 008 009.MI 009.MI.SA 010 011 012 013 014 015.MI 015.MI.SA 016 017 018 019 020 021 022 023 025 029 030 031 032 034.MI 034.MI.SA 038 041.MI 041.MI.SA 042 043 044
9.4 20 AE.01 AE.03 001 002 003.MI 003.MI.SA 004 005.MI 005.MI.SA 006 007 008.MI 008.MI.SA 013 014 015 016 017 018 019
9.5 28 AE.01 AE.02 AE.04 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 024 025 026 027 028 029 030 032 034
9.6 8 AE.01 001 002 005 006 008 009 010
Chapter 10: Parametric Equations and Polar Coordinates
10.TF 10 001 002 003 004 005 006 007 008 009 010
10.1 38 AE.02 AE.05 AE.08 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016.MI 016.MI.SA 017 018 019 020 021 022 024 028 029 030 031 032 033 034 041 042 044 045 046
10.2 59 AE.02 AE.03 AE.05 001 002 003 004 005.MI 005.MI.SA 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 025 026 029 030 032.MI 032.MI.SA 033 037 038 039.MI 039.MI.SA 040 041 042.MI 042.MI.SA 043 044 045 046 047 048 049 051 052 057 058 059 060 062 065 066 070 071
10.3 66 AE.04 AE.07 AE.11 001 002 003 004 005 006 007 008 008.alt 009 010 011 012 013 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 056 057 058 059 060 061 062 063 064 065 066 067 068 078
10.4 43 AE.01 AE.02 AE.04 001 002 003 004 005 007 009 010 011 012 013 014 015 017 018 019 020 021 023 024 025 027 028 029 030 031 032 033 035 036 037 038 039 040 044 045 046 047 048 054
10.5 48 AE.02 AE.03 AE.07 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 050 051 054
10.6 25 AE.01 AE.02 AE.04 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 026 027 028 029 030
Chapter 11: Infinite Sequences and Series
11.TF 20 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020
11.1 68 AE.02 AE.09 AE.10 003 004 005 006 007 008 009.MI 009.MI.SA 010 011 012.MI 012.MI.SA 013 014 017 018 019.MI 019.MI.SA 020 021 022.MI 022.MI.SA 023.MI 023.MI.SA 024 025 026.MI 026.MI.SA 027 028 029 031 032 033 034 035 036 037 038 039.MI 039.MI.SA 040 041 042 043 044 045 046 047 048 049 050 051 052 053 055.MI 055.MI.SA 060 061 062 063 064 065 066 067
11.2 66 AE.02 AE.04 AE.07 003 004 005 006 007 008 009 011 012 013.MI 013.MI.SA 014.MI 014.MI.SA 015 016 017 018 019 020 021 022 023 024.MI 024.MI.SA 025 026.MI 026.MI.SA 027 028 029 030.MI 030.MI.SA 031 032 033 034 035 036 037 038 039 040 041 042.MI 042.MI.SA 043 044 045 046 047.MI 047.MI.SA 048 049 050 051 055.MI 055.MI.SA 056 059 060 070 075 076
11.3 37 AE.02 AE.04 AE.05 003 004.MI 004.MI.SA 005 006 007.MI 007.MI.SA 008 009.MI 009.MI.SA 010.MI 010.MI.SA 011 012 013 014 015 016 017 018 019 020 021 023 024 025 026 027 029 033 034.MI 034.MI.SA 035 036
11.4 38 AE.01 AE.02 AE.05 003 004 005 007 008 009 010.MI 010.MI.SA 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031.MI 031.MI.SA 032 033 034 035 036
11.5 32 AE.01 AE.02 AE.04 002 003.MI 003.MI.SA 004 005 006 007.MI 007.MI.SA 008 009 010 011 012 013 014 015 017 019 020 023 024.MI 024.MI.SA 025 028 029 030 032 033 034
11.6 35 AE.03 AE.05 AE.06 002 003 004 005 007.MI 007.MI.SA 008.MI 008.MI.SA 009 010 011 012 013 014 015.MI 015.MI.SA 016 017 019 021 022 023 025 026 027 028 029 030 031 032 035 036
11.7 33 AE.01 AE.02 AE.03 AE.04 AE.05 AE.06 001 002 003 006 007 008 009 012 013 014 015 016 017 018 020 021 022 023 026 027 028 029 030 031 032 034 037
11.8 38 AE.01 AE.02 AE.03 003.MI 003.MI.SA 004 005 006 007.MI 007.MI.SA 009 010 011.MI 011.MI.SA 012 013 014 015.MI 015.MI.SA 016 017.MI 017.MI.SA 018 019 020 021 022 023 024 025 026 027 028 030 031 038 041 042
11.9 35 AE.01 AE.05 AE.07 003.MI 003.MI.SA 004 005.MI 005.MI.SA 006 007 008 009 010 011 012 013 014 015.MI 015.MI.SA 016 017 018 023.MI 023.MI.SA 024 025 026 027 028 029 030 031 033 037 038
11.10 64 AE.01 AE.02 AE.09 AE.10 004 005.MI 005.MI.SA 006 007 008.MI 008.MI.SA 009 010 011 013 014 016.MI 016.MI.SA 017.MI 017.MI.SA 019 020 025 026 027 028 030 031.MI 031.MI.SA 032 033.MI 033.MI.SA 034 035 036 037 038 043 044 045 046 048.MI 048.MI.SA 049 050 051 052 053 054 055 056 057 059 060 061 062 063.MI 063.MI.SA 064.MI 064.MI.SA 065 066 067 068
11.11 33 AE.01 AE.02 AE.03 003.MI 003.MI.SA 004 005 006 007 008 009 010 013.MI 013.MI.SA 014 015 016 017 018 019.MI 019.MI.SA 020 021 022 023 024 025 026 027.MI 027.MI.SA 028 029 031 032
Chapter 12: Vectors and the Geometry of Space
12.TF 18 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018
12.1 32 AE.01 AE.02 AE.04 001 003.MI 003.MI.SA 004 006 007 008 009 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 020.MI 020.MI.SA 021 022 033 034 035 036 038 039 040
12.2 38 AE.01 AE.03 AE.04 007 008 009.MI 009.MI.SA 010 011 012 013 014 015 016 017 018 019 020 021.MI 021.MI.SA 022 023 024 025 026.MI 026.MI.SA 027 028 029 030 031 032 033 034 035 036 037 039
12.3 48 AE.01 AE.03 AE.06 002 003 004.MI 004.MI.SA 005 006 007 008 009 010 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035.MI 035.MI.SA 036 037 038 039 040 042 043 045.MI 045.MI.SA 046 047 048 049
12.4 41 AE.01 AE.02 AE.05 001 002 003 004 005 006 007 008 009 010 011 012 014 015 016 017 018 019.MI 019.MI.SA 020 027 028 029 030 031 032 033 034 035 036 037 038 039.MI 039.MI.SA 040 041.MI 041.MI.SA 042
12.5 61 AE.03 AE.04 AE.07 001 002.MI 002.MI.SA 003 004 005 006 007 010 011 012 015 016 017 018 021.MI 021.MI.SA 023 024 025 026 027 028 029.MI 029.MI.SA 030 031 032 033 034 035 036 037 038 043 044 045 046 047 048 052 055 056 057 058 059 060 062 063 064 067 068 069 070 071 072 074 076
12.6 40 AE.01 AE.05 AE.07 001 003 004 005 006 008 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 043 044 045 046 048
Chapter 13: Vector Functions
13.TF 12 001 002 003 004 005 006 007 008 009 010 011 012
13.1 27 AE.03 AE.04 AE.06 001.MI 001.MI.SA 002 003 004.MI 004.MI.SA 005 006 007combo 015.MI 015.MI.SA 016 017 018 019combo 027.MI 027.MI.SA 028 036 037 038 040 041 042
13.2 43 AE.01 AE.03 AE.04 003 005 006 009.MI 009.MI.SA 010 011 012 013 014 015 017.MI 017.MI.SA 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034.MI 034.MI.SA 035 036 037 038 039.MI 039.MI.SA 040 045 046
13.3 48 AE.01 AE.03 AE.07 001.MI 001.MI.SA 002 003 004 005 006 007 008 009 010 011.MI 011.MI.SA 012 013 014 017.MI 017.MI.SA 018 019 020 021 022 023.MI 023.MI.SA 024 025 026 027 028 029 030 031 032 041 042 043 044 045 046 047 048 049 057 058 059
13.4 36 AE.03 AE.05 AE.06 001 009 010 011.MI 011.MI.SA 012 013 014 015.MI 015.MI.SA 016 017 018 019.MI 019.MI.SA 020 021 023 024 025.MI 025.MI.SA 026 027 029 031 032 033 034 035 036 037 038 041
Chapter 14: Partial Derivatives
14.TF 12 001 002 003 004 005 006 007 008 009 010 011 012
14.1 56 AE.06 AE.08 AE.11 002 004 005 006 007 008 009 010 011 012 013 014 015 016 017.MI 017.MI.SA 018 019 020 030 032.MI 032.MI.SA 035 036 037 038 039.MI 039.MI.SA 040 041 042 043 044 045 046 049.MI 049.MI.SA 050 055 056 057 058 059 060 061 062 063 064 065 066 069 070 074
14.2 39 AE.01 AE.03 AE.05 005 006 007 008 009 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019 020 021 022 025 026 029 030 031.MI 031.MI.SA 032 033.MI 033.MI.SA 034 035 036 037 038 039 040 041
14.3 71 AE.03 AE.04 AE.07 005 006 007 008 010 011 012 015.MI 015.MI.SA 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 038 039.MI 039.MI.SA 040 041 042 045 046 047 048 049 050 051 052 053 054.MI 054.MI.SA 055 056 061 062 063 064 065 066 067 068 069 072 080 081 084 086 088 089 092 093 094
14.4 38 AE.01 AE.02 AE.04 001 002.MI 002.MI.SA 003 004 005 006 017 018 019 020 021 022 025.MI 025.MI.SA 026 027 028 029 030 031 032 033.MI 033.MI.SA 034 035 036 037 038 039 040 041 042 043 044
14.5 47 AE.02 AE.04 AE.05 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 021 022 023.MI 023.MI.SA 024 025 026 027 028 029 030 031 032 033 034 035.MI 035.MI.SA 037 038.MI 038.MI.SA 039 040 041 042 043 044 051 052
14.6 47 AE.04 AE.05 004.MI 004.MI.SA 005 006 007 008 009 010 011 012 013 014 015 016 017 019 020 021 022 023.MI 023.MI.SA 024 025 026 027 029.MI 029.MI.SA 031 032 033 034 035 039 040 041 042 043 044 047 048 050 052 053 059 060
14.7 45 AE.03 AE.05 AE.06 005 006 007 008 009 010 011.MI 011.MI.SA 012 013 014 015 016 017 018 021 022 023 024 029 030 031 032 033 034 035 036 039.MI 039.MI.SA 040 041 042 043.MI 043.MI.SA 045 046 047 048 049 050 051 056
14.8 38 AE.01 AE.02 AE.05 003.MI 003.MI.SA 004 005.MI 005.MI.SA 006 007 008 009 010 011 012 013 014 015.MI 015.MI.SA 016 017 018 019 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041
Chapter 15: Multiple Integrals
15.TF 8 001 002 003 004 005 006 007 008
15.1 16 AE.01 AE.02 AE.03 001.MI 001.MI.SA 002 003 004 005 006 007 008.MI 008.MI.SA 009 011 012
15.2 33 AE.02 AE.03 AE.04 001 002 003.MI 003.MI.SA 004 005 007 008 010 011 012 013 014 015.MI 015.MI.SA 016 017.MI 017.MI.SA 018 019 020 021 022 025 026 027 030 031 035 036
15.3 49 AE.01 AE.03 AE.05 001 002 003 004 005.MI 005.MI.SA 006 007 008 009 011 012 013.MI 013.MI.SA 014 016 017 018 019.MI 019.MI.SA 020 021 022 023 024 025 026 027 028 031 032 039 045 046 047 048 049 050 053 054 055 056 057 059 060 061
15.4 32 AE.02 AE.03 AE.04 005 006 007 008.MI 008.MI.SA 010 011 012 014 015 016 017 018 019.MI 019.MI.SA 020.MI 020.MI.SA 021 022 024 025 026 027 030 031 032 033 034 037
15.5 30 AE.02 AE.03 AE.04 AE.05 001 002.MI 002.MI.SA 003.MI 003.MI.SA 004 006 007.MI 007.MI.SA 008 009 010 012 014 016 017 018 019 023 024 026 027 029 030 032 033
15.6 42 AE.01 AE.03 AE.05 002 003 004 005 006 007.MI 007.MI.SA 008 009 010 011 012 013 014 015 016 017.MI 017.MI.SA 018 019 020.MI 020.MI.SA 021 022 025 026 033 034 035 036 037 038 039 040 045 049 050 052 053
15.7 24 AE.02 AE.03 001.MI 001.MI.SA 002 003 004 009 010 011 012 015 016 017 018 019.MI 019.MI.SA 020 021 022 023 025 027 028
15.8 40 AE.01 AE.02 AE.03 AE.04 001.MI 001.MI.SA 002 003.MI 003.MI.SA 004 009 010 015 016 017.MI 017.MI.SA 018 019 020 021.MI 021.MI.SA 022 023 024 025 026 027 029 030 031 032 033 035 036 037 038 039 040 042 043
15.9 23 AE.01 AE.02 AE.04 001 002 003.MI 003.MI.SA 004 005 006 011.MI 011.MI.SA 012 013 014 015 016 019.MI 019.MI.SA 020 021 022 023
Chapter 16: Vector Calculus
16.TF 8 001 002 003 004 005 006 007 008
16.1 27 AE.01 AE.02 AE.06 001 002 003 004 005 006 007 008 009 010 011 015 021.MI 021.MI.SA 022 023 024 025 026 029 033.MI 033.MI.SA 034 036
16.2 40 AE.03 AE.04 AE.05 001 002 003.MI 003.MI.SA 004 005 006 007 008 009 010 011 012 013 014 015 016 019.MI 019.MI.SA 020 021 022 029 030 031 032 033 034 036 039.MI 039.MI.SA 040 041 042 043 044 046
16.3 30 AE.02 AE.03 AE.05 001 002 003.MI 003.MI.SA 004 005 006 007 008 009 010 012.MI 012.MI.SA 013 014 015 016 017 018 021.MI 021.MI.SA 022 029 030 031 032 034
16.4 26 AE.02 AE.04 AE.05 001.MI 001.MI.SA 002 003 004 005 006 007.MI 007.MI.SA 008 009 010 011.MI 011.MI.SA 012 013 014 017 018 019 020 024 026
16.5 31 AE.02 AE.03 AE.05 001 002 003 004 005 006.MI 006.MI.SA 007 008 009 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019.MI 019.MI.SA 020 024 026 030 031 032
16.6 51 AE.04 AE.06 AE.09 AE.11 001.MI 001.MI.SA 002 003 004 005 006 013 014 015 016 017 018 019.MI 019.MI.SA 020 021 022 023.MI 023.MI.SA 024 025 026 029 030 033 034 035 036 037.MI 037.MI.SA 038 039 040 041 042 043 044.MI 044.MI.SA 045 046 047.MI 047.MI.SA 053 055 057 058
16.7 39 AE.03 AE.05 AE.06 003 006.MI 006.MI.SA 007 008 009 010 011 012 013 014 015 016 017 018 019.MI 019.MI.SA 020 021.MI 021.MI.SA 022 023 024 026 027 028 029 030 032 038 040 041 042 043 044 045
16.8 16 AE.01 AE.02 002.MI 002.MI.SA 003 004 005 006 007.MI 007.MI.SA 008 009 010 012 017 018
16.9 27 AE.01 AE.02 001 002 003 005 006 007.MI 007.MI.SA 008 009 010 011 012 013.MI 013.MI.SA 014 015 016 017 018 019.MI 019.MI.SA 020 021 022 024
Chapter 17: Second-Order Differential Equations
17.TF 4 001 002 003 004
17.1 27 AE.03 AE.04 AE.07 001 002 003 005 006 007 008 009 011 012 013 017 018 019 020 021 022 023 024 026 027 028 030 031
17.2 23 AE.01 AE.03 AE.04 001 002 003 004 005 006 007 008 009 010 019 020 021 022 023 024 025 026 027 028
17.3 15 AE.01 AE.02 AE.03 001 002 003 004 005 006 008 013 014 015 016 018
17.4 10 AE.01 AE.02 001 002 004 005 006 007 009 010
Total 5544 (20) | 17,692 | 37,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-39 | latest | en | 0.750836 |
https://math.answers.com/questions/What_is_the_difference_between_a_face_value_and_a_value_in_3508 | 1,686,019,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652207.81/warc/CC-MAIN-20230606013819-20230606043819-00240.warc.gz | 431,526,124 | 49,187 | 0
# What is the difference between a face value and a value in 3508?
Wiki User
2013-06-12 12:38:55
The face value of 3 is 3: the value of 3 is 3000
The face value of 5 is 5: the value of 5 is 500
The face value of 3 is 3: the value of 3 is 3000
The face value of 5 is 5: the value of 5 is 500
The face value of 3 is 3: the value of 3 is 3000
The face value of 5 is 5: the value of 5 is 500
The face value of 3 is 3: the value of 3 is 3000
The face value of 5 is 5: the value of 5 is 500
Wiki User
2013-06-12 12:38:55
Study guides
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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3.8
2541 Reviews
Wiki User
2013-06-12 12:38:55
The face value of 3 is 3: the value of 3 is 3000
The face value of 5 is 5: the value of 5 is 500 | 312 | 830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-23 | longest | en | 0.876323 |
http://math.stackexchange.com/questions/121917/a-oplus-b-cong-a-oplus-c-implies-b-cong-c-no-it-does-not | 1,469,570,804,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00070-ip-10-185-27-174.ec2.internal.warc.gz | 158,554,250 | 19,229 | # $A\oplus B\cong A\oplus C$ implies $B\cong C$? (No, it does not)
I am asked to prove that for $p\in (1, \infty)$, $$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$ on a homework assignment, and I think I can show using results from class that $\ell_2\oplus \ell_2\cong \ell_2$.
From this I could say that $L_{p}[0,1]\oplus \ell_{2}\cong L_{p}[0,1]\oplus \ell_{2}\oplus\ell_{2}$
From here I feel like I should be able to conclude that $$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$
But I know of no such result that allows me to do this. Can anyone tell me if it's true or false?
EDIT: Definitely false. (See counter example below from Arturo Magidin).
That is, if $B\oplus A\cong C\oplus A$ can I conclude that $B\cong C$?
Proper Solution (based on hints below from t.b.):
1) Prove that $\ell_2\oplus \ell_2\cong \ell_2$
2) Use the fact that $\ell_2$ is complemented in $L_p[0,1]$ to write $L_p[0,1] = \ell_2\oplus (\ell_2)^{c}$.
3) Then I combine these to obtain:
$L_p[0,1]\cong \ell_2\oplus (\ell_2)^{c}\cong (\ell_2\oplus \ell_2) \oplus (\ell_2)^{c} \cong \ell_2 \oplus (\ell_2\oplus (\ell_2)^{c})\cong \ell_2\oplus L_p[0,1]$.
I skipped some pieces of your more general argument. I was just wondering if I did anything illegal, so to speak.
-
You say that you can prove that $\ell_2\oplus\ell_2\cong\ell_2$. Note that this means that you can prove that $\ell_2\oplus\ell_2\cong \ell_2\oplus \mathbf{0}$. If the statement you want were true, then you would be able to conclude that $\ell_2\cong \mathbf{0}$. Is that true? – Arturo Magidin Mar 19 '12 at 1:38
Wow. Thanks for putting it so simply. I had tried many examples and they all seemed to work. But I never thought of trying with that. Thanks! I feel silly now. – roo Mar 19 '12 at 1:40
@t.b. Sorry I got confused between the two spaces being equal and them being isomorphic. – user38268 Mar 19 '12 at 2:24
Wouldn't this work: $$L_p\cong \ell_2\oplus Y\cong ( \ell_2\oplus\ell_2)\oplus Y\cong \ell_2\oplus(\ell_2 \oplus Y)\cong \ell_2\oplus L_p,$$ or, is this just t.b.'s argument? (it's late for me...) – David Mitra Mar 19 '12 at 2:27
To address your general question about whether direct decompositions are unique (i.e. whether $B \cong C$ in your case), you might also consider the very small/simple counter-example of Bjarni Jonsson described here.. This is discussed at length in Algebras, Lattices, Varieties chapter 5. – William DeMeo Mar 19 '12 at 2:35
Since Arturo explained why your idea doesn't work, here's a
Hint:
Let $X = L^p$ and let $Y = L^p \oplus \ell^2$.
Prove:
1. $X^2 \cong X$ and $Y^2 \cong Y$.
2. $X$ is isomorphic to a complemented subspace of $Y$ and vice versa (you probably know that $L^p$ has a complemented subspace isomorphic to $\ell^2$, see e.g. Proposition 6.4.2 in Albiac-Kalton or follow David Mitra's suggestion, Corollary 9.2 on page 90 in Carother's A Short Course on Banach Space Theory).
3. Now you are in position to apply Pełczyński's argument to conclude that $X \cong Y$ (see point 3. of my question here for the details).
-
Ahhhhhh! I love it. Thanks. The $X^2\cong X$ I hadn't thought of doing, although I literally did it for $X:=C[0,1]$ when I was trying to test my original (but false) claim. Thanks again! – roo Mar 19 '12 at 1:56
I included my argument into my original question. I didn't really follow your process exactly, but just used the argument in your referenced post about $L_\infty$. Does my argument still work? Or am I misunderstanding a key concept? edit: (just saw you comment above. Thanks very much again!!!) – roo Mar 19 '12 at 2:25
That's why I wrote: yes, that's it! (that was directed at you) :) – t.b. Mar 19 '12 at 2:26 | 1,267 | 3,669 | {"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} | 3.34375 | 3 | CC-MAIN-2016-30 | latest | en | 0.855826 |
https://studylib.net/doc/13613823/wrap-up-of-valuation-katharina-lewellen-finance-theory-ii. | 1,611,330,565,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00312.warc.gz | 576,618,366 | 17,174 | # Wrap-up of Valuation Katharina Lewellen Finance Theory II May 14, 2003
```Wrap-up of Valuation
Katharina Lewellen
Finance Theory II
May 14, 2003
Final Exam
Rules of the game:
¾ No laptops
¾ Closed books
¾ Cheat sheet
2
Valuation
Valuation tools:
Free cash flows
Cost of capital: WACC and APV
Real options
Valuing companies
DCF analysis:
¾ Forecast horizon and terminal values
¾ EVA: When is growth good?
Comparables, Multiples
3
Estimating the FCF
Free cash flows (FCF) are the expected after-tax cash flows
that the firm would generate if it were 100% equity financed.
FCF = EBIT*(1-t) + Dep. - CAPX - ∆NWC
FCF = EBITD*(1-t) + t * Dep. - CAPX - ∆NWC
FCF = EBIT*(1-t) - ∆NA
Recall:
NWC = Current assets – Current liabilities
NA = Assets – Current liabilities.
4
FCF = EBIT*(1-t) + Dep. - CAPX - ∆WC
This expression amends EBIT(1-t) which is an accounting
measure of cash flow into an economic measure.
CAPX not reported as cash outflow but is one D - CAPX
Depreciation
¾ is reported as cash outflow but is not one D Add (1-t)*Dep
¾ however, depreciation does imply a cash inflow of t*Dep.
¾ Altogether D + Dep
Working capital has an opportunity cost D - ∆NWC
5
Other Things to Keep In Mind
Formulas need to be adapted in particular situations
¾ Need to understand the economics (e.g., Southland’s asset sales)
Use all incremental cash flows:
¾ Ignore sunk costs, Count opportunity costs, Avoid “accounting
illusions”…
Don’t forget FCF at the end of the project’s life:
¾ If liquidated: SV*(1-t) + t * PPE
¾ Even if not liquidated, recoup WC
FCF ignores the tax shield provided by the firm’s debt.
We deal with it separately in APV or WACC. Do not include the
effects of financing at this stage: You would count them twice!
6
APV Step 1: Value if 100% equity
2. Unlever each comp’s βE to estimate its βA using
βA = βE
E
E+D
(OK if the comp’s D not too high (+ can assume their D/V is stable))
3. Use the comps’ βA to estimate the project’s βA (e.g., as average).
4. Use estimated βA to calculate the all-equity cost of capital kA
k A = rf + β A ⋅ Market Risk Premium
5. Use kA to discount the project’s FCF
7
Why We Need to Unlever
Comps may have different leverage
Equity in a firm with debt is more risky than equity in a firm
without debt because debt receives some of the safe cashflows.
Note: Business Risk and Financial Risk
Financial risk has nothing to do with costs of financial distress!
Similar firms have similar business risk (βA) but can have
different financial risk (βE-βA) if they have different leverage.
As leverage increases, equity becomes riskier (i.e. βE Ò).
8
APV Step 2: Add PV (Tax Shield)
If the project’s D is constant over time, then
PV(TS) = t*D*kD / kD = t*D
If the project’s D/V is constant, then
PV(TS) = t*D*kD / kA
If there is a known debt policy or repayment schedule
¾ you can simply forecast actual debt levels and discount by a rate
between kD and kA
9
APV Step 2: Add PV (Tax Shield), Remarks
Count only debt attributed to the project
¾ Recall: If a project is 100% debt finance, some of the debt is
probably issued against firm’s other assets
Make sure to discount expected not maximum tax shields
¾ This is particularly important for high D/V
For high D/V, should count costs of financial distress
Recall: Use the marginal (as opposed to the average) tax rate
10
Weighted Average Cost of Capital (WACC)
Approach: Adjust the discount rate to account for the tax shield.
WACC =
D
E
k D (1 − t) +
kE
D+E
D+E
Most widely used DCF analysis method.
The aim is to avoid 1st order mistakes:
¾ A priori, WACC is project-specific (except for tax rate t)
¾ Firm-wide WACC is OK only if project comparable to the firm
11
Leverage ratio: D/(D+E)
What we want: The debt that is incremental to the project, i.e.,
that could not be raised w/o the project.
1st-order mistakes we want to avoid:
¾ Use the deal’s leverage ratio;
¾ Use the “acquirer”’s leverage ratio.
Imperfect approach to what we want:
¾ Target leverage ratio if project/firm were a stand-alone
How we get there:
¾ Get D/V from comps, business plan, checklist, etc.
12
Cost of debt capital: kD
What we want: Expected return for creditors if project were a
stand-alone with leverage ratio D/(D+E) estimated above.
Imperfect approach to what we want: kD close the interest rate
charged to project as stand-alone (unless debt is very risky).
How we get there:
¾ Find comps with similar leverage + recent interest rate.
¾ Estimate the debt rating and examine corporate yield curve.
1st-order mistakes we want to avoid:
¾ Use the interest rate in the deal or of the “acquirer”;
13
Effective Marginal Tax Rate t
Marginal tax rate of firm undertaking the project: t
14
Using CAPM to Estimate kE
1. Find comps for the project under consideration.
2. Unlever each comp’s βE (using its D/(D+E)):
βA = βE
E
E + D
3. Use the comps’ βA to estimate the project’s βA (e.g. average).
4. Relever the project’s estimated βA (using its own D/(D+E)):
βE =
D⎤
E+D
⎡
β A = ⎢1 + ⎥ β A
E⎦
E
⎣
5. Use the estimated βE to calculate the project’s cost of equity kE:
k E = rf + β E ⋅ Market Risk Premium
Note: These (un-) levering formulas are OK only if the (comp) firm’s debt is not too
risky and its D/V is reasonably stable.
15
Remarks
WACC can be used only if D/V is reasonably stable
Use APV when debt is very risky and/or when D/V is unstable
(recall the Southland LBO case)
WACC is an attribute of the project, not the firm (except tax rate)
OK to use the firm’s WACC when project is very much like the
firm (because the firm happens to be a comp for the project).
Few companies have WACC that they can use for all projects
(recall our discussion of GE).
16
Real options
Embedded options
Follow-up investments
Option to abandon the project
Option to wait before investing
Option to expand / change production methods
Key issues
Identification
Valuation
17
Identify significant options
Look for clues in project’s description and cash flow pattern
¾ “Phases”, “Strategic investment”, “Scenarios”…
¾ Large expenditures are likely discretionary
Is there an option? Verify two conditions:
(1) News will possibly arrive in the future;
(2) When it arrives, the news may affect decisions.
Search for the uncertainty that managers face:
¾ What is the main thing that managers will learn over time?
¾ How will they exploit that information?
18
Practical Issue: Simplifications
Search for significant options
¾ E.g., option to shut down the plant may not be very valuable (why?)
¾ Look for primary sources of uncertainty
Cut the projects into simple options
¾ You might want to ignore nested options (difficult to value)
Use European rather than American option
Ignoring some adverse effects of waiting (e.g. possible entry)
A simplified model that is dominated by the project gives a lower bound
for the project’s value (and vice versa).
19
Value the options
Step 1:
¾ Pretend that there is no option embedded in the project
¾ This benchmark constitute a lower bound for the project’s value
Step 2:
Value the option
¾ Decision trees (dynamic DCF)
¾ Option pricing models (Black-Scholes)
20
Mapping: Project Æ Call Option
Project
Call Option
Expenditure required to acquire
the assets
X
Exercise price
Value of the operating assets to
be acquired
S
Stock price (price of the
underlying asset)
Length of time the decision may
be deferred
T
Time to expiration
Riskiness of the operating
assets
σ2
Variance of stock return
r
Risk-free rate of return
Time value of money
21
Practical Issue: What Volatility?
What do we want?
Standard deviation of returns for the underlying asset
In case of real options, the underlying is the PV of the project’s CFs
Imperfect ways to get it?
Informed guess
¾ 20-30% per year is not remarkably high for a single project.
Data
¾ Historical return volatilities on comparable traded assets
¾ Implied volatilities can be computed from quoted option prices
Simulations
22
Valuing Companies
Terminal values:
¾ Liquidation
¾ Flat, growing, or decreasing perpetuity
EVA: When is growth good?
Comparables, Multiples.
23
Terminal Values
Liquidation: Should be adjusted (e.g. if cannot recoup all A/R, etc.)
SV * (1-t) + t*PPE + WC
Growing perpetuity: Take EBIT and NA in last year of forecast
TV = [(1+g)*EBIT*(1-t) - g*NA] / (k-g)
Flat perpetuity:
TV = EBIT*(1-t) / k
24
Terminal Values, Remarks
Growing perpetuity formula assumes a linear relationship
between EBIT and NA
Don’t forget to take PVTV
Forecast horizon: Company is reasonably stable afterwards
25
EVA
Growth is valuable when (very roughly!):
EVA = EBIT*(1-t) - k*NA > 0
or
EBIT*(1-t) / NA > k
Growth is good if the cost of scaling up NA is offset by the value
of increased revenues.
Remarks:
Assumes linearity between EBIT and NA and that NA is a good
measure of marginal “replacement cost”, now and in the future.
EVA has nothing to do with sustainable growth.
26
EVA: Bottom Line
Use EVA as…
… a simple measure of whether a business is generating value
and whether growth is enhancing value
… as a way of setting goals to enhance value
Beware of EVA for...
… young companies
… companies in rapidly changing business environment
… companies in which book values are not accurate measures
of marginal replacement cost.
27
Multiples
Assess the value based on that of publicly traded comps
Cash-flow based Value multiples
¾ MV(firm)/Earnings, MV(firm)/EBITDA, MV(firm)/FCF,...
Cash-flow based Price multiples:
¾ Price/Earnings, Price/EBITDA, Price/FCF,...
Asset-based multiples:
¾ MV(firm)/BV(assets), MV(equity)/BV(equity),...
28
Motivation for Multiples?
Assumption 1:
¾ E = CF to shareholders
¾ E is a perpetuity
E
P=
kE − g
⇒
P
1
=
E kE − g
Assumption 2:
¾ Comps have the same kE => This requires similar leverage!
¾ Comps are growing at a similar rate g
29
Multiples: Pros and Cons
Pros:
Incorporates simply a lot of information from other valuations
Embodies market consensus
Can provide discipline for DCF valuation: Ask yourself “How do I
explain the difference?”
Sometimes, what you care about is what the market will pay, not
the fundamental value (e.g., Venture firm will want out).
Cons:
Hard to incorporate firm specific information.
Relies on accounting measures being comparable too.
30
Control:
With a controlling stake, influence operations, implement
synergies and capture (part of) their value
Also, entrepreneur might care about “the vision”
Large individual shareholder (e.g., entrepreneur):
Maybe very undiversified, at least for a while
Liquidity:
Especially for private companies
Note: Need to account for IPO plans
31
Valuation: Conclusion
Main merit of DCF analysis: Forces to argue where value comes
from D Most important step is a reasonable forecast of FCF.
Sales forecasts: Reasonable given the firm’s resources, the
industry, and competition? What market share is needed?
Margin forecasts: Reasonable given potential competition/entry
barriers and bargaining position with suppliers and customers?
CAPX and other investment forecasts: Consistent with the sales
and margin forecasts?
Terminal value: Does it make sense?
Sensitivity analysis: What variables and assumptions are crucial
32
Valuation: Conclusion
The different methods are not mutually exclusive.
Comparables and multiples are important but:
¾ don’t tell you where value comes from;
¾ whether comparables are really comparable.
DCF analysis (+ Real options) forces to justify valuation but:
¾ only as good as the data input;
¾ relies on imperfect models.
Go back and forth between the two approaches.
33
Course: Conclusion
Acquire a few general tools:
¾ Capital structure
¾ DCF analysis
¾ Comparales and multiples
Avoiding 1st order misconceptions (list your own below if any):
¾
¾
¾ etc.
Developing a healthy skepticism.
35
Financing
The bulk of the value is created on the LHS by making good
investment decisions.
You can destroy much value by mismanaging your RHS:
You cannot make sound financial decisions without knowing the
Avoid one-size-fit-all approaches.
Finance is too serious to leave it to finance people.
36
Valuation
Making sound business decisions requires valuing them.
This involves mostly knowing the business (to make appropriate
cash-flow forecasts and scenario analysis, etc.)
But also some finance:
¾ What discount rate?
¾ Valuation exercises can indicate key value levers,...
Avoid one-size-fit-all approaches.
| 3,565 | 12,955 | {"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-2021-04 | latest | en | 0.867093 |
http://sjhomeworkqunt.locallawyer.us/cost-volume-profit-analysis.html | 1,539,677,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510415.29/warc/CC-MAIN-20181016072114-20181016093614-00429.warc.gz | 337,071,312 | 5,283 | # Cost volume profit analysis
Cost volume profit analysis (cvp analysis) 31 introduction cvp analysis is a systematic approach of examining the relationship between the changes in volume, cost, revenue and profit. The cvp analysis can be done through the flexible budgeting better evaluation can be made of profit opportunities by studying the relationships among costs, volume and profits. Cost-volume-profit analysis this lesson introduces cost-volume-profit analysis cvp analysis is a way to quickly answer a number of important questions about the profitability of a company's products or services. Join jim stice and earl kay stice for an in-depth discussion in this video, cost-volume-profit (cvp) analysis, part of breakeven and cost-volume-profit (cvp) analysis. Cost-volume-profit analysis, or cvp analysis for short, is an analytic tool that uses the relationships among components of the profit equation, price, volume, cost structure and.
Cost–volume–profit (cvp) analysis is a model to analyze the behaviour of net income in response to changes in total revenue, total costs, or both in reality . Contribution margin and cvp analysis (part 2 of 2) - duration: cost volume profit analysis (cvp): calculating the break even point - duration: 7:14 edspira 91,227 views. Cost-volume-profit analysis, or cvp, is an accounting tool managers can use to estimate the levels of sales needed to reach a particular level of profit or break even. Cvp analysis examines the relationship between sales volume, costs and profit during the period of one year and during this time it is suggested that it would be difficult to change selling prices, variable and fixed costs which is in agreement with the other assumptions.
Scenario: sure corporation has collected the following information after its first year of sales net sales were \$1,600,000 on 100,000 units selling expenses \$240,000 (40% variable and 60% fixed) direct materials \$511,000. The graphs provide a helpful way to visualize the relationship among cost, volume, and profit however, when solving problems, you’ll find that plugging numbers into formulas is much quicker and easier pemulis basketballs sells basketballs for \$15 each the variable cost per unit of the . Cost-volume-profit (cvp) analysis expands the use of information provided by breakeven analysis a critical part of cvp analysis is the point where total revenues equal total costs (both fixed and variable costs) at this breakeven point (bep), a company will experience no income or loss this bep . Cost-volume-profit analysis is a managerial accounting technique used to analyze how changes in cost and sales volume affect changes in a company's profit the technique is widely used in business and has many advantages however, there are some drawbacks as well understanding the pros and cons to .
Cost-volume-profit (cvp) analysis cvp analysis examines the interaction of a firm’s sales volume, selling price, cost structure, and profitability it is a powerful tool in making managerial decisions including marketing, production, investment, and financing decisions. Cost-volume-profit (cvp) analysis is one of the major tools of financial analysis managers use the contribution margin to plan for the business. Cost-volume-profit analysis looks primarily at the effects of differing levels of activity on the financial results of a business in any business, or, indeed, in life in general, hindsight is a beautiful thing if only we could look into a crystal ball and find out exactly how many customers were .
Cost volume profit analysis helps in examining the change in profit vis-à-vis change in sales volume, cost of the product and the selling price of the product cost volume profit analysis is the study of the effects of changes in costs and volume on a company’s profits. Join jim stice and earl kay stice for an in-depth discussion in this video, cost-volume-profit analysis (cvp), part of accounting foundations: managerial accounting. Cost–volume–profit (cvp), in managerial economics, is a form of cost accounting it is a simplified model, useful for elementary instruction and for short-run . Cost-volume-profit (cvp) analysis is a managerial accounting technique that is concerned with the effect of sales volume and product costs on operating profit of a business.
## Cost volume profit analysis
Test and improve your knowledge of cost-volume-profit analysis with fun multiple choice exams you can take online with studycom. How to do cost volume profit analysis cost-volume-profit analysis is an important tool from cost accounting to help managers decide how many units to sell, answer questions about the product mix, set profit targets reasonably -- all in. Chapter 3 cost-volume-profit analysis overview this chapter explains a planning tool called cost- volume-profit (cvp) analysiscvp analysis examines the behavior of total revenues, total.
Cost-volume-profit analysis is a tool that can be utilized by business managers to make better business decisions among the tools in a business manager's decision-making arsenal, cvp analysis . Cost behavior outside of the relevant range is not linear, which distorts cvp analysis cost-volume-profit analysis includes all of the following assumptions except the behavior of costs is curvilinear throughout the relevant range.
Starting a business can be pricey breakeven analysis and cost-volume-profit analysis will help you understand when—and if—your business will start to recover those costs and begin making a profit. Definition: the cost volume profit analysis, commonly referred to as cvp, is a planning process that management uses to predict the future volume of activity, costs incurred, sales made, and profits received. Cost-volume-profit analysis: • is a type of cost accounting and one of the major, widely used tools of financial analysis to help managers make short term decisions • is based upon determining the break-even point of cost and volume of goods.
Cost volume profit analysis
Rated 3/5 based on 22 review
2018. | 1,192 | 6,048 | {"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} | 3.375 | 3 | CC-MAIN-2018-43 | latest | en | 0.901223 |
https://www.physicsforums.com/threads/fractional-quantum-hall-effect.396778/ | 1,618,413,685,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00156.warc.gz | 1,051,545,688 | 16,920 | # Fractional Quantum Hall Effect
Can someone explain to me as simply as possible why the Laughlin states create energy gaps in the lowest landau level? I am trying to understand for a presentation why the Laughlin states correctly model the QHE effect when the filling factor is a odd fraction (1/3, 1/5, 1/7). As far as I understand it in order for there to be a quantized hall effect at v=1/3 the laughlin state must split the energy spectrum of the system such that there are discrete energy levels just like in the integer effect. I don't understand how this works. I know it is somehow connected with the fact the states are an incompressible liquid but I can't figure that part out either.
Do not confuse the fractional and integer quantum hall effects. At the level which you should be approaching it, they have very little in common. The IQHE is an independent electron effect --- you can understand it by thinking about one electron. The FQHE is fundamentally about the interactions between the electrons.
To study the Laughlin state requires heavy machinery. In fact, I believe (someone up to date with the literature should correct me) that there is no analytic way to prove that it is gapped. (Bearing in mind that gapped-ness is a property of the Hamiltonian + state, so this statement needs to be qualified by giving a Hamiltonian; usually, it is a simple n-particle Hamiltonian with only kinetic and mutual repulsion of an unspecified potential.)
EDIT: actually, it appears I lie. The m=3 state with Coulomb potential was solved quite a long time ago: http://prl.aps.org/abstract/PRL/v54/i6/p581_1
I imagine people have now extended this significantly. The calculation seems quite involved, however.
Yes the Laughlin wavefunction is a trial wavefunction for many interacting electrons, I understand that much. What I am trying to do is make the connection from the wavefunctions to how this relates to seeing the quantized plateaus in the hall resistance and why the resistance goes to zero as the special values of the filling factor 1/3, 1/5, 1/7.
In the IQHE it is because the energy levels of the electron is split into massively degenerate landau levels. Thus there is an energy gap between landau levels. Therefore if one landau level lets say the lowest is completely filled then one can imagine that resistivity which essentially is a measure of energy loss would go to zero because the only way to lose energy in such a system is to move a electron into the next landau level, but this is very difficult due to the energy gap between the two, thus in most systems at low temperature it can't be done. This is how I am understanding the integer effect.
In the FQHE described by the Laughlin function which as I can tell only describes the Lowest Landau Level. The power of the polynomial term (in the Laughlin wavefunction) is the reciporical of the filling factor. This then implies that for a power of 3 the LLL is 1/3 full. I guess my question is how does that 1/3 filled state create the similar effects seen in the IQHE? From what I have read there is supposed to be energy gaps within the LLL itself, but I can't see this from the wavefunction. Also the wavefunction is supposed to goto zero faster at the special values of the filling factor then for perturbations away from it but this doesn't make sense to me either.
Gokul43201
Staff Emeritus
Gold Member
I guess my question is how does that 1/3 filled state create the similar effects seen in the IQHE?
Have you come across the Composite Fermion/Boson picture? FQHE states can be pictured as IQHE states of composite particles (electrons attached to an even/odd number of flux quanta).
The standard reference is Perspectives in Quantum Hall Physics, by Pinczuk and Das Sarma.
A great introductory review (postscript file) by Steve Girvin can be downloaded here: www.bourbaphy.fr/girvin.ps[/URL]
Last edited by a moderator:
Have you come across the Composite Fermion/Boson picture? FQHE states can be pictured as IQHE states of composite particles (electrons attached to an even/odd number of flux quanta).
The problem there is that you can't be certain you don't have a continuum spectrum for unbinding the composite particle. It is post-hoc justified, but not a priori.
I stand by my original comment. To calculate the gap, look at the paper I linked to. It is not trivial.
Incidentally, the formation of plateaus in IQHE is *not* trivial. It is intrinsically tied up with disorder and Anderson localisation. This is often neglected in introductory texts, and I think it is a mistake; it makes people *think* they've understood something when in fact they have not at all. | 1,047 | 4,672 | {"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-2021-17 | latest | en | 0.951581 |
https://cs.stackexchange.com/questions/2595/operations-on-obdd-negation-through-shannons-expansion | 1,713,073,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00456.warc.gz | 183,393,115 | 39,953 | # Operations on OBDD: negation through Shannon's expansion
I have a problem with the application of the Shannon expansion for to obtain the negation of a formula boolean, than will need for implement the negation operator on OBDD (Order Binary Decision Diagram) that is, show that:
$\qquad \displaystyle \neg f(x_1,\ldots,x_n) = (\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1})$
where $f|_{x_i=b}$ is the function boolean in which replaces $x_i$ with b, that is:
$\qquad \displaystyle f|_{x_i=b}(x_1,\ldots,x_n)=f(x_1,\ldots,x_{i-1},b,x_{i+1},\ldots,x_n)$.
The proof says:
$\qquad \displaystyle\neg f(x_1,\ldots,x_n) = \neg((\neg x_1 \wedge f|_{x_1=0}) \vee (x_1 \wedge f|_{x_1=1}))$.
Applying the negation (skip the intermediate steps), we get:
$\qquad \displaystyle (x_1 \wedge \neg x_1) \vee (\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1}) \vee (\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$.
Now $(x_1 \wedge \neg x_1)= \mathrm{false}$ can be dropped, which leads to
$\qquad \displaystyle (\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1}) \vee (\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$
which in turn is, finally, equal to
$\qquad \displaystyle (\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1})$.
Why does this hold?
Maybe it is better to understand if you make a truth table for both functions. You can see that if $(\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ is true then $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1})$ is also true, because both $f$ terms are true and either $x_1$ is true or $\neg x_1$ is true. And if $(\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ is false then you have to do a case study:
• If both $f$ terms are false then is $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1})$ also false
• If exactly one $f$ term (w.l.o.g. $\neg f|_{x_1=0}$) is false then it is equivalent to $(x_1 \wedge \neg f|_{x_1=1})$
Therefore, both formulas $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1}) \vee (\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ and $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee (x_1 \wedge \neg f|_{x_1=1})$ are equivalent. | 922 | 2,185 | {"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} | 3.609375 | 4 | CC-MAIN-2024-18 | latest | en | 0.557767 |
http://math.stackexchange.com/questions/12408/is-a-regular-sequence-ordered | 1,469,647,133,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00069-ip-10-185-27-174.ec2.internal.warc.gz | 154,162,466 | 18,490 | # Is a regular sequence ordered?
A regular sequence is an $n$-fold collection $\{r_1, \cdots, r_n\} \subset R$ of elements of a ring $R$ such that for any $2 \leq i \leq n$, $r_i$ is not a zero divisor of the quotient ring $$\frac R {\langle r_1, r_2, \cdots, r_{i-1} \rangle}.$$
Does the order of the $r_i$'s matter? That is, is any permutation of a regular sequence regular?
-
Not in general. A standard example is $R=k[x,y,z]$, where $k$ is a field. Then, $x,y(1-x),z(1-x)$ is regular but $y(1-x),z(1-x),x$ is not.
On the bright side, if $R$ is Noetherian, local then every permutation of a regular sequence is regular.
In fact more is true. We can extend this notion to modules over rings analogously. Then, if $M$ is a finitely generated module over a Noetherian, local ring, then every permutation of a regular sequence is regular.
-
You beat me by one minute! – Akhil Mathew Nov 29 '10 at 23:46
@Akhil: and with the same example :) – Timothy Wagner Nov 29 '10 at 23:52
In general, no. If $R$ is a local noetherian ring, then yes, though.
A counterexample is if $R = k[x,y,z]/(x-1)z$ over a field $k$, and the elements $x, (x-1)y$. This is a regular sequence even though $(x-1)y$ is a zerodivisor (so the swapped sequence is not regular).
It is true under local hypotheses by the Krull intersection theorem; see
http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/
-
If $R$ is a local non-Noetherian integral domain the answer is still no! See M. Hochster - Pathological maximal R-sequences in quasilocal domains, Portugaliae Mathematica V. 38, Fasc. 3-4 (1979), 33-36. – user26857 Jun 17 '12 at 0:17
Here is a general result for when any permutations of elements of a regular sequence forms a regular sequence:
Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module. If $x_1,...,x_n$ be an $M$-sequence s.t. $x_i \in J(A)$ for $1 \leq i \leq n$, where $J(A)$ is the Jacobson radical of $A$, then any permutation of $x_1,...,x_n$ becomes an $M$-sequence.
- | 639 | 2,012 | {"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} | 3.421875 | 3 | CC-MAIN-2016-30 | latest | en | 0.77442 |
https://www.roulettelife.com/index.php/topic,1052.0.html | 1,571,592,702,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986717235.56/warc/CC-MAIN-20191020160500-20191020184000-00150.warc.gz | 1,048,019,250 | 12,893 | ### Author Topic: Mathematical questions (Read 4671 times)
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#### BlueAngel
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##### Mathematical questions
« on: July 14, 2016, 07:52:00 PM »
Everyone is welcome to contribute in this topic, but I'd be particulary interested for "Bayes" response.
So here we go:
1) If we could win with 12 numbers VS 7 numbers which make us lose, would that ratio be better than 24 numbers to win VS 13 numbers to lose?
The payout for both situations remains the same, 2 units at risk with 1 unit profit.
2) About EC's, if we could avoid all losing streaks, from 2 loses in a row and more, in exchange for every 1st win, would that condition be an improvement?
Does it make any difference from the mathematical point of view, or it's about the same?
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#### BlueAngel
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##### Re: Mathematical questions
« Reply #1 on: July 14, 2016, 08:11:51 PM »
Second part of the mathematical questions:
If we assume that the minimum for any EC bet is 25 hits out of 100 outcomes, then would it be correct to assume that by adding 2 more EC bets, a total of 3 EC's simultaneously for 100 outcomes would reduce the deviation?
Example, 25 appearances in 100 results is 25%, deduct that from 48.65% which is the average expectation and we arrive to 23.65% deviation.
Round it up to 24% and divide it by 3 (3 EC's) and we arrive to 8%.
8% multiplied by 300 bets equals 24 more losing bets than the winning ones, in other words 138 wins VS 162 loses
If the above statement has no flaws, then it could be possible to create a progression based on this minimum expectation , thus enabling us to be permanent winners even when the worst happens.
Is it valid according your point of view or not?
#### kav
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##### Re: Mathematical questions
« Reply #2 on: July 14, 2016, 09:57:00 PM »
Your way of thinking/calculating on the second post is wrong.
What I did find interesting is your first question
Quote
If we could win with 12 numbers VS 7 numbers which make us lose, would that ratio be better than 24 numbers to win VS 13 numbers to lose?
The payout for both situations remains the same, 2 units at risk with 1 unit profit.
This is a question I pondered while designing my new bet selection.
The answer is that at first sight 12 numbers vs 7 numbers is actually worse than 24 vs 13 numbers.
Let's calculate the expected value.
12 - (7x2) = -2 24 - (13x2) = -2
Both have an expected value of -2, but in the first case the disadvantage is 1 number in 19 numbers, or 5%, while in the second case it is 1 number in 37 numbers, or 2,7%.
However there is some benefit in the 12 vs 7 numbers case. If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
So the 12 vs 7 numbers approach is worse in expected value but better in variance. To put it in another way, if two players played the 2 versions for 100 spins, then the one who played the 12 vs 7 has more chances to be in loss than the other player, but he also has more chances that his loss will be lower that the other player's loss.
#### BlueAngel
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##### Re: Mathematical questions
« Reply #3 on: July 14, 2016, 10:06:43 PM »
Quote
Your way of thinking/calculating on the second post is wrong.
Why?
Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?
#### kav
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##### Re: Mathematical questions
« Reply #4 on: July 14, 2016, 10:18:26 PM »
Quote
Your way of thinking/calculating on the second post is wrong.
Why?
Because of this:
"divide it by 3 (3 EC's) and we arrive to 8%"
It doesn't work that way.
Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?
I don't understand the connection between my quote and your conclusion.
#### BlueAngel
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##### Re: Mathematical questions
« Reply #5 on: July 14, 2016, 10:40:23 PM »
Quote
Your way of thinking/calculating on the second post is wrong.
Why?
Because of this:
"divide it by 3 (3 EC's) and we arrive to 8%"
It doesn't work that way.
Quote
If the rest 18 numbers (37-19=18) produce neither profit nor loss, then overall this scenario will offer lower variance than the 24 vs 13 numbers case.
If so, then it would be sound to devise a progression which puts you on the positive by aiming equal amount of wins VS loses.
Wouldn't it?
I don't understand the connection between my quote and your conclusion.
About the first leg of my question, you are of the opinion that the worst EC situation (25/100) could happen on all 3 EC's at the same time?!
Sorry but it doesn't make sense if this is what you mean!
You said that 12 VS 7 ratio has lower variance in comparison with 24 VS 13, right?
So a progression which aims in equal proportion of wins VS loses would stand better chance rather than a progression which tries to handle greater variance like 24 VS 13.
Especially when the aim of the progression is EQUAL proportion of wins VS loses while the winning situations are NOT EQUAL (12 VS 7) which means roughly 65% win chance against 35% losing chance.
What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!
#### kav
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##### Re: Mathematical questions
« Reply #6 on: July 15, 2016, 12:17:34 AM »
Quote
What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!
BA,
I think there is no point to continue the discussion because you lack some basic understanding.
#### BlueAngel
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##### Re: Mathematical questions
« Reply #7 on: July 15, 2016, 12:40:26 AM »
Quote
What you have to realize is that is one thing to say 1 out of 2 and completely another 50 out of 100 possibilities, proportion may be the same BUT variance increases as the total does!
BA,
I think there is no point to continue the discussion because you lack some basic understanding.
If you believe otherwise just let us hear it.
Personally, I'm reinforcing my objections with undeniable evidence, what do you have to deny my point?
Nothing, it's just your word against mine.
So why don't you enlighten me??
After all you could be correct and me wrong, BUT that has to be proved!
Where is the epicentre of your negativity towards my theory??
#### Bayes
##### Re: Mathematical questions
« Reply #8 on: July 15, 2016, 08:04:10 AM »
Hi BA,
I agree with Kav regarding the first part of your question, and as to the second part, your analysis is a little simplistic because variance is a non-linear function, so you can't just subtract and divide as you've done.
You could find the number of wins in 100 bets, given the same deviation and betting on all 3 EC's, like this:
First find the standard deviation for 25 wins in 100 bets on one EC, using the formula for the z-score:
Z = (w - np) / SQRT(npq)
w = no. of wins
n = no. of bets
p = probability
q = 1 - p
So for a single EC, Z = (25 - 0.4865 * 100) / SQRT(100 * 0.4865 * 0.5135) = -4.73 standard deviations below the mean
Now re-arrange the formula because we want the number of wins for the same Z score when betting on all 3 EC's (27 numbers). The probability for 27 numbers is 27/37 ~ 0.73
w = np + Z * SQRT(npq)
= 100 * 0.73 - 4.73 * SQRT(100 * 0.73 * 0.27) = 52.
So this would be the "worst case" number of wins if betting on all 3 EC's.
But as you know, there's always a trade-off between the number of wins and the amount you have to put down in order to recoup any losses, so betting more numbers doesn't really help, and it just increases your exposure to the house edge.
Quote
2) About EC's, if we could avoid all losing streaks, from 2 loses in a row and more, in exchange for every 1st win, would that condition be an improvement?
This is ambiguous, can you clarify? What do you mean by "in exchange for every first win"?
« Last Edit: July 15, 2016, 08:27:33 AM by Bayes »
The following users thanked this post: kav, december, BlueAngel
#### BlueAngel
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##### Re: Mathematical questions
« Reply #9 on: July 15, 2016, 11:33:38 AM »
Quote
This is ambiguous, can you clarify? What do you mean by "in exchange for every first win"?
I mean to use stops after every loss and restarting after every 1st virtual win.
« Last Edit: July 15, 2016, 11:38:33 AM by BlueAngel »
#### Bayes
##### Re: Mathematical questions
« Reply #10 on: July 15, 2016, 12:26:38 PM »
Ok, so does this mean that you stop after every loss and only restart on the first "virtual" win? I.e. betting on red:
R (vw) bet on R
R W
R W
B L (wait for vw)
B (no bet)
R (vw. bet on R)
B L (wait for vw)
R (vw, bet on R)
B L
etc
#### BlueAngel
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##### Re: Mathematical questions
« Reply #11 on: July 15, 2016, 12:41:56 PM »
Ok, so does this mean that you stop after every loss and only restart on the first "virtual" win? I.e. betting on red:
R (vw) bet on R
R W
R W
B L (wait for vw)
B (no bet)
R (vw. bet on R)
B L (wait for vw)
R (vw, bet on R)
B L
etc
yes
#### Bayes
##### Re: Mathematical questions
« Reply #12 on: July 15, 2016, 07:33:13 PM »
The problem with the scheme is that although you catch every streak of 2 or more, these only account for 50% of the total EC outcomes, the rest being singles, and that's where you get clobbered, because every choppy sequence will result in continuous losses.
Generally speaking you can't expect any virtual wins or losses to either give you extra wins or eliminate losses because what you gain on the swings you lose on the roundabouts (something that's usually overlooked by adherents). Like any other bet selection it can be worth trying for a while to mix things up, but all it really does is to redistribute your wins and losses to different patterns, it doesn't give you something for nothing (or at least, nearly nothing. ).
#### BlueAngel
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##### Re: Mathematical questions
« Reply #13 on: July 15, 2016, 09:32:24 PM »
The problem with the scheme is that although you catch every streak of 2 or more, these only account for 50% of the total EC outcomes, the rest being singles, and that's where you get clobbered, because every choppy sequence will result in continuous losses.
Generally speaking you can't expect any virtual wins or losses to either give you extra wins or eliminate losses because what you gain on the swings you lose on the roundabouts (something that's usually overlooked by adherents). Like any other bet selection it can be worth trying for a while to mix things up, but all it really does is to redistribute your wins and losses to different patterns, it doesn't give you something for nothing (or at least, nearly nothing. ).
I was expecting to hear this, in theory every sequence has the same probability but from my empirical viewpoint, please explain me why I'm under the impression that this sequence: R B R B R B R B R B R B is less frequent than this one: R R R R R R R R R R R R or that one: B B B B B B B B B B B B.
Is there any sensible explanation or it's just my idea that alternating EC's are more rare than series of the same EC ??
Another one of my empirical fallacies is that I've observed columns to be more "choppy" than dozens which I consider them more streaky.
Again in theory it doesn't have any difference, 12 numbers one side vs 12 numbers the other, but does anybody else can confirm such observations?
Regarding the comparison, a fact is that 2 out of 3 dozens contain only "low" or "high" numbers, on the other side, none of the 3 columns contain only numbers from 1 EC without its opposing EC.
This could be an explanation, but why the same group of numbers (EC's) to be more streaky, i theory every number has the same chances, yet again we have witness again and again what we call unequal distribution, in other word some numbers to appear more than their probability while some others less.
I believe probability theory cannot satisfy all the questions completely, or perhaps an explanation regarding unequal distribution lays on the wheel's layout.
I'm not affirmative, just assuming that this could be an explanation.
Last but not least, please could you answer me the following: according my calculations I've concluded that 30 numbers can never be less than 72 appearances in 100 outcomes, could you confirm this?
#### scepticus
##### Re: Mathematical questions
« Reply #14 on: July 16, 2016, 04:33:55 AM »
You are right BA to think that “chops” are rarer than non- chops. Consider R/B s as Sets of three.
RB…RB….RB…RB
RB…RB….BR…BR
RB…BR….BR…RB
Your “CHOPS” occur in only two of the possible 8 .
We still have the problem of making use of this “information” though .
Each of the 8 have an equal chance over 3 spins.
Which to choose ?
Murphy’s Law states that whichever one you choose it is the wrong one !
You could even choose BOTH Streaks and Chops !
« Last Edit: July 16, 2016, 04:36:05 AM by scepticus » | 3,872 | 14,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-43 | latest | en | 0.91518 |
https://stats.stackexchange.com/questions/412968/single-group-pre-post-mcnemar-table-set-up | 1,597,439,210,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739370.8/warc/CC-MAIN-20200814190500-20200814220500-00520.warc.gz | 466,296,375 | 34,870 | # Single Group Pre-Post McNemar Table Set Up
I am looking to evaluate if the change in proportion for a group changed pre-to-post intervention.
Essentially, is the ~7% increase from pre (~2%) to post (~9%) significant. I am running the McNemar test in SAS 9.3 as follow:
ODS HTML CLOSE; ODS HTML;
PROC FREQ DATA = AGE0_10_SUM ORDER = DATA;
WEIGHT MEMCOUNT;
TABLES AGE0_10*SRC / AGREE;
TITLE "AGE0_10";
RUN;
;QUIT;
This is resulting in a p-value less than 0.0001. I think the change is significant but not at this magnitude, also, when running other age groups the results for a 0.02% increase was also <0.0001. Something is wrong.
How should the table be set-up to run the McNemar test with AGE0_10 as being the repeated subject pre and post.
Thank you.
You have very large denominators, so you should not be surprised if quite small differences give low p-values.
One should be careful to not confuse p$$\leq$$0.05 with practical relevance or the magnitude of an effect. Differences that are completely irrelevant for any practical purposes can be "statistically significant" with enough data.
The language you use of "repeated subject" does not seem to make a lot of sense in connection to a variable that seems to have two levels and does not sound like it would indicate a particular individual person/experimental unit.
For McNemar's test, your data should be structured like this with each row giving the number of units (Number) with result of assessment 1 = Result1 and the result of assessment 2 = Result2):
Result1 Result2 Number
0 0 900
0 1 100
1 0 100
1 1 900
Other note: pre-/post-tests are typically irrelevant for (almost) any question a researcher truly wants to answer (particularly, if they are to see whether some intervention/change/something had an effect).
• Thank you. You are right. The variable isn’t repeated. There is no way the data can be structured in the table you describe. What is a more appropriate way to understand if the change in distribution of that group is different in the post period? Should I take the difference and test if the difference is equal to 0? – Ronald Sanchez Jun 14 '19 at 11:53
• If you do not have the necessary information, then you cannot use McNemar's test, in fact you might have a bit of a problem, because if you assess the same subjects at two time points, you will have a correlation between the assessments, which you have trouble taking into account, if you do not know how the observations line up within subject. There might be some solution for that, but it's not immediately obvious to me. On the other hand, if these are independent records from totally separate subjects, this would be easier. – Björn Jun 14 '19 at 11:53
• Hmmm, ok. This is the “intervention”. We placed a social worker inside a doctors office. We are evaluating the change in volume of mental health and behavioral health services received at the clinic. First, we need to understand if the demographics (e.g. age groups at different bands) changed before and after the social worker was hired. So there will be patients In both time periods and then new patients in the post period. Thoughts in assessing the change in the post period for age groups? – Ronald Sanchez Jun 14 '19 at 12:30
• Another thought, could I compare the percent change in the total population (171% increase) to the target group (92%) to assess if the subgroup increased proportionally with the general population, the H-null would be that 171% is not different than 92%. Thoughts? – Ronald Sanchez Jun 14 '19 at 13:24
• a) There is uncertainty around 171% and 92%, and the two are presumably not independent (overlap in populations). And you do not really know the uncertainty around a change, because you do not know how much is existing patients changing vs. different people being in the population (at least it seems so from your comment that you cannot get the data into the format suggested). – Björn Jun 14 '19 at 13:53 | 942 | 4,038 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.926129 |
https://chestofbooks.com/architecture/Building-Construction-V4/Roofing-Continued.html | 1,702,272,046,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00723.warc.gz | 185,796,082 | 5,241 | ## 27. Slating
Slating. In measuring slating, the method of determining the number of slates required per square is similar to that given for shingling; but in slating, each course overlaps but two of the course below, instead of three, as in shingling. The usual lap, or cover of the lowest course of slate by the uppermost of the three overlapping courses, is 3 inches; hence, to find the exposed length, deduct the lap from the length of the slate, and divide the remainder by 2.
The exposed area is the width of the slate multiplied by this exposed length, and the number of slates required per square is found by dividing 14,400 by the exposed area of 1 slate.
Thus, if 14"x20" slate are to be used, the exposed length will be 20-3/2 = 8 1/2 inches; the exposed area will be 14 X 8 1/2
= 119 square inches, and the number per square will be 14,400÷119 = 121 slates.
The following rules should be observed in measuring slating: Eaves, hips, valleys, and cuttings against walls are measured extra, 1 foot wide by their whole length-the extra charge being made for waste of material and the increased labor required in cutting and fitting. Openings less than 3 square feet are not deducted, and all cuttings around them are measured extra. Extra charges are also made for borders, figures, and any change in color of the work, and for steeples, towers, and perpendicular surfaces.
The following table, based on 3 inches lap, gives the sizes of the American slates, and the number of pieces required per square: | 366 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.956349 |
https://mql4tradingautomation.com/pip-pipette-forex-trading/ | 1,600,733,526,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00139.warc.gz | 535,839,858 | 27,442 | # What is a PIP (and a PIPETTE) in Forex Trading?
When you start learning about Forex Trading one of the first concepts you will see is the concept of PIP. In this article we are going to see what is a PIP and what is a PIPETTE in Forex Trading. We will also see when to use these terms and how to caluclate the PIP Value. Last but not least we will see in some useful MQL4 function related to PIPS.
### What is a PIP, and a PIPETTE, in Forex Trading
By definition the PIP is the Price Interest Point. A PIP is the unit of measure for the change of value in the exchange rate of two currencies. For currency pairs with 4 decimals 1 pip = 0.0001 while for YEN based pair it is 1 pip = 0.01. Complex? A couple of examples will make it clear.
Assume that the EURUSD is exchanged at 1.0799 meaning that with 1 EUR you buy 1.0799 USD, if the rate changes to 1.0795 it means that there is a difference of 1.0799-1.0795=0.0004 that is 4 pips. If the USDJPY is quoted 107.38, meaning that with 1 USD you can buy 107.38 Yen, and then it changes to 107.40, this results in a price change of 107.40-107.38=0.02 which is 2 pips.
Due to the evolution of technology, increased popularity of online trading platforms and higher volume of trades, Brokers have started to present the exchange rates with 5 decimals for normal pair and 3 decimals for JPY pairs. This has resulted in the introduction of a new term: PIPETTE. A Pipette is a 1/10 of a PIP and usually the last decimal (the 5th for normal pairs and 3rd for JPY pairs) in the exchange rate. 10 Pipettes=1Pip.
### When to use the terms PIP and PIPETTE
PIP is still the most used in the daily Forex Trading jargon. Usually PIP and PIPETTE are used to:
• Express the spread, example, “the spread is 3 pips”, meaning the difference between the ask and the bid price is 3 pips
• Express a price change, “the price has dropped of 120 pips”
• Express a gain or profit, “I made 40 pips with that trade”
• Express a loss, “I lost 50 pips”
• Express the distance between open price and take profit or stop loss price, “Stop loss 30 pips and take profit 60 pips”, meaning the stop loss price will be 30 pips from the open price and the take profit 60 pips away
### How to Calculate the PIP value
Calculate the PIP value is a necessary step to understand the profit/loss of a trade and to implement strategies of risk management.
The PIP value is always calculated for a size or volume, usually a standard lot (100.000 units), a mini lot (10.000 units) or micro lot (1.000 units). The first step to calculate the pip value is to multiply the volume for 0.0001 (0.01 in case of JPY pairs), this will give you the value in the QUOTE currency. Example, in a trade of 20.000 EURUSD I multiply 20.000×0.0001=2 which is the PIP Value in USD, in other words, 1 pip of change in the exchange rate is a profit/loss of 2 USD. In case of 30.000 of USDJPY the PIP value would be 30.000*0.01=300 JPY. Having the pip value in the quote currency is not always useful, usually you want to have the value expressed in your base currency of the account . Once you have the PIP value in a currency you just need to convert in your currency, so in the second example the 300 JPY of PIP value can be converted in USD knowing the USDJPY exchange rate (assume USDJPY 107.40), in fact 300 JPY / 107.40 USDJPY = 2.79 USD per pip.
You can see the pip value in use in the article about the position size.
You can also download a simple script to quickly see the Default lot size and PIP value for the selected Pair
[wpdm_package id=’832′]
### MQL4 and PIP Value, MODE_TICKVALUE – MODE_TICKSIZE – MODE_TICKVALUE
If you are learning about MQL4 and trading automation with MetaTrader 4 you will be happy to know that MQL4 has some native functions to retrieve the PIP Value and the default lot size.
We already introduced the concept of TICK in this article. Basically the tick is the smallest change that can have a price. MQL4 includes the following information natively:
• MarketInfo(Symbol(), MODE_LOTSIZE) gives you the default lot size, so you know what is the tick value referred to
• MarketInfo(Symbol(), MODE_TICKVALUE) is the tick value for the selected currency (Symbol()) in the account currency
• MarketInfo(Symbol(), MODE_TICKSIZE) is the size of the tick, in all modern brokers it is a pipette
So if we run the following code we will get all the information we need. Pay attention as it shows the TICK value and this is not necessarily equal to the PIP value. If the Tick size is a PIPETTE then the Tick Value will be the value of a pipette.
### Conclusion
Understanding the concepts of PIP and PIPETTE is required if you want to trade Forex. You will need to understand the meaning, the common jargon and the importance of PIP value.
If anything is not clear or need any help please feel free to comment or contact us.
I would really appreciate also suggestions for improvements and reports of any bug found. Thank you!
### 3 thoughts on “What is a PIP (and a PIPETTE) in Forex Trading?”
1. HEH u rock Man..this article helped me a lot as i m a newbee on forex..would like to kw wht exacly is 4 & 5 here which u hwe specified..
string TickValue=DoubleToStr(MarketInfo(Symbol(),MODE_TICKVALUE),4);
string TickSize=DoubleToStr(MarketInfo(Symbol(),MODE_TICKSIZE),5);
& if i dont specify 4 & 5 wht happens??
• Thank you for your comment tasaoirse!
DouebleToStr basically converts a variable of double type to string, 4 and 5 are the number of digits to keep after the decimal point, to give you an example
void OnStart(){
Print(DoubleToStr(12.123654789,4)); //returns 12.1237 (because it rounds it)
Print(DoubleToStr(12.123654789,5)); //returns 12.12365
Print(DoubleToStr(12.123654789)); //returns 12.12365479 (the default parameter if not specified is 8)
}
Does that make sense? | 1,545 | 5,828 | {"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} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.932477 |
https://www.nctm.org/Classroom-Resources/Illuminations/Lessons/Puzzling-Relationships/ | 1,726,100,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00331.warc.gz | 847,886,844 | 33,310 | Puzzling Relationships
• ## Puzzling Relationships
Periods: 2
Author: Carol Midgett
### Instructional Plan
Display the Tangram Puzzle Activity Sheet for the whole class to see and ensure that each student has access to the activity sheet and tangrams (cardstock is recommended).
Model how to fill the puzzles with the tangram pieces. For example, the Tangram School can be filled as shown below.
Discuss how it is necessary to turn over, rotate, or slide pieces to complete a puzzle by attending to precise vocabulary.
Have the students work in pairs to complete the tangram puzzles. Encourage them to discuss with each other what needs to be done to fit the pieces to match the outline of the puzzle. (This exercise focuses students' attention on the transformations required to make pieces fit. A turnover, rotation, or slide of pieces may be necessary to complete the puzzle.) This discourse facilitates students' problem solving and reasoning about the process. It connects actions with vocabulary and assists the students in relating what happens with concrete and virtual tools.
As a wrap up activity, pairs of students can share their work (what was most difficult, what feature they used the most, how long it took them, etc.).
### Assessments and Extensions
Assessment Options
1. By completing tangram puzzles, the students should demonstrate the ability to do the following:
• Interpret relative positions in space and apply ideas about relative position.
• Describe and interpret direction and distance in navigating space, and apply ideas about direction and distance.
2. Making notes about what the students do and do not understand is helpful when planning future instruction for individuals, groups of students, and the class as a whole. Collecting this data for future reference helps you understand the impact of your teaching strategies on students' learning. You may use the Class Notes recording sheet for this purpose.
Extensions
The students who find it easy to complete the puzzles could create new ones for the class to solve.
### Questions and Reflections
Questions for Students
1. What did you find when you tried to place the puzzle pieces in place?
2. What happens when you turn (or rotate) a puzzle piece?
3. What happens when you turn over a puzzle piece to fit it into place?
4. How many different ways did you find to fill the puzzle? What did you do to fill the puzzles in different ways? (This question focuses students' attention on different orientations of pieces and whether pieces must be turned over, rotated, or slid to fill the puzzle.)
5. What did you learn from other students as you watched them put their puzzle together?
6. If you were teaching this lesson, what would you tell your classmates about how to put the puzzle together?
7. Why is it necessary to turn over, rotate, or slide the tangram pieces to make them fit?
8. How did you decide which triangle would fit into the vacant space?
9. Which shape(s) caused you the most trouble? Why?
Teacher Reflection
• Which students demonstrate an understanding of how to solve puzzles by turning over, rotating, and sliding puzzle pieces? Which did not? What caused them difficulty?
• What other learning opportunities do these students need? What instructional strategies would provide these learning experiences?
• Were some students able to use more than one strategy to complete the puzzles?
• Which students were able to interpret and use relative position in space as a problem-solving strategy?
• Were all students able to describe how they solved puzzles? If not, what would help them do so?
• Which students demonstrate an understanding of the relative sizes among the tangram pieces by identifying the varying sizes of the triangles as small, medium, or large? For those who do not, what instructional experiences are needed?
• Which students do not realize that rotating or sliding changes the orientation of shapes? What experiences would help them learn this?
• Which students realize that turning over pieces can change the orientation of shapes? If not, what experiences would help them learn this?
### Objectives and Standards
Students will:
• Interpret relative positions in space and apply ideas about relative position.
• Describe and interpret direction and distance in navigating space and apply ideas about direction.
• Use tangram puzzles to experience turns, rotations, and slides.
Common Core State Standards – Mathematics
9th to 12th
• Stats & Probability
• CCSS.Math.Content.HSS-ID.A
Summarize, represent, and interpret data on a single count or measurement variable
Pre K to 2nd
• Kindergarten
• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure.
Pre K to 2nd
• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure.
Pre K to 2nd
• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure.
Common Core State Standards – Practice
• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure. | 1,212 | 5,715 | {"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} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.902021 |
https://forum.yoyogames.com/index.php?threads/is-there-a-faster-way-to-draw-a-thick-line-when-using-draw_path.15204/ | 1,586,372,064,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00344.warc.gz | 477,838,835 | 9,790 | # Is there a faster way to draw a thick line, when using draw_path();?
O
#### Olivebates
##### Guest
Hi!
I want to draw a thick line as a path.
Right now I'm using the following code.
Is there any way to optimize this?
I think it's making my game lag to draw this many paths, and it also looks bad.
Code:
``````// Draw line
draw_set_color(c_red);
draw_path(ip, ix, iy, false);
draw_path(ip, ix+.25, iy, false);
draw_path(ip, ix-.25, iy, false);
draw_path(ip, ix, iy+.25, false);
draw_path(ip, ix, iy-.25, false);
draw_path(ip, ix+.5, iy, false);
draw_path(ip, ix-.5, iy, false);
draw_path(ip, ix, iy+.5, false);
draw_path(ip, ix, iy-.5, false);
draw_path(ip, ix+.75, iy, false);
draw_path(ip, ix-.75, iy, false);
draw_path(ip, ix, iy+.75, false);
draw_path(ip, ix, iy-.75, false);
draw_path(ip, ix+1, iy, false);
draw_path(ip, ix-1, iy, false);
draw_path(ip, ix, iy+1, false);
draw_path(ip, ix, iy-1, false);
//Draw outline
draw_set_color(c_black);
draw_path(ip, ix+1.25, iy, false);
draw_path(ip, ix-1.25, iy, false);
draw_path(ip, ix, iy+1.25, false);
draw_path(ip, ix, iy-1.25, false);
draw_path(ip, ix+1.5, iy, false);
draw_path(ip, ix-1.5, iy, false);
draw_path(ip, ix, iy+1.5, false);
draw_path(ip, ix, iy-1.5, false);``````
Thanks!
A
#### anomalous
##### Guest
A few ways.
In general, you should look at the path functions for getting the path points.
Loop through each point, and draw a line between each path point. This is the core code.
You can then draw the line using draw_line_width in one pass, to draw a thicker line.
Also, draw_line_width_colour can get you some interesting gradient effects.
For a fancier look and additional optimization, you can create a sprite that can be stretched without changing how it looks.
Then draw this stretched between those two points, its likely a bit faster than the draw primitive line.
If the paths do not change, for each path you can create a surface and draw the sprites/lines to the surface, then draw the surface to the screen. This has tradeoffs/overhead, but if its a lot of paths, and they don't update every step, this could save even more.
#### Kubawsky
##### Member
Anomalous, what if the path is not a straight line and also changes shape (possibly at every frame).
Any ideas?
#### Kubawsky
##### Member
For a fancier look and additional optimization, you can create a sprite that can be stretched without changing how it looks.
Can you explain what you mean by this?
Such a solution would work for me - other solutions are too heavy as I'm scaling and shifting the path every step and then drawing it multiple times, every step
I made a test with a sprite that is shaped like my path, it is a little CPU-heavy but doesn't lag the game. But obviously stretching it disproportionately looks bad - that's why my interest in your comment
If draw_path had a width parameter, that would solve my problem... | 779 | 2,890 | {"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-2020-16 | latest | en | 0.77666 |
https://madanswer.com/8037/The-g-c-d-of-1-08-0-36-and-0-9-is | 1,618,084,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00033.warc.gz | 478,126,477 | 8,888 | # The G.C.D. of 1.08, 0.36 and 0.9 is:
in Other
Q:
The G.C.D. of 1.08, 0.36 and 0.9 is:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18. | 100 | 195 | {"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-2021-17 | latest | en | 0.823419 |
http://pawfleet.com/site/bull-country-ashiy/article.php?id=2ccdde-how-to-find-the-range-of-a-quadratic-function | 1,623,558,840,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487600396.21/warc/CC-MAIN-20210613041713-20210613071713-00536.warc.gz | 41,844,162 | 17,998 | # how to find the range of a quadratic function
by Mometrix Test Preparation | Last Updated: March 20, 2020. For example: $$fx=a(x-b)(x-c)$$. Graphs can be helpful, but we often need algebra to determine the range of quadratic functions. For example, say you want to find the range of the function $$f(x) = x + 3$$. This is a property of quadratic functions. x cannot be 0 because the denominator of a fraction cannot be 0). Quadratic function has exactly one y-intercept. RANGE OF A FUNCTION. In order to find a quadratic equation from a graph using only 2 points, one of those points must be the vertex. Example 1. Physics. Once we know the location of the vertex – the x-coordinate – all we need to do is substitute into the function to find the y-coordinate. For example, consider this function: $$\frac{-b}{2a}=\frac{-8}{2(-2)}=\frac{-8}{-4}=2$$. When "a" is negative the graph of the quadratic function will be a parabola which opens down. We’re going to plug it into our original equation: $$f(-1)=-23-3=18$$. If a >0 a > 0, the parabola opens upward. To see the domain, let’s move from left-to-right along the x-axis looking for places where the graph doesn’t exist. We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best. The quadratic parent function is y = x2. Since a is negative, the range of all real numbers is less than or equal to 18. This topic is closely related to the topic of quadratic equations. y = ax2 + bx + c, we have to know the following two stuff. Sometimes, we are only given an equation and other times the graph is not precise enough to be able to accurately read the range. Example 2: Find the inverse function of f\left( x \right) = {x^2} + 2,\,\,x \ge 0, if it exists.State its domain and range. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. We would say the range is all real numbers greater than or equal to 0. Other Strategies for Finding Range of a function . How to find the range of a rational function How to sketch the graph of quadratic functions 4. Example $$\PageIndex{4}$$: Finding the Domain and Range of a Quadratic Function. y-intercept for this function . On the other hand, functions with restrictions such as fractions or square roots may have limited domains and ranges (for example $$fx=\frac{1}{2x}$$. Let’s see how the structure of quadratic functions defines and helps us determine their domains and ranges. To find the x-coordinate use the equation x = -b/2a. If a quadratic function opens up, then the range is all real numbers greater than or equal to the y-coordinate of the range. not transformed in any way). Determine whether $a$ is positive or negative. The domain of a function is the set of all real values of x that will give real values for y . Donate or volunteer today! We need to determine the maximum value. To determine the domain and range of any function on a graph, the general idea is to assume that they are both real numbers, then look for places where no values exist. The range is all the y-values for which the function exists. When x = − b 2 a, y = c − b 2 4 a. Since domain is about inputs, we are only concerned with what the graph looks like horizontally. Email. In fact, the domain of all quadratic functions is all real numbers! Chemistry. Let’s generalize our findings with a few more graphs. As with the other forms, if a is positive, the function opens up; if it’s negative, the function opens down. Our mission is to provide a free, world-class education to anyone, anywhere. We’ll use a similar approach, but now we are only concerned with what the graph looks like vertically. This same quadratic function, as seen in Example 1, has a restriction on its domain which is x \ge 0.After plotting the function in xy-axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. Let us see, how to know whether the graph (parabola) of the quadratic function is … Because $$a$$ is negative, the parabola opens downward and has a maximum value. If a quadratic function opens down, then the range is all real numbers less than or equal to the y-coordinate of the range. To know the range of a quadratic function in the form. The range of a quadratic function is either from the minimum y-value to infinity, or from negative infinity to the maximum v-value. Learn how to graph quadratics in standard form. This is basically how to find range of a function without graphing. As with standard form, if a is positive, the function opens up; if it’s negative, the function opens down. The Basic of quadratic functions 2. 03:57. Quadratic functions generally have the whole real line as their domain: any x is a legitimate input. Finding the roots of higher-degree polynomials is a more complicated task. The domain of this function is all real numbers. The x-intercepts are at -4 and 2 and the vertex is located at $$\frac{-4+2}{2}=-1$$ (simply take the “average” of the x-intercepts). range f ( x) = cos ( 2x + 5) $range\:f\left (x\right)=\sin\left (3x\right)$. Graphs can be helpful, but we often need algebra to determine the range of quadratic functions. There are three main forms of quadratic equations. range y = x x2 − 6x + 8. (c) Find the range of values of y for which the value x obtained are real and are in the domain of f (d) The range of values obtained for y is the Range of the function. When quadratic equations are in standard form, they generally look like this: fx = ax2 + bx + c. If a is positive, the function opens up; if it’s negative, the function opens down. As you can see, outputs only exist for y-values that are greater than or equal to 0. The general form of a quadratic function presents the function in the form. To find the range you need to know whether the graph opens up or down. f (x)= ax2 +bx+c f ( x) = a x 2 + b x + c. where a , b, and c are real numbers and a ≠0 a ≠ 0. Solve the inequality x2 – x > 12. You can plug any x-value into any quadratic function and you will find a corresponding y-value. range f ( x) = √x + 3. The graph of a quadratic function is a parabola. The range of a function is the set of all real values of y that you can get by plugging real numbers into x. Hi, and welcome to this video about the domain and range of quadratic functions! To find y-intercept we put x =0 in the function we get. If you're seeing this message, it means we're having trouble loading external resources on our website. And finally, when looking at things algebraically, we have three forms of quadratic equations: standard form, vertex form, and factored form. Horizontally, the vertex is halfway between them. Specifically, Specifically, For a quadratic function that opens upward, the range consists of all y greater than or equal to the y -coordinate of the vertex. The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. Learn how you can find the range of any quadratic function from its vertex form. Domain and range of quadratic functions (video) | Khan Academy As you can see, there are no places where the graph doesn’t exist horizontally. $range\:f\left (x\right)=\cos\left (2x+5\right)$. When quadratic equations are in standard form, they generally look like this: fx = ax2 + bx + c. Graphing nonlinear piecewise functions (Algebra 2 level). It means that graph is going to intersect at point (0,-5) on y-axis. x-intercept: x-intercept is the point where graph meets x-axis. They are, (i) Parabola is open upward or downward. Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. Quadratic functions together can be called a family, and this particular function the parent, because this is the most basic quadratic function (i.e. Therefore the maximum or minimum value of the quadratic is c − b 2 4 a. Mechanics. Video Transcript. 1. 1) Find Quadratic Equation from 2 Points. In this video, we will explore: How the structure of quadratic functions defines their domains and ranges and how to determine the domain and range of a quadratic function. Determining the range of a function (Algebra 2 level) Domain and range of quadratic functions. Calculate x-coordinate of vertex: x = -b/2a = -6/(2*3) = -1 Video: Finding the Range of Quadratic Functions If : {−4, −1, 4, −2} [6, 25] and () = ² + 5, find the range of . a is negative, so the range is all real numbers less than or equal to 5. The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. The domain of any quadratic function as all real numbers. Learn More... All content on this website is Copyright © 2020. Google Classroom Facebook Twitter. A quadratic equation is an equation whose highest exponent in the variable(s) is 2. Continue to Page 2 (Find quadratic Function given its graph) Continue to Page 3 (Explore the product of two linear functions) More on quadratic functions and related topics Find Vertex and Intercepts of Quadratic Functions - Calculator: An applet to solve calculate the vertex and x and y intercepts of the graph of a quadratic function. Maximum Value of a Quadratic Function. If $a$ is negative, the parabola has a maximum. We can use this function to begin generalizing domains and ranges of quadratic functions. Let’s talk about domain first. The range for this graph is all real numbers greater than or equal to 2, The range here is all real numbers less than or equal to 5, The range for this one is all real numbers less than or equal to -2, And the range for this graph is all real numbers greater than or equal to -3. If a < 0 a < 0, the parabola opens downward. range f ( x) = sin ( 3x) a is positive and the vertex is at -4,-6 so the range is all real numbers greater than or equal to -6. In other words, there are no outputs below the x-axis. Our goals here are to determine which way the function opens and find the y-coordinate of the vertex. Find the vertex of the function if it's quadratic. The range of a function is the set of output values when all x-values in the domain are evaluated into the function, commonly known as the y-values.This means I need to find the domain first in order to describe the range.. To find the range is a bit trickier than finding the domain. Learn how you can find the range of any quadratic function from its vertex form. As we saw in the previous example, sometimes we can find the range of a function by just looking at its graph. $range\:y=\frac {x} {x^2-6x+8}$. Introduction to Rational Functions . If a quadratic function opens up, then the range is all real numbers greater than or equal to the y-coordinate of the range. Here’s the graph of fx = x2. The parabola can either be in "legs up" or "legs down" orientation. Finding the range of a quadratic by using the axis of symmetry to find the vertex. Chemical ... Quadratic Equations Calculator, Part 2. Example, we have quadratic function . How to find the range of values of x in Quadratic inequalities. The range of a function is the set of all real values of y that you can get by plugging real numbers into x. Rational functions are fractions involving polynomials. Graphical Analysis of Range of Quadratic Functions The range of a function y = f(x) is the set of values y takes for all values of x within the domain of f. The graph of any quadratic function, of the form f(x) = a x 2 + b x + c, which can be written in vertex form as follows f(x) = a(x - h) 2 + k , where h = - … This is the currently selected item. The maximum value is "y" coordinate at the vertex of the parabola. Determine max and min values of quadratic function 3. Domain and Range As with any function, the domain of a quadratic function f(x) is the set of x -values for which the function is defined, and the range is the set of all the output values (values of f). Lets see fee examples with various type of functions. The other is the direction the parabola opens. The vertex is given by the coordinates (h,k), so all we need to consider is the k. For example, consider the function $$fx=3(x+4)^2-6$$. The domain of a quadratic function in standard form is always all real numbers, meaning you can substitute any real number for x. So, let’s look at finding the domain and range algebraically. For example, consider the function $$fx=-2(x+4)(x-2)$$. If $a$ is positive, the parabola has a minimum. (ii) y-coordinate at the vertex of the Parabola . Khan Academy is a 501(c)(3) nonprofit organization. If you're working with a straight line or any function … How to Find a Quadratic Equation from a Graph: In order to find a quadratic equation from a graph, there are two simple methods one can employ: using 2 points, or using 3 points. The range of quadratic functions, however, is not all real numbers, but rather varies according to the shape of the curve. Sometimes quadratic functions are defined using factored form as a way to easily identify their roots. Now for the range. For example, find the range of 3x 2 + 6x -2. When quadratic equations are in vertex form, they generally look like this: $$fx=a(x-h)^2+k$$. The graph of this function is shown below. Solution. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. One way to use this form is to multiply the terms to get an equation in standard form, then apply the first method we saw. Range of quadratic functions. We can also apply the fact that quadratic functions are symmetric to find the vertex. Some functions, such as linear functions (for example fx=2x+1), have domains and ranges of all real numbers because any number can be input and a unique output can always be produced. Domain is the set of input values, while range is the set of output values. Using the quadratic formula and taking the average of both roots, the x -coordinate of the stationary point of any quadratic function a x 2 + b x + c (where a ≠ 0) is given by x = − b 2 a. We will discuss further on 4 subtopics below: 1. As with any quadratic function, the domain is all real numbers. We know the roots, and therefore, the locations of the x-intercepts. In this form, the y-coordinate of the vertex is found by evaluating $$f(\frac{-b}{2a})$$. As you can see, the turning point, or vertex, is part of what determines the range. To write the inequality in standard form, subtract both sides of the … Find the domain and range of $$f(x)=−5x^2+9x−1$$. If a quadratic function opens down, then the range is all real numbers less than or equal to the y-coordinate of the range. The structure of a function determines its domain and range. The structure of a function determines its domain and range. Before we begin, let’s quickly revisit the terms domain and range. We know that a quadratic equation will be in the form: y = ax 2 + bx + c. Our job is to find the values of a, b and c after first observing the graph. The range of a quadratic function written in standard form $$f(x)=a(x−h)^2+k$$ with a positive $$a$$ value is $$f(x) \geq k;$$ the range of a quadratic function written in standard form with a negative $$a$$ value is $$f(x) \leq k$$. range f ( x) = 1 x2. This equation is a derivative of the basic quadratic function which represents the equation with a zero slope (at the vertex of the graph, the slope of the function is zero). A rational function f(x) has the general form shown below, where p(x) and q(x) are polynomials of any degree (with the caveat that q(x) ≠ 0, since that would result in an #ff0000 function). How To: Given a quadratic function, find the domain and range. The calculator, helps you finds the roots of a second degree polynomial of the form ax^2+bx+c=0 where a, b, c are constants, a\neq 0.This calculator is automatic, which means that it … Ok, let’s do a quick review before we go. This quadratic function calculator helps you find the roots of a quadratic equation online. The graph is shown below: The quadratic function f(x) = ax 2 + bx + c will have only the maximum value when the the leading coefficient or the sign of "a" is negative. Determining the range of a function (Algebra 2 level). $range\:f\left (x\right)=\sqrt {x+3}$. Look like this: \ ( a\ ) is 2 legs up '' or up... Below: 1 minimum y-value to infinity, or vertex, is not real. Without graphing meets x-axis \ ): finding the domain of any quadratic function is the set of values! Any x-value into any quadratic function is all real numbers greater than or equal to the topic of functions! Algebra to determine which way the function \ ( \PageIndex { 4 } \ ) get... All quadratic functions defines and helps us determine their domains and ranges of functions... A 501 ( c ) ( x-c ) \ ): finding the domain the! With various type of functions 3x 2 + 6x -2 functions are defined using factored form as a way easily. Need Algebra to determine the range is the set of all real numbers is less than equal... Re going to plug it into our original equation: \ ( fx=a x-h... In Standard form is always all real numbers greater than or equal to 18 x-intercept is the of..Kastatic.Org and *.kasandbox.org are unblocked of those points must be the vertex is at -4 -6! Be how to find the range of a quadratic function parabola function in the form meaning you can see, outputs only exist for that... *.kasandbox.org are unblocked 4 } \ ): finding the domain of all real numbers x. Behind a web filter, please enable JavaScript in your browser a < 0 a > 0, the point... The equation x = -b/2a '' coordinate at the vertex 2 + 6x -2 by Test. Opens up, then the range of quadratic functions generally have the real... The graph of the parabola opens upward with what the graph doesn ’ t exist horizontally into any function... Fx=A ( x-h ) ^2+k\ ) /latex ] is positive and the.. X-Intercept: x-intercept is the set of output values education to anyone,.. To log in and use all the y-values for which the function we get 0 a >,... Mode Order minimum maximum Probability Mid-Range range Standard Deviation Variance Lower Quartile Upper Quartile range... See fee examples with various type of functions JavaScript in your browser to... Academy is a 501 ( c ) ( x-c ) \ ) ^2+k\.. On this website is Copyright © 2020 to begin generalizing domains and ranges of quadratic generally... Be helpful, but rather varies according to the maximum v-value x in! The form will be a parabola which opens down, then the range of quadratic... Roots of higher-degree polynomials is a 501 ( c ) ( x-c ) \:... Algebra to determine the range is the set of all real numbers greater than equal... Academy is a more complicated task as their domain: any x is a legitimate input while is. The x-coordinate use the equation x = − b 2 a, y = c − b 2 a. ) nonprofit organization the whole real line as their domain: any how to find the range of a quadratic function is a legitimate input highest in!, they generally look like this: \ ( f ( x ) = (! Calculator helps you find the x-coordinate use the equation x = − b 2 a, y c. In and use all the features of Khan Academy, please enable JavaScript in your browser way to identify. And range of a quadratic function in the variable ( s ) is negative the graph doesn ’ t.., y = c − b 2 4 a or negative Academy, please enable JavaScript in browser. Are only concerned with what the graph of the parabola opens upward be the vertex are in form! Range is all real numbers rather varies according to the y-coordinate of the range of a (! X = -b/2a { x+3 } $is c − b 2 4 a real number x... Revisit the terms domain and range and therefore, the domain, ’... We would say the range of what determines the range is all real numbers greater or. Which the function exists ( 3x\right )$ range\: f\left ( x\right ) =\sin\left ( 3x\right ).! Maximum or minimum value of the curve varies according to the topic of quadratic functions is real! Are, ( i ) parabola is open upward or downward of input values, while is! Revisit the terms domain and range of quadratic functions is all real numbers is less or! The form this is basically how to find the y-coordinate of the function \ ( (. Presents the function \ ( f ( x ) =−5x^2+9x−1\ ) which the function \ ( fx=-2 x+4. On 4 subtopics below: 1 they are, ( i ) parabola is open or... S look at finding the domain is the point where graph meets x-axis Algebra to determine which way the in... Output values maximum value is y '' coordinate at the vertex of the curve exponent in previous... Please enable JavaScript in your browser find y-intercept we put x =0 in previous! Value of the range | Khan Academy find the range you need to the... Since domain is all real numbers, meaning you can find the range of all real numbers -6! Topic of quadratic functions are symmetric to find y-intercept we put x =0 in the function opens up then! '' orientation of y that you can plug any x-value into any quadratic function opens down Academy is a (. Copyright © 2020 learn more... all content on this website is Copyright © 2020 … to the! Either be in legs up '' or legs down '' orientation domains and ranges y '' at! X+4 ) ( 3 ) nonprofit organization whose highest exponent in the variable ( )... Since domain is the point where graph meets x-axis to sketch the graph doesn ’ t horizontally! Quick review before we go Standard Deviation Variance Lower Quartile Upper Quartile range... The denominator of a function is all real numbers less than or equal to the y-coordinate of parabola. Function determines its domain and range algebraically is to provide a free, world-class education to anyone anywhere...: x-intercept is the set of all quadratic functions the topic of quadratic functions ( video |! Terms domain and range which way the function if it 's quadratic move left-to-right. Any function … to find the vertex is at -4, -6 so the range of values of x quadratic! Lower Quartile Upper Quartile Interquartile range Midhinge function in Standard form is always all real numbers than. Bx + c, we are only concerned with what the graph looks horizontally!, but we often need Algebra to determine the range web filter, please enable JavaScript in your browser looks. Features of Khan Academy, please enable JavaScript in your browser from left-to-right along the x-axis ( s ) negative! We go a more complicated task parabola opens downward and has a minimum for x can get by real! Its vertex form, they generally look like this: \ ( f ( x =! ( ii ) y-coordinate at the vertex is at -4, -6 so range. F ( x ) = cos ( 2x + 5 ) $range\: f\left ( x\right =\sin\left! Your browser to 18 ii ) y-coordinate at the vertex is at -4, -6 so range! Ll use a similar approach, but we often need Algebra to determine the range is all real into! Variable ( s ) is negative, so the range is all real numbers greater than or equal 5! The vertex is at -4, -6 so the range of the parabola legs down orientation... As a way to easily identify their roots function 3 similar approach, but now we are only concerned what. Often need Algebra to determine which way the function if it 's quadratic be a parabola which opens down then! 2X + 5 )$ real numbers, meaning you can find the vertex the! Opens downward and has a maximum value is y '' coordinate at the vertex is -4! S do a quick review before we go we begin, let ’ s at! Generally have the whole real line as their domain: any x is a legitimate input s the looks... | 5,665 | 23,567 | {"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": 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} | 4.59375 | 5 | CC-MAIN-2021-25 | latest | en | 0.89327 |
https://pt3mathematics.blog.onlinetuition.com.my/2018/07/9-2-1-loci-in-two-dimensions-pt3-focus-practice-3.html | 1,719,225,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00891.warc.gz | 417,233,234 | 11,149 | # 8.2.3 Loci in Two Dimensions, PT3 Focus Practice
Question 5:
Diagram below in the answer space shows a circle with centre O drawn on a grid of equal squares with sides of 1 unit. POQ is a diameter of the circle.
W, X and Y are three moving points inside the circle.
(a) W is the point which moves such that it is constantly 4 units from the point O. Describe fully the locus of W.
(b) On the diagram, draw,
(i) the locus of the point X which moves such that its distance is constantly 3 units from the line PQ,
(ii) the locus of the point Y which moves such that it is equidistant from the point P and the point Q.
(c) Hence, mark with the symbol the intersection of the locus of X and the locus of Y.
Answer:
(b)(i),(ii) and (c)
Solution:
(a) The locus of W is a circle with the centre O and a radius of 4 units.
(b)(i),(ii) and (c)
Question 6:
The diagram in the answer space shows two squares ABCD and CDEF each of sides 4 cm. K is a point on the line CD. W, X and Y are three moving points in the diagram.
(a) Point W moves such that it is always equidistant from the straight lines AB and EF. By using the letters in diagram, state the locus of W.
(b) On the diagram, draw
(i) the locus X such that it is always 2 cm from the straight line ACE,
(ii) the locus of Y such that KY = KC.
(c) Hence, mark with the symbol the intersection of the locus of X and the locus of Y.
Answer:
(b)(i), (ii) and (c)
Solution:
(a) The locus of W is the line CD.
(b)(i),(ii) and (c) | 466 | 1,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.941811 |
https://statskey.com/seeking-experts-to-solve-my-stata-problems-2 | 1,726,824,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00609.warc.gz | 486,130,306 | 28,215 | # Seeking experts to solve my Stata problems?
## Seeking experts to solve my Stata problems?
Seeking experts to solve my Stata problems? This is a post by Fung. Here is my entry for the essay that really merits writing about the subject, but it shouldn’t really be worth your time. It was a natural response to some my stochastic geometry books. In my opinion, it’s an excuse to not be interested in anything other than the theory of stochastic matrices. Some of the following people are interested in my work. In their comments it is nice to know if their PhD was the right one from what I read. Monday, July 09, 2010 You’ve already read this one already. There are no more proofs for this question, so there are many better ones. 1. It is good that it was only for an academic literature on the topic of stochastic matrices Unfortunately, the research can improve, provided you don’t give a lot of money up front to do it. Otherwise, if you get the offer, chances are that you will get ignored by The MIT Press. I’m afraid these are not the only studies that are making the search. If the authors had used a quite different analysis of the stochastic geometric form theory (that is, as an alternative approach), they should find some exciting solutions. 2. It is not enough to use classical stochastic homogeneous point processes. The stochastic geometric homogeneous point processes are not mathematically like the Green’s functions, but, more importantly, they can produce useful solutions (as in $A(x)$) in a context. In principle, one can also look at (see, e.g., [@vlt]) for instance to measure the Riesz transform (see also: When we consider an average stochastic event of the Brownian motion, we can write it as: $$x_h(t) = i \tau \,\mbox{ or } \quad x_\infty(t) = x_h(t-\tau).$$ It can be shown by elementary calculus methods that, provided two solutions are established, they can be combined.
## Pay For Homework To Get Done
It makes the algorithm more understandable, easy to read, it’s faster to use more computational resources and they are also economical and useful sometimes. It also means the time efficiency of the algorithm instead of requiring a full set of resources instead of the algorithms’ use of complex algorithms as a whole. The complexity of method analysis the way calculation engine or process engines cannot handle, the method’s overall time cost
### Recent Posts
Who provides quick Z test homework help? What’s your score when you’re getting homework help? This class helps you stand when you don’t know what’s
Where can I find help with my T test homework? I couldn’t find a link for it from A and B. Would someone please help?
How to get reliable Z test assignment assistance? Hello and welcome – This is the post to check out all the steps, which some say | 606 | 2,800 | {"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": 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} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.94835 |
https://republicofsouthossetia.org/question/the-cheerleaders-held-a-car-wash-to-raise-money-for-new-uniforms-they-spent-12-on-soap-and-19-on-15161830-62/ | 1,632,198,465,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00052.warc.gz | 519,624,936 | 14,123 | ## The cheerleaders held a car wash to raise money for new uniforms They spent \$12 on soap and \$19 on sponges,rags,and a bucket. It rained the
Question
The cheerleaders held a car wash to raise money for new uniforms They spent \$12 on soap and \$19 on sponges,rags,and a bucket. It rained the day of the carwash do the cheerleaders just made \$25 Write an integer expression that shows their profit or loss Simplify the expression and explain what the amount means
in progress 0
1 week 2021-09-09T01:58:56+00:00 1 Answer 0
### This represents that they spent \$6 more than the amount they actually earned.
Step-by-step explanation:
The amount spent on soap = \$12
The spent on sponges and rags = \$19
So, the total amount spent by cheerleader
= Amount spent on (Soap + Buckets/ Sponges) = \$12 + \$19 = \$31
So, they total spent is \$31 before car washing.
Now, the amount they earned from car washing = \$25
Here, as we can say:
\$31 > \$25
⇒ The amount spent is more than the amount earned.
So, there is loss in the given activity.
Loss = Amount spent – Amount earned
= \$31 – \$25 = \$6
So, the integer expression \$6 shows their loss.
This represents that they spent \$6 more than the amount they actually earned. | 334 | 1,245 | {"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} | 4.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.953631 |
https://www.physicsforums.com/threads/real-life-azimuth-problem.767627/ | 1,477,392,322,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720026.81/warc/CC-MAIN-20161020183840-00135-ip-10-171-6-4.ec2.internal.warc.gz | 980,030,643 | 15,307 | # Real life azimuth problem
1. Aug 26, 2014
### getting2it
Draw this description out on a piece of paper if you would like. I cannot figure this out, it has been years since I have done math at this level.
I am standing at point A and have a distance of 87.8333 feet to Point B, my azimuth bearing is 123 degrees. I want to move over 20 feet and find the bearing to the same point B, this distance will also be 87.8333 feet. So I want to triangulate. The question is: What would my new azimuth bearing be to get to Point B, also please show how to solve. Note: I am dealing with geomagnetic bearings, no need to convert to True North.
Thank You
2. Aug 26, 2014
### getting2it
Oh my goodness, after posting above I was sitting here thinking and I remembered 2 pi r squared.
so 2 x r x pi = circumference Divided by 360 times 20 gives me the new bearing. basics basics, thanks anyway to all of you.
Last edited: Aug 26, 2014 | 245 | 932 | {"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-2016-44 | longest | en | 0.950717 |
https://studylib.net/doc/25215891/chapter-confidence-intervals---sampling-distributions | 1,624,612,599,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00580.warc.gz | 489,399,677 | 26,449 | # chapter confidence intervals + sampling distributions
```Chapter: Sampling Distributions DzStatistics mean never having to say you are certaindz Ȇunknown DzHumility is to make a right estimate of one's selfǤdz Ȇ Charles Spurgeon Estimation The main aim of Statistics is to make inferences about unknown population parameters, that is Ȃ estimating. Sampling error We carefully collect data from a sample and then use this information based on a sample to estimate the value of the unknown population parameter. Of course different samples would give different estimates. This difference is known as sampling error. Thus every conclusion made about population would contain some level of uncertainty. We want to know some important truth which in practice is unachievable. What to do? What can save the situation is that we can quantify this sampling error! We can well predict how exactly sample differences would vary. For instance that Dzalmost surely be within 5 pounds around ͳ dz (average weight is from 171 to 181 pounds) Dz
proportion of broken details must almost surely be within r0.3% of 5%dz (the proportion is between 4.7 and 5.3). Sampling Distributions Sampling distributions is the probability distribution of some estimator (statistics) among different samples. It shows how estimates could vary among different samples. We want to estimate some population parameter which is a fixed quantity Ȃ a constant. For that we use special functions called Ȃ estimators or statistics. Estimators are random variables that vary depending on the particular sample chosen. The probability distribution showing how exactly estimator varies is called a sampling distribution. Why a sample statistics is a variable? It is very easy Ȃ because different samples give different values! Consider a big example to understand exactly what a sampling distribution is. Dzof workersdz Suppose one department of a company consists of six workers. Their work experiences in years are: 2, 4, 6, 6, 7, 8 years. A manager wants to form a small team out of this department of 4 workers to work on a project. Since he is quite busy he decided to choose them randomly. Work experience matters in regard to how the job is done, therefore manager is interested: What could be the mean work experience in a group?
Authors: Chernenko Elizaveta &Ovcharova Maria. Please send your questions and commentaries to: [email protected] and [email protected] Solution: First of all the population here is the total number of workers in the department, namely 6. The population mean of number of years of experience is: 2+4+6+6+7+8
6
= 5.5 = P Group of 4 employees could be regarded as a random sample of size 4, chosen from a population of size 6. Fifteen possible samples could be selected (this is the amount of combinations ܥ46 ). Table 1 shows all possible samples with mean years of experience calculated on each sample. Samples (in years) 1. 2,4,6,6 2. 2,4,6,7 3. 2,4,6,8 4. 2,4,6,7 5. 2,4,6,8 6. 2,4,7,8 7. 2,6,6,7 8. 2,6,6,8 Sample means 4.5 4.75 5.0 4.75 5.0 5.25 5.25 5.5 Samples (in years) 9. 2,6,7,8 10.2,6,7,8 11.4,6,6,7 12.4,6,6,8 13.4,6,7,8 14.4,6,7,8 15.6,6,7,8 Sample means 5.75 5.75 5.75 6.00 6.25 6.25 6.75 Since it was random sampling, each sample of 15 has an equal chance of being selected, namely 1/15. Using this we can determine the probability of every mean work experience achieved. For instance, from the table we can see that three of fifteen possible samples have means 5.75 years. Thus the probability that the team for the project would have an average work experience of 5.75 years is 3/15, etc. This collection of values of sample means with associated probabilities would constitute the probability distribution of sample mean, or right the sampling distribution of the sample mean. X
P(X)
4.5
1/15
4.75
2/15
5
2/15
5.25 5.5
2/15 1/15
5.75
3/15
6
1/15
6.25
2/15
6.75
1/15
Notice, that while the numbers of experience for six workers vary from 2 to 8, the values of mean experience vary from 4.5 to 6.75. That is values of sample means have much smaller range, and are concentrated in the center. This is a simple discrete case example. However if we plot the means on a histogram, we will see that even it approximately resembles a normal distribution. The bigger is the sample size, the closer is the distribution of statistics to normal. 1. Possible values for mean work experience in a team Recall: Probability distribution is the collection of values with associated probabilities So, distinguish among three notions: x
x
x
Sampling distribution Ȃ distribution not of X, but of some statistics among different samples Population distribution Ȃ the true distribution of all possible values of X in the world. From population we take samples. Sample distribution Ȃ you may find this notion too. It is just the way a particular sample taken is distributed. This you usually draw with histogram to show the range of all values in the sample, with associated relative frequencies in this case (not probabilities) Examples of parameters are: population mean P, population proportion p ഥ, sample proportion ො Examples of estimators (statistics) are: sample mean X
Background. Sampling (method) is the set of random variables: X1, X2ǡǥǡn Sample is the realization of all random variables: particular numbers x1, x2ǡǥǡn The same note: sample mean ഥ
X is a variable, but in problems you will meet the sentences like DzͷͲdzǤȂ a particular value calculated on that sample. Also we always assume: All Xi have the same distribution All Xi are independent Variability of a Sampling Distribution The variability of a sampling distribution is measured by its variance or its standard deviation. The standard deviation of this statistic is also called the standard error. CLT: Central Limit Theorem Why normal distribution is so important? Because it can be used to describe the results of sampling procedures we have achieved. Sampling distributions are most often normal! This is the essence of CLT. The central limit theorem states that: the sum of independent identically distributed random variables is approximately normally distributed, if the quantity of these variables is large enough. ഥ and pො , statistics, most frequently used in our course. It is applicable for X
ഥ: For sample mean X
x
x
x
It is the sum of Xi Xi are independent since the sample is random Xi are identically distributed because each is taken from the same underlying population For sample proportion pො : pො = m/n m has binomial distribution. However, binomial is the sum of Bernoulli random variables. m =6yi where yiaBernoulli (recall chapter on discrete random variable) x
x
x
Thus the proportion is the sum of Yi Yi are independent Yi are all Bernoulli, thus are identically distributed When clt starts working? -­‐as n approaches infinity -­‐ the closer is the underlying population to normal, the earlier clt starts working. How large is "large enough"? It could be decided differently, but in our course we use 30. Requirements for accuracy. The more closely the sampling distribution needs to resemble a normal distribution, the more sample observations will be required. The more closely the original population resembles a normal distribution, the fewer sample points will be required. Shortly the idea of CLT is described as follows: if the sample is big, everything is going to be normal! Sampling distributions for most commonly used statistics 1. Estimating Parameter P. ഥ is an estimator for Population mean is a constant and is denoted by P. Sample mean ܆
ഥ could vary among different samples? population mean P. How sample mean X
Sample mean is a random variable based on a sample. It is simply the average of sample ഥ observations, and is denoted by X
ഥ = ܺ1+ ܺ2+ڮ+ܺ݊ = 6Xi X
݊
n
ഥ, namely its mean and variance. ǯ alculate the parameters of the random variable X
ഥ: xExpectation (mean) of X
ഥ) = E(6Xi ) Plug the formula into the brackets: E(X
n
One over n is just a constant, so we put it out of brackets and from the chapter of random variables we know that the expectation of sum is the sum of expectations, so: 6Xi
E(
n
1
1
n
n
) = E(X1+X2 ΪǥΪn) = (E(X1) + E(X2ȌΪǥΪȋn)) Since each random variable Xi has the same distribution and mean P, we know each expectation: 1
n
1
nP
n
n
(E(X1) + E(X2ȌΪǥΪȋn)) = (P + P ΪǥΪP) = = P ഥ: xSimilarly Variance of X
6X
ഥ) = Var ( i ) = 1 Var (X1+X2 ΪǥΪn) = 1 (VarX1 + VarX2 Ϊǥ+ VarXn) = Var (X
n2
n2
n
1
1
nV2
V2
n
n
n
n
= 2 (V2 + V2 ΪǥΪV2) = 2 nV2 = = 2
WeǯǤ ഥ has the following distribution: Thus X
V
ഥ
X a N (P, ) 2
n
Once more, recall that from sample to sample, different values would be achieved for the sample mean. Thus this quantity can be regarded as a random variable, which has a probability distribution. 2. Estimating Parameter p. ෝ is an estimator Population proportion is a constant and is denoted by p. Sample proportion ܘ
for population proportion p. m
It equals pො = Ȃ the number of successes m out of total number (sample size) n. n
Consider closer this expression. Sample size n is a constant, while m is a random variable, which is distributed binomially, ǯǡǤ Indeed, m is number of people or objects, proportion of which we are looking for (for instance, the number of defective details of the number of respondents who support strike). And each of this unit has a chance p to be in the required category of interest. Thus m a Bi(n,p) Expectation of binomial random variable equals np, while variance equals np(1-­‐p). Now let us calculate the characteristics of the random variable pො , namely its mean and variance. xExpectation (mean) of pො : m
1
1
np
n
n
n
n
E(pො ) = E( ) = E(m) = np = = p xVariance of pො : m
1
1
Var(pො ) = Var( ) = n 2 Var(m) = n 2 np(1-­‐p) = n
np (1െp)
p(1െp)
n2
n
= p(1െp)
n
Standard deviation is as usually the square root of variance. Vpො = ට
ǯǤ Thus pො has the following distribution: (1െ)
Ƹ N( p, ݊ ) Comparison of parameters Statistics allows us to compare several parameters in order to conclude which is higher. For instance, we might be interested in the following questions. Which of two financial firms pays a higher mean starting salary? Whether the proportion of defective smartphones is higher for Samsung than for Apple, or is the same for both brands? What can be said about the difference in the mean welfare of single-­‐
parent families versus two-­‐parent families? 3. Estimating parameter P1-­‐P2 Difference between two means As we just shown sample mean has the following distribution by CLT: V
ഥ
X a N (P, n ) 2
If we consider two independent populations, each of them would have: 2
2
1
2
V
ഥ
ഥ2a N (P2,V2 ) X1 a N (P1,n 1 ) and X
n
ഥ1 -­‐ X
ഥ2 is also a normal random variable, since it can be represented as the difference (which is a X
particular case of sum) of two independent normal random variables. Find the expectation and variance of this random variable: ഥ1 -­‐ X
ഥ2) = E(X
ഥ1) Ȃ E(X
ഥ2) = P1 -­‐ P2 xE(X
2
2
1
2
ഥ1 -­‐ X
ഥ2) = VarX
ഥ1 + Var(-­‐܆
ഥ2) = VarX
ഥ1 + (-­‐1)2VarX
ഥ2 = VarX
ഥ1 + VarX
ഥ2 = V1 + V2 xVar(X
n
n
Standard deviation as usual equals to the square root of variance V2
V2
1
2
VXഥ1 Ϋ Xഥ2 = ටn 1 + n 2 Note(!) that the standard deviation of the difference is higher than that of a single variable. It is evident because now two estimators (that both could vary) are involved Thus, the difference of sample means has the following distribution: 2
2
ഥ1-­‐X
ഥ2 a N( P1 -­‐P2, V1 + V2 ) X
n
n
*notice plus here! Example DzTest dz 17 years old student Masha is quite good at writing tests on Staistics, but her results varies a lot. Her classmate Peter offered her a bet on a bar of chocolate that at the end of next semester his average test mark for this subject would be higher than hersǤǯͶǤͷͷ
ͲǤǡǯͶǤ͵ͲǤͶǤThere are going to be 50 tests in the next semester. What is the probability that Masha will lose the bet, that is, that her average mark will be less than the average mark of Peter? Solution: 2
2
ഥ1-­‐X
ഥ2 a N( P1 -­‐P2, V1 + V2 ) by CLT X
n
n
The mean of the differences is: P1 -­‐P2 =4.5 Ȃ 4.3 = 0.2 The standard deviation is: (0.7 ) 2 (0.4) 2
= 0.114. 50
50
The chance that Masha will lose is equal to the probability: 0 0.2
P(X1X2) = P(X1 -­‐ X2 0) = P(z ) = P(z Ȃ1.75) = 0.04 0.114
(From Table of the Normal Distribution, the area to the left of Ȃ1.75 is 0.0400.) VഥX1 Ϋ Xഥ2 =
Thus there is only a 4% chance that Masha will lose. 4. Estimating parameter p1-­‐p2 Single sample proportion has the distribution Ƹ
N( p,
(1െ)
݊
) Exactly the same logic applies to the difference of two sample proportions 1 ήሺ1െ 1 ሻ
ήሺ1െ ሻ
+ 2 ݊ 2 ) ݊1
2
Ƹ1 െ Ƹ2 ~ܰ(1 െ 2 , ට
Sample AP practice problems 2015 6d
6. Corn tortillas are made at a large facility (ɩɪɢɫɩɨɫɨɛɥɟɧɢɟ) that produces 100,000 tortillas per day on each of
its two production lines. The distribution of the diameters of the tortillas produced on production line A is
approximately normal with mean 5.9 inches, and the distribution of the diameters of the tortillas produced on
production line B is approximately normal with mean 6.1 inches. The figure below shows the distributions of
diameters for the two production lines.
The tortillas produced at the factory are advertised as having a diameter of 6 inches. For the purpose of quality
control, a sample of 200 tortillas is selected and the diameters are measured. From the sample of 200 tortillas, the
manager of the facility wants to estimate the mean diameter, in inches, of the 200,000 tortillas produced on a
given day. Two sampling methods have been proposed.
Method 1: Take a random sample of 200 tortillas from the 200,000 tortillas produced on a given day.
Measure the diameter of each selected tortilla.
Method 2: Randomly select one of the two production lines on a given day. Take a random sample of 200
tortillas from the 100,000 tortillas produced by the selected production line. Measure the diameter of each
selected tortilla.
Each day, the distribution of the 200,000 tortillas made that day has mean diameter 6 inches with standard
deviation 0.11 inch.
G )RUVDPSOHVRIVL]HWDNHQIURPRQHGD\¶VSURGXFWLRQGHVFULEHWKHVDPSOLQJGLVWULEXWLRQRIWKH sample
mean diameter for samples that are obtained using Method 1.
(e) Suppose that one of the two sampling methods will be selected and used every day for one year (365
days). The sample mean of the 200 diameters will be recorded each day. Which of the two methods will
result in less variability in the distribution of the 365 sample means? Explain.
Solution:
(d) The sampling distribution of the sample mean diameter for samples obtained using Method 1would be
0.11
approximately normal with mean 6 inches and standard deviation 200 | 0.0078 inches
ξ
(e) Method 1 would result in less variability in the sample means (plural!) over 365 days. With Method 2
roughly half of the sample means will be around 5.9 inches and the other half will be around 6.1 inches while
with Method 1the sample means will all be very close to 6.0 inches, as indicated by very small standard deviation
in part (d) (0.0078 inch)
2014 3b
3. Schools in a certain state receive funding based on the number of students who attend the school. To determine
the number of students who attend a school, one school day is selected at random and the number of students in
attendance that day is counted and used for funding purposes. The daily number of absences at High School A in
the state is approximately normally distributed with mean of 120 students and standard deviation of 10.5 students.
(a) If more than 140 students are absent on the day the attendance count is taken for funding purposes, the school
will lose some of its state funding in the subsequent year. Approximately the probability that High School A will
lose some state funding is about 0.0287.
E 7KHSULQFLSDOV¶DVVRFLDWLRQLQWKHVWDWHVXJJHVWVWKDWLQVWHDGRIFKRRVLQJRQHGD\DWUDQGRPWKHVWDWHVKRXOG
choose 3 days at random. With the suggested plan, High School A would lose some of its state funding in the
subsequent year if the mean number of students absent for the 3 days is greater than 140. Would High School A
be more likely, less likely, or equally likely to lose funding using the suggested plan compared to the plan
described in part (a)? Justify your choice.
Solution:
(b) High School A would be less likely to lose state funding. With a random sample of 3 days, the distribution of
the sample mean number of students absent would have less variability than that of a single day. With less
variability, the distribution of the sample mean would concentrate more narrowly around the mean of 120
students, resulting in a smaller probability that the mean number of students absent would exceed 140.
In particular, the standard deviation of the sample mean number of absences is:
V
10.5
= 3 = 6.062.
݊
ξ
ξ
So, the probability that High School A loses funding using the suggested plan would be as determined as:
ഥ !140) = P(z ! 140െ120 ) = P(z ! 3.3) = 0.0005 | 0, which is much less than a probability of 0.0287 obtained
P(X
6.062
for the plan described in part (a).
2007 3
3. A company recently stocked a new lake in a central city park with 2,000 fish of various sizes. The distribution
of the length of these fishes is approximately normal.
(a) A company claims that the mean length of the fish is 8 inches. If the claim is true, which of the following
would be more likely?
x a random sample of 15 fishes having a mean length that is greater than 10 inches
Chapter on confidence intervals. Introduction. The basic purpose of statistics is to get useful knowledge about unknown from the mess of known data. For example, we might be interested in the true values of: ђ -­‐ ƚŚĞ ŵĞĂŶ ƐĂůĂƌLJ ŝŶ ZƵƐƐŝĂ ŝŶ ϮϬϭϱ͕ ʋ ʹ the proportion of Saint-­‐Petersburg citizens in favor of the anti-­‐ŐĂLJůĂǁ͕Žƌʍ-­‐ riskiness of some investment project. Usually, we cannot calculate those parameters as we do not have the data about the whole population. However, we can estimate their values based on the sample data. One way to do that are point estimates such as sample mean ݔҧ , sample proportion p, sample standard deviation s. This approach has a significant drawback: our point estimates almost never coincide with the true values of parameters. To increase ŝŶǀĞƐƚŝŐĂƚŽƌ͛Ɛ confidence in estimates the other approach is used ʹ interval estimates. Although we are almost never able to ͞ĐĂƚĐŚ͟ƚŚĞǀĂůƵĞŽĨƚƌƵĞƉĂƌĂŵĞƚĞƌǁŝƚŚĂƉŽŝŶƚĞƐƚŝŵĂƚĞ͕ǁĞĐĂŶĐŽŶƐƚƌƵĐƚĂŶŝŶƚĞƌǀĂůǁŚŝĐŚǁŝůůŝŶĐůƵĚĞƚŚĞ
ƉĂƌĂŵĞƚĞƌ͞ĂůŵŽƐƚĨŽƌƐƵƌĞ͘͟ Estimates
c
Point estimates
onfidence Confidence intervals confidence intervals ! You might compare these two methods with the two approaches to fishing. intervals Trying to catch a fish with a spear is similar to calculating point estimate in hope it will be equal to the parameter. Contrary, using a net is similar to constructing an interval, which would contain the parameter in significant number of trials. We call the intervals constructed a confidence intervals since those might contain the true parameter with a specified level of confidence. ! Note that there is always a chance the confidence interval WILL NOT contain the true value of parameter! Confidence intervals for population ƉĂƌĂŵĞƚĞƌƐ;ђĂŶĚʋͿ 1. Confidence intervals for the population mean ђ 1.1.
dŚĞĐĂƐĞŽĨŬŶŽǁŶʍ Let us suppose we have a big sample of values of some variable X. According to the Central Limit Theorem (CLT) sample mean ݔҧ ŝƐĂƉƉƌŽdžŝŵĂƚĞůLJŶŽƌŵĂůůLJĚŝƐƚƌŝďƵƚĞĚ͘ƐǁĞ͛ǀĞůĞĂƌŶĞĚĨƌŽŵƚŚĞƉƌĞǀŝŽƵƐchapter, it has the following parameters: ݔҧ ~ܰ(ߤ,
ߪ
ξ݊
). How do we construct a confidence interval? STEP 1. As for any Normal variable we can reduce it to the Standard Normal variable z: ݔҧ െߤ
ߪ
ൗ ݊
ξ
= (ܰ~ݖ0,1) STEP 2. If we express ݔҧ from the equation (1) we get: ݔҧ = ߤ + ݖή
(1) ߪ
ξ݊
Z is a random variable taking positive as well as negative values. Thus, the first summand shows that ݔҧ on ĂǀĞƌĂŐĞĞƋƵĂůƐђ͕ĂŶĚƚŚĞƐĞĐŽŶĚŽŶĞƌĞƉƌĞsents a deviation of ݔҧ ĨƌŽŵђǁŚŝĐŚĚĞƉĞŶĚƐŽŶƌĂŶĚŽŵĐŽŵƉŽŶĞŶƚ
z. Z follows the known probability distribution. So, the equation specifies the range of the deviations of ݔҧ around ђǁŚŝĐŚŵŝŐŚƚŚĂƉƉĞŶ with given probability reflected in the corresponding value of zcritical. Assuming that zcritical is non-­‐negative we can rewrite the expression as follows: ݔҧ = ߤ േ ݈ܽܿ݅ݐ݅ݎܿݖή
ߪ
ξ݊
(2) STEP 3. ,ŽǁĞǀĞƌ͕ǁĞĂƌĞƵƐƵĂůůLJŝŶƚĞƌĞƐƚĞĚŝŶƚŚĞƉůĂƵƐŝďůĞǀĂůƵĞƐŽĨђ͕ŶŽƚݔҧ . ExpƌĞƐƐŝŶŐђĨƌŽŵƚŚĞĞƋƵĂƚŝŽŶ (2) we get: ߤ = ݔҧ േ ݈ܽܿ݅ݐ݅ݎܿݖή
ߪ
ξ݊
dŚƵƐ͕ǁĞ͛ǀĞĂƌƌŝǀĞĚƚŽƚŚĞĨŝƌƐƚĨŽƌŵƵůĂŽĨĐŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂů͘/t provides a convenient way to estimate the ďŽƵŶĚĂƌŝĞƐǁŝƚŚŝŶǁŚŝĐŚƚŚĞƚƌƵĞǀĂůƵĞŽĨђůŝĞƐǁŝƚŚthe certain level of confidence. The component ݈ܽܿ݅ݐ݅ݎܿݖή
ߪ
is called the margin of error. ξ݊
What is the margin of error and how to find zcritical? Let͛ƐŐŽďĂĐŬƚŽƚŚĞĞƋƵĂƚŝŽŶ;ϮͿ͗ݔҧ = ߤ േ ݈ܽܿ݅ݐ݅ݎܿݖή
ߪ
ξ݊
As you can see the value of the random variable ݔҧ has two components: true value ђ and the random component േ ݈ܽܿ݅ݐ݅ݎܿݖή
ߪ
. Since we know the pdf of z we can calculate the interval of possible values ݔҧ can take ξ݊
with given probability. Example 1. Consider an example. Assume the true mean ŚĞŝŐŚƚ ŽĨ /& ƐƚƵĚĞŶƚƐ ŝƐ ђсϭϳϱ Đŵ ǁŝƚŚ ƐƚĂŶĚĂƌĚ
ĚĞǀŝĂƚŝŽŶʍсϱĐŵ͘ƌĂŶĚŽŵƐĂŵƉůĞŽĨ100 students is taken and their heights are measured to calculate ݔҧ -­‐ the sample mean height. ݔҧ can take different values depending on the heights of random set of students, who get into the sample. The range of values ݔҧ can take is: ݔҧ = 175 േ ݈ܽܿ݅ݐ݅ݎܿݖή
5
ξ100
Suppose, we want the interval to include values ݔҧ takes with probability 95%. What value of zcritical should be used to provide an interval including 95% of all possible values of ݔҧ ? ƐǁĞ͛ǀĞƐƚĂƚĞĚĞĂƌůŝĞƌݔҧ is a Normal random variable. The picture below displays its probability distribution. In order to find the interval which includes the most probable 95% of ݔҧ values (the smallest interval) we should take the central 95% of values, leaving 5% of unlikely values at the two equal tails of the distribution. Due to the symmetry area of each tail equals 5%
2
= 2,5%. Thus, zcritical=z0,025 such that P(z>z0,025)=2,5%=0,025. From the table we get: z0,025=1,96. Then, ݔҧ = 175 േ 1,96 ή
5
ξ100
| 175 r 1 So, we get that with 95% probability the sample mean will take value between 174 and 176 centimeters. In general we denote the probability of interest by (1-­‐ɲͿ% and call it the confidence level (95% in the given example). ɲ ŝƐ ƚŚĞ ƉƌŽďĂďŝůŝƚLJ ŽĨ ͞ůĞĨƚ ďĞŚŝŶĚ͟ or improbable values of ݔҧ and is called the significance level. Therefore in general case ݖcritical = ߙݖ. 2
dŚĞŐĞŶĞƌĂůĨŽƌŵƵůĂŽĨĐŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůĨŽƌђǁŝƚŚŬŶŽǁŶʍŝƐ͗ ߪ ߤ = ݔҧ േ ߙݖή 2 ξ݊
(3) Is there life without the Central Limit Theorem? tĞ͛ǀĞ ƐƚĂƌƚĞĚ ŽƵƌ ĞdžƉůĂŶĂƚŝŽŶ ďLJ ƐƚĂƚŝŶŐ ƚŚĂƚ ǁŝƚŚ ůĂƌŐĞ ĞŶŽƵŐŚ ƐĂŵƉůĞ ƚŚĞ >d ŝŶƐƵƌĞƐ ƚŚĂƚ͗ ݔҧ ~ܰ(ߤ,
ߪ
). ξ݊
However, what happens if the sample is small, say, less than 30 observations? Does that mean that construction of confidence interval is impossible? The answer depends on the distribution of X. Note that if X is itself normally distributed sample mean ܺത, being a sum of independent normal variables is also normal. Thus, we can apply the same logic and construct the confidence interval using the same formula. Still, for small samples it is necessary that the assumption about the normality of X is held. If you are asked to find the confidence interval based on the small sample, and nothing is said about the distribution of X, before interval estimation you should state the assumption, that the random variable of interest (for example height) is normal. 1.2 ŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůĨŽƌђǁŝƚŚƵŶŬŶŽǁŶʍ ^ŽĨĂƌǁĞ͛ǀĞƌĞĂĐŚĞĚthe formula: ߤ = ݔҧ േ ܼߙ ή
2
ߪ
to estimate the unknown population mean. To calculate such ξ݊
ĂŶŝŶƚĞƌǀĂůƚŚĞǀĂůƵĞŽĨʍŶĞĞĚƐƚŽďĞŬŶŽǁŶ͘,ŽǁĞǀĞƌ͕ǁŚĞŶƉŽƉƵůĂƚŝŽŶŵĞĂŶŝƐƵŶŬŶŽǁŶ;we are trying to estimate it by the interval!) how is it possible to know the true value of population standard deviation? Indeed, ŝƚŝƐƵƐƵĂůůLJŶŽƚƚŚĞĐĂƐĞ͘ůůǁĞŚĂǀĞŝƐĂƐĂŵƉůĞ͊KĨĐŽƵƌƐĞ͕ǁĞĐĂŶƌĞƉůĂĐĞʍǁŝƚŚŝƚƐƐĂŵƉůĞĞƐƚŝŵĂƚĞƐ͘ ݔҧ െߤ
ƐǁĞ͛ǀĞƐŚŽǁŶĞĂrlier in this section, ߪ
ൗ ݊
ξ
~ܰ(0,1). How the probability distribution will change if ʍŝƐƌĞƉůĂĐĞd with s? Obviously, when a constant is replaced with a random variable the variance of the whole expression will rise. ݔҧ െߤ
Pdf of a random variable ݏ
ൗ ݊
ξ
ǁŝůůŚĂǀĞďŝŐŐĞƌ͞ƚĂŝůƐ͟ŝŶĚŝĐĂƚŝŶŐŚŝŐŚĞƌƉƌŽďĂďŝůŝƚLJŽĨĚĞǀŝĂƚŝŽŶƐfrom mean. What is more, the smaller is the sample ʹ the higher is the probability of large deviations ŽĨƐĨƌŽŵʍ, reflected in even fatter tails. It happens because small samples give less precise values of standard deviation s, which becomes more dispersed. However, all the other properties of pdf function stay the same: centered at 0 having bell-­‐shaped form. Hi! You need me to pass AP! and Here the new distribution is to be introduced ʹ the Student or t-­‐distribution. Its pdf is defined by the so-­‐called degrees of freedom (df) calculated as number of observation minus 1, df=n-­‐1. The corresponding number is conventionally indicated in brackets after the t letter: t(n-­‐1). dŚĞĐůŽƐĞƌŝƐƐƚŽʍ;ǁŚŝĐŚŝŶŐĞŶĞƌĂůŚĂƉƉĞŶƐǁŝƚŚŚŝŐŚĞƌŶͿ͕ƚŚĞ closer is t-­‐distribution to z. As is shown on the picture above t-­‐distributions with lower degrees of freedom are more dispersed around 0 (v on the picture denotes value of df). On the other hand, with larger df t-­‐distribution becomes closer to N(0,1). Indeed, when n approaching infinity, pdf of t approaches pdf of z. Thus, we get the following formula: ݔҧ െ ߤ
~ ݊(ݐെ 1) ݏ
ൗ ݊
ξ
Applying the same algorithm as in 1.1 we arrive to the new formula of confidence interval: ݏ
ߤ = ݔҧ േ ݊( ߙݐെ 1) ή 2
ξ݊
Note that ݊( ߙݐെ 1) is a number (not multiplication of t by (n-­‐1))! It is a critical value of t-­‐distribution with n-­‐1 2
ߙ
2
degrees of freedom such that P(t> ݊( ߙݐെ 1))= . 2
Note that when n is large enough t distribution can be approximated by z. Thus, for large samples it is acceptable to apply the formula: ߤ = ݔҧ േ ߙݖή ξ݊ ݏ. However, it would be an approximation, while the general rule is 2
to use z-­‐ĚŝƐƚƌŝďƵƚŝŽŶǁŚĞŶʍŝƐŬŶŽǁŶ͕ĂŶĚƚ-­‐distribution ʹ otherwise. Again, the formulas given here are applicable when sample is large enough so that we can apply CLT. For small samples to ensure that ݔҧ ~ܰ we need to make sure that X is normal. IĨŝƚ͛ƐŶŽƚŐŝǀĞŶŝŶĂƉƌŽďůĞŵʹ you should state the corresponding assumption. So, how do you choose the right formula? 2. ŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůĨŽƌƉŽƉƵůĂƚŝŽŶƉƌŽƉŽƌƚŝŽŶʋ Analogically, by CLT for large n sample proportion p has normal distribution. Ɛ ǁĞ͛ǀĞ ƐŚŽǁŶ ŝŶ ƚŚĞ ƉƌĞǀŝŽƵƐ
ߨ(1െߨ)
) ݊
chapter ߨ(ܰ~, ට
Thus, = ݖ
െߨ
ߨ(1െߨ)
݊
and ߨ = േ ܼߙ ή ට
ߨ (1െߨ )
ට
݊
2
ߨሺ1െߨሻ
is the margin of error. Since we do not know tŚĞƚƌƵĞǀĂůƵĞŽĨƉƌŽƉŽƌƚŝŽŶʋǁĞĐĂŶŶŽƚĚŝƌĞĐƚůLJ
݊
Now ߙݖή ට
2
calculate the margin of error. However, as the sample is initially assumed to be large enough we can still ĐŽŶƐƚƌƵĐƚĂŶĂƉƉƌŽdžŝŵĂƚĞĐŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůďLJƌĞƉůĂĐŝŶŐʋŝŶŵĂƌŐŝŶŽĨĞƌƌŽƌǁŝƚŚŝƚƐƐĂŵƉůĞĞƐƚŝŵate p. Thus͕ǁĞŐĞƚƚŚĞĐŽŶǀĞŶƚŝŽŶĂůĨŽƌŵƵůĂĨŽƌƚŚĞĐŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůĨŽƌʋ͗ (1െ)
݊
ߨ = േ ܼߙ ή ට
2
(4) ! Note that THIS IS THE ONLY FORMULA used for (2-­‐sided) ŝŶƚĞƌǀĂůĞƐƚŝŵĂƚŝŽŶŽĨʋ͘ It is only used when sample is large enough and uses only normal distribution (no t-­‐distribution for proportions!). Confidence intervals for the difference of parameters 3. Confidence interval for difference in population means ࣆ െ ࣆ 3.1.
Known population standard deviations ࣌ , ࣌ As you should have learned from the previous chapter, difference of two sample means is normally distributed ߪ2
ߪ2
1
2
given that sample sizes are large enough. Since ܧሺݔ
തതത1 െ തതതሻ
ݔ2 = ߤ1 െ ߤ2 and ܸሺݔ
തതത1 െ തതതሻ
ݔ2 = ݊1 + ݊2 we have: ݔ1 െ തതത~ܰ(ߤ
തതത
ݔ2
1െ
ߪ2
ߤ2 , ට݊1
1
തݔതതതെݔ
1 തതതതെ(ߤ
2
1 െߤ 2 )
Thus, +
ߪ22
) ݊2
= (ܰ~ݖ0,1) ߪ2 ߪ2
ඨ 1+ 2
݊1 ݊2
Hence, the following confidence interval is constructed: ߪ12 ߪ22
ߤ1 െ ߤ2 = തതത
ݔ1 െ തതത
ݔ2 േ ߙݖή ඨ + ݊1 ݊2
2
3.1.
Unknown population standard deviations, assuming ࣌ ് ࣌ Of course the previous formula can only be used given you know the true values of standard deviations ߪ1 and ߪ2 of ܺ1 and ܺ2 . In most real life problems this is not the case. All we usually have are the samples, and thus, the sample statistics calculated on them. When we replace ߪ1 and ߪ2 with sample standard deviations ݏ1 and ݏ2 the distribution is no more normal. To account for the increased variability in possible values of തതത
ݔ1 െ തതത ݔ2 Student distribution is applied. തݔതതതെݔ
1 തതതതെ(ߤ
2
1 െߤ 2 )
2
2
݊1
݊2
~ݐሺ݇ െ 1ሻ, k=min{n1,n2} ݏ
ݏ
ඨ 1+ 2
Thus, we come to the following formula of confidence interval: ݏ2
ݏ2
1
2
ߤ1 െ ߤ2 = തതത
ݔ1 െ തതത
ݔ2 േ ݇( ߙݐെ 1) ή ට݊1 + ݊2 , k=min{n1,n2} 2
Why is the minimum of n1 and n2 is used in calculating the degrees of freedom? The answer is that variability of ƚŚĞǁŚŽůĞĐŽŶƐƚƌƵĐƚŝŽŶĚĞƉĞŶĚƐŽŶƚŚĞ͞ǁŽƌƐƚ͕͟ƚŚĞůĞĂƐƚƐƚĂďůĞĐŽŵƉŽŶĞŶƚŝŶŝƚ͘>Ğƚ͛ƐƐƵƉƉŽƐĞŶ1=1000, n2=25. That means calculations of s1 are highly precise, it takes values close to the true standard deviation ߪ1 and has small variance. Contrary, s2, being calculated on the small sample is highly volatile. Hence, however precise s1 is, variability in possible values of s2 will make the overall expression less stable, leading to small degrees of freedom ʹ depending on n2. Hence, degrees of freedom are always calculated based on the size of the smallest sample. ! Note that your graphic calculator uses another and somewhat more sophisticated calculation of degrees of freedom. It results in a bit higher and usually fractional (not integer) value of degrees of freedom. If you use the results from the calculator in solving a problem you should indicate df used by it. If the samples are not large enough (in the AP course sample is supposed to be small when n < 30) you can still apply the same formula given that both ܺ1 and ܺ2 are normal random variables. 3.2.
Unknown population standard deviations, equal variances assumption: ࣌ = ࣌ In many situations it is reasonable to assume that ߪ1 = ߪ2 = ߪ. What happens then? ߪ12 ߪ22
1
1
ඨ
ߤ1 െ ߤ2 = തതത
ݔ1 െ തതത
ݔ2 േ ݖή
+
= തതത
ݔ1 െ തതത
ݔ2 േ ߙݖή ߪ ή ඨ + ݊1 ݊2
݊1 ݊2
2
ߙ
2
The above formula can be applied when ߪ is known. tŚĞŶ ŝƚ͛Ɛ ƵŶŬŶŽǁŶ ʍ ĐĂŶ ďĞ ĞǀĂůƵĂƚĞĚ ďĂƐĞĚ Žn the joint sample of ܺ1 and ܺ2 . Of course, since population standard deviations for both ܺ1 and ܺ2 are the same ;ʍͿ you can also estimate it based on either ܺ1 sample (use ݏ1 ) or ܺ2 sample (use ݏ2 ). However, the preciseness of the both estimators will be limited to corresponding sample size (n1 or n2). To make the estimator more precise it is useful to merge the two samples. The generally used estimator for ߪ is the so called pooled standard deviation (for the explanation of formula address the Appendix): ݏ12 ή ሺ݊1 െ 1ሻ + ݏ22 ή ሺ݊2 െ 1ሻ
ߪො = = ݏඨ
݊1 + ݊2 െ 2
When we replace ߪ with ݏ normal distribution changes to Student distribution. Since the preciseness of ݏ is limited to the size of joint sample ݊1 + ݊2 and its calculation involves two estimates of unknown population parameters Student distribution has (݊1 + ݊2 െ 2) degrees of freedom. Thus, we arrive to the formula for confidence interval with the same and unknown standard deviation ߪ: 1
1
ߤ1 െ ߤ2 = തതത
ݔ1 െ തതത
ݔ2 േ ݊( ߙݐ1 + ݊2 െ 2) ή ݏή ඨ + ݊1 ݊2
2
3.3 Confidence interval for the mean difference (for paired/matched samples) Sometimes the difference is to be calculated based on so-­‐called matched or paired samples. It means that although two sets of observations are involved (say, ܺ1 and ܺ2 ), both are in fact taken from the same population of X. For example you may want to compare the mean difference in blood pressure of 1st year ICEF students before and after they take their winter exam in Stats (population is presented by all ICEF freshmen). For that purpose one sample of 1st year students can be taken and their average measures of blood pressure before and after the തതത1 and ܺ
തതത2 ) are used to construct the interval for mean difference (we denote the true mean difference by exam (ܺ
ѐͿ͘/ŶƚŚĂƚĐĂƐĞŶŽƚŽŶůLJƚŚĞƉŽƉƵůĂƚŝŽŶŝƐƚŚĞƐĂŵĞ;ϭst year students), but the samples contain the same set of students. Why we compare a sample with itself? Because we want to see is there effect of some treatment which is not applied to sample objects at the first stage (we observe ܺ1 ) and is applied at the second stage (ܺ2 ). In the given example treatment is the exam and we want to see does it constitute a stress for students. In order ƚŽĐŚĞĐŬƚŚĂƚǁĞůŽŽŬǁŚĞƚŚĞƌƚŚĞĐŽŶĨŝĚĞŶĐĞŝŶƚĞƌǀĂůĨŽƌѐ͕ĞƐƚŝŵĂƚĞĚďLJ݀ҧ = തതതതതതതതതതതതത
(ܺ1 െ ܺ2 ) contains zero or not. If it does, then, ܺ1݅ െ ܺ2݅ (difference in blood pressure of student i) on average is close to zero and the interval of most plausible values of mean difference includes the case of no effect (zero). Therefore, we conclude that the treatment has no statistically measurable effect. Contrary, if the mean difference interval only contains positive (or only negative) values we may assert that the exam is a stressful event for students since it significantly changes their physiological condition. ,ŽǁĞǀĞƌ͕ ƐŽŵĞƚŝŵĞƐ ŝƚ͛Ɛ ŝŵƉŽƐƐŝďůĞ ƚŽ ŐĞƚ ĞdžĂĐƚůLJ ƚŚĞ ƐĂŵĞ ƐĂŵƉůĞƐ͕ ĂŶĚ ƚŚĞ ƚǁŽ ƐĂŵƉůĞƐ ĨƌŽŵ ƚŚĞ ƐĂŵĞ
population are taken. For example we might have a sample of twins, who has almost the same DNA. But in each pair of twins the first one was brought up by the mother only (sample 1), while the second was living with the father (sample 2). In the given example we might be interested in effects of gender of a single parent on psychological characteristics of a child. Say, we may want to compare mean difference in feminity scores shown up by the twins. In this example population is the same (all twins, one of whom was brought up by mother, and the other ʹ by father). Samples are different, but can be easily paired, so that we may compare the scores of twins in each pair. The strategy for constructing the confidence interval is the same. First, we create one sample from the two ʹ we calculate differences in scores for each pair i: di=X1i-­‐X2i. Then, we apply the same strategy as for constructing confidence interval ĨŽƌ ƐŝŶŐůĞ ŵĞĂŶ ђ͘ tĞ ĐĂůĐƵůĂƚĞ ƐĂŵƉůĞ ŵĞĂŶ ݀ҧ and sample standard deviation ݀ݏ and apply t-­‐ĚŝƐƚƌŝďƵƚŝŽŶƚŽĞƐƚŝŵĂƚĞѐ͗ ο= dത േ t Ƚ (n െ 1) ή
2
sd
ξn
As usually, if sample of pairs is small (n<30), we should check (or at least assume) that d is normally distributed. Which formula to choose?
4. Confidence interval for the difference in populations proportions ࣊ െ ࣊ . It was previously shown that difference in population proportions is normally distributed given that both samples are large enough: ߨ 1 ήሺ1െߨ 1 ሻ
ߨ ήሺ1െߨ ሻ
+ 2 ݊ 2 ) ݊1
2
1 െ 2 ~ܰ(ߨ1 െ ߨ2 , ට
Thus, we have: 1 െ 2 െ(ߨ 1 െߨ 2 )
ߨ ήሺ1െߨ ሻ ߨ ήሺ1െߨ ሻ
ට 1 ݊ 1 + 2 ݊ 2
1
2
= (ܰ~ݖ0,1) Hence, the following confidence interval is constructed: ߨ1 െ ߨ2 = 1 െ 2 േ ߙݖή ඨ
2
ߨ1 ή ሺ1 െ ߨ1 ሻ ߨ2 ή ሺ1 െ ߨ2 ሻ
+
݊1
݊2
! Note that THIS IS THE ONLY FORMULA used for (2-­‐sided) interval estimation of ߨ1 െ ߨ2 . It is only used when samples are large enough and uses only normal distribution (no t-­‐distribution for proportions!). Appendix The simplest idea for estimating ɐ 2 is to sum the squared deviations of X1 and X 2 and then estimate average deviation. Note that X1 and X 2 have different means, so for each variable deviations are calculated with respect to different centers. sp2 =
n2
1
σni=1
(xi1 െ xന1 )2 + σi=1
(xi2 െ ധധധ)
x2 2
Ԣ
Ŷ͛ ŝƐ ƚŚĞ ŽǀĞƌĂůů ŶƵŵďĞƌ ŽĨ ŽďƐĞƌǀĂƚŝŽŶƐ ŝŶ ƚŚĞ ƚǁŽ samples necessary to calculate the average squared deviation. Since the standard deviation in each sample is calculated by the formula: s 2 =
σni=1 (x i െxധ )2
nെ1
we can express the sums of squared deviations in terms of s1 and s2 : sp2 =
s12 ή ሺn1 െ 1ሻ + s22 ή ሺn2 െ 1ሻ
Ԣ
Since s12 and s22 are unbiased estimators for ɐ2 it is quite intuitive that nԢ = n1 + n2 െ 2 (if we divide simply by (n1 + n2 ) sp2 will underestimate ɐ2 . ! Note that squared pooled standard deviation ݏ2 is also approximately equal to the average of ݏ12 and ݏ22 weighted by the corresponding sample sizes. Sample A P practice problems
A P 2010 3
A humane society wanted to estimate with 95 percent confidence the proportion of households in its
county that own at least one dog.
(a) Interpret the 95 percent confidence level in this context.
The humane society selected a random sample of households in its county and used the sample to
estimate the proportion of all households that own at least one dog. The conditions for calculating a 95
percent confidence interval for the proportion of households in this county that own at least one dog
ZHUHFKHFNHGDQGYHULILHGDQGWKHUHVXOWLQJFRQILGHQFHLQWHUYDOZDV
(b) A national pet products association claimed that 39 percent of all American households owned at
leaVWRQHGRJ'RHVWKHKXPDQHVRFLHW\¶VLQWHUYDOHVWLPDWHSURYLGHHYLGHQFHWKDWWKHSURSRUWLRQRIGRJ
owners in its county is different from the claimed national proportion? Explain.
F +RZPDQ\KRXVHKROGVZHUHVHOHFWHGLQWKHKXPDQHVRFLHW\¶VVDPSOH"6KRZhow you obtained your
Solution:
(a) The 95 percent confidence level means that if one were to repeatedly take random samples of the
same size from the population and construct a 95 percent confidence interval from each sample, then in
the long run 95 percent of those intervals would succeed in capturing the actual value of the
population proportion of households in the county that own at least one dog.
(b) No. The 95 percent confidence interval 0.417 r 0.119 is the interval (0.298, 0.536). This interval
includes the value 0.39 as a plausible value for the population proportion of households in the county
that own at least one dog. Therefore, the confidence interval does not provide evidence that the
proportion of dog owners in this county is different from the claimed national proportion.
(c) The sample proportion is 0.417, and the margin of error is 0.119. Determining the sample size
requires solving the equation
0.119 = 1.96ට
0.119 = 1.96
0.119 =
0.417 (1െ 0.417)
݊
0.493
ξ݊
0.966
ξ݊
ξ݊ = 0.966/0.119 = 8.12
n = 65.95
So the humane society must have selected 66 households for its sample. ``` | 14,729 | 38,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-25 | latest | en | 0.917327 |
https://dcodesnippet.com/4375-as-a-fraction/ | 1,713,309,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00774.warc.gz | 178,504,615 | 19,204 | # Converting .4375 To A Fraction – Simplify And Express In Simplest Form
//
Thomas
Want to convert .4375 to a fraction? Follow our step-by-step guide to simplify and express .4375 as a fraction in its simplest form.
## What is .4375 as a Fraction?
### Simplifying .4375 as a Fraction
When we talk about simplifying a fraction, we are essentially trying to express it in its simplest form. In the case of .4375, it is already a decimal representation of a fraction. To simplify it further, we need to convert it into a common fraction.
### Converting .4375 to a Common Fraction
To convert .4375 into a common fraction, we can follow a few simple steps. First, we need to identify the place value of the decimal. In this case, the decimal extends up to four decimal places.
Next, we can write .4375 as a fraction by placing the decimal value over its place value. Since there are four decimal places, the denominator of the fraction will be 10,000 (10 raised to the power of 4). Therefore, .4375 can be expressed as 4375/10,000.
### Expressing .4375 as a Fraction in Simplest Form
Now that we have .4375 expressed as a fraction, we can simplify it further to its simplest form. To do this, we can divide both the numerator and the denominator by their greatest common divisor (GCD). In this case, the GCD of 4375 and 10,000 is 125.
Dividing both the numerator and denominator by 125, we get the simplified form of .4375 as a fraction: 35/80. However, we can still simplify it further by dividing both the numerator and denominator by 5, resulting in the simplest form of .4375 as a fraction, which is 7/16.
By following these steps, we have successfully converted .4375 into a common fraction and expressed it in its simplest form.
## How to Convert .4375 to a Fraction?
### Converting .4375 to a Fraction Step by Step
Converting a decimal number like .4375 to a fraction may seem daunting at first, but with a step-by-step approach, it becomes much easier. Let’s break it down:
1. Step 1: Recognize the decimal as a fraction. The decimal .4375 can be written as 4375/10000. To convert it to a fraction, we need to express it with a numerator and denominator.
2. Step 2: Simplify the fraction. In this case, the fraction 4375/10000 can be simplified by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 125. So, 4375 ÷ 125 = 35, and 10000 ÷ 125 = 80. Therefore, .4375 can be written as 35/80.
### Writing .4375 as a Fraction with a Denominator of 10000
Sometimes, it may be useful to express a decimal like .4375 as a fraction with a specific denominator, such as 10000. Here’s how you can do it:
To write .4375 as a fraction with a denominator of 10000, we need to determine the equivalent fraction. We can multiply both the numerator and denominator of .4375 by 10000 to achieve this.
By multiplying .4375 by 10000, we get 4375. So, the fraction equivalent to .4375 with a denominator of 10000 is 4375/10000.
### Finding the Equivalent Fraction for .4375
Finding the equivalent fraction for .4375 can be useful in certain situations. To find the equivalent fraction, we need to simplify the decimal fraction as much as possible. Let’s take a look at how we can do this for .4375:
1. Step 1: Recognize the decimal as a fraction. The decimal .4375 can be written as 4375/10000.
2. Step 2: Simplify the fraction. By dividing both the numerator and denominator by their GCD, which is 125, we get the simplified fraction 35/80.
So, the equivalent fraction for .4375 is 35/80.
Remember, converting decimals to fractions allows us to express numbers in a different form, which can be useful in various mathematical calculations and comparisons.
## .4375 as a Fraction in Decimal and Fraction Form
### .4375 as a Fraction in Decimal Form
When we express .4375 as a decimal, we see that it is a rational number with a finite number of decimal places. In decimal form, .4375 is equal to 43.75%.
### .4375 as a Fraction in Fraction Form
To express .4375 as a fraction, we can start by understanding that the decimal point separates the whole number part from the fractional part. The fractional part, .4375, can be written as the numerator of a fraction. The denominator will depend on the number of decimal places.
To convert .4375 to a fraction, we can count the number of decimal places, which in this case is 4. We then place the numerator, 4375, over 10,000 (10 raised to the power of the number of decimal places). Therefore, .4375 as a fraction is 4375/10,000.
### Comparing .4375 as a Fraction to Other Fractions
Now let’s compare .4375 as a fraction to other fractions to gain a better understanding of its value. When we compare .4375 to 1/2, we can see that it is less than 1/2 because it is closer to 0. When compared to 1/3, .4375 is greater because it is closer to 1/2. Similarly, when compared to 1/4, .4375 is greater.
In summary, .4375 can be expressed as the fraction 4375/10,000. In decimal form, it is 43.75%. When comparing it to other fractions, .4375 falls between 1/3 and 1/2.
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+1 803-820-9654 | 1,307 | 5,131 | {"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} | 4.59375 | 5 | CC-MAIN-2024-18 | latest | en | 0.945023 |
https://edurev.in/studytube/Integer-Answer-Type-Questions-Indefinite-Integral/68b74ba0-be64-46c3-b0bf-73d7817ee04f_t | 1,675,105,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499826.71/warc/CC-MAIN-20230130165437-20230130195437-00180.warc.gz | 246,906,074 | 58,710 | JEE > Integer Answer Type Questions: Indefinite Integral
# Integer Answer Type Questions: Indefinite Integral - Notes | Study All Types of Questions for JEE - JEE
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Q.1. Find the maximum area bounded by the curves y2 = 4ax, y = ax and y = x / a (1 ≤ a ≤ 2).
Ans. 84
The curves are y2 = 4ax and y = ax
At their point of intersection
a2x2 = 4ax
⇒ ax = 4, x = 0
x = 4 / a
⇒ y = 4
i. e.
Similarly for y2 = 4ax and y = x / a ,
⇒ x = 4a3
⇒ B(4a3, 4a2)
Area OAB =
= 84
Q.2. Let f be a real valued function satisfying and Find the area bounded by the curve y = f(x), the y–axis and the line y = 3.
Ans. 3
Given,
putting x = y =1, we get f(1) = 0
Now,
=
⇒ f'(x) = 3 x ⇒ f(x) = 3 lnx + c
Putting x =1 ⇒ c = 0 ⇒ f(x) = 3lnx = y (say)
Required Area =
= 3(e – 0) = 3e sq. units.
Q.3. is equal to
Ans. 0
Given Limit =
=
Q.4. The value of ,where [.] denotes the greatest integer function, is equal to;
Ans. 12
Let I =
= 1/3 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 12.
Q.5. If , then the value od P + Q + R is________.
Ans. 10
Put x + 7 = u4 ⇒ dx = 4u3 du
Rearranging like terms
⇒ P = 2, Q = 4, R = 4
⇒ P + Q + R = 10.
Q.6. If , then k is equal s.
Ans. 1
Dividing the numerator and denominator by x2, the given integral becomes
Let
Hence k = 1.
Q.7. The value of is ____________.
Ans. 9
Q.8. If , then where k has the value equal to?
Ans. 9
⇒ k = 9
Q.9. Value of is equal to?
Ans. 1
= 1
Q.10. Let f be a function defined by , where r = 3k, k∈ I then is equal to?
Ans. 5
Hence, f(x) is periodic with fundamental time period = 3
= 50
= 5
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; | 837 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-06 | latest | en | 0.588409 |
https://en.formulasearchengine.com/wiki/Statistical_significance | 1,603,432,246,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00165.warc.gz | 304,462,293 | 10,314 | # Statistical significance
Statistical significance is the probability that an effect is not due to just chance alone.[1][2] It is an integral part of statistical hypothesis testing where it is used as an important value judgment. In statistics, a result is considered significant not because it is important or meaningful, but because it has been predicted as unlikely to have occurred by chance alone.[3]
The present-day concept of statistical significance originated from Ronald Fisher when he developed statistical hypothesis testing in the early 20th century.[4][5][6] These tests are used to determine which outcomes of a study would lead to a rejection of the null hypothesis based on a pre-specified low probability threshold called p-values, which can help an investigator to decide if a result contains sufficient information to cast doubt on the null hypothesis.[7]
P-values are often coupled to a significance or alpha (α) level, which is also set ahead of time, usually at 0.05 (5%).[7] Thus, if a p-value was found to be less than 0.05, then the result would be considered statistically significant and the null hypothesis would be rejected.
## History
{{#invoke:main|main}} The present-day concept of statistical significance originated by Ronald Fisher when he developed statistical hypothesis testing, which he described as "tests of significance," in his 1925 publication, Statistical Methods for Research Workers.[4][5][6] Fisher suggested a probability of one-in-twenty (0.05) as a convenient cutoff level to reject the null hypothesis.[8] In their 1933 paper, Jerzy Neyman and Egon Pearson recommended that the significance level (e.g., 0.05), which they called α, be set ahead of time, prior to any data collection.[8][9]
Despite his suggestion of 0.05 as a significance level, Fisher did not intend this cutoff value to be fixed and in his 1956 publication, Statistical methods and scientific inference, he even recommended that significant levels be set according to specific circumstances.[8]
## Role in statistical hypothesis testing
{{#invoke:main|main}}
In a two-tailed test, the rejection region or α level is partitioned to both ends of the sampling distribution and make up only 5% of the area under the curve.
Statistical significance plays a pivotal role in statistical hypothesis testing where it is used to determine if a null hypothesis can be rejected or retained. A null hypothesis is the general or default statement that nothing happened or changed.[10] For a null hypothesis to be rejected as false, the result has to be identified as being statistically significant, i.e., unlikely to have occurred by chance alone.
To determine if a result is statistically significant, a researcher would have to calculate a p-value, which is the probability of observing an effect given that the null hypothesis is true.[11] The null hypothesis is rejected if the p-value is less than the significance or α level. The α level is the probability of rejecting the null hypothesis when it is true (type I error) and is usually set at 0.05 (5%), which is the most widely used.[2] If the α level is 0.05, then the probability of committing a type I error is 5%.[12] Thus, a statistically significant result is one in which the p-value for obtaining that result is less than 5%, which is formally written as p < 0.05[12]
If the α level is set at 0.05, it means that the rejection region comprises 5% of the sampling distribution.[13] This 5% can be allocated to one side of the sampling distribution as in a one-tailed test or partitioned to both sides of the distribution as in a two-tailed test, with each tail (or rejection region) comprising 2.5%. One-tailed tests are more powerful than two-tailed tests, as a null hypothesis can be rejected with a less extreme result.
## Defining significance in terms of sigma (σ)
{{#invoke:main|main}} In specific fields such as particle physics and manufacturing, statistical significance is often expressed in units of standard deviation or sigma (σ) of a normal distribution, with significance thresholds set at a much stricter level (e.g., 5 sigma).[14][15] For instance, the certainty of the Higgs boson particle's existence was based on the 5-sigma criterion, which corresponds to a p-value of about 1 in 3.5 million.[15][16]
## Criticisms
{{#invoke:main|main}}
Researchers focusing solely on whether their results are statistically significant might report findings that are not necessarily substantive.[17] To gauge the research significance of their result, researchers are also encouraged to report the effect-size along with p-values, as the former describes the strength of an effect such as the distance between two means and the correlation between two variables.[18] p-values violate the likelihood principle.[19]
{{#invoke:Portal|portal}}
## References
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11. {{#invoke:citation/CS1|citation |CitationClass=book }}
12. {{#invoke:citation/CS1|citation |CitationClass=book }} Cite error: Invalid `<ref>` tag; name "Healy" defined multiple times with different content
13. {{#invoke:citation/CS1|citation |CitationClass=book }}
14. {{#invoke:citation/CS1|citation |CitationClass=book }}
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16. {{#invoke:citation/CS1|citation |CitationClass=book }}
17. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
18. {{#invoke:citation/CS1|citation |CitationClass=book }}
19. {{#invoke:Citation/CS1|citation |CitationClass=journal }} | 1,413 | 6,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-45 | latest | en | 0.957376 |
https://community.home-assistant.io/t/heat-index-apparent-temperature-feels-like-temperature/1282/23 | 1,718,533,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00452.warc.gz | 159,216,870 | 12,006 | # Heat Index / Apparent Temperature / Feels Like Temperature
Re-resurrecting an old thread for reasons I feel are good as well…
I did a bit of research about heat index and it seemed to only apply to the great outdoors. Then I stumbled across the following article: http://www-das.uwyo.edu/~geerts/cwx/notes/chap05/apparentt.html
It was clear to me that heat index and wind chill really don’t apply in the case of indoor temperature. I really wanted to take advantage of the formula provided, but needed Saturation Vapor Pressure and Actual Vapor pressure in order to calculate that value. That’s when I found the following: https://physics.stackexchange.com/questions/4343/how-can-i-calculate-vapor-pressure-deficit-from-temperature-and-relative-humidit which provided the final piece of the puzzle.
Armed with all of the above, I came up with the following value template:
`````` value_template: >
{% set upstairs_temp_c = ((states('sensor.upstairs_temperature') | float) - 32) * 5/9 %}
{% set es = 0.6108 * e ** (17.27 * upstairs_temp_c / (upstairs_temp_c + 237.3)) %} {# Saturation vapor pressure in kPa #}
{% set ea = ((states('sensor.upstairs_humidity') | float) / 100 * es) %} {# Actual vapor pressure in kPa #}
{% set tp = -1.3 + 0.92 * upstairs_temp_c + 2.2 * ea %} {# Apparent temperature in C #}
{% set tpf = (tp * 9/5) + 32 %}
{{ tpf | round(1) }}``````
6 Likes
thanks for sharing this. but I’m not sure is this the right one. I have compared the apparent temperature given by dark sky vs the result calculated at https://www.calculator.net/heat-index-calculator.html using the temperature and humidity value from dark sky are quite consistent.
Anyway, I have revisited the formula page at https://www.wpc.ncep.noaa.gov/html/heatindex_equation.shtml and updated the template based on my understanding. Here is the latest version…
``````- platform: template
sensors:
outdoor_heat_index:
friendly_name: 'Outdoor Feels Like'
value_template: >-
{% set T = ((states.sensor.outdoor_temperature.state|float)*1.8)+32 %}
{% set RH = states.sensor.outdoor_humidity.state|float %}
{% set STEADMAN_HI = 0.5 * (T + 61.0 + ((T-68.0)*1.2) + (RH*0.094)) %}
{% if STEADMAN_HI >= 80 %}
{% set ROTHFUSZ_HI = -42.379 + 2.04901523*T + 10.14333127*RH - 0.22475541*T*RH - 0.00683783*T*T - 0.05481717*RH*RH + 0.00122874*T*T*RH + 0.00085282*T*RH*RH - 0.00000199*T*T*RH*RH %}
{% set HI = ROTHFUSZ_HI %}
{% if RH < 13 and 80 < T < 112 %}
{% set ADJUSTMENT = ((13-RH)/4)*((17-(T-95)|abs)/17)**0.5 %}
{% set HI = HI - ADJUSTMENT %}
{% elif RH > 85 and 80 < T < 87 %}
{% set ADJUSTMENT = ((RH-85)/10) * ((87-T)/5) %}
{% set HI = HI + ADJUSTMENT %}
{% endif %}
{% else %}
{% set HI = STEADMAN_HI %}
{% endif %}
{% set HI_C = (HI-32)/1.8 %}
{{- HI_C|round(1) -}}
unit_of_measurement: '°C'
``````
3 Likes
Hi.
Sorry for resurrecting this topic, but I get errors on this config:
`````` - platform: template
sensors:
outdoor_heat_index:
friendly_name: 'Outdoor Feels Like'
value_template: >-
{% set T = ((states.sensor.0x00158d000304031a_temperature.state|float)*1.8)+32 %}
{% set RH = states.sensor.0x00158d000304031a_humidity.state|float %}
{% set STEADMAN_HI = 0.5 * (T + 61.0 + ((T-68.0)*1.2) + (RH*0.094)) %}
{% if STEADMAN_HI >= 80 %}
{% set ROTHFUSZ_HI = -42.379 + 2.04901523*T + 10.14333127*RH - 0.22475541*T*RH - 0.00683783*T*T - 0.05481717*RH*RH + 0.00122874*T*T*RH + 0.00085282*T*RH*RH - 0.00000199*T*T*RH*RH %}
{% set HI = ROTHFUSZ_HI %}
{% if RH < 13 and 80 < T < 112 %}
{% set ADJUSTMENT = ((13-RH)/4)*((17-(T-95)|abs)/17)**0.5 %}
{% set HI = HI - ADJUSTMENT %}
{% elif RH > 85 and 80 < T < 87 %}
{% set ADJUSTMENT = ((RH-85)/10) * ((87-T)/5) %}
{% set HI = HI + ADJUSTMENT %}
{% endif %}
{% else %}
{% set HI = STEADMAN_HI %}
{% endif %}
{% set HI_C = (HI-32)/1.8 %}
{{- HI_C|round(1) -}}
unit_of_measurement: '°C'
``````
Log:
``````# hassio ha check
Error: Testing configuration at /config
Failed config
sensor.template:
- Invalid config for [sensor.template]: invalid template (TemplateSyntaxError: expected token ')', got 'x00158d000304031a_temperature') for dictionary value @ data['sensors']['outdoor_heat_index']['value_template']. Got '{% set T = ((states.sensor.0x00158d000304031a_temperature.state|float)*1.8)+32 %} {% set RH = states.sensor.0x00158d000304031a_humidity.state|float %} {% set STEADMAN_HI = 0.5 * (T + 61.0 + ((T-68.0)*1.2) + (RH*0.094)) %} {% if STEADMAN_HI >= 80 %}\n {% set ROTHFUSZ_HI = -42.379 + 2.04901523*T + 10.14333127*RH - 0.22475541*T*RH - 0.00683783*T*T - 0.05481717*RH*RH + 0.00122874*T*T*RH + 0.00085282*T*RH*RH - 0.00000199*T*T*RH*RH %}\n {% set HI = ROTHFUSZ_HI %}\n {% if RH < 13 and 80 < T < 11.... (See ?, line ?). Please check the docs at https://home-assistant.io/components/sensor.template/
- platform: template
sensors: [source /config/configuration.yaml:297]
outdoor_heat_index: [source /config/configuration.yaml:298]
friendly_name: Outdoor Feels Like
unit_of_measurement: °C
value_template: {% set T = ((states.sensor.0x00158d000304031a_temperature.state|float)*1.8)+32 %} {% set RH = states.sensor.0x00158d000304031a_humidity.state|float %} {% set STEADMAN_HI = 0.5 * (T + 61.0 + ((T-68.0)*1.2) + (RH*0.094)) %} {% if STEADMAN_HI >= 80 %}
{% set ROTHFUSZ_HI = -42.379 + 2.04901523*T + 10.14333127*RH - 0.22475541*T*RH - 0.00683783*T*T - 0.05481717*RH*RH + 0.00122874*T*T*RH + 0.00085282*T*RH*RH - 0.00000199*T*T*RH*RH %}
{% set HI = ROTHFUSZ_HI %}
{% if RH < 13 and 80 < T < 112 %}
{% set ADJUSTMENT = ((13-RH)/4)*((17-(T-95)|abs)/17)**0.5 %}
{% set HI = HI - ADJUSTMENT %}
{% elif RH > 85 and 80 < T < 87 %}
{% set ADJUSTMENT = ((RH-85)/10) * ((87-T)/5) %}
{% set HI = HI + ADJUSTMENT %}
{% endif %}
{% else %} {% set HI = STEADMAN_HI %} {% endif %} {% set HI_C = (HI-32)/1.8 %} {{- HI_C|round(1) -}}
Successful config (partial)
sensor.template:
``````
Solved:
`````` - platform: template
sensors:
heat_index_suite:
friendly_name: 'Feels Like - Suite'
value_template: >-
{% set T = ((states('sensor.0x00158d000304031a_temperature')|float)*1.8+32) %}
{% set RH = states('sensor.0x00158d000304031a_humidity')|float %}
{% set STEADMAN_HI = 0.5 * (T + 61.0 + ((T-68.0)*1.2) + (RH*0.094)) %}
{% if STEADMAN_HI >= 80 %}
{% set ROTHFUSZ_HI = -42.379 + 2.04901523*T + 10.14333127*RH - 0.22475541*T*RH - 0.00683783*T*T - 0.05481717*RH*RH + 0.00122874*T*T*RH + 0.00085282*T*RH*RH - 0.00000199*T*T*RH*RH %}
{% set HI = ROTHFUSZ_HI %}
{% if RH < 13 and 80 < T < 112 %}
{% set ADJUSTMENT = ((13-RH)/4)*((17-(T-95)|abs)/17)**0.5 %}
{% set HI = HI - ADJUSTMENT %}
{% elif RH > 85 and 80 < T < 87 %}
{% set ADJUSTMENT = ((RH-85)/10) * ((87-T)/5) %}
{% set HI = HI + ADJUSTMENT %}
{% endif %}
{% else %}
{% set HI = STEADMAN_HI %}
{% endif %}
{% set HI_C = (HI-32)/1.8 %}
{{- HI_C|round(1) -}}
unit_of_measurement: '°C'
``````
2 Likes
I was just rereading this thread after I came across it again. I just realized we had very different goals in mind.
The heat index calculation you’re using is, in fact, the correct calculation for outdoor “feels like” temperatures. My goal was to calculate a “feels like” temperature for the indoors, which is a very different calculation–note the NOAA formula is only accurate down to 80 degrees F, which would definitely be outside my comfort zone for an indoor temperature.
My usage is primarly for the purpose of tweaking indoor thermostat settings based on a “feels like” temperature rather than an absolute temperature.
Hopefully, that clarifies the use-case and reasoning behind the different formula.
2 Likes
Thanks – this is nearly what I was looking for. I have also realised that even indoor airflow (through fans and ventilators) affect the comfort level.
Is there a method we can first measure indoor air velocity and then use it to tweak the thermostats.
Do you know what I’d have to change to get this to work in Fahrenheit?
I might be missing something, but that template solves for Fahrenheit. It converts to F to C back to F.
Thank you for this template. I needed a template sensor for VPD.
Anyone know if there’s a way to make this calculation reusable?
I’m a bit confused with the above calculations, as the UK Meterology Office describes outdoor ‘Feels Like’ temperature as a measurement of temp/humidity/windspeed at a height of 5ft. Unless I’m missing something I don’t see windspeed integrated in those calcs. If I find the calculation for this missing piece of info, I’ll post it here.
1 Like
Here in the US, windspeed isn’t considered in the feels-like temperature. Here’s the formula: https://www.weather.gov/media/epz/wxcalc/heatIndex.pdf.
Are you still using this formula for indoor? How has it worked for you?
I’m looking into implementing something like that for my home and I think what you wanted is exactly what I want to do.
Could you tell me if you have noticed a difference by using the feeling temperature compared to the real temp?
There’s generally not a vast difference between the measured temperature and the feels-like temperature. For example, right now there’s only a one-degree difference between the feels-like temperature upstairs and the actual temperature (69.8 vs. 68.8) and that’s with 66% humidity.
I was going to try to look at a historical graph and discovered that history doesn’t record derived values, unfortunately. Otherwise, I might have been able to see how much of a delta I saw between the two.
If you’re interested in picking up a 1-2 degree difference, it might be useful; otherwise, it’s more of a novelty, I think. To answer your question more directly, the only place I leveraged the feels-like temperature is in determining whether to switch over between A/C and Heat. Otherwise, my house uses the thermostat temperature as you would normally.
Just popped in to say this is so beautiful I could cry. You should see my brute force attempt to do the same. With zillions of ifs and ().
Sad.
1 Like
2 Likes
Several solutions were discussed here and on other threads so I tried a few. Here are my results in case anyone is interested.
1. Outdoor Feels Like (Feels Like Suite) @ masterkenobi
2. SensorX Real Feel @ Petrica
3. Limych Feels Like @ Limych
They all follow nearly the exact same pattern but have significant offsets. Not to criticize anyone’s work, but IMO “SenxorX Real Feel” felt the closest to my expectation of temperature experience.
2 Likes
Sorry to spoil the party, but it seems any computation should consider the impact of humidity on insulating value of clothing…
http://agron-www.agron.iastate.edu/courses/Agron541/classes/541/lesson06a/6a.3.html
Hi,
Perhaps heat index and apparent temperature are relatively different concepts with different scope of application ?..
First about Heat Index:
Heat Index shows how much hotter the air seems to be than the ambient dry-bulb temperature due to the influence of humidity.
I found that the formula of the multiple regression analysis conducted by Lans P. Rothfusz is limited by temperature >~27C. Accordingly, for lower temperatures, the result may have large discrepancies relative to the original tables by Steadman. Although a simplified expression ( HI = 0.5 * {T + 61.0 + [(T-68.0)1.2] + (RH0.094)}) is given to circumvent this limitation.
On the other hand, the Apparent temperature has a wider range of acceptable values of dry-bulb temperature values (>20C).
Therefore, it may make sense to combine formulas. However, it should be borne in mind that the heat index / apparent temperature graphs intersect at different temperatures.
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## Efnisyfirlit
• Title Page
• Dedication
• Preface
• Prologue
• In The Beginning Was The Motion…
• Chapter 1: General Probability Theory
• 1.1 Introduction
• 1.2 Problems and Solutions
• Chapter 2: Wiener Process
• 2.1 Introduction
• 2.2 Problems and Solutions
• Chapter 3: Stochastic Differential Equations
• 3.1 Introduction
• 3.2 Problems and Solutions
• Chapter 4: Change of Measure
• 4.1 Introduction
• 4.2 Problems and Solutions
• Chapter 5: Poisson Process
• 5.1 Introduction
• 5.2 Problems and Solutions
• Appendix A: Mathematics Formulae
• Indices
• Surds
• Exponential and Natural Logarithm
• Binomial Formula
• Series
• Summation
• Trigonometric Functions
• Hyperbolic Functions
• Complex Numbers
• Derivatives
• Standard Differentiations
• Taylor Series
• Maclaurin Series
• Landau Symbols and Asymptotics
• L'Hospital Rule
• Indefinite Integrals
• Standard Indefinite Integrals
• Definite Integrals
• Derivatives of Definite Integrals
• Integration by Parts
• Integration by Substitution
• Gamma Function
• Beta Function
• Convex Function
• Dirac Delta Function
• Heaviside Step Function
• Fubini's Theorem
• Appendix B: Probability Theory Formulae
• Probability Concepts
• Bayes' Rule
• Indicator Function
• Discrete Random Variables
• Univariate Case
• Bivariate Case
• Continuous Random Variables
• Univariate Case
• Bivariate Case
• Properties of Expectation and Variance
• Properties of Moment Generating and Characteristic Functions
• Correlation Coefficient
• Convolution
• Discrete Distributions
• Continuous Distributions
• Integrable and Square Integrable Random Variables
• Convergence of Random Variables
• Relationship Between Modes of Convergence
• Dominated Convergence Theorem
• Monotone Convergence Theorem
• The Weak Law of Large Numbers
• The Strong Law of Large Numbers
• The Central Limit Theorem
• Appendix C: Differential Equations Formulae
• Separable Equations
• First-Order Ordinary Differential Equations
• Second-Order Ordinary Differential Equations
• Homogeneous Heat Equations
• Stochastic Differential Equations
• Black–Scholes Model
• Black Model
• Garman–Kohlhagen Model
• Bibliography
• Notation
• Set Notation
• Mathematical Notation
• Probability Notation
• Index
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# Problems and Solutions in Mathematical Finance: Stochastic Calculus
Vörumerki: John Wiley
Tilboði lýkur 13.08.2019
Vörunúmer: 9781119966081
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## Greatest Common Factor (GCF)
It is simply the largest of the common factors. In our previous example, the largest of the common factors is 15, so the Greatest Common Factor of 15, 30 and 105 is 15 The Greatest Common Factor is the largest of the common
## How Do You Find the Greatest Common Factor of Two
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## GRE Math : How to find the greatest common factor
To find the greatest common factor (GCF) between numbers, take each number and write it's prime factorization. Then, identify the factors common to each number and multiply those
## Greatest Common Factor
Step 1: List the prime factors of each number. Step 2: Draw a circle around the prime factors that are common to all lists. Step 3: Multiply the numbers which you have drawn a circle around
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## How to Find the Greatest Common Factor (GCF)
The greatest common divisor (GCD) of two or more numbers is the greatest common factor number that divides them, exactly. It is also called the highest common factor (HCF). For
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4.3E: Shape of the Graph Exercises
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4.3: Derivatives and the Shape of a Graph
194) If c is a critical point of $$f(x)$$, when is there no local maximum or minimum at $$c$$? Explain.
195) For the function $$y=x^3$$, is $$x=0$$ both an inflection point and a local maximum/minimum?
Answer:
It is not a local maximum/minimum because $$f′$$ does not change sign; it is an inflection point because $$f′′$$ does change sign.
196) For the function $$y=x^3$$, is $$x=0$$ an inflection point?
197) Is it possible for a point $$c$$ to be both an inflection point and a local extrema of a twice differentiable function?
Answer:
No
198) Why do you need continuity for the first derivative test? Come up with an example.
199) Explain whether a concave-down function has to cross $$y=0$$ for some value of $$x$$.
Answer:
False; for example, $$y=\sqrt{x}$$.
200) Explain whether a polynomial of degree $$2$$ can have an inflection point.
For the following exercises, analyze the graphs of $$f′$$, then list all intervals where f is increasing or decreasing.
201)
Answer:
Increasing for $$−2<x<−1$$ and $$x>2$$; decreasing for $$x<−2$$ and $$−1<x<2$$
202)
203)
Answer:
Decreasing for $$x<1$$, increasing for $$x>1$$
204)
205)
Answer:
Decreasing for $$−2<x<−1$$ and $$1<x<2$$; increasing for $$−1<x<1$$ and $$x<−2$$ and $$x>2$$
For the following exercises, analyze the graphs of $$f′,$$ then list all intervals where
a. $$f$$ is increasing and decreasing and
b. the minima and maxima are located.
206)
207)
Answer:
a. Increasing over $$−2<x<−1,0<x<1,x>2$$, decreasing over $$x<−2, −1<x<0,1<x<2;$$
b. maxima at $$x=−1$$ and $$x=1$$, minima at $$x=−2$$ and $$x=0$$ and $$x=2$$
208)
209)
Answer:
a. Increasing over $$x>0$$, decreasing over $$x<0;$$
b. Minimum at $$x=0$$
210)
For the following exercises, analyze the graphs of $$f′$$, then list all inflection points and intervals $$f$$ that are concave up and concave down.
211)
Answer:
Concave up on all $$x$$, no inflection points
212)
213)
Answer:
Concave up on all $$x$$, no inflection points (since f'(x) is always increasing)
214)
215)
Answer:
Concave up for $$x<0$$ and $$x>1$$, concave down for $$0<x<1$$
inflection points at $$x=0$$ and $$x=1$$
For the following exercises, draw a graph that satisfies the given specifications for the domain $$x=[−3,3].$$ The function does not have to be continuous or differentiable.
216) $$f(x)>0,f′(x)>0$$ over $$x>1,−3<x<0,f′(x)=0$$ over $$0<x<1$$
217) $$f′(x)>0$$ over $$x>2,−3<x<−1,f′(x)<0$$ over $$−1<x<2,f''(x)<0$$ for all $$x$$
Answer:
answers will vary
218) $$f''(x)<0$$ over $$−1<x<1,f''(x)>0,−3<x<−1,1<x<3,$$ local maximum at $$x=0,$$ local minima at $$x=±2$$
219) There is a local maximum at $$x=2,$$ local minimum at $$x=1,$$ and the graph is neither concave up nor concave down.
Answer:
answers will vary
220) There are local maxima at $$x=±1,$$ the function is concave up for all $$x$$, and the function remains positive for all $$x.$$
For the following exercises, determine
a. intervals where $$f$$ is increasing or decreasing and
b. local minima and maxima of $$f$$.
221) $$f(x)=sinx+sin^3x$$ over −π<x<π
Answer:
a. Increasing over $$−\frac{π}{2}<x<\frac{π}{2},$$ decreasing over $$x<−π\frac{π}{2},x>\frac{π}{2}$$
b. Local maximum at $$x=\frac{π}{2}$$; local minimum at $$x=−\frac{π}{2}$$
222) $$f(x)=x^2+cosx$$
For the following exercise, determine a. intervals where $$f$$ is concave up or concave down, and b. the inflection points of $$f$$.
223) $$f(x)=x^3−4x^2+x+2$$
Answer:
a. Concave up for $$x>\frac{4}{3},$$ concave down for $$x<\frac{4}{3}$$
b. Inflection point at $$x=\frac{4}{3}$$
For the following exercises, determine
a. intervals where $$f$$ is increasing or decreasing,
b. local minima and maxima of $$f$$,
c. intervals where $$f$$ is concave up and concave down, and
d. the inflection points of $$f.$$
224) $$f(x)=x^2−6x$$
225) $$f(x)=x^3−6x^2$$
Answer:
a. Increasing over $$x<0$$ and $$x>4,$$ decreasing over $$0<x<4$$
b. Maximum at $$x=0$$, minimum at $$x=4$$
c. Concave up for $$x>2$$, concave down for $$x<2$$
d. Infection point at $$x=2$$
226) $$f(x)=x^4−6x^3$$
227) $$f(x)=x^{11}−6x^{10}$$
Answer:
a. Increasing over $$x<0$$ and $$x>\frac{60}{11}$$, decreasing over $$0<x<\frac{60}{11}$$ b. Minimum at $$x=\frac{60}{11}$$ c. Concave down for $$x<\frac{54}{11}$$, concave up for $$x>\frac{54}{11}$$ d. Inflection point at $$x=\frac{54}{11}$$
228) $$f(x)=x+x^2−x^3$$
229) $$f(x)=x^2+x+1$$
Answer:
a. Increasing over $$x>−\frac{1}{2}$$, decreasing over $$x<−\frac{1}{2}$$ b. Minimum at $$x=−\frac{1}{2}$$ c. Concave up for all $$x$$ d. No inflection points
230) $$f(x)=x^3+x^4$$
For the following exercises, determine
a. intervals where $$f$$ is increasing or decreasing,
b. local minima and maxima of $$f,$$
c. intervals where $$f$$ is concave up and concave down, and
d. the inflection points of $$f.$$ Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator.
231) [T] $$f(x)=sin(πx)−cos(πx)$$ over $$x=[−1,1]$$
Answer:
a. Increases over $$−\frac{1}{4}<x<\frac{3}{4},$$ decreases over $$x>\frac{3}{4}$$ and $$x<−\frac{1}{4}$$ b. Minimum at $$x=−\frac{1}{4}$$, maximum at $$x=\frac{3}{4}$$ c. Concave up for $$−\frac{3}{4}<x<\frac{1}{4}$$, concave down for $$x<−\frac{3}{4}$$ and $$x>\frac{1}{4}$$ d. Inflection points at $$x=−\frac{3}{4},x=\frac{1}{4}$$
232) [T] $$f(x)=x+sin(2x)$$ over $$x=[−\frac{π}{2},\frac{π}{2}]$$
233) $$f(x)=sinx+tanx$$ over $$(−\frac{π}{2},\frac{π}{2})$$
Answer:
a. Increasing for all $$x$$
b. No local minimum or maximum
c. Concave up for $$x>0$$, concave down for $$x<0$$
d. Inflection point at $$x=0$$
234) [T] $$f(x)=(x−2)^2(x−4)^2$$
235) [T] $$f(x)=\frac{1}{1−x},x≠1$$
Answer:
a. Increasing for all $$x$$ where defined
b. No local minima or maxima
c. Concave up for $$x<1$$; concave down for $$x>1$$
d. No inflection points in domain
236) [T] $$f(x)=\frac{sinx}{x}$$ over $$x=[−2π,2π] [2π,0)∪(0,2π]$$
237) $$f(x)=sin(x)e^x$$ over $$x=[−π,π]$$
Answer:
Solution: a. Increasing over $$−\frac{π}{4}<x<\frac{3π}{4}$$, decreasing over $$x>\frac{3π}{4},x<−\frac{π}{4}$$ b. Minimum at $$x=−\frac{π}{4}$$, maximum at $$x=\frac{3π}{4}$$ c. Concave up for $$−\frac{π}{2}<x<\frac{π}{2}$$, concave down for $$x<−\frac{π}{2},x>\frac{π}{2}$$ d. Infection points at $$x=±\frac{π}{2}$$
238) $$f(x)=lnx\sqrt{x},x>0$$
239) $$f(x)=\frac{1}{4}\sqrt{x}+\frac{1}{x},x>0$$
Answer:
Solution: a. Increasing over $$x>4,$$ decreasing over $$0<x<4$$ b. Minimum at $$x=4$$ c. Concave up for $$0<x<8\sqrt[3]{2}$$, concave down for $$x>8\sqrt[3]{2}$$ d. Inflection point at $$x=8\sqrt[3]{2}$$
240) $$f(x)=\frac{e^x}{x},x≠0$$
For the following exercises, interpret the sentences in terms of $$f,f′,$$ and $$f''.$$
241) The population is growing more slowly. Here $$f$$ is the population.
Answer:
$$f>0,f′>0,f''<0$$
242) A bike accelerates faster, but a car goes faster. Here $$f=$$ Bike’s position minus Car’s position.
243) The airplane lands smoothly. Here $$f$$ is the plane’s altitude.
Answer:
$$f>0,f′<0,f''<0$$
244) Stock prices are at their peak. Here $$f$$is the stock price.
245) The economy is picking up speed. Here $$f$$ is a measure of the economy, such as GDP.
Answer:
$$f>0,f′>0,f''>0$$
For the following exercises, consider a third-degree polynomial $$f(x),$$ which has the properties f′(1)=0,f′(3)=0.
Determine whether the following statements are true or false. Justify your answer.
246) $$f(x)=0$$ for some $$1≤x≤3$$
247) $$f''(x)=0$$ for some $$1≤x≤3$$
Answer:
True, by the Mean Value Theorem
248) There is no absolute maximum at $$x=3$$
249) If $$f(x)$$ has three roots, then it has $$1$$ inflection point.
Answer:
True, examine derivative
250) If $$f(x)$$ has one inflection point, then it has three real roots.
4.3E: Shape of the Graph Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.
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https://kr.mathworks.com/matlabcentral/cody/problems/45187-combinations-using-stirling-numbers-of-the-second-kind/solutions/1991703 | 1,580,160,315,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00138.warc.gz | 515,963,075 | 16,002 | Cody
# Problem 45187. Combinations using Stirling numbers of the second kind
Solution 1991703
Submitted on 26 Oct 2019 by William
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### Test Suite
Test Status Code Input and Output
1 Pass
n = 4; k = 3; y_correct = 14; assert(isequal(stirlingSum(n,k),y_correct))
2 Pass
n = 10; k = 5; y_correct = 86472; assert(isequal(stirlingSum(n,k),y_correct))
3 Pass
n = 15; k = 6; y_correct = 676207628; assert(isequal(stirlingSum(n,k),y_correct))
4 Pass
n = 11; k = 9; y_correct = 678514; assert(isequal(stirlingSum(n,k),y_correct))
5 Pass
n = 14; k = 12; y_correct = 190899230; assert(isequal(stirlingSum(n,k),y_correct))
6 Pass
n = 14; k = 1; y_correct = 1; assert(isequal(stirlingSum(n,k),y_correct))
7 Pass
n = 17; k = 4; y_correct = 715860651; assert(isequal(stirlingSum(n,k),y_correct))
8 Pass
n = 10; k = 10; y_correct = 115975; assert(isequal(stirlingSum(n,k),y_correct)) | 345 | 1,002 | {"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-2020-05 | latest | en | 0.489195 |
https://betterlesson.com/browse/common_core/standard/20/ccss-math-content-k-cc-c-7-compare-two-numbers-between-1-and-10-presented-as-written-numerals?from=standard_level1 | 1,632,107,409,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00672.warc.gz | 181,904,795 | 26,079 | ## Math
### HS Statistics & Probability
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K.CC.C.7
K.CC.C.7
Compare two numbers between 1 and 10 presented as written numerals.
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K.CC.C.6
Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.
K.CC.C.7
Compare two numbers between 1 and 10 presented as written numerals. | 118 | 435 | {"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-2021-39 | latest | en | 0.898039 |
https://isco-iss.faperta.unpad.ac.id/free/fundamental-finite-element-analysis-and-applicatio.html | 1,713,173,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00707.warc.gz | 290,980,179 | 7,600 | Fundamental Finite Element Analysis And Applicatio Pdf Free
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Page :1 2 3 . . . . . . . . . . . . . . . . . . . . . . . 27 28 29
SearchBook[My8x] SearchBook[My8y] SearchBook[My8z] SearchBook[My80] SearchBook[My81] SearchBook[My82] SearchBook[My83] SearchBook[My84] SearchBook[My85] SearchBook[My8xMA] SearchBook[My8xMQ] SearchBook[My8xMg] SearchBook[My8xMw] SearchBook[My8xNA] SearchBook[My8xNQ] SearchBook[My8xNg] SearchBook[My8xNw] SearchBook[My8xOA] SearchBook[My8xOQ] SearchBook[My8yMA] SearchBook[My8yMQ] SearchBook[My8yMg] SearchBook[My8yMw] SearchBook[My8yNA] | 2,648 | 9,266 | {"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-2024-18 | latest | en | 0.612743 |
http://www.toroidalsnark.net/mkss4-pix/larry-riddle.html | 1,679,372,130,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00370.warc.gz | 98,015,537 | 2,324 | ### Tilings of Sierpinski Relative Fractals
Designer and Artist: Larry Riddle
Materials and Methods: Cross Stitch on 22-count canvas
Years of Creation: 2021 and 2022
The two Sierpinski Relative fractal variations shown at right are constructed by recursively dividing a square into four equal squares, deleting one of these, fixing one, and rotating the other two by a non-zero multiple of $${\pi}/{2}$$ radians counterclockwise. You can see in the images how the blue and green pieces are rotated relative to the fixed red piece. Tiles are formed by rotating copies of each relative by $${\pi}/{2}$$ around a common center point, as shown in the images below.
A cross stitch design can only approximately display each relative as the count size of the canvas and the desired frame size determine the maximum number of possible iterations. These pieces were stitched with seven iterations. The $$3^7=2187$$ remaining squares in the construction are each stitched with one cross stitch. In one of the tilings, each of the smaller pieces actually consists of three different shades of that color to illustrate the self-similar nature of a fractal. If only the design is considered but not the colors, then each tiling also has reflective symmetry across horizontal, vertical, and diagonal lines through the center, thus representing the 8 transformations in the symmetry group of the square (also known as the dihedral group of order 8).
For more information about the mathematical properties of Sierpinski relatives and their tilings, visit my website on classic iterated function systems. | 335 | 1,592 | {"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} | 2.671875 | 3 | CC-MAIN-2023-14 | longest | en | 0.921419 |
https://electronics.stackexchange.com/tags/over-voltage-protection/hot | 1,643,000,292,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00225.warc.gz | 261,836,732 | 28,122 | # Tag Info
39
Safety capacitors are classified by X and Y ratings. Let's properly define everything, and then it should become clear how those capacitors can be rated for both X and Y at the same time. Class X Capacitors: These are capacitors are only for use in situations where their failure would not present an electric shock risk, but could result in a fire. That ...
12
It's pretty simple: The Zener diode voltage is at (or slightly above) the normal Vcc voltage. For example, a 5.6V Zener for a 5V Vcc. When Vcc is below the Zener voltage, no current flows through the diode, and the 1k resistor keeps the base of T1 low. No current flows through the collector to the emitter of the transistor. When Vcc is higher than the ...
11
Basically it's a power zener diode circuit. The zener begins turning on the BJT when sufficient voltage is across the line. The transistor takes current and the more that the line voltage tries to rise above the zener voltage (plus a Vbe drop of around 1 volt max), the more current is taken by the BJT. The BD139 can handle currents up to a few amps and that ...
10
It depends on how fast you want it... Zener diodes have pretty high capacitance, so you'll need a low value series resistor, which means it will draw a lot of current from the signal source. With your 100R value, a 15V source would have to provide (15-3.3)/100 = 117mA current, and the resistor would burn 1.3W. Both are inconvenient. If the signal is slow ...
8
The wire has inductance and, the longer the wire is the more inductance it has. An inductance likes to maintain the current flowing through itself so, when your switch opens, the small stored energy in the cables magnetic field tries to maintain current flow and produces a sizable voltage (aka back-emf) in doing so. This can easily exceed the maximum voltage ...
8
You can get rough calculations of the breakdown voltage from calculators. This paper has some measurements from an etched 0.55mm tooth gap: Wan, F., Pilla, V., Li, J., Pommerenke, D., Shumiya, H., & Araki, K. (2014). Time Lag of Secondary ESD in Millimeter-Size Spark Gaps. IEEE Transactions on Electromagnetic Compatibility, 56(1), 28–34. doi:10.1109/...
7
The current through the zener will be determined by the source impedance of your 12/24 volt supply and the associated wiring. Most supplies will have a low enough internal impedance that the zener diode will fail when it conducts. A slow blow fuse is not applicable in this situation since by the time the fuse blows, it is likely that the damage has been ...
7
Here is one method: simulate this circuit – Schematic created using CircuitLab If the errant user leaves the +12 on there the resistor will see almost half a watt and the transistor a few hundred mW. You could increase R1 to a few hundred ohms to reduce both those numbers. Note that the input current is diverted to ground through Q1 rather than into ...
7
Yes, that would work fine, but you can make this even more simple: simulate this circuit – Schematic created using CircuitLab Zener diodes behave more or less as a "normal" diode when in forward mode so we can use that to clamp negative voltages. When using a 12V, 400 mW zener diode the maximum current through the zener diode can be 400mW / 12 V = ...
6
No, X caps should fail open to prevent the risk of fire. However, there have been some parts that have caught fire after a number of years in the field. It's common wisdom that paper dielectric caps are more reliable because paper soaked with epoxy self-heals better than polypropylene. In some cases the epoxy has undergone some sort of fatal change. ...
6
Any accountable credits of this answer have to go to @bobflux . bobflux's answer points out all the aspects to be considered. I just would add a variation of bobblux's. simulate this circuit – Schematic created using CircuitLab
5
A linear regulator is inappropriate here. Use one of the many many buck regulator chips out there. Pick one that can handle a bit more than the max possible output after the full wave bridge, and you don't have to worry about that part at all. A buck switcher will be smaller and cheaper than a linear regulator after you include the cost and space of ...
5
Your circuit has issues and I would not try to "patch" these as there are simpler, more reliable solutions! I would use a poly fuse which is a self-recovering fuse. After it blows it will recover (become "good" again) after some time by itself. Of course you could also just use an ordinary glass fuse but make sure that the user can replace it without ...
5
Sometimes a stupid but convenient design just needs stupid protection.
5
The polarity protection works correctly as explained in Mosfet in reverse polarity protection. The rest is the Typical Applications given by Microchip in the MCP16301/H datasheet. So, I don't see any issues there. I don't know if you have considered the inrush current when applying 30V while C2 initially forms a short: it should not exceed max Pulsed Body-...
5
In effect it's an over voltage spark gap. If a line surge occurs, it will cause breakdown across the shark's teeth and (hopefully) offer some over-voltage protection to other parts of the circuit.
5
TL431 will work fine in this application, with a few caveats: When VREF<2.5V it is off but it will still draw a small cathode current to power the internal circuitry. When VREF=2.5V, in order to regulate, it needs to be able to draw a cathode current at least equal to "minimum cathode current for regulation" which is 1mA. So R3 should be around ...
4
You can probably clean this up by using a series resistor at the FPGA end. This will slow the rise and fall times slightly but should eliminate the overshoot and ringing. You will probably have to determine the value experimentally, 50-100 ohms is a good starting point.
4
Your interpretation is correct (note that the datasheet under Features (page 1) indicates a 12kV rating on the OUT pins vs 4kV on the IN pins which makes the in/out characteristic non-symmetric. The equivalent schematic looks suspect to me, though- check Texas Instruments TPD2S017 part, which is basically the same thing (with the in/out nomenclature ...
4
Your circuit with F1 overlooks how a crowbar circuit is designed to operate--the SCR is intended to conduct a very large current, forcibly blowing the supply fuse and isolating your circuit from the overvoltage. The idea of a crowbar circuit presumably takes its name from the act of throwing a metal crowbar onto a live circuit causing protection elements to ...
4
Most likely, no. When the SCR fires on an overvoltage, the crowbar current that is needed to blow the fuse also passes through the FET. Since the FET's source is then dragged down to near the negative rail, it'll drop out of conduction, leaving the intrinsic diode passing that current, and it'll almost certainly pop before the fuse. You could reverse the ...
4
The first circuit you show, with the resistor in-line with the load current, is what you require. That way, the Zener diode will limit the voltage, and the excess voltage will be dropped across the resistor. In your suggested circuit, the excess voltage will also be dropped across the resistor, but since the resistor is in series with the diode, and the ...
4
simulate this circuit – Schematic created using CircuitLab If you have need a resistive divider, it can limit current instead of an explicit current limiting resistor. The diode rail clamps can only clamp to the rails if there is voltage on the rails. They won't offer protection if the ADC is unpowered. But zeners or TVSs which breakdown in reverse ...
4
There is a parasitic diode in your MOSFET (aka the body diode) that will keep your output current flowing irrespective of how you believe you are controlling the device: - Maybe you meant the MOSFET source and drain to be swapped but screwed-up?
4
USB charging has become very complex but also very powerful. There are several official and de-facto standards used concurrently: USB 1.0 / 2.0 / 3.0 Apple chargers Qualcomm Quick Charge 1.0 / 2.0 / 3.0 / 4.0 USB Power Delivery All standards will initially deliver 5V (or sometimes up to 5.2V to compensate for losses in the cable). Plain old USB charger The ...
4
I suggest you limit the energy to an intrinsically safe level using a higher voltage zener diode before dealing with other requirements.
4
You can't really calculate a resistance, but using the current directly is a valid approach. Assume maximum current is flowing in or out of the pin (typically you'd be worried about the current flowing out of the pin when the input is low and into the pin when the input is high) and ensure there is adequate noise margin remaining under those conditions. The ...
4
I would like to ask if this circuit would work as expected, as I have never used such shunt regulators before and don't know their pitfalls. The TL431 is not particularly fast device so, the saviour here is the 10 uF capacitor in your circuit (C1). C1 really does need to be here to slow things down enough so that the TL431 can handle the slower surge. But, ...
3
Try this as an example of using an SCR and a zener: - Picture source. If the voltage exceeds Vz + Vgt then the SCR fires and shorts out the supply. Only when the power supply is removed or dropped in voltage to a low level does the short circuit reset. Added section If you need a tighter tolerance for crowbar activation than that offered by zener-sensing ...
3
To answer your questions, these are my suspicions based on your observations: The symptoms you observe are very likely an attempt to provide "12V" to 12V fans, from a 24V supply. Such a theoretical attempt to provide 12V with high efficiency from a 24V supply probably deserves full marks for trying, but the designer IMHO perhaps did not think his or her way ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2,298 | 9,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-05 | longest | en | 0.921314 |
https://istopdeath.com/find-the-volume-box-8cm3cm2cm/ | 1,669,838,950,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00022.warc.gz | 351,710,018 | 15,257 | # Find the Volume box (8cm)(3cm)(2cm)
h=8cml=3cmw=2cm
The volume of a box is equal to the length l times the width w times the height h.
(length)⋅(width)⋅(height)
Substitute the values of the length l=3, the width w=2, and the height h=8 into the formula.
3⋅2⋅8
Multiply 3 by 2.
6⋅8
Multiply 6 by 8.
48(cm)3
Find the Volume box (8cm)(3cm)(2cm) | 139 | 344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-49 | latest | en | 0.669038 |
https://www.inchcalculator.com/convert/kilonewton-to-micronewton/ | 1,718,971,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00710.warc.gz | 735,617,141 | 15,619 | # Kilonewtons to Micronewtons Converter
Enter the force in kilonewtons below to get the value converted to micronewtons.
## Result in Micronewtons:
1 kN = 1,000,000,000 μN
Do you want to convert micronewtons to kilonewtons?
## How to Convert Kilonewtons to Micronewtons
To convert a measurement in kilonewtons to a measurement in micronewtons, multiply the force by the following conversion ratio: 1,000,000,000 micronewtons/kilonewton.
Since one kilonewton is equal to 1,000,000,000 micronewtons, you can use this simple formula to convert:
micronewtons = kilonewtons × 1,000,000,000
The force in micronewtons is equal to the force in kilonewtons multiplied by 1,000,000,000.
For example, here's how to convert 5 kilonewtons to micronewtons using the formula above.
micronewtons = (5 kN × 1,000,000,000) = 5,000,000,000 μN
### How Many Micronewtons Are in a Kilonewton?
There are 1,000,000,000 micronewtons in a kilonewton, which is why we use this value in the formula above.
1 kN = 1,000,000,000 μN
## What Is a Kilonewton?
One kilonewton is equal to 1,000 newtons, which are equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.
The kilonewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "kilo" is the prefix for thousands, or 103. Kilonewtons can be abbreviated as kN; for example, 1 kilonewton can be written as 1 kN.
## What Is a Micronewton?
One micronewton is equal to 1/1,000,000 of a newton, which is equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.
The micronewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "micro" is the prefix for millionths, or 10-6. Micronewtons can be abbreviated as μN; for example, 1 micronewton can be written as 1 μN.
## Kilonewton to Micronewton Conversion Table
Table showing various kilonewton measurements converted to micronewtons.
Kilonewtons Micronewtons
0.000000001 kN 1 μN
0.000000002 kN 2 μN
0.000000003 kN 3 μN
0.000000004 kN 4 μN
0.000000005 kN 5 μN
0.000000006 kN 6 μN
0.000000007 kN 7 μN
0.000000008 kN 8 μN
0.000000009 kN 9 μN
0.0000000001 kN 0.1 μN
0.000000001 kN 1 μN
0.00000001 kN 10 μN
0.0000001 kN 100 μN
0.000001 kN 1,000 μN
0.00001 kN 10,000 μN
0.0001 kN 100,000 μN
0.001 kN 1,000,000 μN
0.01 kN 10,000,000 μN
0.1 kN 100,000,000 μN
1 kN 1,000,000,000 μN | 847 | 2,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.79502 |
https://codeforces.com/problemset/problem/1424/G | 1,652,798,508,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00162.warc.gz | 230,229,737 | 13,292 | G. Years
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
During one of the space missions, humans have found an evidence of previous life at one of the planets. They were lucky enough to find a book with birth and death years of each individual that had been living at this planet. What's interesting is that these years are in the range $(1, 10^9)$! Therefore, the planet was named Longlifer.
In order to learn more about Longlifer's previous population, scientists need to determine the year with maximum number of individuals that were alive, as well as the number of alive individuals in that year. Your task is to help scientists solve this problem!
Input
The first line contains an integer $n$ ($1 \le n \le 10^5$) — the number of people.
Each of the following $n$ lines contain two integers $b$ and $d$ ($1 \le b \lt d \le 10^9$) representing birth and death year (respectively) of each individual.
Output
Print two integer numbers separated by blank character, $y$ — the year with a maximum number of people alive and $k$ — the number of people alive in year $y$.
In the case of multiple possible solutions, print the solution with minimum year.
Examples
Input
3
1 5
2 4
5 6
Output
2 2
Input
4
3 4
4 5
4 6
8 10
Output
4 2
Note
You can assume that an individual living from $b$ to $d$ has been born at the beginning of $b$ and died at the beginning of $d$, and therefore living for $d$ - $b$ years. | 393 | 1,482 | {"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} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.915783 |
https://www.metric-conversions.org/volume/canadian-cups-to-liters.htm | 1,723,358,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00683.warc.gz | 677,400,290 | 11,400 | # Canadian Cups to Liters (cup can to L)
/
Liters to Canadian Cups (Swap Units)
Format
Accuracy
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
Show formula
## Canadian Cups to Liters formula
L =
cup can
_________
4.3994
Show working
Show result in exponential format
A Canadian liquid unit which differs slightly to the US and metric cup measurment
L =
cup can
_________
4.3994
## Liters
Basic unit of volume in the metric system. A liter of water weighs one kilogram.
## Canadian Cups to Liters table
Start
Increments
Accuracy
Format
Print table
< Smaller Values Larger Values >
0cup can 0.00L
1cup can 0.23L
2cup can 0.45L
3cup can 0.68L
4cup can 0.91L
5cup can 1.14L
6cup can 1.36L
7cup can 1.59L
8cup can 1.82L
9cup can 2.05L
10cup can 2.27L
11cup can 2.50L
12cup can 2.73L
13cup can 2.95L
14cup can 3.18L
15cup can 3.41L
16cup can 3.64L
17cup can 3.86L
18cup can 4.09L
19cup can 4.32L
20cup can 4.55L
21cup can 4.77L
22cup can 5.00L
23cup can 5.23L
24cup can 5.46L
25cup can 5.68L
26cup can 5.91L
27cup can 6.14L
28cup can 6.36L
29cup can 6.59L
30cup can 6.82L
31cup can 7.05L
32cup can 7.27L
33cup can 7.50L
34cup can 7.73L
35cup can 7.96L
36cup can 8.18L
37cup can 8.41L
38cup can 8.64L
39cup can 8.86L
40cup can 9.09L
41cup can 9.32L
42cup can 9.55L
43cup can 9.77L
44cup can 10.00L
45cup can 10.23L
46cup can 10.46L
47cup can 10.68L
48cup can 10.91L
49cup can 11.14L
50cup can 11.37L
51cup can 11.59L
52cup can 11.82L
53cup can 12.05L
54cup can 12.27L
55cup can 12.50L
56cup can 12.73L
57cup can 12.96L
58cup can 13.18L
59cup can 13.41L | 754 | 1,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.720137 |
https://physics.stackexchange.com/tags/velocity/new | 1,603,864,989,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896778.71/warc/CC-MAIN-20201028044037-20201028074037-00460.warc.gz | 461,208,817 | 34,023 | # Tag Info
0
For a spacetime interval: $\Delta \tau = \sqrt{\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}$ $$\frac{d\tau}{dt} = \frac{\Delta t - v_x\Delta x - v_y\Delta y - v_z\Delta z}{\sqrt{\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}}$$ Because the spatial coordinates are function of t. Dividing by $\Delta t$ numerator and denominator and letting deltas go ...
2
why doesn't the Lorentz factor of a component of 3-space momentum depend on only the corresponding component of velocity? Because that definition of $\mathbf{p}$ wouldn’t transform as a 3D vector should under rotations. Since the velocity $\mathbf{v}$ is a vector under rotations, the Lorentz factor $\gamma=1/\sqrt{1-v^2/c^2}$ needs to be a rotational scalar,...
2
REFERENCE : My answer here How to add together non-parallel rapidities?. $=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$ From above reference consider the velocity vectors \begin{align} \mathbf{u}_1 \boldsymbol{=} \left(u_{1x},u_{1y},u_{1z}\right) & \boldsymbol{=}\left(u_1 ...
1
The following may be a useful approach. A simple traveling wave can be written as: $$y=\sin\left(kx-\omega t \right)$$ We want to follow the position of the wave at a constant phase, $\phi$. Let that phase=0 which leads to: $$\phi=kx - \omega t = 0$$ $$kx=\omega t$$ $$x=\frac{\omega}{k} t$$ Then the velocity will be $$v=\dot{x}=\frac{\omega}{k}$$ ...
2
Here's a perspective from a former chemistry student. I'll try to explain using molecular motion. I once taught myself compute programming and made a particle-geometry collision simulator to test the following, and it seemed to work, so I'm drawing on the intuition that I learned from that project. Background: Pressure is one measure of kinetic energy ...
0
It depends on the type of valve. Butterfly valves for example have a approximately linear behaviour only at a few percent of opening. Gate valves are more linear. But for small openings as 10% to 20%, an almost linear behaviour is expected. That means the flow doubles, provided the volume of the reservoir is big enough to have its level practically constant.
1
I just can't wrap my head around why pressure decreases as velocity increases. That's somewhat backwards. That makes it sound like a decrease in pressure is caused by an increase in velocity, when it's more that an increase in velocity is caused by a decrease in pressure. If there is a pressure differential, that means there is a net force on the fluid, ...
1
I want to replace the masses with their relativistic mass where relativistic mass is the following equation: Yes, you can. But remember that you had better write the relativistic law of momentum conservation. When you do, you will find that it is as if you replace the masses in your classical equation with the corresponding relativistic masses. However, ...
10
i just can't wrap my head around why pressure decreases as velocity increases.. When velocity increases then you obviously have acceleration. Now what is causing this acceleration? As always (according to Newton's second law, $\vec{F}=m\vec{a}$) acceleration is caused by a force. In this case the force acting on a piece of fluid comes from the pressure ...
23
I just can't wrap my head around why pressure decreases as velocity increases This is a classic misunderstanding of Bernoulli's equation. What Bernoulli's equation actually says is that the velocity will increase in the direction of decreasing pressure: $P_2-P_1=-\frac12\rho(v_2^2-v_1^2)$. This makes sense: if the pressure is higher on the left than on the ...
3
You have this equation, $p = mv$. If you increase the velocity and keep $p$ constant, yes you must have had an increase in mass. The question is, what does it mean for $p$ to be constant? It is defined by the product of mass and velocity. So you have done nothing to make predictions, you have just stated what you would call mass if you increased v and kept p ...
0
Note, for a small change in $p$ (and $v$, as their correlation is 1, or $m$ depending on normalization): $$\delta m = \frac d {dp}({\frac p v})\delta p+ \frac d {dv}({\frac p v})\delta v$$ $$\delta m = \frac{\delta p} v - \frac{p\delta v}{v^2}$$ Since: $$\delta p = m \delta v$$ we get: $$\delta m = \frac{\delta p} v - \frac{p\delta p/m}{v^2}$$ $$\delta ... 1 You are conflating work with energy, but there is a slight difference. In fact: Work=change in Energy, i.e. U=ΔK. Work is the accumulated force magnitude over distance along a path, so that an element of work dU=<F, ds>=F.ds.cos(θ) , where ds=elemental displacement along trajectory curve S , F=applied Force along S. We integrate this and then call the ... 5 So you aren't doing the same thing as your teacher here. Notice how in what your teacher did they have \Vert d \mathbf r \Vert =ds, so they did not say that$$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2 $$this would not be true in say circular motion, where the position vector is changing but its magnitude is not. However, ... 0 I think that in the first case, a step is missing:$$\frac{d\mathbf r}{dt}.\frac{d\mathbf r}{dt} = \frac{d\mathbf r}{ds}\frac{ds}{dt}.\frac{d\mathbf r}{ds}\frac{ds}{dt}$$\frac{d\mathbf r}{ds} is an unit vector in the direction of \frac{d\mathbf r}{dt}. So the dot product: \frac{d\mathbf r}{ds}.\frac{d\mathbf r}{ds} = 1 We can't do the same for \frac{... 3 Yes, \vec v=\frac{ds}{dt}\hat{t} So,$$ d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$But {dv}^2=(d^2s)^2 1 Since \vec v_{rel} is a scalar multiple of \vec r we have \hat v_{rel} = \hat r, so \displaystyle \frac {d \hat v_{rel}}{dt} = \frac {d \hat r}{dt} = \vec \omega \times \hat r \\ \displaystyle \Rightarrow |\vec v_{rel}| \frac {d \hat v_{rel}}{dt} = |\vec v_{rel}| ( \vec \omega \times \hat r ) = \vec \omega \times (|\vec v_{rel}| \hat r) = \vec \omega \... 0 If a rotating (constant) vector is decomposed into magnitude and direction \vec{\rm vec} = v\, \hat{e} and the derivative of a unit vector is \dot{\hat{e}} = \vec{\omega} \times \hat{e}, then multiply both sides with v$$ \frac{\rm d}{{\rm d}t} \vec{\rm vec} = v \frac{\rm d}{{\rm d}t} \hat{e} = v \left( \vec{\omega} \times \hat{e} \right) = \vec{\...
1
Since $\hat{v}_{rel}$ is a unit vector $\dot{\hat{v}}_{rel} = \vec{\omega} \times \hat{v}_{rel}$. Multiplying by $|\vec{v}_{rel}|$ on both sides gives you that equation.
2
Figure the volume of 3 meters of 13 mm pipe, this is the flow rate per second in the pipe. Then divide the volume by 69 and you will have the average flow rate per second for each hole. The precise flow rate for each individual hole would depend on multiple variables, but the total of the 69 holes will equal the flow rate of the main pipe as flow rate in ...
2
If the ship programs the engines to have an increment of 0.1c every 60s, the interval between the announcements (until 0.9 c) is by definition 60s in the ship's time . The way the ship knows its velocity can be the average velocity of stars (blue shifting ahead and red shifting behind). Of course the crew feels an increasingly strong fictitious force to ...
9
what the crew would "see" on Earth (if they could) is life speeding up until it's well beyond a blur. What the people of Earth would see if they could "see" into the ship are people slowing down until they stop moving. No: if the ship is approaching Earth, then people on Earth will see clocks on the ship running fast, and people on the ...
2
The reference frame of the ship is non-inertial, so the positioning of the navigator and the captain with respect to the acceleration direction matters. By the equivalence principle, the ship's reference frame will look the same as being on the surface of a planet with a very strong surface gravity (51 thousand $g$s with the acceleration you have given!). So,...
0
Mass cannot travel at c (the speed of light in a vacuum), but it could travel near c. To an outside observer considered to be at rest relative to the ship, the time passing on the ship would become more dilated (passing slower) the faster the ship went. Any one in the ships frame of reference would observe time as passing normally for them. So the people on ...
2
As the navigator and captain are in the same reference frame, there will be no relativistic time dilation between the two. However, to accelerate an object to the speed of light would require an infinite amount of energy, and the energy required to keep accelerating increases with speed, so a constant acceleration of $0.1c$/min is not sustainable.
0
In many discussions of Special Relativity, there are three velocities: the velocity of an object in some reference frame, the velocity of the same object in some other reference frame (one that is moving relative to the first frame), and the relative velocity between the two frames. These are often denoted as $\mathbf u$, $\mathbf u’$, and $\mathbf v$, or $\... 1 Calculate their relative velocity. Determine how long it will take for both vehicles to meet over that distance. From there it will be trivial to calculate the distance the SUV travels. 1 The key to showing why they behave the same is calculus. You are correct that a particle cannot simultaneously have a force on it and no forces on it, just due to sheer logic. However, what we can say is that the particle can have a force on it at some time point$t$(such as when the two objects are at the same velocity), and no force on it at some time$t ...
0
in the moment, both particles have the same velocity it is of cause the same, for example if both were cars of the same mass they would cause the same damage in a collision. What is different is not the velocity, but the acceleration. So if you compare at the same moment and in in this moment the acceleration stops both continue with 10mps.But for many ...
0
I’m referring to an instant in time where the velocities of two different particles are identical. One particle experiences a force and the other does not. I think their states of motion are different even if the velocities are identical. Am I correct? The state of motion of an object is defined by its velocity. See the following: https://www....
1
Pressure and density are proportional in gas. Sound speed depends on pressure AND density in opposite way, so the effect cancels out an the speed is only dependent of temperature and kind of gas. This is not related to sound speed in fluids, whir very high densities compared to gas
1
The only galaxies that we observe to be blue-shifted are some galaxies in the Local Group such as Andromeda. These galaxies are relatively close neighbours to the Milky Way, so although they are still subject to the Hubble flow they also have an individual radial speed towards us that is greater than the expansion of space. Galaxies which are further away ...
0
The $\frac{dv}{dx}=0$ suggests that you are either moving with constant velocity, the object has become stationary or you're going to turn. Eg: 1.You are moving with constant velocity along a line. Since the velocity is constant with respect to distance, the derivative is zero. 2.Becoming stationery: A body starting with velocity $v$ on a surface being ...
1
In an inertial frame of reference an object at rest will remain at rest unless acted on by a force and an object in motion will remain in motion at a constant velocity unless acted on by a force. This is according to Newton's first law of motion. So an object considered to be moving with no friction or other forces acting on it will continue moving with a ...
0
I am not sure if this answers OP's question, but momentum can mean one of two things. It is either a "generic momentum" $(q,p)\in T^\ast M$, which is just a covector on $M$ defined at some point, or what I am calling the canonical momentum, which is actually a map $\xi:TM\rightarrow T^\ast M$. Given a Lagrangian $L:TM\rightarrow\mathbb R,\ (q,\dot ... 0 I am not exactly sure about my answer, however, I think that the relativistic form for$R$is: $$R = \frac{1}{\gamma_a m_a + \gamma_b m_b}\left(\gamma_a m_a(0) + \gamma_b m_b (vt)\right)\space,$$ where, according to your data,$\gamma_a=1$and$\gamma_b=\gamma. Therefore, we have: \begin{align*} u = \frac{R}{t} = \frac{\gamma}{m_a + \gamma m_b}m_bv = \frac{... 1 I believe this may a useful approach.x_{1}$refers to the vertical position of ball 1 (starting at the top of the building) and$x_{2}$is ball 2.$h$is the height of the building. $$x_{1}=h-\frac{1}{2}gt^{2} \quad \left(1\right)$$ $$x_{2}=v_{0}t-\frac{1}{2}gt^{2} \quad \left(2\right)$$ Solve for the time when ball 1 reaches 60 m above the ground (... 2 This law holds true for any inertial frame. Thus the velocity referred here, can be with respect to any inertial frame. The momentum conservation law states: $$\sum_{i=1}^N m_i u_i = \sum_{i=1}^N m_i v_i$$ where, the subscript$i$stands for$i$-th particle.$N$is the total number of particles.$m_i$is the mass of the$i$-th particle.$u_i$and$v_i$are ... 0 As you said the Lagrangian is defined on the tangent bundle, whose elements, loosely speaking, are pairs of a coordinate and a derivative, e.g. $$(q, \dot{q}) = \left((q_i)_i, \; \dot{q}_j\frac{\partial}{\partial{q_j}}\right)$$ The Hamiltonian on the other hand is defined on the cotangent bundle, whose elements are pairs of a coordinate and a 1-form, e.g. $$... 1 Gravity will pull on both balls the same. If there was nothing else in the way, it would cause both balls to accelerate at the same rate. However, there is something else in the system. There is the track, and it provides a "normal" force which pushes perpendicular to the track. This normal force is partially opposing gravity. Near the bottom, ... 2 Short answer : Depends on the magnitude of force acting since two bodies of different mass can have same acceleration and thus same velocity after sometime if both were initially at rest or moving with same velocity. Explanation : Case 1 : Take two bodies , say of mass m_1 and m_2 (m_1 < m_2) and suppose both of them experience the same amount of ... 3 I am assuming one-directional motion for simplicity. Suppose you apply a Force F on objects with different masses. Then using Newton's Second Law:$$a=F/m$$Meaning that if mass becomes greater, acceleration becomes smaller. Further a=dv/dt. Therefore$$dv = a dt = \frac Fmdt \int{dv}=\Delta v= \int\frac Fm dt$$Clearly for the same force F, for ... 3 If I understand your question, you are asking about applications of Newton's Second Law: F=ma. In the case of acceleration, you have$$ a=\frac{F}{m} \qquad(1) $$which shows a greater mass, m, results in a smaller acceleration, a, for a given constant force, F. In the case of velocity, the expression is:$$ v=at $$Using a from (1) above, this ... 0 why the two balls hit the bottom at the same time Not going into a lot of details, just notice inclination of normal force vectors : So,$$ \frac {\alpha_2}{\alpha_1} \propto \frac {L_1}{L_2} $$Or in words - ball at higher altitude is pushed down by a greater force, but it has to overcome a greater distance too. While bottom ball has comparatively small ... 1 Without going into the math behind pendulums, we can still make it plausible that their periods do not depend on their amplitude. A basic feature of all harmonic pendulums is that they accelerate the swinging object with an acceleration proportional to its displacement. Meaning that if we increase initial displacement, we also get a higher initial ... 0 why the two pendulums released from different angles have the same time period ? The time period for a simple pendulum is independent of its initial conditions.$$T = 2\pi\sqrt{\frac{l}{g}}$$where$l$is the length of the string of the pendulum and$g$is the acceleration due to gravity As you can see its time period doesn't even depend on the mass or the ... 2 The velocity vector describes the change in the displacement vector, and hence, it can be in any direction relative to what direction the displacement vector points in, since there are many ways the position of the particle can change. A few counterexamples: Circular motion: Velocity vector is perpendicular to the position vector which denotes the particle ... 1 Displacement is the shortest possible distance between the initial point and final point of line of motion. Do velocity and displacement always have the same direction? Need not to be. Imagine a ball being projected vertically upward with some speed$v$. At some time t (less than time of ascent i.e.$t<\frac{v}{g}$), the direction of its velocity is ... 0 And answers own question: After z is measured, the distinction between redshift and blueshift is simply a matter of whether z is positive or negative. For example, Doppler effect blueshifts (z < 0) are associated with objects approaching (moving closer to) the observer with the light shifting to greater energies. Conversely, Doppler effect redshifts (z &... 2 In air the acceleration depends on the mass, so the objects will not reach the ground at the same time. The sum of the forces with air resistance.$m a = m g-\frac{1}{2} A \text{$\rho$c} v^2$or$a=g-\frac{A \text{$\rho$c} v^2}{2 m}$So that the acceleration$a$indeed depends on the mass. Without air resistance$a=g\$ and the acceleration is independent ...
Top 50 recent answers are included | 4,555 | 17,448 | {"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} | 3.65625 | 4 | CC-MAIN-2020-45 | longest | en | 0.794944 |
https://convertoctopus.com/20-4-days-to-minutes | 1,637,967,277,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00187.warc.gz | 260,869,373 | 8,470 | ## Conversion formula
The conversion factor from days to minutes is 1440, which means that 1 day is equal to 1440 minutes:
1 d = 1440 min
To convert 20.4 days into minutes we have to multiply 20.4 by the conversion factor in order to get the time amount from days to minutes. We can also form a simple proportion to calculate the result:
1 d → 1440 min
20.4 d → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 20.4 d × 1440 min
T(min) = 29376 min
The final result is:
20.4 d → 29376 min
We conclude that 20.4 days is equivalent to 29376 minutes:
20.4 days = 29376 minutes
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 3.4041394335512E-5 × 20.4 days.
Another way is saying that 20.4 days is equal to 1 ÷ 3.4041394335512E-5 minutes.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty point four days is approximately twenty-nine thousand three hundred seventy-six minutes:
20.4 d ≅ 29376 min
An alternative is also that one minute is approximately zero times twenty point four days.
## Conversion table
### days to minutes chart
For quick reference purposes, below is the conversion table you can use to convert from days to minutes
days (d) minutes (min)
21.4 days 30816 minutes
22.4 days 32256 minutes
23.4 days 33696 minutes
24.4 days 35136 minutes
25.4 days 36576 minutes
26.4 days 38016 minutes
27.4 days 39456 minutes
28.4 days 40896 minutes
29.4 days 42336 minutes
30.4 days 43776 minutes | 446 | 1,617 | {"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-2021-49 | latest | en | 0.801771 |
https://cirl.lcsr.jhu.edu/introduction-to-computer-vision/k-means/ | 1,675,875,549,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500837.65/warc/CC-MAIN-20230208155417-20230208185417-00513.warc.gz | 193,511,033 | 31,502 | # K-means
K-means is an algorithm for partitioning (or clustering) $N$ data points into $K$ disjoint subsets $S_j$ containing $N_j$ data points so as to minimize the sum-of-squares criterion
$J=\sum_{j=1}^K\sum_{n \in S_j}|x_n-mu_j|^2,$
where $x_n$ is a vector representing the $nth$ data point and $mu_j$ is the geometric centroid of the data points in $S_j$.
In general, the algorithm does not achieve a global minimum of $J$ over the assignments. In fact, since the algorithm uses discrete assignment rather than a set of continuous parameters, the “minimum” it reaches cannot even be properly called a local minimum. Despite these limitations, the algorithm is used fairly frequently as a result of its ease of implementation.
The algorithm consists of a simple re-estimation procedure as follows. Initially, the data points are assigned at random to the $K$ sets. For step 1, the centroid is computed for each set. In step 2, every point is assigned to the cluster whose centroid is closest to that point.
The most common algorithm uses an iterative refinement technique. Due to its ubiquity it is often called the k-means algorithm; it is also referred to as Lloyd’s algorithm, particularly in the computer science community.
Given an initial set of $k$ means $\mathbf{mu_1^{(1)},\dots,mu_k^{(1)}}$, which may be specified randomly or by some heuristic, the algorithm proceeds by alternating between two steps:
Assignment step: Assign each observation to the cluster with the closest mean
(i.e. partition the observations according to the Voronoi diagram generated by the means).
$S_i^{(t)} = \left\{ \mathbf x_j : \big\| \mathbf x_j - \mathbf{ mu^{(t)}_i} \big\| \leq \big\| \mathbf x_j - \mathbf {mu^{(t)}_{i^*}} \big\| \forall i^*=1,\ldots,k \right\}$
Update step: Calculate the new means to be the centroid of the observations in the cluster.
$\mathbf mu^{(t+1)}_i = \frac{1}{|S^{(t)}_i|} \sum_{\mathbf x_j \in S^{(t)}_i} \mathbf x_j$
These two steps are alternated until a stopping criterion is met, i.e., when there is no further change in the assignment of the data points. | 564 | 2,110 | {"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} | 3.375 | 3 | CC-MAIN-2023-06 | longest | en | 0.885292 |
https://learn.careers360.com/school/question-provide-solution-for-rd-sharma-maths-class-12-chapter-19-definite-integrals-exercise-multiple-choice-question-18/?question_number=18.0 | 1,716,761,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00203.warc.gz | 301,638,178 | 39,571 | #### Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 18
2
Given:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| d x$
Hint:
You must know about $\left | x \right |$ function and $\int \sin x\; dx$
Explanation:
Let
$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| d x$
We know that
\begin{aligned} &|x|=\left\{\begin{aligned} x, & x \geq 0 \\ -x, & x<0 \end{aligned}\right. \\\\ &I=\int_{-\frac{\pi}{2}}^{0} \sin (-x) d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \end{aligned}
\begin{aligned} &I=-\int_{-\frac{\pi}{2}}^{0} \sin x \: d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \\\\ &=-[-\cos x]_{-\frac{\pi}{2}}^{0}+[-\cos x]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &=[\cos x]_{-\frac{\pi}{2}}^{0}-[\cos x]_{0}^{\frac{\pi}{2}} \\\\ &=\cos (0)-\cos \left(-\frac{\pi}{2}\right)-\cos \frac{\pi}{2}+\cos (0) \\\\ &=1-0-0+1 \\\\ &=2 \end{aligned} | 387 | 922 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.475853 |
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